Fluid power-data-book
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About This Presentation
fluido
Slide Content
Fluid Power
DATA BOOK
A collection of useful fluid power data, published
in this condensed form for convenient reference.
For expanded educational material on fluid power,
see textbook listings on the back cover.
Ask your Catching Inside Sales Person
for a printed copy
DATA BOOK
A collection of useful fluid power data, published
in this condensed form for convenient reference.
For expanded educational material on fluid power,
see textbook listings on the back cover.
Ask your Catching Inside Sales Person
for a printed copy
Table of Contents
Topic Page — Topic Page
Troubleshooting Design Data (cont)
Pa power graphic bola A;
adorar ; Hydraulic pipe table st
a a Pour mm Oi ow capacity fines =
ee Où posar os hr pipes 39
on Carton sel tio ata ná
laid power formulas 9 „Cooper cabina 40
si pores mal in ns setting data 2
Engine moves BO apacty eating pa
Dre ‘Airline sin m
ri u; Anm =
ehe an een u Km: E
een Pressure lass through fitting 46
Hydraulie cylinder force and A E >
Deere Mo Pinbnsfarindem se
ral der force ‘Nel nd vacoum ow
od epee 6 throug ones a
Prat onder sic Où ow through orcs n
ump 18 AE Mange dimensional data a
Pneumatic cylinder force table 20 = ha ja
omis il 2 Sri thread iting sine so
ston rod column strength 2 ‘Equivalent pipe and tubing sizes 60
Aa 150 anderen «or so
Dane sra pun. 3 Temadfmmsefflidemeden.. BL
Tor PPS aie À nice =
Pump and motor torque table a Lans a
Aingialteanmcon cto 28 Wire selection guide ss
opinas ang Ranger 20
Miscellaneous Formulas and
Required flow for operating Conversions
an ar or hydraulic ein = Table ofequivaents
HP to compress aie 23 Decimal and metric equivalents
Tank pump-down time 3 Conversion between English and
Cooling in hydraulie systems 30 SD Standard units ss
Fete 3 Imerhange between unis so
a "Temperature conversion chart. | 61
Vis rating aptes 3 Tableofstandard wire gauges 62
SU risen variation 38 Denis peiicgraviion and
Sea! compatible mechanical proper of
“commen Quid 26 ‘commun materais ss
Orders ile accepted by the publisher (address below fra minimum of
10 copies. Quantities of 1000 or more booklets can be purchased with your com.
pany name and address printed on the front cover. For ordering and price infor
{nation please visit our website at wirwwomackemachine com
NOTICE! We believe the information inthis booklet to be accurate. But errors
can occur in spite of careful checking. Therefore, neither this company nor
‘Womack Educational Publications will assume any liablty for damage or injury.
nor for the safe and/or satisfactory operation of any machine or system designed
‘rom information inthis booklet.
‘Tenth Edition — Ninth Printing, February 2005
‘©1998 by WOMACK EDUCATIONAL PUBLICATIONS
8020 Wyche Blvd. « Dallas, Texas 75235
Phone: 214-631-7083 + Fax: 214-630-5314
www: womack-machine.com
All Rights Reserved — Printed in the USA,
2
Topic Page — Topic Page
Troubleshooting Design Data (cont)
Pa power graphic bola A;
adorar ; Hydraulic pipe table st
a a Pour mm Oi ow capacity fines =
ee Où posar os hr pipes 39
on Carton sel tio ata ná
laid power formulas 9 „Cooper cabina 40
si pores mal in ns setting data 2
Engine moves BO apacty eating pa
Dre ‘Airline sin m
ri u; Anm =
ehe an een u Km: E
een Pressure lass through fitting 46
Hydraulie cylinder force and A E >
Deere Mo Pinbnsfarindem se
ral der force ‘Nel nd vacoum ow
od epee 6 throug ones a
Prat onder sic Où ow through orcs n
ump 18 AE Mange dimensional data a
Pneumatic cylinder force table 20 = ha ja
omis il 2 Sri thread iting sine so
ston rod column strength 2 ‘Equivalent pipe and tubing sizes 60
Aa 150 anderen «or so
Dane sra pun. 3 Temadfmmsefflidemeden.. BL
Tor PPS aie À nice =
Pump and motor torque table a Lans a
Aingialteanmcon cto 28 Wire selection guide ss
opinas ang Ranger 20
Miscellaneous Formulas and
Required flow for operating Conversions
an ar or hydraulic ein = Table ofequivaents
HP to compress aie 23 Decimal and metric equivalents
Tank pump-down time 3 Conversion between English and
Cooling in hydraulie systems 30 SD Standard units ss
Fete 3 Imerhange between unis so
a "Temperature conversion chart. | 61
Vis rating aptes 3 Tableofstandard wire gauges 62
SU risen variation 38 Denis peiicgraviion and
Sea! compatible mechanical proper of
“commen Quid 26 ‘commun materais ss
Orders ile accepted by the publisher (address below fra minimum of
10 copies. Quantities of 1000 or more booklets can be purchased with your com.
pany name and address printed on the front cover. For ordering and price infor
{nation please visit our website at wirwwomackemachine com
NOTICE! We believe the information inthis booklet to be accurate. But errors
can occur in spite of careful checking. Therefore, neither this company nor
‘Womack Educational Publications will assume any liablty for damage or injury.
nor for the safe and/or satisfactory operation of any machine or system designed
‘rom information inthis booklet.
‘Tenth Edition — Ninth Printing, February 2005
‘©1998 by WOMACK EDUCATIONAL PUBLICATIONS
8020 Wyche Blvd. « Dallas, Texas 75235
Phone: 214-631-7083 + Fax: 214-630-5314
www: womack-machine.com
All Rights Reserved — Printed in the USA,
2
Graphic Symbols
For Use on Fluid Power Drawings
‘These are the more common ANSI graphic symbols from the American Na-
tional Standards Institute for use on Avid power circuit drawings. A more com:
ete list can be obtained from the National Fluid Power Aseos
ayfair Rd. Milwaukee, W1 83222. Write for listing and prices
ER
1 [ 1
E | ]
T T T
Double-Acting Doubie-Acingylinder Single-Acting
Sylinder ‘wth Double End Rod Sindor
HYDRAULIC PUMPS ———
SIO
Fed Fixed Variable ‘Variable Variable Over
Disp Diep. Disp Diepl. Displ,Pres. Center
Umidrect. Brórect Umawect. Over Ent. Compensalor wComp.
HYDRAULIC & ELECTRIC MOTORS ———
ODDS ©
pos Ewa Verano Brea Paral Eee
Bape bast ge en Sasi
Uña. aia, under a
| zusam
Ub «ico Am 0
Raton Bere) Tonalonnat Selector
HX d * DE
Boston 2Postion Posten
Dr ET
-seoot cenrens ron a posmon vawves WHT TT TX RE
Hl M m
co Tandem Fam Sh
E Sie
For Use on Fluid Power Drawings
‘These are the more common ANSI graphic symbols from the American Na-
tional Standards Institute for use on Avid power circuit drawings. A more com:
ete list can be obtained from the National Fluid Power Aseos
ayfair Rd. Milwaukee, W1 83222. Write for listing and prices
ER
1 [ 1
E | ]
T T T
Double-Acting Doubie-Acingylinder Single-Acting
Sylinder ‘wth Double End Rod Sindor
HYDRAULIC PUMPS ———
SIO
Fed Fixed Variable ‘Variable Variable Over
Disp Diep. Disp Diepl. Displ,Pres. Center
Umidrect. Brórect Umawect. Over Ent. Compensalor wComp.
HYDRAULIC & ELECTRIC MOTORS ———
ODDS ©
pos Ewa Verano Brea Paral Eee
Bape bast ge en Sasi
Uña. aia, under a
| zusam
Ub «ico Am 0
Raton Bere) Tonalonnat Selector
HX d * DE
Boston 2Postion Posten
Dr ET
-seoot cenrens ron a posmon vawves WHT TT TX RE
Hl M m
co Tandem Fam Sh
E Sie
—— Graphic Symbols (continued) ——
ACTUATORS FOR VALVES.
r AL fof -{ oa
cron ‘wna “ra On se, Bike
Soret Met IA
“oo Lo L
fowls Seg Pomme Fated Sem Par
AE
cy ch ch
F I h +4
N ¡El
I | wl ut Pressure
ou mets ‘jen ane conte
cet peat, eet
ee Se
Ga fae re
ted WE ote cs Ge
ar u Om
' <> DEKE
pc A ey
Re mn nt etifhe
Bypass Flow Control
a
O Er ES 24 — —
nue Men li Mo
mn ie OS JS de a
ES
Gonna Que A
Outine Lines Lines Lines
Crossing Connecting
Moose
Air To ArorOi Ari Airline AirFiter
Unit iter Lubricator Aegulalor wiDrain
4
ACTUATORS FOR VALVES.
r AL fof -{ oa
cron ‘wna “ra On se, Bike
Soret Met IA
“oo Lo L
fowls Seg Pomme Fated Sem Par
AE
cy ch ch
F I h +4
N ¡El
I | wl ut Pressure
ou mets ‘jen ane conte
cet peat, eet
ee Se
Ga fae re
ted WE ote cs Ge
ar u Om
' <> DEKE
pc A ey
Re mn nt etifhe
Bypass Flow Control
a
O Er ES 24 — —
nue Men li Mo
mn ie OS JS de a
ES
Gonna Que A
Outine Lines Lines Lines
Crossing Connecting
Moose
Air To ArorOi Ari Airline AirFiter
Unit iter Lubricator Aegulalor wiDrain
4
Hydraulic Troubleshooting
Many of the failures in a hydraulic system show similar symptoms: a grad
tal or sudden loss of high pressure, resting in Toes of power or speed ithe
Sslinders. In fact, the lindo i under light loads or may not move at
A. Ofen the los of power e acmpanied by an Inerense in pump noise, espe:
Sally the A o a o,
‘Any major component (pump, Coll valve, directional valve, or elinder)
could be at fault Ina sophisticated system other components could also be at
fault bur thie would require he services ofan experienced tech
By following an organized sterby-step testing procedure inthe order given
problem can be traced o a general area hen i necessary, cach cam
Fat area can be vested or replace
JK F
8 o
IT
‘WORK
CYUNDER
A
6
BASIC HYDRAULIC SYSTEM
STEP 1 - Pump Suction Strainer
Probably the trouble encountered most often is cavitation of the hydraulic
jump inlet caused by restriction due to a dirt build-up on the auction strainer.
ca happen on a new ax well san older agur. produces the symptoms
described above: increased pump noise loss of high pressure and/or =
Tf the strainer is not located in the pump suction line it will be found im-
mersed below the oil level in the reservotr (point A), Some operators of hydra
Tie equipment never give the equipment
fails: Under these conditions, sooner or later, the suetion strainer will probably
become suficientiy restricted to cause a breakdown of the whole aystem and
damage to the pumy
"The suction strainer should be removed for inspection and should be cleaned
before reinstallation. Wire mesh strainers can best be cleaned with an air hose,
blowing from inside out, They can also be washed in a solvent which ie compat:
ible with the reservoir fluid. Kerosene may be used for strainers operating la
roleum base hydraulic où. Do not use gasoline or other explosive or amma
Bie solvents. The strainer should be cleaned even though it may not appear to
fry. Some clogging materials cannot be seen except by close inspection. If
there are holes in the mesh or if there is mechanical damage, the strainer
Should be replaced. When reinstalling the strainer, inspect all joints for possible
fir leaks, particularly at union joints (points B, E, G,H, 9, and K). There must
be ne air leaks in the suction line. Chee Tove to be sure it cov
ers the top of the strainer by at least 3 level, with all evlinders
Extended Tf it does not cover to this depth there ie danger of a vortex forming
ich may allow air to enter the system when the pump is running.
STEP 2 - Pump and Relief Valve
Afcleaning the pump suction strain doesnot correct the trouble sat the
pump and relief valve from the rest of the eircut by disconnecting at point E so
{That only the pump, relietvalve, and pressure gauge remain in the pump circuit
Cap or plug both ends of the plumbing which was disconnected, The pump is
now deadheaded into the relief valve. Start the pump and watch for pressure
build-up on the gauge while tightening the adjustment on the relief valve. If fall
pressure can be developed, obviously the pump and relief valve are operating
Correctly, and the trouble ia to be found further down the Nine. I full pressure
fannot be developed in this test, continue with STEP 3
5
Many of the failures in a hydraulic system show similar symptoms: a grad
tal or sudden loss of high pressure, resting in Toes of power or speed ithe
Sslinders. In fact, the lindo i under light loads or may not move at
A. Ofen the los of power e acmpanied by an Inerense in pump noise, espe:
Sally the A o a o,
‘Any major component (pump, Coll valve, directional valve, or elinder)
could be at fault Ina sophisticated system other components could also be at
fault bur thie would require he services ofan experienced tech
By following an organized sterby-step testing procedure inthe order given
problem can be traced o a general area hen i necessary, cach cam
Fat area can be vested or replace
JK F
8 o
IT
‘WORK
CYUNDER
A
6
BASIC HYDRAULIC SYSTEM
STEP 1 - Pump Suction Strainer
Probably the trouble encountered most often is cavitation of the hydraulic
jump inlet caused by restriction due to a dirt build-up on the auction strainer.
ca happen on a new ax well san older agur. produces the symptoms
described above: increased pump noise loss of high pressure and/or =
Tf the strainer is not located in the pump suction line it will be found im-
mersed below the oil level in the reservotr (point A), Some operators of hydra
Tie equipment never give the equipment
fails: Under these conditions, sooner or later, the suetion strainer will probably
become suficientiy restricted to cause a breakdown of the whole aystem and
damage to the pumy
"The suction strainer should be removed for inspection and should be cleaned
before reinstallation. Wire mesh strainers can best be cleaned with an air hose,
blowing from inside out, They can also be washed in a solvent which ie compat:
ible with the reservoir fluid. Kerosene may be used for strainers operating la
roleum base hydraulic où. Do not use gasoline or other explosive or amma
Bie solvents. The strainer should be cleaned even though it may not appear to
fry. Some clogging materials cannot be seen except by close inspection. If
there are holes in the mesh or if there is mechanical damage, the strainer
Should be replaced. When reinstalling the strainer, inspect all joints for possible
fir leaks, particularly at union joints (points B, E, G,H, 9, and K). There must
be ne air leaks in the suction line. Chee Tove to be sure it cov
ers the top of the strainer by at least 3 level, with all evlinders
Extended Tf it does not cover to this depth there ie danger of a vortex forming
ich may allow air to enter the system when the pump is running.
STEP 2 - Pump and Relief Valve
Afcleaning the pump suction strain doesnot correct the trouble sat the
pump and relief valve from the rest of the eircut by disconnecting at point E so
{That only the pump, relietvalve, and pressure gauge remain in the pump circuit
Cap or plug both ends of the plumbing which was disconnected, The pump is
now deadheaded into the relief valve. Start the pump and watch for pressure
build-up on the gauge while tightening the adjustment on the relief valve. If fall
pressure can be developed, obviously the pump and relief valve are operating
Correctly, and the trouble ia to be found further down the Nine. I full pressure
fannot be developed in this test, continue with STEP 3
5
STEP 3 - Pump or Relief Valve ..
Ti high pressure cannot be obtained in STEP 2 by running the pump against
the relief valve, further testing must be conducted to see whether the fault I
inthe pump or in the relie valve. Proceed ns follow:
‘posable, disconnect the reservoir return line from the relief valve at point
H. Atlach a short length of hose to the relef valve outlet. Hold the open end of
this hose over the reservoir filer opening ao Ihe rate of ol low can be observed.
Start the pump and run the reel valve adjustment up and down while observ
ing the flow through the hose. If the pump is bad. there will probably be a fui
stream of oil when the relief adjustment ie backed off but this flow will dim:
{sh or stop as the adjustment in increased
Fa flowmeter is available, the flow can be measured and compared with the
pump catalog rating. a flowmeter is not available, the rate of How on small
umps.can be measured by discharging the hose into a bucket while timing with
a watch, For example, ia volume of 10 gallons is collected in 15 seconds, the
‘Dumping rate à 40 GPM, ete
the gauge pressure does not rise above a low value, say 100 P*
fou docs not substantially decrease as the reel v
‘ened, the relier valve is probably at fault and should be el
structed in STEP 5. Ifthe oil substantially decreases as the relict valve ad
justment is tightened, and ifonly alow or moderate pressure can be developed,
this indicates trouble inthe pump. Proceed to STEP 4
STEP 4- Pump
Ifa full stream of oi is not obtained in STEP 3, or if he stream diminishes
as the relief valve adjustment e tightened, the pump is probably at fault. Aw
Suming that the suction strainer has already been cleaned and the inlet plumb:
ing has been examined for air leaks, as in STEP À, the ol a nipping across the
1 elements inside the pump. This can means wora-out pump, or Wo
an Oi temperature. High shippage inthe pump will cause the pump to run
considerably hotter than the oil reservoir temperature. In normal operation,
With a good pump, the pump ease will probably run about 20°F above the reer:
oir temperature, greater than this, exce lppaye, uned by wear, may be
‘Check alo for slipping bets, sheared shat pin or ke, broke sha broken
coupling, or loosened net screw
STEP 5 - Relief Valve ...
the test in STEP 3 has indicated the trouble tobe in the relief valve, point
D. the quickest remedy is to replace the valve with one known to be good. The
faulty valve may later be disassembled for inspection and cleaning: Pilot-oper
ated relie valves have small orifices which may be blocked with accumulations
öfdirt Blow out all passages with an air hose and run a small wire through or.
tien Check slo for free movement of the spon. in relief valve with pipe
thread connections in the body the spool may bind pipe ftings are over.
toned: If possible, test the spool for bind before unscrewing threaded connections
{rom the body. or screw in Rttings tightly during inspection ofthe valve
STEP 6 - Cylinder
[the pump will deliver full pressure when operating across the relief val
in STEP 2 both pump and reli valve can b lered good, and the trol
Is further downstream. The cylinder should be tested first for worn-out or detec:
tive packings by the method described on page 7.
STEP 7 - Directional Control Valve
Tithe cylinder has been tested (STEP 6) and found to have reasonably tight
isi seal he wy vale sald be ehecked next Although ex ol ale
pen, an excessively worn valve spool can slip enough al 1 prevent build-a
of maximum pressure: Symptom of this condition are a Tons ot cylinder spect
together with eu sn building upto fall pressure even with the oli valve
‘adjusted toa high setting. This condition would be more likely to vecur with
high pressure pumpe offow volume output, and would develop gradually over a
long period of me, Foursway valves may be tested by the method described on
page
‘Other Components
eck other components such as bypass flow controls, hydr
Solenoid way valten ofthe pilotoperated type with tandem or 0
Spools may not Rave sulticiont ilot pressure fo shit the spon!
Ti high pressure cannot be obtained in STEP 2 by running the pump against
the relief valve, further testing must be conducted to see whether the fault I
inthe pump or in the relie valve. Proceed ns follow:
‘posable, disconnect the reservoir return line from the relief valve at point
H. Atlach a short length of hose to the relef valve outlet. Hold the open end of
this hose over the reservoir filer opening ao Ihe rate of ol low can be observed.
Start the pump and run the reel valve adjustment up and down while observ
ing the flow through the hose. If the pump is bad. there will probably be a fui
stream of oil when the relief adjustment ie backed off but this flow will dim:
{sh or stop as the adjustment in increased
Fa flowmeter is available, the flow can be measured and compared with the
pump catalog rating. a flowmeter is not available, the rate of How on small
umps.can be measured by discharging the hose into a bucket while timing with
a watch, For example, ia volume of 10 gallons is collected in 15 seconds, the
‘Dumping rate à 40 GPM, ete
the gauge pressure does not rise above a low value, say 100 P*
fou docs not substantially decrease as the reel v
‘ened, the relier valve is probably at fault and should be el
structed in STEP 5. Ifthe oil substantially decreases as the relict valve ad
justment is tightened, and ifonly alow or moderate pressure can be developed,
this indicates trouble inthe pump. Proceed to STEP 4
STEP 4- Pump
Ifa full stream of oi is not obtained in STEP 3, or if he stream diminishes
as the relief valve adjustment e tightened, the pump is probably at fault. Aw
Suming that the suction strainer has already been cleaned and the inlet plumb:
ing has been examined for air leaks, as in STEP À, the ol a nipping across the
1 elements inside the pump. This can means wora-out pump, or Wo
an Oi temperature. High shippage inthe pump will cause the pump to run
considerably hotter than the oil reservoir temperature. In normal operation,
With a good pump, the pump ease will probably run about 20°F above the reer:
oir temperature, greater than this, exce lppaye, uned by wear, may be
‘Check alo for slipping bets, sheared shat pin or ke, broke sha broken
coupling, or loosened net screw
STEP 5 - Relief Valve ...
the test in STEP 3 has indicated the trouble tobe in the relief valve, point
D. the quickest remedy is to replace the valve with one known to be good. The
faulty valve may later be disassembled for inspection and cleaning: Pilot-oper
ated relie valves have small orifices which may be blocked with accumulations
öfdirt Blow out all passages with an air hose and run a small wire through or.
tien Check slo for free movement of the spon. in relief valve with pipe
thread connections in the body the spool may bind pipe ftings are over.
toned: If possible, test the spool for bind before unscrewing threaded connections
{rom the body. or screw in Rttings tightly during inspection ofthe valve
STEP 6 - Cylinder
[the pump will deliver full pressure when operating across the relief val
in STEP 2 both pump and reli valve can b lered good, and the trol
Is further downstream. The cylinder should be tested first for worn-out or detec:
tive packings by the method described on page 7.
STEP 7 - Directional Control Valve
Tithe cylinder has been tested (STEP 6) and found to have reasonably tight
isi seal he wy vale sald be ehecked next Although ex ol ale
pen, an excessively worn valve spool can slip enough al 1 prevent build-a
of maximum pressure: Symptom of this condition are a Tons ot cylinder spect
together with eu sn building upto fall pressure even with the oli valve
‘adjusted toa high setting. This condition would be more likely to vecur with
high pressure pumpe offow volume output, and would develop gradually over a
long period of me, Foursway valves may be tested by the method described on
page
‘Other Components
eck other components such as bypass flow controls, hydr
Solenoid way valten ofthe pilotoperated type with tandem or 0
Spools may not Rave sulticiont ilot pressure fo shit the spon!
Cylinder and Valve Testing
On an air system, if air is detected
‘escaping from a Away valve exhaust 7
while the cylinder is stopped, this air is N| cyinder
ither blowing. by worneut piston Fr
I. or is leaking across the spool in —w Law
the Away valve: These two leakage — | 7 ge
paths are shown in the figure to the Le
Fight
Most air cylinders and valves have
soft sete and should be Teeth. Vahe
However, those ‘air valves having a Gyindér
metalto.metal seal between spool and
ody may be expected to have a small POS
amount of leakage.
Sea te, mare key
o be coming through the cylinder than
{rors the valve spool, and the cylinder
should be tested frst.
vane
Spool
Leakage
PS
ehe pin lo one end of ity FE Fe
stroke and leave it stalled in this posi- Ñ
lon under pressure: Crack the Blin Two Leak Paths
tn thesameendofthe clinder to ches
for id lake
After checking, tighten the fitting and run the piston to the opposite end of
the barrel and repeat the text Ocenslonall a elder wil Teak a one point
sedge rato den the rel Check rapes pon
mid stroke by installing a positive stop atthe suspected position and run the
ston Tod against lo! testing. Once In a great white a piston eal may leak
Intermiitemly This is usually caused by a salt packing or O-ring moving slight
Ivor rolling ino different positions on the piston, and is more likely to happen
on cylinders of argo bore. = “
Pug
Presse Diagram for
4 Cylinder Testing
Cyinder
Leakage
When making this test on hydraulic cylinders, the line should be complete
removed from a cylinder port during the test. The open line from the valve
ie plage or cape since fight hack pene inthe tank return ine
‘spill if from the Hoe i not plugged. Pistons with metal ring seals can be
expected to have a small amount of leakage across the rings, and even “leak
tight soft seals may have a amall bypass during new seal break-in or after the
Seals are well worn
4-Way Valve Testing
For testing 4-way valves, either air or hydraulici is necessary to obtain ac
cess to the exhaust or tank return ports so that the amount of leakage can be
Sbserved To make the test, disconnect both cylinder lines and plug these ports
On the valve. Start up the system and shift the valve to one working postion
Any flow out the exhausts or tank return line while the valve fs under pressure
tthe amount of eakage: Repeat the test in all other working positions of the
vs Safe Pump Inlet Vacuum
| Gear Pumps | Vane Pumps | Piston Pumps
Wax Sate et Vacuum, PST EE 203 2
Max Sate InletVacuum.in.Hg | 61010 406 4
aya sucio strainer shold be clean or replaced when inet saca
‚ydraulie pump reaches these values, Sustained operation at these va
may damage the pump. When the suction strainer 1 clean, the inlet vacuum
Should not be more than LA of these values
7
On an air system, if air is detected
‘escaping from a Away valve exhaust 7
while the cylinder is stopped, this air is N| cyinder
ither blowing. by worneut piston Fr
I. or is leaking across the spool in —w Law
the Away valve: These two leakage — | 7 ge
paths are shown in the figure to the Le
Fight
Most air cylinders and valves have
soft sete and should be Teeth. Vahe
However, those ‘air valves having a Gyindér
metalto.metal seal between spool and
ody may be expected to have a small POS
amount of leakage.
Sea te, mare key
o be coming through the cylinder than
{rors the valve spool, and the cylinder
should be tested frst.
vane
Spool
Leakage
PS
ehe pin lo one end of ity FE Fe
stroke and leave it stalled in this posi- Ñ
lon under pressure: Crack the Blin Two Leak Paths
tn thesameendofthe clinder to ches
for id lake
After checking, tighten the fitting and run the piston to the opposite end of
the barrel and repeat the text Ocenslonall a elder wil Teak a one point
sedge rato den the rel Check rapes pon
mid stroke by installing a positive stop atthe suspected position and run the
ston Tod against lo! testing. Once In a great white a piston eal may leak
Intermiitemly This is usually caused by a salt packing or O-ring moving slight
Ivor rolling ino different positions on the piston, and is more likely to happen
on cylinders of argo bore. = “
Pug
Presse Diagram for
4 Cylinder Testing
Cyinder
Leakage
When making this test on hydraulic cylinders, the line should be complete
removed from a cylinder port during the test. The open line from the valve
ie plage or cape since fight hack pene inthe tank return ine
‘spill if from the Hoe i not plugged. Pistons with metal ring seals can be
expected to have a small amount of leakage across the rings, and even “leak
tight soft seals may have a amall bypass during new seal break-in or after the
Seals are well worn
4-Way Valve Testing
For testing 4-way valves, either air or hydraulici is necessary to obtain ac
cess to the exhaust or tank return ports so that the amount of leakage can be
Sbserved To make the test, disconnect both cylinder lines and plug these ports
On the valve. Start up the system and shift the valve to one working postion
Any flow out the exhausts or tank return line while the valve fs under pressure
tthe amount of eakage: Repeat the test in all other working positions of the
vs Safe Pump Inlet Vacuum
| Gear Pumps | Vane Pumps | Piston Pumps
Wax Sate et Vacuum, PST EE 203 2
Max Sate InletVacuum.in.Hg | 61010 406 4
aya sucio strainer shold be clean or replaced when inet saca
‚ydraulie pump reaches these values, Sustained operation at these va
may damage the pump. When the suction strainer 1 clean, the inlet vacuum
Should not be more than LA of these values
7
Replacement of Pump or Motor
Calculating the Theoretical GPM of a Pump by Measuring lts Internal Parts.
‘To select a replacement for a broken or worn out hydraulic pump or motor
which has no nameplate or has na rating marked or
below after making internal physical measurement
When replacing a pump, catalog ratings will usually bo shown in
specified shaft speed. On a motor, catalog ratings will usually be in C.LR. (cubic
Inches displacement per shaft revolution) Formulas are given fr ealeulatingei
ther GPM at 1800 REM or calculating C.LR. Use the formula which is appropri
ate. Make all measurements. in inches, as accurately as possible. Convert
Fractional dimensions nto decimal equivalent for une In Ihe forma
lake sure the catalog pressure rating is adequate for your application, an
in the case of a pump, be sure direction of shaft rotation is correct.
Gear Pumps
1. Measure gear wi
Measure distance across both gear
chambers; this ic.
‘GPM 1800 nen
of 1800 RPA ac he or
other speeds, GPM is propor
tonal to RPM
CLR Dile
Vane Pumps and Motors
(Balanced type; not variable displ.)
1. Measure width of rotor. This is W.
2: Measure shortest distance across
borer this is D.
3: Measure longest distance across
Dore this ie,
A speed of 1800 RPM is used in the for:
a: At other speeds, GPM is propor.
Boni to RPM,
AR Dipacement
m) dD)
VANE PUMP (Balanced Type Only)
Piston Pumps and Motors
1. Find piston area from piston diameter: this is Ain formula,
2 Measure length of stroke; this is Ln formula,
3. Count number of pistons; this is Nin formula.
PNG 1800 RENE = Av Loe N #800231
A speed of 1800 RPM is used in formula. At other speeds, GPM is proportion
alo RPM
CAR Diplcement = ALAN
Ifa pump of higher GPM has to be used, it will require more HP at the same
pressure and eylinders in the system will move faster I one with lower GPL Is
Used, the system will have plenty of power but eylinders will move more slowly
than originally
I a motor with greater displacement is used, it will deliver more torque at a
reduced RPM, but will require no more fluid HP from the pump. It has less
displacement it will rotate faster with less torque,
g
Calculating the Theoretical GPM of a Pump by Measuring lts Internal Parts.
‘To select a replacement for a broken or worn out hydraulic pump or motor
which has no nameplate or has na rating marked or
below after making internal physical measurement
When replacing a pump, catalog ratings will usually bo shown in
specified shaft speed. On a motor, catalog ratings will usually be in C.LR. (cubic
Inches displacement per shaft revolution) Formulas are given fr ealeulatingei
ther GPM at 1800 REM or calculating C.LR. Use the formula which is appropri
ate. Make all measurements. in inches, as accurately as possible. Convert
Fractional dimensions nto decimal equivalent for une In Ihe forma
lake sure the catalog pressure rating is adequate for your application, an
in the case of a pump, be sure direction of shaft rotation is correct.
Gear Pumps
1. Measure gear wi
Measure distance across both gear
chambers; this ic.
‘GPM 1800 nen
of 1800 RPA ac he or
other speeds, GPM is propor
tonal to RPM
CLR Dile
Vane Pumps and Motors
(Balanced type; not variable displ.)
1. Measure width of rotor. This is W.
2: Measure shortest distance across
borer this is D.
3: Measure longest distance across
Dore this ie,
A speed of 1800 RPM is used in the for:
a: At other speeds, GPM is propor.
Boni to RPM,
AR Dipacement
m) dD)
VANE PUMP (Balanced Type Only)
Piston Pumps and Motors
1. Find piston area from piston diameter: this is Ain formula,
2 Measure length of stroke; this is Ln formula,
3. Count number of pistons; this is Nin formula.
PNG 1800 RENE = Av Loe N #800231
A speed of 1800 RPM is used in formula. At other speeds, GPM is proportion
alo RPM
CAR Diplcement = ALAN
Ifa pump of higher GPM has to be used, it will require more HP at the same
pressure and eylinders in the system will move faster I one with lower GPL Is
Used, the system will have plenty of power but eylinders will move more slowly
than originally
I a motor with greater displacement is used, it will deliver more torque at a
reduced RPM, but will require no more fluid HP from the pump. It has less
displacement it will rotate faster with less torque,
g
Fluid Power Formulas
Tongue gat heces Jeans: Eevee fain waking ot
Malis Fama
de = pounds; F is the part of the force
Hydraulic (fluid power) horsepower: which is effective, in pounds; A is
A Be eon ante
ase ieee aati lr
Tia =T2Vy, OF Ta =T2Py Burst pressure of pipe or tubing:
PAVy = Pava
Pj and Vy are initial pressure and Relationship between displacement
volume: Pz and Vz are final condi> Fp PSI 240
Bons Tis torque in fonte D is dis
ia lacemant in eubie inches
ation: PSI is pressure di
‘Area =, or x02 4 across motor: R= 3-14.
Cireumferenee = Zar, or xD
Ea GEES ete Rulesot-Thum>
Bou gilet sad power: recover to are eu
BTU por eur PS GP “RPO AES PH.
Hydraulic cyl. piston travel speed: ‘slant of | GPM @ 1500 PSI can be
“sn ul
ÉD Le nd de A
Bin en sol fee triage ps enden
RP TE er
A shoved
Tree eae ly ints
avi} ey e ay
A e e
FA guage pres Compressibility of water:
perimeter around area to be ‘Bach watt will raise the tempera
ess in inches; PSI is the shear hour.
strength rating of the material
pounds
er sa are Inch Floy velocity in hydraulic lines
Pump auction lines 210.4 feet por
Side load on pump or motor shaft: second: pressure lines up to 500
E = (HP x 83024) > (RPM R) PSI, 10 to 15 foot per see: pressure
F isthe side oad, in pounds, lines 500 0 3000 PSI, 15 to 20 foot
‘against shalt: isthe piteh radius, per sec; pressure lines over 3000
ichesyof sheave on pump shaft PSI, 28 feet por sec. al oil lines in
HP is driving power applied to airóveroll aystem, foot per see.
shal,
Tongue gat heces Jeans: Eevee fain waking ot
Malis Fama
de = pounds; F is the part of the force
Hydraulic (fluid power) horsepower: which is effective, in pounds; A is
A Be eon ante
ase ieee aati lr
Tia =T2Vy, OF Ta =T2Py Burst pressure of pipe or tubing:
PAVy = Pava
Pj and Vy are initial pressure and Relationship between displacement
volume: Pz and Vz are final condi> Fp PSI 240
Bons Tis torque in fonte D is dis
ia lacemant in eubie inches
ation: PSI is pressure di
‘Area =, or x02 4 across motor: R= 3-14.
Cireumferenee = Zar, or xD
Ea GEES ete Rulesot-Thum>
Bou gilet sad power: recover to are eu
BTU por eur PS GP “RPO AES PH.
Hydraulic cyl. piston travel speed: ‘slant of | GPM @ 1500 PSI can be
“sn ul
ÉD Le nd de A
Bin en sol fee triage ps enden
RP TE er
A shoved
Tree eae ly ints
avi} ey e ay
A e e
FA guage pres Compressibility of water:
perimeter around area to be ‘Bach watt will raise the tempera
ess in inches; PSI is the shear hour.
strength rating of the material
pounds
er sa are Inch Floy velocity in hydraulic lines
Pump auction lines 210.4 feet por
Side load on pump or motor shaft: second: pressure lines up to 500
E = (HP x 83024) > (RPM R) PSI, 10 to 15 foot per see: pressure
F isthe side oad, in pounds, lines 500 0 3000 PSI, 15 to 20 foot
‘against shalt: isthe piteh radius, per sec; pressure lines over 3000
ichesyof sheave on pump shaft PSI, 28 feet por sec. al oil lines in
HP is driving power applied to airóveroll aystem, foot per see.
shal,
Fluid Power Formulas
in SI Metric Units
luid power formulas in English units
ternational) unit equivalents of these fo
English Units Metric (SI) Unit
are shown in the left column. SI (In
las are shown in the right column,
Torque, HP, Speed Relations in Hydraulic Pumps & Motors
7 HP 5252 + RPM T= Kw x 9543 RPM
HB ITS RPM. 5282 Ku T2 RPM - 9543
RPM HP 9252 RPM = kw 9843 T
T > Torque, footbs T > Torque, Nm (Newton-meters)
APM = Speed, reva/minute RPM = Speed, reve/minute
HP = Horsepower Kw = Power in kilowatts
Hydraulic Power Flowing through the Pipes
HP = PSI» GPM + 1714 Kw = Bar x dmfimin + 600
HP = Horsepower Kw = Power in kilowatts
PSI = Gauge pressure, Ib/sq. inch S
GPM = Flow, galions por minute
Force Developed by an Air or Hydraulic Cylinder
Fa Ax PSI Ax Bar 10
cylinder force in Newtons
Piston area, sq. centimeters
ae pressure
force or thrust, Ibs.
Picton area, square inches
Gauge prossure,lbvsq.ineh | | Bar
Travel Speed of a Hydraulic Cylinder Piston
“A “A
ravel Speed, inches/minute = Travel Speed, meter
Volume er oilto sl, cu.inimin. | | Y = Oil flow, dm minute
A= Piston area, square inches Piston area, square centimeters
Barlow's Formula — Burst Pressure of Pipe & Tubing
mes ae
fensile strength, pipe mat fensile strength, pipe mat,
© = Outside diameter ol pipe, inches | | © = Outside diameter of pipe, mim
Velocity of Oil Flow in Hydraulic Lines
PM 0.3208 + A dein + 6A
locity, feet per second V = Oil velocity, meters/second
Si low: gallons/minute dim = Oil how, cu. dm/minute
Inside area of pipe, sq. inches | | A’ Inside area of pipe, sq. cm
Recommended Maximum Oil Velocity in Hydraulic Lines
feet per second mps = Meters por second
Pump suction lines - 210 4 fps Pump suction lines - 0.6 10 1.2 mps
Pres ines to 500 PSI — 1010 15 fps | | Pres lines Lo 35 bar- 3 to 4% mps
vs Hines to 3000 PSI 15 020 fps| | Pres. lines to 200 bar- 4¥%to 6 mps
Pres lines over 3000 PSI~25 fps | | Pres, lines over 200 bar~ 7% mps
Oil lines in rfl syatem — 4 tps Of ines in all system = 1% mps
10
tos
in SI Metric Units
luid power formulas in English units
ternational) unit equivalents of these fo
English Units Metric (SI) Unit
are shown in the left column. SI (In
las are shown in the right column,
Torque, HP, Speed Relations in Hydraulic Pumps & Motors
7 HP 5252 + RPM T= Kw x 9543 RPM
HB ITS RPM. 5282 Ku T2 RPM - 9543
RPM HP 9252 RPM = kw 9843 T
T > Torque, footbs T > Torque, Nm (Newton-meters)
APM = Speed, reva/minute RPM = Speed, reve/minute
HP = Horsepower Kw = Power in kilowatts
Hydraulic Power Flowing through the Pipes
HP = PSI» GPM + 1714 Kw = Bar x dmfimin + 600
HP = Horsepower Kw = Power in kilowatts
PSI = Gauge pressure, Ib/sq. inch S
GPM = Flow, galions por minute
Force Developed by an Air or Hydraulic Cylinder
Fa Ax PSI Ax Bar 10
cylinder force in Newtons
Piston area, sq. centimeters
ae pressure
force or thrust, Ibs.
Picton area, square inches
Gauge prossure,lbvsq.ineh | | Bar
Travel Speed of a Hydraulic Cylinder Piston
“A “A
ravel Speed, inches/minute = Travel Speed, meter
Volume er oilto sl, cu.inimin. | | Y = Oil flow, dm minute
A= Piston area, square inches Piston area, square centimeters
Barlow's Formula — Burst Pressure of Pipe & Tubing
mes ae
fensile strength, pipe mat fensile strength, pipe mat,
© = Outside diameter ol pipe, inches | | © = Outside diameter of pipe, mim
Velocity of Oil Flow in Hydraulic Lines
PM 0.3208 + A dein + 6A
locity, feet per second V = Oil velocity, meters/second
Si low: gallons/minute dim = Oil how, cu. dm/minute
Inside area of pipe, sq. inches | | A’ Inside area of pipe, sq. cm
Recommended Maximum Oil Velocity in Hydraulic Lines
feet per second mps = Meters por second
Pump suction lines - 210 4 fps Pump suction lines - 0.6 10 1.2 mps
Pres ines to 500 PSI — 1010 15 fps | | Pres lines Lo 35 bar- 3 to 4% mps
vs Hines to 3000 PSI 15 020 fps| | Pres. lines to 200 bar- 4¥%to 6 mps
Pres lines over 3000 PSI~25 fps | | Pres, lines over 200 bar~ 7% mps
Oil lines in rfl syatem — 4 tps Of ines in all system = 1% mps
10
tos
English/Metric Conversions
Pressure - PSI and Bar
SSBR ER ER
Pst
RER
7250,
32
Inagereuenssgagassg
lpgggagronononanauel
cano ER CE
1 bar = 145 Pst
afro eeerensess
min GPM
1 Hiter/min = 0.2682 GPM
‘GP
min
noseasangs-agsrges|
RRSSRRSBRRRRSERRER
AS
MERBSRELBTELTERERE|
224187 CFM,
émis)
1 eu. dmmee
mts CFM
mils CFM
ERRE
leggesgeesess: RE
N28389833223322382
Beensgerga-gnasenı)
VERRREBTITTREELEER|
FENETRE
(ABS ETS RIT BR ENS |
Erre isis
Banssszagrassaszaa|
anise aostRnBTERLE
glRssesresssssesesss
GP min
|
2
3
E
=
a
E
3
2
5
=
a
[a]
1
3
2
lr
2
E
3
=
cpm
gem Umin
gacsgesearresseesa|
ed
eseegeacergrasare|
less ageegentonr ee)
Ser nee PE RNA]
| sosweeanaseeanase
Air Flow - CFM and Cubic Decimeters per Second
1 CFM = 0.47195 cu, drvsee
ÉRSSFS Ra EE RATE
FRSSABREE SR ERRES
ee
|-sewovgnaaeeaneeR|
11
Pressure - PSI and Bar
SSBR ER ER
Pst
RER
7250,
32
Inagereuenssgagassg
lpgggagronononanauel
cano ER CE
1 bar = 145 Pst
afro eeerensess
min GPM
1 Hiter/min = 0.2682 GPM
‘GP
min
noseasangs-agsrges|
RRSSRRSBRRRRSERRER
AS
MERBSRELBTELTERERE|
224187 CFM,
émis)
1 eu. dmmee
mts CFM
mils CFM
ERRE
leggesgeesess: RE
N28389833223322382
Beensgerga-gnasenı)
VERRREBTITTREELEER|
FENETRE
(ABS ETS RIT BR ENS |
Erre isis
Banssszagrassaszaa|
anise aostRnBTERLE
glRssesresssssesesss
GP min
|
2
3
E
=
a
E
3
2
5
=
a
[a]
1
3
2
lr
2
E
3
=
cpm
gem Umin
gacsgesearresseesa|
ed
eseegeacergrasare|
less ageegentonr ee)
Ser nee PE RNA]
| sosweeanaseeanase
Air Flow - CFM and Cubic Decimeters per Second
1 CFM = 0.47195 cu, drvsee
ÉRSSFS Ra EE RATE
FRSSABREE SR ERRES
ee
|-sewovgnaaeeaneeR|
11
Fluid Power Equivalents
Exact Equivalents 1 PSI = 20416 Hg
1 US, gallon 27.71 water
2 uf cubic inches 0.0689 bar
quarts or 8 pints Uaboeptire
128 oumces Liquid) eye
133.97 ounces weigh) ie be
5.3556 pounds |
23785 Mera DER
1 Imperial gallon = 1.2 US. gal 1 Foot water column = 0.452.PS1
1 Liter = 0.2642 US. gallons 1 Foot ol column = 0.354 PSI
1 Cubic Rat arr ol = 42 gallons
= 7.48 gallons pharrell ae
= 1728 eubie inches 1 Miero-meter um:
624 pounds water) 000001 meter (micron)
1.Cu, A. water weighs 62.4 Ibs Arr
1 Bar at sea level: 25 Microsmeters = 0.001 inch
214804 Pst
08692 atmosphere Approximate Equivalents
33.6 fot water column 1 Pint 22 cups = 32 tablespoons =
41 foot oi column 96 teaspoons = 16.02. = 1 Ib
Approx. 1/2 PSI decrease each 1000 1 Quart = 4 cups = 2 pints
riet of elevation 32 fluid ounces = pounds
Y Hg = 0.490 PSI 1 Gallon = 16 cups = 4 quarts =8 pints
SAS water ‘P18 Non 2 291 cu. ins
1 Horsepower: 1 Cup
6 tablespoons = 48 tsp.
33.000 f Ibs. per minute
1 Tablespoon = 3 tap. = 1/2 uid ox
O per minute 1 Fluid oz. (volume) = 600 drops hy-
2545 BTU Der hour raulic ot
796 watts or 0.148 kw 1 Cobie inch = 330 drops oi.
Fluid Power Abbreviations
shot ax ie) ipm inches per minute
‘Alternating current ips’ inches per second
Bone hardness number ID pound
Erich hema pit | ax Maximum
a center ‘mounted
‘normally closed
formally open
national pipe thread
Arsen! pipe threads
mentor clockwise
tabi fet pr minute
tub feet por second
Kubi inches por revolution
gue inches per open center
ycles per minute pilot operated
Seles per acco Pressure
bie inches per revolution pounds/quare inch
locke psi absolute
Sylinder psi gauge
Are current Pint
liameter Quart
external radis
root mean square
revolutions per minute
‘evolutions per second
degrees Fahrenheit
a
Feet per minute
foot Standard cu’ ft per minute
cl nd
Fillets por minute alt seconds universal
ercury
Hens
inside vacuum
Viscosity index
Exact Equivalents 1 PSI = 20416 Hg
1 US, gallon 27.71 water
2 uf cubic inches 0.0689 bar
quarts or 8 pints Uaboeptire
128 oumces Liquid) eye
133.97 ounces weigh) ie be
5.3556 pounds |
23785 Mera DER
1 Imperial gallon = 1.2 US. gal 1 Foot water column = 0.452.PS1
1 Liter = 0.2642 US. gallons 1 Foot ol column = 0.354 PSI
1 Cubic Rat arr ol = 42 gallons
= 7.48 gallons pharrell ae
= 1728 eubie inches 1 Miero-meter um:
624 pounds water) 000001 meter (micron)
1.Cu, A. water weighs 62.4 Ibs Arr
1 Bar at sea level: 25 Microsmeters = 0.001 inch
214804 Pst
08692 atmosphere Approximate Equivalents
33.6 fot water column 1 Pint 22 cups = 32 tablespoons =
41 foot oi column 96 teaspoons = 16.02. = 1 Ib
Approx. 1/2 PSI decrease each 1000 1 Quart = 4 cups = 2 pints
riet of elevation 32 fluid ounces = pounds
Y Hg = 0.490 PSI 1 Gallon = 16 cups = 4 quarts =8 pints
SAS water ‘P18 Non 2 291 cu. ins
1 Horsepower: 1 Cup
6 tablespoons = 48 tsp.
33.000 f Ibs. per minute
1 Tablespoon = 3 tap. = 1/2 uid ox
O per minute 1 Fluid oz. (volume) = 600 drops hy-
2545 BTU Der hour raulic ot
796 watts or 0.148 kw 1 Cobie inch = 330 drops oi.
Fluid Power Abbreviations
shot ax ie) ipm inches per minute
‘Alternating current ips’ inches per second
Bone hardness number ID pound
Erich hema pit | ax Maximum
a center ‘mounted
‘normally closed
formally open
national pipe thread
Arsen! pipe threads
mentor clockwise
tabi fet pr minute
tub feet por second
Kubi inches por revolution
gue inches per open center
ycles per minute pilot operated
Seles per acco Pressure
bie inches per revolution pounds/quare inch
locke psi absolute
Sylinder psi gauge
Are current Pint
liameter Quart
external radis
root mean square
revolutions per minute
‘evolutions per second
degrees Fahrenheit
a
Feet per minute
foot Standard cu’ ft per minute
cl nd
Fillets por minute alt seconds universal
ercury
Hens
inside vacuum
Viscosity index
Vehicle Drive Calculations
"The force to drive a vehicle is composed of the sum of (1) road resistance, (2)
force necessary to climb a grade, (3) loree needed to accelerate to final velocity
in the allowable time, (4) foree to overcome air resistance, on fast moving vehi
«le Bach of these rs can be cout or estimated from the formulas on
{this page, then added together In selecting an engine, allow enough extra pow-
er to make up for losses in the mechanical transmission system including gear
boxes, clutches, differentials, chain or belt drives
Travel Speed in M.PH. (miles por Acceleration of a ve
APMP. shoeleircumtoreses CT RN, cteerating ores and me.
MPH, = RPM x d + 336, or F=(VxW)= (9x7)
RPM = 396 «MPH. + d Fis accelerating force in pounds.
à sonne nie Vis Anal velocity in feet per second
Wir vehicle weights» pounds
gis gravity acceleration = 82.16
Axle Torque for driving the vehicle Pi fimo in seconds that force act
is found by multiplying drawbar pull Noto: The gravity acceleration sy.
{ox push) Has wel rides bol. £ converts weight into mass
T=Fxron, —
Ts ale torque in inch pounds Grade, in mobile work, is usually
is drawbar pullin pounds ‘expressed in percentage rather than
Tis wheel radius in inches. Ÿ Lap
"degrees. For example, a 10% grade,
has rio of 10 feet In a distance oF
Drawbar Pull to keep the vehicle in 100 feet etc.
steady motion on level ground de: Grade Resistance is the drawbar
pende on the road surface. The follow. pull needed to keep the vehicle in con:
Ing figures are pounds of drawbar pull Stant motion up a grade. This is in ad.
mer 1000 Ihe ol vello weg Allen to the drandar pull overcome
Concrete 10 to 90 lhe, fond resistance as expressed ty amo
Ssphait 131232 he or formula.
Ma Beare Feonxw
eine $510 851s Eis drawbar pullin pounds
ore 3310 STD Gis grade resistance in percent
ha 120% fe written as 030, te)
Me, Win grosa vehicle weight in pounds
Horsepower required on vehicle ‚Air Resistance will be important
A Gate Ria antag moving vehicles (vor 20
HP =T x RPM + 60008 FR 2 0.0025 « MPH?
Tis wheel torque in inch pounds. is additional drawbar pull needed to
NOTE: Additonal HP is required at gvereame ar resistance
he engine to overcome transmission FA Is frontal arca of vehicle in square
stent losses. feet
iH. is vehicle speed, miles per hour.
Onin Fórmula na es
torque, HP and speed. Axe and drive shaft must haven
T2 HP x 63024 + RPM diameter large enough to transmit the
torque without ‘excessive. deflection
‘Torque values are in inch pounds. The angle of deflection for solid
wind ante may be calculated from
Momentum of a vehicle is equiva- this formula
lent to that constant force which a= 889.8 xx L + (OE)
TEE en nd ane af detection dere
IS applied torque in Inch pounds
Momentum - Weight «V + 9 Lis shan length in inches.
eight sin pounds Eis modulus of elsteity of material
VE ect feet per second 12.000.000 fr ice)
Bis gravity ace SG Ds shaft diameter in inches.
Some authorities say that a steel
shaft should be limited to an angular
election of 0.08 degrees por fot of
length to avoid failure
13
"The force to drive a vehicle is composed of the sum of (1) road resistance, (2)
force necessary to climb a grade, (3) loree needed to accelerate to final velocity
in the allowable time, (4) foree to overcome air resistance, on fast moving vehi
«le Bach of these rs can be cout or estimated from the formulas on
{this page, then added together In selecting an engine, allow enough extra pow-
er to make up for losses in the mechanical transmission system including gear
boxes, clutches, differentials, chain or belt drives
Travel Speed in M.PH. (miles por Acceleration of a ve
APMP. shoeleircumtoreses CT RN, cteerating ores and me.
MPH, = RPM x d + 336, or F=(VxW)= (9x7)
RPM = 396 «MPH. + d Fis accelerating force in pounds.
à sonne nie Vis Anal velocity in feet per second
Wir vehicle weights» pounds
gis gravity acceleration = 82.16
Axle Torque for driving the vehicle Pi fimo in seconds that force act
is found by multiplying drawbar pull Noto: The gravity acceleration sy.
{ox push) Has wel rides bol. £ converts weight into mass
T=Fxron, —
Ts ale torque in inch pounds Grade, in mobile work, is usually
is drawbar pullin pounds ‘expressed in percentage rather than
Tis wheel radius in inches. Ÿ Lap
"degrees. For example, a 10% grade,
has rio of 10 feet In a distance oF
Drawbar Pull to keep the vehicle in 100 feet etc.
steady motion on level ground de: Grade Resistance is the drawbar
pende on the road surface. The follow. pull needed to keep the vehicle in con:
Ing figures are pounds of drawbar pull Stant motion up a grade. This is in ad.
mer 1000 Ihe ol vello weg Allen to the drandar pull overcome
Concrete 10 to 90 lhe, fond resistance as expressed ty amo
Ssphait 131232 he or formula.
Ma Beare Feonxw
eine $510 851s Eis drawbar pullin pounds
ore 3310 STD Gis grade resistance in percent
ha 120% fe written as 030, te)
Me, Win grosa vehicle weight in pounds
Horsepower required on vehicle ‚Air Resistance will be important
A Gate Ria antag moving vehicles (vor 20
HP =T x RPM + 60008 FR 2 0.0025 « MPH?
Tis wheel torque in inch pounds. is additional drawbar pull needed to
NOTE: Additonal HP is required at gvereame ar resistance
he engine to overcome transmission FA Is frontal arca of vehicle in square
stent losses. feet
iH. is vehicle speed, miles per hour.
Onin Fórmula na es
torque, HP and speed. Axe and drive shaft must haven
T2 HP x 63024 + RPM diameter large enough to transmit the
torque without ‘excessive. deflection
‘Torque values are in inch pounds. The angle of deflection for solid
wind ante may be calculated from
Momentum of a vehicle is equiva- this formula
lent to that constant force which a= 889.8 xx L + (OE)
TEE en nd ane af detection dere
IS applied torque in Inch pounds
Momentum - Weight «V + 9 Lis shan length in inches.
eight sin pounds Eis modulus of elsteity of material
VE ect feet per second 12.000.000 fr ice)
Bis gravity ace SG Ds shaft diameter in inches.
Some authorities say that a steel
shaft should be limited to an angular
election of 0.08 degrees por fot of
length to avoid failure
13
Hydraulic Cylinder
Force & Speed Calculations
Calculation of Hydraulic Cylinder Force...
EXAMPLE: A certain application requires a cylinder force of 25 ton
What should be the cylinder bore diameter used and sl what yauge
Pr SOLUTION: The required force is 25 tons x 2000 = 50,000 pounds. Refer to
the “Hydraulie Cylinder Force table on pages 19 and 10 which hows several
Combinations of piston diameter and PSÍ pressure which will produce 50,000
Pounds of force or more. For example, inch piston will produce 38,950 pounds
112000 PSI; a7 Inch piston wi roduee 52,725 Ibs at 1360 PSI: an inch piston
‘ll produce 30.285 tbs at 1000 PSI, à 10 ch piston will produce 58.000 los. at
ete So there are many combinations which could be use, and the Anal
a matter of preference or of matching the pressure and flow capability
Stother components. particularly the pump
In practice, choose a combination wie will produce from 10% to 25% mor
than actually required by the load alone. This wil provide a safety allowance
sehen will take are of pressure losses in valves and piping and mechanical
donnes in the eylinder.
EXAMPLE: How many pounds of force will be developed on the exten-
sion stroke of a 3% bore eylinder operating at 1500 PSI? If this cylinder
has a 17. diameter piston rod, how much force will be developed on the
retraction stroke?
SOLUTION! eter to the “Hydraulic Cylinder Force” table on pages 15 and
16. The chart shows 12,444 bs. À solution can also be obtained by bring the pls.
{on area (8.296 square inches) and multiplying by the pressure (1900 PSI
8296 square inches x 1500 PSI = 13.444 bs
On the retraction stroke the amount of force developed on the 2.41 square
inch od area must be subtracted: T2444 3608 = 8836 lbs
EXAMPLE: What PSI gauge pressure ix required for retraction of a
30,000 Ih. load with an finch boro cylinder having a 4 inch diameter
SOLUTION: The net piston area must be
‘minus the rod area. 80:37 (piston area) = 12. a
PSI = 50,000 + 3.7 = 1926 PSL The actual pressure will be slightly greater dut
do fiction of the piston in the barrel.
Calculation of Hydraulic Cylinder Speed...
EXAMPLE: At what speed would the piston of a 4 inch bore cylinder
extend on an oil How of 12 GPM?
SOLUTION: The table of “Hydraulic Cylinder
may be used or the speed figured with
‘equal to the incoming flow of oil in cubie inches per minut
‘Square inch area of the piston”. The speed will bein inches per minute.
‘A low of 12 GPM is 231 x 12 = 2772 eubie inches per minute. The speed ix
2772 (flow rate) + 12.57 (piston area) = 220.5 inches per minute, This checks
‘ery closely with the value ahoven in the table on page 17.
EXAMPLE: Find the GPM flow necessary to cause a 5 inch bore eylin-
der to travel at a rate of 175 inches per minute while extending.
How fast would this cylinder retract on the same oil flow fit had a
2 inch diameter piston rod?
SOLUTION: Few ie determined by multiplying the piston ara in square
inches times the travel rate in inches per minute. This gives How in cubic inches
per minuto. Divide by 231 to convert to GPM: 19,64 (piston area) x 178 = 3497
Cubic inches per minute. 3437 + 251 = 14.88 GPM, This checks very closely with
15 GPM at 174 inches per minute shown on the chart on page 17.
To find the retraction speed on 14.88 GPM, the net piston area must be
found. This is he full piston area minus the rod area: 19,64 (piston area) ~ 6.5.
(rod area) = 16.5 square inches. The flow rate is 3437 cube inches per minuts
vo 14.58 GPAD + 16.5 (net area) = 208 inches per minute Note that
this is faster than the extension speed un the same oll flow
Speeds on pages 17 and 18
€ formula which nays that “speed te
divided by the
14
Force & Speed Calculations
Calculation of Hydraulic Cylinder Force...
EXAMPLE: A certain application requires a cylinder force of 25 ton
What should be the cylinder bore diameter used and sl what yauge
Pr SOLUTION: The required force is 25 tons x 2000 = 50,000 pounds. Refer to
the “Hydraulie Cylinder Force table on pages 19 and 10 which hows several
Combinations of piston diameter and PSÍ pressure which will produce 50,000
Pounds of force or more. For example, inch piston will produce 38,950 pounds
112000 PSI; a7 Inch piston wi roduee 52,725 Ibs at 1360 PSI: an inch piston
‘ll produce 30.285 tbs at 1000 PSI, à 10 ch piston will produce 58.000 los. at
ete So there are many combinations which could be use, and the Anal
a matter of preference or of matching the pressure and flow capability
Stother components. particularly the pump
In practice, choose a combination wie will produce from 10% to 25% mor
than actually required by the load alone. This wil provide a safety allowance
sehen will take are of pressure losses in valves and piping and mechanical
donnes in the eylinder.
EXAMPLE: How many pounds of force will be developed on the exten-
sion stroke of a 3% bore eylinder operating at 1500 PSI? If this cylinder
has a 17. diameter piston rod, how much force will be developed on the
retraction stroke?
SOLUTION! eter to the “Hydraulic Cylinder Force” table on pages 15 and
16. The chart shows 12,444 bs. À solution can also be obtained by bring the pls.
{on area (8.296 square inches) and multiplying by the pressure (1900 PSI
8296 square inches x 1500 PSI = 13.444 bs
On the retraction stroke the amount of force developed on the 2.41 square
inch od area must be subtracted: T2444 3608 = 8836 lbs
EXAMPLE: What PSI gauge pressure ix required for retraction of a
30,000 Ih. load with an finch boro cylinder having a 4 inch diameter
SOLUTION: The net piston area must be
‘minus the rod area. 80:37 (piston area) = 12. a
PSI = 50,000 + 3.7 = 1926 PSL The actual pressure will be slightly greater dut
do fiction of the piston in the barrel.
Calculation of Hydraulic Cylinder Speed...
EXAMPLE: At what speed would the piston of a 4 inch bore cylinder
extend on an oil How of 12 GPM?
SOLUTION: The table of “Hydraulic Cylinder
may be used or the speed figured with
‘equal to the incoming flow of oil in cubie inches per minut
‘Square inch area of the piston”. The speed will bein inches per minute.
‘A low of 12 GPM is 231 x 12 = 2772 eubie inches per minute. The speed ix
2772 (flow rate) + 12.57 (piston area) = 220.5 inches per minute, This checks
‘ery closely with the value ahoven in the table on page 17.
EXAMPLE: Find the GPM flow necessary to cause a 5 inch bore eylin-
der to travel at a rate of 175 inches per minute while extending.
How fast would this cylinder retract on the same oil flow fit had a
2 inch diameter piston rod?
SOLUTION: Few ie determined by multiplying the piston ara in square
inches times the travel rate in inches per minute. This gives How in cubic inches
per minuto. Divide by 231 to convert to GPM: 19,64 (piston area) x 178 = 3497
Cubic inches per minute. 3437 + 251 = 14.88 GPM, This checks very closely with
15 GPM at 174 inches per minute shown on the chart on page 17.
To find the retraction speed on 14.88 GPM, the net piston area must be
found. This is he full piston area minus the rod area: 19,64 (piston area) ~ 6.5.
(rod area) = 16.5 square inches. The flow rate is 3437 cube inches per minuts
vo 14.58 GPAD + 16.5 (net area) = 208 inches per minute Note that
this is faster than the extension speed un the same oll flow
Speeds on pages 17 and 18
€ formula which nays that “speed te
divided by the
14
Hydraulic Cylinder Force
Low Pressure Range - 500 to 1500 PSI - 144"to 14° Bores
Cylinder forces, both extension and retraction, are shown in pounds. The
chart on this page covers cslinder operation inthe pressure range of 300 to 1500
PS and the chart on the next page covers the 2000 to 5000 PSI range. Lines in
bold type show extension foree, using the full piston area, Lines in italie type
Show retraction force with various size piston rods
Remember that force values are theoretical, derived by calculation, Experi-
ence has shown that probably 9%, but cerainy no more than 10% additional
resaure will be required to make up cylinder losses.
Por pressures not shown, the effective piston areas in the third column can
be tised'a power factors. Multiply effective area times (continued on page 16)
Bore) Rod Pressure Differential Across Cylinder Ports
Dias | Di.
| m SOOPSI 7S0PSI 100OPSI 1250PSI_1500PSI
1% | Hans" a VE za 2651
E > To age, 28
7 Fe OO OI Nr
2 | None mE RT AR
Y eer e le E
1 MS IS Of 01 Bas
2% | None" EC 3562 6136
N 2% Me a din
e E E
i E E
3 | None: Pre ee
Y se TE abe
be Bee dé u)
he 288 OM ates 5800
sa | None sg m ms 10370
le Se Se en ‘u
pe E Zag
2 FE Bisa Baas
a | None Pe 18709
"8 E Me Boas
3 Fe u}
bn Spat Ed
5 | Non wo 14726 24.508
x Be 2 Er
Be Beg Wan 78408
3 es Usas 13703
du Soor 50 12918
6 | None: | 28.274) 14137 21206 3528
HER 7 328
HE 36508
1389 Ft
11780 sass
7 28564 49,108
E37 Edad
Bid Er
18 288
16958 3
14:38 38 828
37700 52893 75,399
5 Er 5% Se
2 Gin Sse
772 LE a
235 Bee ear
10.881 ES
10 908 ws 117810
2% BE ‘Boe
BY Be de
Hoge Bie BIB
30 Er
2 110) 55550 84,825 141375 169.850
| S38 870 Me 1208
14 |Noner| 15394 | 76970 115455 192425 230910
Mone’ | NES] 3 1555 FRS 731%
"These figures are for extension force. N
15
o piston rod diameter is involved,
Low Pressure Range - 500 to 1500 PSI - 144"to 14° Bores
Cylinder forces, both extension and retraction, are shown in pounds. The
chart on this page covers cslinder operation inthe pressure range of 300 to 1500
PS and the chart on the next page covers the 2000 to 5000 PSI range. Lines in
bold type show extension foree, using the full piston area, Lines in italie type
Show retraction force with various size piston rods
Remember that force values are theoretical, derived by calculation, Experi-
ence has shown that probably 9%, but cerainy no more than 10% additional
resaure will be required to make up cylinder losses.
Por pressures not shown, the effective piston areas in the third column can
be tised'a power factors. Multiply effective area times (continued on page 16)
Bore) Rod Pressure Differential Across Cylinder Ports
Dias | Di.
| m SOOPSI 7S0PSI 100OPSI 1250PSI_1500PSI
1% | Hans" a VE za 2651
E > To age, 28
7 Fe OO OI Nr
2 | None mE RT AR
Y eer e le E
1 MS IS Of 01 Bas
2% | None" EC 3562 6136
N 2% Me a din
e E E
i E E
3 | None: Pre ee
Y se TE abe
be Bee dé u)
he 288 OM ates 5800
sa | None sg m ms 10370
le Se Se en ‘u
pe E Zag
2 FE Bisa Baas
a | None Pe 18709
"8 E Me Boas
3 Fe u}
bn Spat Ed
5 | Non wo 14726 24.508
x Be 2 Er
Be Beg Wan 78408
3 es Usas 13703
du Soor 50 12918
6 | None: | 28.274) 14137 21206 3528
HER 7 328
HE 36508
1389 Ft
11780 sass
7 28564 49,108
E37 Edad
Bid Er
18 288
16958 3
14:38 38 828
37700 52893 75,399
5 Er 5% Se
2 Gin Sse
772 LE a
235 Bee ear
10.881 ES
10 908 ws 117810
2% BE ‘Boe
BY Be de
Hoge Bie BIB
30 Er
2 110) 55550 84,825 141375 169.850
| S38 870 Me 1208
14 |Noner| 15394 | 76970 115455 192425 230910
Mone’ | NES] 3 1555 FRS 731%
"These figures are for extension force. N
15
o piston rod diameter is involved,
Hydraulic Cylinder Force
High Pressure Range - 2000 to 5000 PSI - 114"10 14° Bores
(continued from page 15) pressure to obtain eylinder force produeed. Values in
{two or more columns can be added for pressure not listed, or, force values can
be obtained by interpolating between the next higher and the next lower pres:
sure columns,
‘Pressure values along the top, of each chart are differential pressures across
the two evlinder ports. This ts the pressure to just balance the load, and not the
resaure which must be produced by the system pump. There will be circuit ow
[ocios in pressure and return lines due to ol flaw, and these will require extra
pressure. When designing a system, be sure to allow sufficient pump pressure,
probably an extra 28% to 30% on the average, to supply both the cylinder and
{satisfy system flow losses.
Pressure Differential Across Cylinder Ports
2000 PSI_2500 PSI 2000 PSI_4000 PSI_5000 PSI
3534 7069 — 8838
E EA
1964 3927 4009
6283 12566 15,708
oe des 1
3313 EN
9087 | 9817 19.595 2454
wore’) 29 | Sear 18493 20017
m | 34298) 6e 13608 17118
m | 2s034| S007 Wor 12517
3 | Noner| 70886 | 14,197 28274 35343
vere’) ee | 1230 Biss Sie
m | 55837 | 15.107 3 27919
me | 2008 2302
3% | None | 82958 | 16.5 41479
me | 6810 | 1382 pr
m [80903 | 11281 238
2° 187882) 10308 27H
4 | Momer| 12567 | 25.134 as
ter’ | 1082| 202 30810
2, [os] 1881
bn | 70589 | 18317
5 | Noner | 19.635 | 39270
16.493 | 32308
= | 17%] 240
Be 35139
Se | 1001 | 20028
6 | Noner| 28.274 | 56548
Dee | 25508 120
3 1% 239
Se || 37509
2 | 1207| Stata
7 | Noner| 38485 | 76,970
| e
3+ | 20804 | 87728
32 | 33818) 21e
de | Bar| Ge
3° |f80| 37700
8 | Noner| 50266 | 100532
32e | 2085| 81290
a. | eos | 75308
de [eso 67
3 | Soest 8122
3% | 26508 | 53016
10 | None: | 78,540 | 157.080
prod | ee) 128272
5 | $8908 | 117810
Su | 54782 | 109564
7° | Zoos | "so
12 | None*| 11310
52° | 59330 | 178678
14 | Noner| 153.98 | 207,880
pone’ | 11826 | 2301920
“These Hares are for extension force, No piston rod diameter is involved.
16
High Pressure Range - 2000 to 5000 PSI - 114"10 14° Bores
(continued from page 15) pressure to obtain eylinder force produeed. Values in
{two or more columns can be added for pressure not listed, or, force values can
be obtained by interpolating between the next higher and the next lower pres:
sure columns,
‘Pressure values along the top, of each chart are differential pressures across
the two evlinder ports. This ts the pressure to just balance the load, and not the
resaure which must be produced by the system pump. There will be circuit ow
[ocios in pressure and return lines due to ol flaw, and these will require extra
pressure. When designing a system, be sure to allow sufficient pump pressure,
probably an extra 28% to 30% on the average, to supply both the cylinder and
{satisfy system flow losses.
Pressure Differential Across Cylinder Ports
2000 PSI_2500 PSI 2000 PSI_4000 PSI_5000 PSI
3534 7069 — 8838
E EA
1964 3927 4009
6283 12566 15,708
oe des 1
3313 EN
9087 | 9817 19.595 2454
wore’) 29 | Sear 18493 20017
m | 34298) 6e 13608 17118
m | 2s034| S007 Wor 12517
3 | Noner| 70886 | 14,197 28274 35343
vere’) ee | 1230 Biss Sie
m | 55837 | 15.107 3 27919
me | 2008 2302
3% | None | 82958 | 16.5 41479
me | 6810 | 1382 pr
m [80903 | 11281 238
2° 187882) 10308 27H
4 | Momer| 12567 | 25.134 as
ter’ | 1082| 202 30810
2, [os] 1881
bn | 70589 | 18317
5 | Noner | 19.635 | 39270
16.493 | 32308
= | 17%] 240
Be 35139
Se | 1001 | 20028
6 | Noner| 28.274 | 56548
Dee | 25508 120
3 1% 239
Se || 37509
2 | 1207| Stata
7 | Noner| 38485 | 76,970
| e
3+ | 20804 | 87728
32 | 33818) 21e
de | Bar| Ge
3° |f80| 37700
8 | Noner| 50266 | 100532
32e | 2085| 81290
a. | eos | 75308
de [eso 67
3 | Soest 8122
3% | 26508 | 53016
10 | None: | 78,540 | 157.080
prod | ee) 128272
5 | $8908 | 117810
Su | 54782 | 109564
7° | Zoos | "so
12 | None*| 11310
52° | 59330 | 178678
14 | Noner| 153.98 | 207,880
pone’ | 11826 | 2301920
“These Hares are for extension force, No piston rod diameter is involved.
16
Hydraulic Cylinder Speeds
Figures in body
ton and rod di
Specds. Lines in italiefype are retraction speeds for the bore and rod diamete
chart are cylinder piston speeds in inches por minute. Pis
iametore are in inches, Horizontal ines in bold type are extension
shown in the frst two columns, using "net" piston area,
For Fluid Flows from 1 to 20 GPM.
Platon | Rod | 1 a 5 8 12 15 20
Diam. | Diam. | GPM Gem GPM GPM GPM pm apm
1% Toner] 131 -
E =
i 238 E
2 [None] 7a
sa | be
1 E
| 19 2
2% | None | 47 E
1 56 220 3
# | à :
| 9 de E
3 [None] 33 163 5
1 S 1 735
me | # 208 Ed
> 5 204 :
3% | None | 28 13 557
te | 32 10 ee
m | & 196 fes
2 de ts 22 30 896
SA [Non] A 72 120 19% 480
| 2% as 1% 20 =
m | B D 16 28 Go
2 % 10 1 2% 713
da 195 222 86 890,
7 CES 28
2% 6 1 1e 207
2 6 14 1& des
a A 18 16 290
D 0 m 24 608
35 108 Va 2 697
5 235 5994 2
% 3 8 10 250
a 8 © Me 280
nz m 125 3
19 u d 1 Ses
3 8 16 18 Ei
y RE EE | 163
5 À 4 7 LA
D D 8 7% 198
M D Ba 218
2 7 & % Sas
5 A 18 298
7 60183048 m
% Be 8 17
do À do 6 160
3 ss 7 e 7 178
is mo 3 e 208
5 12 37 or. 28
3} None} 46 1 23 97 9
| so 1 2 4 6
81 eB 123
de | 67 2 D 8% 1
3 A % © 151
de | a7 2% 4 % 178
vo [Non] 29 86 15 A EJ
E E] FA]
5 an 2 à 7
ge | 2 8 À & 3
2 56 17 8 % its
"These figures are for extension speed, No piston rod diameter is involved.
17
Figures in body
ton and rod di
Specds. Lines in italiefype are retraction speeds for the bore and rod diamete
chart are cylinder piston speeds in inches por minute. Pis
iametore are in inches, Horizontal ines in bold type are extension
shown in the frst two columns, using "net" piston area,
For Fluid Flows from 1 to 20 GPM.
Platon | Rod | 1 a 5 8 12 15 20
Diam. | Diam. | GPM Gem GPM GPM GPM pm apm
1% Toner] 131 -
E =
i 238 E
2 [None] 7a
sa | be
1 E
| 19 2
2% | None | 47 E
1 56 220 3
# | à :
| 9 de E
3 [None] 33 163 5
1 S 1 735
me | # 208 Ed
> 5 204 :
3% | None | 28 13 557
te | 32 10 ee
m | & 196 fes
2 de ts 22 30 896
SA [Non] A 72 120 19% 480
| 2% as 1% 20 =
m | B D 16 28 Go
2 % 10 1 2% 713
da 195 222 86 890,
7 CES 28
2% 6 1 1e 207
2 6 14 1& des
a A 18 16 290
D 0 m 24 608
35 108 Va 2 697
5 235 5994 2
% 3 8 10 250
a 8 © Me 280
nz m 125 3
19 u d 1 Ses
3 8 16 18 Ei
y RE EE | 163
5 À 4 7 LA
D D 8 7% 198
M D Ba 218
2 7 & % Sas
5 A 18 298
7 60183048 m
% Be 8 17
do À do 6 160
3 ss 7 e 7 178
is mo 3 e 208
5 12 37 or. 28
3} None} 46 1 23 97 9
| so 1 2 4 6
81 eB 123
de | 67 2 D 8% 1
3 A % © 151
de | a7 2% 4 % 178
vo [Non] 29 86 15 A EJ
E E] FA]
5 an 2 à 7
ge | 2 8 À & 3
2 56 17 8 % its
"These figures are for extension speed, No piston rod diameter is involved.
17
Cylinder Speeds for Fluid Flows From 25 to 100 GPM
Piston | Rod | 25 30 40 50 60 75 100
Diam. | Diem. | GPM GPM GPM __GPM__GPM__GPM_ GPM
3 | Noner 87 DE
1 9 2
3% 596
End
880
Es 600
E
Edo -
ES
7 460
508
el
2 |
2 | am -
LE
we | aes
2 50
e
3 360
Se | 57 :
LE: ss
ee | es m de 70
2 7 Bor Ss 7
3 E as 7
3s | 30 Se 40% B19 7 %
3 dei Ses 7 32 +
T Tone | 150 180 240 300 360 450
3 Ta OZ 2 OS HI 1
Bs E 10 40 do
3 es 2er 7 He 29 0
de 307 dos SI 4 7%
5 0 3es 30 Bi 7 919
ST Were} “115138 18 20 20 345
gee | tae m 27 24 HI Saw
3 1 leg AS 300 de 260 83
dre | 188 m m 39 10 ns 62
3 109 28 NE
Se | 218 21 m 406 53 es am
TO [Nom] 74 88 119 147 Ve 21 24
ge) & M as M 21 27 3
5 ig ler 6 23 24 3%
Se | 108 12 le m 23 Ste 4
7 iat 179 21 Zu 3 35 57
12 [ion] Si 61 2102123153204
ge | & 78 103 1% 188 In 2
2 mn D 4 m & 2% 30
es | 102 15 16 208 26 307 “o
va [ome] 38 45 6 78 99 113. 180
7 3 6 9 100 120 10 20
as | 3 7 2 m9 4 Te 2%
10 7 2 123 © 10 20 3%
"These figures are for extension speed. No piston rod diameter is involved.
Interpolation of Cylinder Speed Charts
Cylinder speed is directly proportional to GPM. To find
shown in charts, add speed in two columns Example: Spec
the sum of speeds in the 5 GPM and 30 GPM col
Calculation of Cylinder Speed
‘These charts were calculated from the formula: 5 = CIM + A, in which S it
iston speed in inches per minute; CIM is flow in cubic inches per minute: and
Ris ross aoctional area of cavity being filled in square inches. GPM Hows must
à at a flow not
36 GPM is
‘be converted to cubic inches per minute. Multiply GPM times 231
Esto
ion speeds are calculated with full piston area, retraction speeds with
‘whichis piston area minus rod arca. (See Hydraulle Cylinder Force)
18
Piston | Rod | 25 30 40 50 60 75 100
Diam. | Diem. | GPM GPM GPM __GPM__GPM__GPM_ GPM
3 | Noner 87 DE
1 9 2
3% 596
End
880
Es 600
E
Edo -
ES
7 460
508
el
2 |
2 | am -
LE
we | aes
2 50
e
3 360
Se | 57 :
LE: ss
ee | es m de 70
2 7 Bor Ss 7
3 E as 7
3s | 30 Se 40% B19 7 %
3 dei Ses 7 32 +
T Tone | 150 180 240 300 360 450
3 Ta OZ 2 OS HI 1
Bs E 10 40 do
3 es 2er 7 He 29 0
de 307 dos SI 4 7%
5 0 3es 30 Bi 7 919
ST Were} “115138 18 20 20 345
gee | tae m 27 24 HI Saw
3 1 leg AS 300 de 260 83
dre | 188 m m 39 10 ns 62
3 109 28 NE
Se | 218 21 m 406 53 es am
TO [Nom] 74 88 119 147 Ve 21 24
ge) & M as M 21 27 3
5 ig ler 6 23 24 3%
Se | 108 12 le m 23 Ste 4
7 iat 179 21 Zu 3 35 57
12 [ion] Si 61 2102123153204
ge | & 78 103 1% 188 In 2
2 mn D 4 m & 2% 30
es | 102 15 16 208 26 307 “o
va [ome] 38 45 6 78 99 113. 180
7 3 6 9 100 120 10 20
as | 3 7 2 m9 4 Te 2%
10 7 2 123 © 10 20 3%
"These figures are for extension speed. No piston rod diameter is involved.
Interpolation of Cylinder Speed Charts
Cylinder speed is directly proportional to GPM. To find
shown in charts, add speed in two columns Example: Spec
the sum of speeds in the 5 GPM and 30 GPM col
Calculation of Cylinder Speed
‘These charts were calculated from the formula: 5 = CIM + A, in which S it
iston speed in inches per minute; CIM is flow in cubic inches per minute: and
Ris ross aoctional area of cavity being filled in square inches. GPM Hows must
à at a flow not
36 GPM is
‘be converted to cubic inches per minute. Multiply GPM times 231
Esto
ion speeds are calculated with full piston area, retraction speeds with
‘whichis piston area minus rod arca. (See Hydraulle Cylinder Force)
18
Pneumatic Cylinder Air Consumption
‚The purpose of estimating air consumption ofa cylinder is usually to find the
HP capacity which must be available from the air compressor to operate the cp
inder on a continuous eyeling application
‘Air consumption can be estimated from the table below The consumption
can then he converted into compressor HP.
Using the Table to Calculate Air Consumption
Figures in the body of the table are air consumptions for elinders with stan
dard diameter piston roda. The saving of air for eslinders with larger diameter
rods is negligible for most calculations
‚Air consumption was calculated assuming the eylinder piston will be allowed
to stall at least momentarily, at each end of its stroke, giving it time to fil up
With ai to the pressure regulator setting. IP revorsed at either end of a stroke
Before full stall occurs, air consumption will be less than shown in the table
“The frst step inthe calculation a tobe sure that the bore size of the selected.
cylinder will just Balance the load at a pressure of 75% or less of the maximum
pressure available to the system. This leaves about 25% of available pressure
{whieh can be used to overcome flow losses through piping and valving. his sur
plus pressure must be available or the eylinder cannot travel at normal speed.
‘Determine the exact air pressure needed to just balance the load resistance.
Add about 25% for How lostes and set the system regulator to this pressure.
‘This is also the pressure figure which should be used when going into the tabl
‘Aller determining the Feyulator pressure, go into the proper column of the
table, The figure shown in the table fe the ait consumption fora J-inch stroke,
forward and return. Take this figure and multiply times the number of inches
of troke and by the number of eomplete cycles, forward and back which the yi
Inder is expected to make in one minute. This gives the SCFM for the applica:
tion
‘Cylinder Air Consumption per 1-inch Stroke, Forward and Return
Regulator Outlet PSI (At Least 25% Above Load Balance PSI)
Go 70 80 90 100 110 120 130 140 150
PSI PSI PSI PSI PSI PSI PSI PSI PSI PSI
150 009.010 012 013 015 016 OI 018 020 02
200 018 020 022 .025 .027 029 032 (05 026 039
250 028 032 035 039 .043 047 050 054 058.062
300 039 044 050 055 060 066 070 078 081 087
325 046 053 (059 065 .071 .078 084 090 .09 102
400 072 081 091 100 110 19 129 199 148 188
500 113 128 143 169 174 189 204 219 29% 249
600 162 164 208 227 249 270 202 914 995 367
800 201 330 389 408 447 486 525 564 502 602
100 455 516 576 007 698 759 820 881 940 100
120 656 744 831 919 101 109 118 127 196 145
140 800 101 113 125 197 149 161 172 184 19
Converting SCFM Into Compressor HP.
Compression of air is an inefficient process because part of the energy is lost
‘as heat of compression and can never be recovered. By over-compressing the air
‘and then reducing it to a lower pressure through a regulator, the system losses
fre increased. The amount of this loss is nearly impossible Lo calculate, but on
the average system may amount to 5 or 10%. Also, there is à small loss due to
flow resistance through the regulator.
‘after finding the SCFM o operate the cinder, refer to the tales on page
30. Convert into HP according to the kind of compressor used. Add 5 to 10% for
the miscellancous losses deseribed above, This should be very lose to the actual
HP capacity needed,
EXAMPLE: Find compressor HP needed to cycle an air cylinder through a 28-
inch stroke, 11 times a minute, against a load resistance of 1000 Ibs,
‘SOLUTION: À 4 bore cvlinder working at 80 PSI would balance the 1000 lb
load Add 25% more pressure (20 PSI) and set the pressure regulator at 100 PSI
From the above table, air consumption would be
0.110 x 28 (stroke) x 11 (times a minute) = 33.88 SCFM
Refer to page 30. Assume a 2-staxe compressor. At 100 PSI, 0.164 HP is re.
‚ired for each 1 SCFM. Total HP = 0.163 33.98 = 5.57 HP. Add 5% (or 0.278
IP) for miscellaneous losses. Total compressor HP = 5.97 + 0.278 = 5.848 HP.
19
‚The purpose of estimating air consumption ofa cylinder is usually to find the
HP capacity which must be available from the air compressor to operate the cp
inder on a continuous eyeling application
‘Air consumption can be estimated from the table below The consumption
can then he converted into compressor HP.
Using the Table to Calculate Air Consumption
Figures in the body of the table are air consumptions for elinders with stan
dard diameter piston roda. The saving of air for eslinders with larger diameter
rods is negligible for most calculations
‚Air consumption was calculated assuming the eylinder piston will be allowed
to stall at least momentarily, at each end of its stroke, giving it time to fil up
With ai to the pressure regulator setting. IP revorsed at either end of a stroke
Before full stall occurs, air consumption will be less than shown in the table
“The frst step inthe calculation a tobe sure that the bore size of the selected.
cylinder will just Balance the load at a pressure of 75% or less of the maximum
pressure available to the system. This leaves about 25% of available pressure
{whieh can be used to overcome flow losses through piping and valving. his sur
plus pressure must be available or the eylinder cannot travel at normal speed.
‘Determine the exact air pressure needed to just balance the load resistance.
Add about 25% for How lostes and set the system regulator to this pressure.
‘This is also the pressure figure which should be used when going into the tabl
‘Aller determining the Feyulator pressure, go into the proper column of the
table, The figure shown in the table fe the ait consumption fora J-inch stroke,
forward and return. Take this figure and multiply times the number of inches
of troke and by the number of eomplete cycles, forward and back which the yi
Inder is expected to make in one minute. This gives the SCFM for the applica:
tion
‘Cylinder Air Consumption per 1-inch Stroke, Forward and Return
Regulator Outlet PSI (At Least 25% Above Load Balance PSI)
Go 70 80 90 100 110 120 130 140 150
PSI PSI PSI PSI PSI PSI PSI PSI PSI PSI
150 009.010 012 013 015 016 OI 018 020 02
200 018 020 022 .025 .027 029 032 (05 026 039
250 028 032 035 039 .043 047 050 054 058.062
300 039 044 050 055 060 066 070 078 081 087
325 046 053 (059 065 .071 .078 084 090 .09 102
400 072 081 091 100 110 19 129 199 148 188
500 113 128 143 169 174 189 204 219 29% 249
600 162 164 208 227 249 270 202 914 995 367
800 201 330 389 408 447 486 525 564 502 602
100 455 516 576 007 698 759 820 881 940 100
120 656 744 831 919 101 109 118 127 196 145
140 800 101 113 125 197 149 161 172 184 19
Converting SCFM Into Compressor HP.
Compression of air is an inefficient process because part of the energy is lost
‘as heat of compression and can never be recovered. By over-compressing the air
‘and then reducing it to a lower pressure through a regulator, the system losses
fre increased. The amount of this loss is nearly impossible Lo calculate, but on
the average system may amount to 5 or 10%. Also, there is à small loss due to
flow resistance through the regulator.
‘after finding the SCFM o operate the cinder, refer to the tales on page
30. Convert into HP according to the kind of compressor used. Add 5 to 10% for
the miscellancous losses deseribed above, This should be very lose to the actual
HP capacity needed,
EXAMPLE: Find compressor HP needed to cycle an air cylinder through a 28-
inch stroke, 11 times a minute, against a load resistance of 1000 Ibs,
‘SOLUTION: À 4 bore cvlinder working at 80 PSI would balance the 1000 lb
load Add 25% more pressure (20 PSI) and set the pressure regulator at 100 PSI
From the above table, air consumption would be
0.110 x 28 (stroke) x 11 (times a minute) = 33.88 SCFM
Refer to page 30. Assume a 2-staxe compressor. At 100 PSI, 0.164 HP is re.
‚ired for each 1 SCFM. Total HP = 0.163 33.98 = 5.57 HP. Add 5% (or 0.278
IP) for miscellaneous losses. Total compressor HP = 5.97 + 0.278 = 5.848 HP.
19
Pneumatic Cylinder Force
Extension and Retraction - 60 to 130 PSI Pressure Range
Ccytinder forces are shown in pounds for bath extension and retracton
in bold type show extension forces, using the full pston reo, Lines in ae
type show retraction forces with various Sie piston rods, Remember that force
Fe re ain rived by clean i
essor along the top ofthe chart do not represent air supply pressure:
they are differential pressures across te two eyinder pots fn practice the air
Supply line must supply another Bf of pressure to make up for cylinder lo
‘ind must supply an estimated 29 to 50% additional pressure to make up for low
Fosses in tines and valving o the cylinder will have aulicien travel speed
or good design and highest circuit slcioney open theeslinder speed control
volves da wide as practical and reduce the pressure regulator ating lo ar low
pressure as vil iv star) eyiner force and sped.
For pressures not shown, une the effective arean in the third column as pow
cer factors, Multiply effective area times differential pressure to obtain theare.
{cal eplinder fore
Piston Rod Etec.
Dia, Die. Am] 60 70 99 90 100 110 120 190
GENES PS PSI PSI PSI
1% Wone 177 | 108124142 19 17 195 A7 230
Se” tae | "se 102 117 182 lo 161 176 190
NI u Bu 18 12
1% None 241 | 144 168 192 216 201 205 200 319
A EEE 20
im Tie | 7 a 10 18 10 142 14
2 None ata | 198 220 251 28% 314 248 37 40
Se 205 | Vo fe 27 E EEE
7° 353 | 141 165 168 212 23 259 283 306
2% None 491 | 295 mu 3093 442 491 540 589 63%
Se" Ye | 2e de Gee 14 40 00 582 S00
i) 493 | 27 ze 390 m 2 a 46 20
im 343 | 26 20 24 6 u 57 a 4
a None 707 | 424 495 565 5% 707 778 am 91
Wore G25 | Sr MO S03 Ses Gee or 74 BIT
hu 460 | 200 320 373 420 486 91 50 606
3% None 820 | 498 Set 668 7a7 80 919 996 107
wore st | 41 S26 G01 67% 751 827 902 97
1m Ge | Soo 377 Sip 61 61 70 818806
Fi Sao | de dis 472 a Se Gee 707 706
3% None 962 | 577 era 770 086 962 1050 1155 1251
wore Bes | S20 Sie 707 795 ass 972 1000 1149
4 None 1257 | 754 990 1008 1131 1257 1982 1508 1634
one al 7or gas DS JO! 1178 1296 1418 1802
me 1100| des 776 807 "G09 1100 1219 1990 1941
i 1010| 610 712 813 918 101 1118 1220 1321
S Nono 1964) 1178 1975 1871 1768 1968 2160 2897 2058
ore 188 | vat 1320 1208 1097 1885 2074 2289 2401
han 1816 1089 1271 1482 183% 1919 1997 2179 2960
6 None 2827 | 1696 1979 2262 2544 2827 3110 3992 3675
Vu 2870| 1807 1075 2143 2411 2079 2048 3214 ME
Fi 2390| 1982 101) 2009 Zus 2008 2045 310% 3002
7 None 3640/2900 2698 3079 3464 3849 4234 4619 S004
Vue 3701| 3220 2590 2060 3801 SOI 4071 aa dt
8 None 5027| 3018 3519 022 4524 5027 5530 6032 6535
Vue ao | 207 HIS 3009 Ser 7 S300 a Ge
m 3700| 2072 3351 3029 4308 S706 020 Va 0222
10 None 7854712 5498 6289 7059 7858 0539 9425 10210
17° 7ora| As 5920 6091 Ge ram 237 9195 cas
2" 7800| 4524 3278 0e 6766 7510 200 908 0002
12 None 113: | 6786 7017 9048 10179 11910 12481 13572 14709
Done ño seas 7897 2797 906 10009 12095 13105 11205
Zu 1082| 6401 7573 8655 8737 10819 11901 12983 14075
14 None 1530| 9204 10779 12912 13881 15900 16829 18468 20007
DA 1490| 8080 10420 11910 13400 1490 16389 17879 19360
5° 1468| 8810 10278 11747 12215 14083 16151 17620 19088
20
Extension and Retraction - 60 to 130 PSI Pressure Range
Ccytinder forces are shown in pounds for bath extension and retracton
in bold type show extension forces, using the full pston reo, Lines in ae
type show retraction forces with various Sie piston rods, Remember that force
Fe re ain rived by clean i
essor along the top ofthe chart do not represent air supply pressure:
they are differential pressures across te two eyinder pots fn practice the air
Supply line must supply another Bf of pressure to make up for cylinder lo
‘ind must supply an estimated 29 to 50% additional pressure to make up for low
Fosses in tines and valving o the cylinder will have aulicien travel speed
or good design and highest circuit slcioney open theeslinder speed control
volves da wide as practical and reduce the pressure regulator ating lo ar low
pressure as vil iv star) eyiner force and sped.
For pressures not shown, une the effective arean in the third column as pow
cer factors, Multiply effective area times differential pressure to obtain theare.
{cal eplinder fore
Piston Rod Etec.
Dia, Die. Am] 60 70 99 90 100 110 120 190
GENES PS PSI PSI PSI
1% Wone 177 | 108124142 19 17 195 A7 230
Se” tae | "se 102 117 182 lo 161 176 190
NI u Bu 18 12
1% None 241 | 144 168 192 216 201 205 200 319
A EEE 20
im Tie | 7 a 10 18 10 142 14
2 None ata | 198 220 251 28% 314 248 37 40
Se 205 | Vo fe 27 E EEE
7° 353 | 141 165 168 212 23 259 283 306
2% None 491 | 295 mu 3093 442 491 540 589 63%
Se" Ye | 2e de Gee 14 40 00 582 S00
i) 493 | 27 ze 390 m 2 a 46 20
im 343 | 26 20 24 6 u 57 a 4
a None 707 | 424 495 565 5% 707 778 am 91
Wore G25 | Sr MO S03 Ses Gee or 74 BIT
hu 460 | 200 320 373 420 486 91 50 606
3% None 820 | 498 Set 668 7a7 80 919 996 107
wore st | 41 S26 G01 67% 751 827 902 97
1m Ge | Soo 377 Sip 61 61 70 818806
Fi Sao | de dis 472 a Se Gee 707 706
3% None 962 | 577 era 770 086 962 1050 1155 1251
wore Bes | S20 Sie 707 795 ass 972 1000 1149
4 None 1257 | 754 990 1008 1131 1257 1982 1508 1634
one al 7or gas DS JO! 1178 1296 1418 1802
me 1100| des 776 807 "G09 1100 1219 1990 1941
i 1010| 610 712 813 918 101 1118 1220 1321
S Nono 1964) 1178 1975 1871 1768 1968 2160 2897 2058
ore 188 | vat 1320 1208 1097 1885 2074 2289 2401
han 1816 1089 1271 1482 183% 1919 1997 2179 2960
6 None 2827 | 1696 1979 2262 2544 2827 3110 3992 3675
Vu 2870| 1807 1075 2143 2411 2079 2048 3214 ME
Fi 2390| 1982 101) 2009 Zus 2008 2045 310% 3002
7 None 3640/2900 2698 3079 3464 3849 4234 4619 S004
Vue 3701| 3220 2590 2060 3801 SOI 4071 aa dt
8 None 5027| 3018 3519 022 4524 5027 5530 6032 6535
Vue ao | 207 HIS 3009 Ser 7 S300 a Ge
m 3700| 2072 3351 3029 4308 S706 020 Va 0222
10 None 7854712 5498 6289 7059 7858 0539 9425 10210
17° 7ora| As 5920 6091 Ge ram 237 9195 cas
2" 7800| 4524 3278 0e 6766 7510 200 908 0002
12 None 113: | 6786 7017 9048 10179 11910 12481 13572 14709
Done ño seas 7897 2797 906 10009 12095 13105 11205
Zu 1082| 6401 7573 8655 8737 10819 11901 12983 14075
14 None 1530| 9204 10779 12912 13881 15900 16829 18468 20007
DA 1490| 8080 10420 11910 13400 1490 16389 17879 19360
5° 1468| 8810 10278 11747 12215 14083 16151 17620 19088
20
‘This table
of the table are internal
Internal Fluid PSI on Tubing
for use in selecting wall thickness of tui
fluid pressures in PSI that wi
Figures inthe body
Produce a fiber stress
{10,000 PSI along the circumference, tending to rupture the tabing. fa tube
iS made of steel with an ultimate strength of 40,000 PSI, the safety factor would
bel
ina Thickness of Tubing Wall, Inches
0. |.120 . 219 250 319 375 500 .625 750 875 1.000]
“1% [1600 2080 2493 2920 9333 4173 5000 6667 8333
10% [1371 1783 2197 2503 2857 3577 4286 5714 7143
2 [1200 1560 1870 2190 2500 3130 3750 5000 6250
2% [1067 1387 1662 1947 2222 2782 3393 4444 5556
21 | 960 1248 1496 1752 2000 2504 3000 4000 5000
2% | 873 1135 1360 1598 1818 2276 2727 3636 4545
3 | 800 1040 1247 1460 1667 2087 2500 3339 4167
3% | 738 960 1151 1348 1538 1926 2308 3077 3848
an | 686 891 1060 1251 1429 1789 2143 2867 3571
3% | 640 832 997 1168 1333 1669 2000 2667 3333 4000 4867 5333,
4 | 600 780 935 1095 1250 1565 1875 2500 3125 3750 4375 5000
an | 565 734 880 1031 1176 1473 1765 2353 2941 3529 4118 4706
4% | 593 698 891 973 1111 1991 1667 2222 2778 3933 3889 4044
am | 505 657 787 922 1053 1918 1579 2105 2692 3158 3684 4211
s | 480 624 748 876 1000 1252 1500 2000 2500 3000 3500 4000
ES 796 909 1198 1364 1818 2273 2727 3182 3696
6 730 833 1043 1250 1667 2088 2500 2017 3893
or 674 769 963 1154 1538 1923 2308 2692 3077
7 626 714 804 1071 1429 1786 2143 2500 2857
m 584 667 835 1000 1333 1667 2000 2333 2667
8 548 625 783 908 1250 1563 1875 2188 2500
ES 515 588 736 882 1176 1471 1765 2059 2353
9 696 893 1111 1389 1667 1944 2222
9% 659 789 1052 1916 1579 1842 2105
10 626 750 1000 1250 1500 1750 2000
10% 506 714 962 1190 1429 1667 1905
‚The above table is calculated by Barlow's formula: P = 21x S + O in which P
tis wall thickness of tubing in inches, Sis the
Über stress allowable in the tubin
is the internal pressure in
ins
à value of 10,000 PS
(O's the outside diameter of the tubing in inches,
21
sed in the
of the table are internal
Internal Fluid PSI on Tubing
for use in selecting wall thickness of tui
fluid pressures in PSI that wi
Figures inthe body
Produce a fiber stress
{10,000 PSI along the circumference, tending to rupture the tabing. fa tube
iS made of steel with an ultimate strength of 40,000 PSI, the safety factor would
bel
ina Thickness of Tubing Wall, Inches
0. |.120 . 219 250 319 375 500 .625 750 875 1.000]
“1% [1600 2080 2493 2920 9333 4173 5000 6667 8333
10% [1371 1783 2197 2503 2857 3577 4286 5714 7143
2 [1200 1560 1870 2190 2500 3130 3750 5000 6250
2% [1067 1387 1662 1947 2222 2782 3393 4444 5556
21 | 960 1248 1496 1752 2000 2504 3000 4000 5000
2% | 873 1135 1360 1598 1818 2276 2727 3636 4545
3 | 800 1040 1247 1460 1667 2087 2500 3339 4167
3% | 738 960 1151 1348 1538 1926 2308 3077 3848
an | 686 891 1060 1251 1429 1789 2143 2867 3571
3% | 640 832 997 1168 1333 1669 2000 2667 3333 4000 4867 5333,
4 | 600 780 935 1095 1250 1565 1875 2500 3125 3750 4375 5000
an | 565 734 880 1031 1176 1473 1765 2353 2941 3529 4118 4706
4% | 593 698 891 973 1111 1991 1667 2222 2778 3933 3889 4044
am | 505 657 787 922 1053 1918 1579 2105 2692 3158 3684 4211
s | 480 624 748 876 1000 1252 1500 2000 2500 3000 3500 4000
ES 796 909 1198 1364 1818 2273 2727 3182 3696
6 730 833 1043 1250 1667 2088 2500 2017 3893
or 674 769 963 1154 1538 1923 2308 2692 3077
7 626 714 804 1071 1429 1786 2143 2500 2857
m 584 667 835 1000 1333 1667 2000 2333 2667
8 548 625 783 908 1250 1563 1875 2188 2500
ES 515 588 736 882 1176 1471 1765 2059 2353
9 696 893 1111 1389 1667 1944 2222
9% 659 789 1052 1916 1579 1842 2105
10 626 750 1000 1250 1500 1750 2000
10% 506 714 962 1190 1429 1667 1905
‚The above table is calculated by Barlow's formula: P = 21x S + O in which P
tis wall thickness of tubing in inches, Sis the
Über stress allowable in the tubin
is the internal pressure in
ins
à value of 10,000 PS
(O's the outside diameter of the tubing in inches,
21
sed in the
Piston Rod Column Strength
Long, slim piston rods may buckle if subjected to too heavy a push load. The
table below suggests the minimum diameter piston rod to use under various
conditions of load and unsupported rod length and is to be used in accordance
‘with the instructions in the next paragraph. There must be no side load or bend:
ing stress at any point along the rod.
How to Use the Table. Exposed rod length is shown along the top ofthe table
"This 1s usually somewhat longer than the actual stroke of the evlinder The ver.
tical scale, column 1, shows the load on the eslinder and is expressed in English
tons (1 ton = 2000 ls. IF both the end of the rod and the FRONT end of the
cylinder barrel are rigidly supported, a smaller rod will have sufheient column
Strength, and you may ute, as Exposed Length of Piston Rod, one-half of the
factual rod length.
For example, ifthe actual length is 80”, and ifthe cylinder barrel and rod
end are supported as described, you could enter the table in the column marked
40. On the other hand it hinge mounting is used on both cylinder and rod (pin
to-pin), you may not be safe in using actual exposed rod length, and should use
about twice the actual length, For example, if the actual length is 20°, You
Should enter the table in the 40° column,
Minimum Piston Rod Diameter
Figures in body of chart are suggested minimum rod diameters, in inches.
Exposed Length of Piston Rod, Inches
Tons [102040607080 100 120
ve wat
EN wis 1%
1 sm 78 ‘eH.
m A6 1616 mee
2 sa 1 % ve
3 | 136 m m m %
4 Lise 1 ee m 24
s |: Wa THe m 2
78 | m mom m 2 24
| we mo 2 2 zu
15 | 1% Ye 2% 2m 3
2 |2 2h mm 24 24 Er
30 | 2% 2% mm 2 3%
“| Mm 3 3 m EN
50 [34 3% SA 34 34 34 4
75 [su m m am a
ml mm dm 44 5
150 | 5% 5% 54 54 56 54 sn 6
CAUTION: Hinge mounted cylinders, when mounted horizontally or at an
angle other than serial create a bending stress onthe rod when extended de
the trunnion mounting rather than tang or levis mounting should be used, and
innion should De located in a position which will balance the eslinder
en extended.
22
Long, slim piston rods may buckle if subjected to too heavy a push load. The
table below suggests the minimum diameter piston rod to use under various
conditions of load and unsupported rod length and is to be used in accordance
‘with the instructions in the next paragraph. There must be no side load or bend:
ing stress at any point along the rod.
How to Use the Table. Exposed rod length is shown along the top ofthe table
"This 1s usually somewhat longer than the actual stroke of the evlinder The ver.
tical scale, column 1, shows the load on the eslinder and is expressed in English
tons (1 ton = 2000 ls. IF both the end of the rod and the FRONT end of the
cylinder barrel are rigidly supported, a smaller rod will have sufheient column
Strength, and you may ute, as Exposed Length of Piston Rod, one-half of the
factual rod length.
For example, ifthe actual length is 80”, and ifthe cylinder barrel and rod
end are supported as described, you could enter the table in the column marked
40. On the other hand it hinge mounting is used on both cylinder and rod (pin
to-pin), you may not be safe in using actual exposed rod length, and should use
about twice the actual length, For example, if the actual length is 20°, You
Should enter the table in the 40° column,
Minimum Piston Rod Diameter
Figures in body of chart are suggested minimum rod diameters, in inches.
Exposed Length of Piston Rod, Inches
Tons [102040607080 100 120
ve wat
EN wis 1%
1 sm 78 ‘eH.
m A6 1616 mee
2 sa 1 % ve
3 | 136 m m m %
4 Lise 1 ee m 24
s |: Wa THe m 2
78 | m mom m 2 24
| we mo 2 2 zu
15 | 1% Ye 2% 2m 3
2 |2 2h mm 24 24 Er
30 | 2% 2% mm 2 3%
“| Mm 3 3 m EN
50 [34 3% SA 34 34 34 4
75 [su m m am a
ml mm dm 44 5
150 | 5% 5% 54 54 56 54 sn 6
CAUTION: Hinge mounted cylinders, when mounted horizontally or at an
angle other than serial create a bending stress onthe rod when extended de
the trunnion mounting rather than tang or levis mounting should be used, and
innion should De located in a position which will balance the eslinder
en extended.
22
Horsepower to Drive a Pump
Figen he ny of hia shred dr
nn. er see
ne ren
ER N leere
ea susan sal ese ce
Using ne rable
RS 180 Pa cor most trae tems, but power e
pi BOL e
ran ar SEES
REN a ERP Ua OU
Tea ieee STE ces oe de
rns IS. pompa tendo came in.
antennae et a
Mule Of Thumb
na PE
TH enue foreach GPM 10001
For ese GPM pampering a 120 PS nal eed MP o at
so Canaries APN Pa an at
SE a al Lenk alo ee
See ed eae ES pump masa rated
noir etcquge ithe any ey anh Pa ee
SOSA at PER E
A gant à pr counele
8 can be figured with simple mental arith-
Figures in table are HP's required to drive a hydraulic pump.
500 750 1000 1250 1500 1760 2000 2500 3000 5000
cpm| PSI PSI PSI PSI Psi psi PSI PSI PSI PSI
We | 172257 343 428 518 600 688 858 103 172
{| 343 515 6e 858 103 120 197 172 206 343
ws | 515 772 103 129 158 180 206 257 309 518
2 | 686 103 137 172 206 240 275 343 412 686
2% | 868 120 172 214 297 300 343 420 518 858
3 | 103 154 206 257 309 360 412 518 618 103
au | 120 180 240 300 360 420 490 600 721 120
4 | 137 206 275 343 412 480 S49 686 82% 137
5 | 172 257 343 420 515 600 685 858 103 172
5 | 206 309 412 515 618 721 824 103 124 206
7 | 240 360 480 600 721 841 961 120 144 240
8 | 275 412 509 68 82 961 110 197 165 275
9 | 309 463 618 772 927 108 124 154 185 309
10 | 343 518 8886 858 103 120 137 172 208 343
12 | 412 618 824 103 124 144 105 206 247 412
15 | 515 772 103 129 154 180 208
2
25
30
35
257 309 515
686 103 137 172 206 240 275 343 412 688
858 129 172 214 257 300 343 420 515 858
103 154 206 257 309 960 412 515 618 103
120 180 280 300 380 420 480 800 721 120
40 | 137 206 275 343 212 460 549 686 824 197
4 | 154 232 309 386 463 541 618 772 927 154
50 | 172 257 343 429 513 600 6809 656 103 172
55 | 189 283 378 472 866 661 788 94 13 189
60 | 206 309 412 515 618 721 824 103 12% 206
65 | 223 335 446 588 609 781 092 112 1% 22
70 | 240 360 480 600 721 801 91 120 144 240
75 | 257 386 515 643 772 901 103 129 154 257
go | 275 412 549 686 824 961 110 197 165 275
85 | 202 498 583 729 875 102 117 146 175 22
go | 309 463 618 772 927 108 124 154 185 309
9 | 326 489 692 815 978 114 190 163 19 328
100 | 343 515 686 658 103 120 197 172 206 38
23
Figen he ny of hia shred dr
nn. er see
ne ren
ER N leere
ea susan sal ese ce
Using ne rable
RS 180 Pa cor most trae tems, but power e
pi BOL e
ran ar SEES
REN a ERP Ua OU
Tea ieee STE ces oe de
rns IS. pompa tendo came in.
antennae et a
Mule Of Thumb
na PE
TH enue foreach GPM 10001
For ese GPM pampering a 120 PS nal eed MP o at
so Canaries APN Pa an at
SE a al Lenk alo ee
See ed eae ES pump masa rated
noir etcquge ithe any ey anh Pa ee
SOSA at PER E
A gant à pr counele
8 can be figured with simple mental arith-
Figures in table are HP's required to drive a hydraulic pump.
500 750 1000 1250 1500 1760 2000 2500 3000 5000
cpm| PSI PSI PSI PSI Psi psi PSI PSI PSI PSI
We | 172257 343 428 518 600 688 858 103 172
{| 343 515 6e 858 103 120 197 172 206 343
ws | 515 772 103 129 158 180 206 257 309 518
2 | 686 103 137 172 206 240 275 343 412 686
2% | 868 120 172 214 297 300 343 420 518 858
3 | 103 154 206 257 309 360 412 518 618 103
au | 120 180 240 300 360 420 490 600 721 120
4 | 137 206 275 343 412 480 S49 686 82% 137
5 | 172 257 343 420 515 600 685 858 103 172
5 | 206 309 412 515 618 721 824 103 124 206
7 | 240 360 480 600 721 841 961 120 144 240
8 | 275 412 509 68 82 961 110 197 165 275
9 | 309 463 618 772 927 108 124 154 185 309
10 | 343 518 8886 858 103 120 137 172 208 343
12 | 412 618 824 103 124 144 105 206 247 412
15 | 515 772 103 129 154 180 208
2
25
30
35
257 309 515
686 103 137 172 206 240 275 343 412 688
858 129 172 214 257 300 343 420 515 858
103 154 206 257 309 960 412 515 618 103
120 180 280 300 380 420 480 800 721 120
40 | 137 206 275 343 212 460 549 686 824 197
4 | 154 232 309 386 463 541 618 772 927 154
50 | 172 257 343 429 513 600 6809 656 103 172
55 | 189 283 378 472 866 661 788 94 13 189
60 | 206 309 412 515 618 721 824 103 12% 206
65 | 223 335 446 588 609 781 092 112 1% 22
70 | 240 360 480 600 721 801 91 120 144 240
75 | 257 386 515 643 772 901 103 129 154 257
go | 275 412 549 686 824 961 110 197 165 275
85 | 202 498 583 729 875 102 117 146 175 22
go | 309 463 618 772 927 108 124 154 185 309
9 | 326 489 692 815 978 114 190 163 19 328
100 | 343 515 686 658 103 120 197 172 206 38
23
Torque/HP/Speed Relations
‘This chart can be used to find the torque, horsepower or speed of any kind of
drive (electric motor, hydraulic or pneumatic motor, engine, rotary actuator,
ete. if two of those three values are known, The chart i a tabular solution to
the following basic formulas:
To ind Horsepower (HP), use the formula: HP = IT RPM) + 5252
To find Torque in fe, Ibs. (7), use the formula: 1 = (HP x 8282) > RPM
To find Speed (RPM), uso the formula: APM = (HP x 5282) + Y
‘The figures in the body of the chart are torque values in foot/pounds
Pw (Revolutions Per Minuto)
MP | 100 500 750 1000 1200 18002400 3000 3600|
IS
VS | 175 350 234 175 148 117 (972 730 ‘see 488
1R| 263 525 350 263 220 175 146 110 875 730
ga] 394 787 524 394 328 262 218 164 131 109
1| 525 105 700 525 238 350 292 219 175 147
m| 788 187 105 788 656 528 438 328 263 219
2| 105 210 mo 105 876 700 ses am 350 282
3| 158 315 210 188 131 108 657 525 438
5| 263 525 350 283 220 178 148 110 87 730
m| ges 788 532 394 328 263 218 164 131 109
| 525 105 700 525 438 350 202 219 175 148
15| 788 158 105 788 656 526 438 328 265 219
20| 1050 210 140 105 876 700 584 438 350 292
25| 13138 263 178 181 110 877 730 518 438 385
30 | 1578 315 210 158 191 105 874 657 526 437
40] 2100 420 280 210 175 10 116 875 700 582
SO | 22% 523 350 283 220 1/5 lé 110 875 728
BO | 101 630 420 1 262 210 173 13% 103 874
75| 3940 788 525 398 328 262 218 164 191 109
100| S280 1060 700 525 398 350 202 219 175 148
125| 6565 1313 878 657 548 497 364 274 218 182
150 | 7878 1580 1050 788 656 526 408 328 265 219
200 19500 2100 100 1050 ae 70D sa A 30 2%
250 | 13130 1750 1910 1100 877 365
Pump and Motor Torque
‘This chart can be applied either toa hydraulic motor or pump. Figures in the
chart are theoretical torque values, in foot pounds, required to turn the shaft of
‘hydraulic pump or that wil be placed on Ihe ahaft ofa hydraulic motor Chart
Values were calculated from the formula
T2 D x PSI» 245
Where: T is torque in foot pounds, D is displacement in cubic inches per
revolution (G.I. A.) PSI is pressure across pump or motor,
andre 314
“The figures in the body of the chart are torque values in foot pounds
cpm
€ 1200) is | PSI (Pounds Per Square men)
RPM [CLR] 250 500 780 2500 3000
Se ee ee
5| 0062 | 319 635 956 127 159 191 285 319 382
a] 183] 522 104 167 209 201 313 418 522 627
10| 192| 637 127 191 255 319 382 510 637 765
12| 231| 704 183 220 300 382 258 811 764 918
18| 346 114 229 344 459 874 688 917 118 138
25| 481| 159 319 479 699 797 957 127 189 191
30 | 770| 285 511 765 102 127 183 204 256 308
Sol 962| 319 638 956 127 159 191 255 319 382
75| 14:43 | 478 956 144 191 239 287 383 478 57%
85 | 1649| 542 108 183 217 271 325 20 542 651
100| '102| 657 127 191 255 S19 30 510 637 765
24
‘This chart can be used to find the torque, horsepower or speed of any kind of
drive (electric motor, hydraulic or pneumatic motor, engine, rotary actuator,
ete. if two of those three values are known, The chart i a tabular solution to
the following basic formulas:
To ind Horsepower (HP), use the formula: HP = IT RPM) + 5252
To find Torque in fe, Ibs. (7), use the formula: 1 = (HP x 8282) > RPM
To find Speed (RPM), uso the formula: APM = (HP x 5282) + Y
‘The figures in the body of the chart are torque values in foot/pounds
Pw (Revolutions Per Minuto)
MP | 100 500 750 1000 1200 18002400 3000 3600|
IS
VS | 175 350 234 175 148 117 (972 730 ‘see 488
1R| 263 525 350 263 220 175 146 110 875 730
ga] 394 787 524 394 328 262 218 164 131 109
1| 525 105 700 525 238 350 292 219 175 147
m| 788 187 105 788 656 528 438 328 263 219
2| 105 210 mo 105 876 700 ses am 350 282
3| 158 315 210 188 131 108 657 525 438
5| 263 525 350 283 220 178 148 110 87 730
m| ges 788 532 394 328 263 218 164 131 109
| 525 105 700 525 438 350 202 219 175 148
15| 788 158 105 788 656 526 438 328 265 219
20| 1050 210 140 105 876 700 584 438 350 292
25| 13138 263 178 181 110 877 730 518 438 385
30 | 1578 315 210 158 191 105 874 657 526 437
40] 2100 420 280 210 175 10 116 875 700 582
SO | 22% 523 350 283 220 1/5 lé 110 875 728
BO | 101 630 420 1 262 210 173 13% 103 874
75| 3940 788 525 398 328 262 218 164 191 109
100| S280 1060 700 525 398 350 202 219 175 148
125| 6565 1313 878 657 548 497 364 274 218 182
150 | 7878 1580 1050 788 656 526 408 328 265 219
200 19500 2100 100 1050 ae 70D sa A 30 2%
250 | 13130 1750 1910 1100 877 365
Pump and Motor Torque
‘This chart can be applied either toa hydraulic motor or pump. Figures in the
chart are theoretical torque values, in foot pounds, required to turn the shaft of
‘hydraulic pump or that wil be placed on Ihe ahaft ofa hydraulic motor Chart
Values were calculated from the formula
T2 D x PSI» 245
Where: T is torque in foot pounds, D is displacement in cubic inches per
revolution (G.I. A.) PSI is pressure across pump or motor,
andre 314
“The figures in the body of the chart are torque values in foot pounds
cpm
€ 1200) is | PSI (Pounds Per Square men)
RPM [CLR] 250 500 780 2500 3000
Se ee ee
5| 0062 | 319 635 956 127 159 191 285 319 382
a] 183] 522 104 167 209 201 313 418 522 627
10| 192| 637 127 191 255 319 382 510 637 765
12| 231| 704 183 220 300 382 258 811 764 918
18| 346 114 229 344 459 874 688 917 118 138
25| 481| 159 319 479 699 797 957 127 189 191
30 | 770| 285 511 765 102 127 183 204 256 308
Sol 962| 319 638 956 127 159 191 255 319 382
75| 14:43 | 478 956 144 191 239 287 383 478 57%
85 | 1649| 542 108 183 217 271 325 20 542 651
100| '102| 657 127 191 255 S19 30 510 637 765
24
Mechanical Transmission Efficiency
‘A hydraulic motor coupled to a load through mechanical transmission items
must have additional power to supply transmission losses, Values given below.
fare average, and where calculations must be precise, olicieney ratings should
be requested from the transmission item manufacturer. Losses must be figured
progressively through each stage of mechanical transmission, For exam
Start with the Anal stage. Figure loss through this stage, Add the extra power
required, Then proceed to the next stage upstream on the power flow. Figure ex
tra power for this stage, add it to the total, then proceed Lo the next stage, ete,
Lup to the driving motor or engine.
The following chart shows typical efficiencies for various
‘mechanical transmission items.
Typical Power Transmission Efficiencies
Machine ‘Typical Eticiency
Vitel dives 95%
Timing bet drives or
Poly-V or ribbed bel ives 97%
Flat belt rives, leather or rubber 98%
Nylon core 98% to 99%
‘Variable speed, spring loaded, wide range
Vibok drives. 20% to 90%
Compound dive 78% to 90%
‘Cam-reacton deve 95%
Helical gear reducer
‘Single-siage 98%
Two-stage 96%
Worm gear reducer
10:1 ratio 86%
25:1 ratio sex
60:1 ratio 66%
Roller chain E
Leadscrew, 60 deg. helix angle 65% to 5%
Flexible coupling, shear-ype E
25
‘A hydraulic motor coupled to a load through mechanical transmission items
must have additional power to supply transmission losses, Values given below.
fare average, and where calculations must be precise, olicieney ratings should
be requested from the transmission item manufacturer. Losses must be figured
progressively through each stage of mechanical transmission, For exam
Start with the Anal stage. Figure loss through this stage, Add the extra power
required, Then proceed to the next stage upstream on the power flow. Figure ex
tra power for this stage, add it to the total, then proceed Lo the next stage, ete,
Lup to the driving motor or engine.
The following chart shows typical efficiencies for various
‘mechanical transmission items.
Typical Power Transmission Efficiencies
Machine ‘Typical Eticiency
Vitel dives 95%
Timing bet drives or
Poly-V or ribbed bel ives 97%
Flat belt rives, leather or rubber 98%
Nylon core 98% to 99%
‘Variable speed, spring loaded, wide range
Vibok drives. 20% to 90%
Compound dive 78% to 90%
‘Cam-reacton deve 95%
Helical gear reducer
‘Single-siage 98%
Two-stage 96%
Worm gear reducer
10:1 ratio 86%
25:1 ratio sex
60:1 ratio 66%
Roller chain E
Leadscrew, 60 deg. helix angle 65% to 5%
Flexible coupling, shear-ype E
25
Pump/Motor Shafts € Flanges
ANSI Shaft and flange code assignments for Auid power pumps and motors.
‘Complete dimensions may be obtained from ANSI B98.6-1972,
ansı shor Long ge pag
SM chant Shaft Shak Kay Shat Sy
code Diam Lg tg Win Gove | | [==
er AA] one
E E dE dE à]
Bi Ge Ve 300 | ©
Bi fo 1 280 M | 7
Hi 1388 D ado da ce 7,
a 28 dal da GR] straight shat
Fr saz; (Without Threads)
Shaft Shan Shaft Spine Sha ES
duos Dim. Egin Specicatons Gove) | [E]
Va DÍ gr 2040 DP AA
dl IRE à See
224 878 1312 191, 18/32 DP
Bi le 10 3
Fri
CR 28
Pr
di Je 3
ansı su
Shot nan Shan
dote Slam. Egin.
192 9300 0750
13 99 qu
HB di 838
3 de 1
Bo a ln
Eee
23 1383 368
ANS! To ma f
Sou Dam. Lom Lgm Size Min =
cae Blam ie ss
oa ES
| 7 7
D SU au A
IRB IA Ne 38) tapered shatt
da 15% les Tae lee 35] GWintireade)
HS ee à
ds 2083 DR tes in e
SRE Mount Mount Pilot
Ber ok" Hel" Bam.
Fung Eirie_Dlam. In
UE
HE dl vou
unting Flange
Bo 3063
23 12500 108 880 ‘(Iwo Bolt)
mt Loe 98
ANS! (SAE, SAE Mount Mount Pit
I ek" Hole" Diem.
Code. Code Rating Cireie Diam. In.
‘is BBs 5000 5605 400
Bi À à E
Be 803 den
Wed À M 288 dE 88
Fa DE 68 Nam hes Sao | ee ge
(Four Bol)
ANSI Shaft and flange code assignments for Auid power pumps and motors.
‘Complete dimensions may be obtained from ANSI B98.6-1972,
ansı shor Long ge pag
SM chant Shaft Shak Kay Shat Sy
code Diam Lg tg Win Gove | | [==
er AA] one
E E dE dE à]
Bi Ge Ve 300 | ©
Bi fo 1 280 M | 7
Hi 1388 D ado da ce 7,
a 28 dal da GR] straight shat
Fr saz; (Without Threads)
Shaft Shan Shaft Spine Sha ES
duos Dim. Egin Specicatons Gove) | [E]
Va DÍ gr 2040 DP AA
dl IRE à See
224 878 1312 191, 18/32 DP
Bi le 10 3
Fri
CR 28
Pr
di Je 3
ansı su
Shot nan Shan
dote Slam. Egin.
192 9300 0750
13 99 qu
HB di 838
3 de 1
Bo a ln
Eee
23 1383 368
ANS! To ma f
Sou Dam. Lom Lgm Size Min =
cae Blam ie ss
oa ES
| 7 7
D SU au A
IRB IA Ne 38) tapered shatt
da 15% les Tae lee 35] GWintireade)
HS ee à
ds 2083 DR tes in e
SRE Mount Mount Pilot
Ber ok" Hel" Bam.
Fung Eirie_Dlam. In
UE
HE dl vou
unting Flange
Bo 3063
23 12500 108 880 ‘(Iwo Bolt)
mt Loe 98
ANS! (SAE, SAE Mount Mount Pit
I ek" Hole" Diem.
Code. Code Rating Cireie Diam. In.
‘is BBs 5000 5605 400
Bi À à E
Be 803 den
Wed À M 288 dE 88
Fa DE 68 Nam hes Sao | ee ge
(Four Bol)
Required Flow for Operating
an Air Cylinder
€ Ares Stoke XA Xf
Time X29
“Area = x dis? or see Table 2 (ea. in.)
Stroke = Travel in.)
pressure drop constant see Table 1
CT Compression Factor see Table 1
Time = seconds
Table 1 - Compression Factors
"A" Constante
Note: Use “A” constant at 6 PSI AP for most applications. On very critical appli
Cations, use À at 2 PSL AP. You will ind in many cases, a 10 PSI AP is not det:
‘imental, and can save money and mounting space,
Required Flow for Operating
a Hydraulic Cylinder
Area x Stroke X60
OPM = Time x231
Area = x x la2/4 or see Table 2 (sa. in.)
Stroke = Travel (in)
Time = seconds.
27
an Air Cylinder
€ Ares Stoke XA Xf
Time X29
“Area = x dis? or see Table 2 (ea. in.)
Stroke = Travel in.)
pressure drop constant see Table 1
CT Compression Factor see Table 1
Time = seconds
Table 1 - Compression Factors
"A" Constante
Note: Use “A” constant at 6 PSI AP for most applications. On very critical appli
Cations, use À at 2 PSL AP. You will ind in many cases, a 10 PSI AP is not det:
‘imental, and can save money and mounting space,
Required Flow for Operating
a Hydraulic Cylinder
Area x Stroke X60
OPM = Time x231
Area = x x la2/4 or see Table 2 (sa. in.)
Stroke = Travel (in)
Time = seconds.
27
HP to Compress Air
e approximate HP roquired to compress 1 SCEM (st
dard cubie Toot per minute) of air from atmospheric pressure of O PSH to the
ressures show sn the tables. Since isothormal and adlabatie compression are
th theoretical conditions, these tables were calculated for compression condi
lon about halfway between these two theoretical extremes. niet ar e as.
Sumed to be about room temperature.
Tables are shown lor single-stage, ota
compressors, assuming thet íficieney to be abo
Sng Water was not considered
“The tables were prepared from information published in Machinery's Hand:
book, Please refer to your copy ofthe handbook for additional information on the
compression of ar
HP to Operate Air Cylinders
One important use for these tables is to estimate the compressor HP capac:
ity needed to operate an air cylinder: The SCFM required by the cylinder under
stated operating conditions must frst be calculated by the method on page 10,
‘Then the appropriate column in the table below can be used to convert SCFM
into MP. For example, cylinder consumption has been calculated to be 24
M and if the compressor la a 2:stage model, the HP needed at 90 PST will
be: HP = 24% 0.156 = 371.
Power Loss Through a Pressure Regulator
Air compressor pers wasted by compretsingair ton pressure higher than
ecossary then reducing it through a regulator. The power wasted cannot be
sly calculated because accurate data cannot be obtained. But it can be est
‘mated with sufficient accuracy with this method.
Use the chart below for your kind of compressor. Caleulate the HP to com-
press 1 SCFM of air to the regulator inlet pressure. Then calculate the HP to
and three-stage piston-type
85%. The effet of acket cook.
Eompress 1 SCFM to the regulator outlet or reduced pressure. Subtract the two
‘This will show the HP wasted for every 1 SCFM passing through the regulator.
Multiply times the SCFM air flow through the cireut.
Horsepower for Compressing Air
ficiency of All Compressors is Assumed to Bo 85%
1-Stage Compressor 2-Stage Compressor 3-Stage Compressor
PSI "PHP" Ps "HP" PS" PHP"
5 où 16 100 159
1 00 & 1 150 180
13056 138 20 212
2% 07 Fe 30 330
2 0 90 Tse CR
5 0 100 ies So 2%
5 m m A 400369
4 107 wo 178 30 2
& ne 130 18 So 289
5 1 10 10 So bor
810 150 10 0 208
a 18 fo 20 es sn
es a 70 206 mo 3
7 18 wu mo
m 188 19 21 800 9
% 160 20 320 Bo “35
8 18 210 2 340
% 10 20 2 90 as
5 18 28 1000 350
100 18 20 335 100 4
no tas 550 23 non 388
120196 20 24 150 3e
10 20 bro 26 120 586
ww tt 20 20 1250 30
10 218 300 23 1300 37a
160 25 S00 385 1550 37
17 2 50 1400 380
10 2% Fe 10 388
190 2 a0 2% 1500 386
200 25 500 1850 390
“HP to comprass Y SCFM Tom 0 PSITo he values shown
NOTE: The power required from other types of compressors ofthe same nu
ber of sages WI be related to these Van ay the efficiency ofthe other com.
pressor isto the assumed 85% efficiency uned for these tables.
28
e approximate HP roquired to compress 1 SCEM (st
dard cubie Toot per minute) of air from atmospheric pressure of O PSH to the
ressures show sn the tables. Since isothormal and adlabatie compression are
th theoretical conditions, these tables were calculated for compression condi
lon about halfway between these two theoretical extremes. niet ar e as.
Sumed to be about room temperature.
Tables are shown lor single-stage, ota
compressors, assuming thet íficieney to be abo
Sng Water was not considered
“The tables were prepared from information published in Machinery's Hand:
book, Please refer to your copy ofthe handbook for additional information on the
compression of ar
HP to Operate Air Cylinders
One important use for these tables is to estimate the compressor HP capac:
ity needed to operate an air cylinder: The SCFM required by the cylinder under
stated operating conditions must frst be calculated by the method on page 10,
‘Then the appropriate column in the table below can be used to convert SCFM
into MP. For example, cylinder consumption has been calculated to be 24
M and if the compressor la a 2:stage model, the HP needed at 90 PST will
be: HP = 24% 0.156 = 371.
Power Loss Through a Pressure Regulator
Air compressor pers wasted by compretsingair ton pressure higher than
ecossary then reducing it through a regulator. The power wasted cannot be
sly calculated because accurate data cannot be obtained. But it can be est
‘mated with sufficient accuracy with this method.
Use the chart below for your kind of compressor. Caleulate the HP to com-
press 1 SCFM of air to the regulator inlet pressure. Then calculate the HP to
and three-stage piston-type
85%. The effet of acket cook.
Eompress 1 SCFM to the regulator outlet or reduced pressure. Subtract the two
‘This will show the HP wasted for every 1 SCFM passing through the regulator.
Multiply times the SCFM air flow through the cireut.
Horsepower for Compressing Air
ficiency of All Compressors is Assumed to Bo 85%
1-Stage Compressor 2-Stage Compressor 3-Stage Compressor
PSI "PHP" Ps "HP" PS" PHP"
5 où 16 100 159
1 00 & 1 150 180
13056 138 20 212
2% 07 Fe 30 330
2 0 90 Tse CR
5 0 100 ies So 2%
5 m m A 400369
4 107 wo 178 30 2
& ne 130 18 So 289
5 1 10 10 So bor
810 150 10 0 208
a 18 fo 20 es sn
es a 70 206 mo 3
7 18 wu mo
m 188 19 21 800 9
% 160 20 320 Bo “35
8 18 210 2 340
% 10 20 2 90 as
5 18 28 1000 350
100 18 20 335 100 4
no tas 550 23 non 388
120196 20 24 150 3e
10 20 bro 26 120 586
ww tt 20 20 1250 30
10 218 300 23 1300 37a
160 25 S00 385 1550 37
17 2 50 1400 380
10 2% Fe 10 388
190 2 a0 2% 1500 386
200 25 500 1850 390
“HP to comprass Y SCFM Tom 0 PSITo he values shown
NOTE: The power required from other types of compressors ofthe same nu
ber of sages WI be related to these Van ay the efficiency ofthe other com.
pressor isto the assumed 85% efficiency uned for these tables.
28
Tank Pump-Down Time
For Large Vacuum Tanks
Use this chart or the formula atthe foot ofthis page to estimate the time to
rasante tank 10 desired degre of vacuum staring eter from atmosph
"The chart is for a amall vane or piston type vacuum pump which will dead
head (with inlet blocked) at a vacuum of 27 oF 28 Hg (when the barometer de at
30° Hg), using a vacuum pump much different from this, caleulate running
time form the formula
“he char shows rnnin time in minutes, fora vacuum pump witha free
running displacement (both ports open to atmosphere) of 1 SCEM. For pump
wrth a diferent displacement the running time must be adjusted by dividing
Shae value by the actual pump placement um a
tunning time values are approximate because the efficiency will var
tween pumps of diferent manufacture iá
Running Time on a 1 SCFM Vacuum Pump
di AR
‘Vac "Hg Time, in Minutes, io Evacuate Tank
EIS GED I) Go Bae
|i Gia abs Bt ab A ie
duos
S| ui Ba ese eB
reee aa
allons, are shown along the top of
chart Degree of vacuum ls shown in the left column.
‘EXAMPLE of running time starting with atmospheric pressure: Estimate
pumping time on a 300 gallon tank to a vacuum leve) of20-g using a vacuum
bmp having fe running dsplacóment ol SO
SOLUTION: From the chart And the running time of 50.1 minutos fora pump
having a free running displacement of 1 SCFM.
Time = 50.1 + 9= 5.67 minutes
EXAMPLE of running time starting with a partial vacuum: Estimate porn
ing time to develop a 24° Hg vacuum mn a 25-cubic foot tank, starting with 15°
Hig vacuum and using a 6 SCFM vacuum pump:
‘SOLUTION: First estimate time from atmosphere up to the present 12° Hg
vacuum, then estimate time from atmosphere to the new Vacuum of 24. Hg
‘Then subtract these times to find the running time between 12" Hg and 247
‘Atmosphere to 12°Hg = 14.0 minutes per 1 SCFM ca
‘Atmosphere to 24° Hg = 48.6 minutes per 1 SCFM capacity
$018.0 94.6 minutes port SCFM capacty
‘Adjust for 6 SCFM pump: Time = 48.6 - 65.77 minutes
Formula for Any Vacuum Pump
This formula, published by Gast, should give a close estimate for any vacu-
um pump, used on any tank, and to any degree of vacuum up to 27" Hg
T= [VD] Loge (A (A-B))
Tis pumping time, in minutes: V is tank volume, in cubic fet; D is free run.
ning displacement, in SCFM: À is deadhead rating of pump in Hg (with inlet
Blocked: B is the desired level of vacuum in tank, in Hg:
29
For Large Vacuum Tanks
Use this chart or the formula atthe foot ofthis page to estimate the time to
rasante tank 10 desired degre of vacuum staring eter from atmosph
"The chart is for a amall vane or piston type vacuum pump which will dead
head (with inlet blocked) at a vacuum of 27 oF 28 Hg (when the barometer de at
30° Hg), using a vacuum pump much different from this, caleulate running
time form the formula
“he char shows rnnin time in minutes, fora vacuum pump witha free
running displacement (both ports open to atmosphere) of 1 SCEM. For pump
wrth a diferent displacement the running time must be adjusted by dividing
Shae value by the actual pump placement um a
tunning time values are approximate because the efficiency will var
tween pumps of diferent manufacture iá
Running Time on a 1 SCFM Vacuum Pump
di AR
‘Vac "Hg Time, in Minutes, io Evacuate Tank
EIS GED I) Go Bae
|i Gia abs Bt ab A ie
duos
S| ui Ba ese eB
reee aa
allons, are shown along the top of
chart Degree of vacuum ls shown in the left column.
‘EXAMPLE of running time starting with atmospheric pressure: Estimate
pumping time on a 300 gallon tank to a vacuum leve) of20-g using a vacuum
bmp having fe running dsplacóment ol SO
SOLUTION: From the chart And the running time of 50.1 minutos fora pump
having a free running displacement of 1 SCFM.
Time = 50.1 + 9= 5.67 minutes
EXAMPLE of running time starting with a partial vacuum: Estimate porn
ing time to develop a 24° Hg vacuum mn a 25-cubic foot tank, starting with 15°
Hig vacuum and using a 6 SCFM vacuum pump:
‘SOLUTION: First estimate time from atmosphere up to the present 12° Hg
vacuum, then estimate time from atmosphere to the new Vacuum of 24. Hg
‘Then subtract these times to find the running time between 12" Hg and 247
‘Atmosphere to 12°Hg = 14.0 minutes per 1 SCFM ca
‘Atmosphere to 24° Hg = 48.6 minutes per 1 SCFM capacity
$018.0 94.6 minutes port SCFM capacty
‘Adjust for 6 SCFM pump: Time = 48.6 - 65.77 minutes
Formula for Any Vacuum Pump
This formula, published by Gast, should give a close estimate for any vacu-
um pump, used on any tank, and to any degree of vacuum up to 27" Hg
T= [VD] Loge (A (A-B))
Tis pumping time, in minutes: V is tank volume, in cubic fet; D is free run.
ning displacement, in SCFM: À is deadhead rating of pump in Hg (with inlet
Blocked: B is the desired level of vacuum in tank, in Hg:
29
Cooling in Hydraulic Systems
Heat Generation
Heat is generated in a hydraulic system whenever ol dumps from a higher
to lower pressure without producing a mechanical work output. Examples ae
sit Howing to tank through à relief valve, owing through a flow control or pres:
Sure reducing valve, or simply fowing through small piping, Pressure drops
{being converted to work output a bef
Yo lower power hydraulie aystems this waste heat is radiated by the walls of
the reservoir In larger systems a heat exchanger must be added. Oil tempera:
ture should be held to 180° to 140°F in an industrial system, but on moving
equipment where heat removal is dificult, the temperature à sometimes al
{owed to reach 200°F although this is not desirable as itis destructive tothe oil
and to components. At high temperatures various chemical reactions produce
sludges which interfere with system operation by clogging orifices and produc
Ing excessive wear in moving parts.
Heat Generation Formulas
Heat generated by oll flow through a valve, piping. or through a relief valve
can do salenlate with the formula Below ithe Pal pressure diffrence genen
the device is known or can be measured, and if the GPM flow through iis
known. Formulas are given for converting heat into other units
1 HP = 2545 BTUMhr = 42.4 BTUlmin = 33,000 ft. Ibs/min = 746 watts
HP = PSI » GPM + 1714 or, BTUIRr = 1% x PSI x GPM
1 BTUMhr = .0167 BTU/min = 00039 HP
EXAMPLE: 12 GPM bypassing a relief valve at a pressure drop of 500 PSI
generates 8): HP of heat, most of which is carried back to the tank
"NOTE: Heat is generated only when no mechanical work is produced
Estimating Heat Build-Up
In most systems the main source of heat may be from the relief valve. this
valve fs in action for only a part of every eyele, End the heat generated while
ia by the formula above. Then average this forthe entire evel. For
example if ol à passing the rehef for about 1 ofthe time in each cycle and
generating 3 HP heat while flowing, hen the average rate of heat generation 1.
far
‘Another source of heat is low control valves used to regulate speed of hy
draulie cylinders or motors. The metered oil generates heat when the cylinder
ar motor 1s running unloaded or ighty loaded: In addition, any ol forced across
the tem relief valve because of metering the cylinder or motor alas gener-
trate less heat than series connected flow controle nn
Pressure reducing valves generate heat during the time oil is flowing
through them and during the ime when pressure differences greatest between
their inlet and outlet ports
ifthe hydraulic syatem is plumbed with pipe sizes adequate to carry the flow
at recommended vit heat generated hy sos through te nes vl un,
ally be small compared with other sources of heat in the system.
‘im addition to these major causes for overheating there wil be heat generat
sg from mechanical owes, mainly in the pump, rin a hydraulic mota About
15% of the input power will go into heat for each pump or motor Ax a rule-o.
thumb, an allowance of 25% of che input power will usually be adequate to take
‘are of all miscellaneous Tosses (and heat) including flow loss through day
Valves, piping. and mechanical loss in one pump. If there is a hydraulic motor,
dr more than one pump in the system, the loses wil be somewhat higher: Then,
{6 this 25%, add the lsses through reli reducing, and How control valves, 14
“any, and this will be a good approximation ofthe heat generated in the system
Cooling Capacity of Steel Reservoirs
‘After estimating the HP or BTU heat generation in your hydraulic system
the next step is to decide whether this heat can be radiated entirely from the
‘walls of the il tank or whether a heat exchanger is needed.
In many systems about 1/8 the heat 1 radiated from walls of eyinder,
pumpe, fluid motors, valves, and plumbing. The remainder is radiated from the
Side walls and top of the reservoir, Radiation from the bottom of the reservoir
¿an be counted if the reservoir is elevated at least 6 inches from the floor. The
‘Amount of heat which can be radiated from the surface of steel tanks can be cal.
ulated from this formula:
30
Heat Generation
Heat is generated in a hydraulic system whenever ol dumps from a higher
to lower pressure without producing a mechanical work output. Examples ae
sit Howing to tank through à relief valve, owing through a flow control or pres:
Sure reducing valve, or simply fowing through small piping, Pressure drops
{being converted to work output a bef
Yo lower power hydraulie aystems this waste heat is radiated by the walls of
the reservoir In larger systems a heat exchanger must be added. Oil tempera:
ture should be held to 180° to 140°F in an industrial system, but on moving
equipment where heat removal is dificult, the temperature à sometimes al
{owed to reach 200°F although this is not desirable as itis destructive tothe oil
and to components. At high temperatures various chemical reactions produce
sludges which interfere with system operation by clogging orifices and produc
Ing excessive wear in moving parts.
Heat Generation Formulas
Heat generated by oll flow through a valve, piping. or through a relief valve
can do salenlate with the formula Below ithe Pal pressure diffrence genen
the device is known or can be measured, and if the GPM flow through iis
known. Formulas are given for converting heat into other units
1 HP = 2545 BTUMhr = 42.4 BTUlmin = 33,000 ft. Ibs/min = 746 watts
HP = PSI » GPM + 1714 or, BTUIRr = 1% x PSI x GPM
1 BTUMhr = .0167 BTU/min = 00039 HP
EXAMPLE: 12 GPM bypassing a relief valve at a pressure drop of 500 PSI
generates 8): HP of heat, most of which is carried back to the tank
"NOTE: Heat is generated only when no mechanical work is produced
Estimating Heat Build-Up
In most systems the main source of heat may be from the relief valve. this
valve fs in action for only a part of every eyele, End the heat generated while
ia by the formula above. Then average this forthe entire evel. For
example if ol à passing the rehef for about 1 ofthe time in each cycle and
generating 3 HP heat while flowing, hen the average rate of heat generation 1.
far
‘Another source of heat is low control valves used to regulate speed of hy
draulie cylinders or motors. The metered oil generates heat when the cylinder
ar motor 1s running unloaded or ighty loaded: In addition, any ol forced across
the tem relief valve because of metering the cylinder or motor alas gener-
trate less heat than series connected flow controle nn
Pressure reducing valves generate heat during the time oil is flowing
through them and during the ime when pressure differences greatest between
their inlet and outlet ports
ifthe hydraulic syatem is plumbed with pipe sizes adequate to carry the flow
at recommended vit heat generated hy sos through te nes vl un,
ally be small compared with other sources of heat in the system.
‘im addition to these major causes for overheating there wil be heat generat
sg from mechanical owes, mainly in the pump, rin a hydraulic mota About
15% of the input power will go into heat for each pump or motor Ax a rule-o.
thumb, an allowance of 25% of che input power will usually be adequate to take
‘are of all miscellaneous Tosses (and heat) including flow loss through day
Valves, piping. and mechanical loss in one pump. If there is a hydraulic motor,
dr more than one pump in the system, the loses wil be somewhat higher: Then,
{6 this 25%, add the lsses through reli reducing, and How control valves, 14
“any, and this will be a good approximation ofthe heat generated in the system
Cooling Capacity of Steel Reservoirs
‘After estimating the HP or BTU heat generation in your hydraulic system
the next step is to decide whether this heat can be radiated entirely from the
‘walls of the il tank or whether a heat exchanger is needed.
In many systems about 1/8 the heat 1 radiated from walls of eyinder,
pumpe, fluid motors, valves, and plumbing. The remainder is radiated from the
Side walls and top of the reservoir, Radiation from the bottom of the reservoir
¿an be counted if the reservoir is elevated at least 6 inches from the floor. The
‘Amount of heat which can be radiated from the surface of steel tanks can be cal.
ulated from this formula:
30
HP (heat) = 0,001 x TD x À
A is the surface area in square feet; TD is the temperature difference in de
gros fahrenheit between surrounding air and oil temperature inside the tank
‘Oil tanks should be installed where there is free air circulation around al
sides and under the tank. A forced blast of air directed on the side of the tank
San increase the radiation capacity as much as 50%
Cooling Capacity of Standard Oi Tanks
Pa dae shots hen edite capac o comercial availble tal y.
date i resrsois having a ch pase underneath and re air cation
Sal side: Euros in the body of tablon HB radiating ably of tanks rom
1010 S00 gall capacity at vacioun temperature diteresces Between ll om
perature and surrounding ai temperature
EXAMPLE: ifs Lei seri tale i a room with 70° ambient
temperature andthe desire maximum olf temperature 180° the tempera
{rierence 1 50 and he heat radiating aby 3.6 HP according to the
Figures in body of chart are heat radiation capacities in HP
Sa, Ft ‘Temperature Difference - Oil to Air °F
Suriace|
Area | 30 40 50 60 70 80 90 100
Del E E & 7% it
128 | 33 Si a m © 10 12 13
140 | 4 56 70 m wo 11 19 14
161 | A6 (64 81 97 11 13 18 16
243 | 73 97 12 18 17 19 22 24
202 | @ 12 15 18 20 23 26 29
316 | 95 13 18 19 22 25 29 32
4o2| 12 16 20 24 28 32 36 ao
a7a | 14 18 22 27 31 36 40 44
529 | 16 21 26 32 37 22 48 53
554 | 17 22 28 33 39 44 50 85
ses | 21 28 35 42 49 56 63 70
‘To Reduce Heat Build-Up
1. Únlcad the pump during intervals when pressure is not required
2. On presses where a high activo pressure must be held for Ton time, an
sir operated pressure intensifier or aD secumulator may be used
‘3, Use as large a reservoir as practical, witha large surface area to dissipate
heat
"4. Pressure compensated flow control valves, if used, should be connected as
“bypass instead of series” control if possible
BSc te main pump rel valve tothe lowest pressure that wil do the
D es open en i ir ao
e thy ot ne oe i ceca
A A
Ea a Meat
Sizing Shell and Tube Heat Exchangers
Surface Area Required. On shell and tube heat exchangers there must be
at least 0.46 square feet of heat transfer surface for each I HP heat load. 2.16
Hp heat load can be removed for every square foot of heat transfer surface un
der the following conditions of usage:
4. Hydraulic o in the shell side at entering temperature of 160°R, leaving
temperature of 140%%
Water in the tube side with a flow equal to 1/2 the oil flow, and at a tem-
peraiure not over 90°F,
‘Correct low velocity in oil and water to obtain optimum heat transfer.
Battles. Bale spacing should bo arranged to give a velocity of 3 N. por see.
in the ol but not outside the range of 2 to 6 feet per second.
Passes. End bonnets should have the correct number of passes on the water
side to give 3 fect por second velocity, but not outside the range of2 to 5 feet per
Second
31
A is the surface area in square feet; TD is the temperature difference in de
gros fahrenheit between surrounding air and oil temperature inside the tank
‘Oil tanks should be installed where there is free air circulation around al
sides and under the tank. A forced blast of air directed on the side of the tank
San increase the radiation capacity as much as 50%
Cooling Capacity of Standard Oi Tanks
Pa dae shots hen edite capac o comercial availble tal y.
date i resrsois having a ch pase underneath and re air cation
Sal side: Euros in the body of tablon HB radiating ably of tanks rom
1010 S00 gall capacity at vacioun temperature diteresces Between ll om
perature and surrounding ai temperature
EXAMPLE: ifs Lei seri tale i a room with 70° ambient
temperature andthe desire maximum olf temperature 180° the tempera
{rierence 1 50 and he heat radiating aby 3.6 HP according to the
Figures in body of chart are heat radiation capacities in HP
Sa, Ft ‘Temperature Difference - Oil to Air °F
Suriace|
Area | 30 40 50 60 70 80 90 100
Del E E & 7% it
128 | 33 Si a m © 10 12 13
140 | 4 56 70 m wo 11 19 14
161 | A6 (64 81 97 11 13 18 16
243 | 73 97 12 18 17 19 22 24
202 | @ 12 15 18 20 23 26 29
316 | 95 13 18 19 22 25 29 32
4o2| 12 16 20 24 28 32 36 ao
a7a | 14 18 22 27 31 36 40 44
529 | 16 21 26 32 37 22 48 53
554 | 17 22 28 33 39 44 50 85
ses | 21 28 35 42 49 56 63 70
‘To Reduce Heat Build-Up
1. Únlcad the pump during intervals when pressure is not required
2. On presses where a high activo pressure must be held for Ton time, an
sir operated pressure intensifier or aD secumulator may be used
‘3, Use as large a reservoir as practical, witha large surface area to dissipate
heat
"4. Pressure compensated flow control valves, if used, should be connected as
“bypass instead of series” control if possible
BSc te main pump rel valve tothe lowest pressure that wil do the
D es open en i ir ao
e thy ot ne oe i ceca
A A
Ea a Meat
Sizing Shell and Tube Heat Exchangers
Surface Area Required. On shell and tube heat exchangers there must be
at least 0.46 square feet of heat transfer surface for each I HP heat load. 2.16
Hp heat load can be removed for every square foot of heat transfer surface un
der the following conditions of usage:
4. Hydraulic o in the shell side at entering temperature of 160°R, leaving
temperature of 140%%
Water in the tube side with a flow equal to 1/2 the oil flow, and at a tem-
peraiure not over 90°F,
‘Correct low velocity in oil and water to obtain optimum heat transfer.
Battles. Bale spacing should bo arranged to give a velocity of 3 N. por see.
in the ol but not outside the range of 2 to 6 feet per second.
Passes. End bonnets should have the correct number of passes on the water
side to give 3 fect por second velocity, but not outside the range of2 to 5 feet per
Second
31
Accumulator Sizing
‘Accumulator Ratings
“Accumulators are cotalog.rted by the gas volume when all uid has been
discharged, and are oly fated Bop measure un arts, and ga
dons, U ÈS gallon = 231 cube inches). Fhe amount of uid which can be stored
In an accumulator is always less than is total gas volume. Only part of the
stored fluid can be used each cycle because the fuld pressure deereates as Auid
Is discharged and when the Auid pressure decreases to the minimum usable
‘alu which wil perform the work inthe hydrauife circuit, no more fluid can be
Uisplaced from the accumulator The actual amount is determined by the ratio
Sf maximum sytem pressure divided by” precharge pressure (example
30007780 = 4:1, maximum Buid volume inthis example is 75% of accumulator
‘olume). In summary: the change in pressure controle the change in volume
“The problem im sciectingan accumulator isto select aise which wil deliver
a sant a eh cycle without the system pressure dropping too
iow Selection e derbe in et later on hl page. Heard information for
Selection; aystem minimum pressure, apstem maximum pressure, id volume,
tobe discharged, and Huid volume discharge time
Precharge Pressure
Gas Charging
A CAUTION SE moe ens on
Precharge new or repaired accumulators with dry nitrogen gas to the recom-
mended gas precharge pressure (PO) listed below, prior to applying hydraulic
system pressure.
For Energy Storage Po=0.9xP;
For Shock Absorption Po = (0.6 to 0.9) x Pry
For Pulsation Dampening Po = (0.6 to 0.8) x Pry
P, = minimum working pressure _P, = median working pressure
Having the precharge pressure set below the minimum system pressure al
lows a small amount of fud to remain in the accumulator, thus preventing the
elastomer from chafing against the valve on each eyelo.
Ett Fd qurements
tana dpa
ial eae tak non ig i en ee e
Ee one ea ne ey aren
yaya out st et
ee
Ft abort, una anise
a
Se
or laminating operation, If the cyl-
32
‘Accumulator Ratings
“Accumulators are cotalog.rted by the gas volume when all uid has been
discharged, and are oly fated Bop measure un arts, and ga
dons, U ÈS gallon = 231 cube inches). Fhe amount of uid which can be stored
In an accumulator is always less than is total gas volume. Only part of the
stored fluid can be used each cycle because the fuld pressure deereates as Auid
Is discharged and when the Auid pressure decreases to the minimum usable
‘alu which wil perform the work inthe hydrauife circuit, no more fluid can be
Uisplaced from the accumulator The actual amount is determined by the ratio
Sf maximum sytem pressure divided by” precharge pressure (example
30007780 = 4:1, maximum Buid volume inthis example is 75% of accumulator
‘olume). In summary: the change in pressure controle the change in volume
“The problem im sciectingan accumulator isto select aise which wil deliver
a sant a eh cycle without the system pressure dropping too
iow Selection e derbe in et later on hl page. Heard information for
Selection; aystem minimum pressure, apstem maximum pressure, id volume,
tobe discharged, and Huid volume discharge time
Precharge Pressure
Gas Charging
A CAUTION SE moe ens on
Precharge new or repaired accumulators with dry nitrogen gas to the recom-
mended gas precharge pressure (PO) listed below, prior to applying hydraulic
system pressure.
For Energy Storage Po=0.9xP;
For Shock Absorption Po = (0.6 to 0.9) x Pry
For Pulsation Dampening Po = (0.6 to 0.8) x Pry
P, = minimum working pressure _P, = median working pressure
Having the precharge pressure set below the minimum system pressure al
lows a small amount of fud to remain in the accumulator, thus preventing the
elastomer from chafing against the valve on each eyelo.
Ett Fd qurements
tana dpa
ial eae tak non ig i en ee e
Ee one ea ne ey aren
yaya out st et
ee
Ft abort, una anise
a
Se
or laminating operation, If the cyl-
32
Discharge Flow Rate ...
A discharge flow rate of 2 pm or les is considered slow, the precharge gas
doce not onde significant heat meaning that the gos precharge presse re
maine constant, and o does the dd discharge volume Gea! condition, but not
always eal word) This s considered an Lofhermal exchange
Ralscharge low rate of2 gpm or more ia considered rapid, the precharge
significant heat, meaning thatthe gas precharge pressure chan
‘van the precharge gas temperature changes and so dace Ihe fluid discharge
‘jac lo rapid compression and expansion of the gas (most like Tea
‘This is considered an adlabate exchange, The volume of usable uid It
Tung with a rapid (adiabatic) exchange
“The selection chart below has been calculated based on the rapid (adiabatic)
design pring
‘Summary...
The sing information is intended to allow the hydraulic system designer
the ability to estimate the approtimate accumulator size for a given applica,
ton: There are many factors that yo into the Anal selection a an accom
for that reason iti suggested that you consult Womack: Always design cy
ders or hydraulic motor of sulKcient size todo satisfactory job ak ts decreased
pressure
Using the Selection Chart.
STEP 1. Calculate or estimate the fluid volume, in eubie inches, which will
be required on every discharge of the accumulator. Consider the design param:
ergy pretty deri
‘STEP 2, Decide on an acceptable decrease in system pressure when the ac-
‘cumulator has discharged the volume of Nuid estimated in Step 1. System pre
Sure always deereases when à charged accumulator delivers a How of
When using the chart, the minimum acceptable system pressure is listed in the
St column he the fully charged accumulator (and system) pressure ts
shown along the to
STEP 3. With the data from Steps 1 and 2 enter the chart in the column
headed by Maximum System Pressure. Go down this column to the line showing
tthe minimum acceptable system pressure. The figure shown is the number of
cubie inches of oil delivered from a “Igallon" accumulator during its discharge
from fully charged to minimum acceptable system pressure, À 5-gallon aecumts
ator will deliver approximately Ave times (within 90%) this amount and a 10.
gallon accumulator wall deliver approximately ten tims (within, 90%) this
Amount, etc. From this information an accumulator of sufficient gallonage can.
be selected which will do the Job.
Minimum Maximum System Pressure (PSI)
Acgapana Sytem
Présure (PS) |3000 2750 2500 2250 2000 1750 1500 1250 1000_750
Fr ee
208 À Gui inches of ci otero by 3"1-goon came
Fd a & rare
BB BS q
Be Bas E
E BG
Es EA
= BR ES E
in BBB Ss
8 £22 3
1 £282 %
i BEB SE
ES 233
EE 22%
a &
E
ge:
285888:
NOTE 1: 231 cubic inches = LUS. gallon
NOTE 2: This chart is calculated using a precharge pressure equal to 90% of
w minimo acceptable system presure with an ambient temperature range
tween 50°F and 120°.
33
A discharge flow rate of 2 pm or les is considered slow, the precharge gas
doce not onde significant heat meaning that the gos precharge presse re
maine constant, and o does the dd discharge volume Gea! condition, but not
always eal word) This s considered an Lofhermal exchange
Ralscharge low rate of2 gpm or more ia considered rapid, the precharge
significant heat, meaning thatthe gas precharge pressure chan
‘van the precharge gas temperature changes and so dace Ihe fluid discharge
‘jac lo rapid compression and expansion of the gas (most like Tea
‘This is considered an adlabate exchange, The volume of usable uid It
Tung with a rapid (adiabatic) exchange
“The selection chart below has been calculated based on the rapid (adiabatic)
design pring
‘Summary...
The sing information is intended to allow the hydraulic system designer
the ability to estimate the approtimate accumulator size for a given applica,
ton: There are many factors that yo into the Anal selection a an accom
for that reason iti suggested that you consult Womack: Always design cy
ders or hydraulic motor of sulKcient size todo satisfactory job ak ts decreased
pressure
Using the Selection Chart.
STEP 1. Calculate or estimate the fluid volume, in eubie inches, which will
be required on every discharge of the accumulator. Consider the design param:
ergy pretty deri
‘STEP 2, Decide on an acceptable decrease in system pressure when the ac-
‘cumulator has discharged the volume of Nuid estimated in Step 1. System pre
Sure always deereases when à charged accumulator delivers a How of
When using the chart, the minimum acceptable system pressure is listed in the
St column he the fully charged accumulator (and system) pressure ts
shown along the to
STEP 3. With the data from Steps 1 and 2 enter the chart in the column
headed by Maximum System Pressure. Go down this column to the line showing
tthe minimum acceptable system pressure. The figure shown is the number of
cubie inches of oil delivered from a “Igallon" accumulator during its discharge
from fully charged to minimum acceptable system pressure, À 5-gallon aecumts
ator will deliver approximately Ave times (within 90%) this amount and a 10.
gallon accumulator wall deliver approximately ten tims (within, 90%) this
Amount, etc. From this information an accumulator of sufficient gallonage can.
be selected which will do the Job.
Minimum Maximum System Pressure (PSI)
Acgapana Sytem
Présure (PS) |3000 2750 2500 2250 2000 1750 1500 1250 1000_750
Fr ee
208 À Gui inches of ci otero by 3"1-goon came
Fd a & rare
BB BS q
Be Bas E
E BG
Es EA
= BR ES E
in BBB Ss
8 £22 3
1 £282 %
i BEB SE
ES 233
EE 22%
a &
E
ge:
285888:
NOTE 1: 231 cubic inches = LUS. gallon
NOTE 2: This chart is calculated using a precharge pressure equal to 90% of
w minimo acceptable system presure with an ambient temperature range
tween 50°F and 120°.
33
Viscosity Rating Systems
Kinematic Viscosity expresses total resistance to Auid flow including inter
nal ui fiction plus effect of mass or weight ofthe uid. Ie» measured in ey.
raf systems, with equivalent values shown in the chart compared to SUS
‘ating in the first column. All these systems are based on the Lime for a quan:
{ity of Mud to flow through a standard orifice under specihed conditions, In the
US the Saybolt Universal Second (SUS) rating is mont often used. I ia derived
{rom English units. The Centistoke isthe standard for international fuid pow:
cr Its derived from metric units (1 Stoke = 100 Centistokes).
Absolute Viscosity is an expression only ofthe internal id fiction with.
cout taking Into account the effect of the mass or weight ofthe uid. À statement
‘f absolute viscosity must also include a statement ofthe specie gravity ofthe
fluid: The international standard unit for absolute viscosity is Che Poise or Cen:
Tipoise (1 Polse = 100 Gentipoise) t's derived from metric units. la the En.
ql aya the unt i the Ream Conte vacates he at column af
fhe chart are for any fluid including standard hydraulic il, which has a specific
gravity of 0.9. The Centipoise is related tothe Centistoke. Any value of kinemat-
dE vinci in Centntken can be converted o absolute Visconti Conioine
by multiplying Centistokes times the specific gravity. Thus water with specie
ravi al 1.0 bas the same kinematic and absolute viscosity ratings.
‘While absolute visconty is important In seentif processes it nf litle val.
e in fluid power because viscosity effects such as pump cavitation, pressure
Fes invalid piping are pre not ols by internal id ction but
by the weight (specie gravity) ofthe Huid as well. Thus, we express visco:
Kinematic SUS values almost entirely. in the USA.
A
we ese Pen
ES naa, Las Br
Er a
O oe D E Ez
in © mass
o © km =
u an Sst
e u u
E
2 2 SN e >= y
= DE
|
Ft
ER À D KR O»
RE
EN a cH
SB 2 e > & à
a 2D 8 8 8
u a — à
= + à À E à =
a 2s 2 2 #8 8 =
2 + à à y Y Eu
ee
NES SE D:
s = m: 2 = |#
ess 8 8 > |&
+ ==
ge:
a == BE # #8 == |%
S = 8 3 8 == 5
2 = be |:
For Redwood No, 2 Admiralty Seconds viscosity, divide values in this column
by 10,
‘For Engler viscosity values in seconds, multiply values in this column by 50.
Sole acon in Contienen related to the specie gravity of the Hu
Values in this column are for hydraulie 0 of 0.9 specihe gravity. For ids with
other values of specific gravity. Centipoise viscosity is found by multiplying val
‘dew in Centistokes column by specific gravity of Hud,
34
Kinematic Viscosity expresses total resistance to Auid flow including inter
nal ui fiction plus effect of mass or weight ofthe uid. Ie» measured in ey.
raf systems, with equivalent values shown in the chart compared to SUS
‘ating in the first column. All these systems are based on the Lime for a quan:
{ity of Mud to flow through a standard orifice under specihed conditions, In the
US the Saybolt Universal Second (SUS) rating is mont often used. I ia derived
{rom English units. The Centistoke isthe standard for international fuid pow:
cr Its derived from metric units (1 Stoke = 100 Centistokes).
Absolute Viscosity is an expression only ofthe internal id fiction with.
cout taking Into account the effect of the mass or weight ofthe uid. À statement
‘f absolute viscosity must also include a statement ofthe specie gravity ofthe
fluid: The international standard unit for absolute viscosity is Che Poise or Cen:
Tipoise (1 Polse = 100 Gentipoise) t's derived from metric units. la the En.
ql aya the unt i the Ream Conte vacates he at column af
fhe chart are for any fluid including standard hydraulic il, which has a specific
gravity of 0.9. The Centipoise is related tothe Centistoke. Any value of kinemat-
dE vinci in Centntken can be converted o absolute Visconti Conioine
by multiplying Centistokes times the specific gravity. Thus water with specie
ravi al 1.0 bas the same kinematic and absolute viscosity ratings.
‘While absolute visconty is important In seentif processes it nf litle val.
e in fluid power because viscosity effects such as pump cavitation, pressure
Fes invalid piping are pre not ols by internal id ction but
by the weight (specie gravity) ofthe Huid as well. Thus, we express visco:
Kinematic SUS values almost entirely. in the USA.
A
we ese Pen
ES naa, Las Br
Er a
O oe D E Ez
in © mass
o © km =
u an Sst
e u u
E
2 2 SN e >= y
= DE
|
Ft
ER À D KR O»
RE
EN a cH
SB 2 e > & à
a 2D 8 8 8
u a — à
= + à À E à =
a 2s 2 2 #8 8 =
2 + à à y Y Eu
ee
NES SE D:
s = m: 2 = |#
ess 8 8 > |&
+ ==
ge:
a == BE # #8 == |%
S = 8 3 8 == 5
2 = be |:
For Redwood No, 2 Admiralty Seconds viscosity, divide values in this column
by 10,
‘For Engler viscosity values in seconds, multiply values in this column by 50.
Sole acon in Contienen related to the specie gravity of the Hu
Values in this column are for hydraulie 0 of 0.9 specihe gravity. For ids with
other values of specific gravity. Centipoise viscosity is found by multiplying val
‘dew in Centistokes column by specific gravity of Hud,
34
4.001 0
ooieu'sng avs Bet ans ave
‘s6uney 110 AVS Á8 paian0O abuey SAS
35
3002 3.001 3091 4.051 Orr 400) 402 AO AO 406 4.08 409 4.09 402 40
"Simeiodue 10.
24NIDLIAUAL, YIIN UORDLIDA 71809814 SNS
ooieu'sng avs Bet ans ave
‘s6uney 110 AVS Á8 paian0O abuey SAS
35
3002 3.001 3091 4.051 Orr 400) 402 AO AO 406 4.08 409 4.09 402 40
"Simeiodue 10.
24NIDLIAUAL, YIIN UORDLIDA 71809814 SNS
me
pee
Seal Compatibility with Common Fluids
[GRA
esmero
cord
o) Ingram) Nese Nyon
D» coho eee ooo
Po»
CRE ETS
Te bors spp» coco DE pepe CCE bape mene DDR
ee bebe ebb PE peer PA op pepe DE]
pee
Seal Compatibility with Common Fluids
[GRA
esmero
cord
o) Ingram) Nese Nyon
D» coho eee ooo
Po»
CRE ETS
Te bors spp» coco DE pepe CCE bape mene DDR
ee bebe ebb PE peer PA op pepe DE]
Hydraulic Pipe Table
Physical Dimensions and Pressure Ratings
Schedule 40 (Standard Weight) Pipe
Ps emer” wee a
TS ZE E E Y iQ
# E 8 E M Be
da os a tee
Y E D 8 E E à
1% 1900 1810 145 2038 1017 8105
E JM Hé Œ
Î ee EE # #
Schedule 80 (Extra Strong Weight) Pipe
em Sa Meta Ba
Bop in mts M Me Sa
es E à 8 M Ie
¿331330
2 à à E # à
(ifs # u if
De à 2 à e@
aoe D # & E ©
¿HD
UE i i BE 5
Pi Burst
sie 00. Pst
Een E ON JU E le
E Ou 20 1320
a 1800 1H 2 10 208 12
1 JO JS 21 140 O2 11e
5 2 Je Où 2% Ji oo
26 DR 2 u u 1% 194%
3 3803 fé 488 ódos 168 don
"Working PSI at a safety actor o 6:1.
‘The above charts are for welded and seamless wrought steel pipe. Wall thick
ness on wrought iron pipe i slightly greater than for tee pipe, and the inside
fea is, therefore, slightly smaller. Burst strength is about the same.
‘Schedule 40s the same as “standard wall" up to 10 inch size, Schedule 80 is
the came as “extra strong” up to inch size. There is no schedule number for
“double extra strong". Schedule 160 is lighter than “double extra strong" and
heavier than “extra strong”
Pressure Ratings
Burst strength has bcen figured on a tensile strength of 40,000 PST for butt
welded steel pipe. Lap welded steel pipe has a strength of 50,000 PSI, and will
Stand 20% more pressure than shown in the tables. Burst strength is by Bar
fows Formula: P= 24 x 8 + O in which P is bursting pressure in PSL, Us wall
thickness in inches, $ is tensile strength of material in PSI, and O is outside di
meter of pipe in inches,
Safety Factor ...
‘The working pressure ratings in the next to last column are figured with a
safety factor of 8 In the usual hydraulic system a factor of at least 6 should be
used: However, to find working pressure at another safety factor take the burst
pressure rating and divide by the desired safety factor,
37
Physical Dimensions and Pressure Ratings
Schedule 40 (Standard Weight) Pipe
Ps emer” wee a
TS ZE E E Y iQ
# E 8 E M Be
da os a tee
Y E D 8 E E à
1% 1900 1810 145 2038 1017 8105
E JM Hé Œ
Î ee EE # #
Schedule 80 (Extra Strong Weight) Pipe
em Sa Meta Ba
Bop in mts M Me Sa
es E à 8 M Ie
¿331330
2 à à E # à
(ifs # u if
De à 2 à e@
aoe D # & E ©
¿HD
UE i i BE 5
Pi Burst
sie 00. Pst
Een E ON JU E le
E Ou 20 1320
a 1800 1H 2 10 208 12
1 JO JS 21 140 O2 11e
5 2 Je Où 2% Ji oo
26 DR 2 u u 1% 194%
3 3803 fé 488 ódos 168 don
"Working PSI at a safety actor o 6:1.
‘The above charts are for welded and seamless wrought steel pipe. Wall thick
ness on wrought iron pipe i slightly greater than for tee pipe, and the inside
fea is, therefore, slightly smaller. Burst strength is about the same.
‘Schedule 40s the same as “standard wall" up to 10 inch size, Schedule 80 is
the came as “extra strong” up to inch size. There is no schedule number for
“double extra strong". Schedule 160 is lighter than “double extra strong" and
heavier than “extra strong”
Pressure Ratings
Burst strength has bcen figured on a tensile strength of 40,000 PST for butt
welded steel pipe. Lap welded steel pipe has a strength of 50,000 PSI, and will
Stand 20% more pressure than shown in the tables. Burst strength is by Bar
fows Formula: P= 24 x 8 + O in which P is bursting pressure in PSL, Us wall
thickness in inches, $ is tensile strength of material in PSI, and O is outside di
meter of pipe in inches,
Safety Factor ...
‘The working pressure ratings in the next to last column are figured with a
safety factor of 8 In the usual hydraulic system a factor of at least 6 should be
used: However, to find working pressure at another safety factor take the burst
pressure rating and divide by the desired safety factor,
37
Oil Flow Capacity of Pipes
Schedule 40 (Standard Weight) Pipe
2 4 os mw
née Fée fie nie Fôe Fle.
pooner "Gu Mr a Ma Ta am
18 38 on IN 38 14 44
Y E 318 14 u + à
ue Ko ee © Be me; Be zu
1e 189 37 247 M2 JM 241
Ed E de de 48 3% Ly
i 3% de 18 ni Se de
ty RE m 0805 1898
ik ¿ BS SE Be Be E
2 E E]
ES 2905 SOTO MON ZM ze az
3 RS Sn 1104 2e Te e
Schedule 80 (Extra Strong Weight) Pipe
2 4 m0 2
née née fie «Fee, Pie se.
pooner “com Mor m Tu ‘ot am
TR FT Ton IN 7 34
ia te 88 3 E E OR
ae bh 35 18 85 4% ni
a of i ge Sl Be ag
; i iv 22 85 un 02
me 800 3599 299% SM 97 jes
y ¿e 28 $8 88 dy ise
2 Bi HN Re li Des
2 Re SM Jen 1017 242 en
3 RB 8% EN MR
Schedule 160 Pipe
2 4 10 15 a x»
O
Pioenet Tr To Mort m GP “hit
% 1520 E 7 WA IN
ES 18 Où $57 de Du Za
y 38 60 x AS 9 dede
e 131 24 en
ha Se JE EN gén ES 18
2 Bee 200
2% Bi 4492110556583 22110 3316
3 Bn de 185 38 Br
Pipe size should be selected on the basis o il low velocity. Undersizing re-
sults Im a high pressure and power loss and system overheating, Oversizing re
¿luces pressure and power losses but may be unnecessarily expensive to plus,
Pump Suction Lines .
Schedule 40 pipe should be used and a size chosen which will keep oil veloc-
ty within the range of 2 to 4 feet per second.
Oil Return Lines
Schedule 40 pipe should be used and a size chosen which will keep oil veloc-
ity within the range of 10 to 15 feet per second
Medium Pressure Lines ...
In those lines carrying 500 (0 2000 PSI, flow velocity should be kept at 15 to
20 feet per second. Use Schedule 80 or 160 pipe, or use steel tubing as listed on
the next page.
igh Pressure Lines ...
Flow velocity may be allowed up to 30 feet per second in lines carrying 3000
10 5000 PSI- Normally tel tubing is vsd, but the abla may be used for find
Ing pipe size, then tubing ahould be selected with the same inside area.
38
Schedule 40 (Standard Weight) Pipe
2 4 os mw
née Fée fie nie Fôe Fle.
pooner "Gu Mr a Ma Ta am
18 38 on IN 38 14 44
Y E 318 14 u + à
ue Ko ee © Be me; Be zu
1e 189 37 247 M2 JM 241
Ed E de de 48 3% Ly
i 3% de 18 ni Se de
ty RE m 0805 1898
ik ¿ BS SE Be Be E
2 E E]
ES 2905 SOTO MON ZM ze az
3 RS Sn 1104 2e Te e
Schedule 80 (Extra Strong Weight) Pipe
2 4 m0 2
née née fie «Fee, Pie se.
pooner “com Mor m Tu ‘ot am
TR FT Ton IN 7 34
ia te 88 3 E E OR
ae bh 35 18 85 4% ni
a of i ge Sl Be ag
; i iv 22 85 un 02
me 800 3599 299% SM 97 jes
y ¿e 28 $8 88 dy ise
2 Bi HN Re li Des
2 Re SM Jen 1017 242 en
3 RB 8% EN MR
Schedule 160 Pipe
2 4 10 15 a x»
O
Pioenet Tr To Mort m GP “hit
% 1520 E 7 WA IN
ES 18 Où $57 de Du Za
y 38 60 x AS 9 dede
e 131 24 en
ha Se JE EN gén ES 18
2 Bee 200
2% Bi 4492110556583 22110 3316
3 Bn de 185 38 Br
Pipe size should be selected on the basis o il low velocity. Undersizing re-
sults Im a high pressure and power loss and system overheating, Oversizing re
¿luces pressure and power losses but may be unnecessarily expensive to plus,
Pump Suction Lines .
Schedule 40 pipe should be used and a size chosen which will keep oil veloc-
ty within the range of 2 to 4 feet per second.
Oil Return Lines
Schedule 40 pipe should be used and a size chosen which will keep oil veloc-
ity within the range of 10 to 15 feet per second
Medium Pressure Lines ...
In those lines carrying 500 (0 2000 PSI, flow velocity should be kept at 15 to
20 feet per second. Use Schedule 80 or 160 pipe, or use steel tubing as listed on
the next page.
igh Pressure Lines ...
Flow velocity may be allowed up to 30 feet per second in lines carrying 3000
10 5000 PSI- Normally tel tubing is vsd, but the abla may be used for find
Ing pipe size, then tubing ahould be selected with the same inside area.
38
Oil Pressure Loss Through Pipes
‘Table 1 shows the pressure loss per 100 feet of Schedule 40 pipe, It is for
standard hydraulic oil of 0.9 specific gravity and 220 SUS viscosity: For other
Speciffe gravities and viscosities se information at the bottom of this page.
Table 1. Pressure Loss Per 100 Feet of Schedule 40 Pipe
Pipe, Pres" Flow Pipe, Pres” Flow Pipe, Pres" Flow
[GPM Site: Drop Velt | GPM Size" Drop Volt | GPM Size” Drop Verf
TO 38 185 17 | 35 12 249 38 | 70 SA 25 42
eB ht oa ea ES
dá ¿a go Y 43 M À
1 $0 37 ho 37 wR OR
1 30 22 14 or 88 2 42 ar
15 12 10 16 | 40 a 95 2 | 80 1 76 31
di 0% où Y 8 nah
Y 14 88 % 2 es we sg
1 48 33 gs 83 2 48 77
PRES 2 24 38 de 23 54
20 12 148 21 | 45 sa 106 27 | 00 1 go 3
a 7 Pr ae M 2%
Y 18 78 dde of vé a
Mm 89 43 we 44 71 2 34 86
m 32 32 2 27 43 de 26 80
25 12 190 26 | so sa 12 91 | 100 1 gp 38
ah ee iS Y 8 io M 22
1 23 93 1 38 ve 18 36
de fe 54 14 at 78 2 89 98
18 40 39 2 30 48 de 28 87
30 ye 219 91 | 60 sa 12 36 | a
an v8 2 M US 2
von he À 6 “2%
1 $0 64 De 93 2 fs 2
14 48 47 2 38 87 de 98 8a
Table 2. Conversion Factors for Tubing
For pressure loss per 100 feet of tubing find tubing 1.D, in table below. The
next larger NPT size de shown in Column 2. Refer back to Table 1 for pressure
Fons for this pipe size. Multiply times factor in Column 3 of Table 2:
EXAMPLE! For 50 GPM flow through a tube with 1310 1D», Column 2
AU NPT tobe the next larger pipe size. From Tabl
for 1% pipe. Multiply this times the factor trom column 3 of Table 2: 15 PSI x
Win 218.65 (or 19) PSI pressure loss per 100 fet
For other schedules of pipe or for hose low loss will be
n proportion to the
inside area of pipe compared to one of the pipe sizes in Table
Tube Use Mult. | Tube Use Mu | Tubo Use Mult
10. NPT by Vor NPT by too NET by
Ta 382g 79 TRE IE
q er 78 I [TE 18
856 à 17 816 dos | 1280 12
r 1 tse | tae 1 116
e 518 E E
46 de 12 876 18 [18% 115
13 ges var | 135 me 10
e Y E 18 BE 414
0360 12 123 | 1010 ww | 1782 2 1e
1e Ya 1092 so | 4760 2 12
Bee 1210 | 1885 18 (18 31%
05% 12 ro | 108% 162 | 180 21%
ga a4 16 | 1108 156 | 182 2 127
see 54 1e | 113 18 Es 3 1%
For Flows Not Shown: Pressure loss increas
to rea nk neos tera ae
¡sing for Other Viscosities: Pressure loss through a pipe it directly pro:
nal to Aid viscosity (on SUS of 100 and above) A440 SUS Avid would
lave approximately twice the pressure loss shown in the tables.
‘Adjusting for Specific Gravity: Pressure loss is directly proportional to spe-
cie gravity, Water emulsions will have 7% higher, waterigiyeo! vids wil
have 14%, and phosphate ester fluids will have 22% higher pressure loss than
calculated fromthe tables.
"Schedule 40 pipe. “PSI loss por 100 fect. Oil flow velocity, (second.
‘approximately ia proportion
39
‘Table 1 shows the pressure loss per 100 feet of Schedule 40 pipe, It is for
standard hydraulic oil of 0.9 specific gravity and 220 SUS viscosity: For other
Speciffe gravities and viscosities se information at the bottom of this page.
Table 1. Pressure Loss Per 100 Feet of Schedule 40 Pipe
Pipe, Pres" Flow Pipe, Pres” Flow Pipe, Pres" Flow
[GPM Site: Drop Velt | GPM Size" Drop Volt | GPM Size” Drop Verf
TO 38 185 17 | 35 12 249 38 | 70 SA 25 42
eB ht oa ea ES
dá ¿a go Y 43 M À
1 $0 37 ho 37 wR OR
1 30 22 14 or 88 2 42 ar
15 12 10 16 | 40 a 95 2 | 80 1 76 31
di 0% où Y 8 nah
Y 14 88 % 2 es we sg
1 48 33 gs 83 2 48 77
PRES 2 24 38 de 23 54
20 12 148 21 | 45 sa 106 27 | 00 1 go 3
a 7 Pr ae M 2%
Y 18 78 dde of vé a
Mm 89 43 we 44 71 2 34 86
m 32 32 2 27 43 de 26 80
25 12 190 26 | so sa 12 91 | 100 1 gp 38
ah ee iS Y 8 io M 22
1 23 93 1 38 ve 18 36
de fe 54 14 at 78 2 89 98
18 40 39 2 30 48 de 28 87
30 ye 219 91 | 60 sa 12 36 | a
an v8 2 M US 2
von he À 6 “2%
1 $0 64 De 93 2 fs 2
14 48 47 2 38 87 de 98 8a
Table 2. Conversion Factors for Tubing
For pressure loss per 100 feet of tubing find tubing 1.D, in table below. The
next larger NPT size de shown in Column 2. Refer back to Table 1 for pressure
Fons for this pipe size. Multiply times factor in Column 3 of Table 2:
EXAMPLE! For 50 GPM flow through a tube with 1310 1D», Column 2
AU NPT tobe the next larger pipe size. From Tabl
for 1% pipe. Multiply this times the factor trom column 3 of Table 2: 15 PSI x
Win 218.65 (or 19) PSI pressure loss per 100 fet
For other schedules of pipe or for hose low loss will be
n proportion to the
inside area of pipe compared to one of the pipe sizes in Table
Tube Use Mult. | Tube Use Mu | Tubo Use Mult
10. NPT by Vor NPT by too NET by
Ta 382g 79 TRE IE
q er 78 I [TE 18
856 à 17 816 dos | 1280 12
r 1 tse | tae 1 116
e 518 E E
46 de 12 876 18 [18% 115
13 ges var | 135 me 10
e Y E 18 BE 414
0360 12 123 | 1010 ww | 1782 2 1e
1e Ya 1092 so | 4760 2 12
Bee 1210 | 1885 18 (18 31%
05% 12 ro | 108% 162 | 180 21%
ga a4 16 | 1108 156 | 182 2 127
see 54 1e | 113 18 Es 3 1%
For Flows Not Shown: Pressure loss increas
to rea nk neos tera ae
¡sing for Other Viscosities: Pressure loss through a pipe it directly pro:
nal to Aid viscosity (on SUS of 100 and above) A440 SUS Avid would
lave approximately twice the pressure loss shown in the tables.
‘Adjusting for Specific Gravity: Pressure loss is directly proportional to spe-
cie gravity, Water emulsions will have 7% higher, waterigiyeo! vids wil
have 14%, and phosphate ester fluids will have 22% higher pressure loss than
calculated fromthe tables.
"Schedule 40 pipe. “PSI loss por 100 fect. Oil flow velocity, (second.
‘approximately ia proportion
39
Carbon Steel Tubing Data
Steel tubing is called out by outside diameter and wall thickness, For hy-
draulic plumbing a low carbon seamless steel tubing should be used which can
be bent and flared without cracking. Order “hydraulke grade” tubing.
Pressure ratings in this table are based o a tubing with tose strength of
55,000 PSI, and were ealeulated by Barlow's formula: P = 208 + O, in which P
= burat strength in PSL t= wall thickness, § = tensile strength in PSI, and O =
outside diameter. This formula may be used to calculate tubing sizes not listed
All dimensions in the table are in inches.
For hydraulic plumbing, a safety factor of at least six should be used and rat
¿nan for thin factor age sh In the table For pressure rating at other safety
So aa) a a lee en
Ee OX ie On foin MS
PR ST
E E EE il 14
- 035, 055 ‚0024 30.800 5133 3.8507
e toe ;
EME, NS ec E
E A A La
“043 152 ‚9181 21560 3593 2,695
Sd Mos à à
m amt —
HO E a 2 je
# e ela ss
a e ss sn y à
e u oe
ON E a a |
HO À D ie à à
a 2 El ao
2 e $ y 4 de
O
HO À dd de à là
E E Ee dé à
072 286 COR 2 1%
2 33 Hi HA e le
oo A ne u ee,
E E 3 E là 1
A là ue
E 3 la ig là
e E ON lé a
S À llas
E dé ús ds 1
Pe
3 E E E E e
Ss E gf da là ie
083 584 2677 12173 2028 1522
E À E he e le
e E ls ly del la
gee ee i
E E 3 ie y
072 731 4195 9.051 1809 i
E E E $ là 8
109 657 3388 13.703 2.284 1713
TE E M mu A
Ss E u là E
e E Bee à
E E te 8 ig JB
E E Ee la là le
a N SE 5 8
ER ole là 8
corp a
Steel tubing is called out by outside diameter and wall thickness, For hy-
draulic plumbing a low carbon seamless steel tubing should be used which can
be bent and flared without cracking. Order “hydraulke grade” tubing.
Pressure ratings in this table are based o a tubing with tose strength of
55,000 PSI, and were ealeulated by Barlow's formula: P = 208 + O, in which P
= burat strength in PSL t= wall thickness, § = tensile strength in PSI, and O =
outside diameter. This formula may be used to calculate tubing sizes not listed
All dimensions in the table are in inches.
For hydraulic plumbing, a safety factor of at least six should be used and rat
¿nan for thin factor age sh In the table For pressure rating at other safety
So aa) a a lee en
Ee OX ie On foin MS
PR ST
E E EE il 14
- 035, 055 ‚0024 30.800 5133 3.8507
e toe ;
EME, NS ec E
E A A La
“043 152 ‚9181 21560 3593 2,695
Sd Mos à à
m amt —
HO E a 2 je
# e ela ss
a e ss sn y à
e u oe
ON E a a |
HO À D ie à à
a 2 El ao
2 e $ y 4 de
O
HO À dd de à là
E E Ee dé à
072 286 COR 2 1%
2 33 Hi HA e le
oo A ne u ee,
E E 3 E là 1
A là ue
E 3 la ig là
e E ON lé a
S À llas
E dé ús ds 1
Pe
3 E E E E e
Ss E gf da là ie
083 584 2677 12173 2028 1522
E À E he e le
e E ls ly del la
gee ee i
E E 3 ie y
072 731 4195 9.051 1809 i
E E E $ là 8
109 657 3388 13.703 2.284 1713
TE E M mu A
Ss E u là E
e E Bee à
E E te 8 ig JB
E E Ee la là le
a N SE 5 8
ER ole là 8
corp a
Carbon Steel Tubing Data (cont.)
Tube Wall Tue de Burst Working Work
OD mex iD Are PS poe PSI 8.
CI NW JON 43 7
BEE E E E
E O1 O]
E ne 06
Eu re I 1
ee de de le 1%
F0 100 08 100 1 129
Tees 0 JM ES JM 2%
oe I 3 32 E
u: In dé nn à
“SO I Ov IA 100
V0 1200 12 a 1 1108
EE DNS DE DNS NS:
BB 2 1 ON ne
M "u A 0 3
oe ee 2 ds e
SO IE UM M 1e 4
FH IN jee GER BR
de a ra 6e 1a 108
z Ds TRIO 3 7451 38 7
Te u 3% 0 de
M lt JU 28 A di
Oe No ra: ee a
OI 24 2 100 1
ES E ae
e 178 88 10 18 3
Copper Tubing Data
Burst pressures are calculated by Barlow s formula: P = 21 S + O in which
P ie burst pressure PSI: tis tubing wall thickness; is ultimate strength of ma:
terial (32,000 PSI for coppen); and O ls outside diameter of tubing
Met Tee at M var
Bl Mar Me Men js ee
50 ee
Pr ee ee ee Eu;
5 H tm Ge ue
ae D me ig 1
Ae à à D |
EEE # à
ae 6 de à
oe @ 4 4 à
2 à © i à à
oo y D D à à
SE we 8
#2 u 5 à
«8 u u Y
‘Ss E $ E E à
2 8 i @ à
o 8 BE BE à à
E 8 E E ROS
VS D Oo 0 D
u i us à Où ON à
ee lm ta 2 © à
a a o o
RAR ET Satay aso TET
‘These are stands retigeaton sits ala aa mal supply houses
41
Tube Wall Tue de Burst Working Work
OD mex iD Are PS poe PSI 8.
CI NW JON 43 7
BEE E E E
E O1 O]
E ne 06
Eu re I 1
ee de de le 1%
F0 100 08 100 1 129
Tees 0 JM ES JM 2%
oe I 3 32 E
u: In dé nn à
“SO I Ov IA 100
V0 1200 12 a 1 1108
EE DNS DE DNS NS:
BB 2 1 ON ne
M "u A 0 3
oe ee 2 ds e
SO IE UM M 1e 4
FH IN jee GER BR
de a ra 6e 1a 108
z Ds TRIO 3 7451 38 7
Te u 3% 0 de
M lt JU 28 A di
Oe No ra: ee a
OI 24 2 100 1
ES E ae
e 178 88 10 18 3
Copper Tubing Data
Burst pressures are calculated by Barlow s formula: P = 21 S + O in which
P ie burst pressure PSI: tis tubing wall thickness; is ultimate strength of ma:
terial (32,000 PSI for coppen); and O ls outside diameter of tubing
Met Tee at M var
Bl Mar Me Men js ee
50 ee
Pr ee ee ee Eu;
5 H tm Ge ue
ae D me ig 1
Ae à à D |
EEE # à
ae 6 de à
oe @ 4 4 à
2 à © i à à
oo y D D à à
SE we 8
#2 u 5 à
«8 u u Y
‘Ss E $ E E à
2 8 i @ à
o 8 BE BE à à
E 8 E E ROS
VS D Oo 0 D
u i us à Où ON à
ee lm ta 2 © à
a a o o
RAR ET Satay aso TET
‘These are stands retigeaton sits ala aa mal supply houses
41
Stainless Steel Tubing Data
Stainless sec tubing is sometimes employed ether to handle corrosive 1
ids or higher pressures. If assembled with fare-type ftings, great care must be
Used not to crack the tubing while faring.
Pressure ratings are based on a maximum strength of 75.000 PSL typical of
‘Types 302, 303, 301, 309, 310, $16, 321, and 416. Types 202 and 440€ have
100.000 PSI while Types 410 and 430 have only 60.000 PSI maximum.
‘in hydraulic systems, a safety factor of atleast aix (6) should be used if there
is likely to be any shock in the system. To calculate working pressure at any
Safety factor take burst strength and divide by desired safety factor
Pressure ratings were calculated by Barlow's formula: P = 21% 8 + O, in
which P is burst pressure in PSL tis tubiny
‘Strength of tube material in PSI, © is tube O.D. All dimensions are in inches.
‘abe, ot te get Mine Neg
BB mille CRETE
= à = ro
EE EE.
7 | 13 38 |
LE | CE me
E CRE e
mE E —
& an 10
& a su a de
ST
da EE da à
== $ a es le
"E E 3 1% 8 le
a E la ee
# à E &
E a à
& e e $e
u u O ee
E E 8 3 12
INE à iB
E E i Be à À
g à ee
4 E E BUS là
E 8 ua le
5 à à Ee 2 le
À à à à a
oe ee
ee ae
e @ À 1 if ig
ES o ae
2 als à
& e la de la
E 8 DE E:
—e E TS ur
E E E id à là
SI à à à ie us
e E ue ie la
SH à oe ie |
ae oe la
ed oda a yl
"E 78 E 32 8 E
109 1.032 8360 13.080 2180 1535
5 1% de le e la
m i a a
"E D EC
HN à IE NE LE à
y 16 m la ls 1
Stainless sec tubing is sometimes employed ether to handle corrosive 1
ids or higher pressures. If assembled with fare-type ftings, great care must be
Used not to crack the tubing while faring.
Pressure ratings are based on a maximum strength of 75.000 PSL typical of
‘Types 302, 303, 301, 309, 310, $16, 321, and 416. Types 202 and 440€ have
100.000 PSI while Types 410 and 430 have only 60.000 PSI maximum.
‘in hydraulic systems, a safety factor of atleast aix (6) should be used if there
is likely to be any shock in the system. To calculate working pressure at any
Safety factor take burst strength and divide by desired safety factor
Pressure ratings were calculated by Barlow's formula: P = 21% 8 + O, in
which P is burst pressure in PSL tis tubiny
‘Strength of tube material in PSI, © is tube O.D. All dimensions are in inches.
‘abe, ot te get Mine Neg
BB mille CRETE
= à = ro
EE EE.
7 | 13 38 |
LE | CE me
E CRE e
mE E —
& an 10
& a su a de
ST
da EE da à
== $ a es le
"E E 3 1% 8 le
a E la ee
# à E &
E a à
& e e $e
u u O ee
E E 8 3 12
INE à iB
E E i Be à À
g à ee
4 E E BUS là
E 8 ua le
5 à à Ee 2 le
À à à à a
oe ee
ee ae
e @ À 1 if ig
ES o ae
2 als à
& e la de la
E 8 DE E:
—e E TS ur
E E E id à là
SI à à à ie us
e E ue ie la
SH à oe ie |
ae oe la
ed oda a yl
"E 78 E 32 8 E
109 1.032 8360 13.080 2180 1535
5 1% de le e la
m i a a
"E D EC
HN à IE NE LE à
y 16 m la ls 1
GPM flow capacities of tubing, and were calculated
V x A 4.0.9208, In which V = velocity of low in feet per
second, and A is inside square inch area of tubing
Figures in body of chart are GPM flows.
Oil Flow Capacity of Tubing
Figures in the chart are
from the formula: GPM
esse uses
SEEN REES
[2887888 25938928
Bere
¡5083338 58N3E9888
SENSES 53858988
8232088 83382033 3885383 3885885 38853328
ve
58
PARSONS 1088887 BHBRDSTS SSASERLE
RESCESS RERRRRa 20000 SANS
SESQNTS ARRZESE RANRRENR 36837580
Soadars 2APPNTS SERTESLE IEBERRER
a EEE
RER RUN BEER BEER
38888888
sa
78
‘
832395)
BESSER PRRPSSS EBBSAIS
aaa Rae SETS
882888 8538223 8538223
43
V x A 4.0.9208, In which V = velocity of low in feet per
second, and A is inside square inch area of tubing
Figures in body of chart are GPM flows.
Oil Flow Capacity of Tubing
Figures in the chart are
from the formula: GPM
esse uses
SEEN REES
[2887888 25938928
Bere
¡5083338 58N3E9888
SENSES 53858988
8232088 83382033 3885383 3885885 38853328
ve
58
PARSONS 1088887 BHBRDSTS SSASERLE
RESCESS RERRRRa 20000 SANS
SESQNTS ARRZESE RANRRENR 36837580
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BESSER PRRPSSS EBBSAIS
aaa Rae SETS
882888 8538223 8538223
43
Air Line Pipe Size
Figures in body of chart are pipe izes on a 100 PSI ar system wo cary air
at about a 1 PSI per 100 fect prescure loss. When measuring piping distances,
to be conservative, count ench pipe Biting as equal to ect ul pipe, At other
than 100 PSI, How capacity will bein inverse proportion to pressure (as based
‘on PSIA (absolute) pressure level and calculated by Boyle's Law.
Serm Length of Run - Feet Comp.
Flow | 25 so 75 100 150 200 300 500 1000| "Hi
ol m ve v2 12 34 aa] 1
mr 12 EN FA fe del
30 | 3% EN i mm 1m] 8
45 | se sa 1 1 m m| m
gs À 1 Me m m| 10
CRE m ve 6 2 |
w|ı m m ve 2 2 | 2
150 | 1% m va 2 2 | 2
130 | 1% m ve 2 2 2 | %
200 | mm 2 2 m 3 | 4
So | 2 2 En 3 3 | %
So | te 2 2 De 3 3 | &
so |2 2 mm 9 3. 36 | 7
Goo | 2 3% 2m Bf 5° 3 3 3 4 | 109
ml 2 os 3° 3 3 m 3% à | ia
Air Pressure Loss
Figures in this table are approximate PSI compressed air pressure losses for
every 100 feet of clean commercial steel pipe, Schedule 40.
‘Nominal Pipe Diameter
crm | WINCH | samc | sich | ranch
Free | 80 125] 80 125| 90 128| 90 125
‘air | PS PSI] psı PSI] PSI PSI| psı PSI
410] 175 120
515| 225 150| 5
mo] as alu 08] 04 02
20| 176 115| 40 28/15 08
30| 385 255| 90 60/ 30 20
40] 695 455] 155 105] 45 30
so] 105 700| 240 160| 75 50
© 235| 100 70
70 315| 135 90
E
El
100 635 | 270 180| .65
125 9so| 420 280| 105
150 145| 575 400| 145
5as | 200
7,0 | 260 115 0
180 120
255 170
790 355 23
103 455 305
--- 580 380
= 710 470
Figures in body of chart are pipe izes on a 100 PSI ar system wo cary air
at about a 1 PSI per 100 fect prescure loss. When measuring piping distances,
to be conservative, count ench pipe Biting as equal to ect ul pipe, At other
than 100 PSI, How capacity will bein inverse proportion to pressure (as based
‘on PSIA (absolute) pressure level and calculated by Boyle's Law.
Serm Length of Run - Feet Comp.
Flow | 25 so 75 100 150 200 300 500 1000| "Hi
ol m ve v2 12 34 aa] 1
mr 12 EN FA fe del
30 | 3% EN i mm 1m] 8
45 | se sa 1 1 m m| m
gs À 1 Me m m| 10
CRE m ve 6 2 |
w|ı m m ve 2 2 | 2
150 | 1% m va 2 2 | 2
130 | 1% m ve 2 2 2 | %
200 | mm 2 2 m 3 | 4
So | 2 2 En 3 3 | %
So | te 2 2 De 3 3 | &
so |2 2 mm 9 3. 36 | 7
Goo | 2 3% 2m Bf 5° 3 3 3 4 | 109
ml 2 os 3° 3 3 m 3% à | ia
Air Pressure Loss
Figures in this table are approximate PSI compressed air pressure losses for
every 100 feet of clean commercial steel pipe, Schedule 40.
‘Nominal Pipe Diameter
crm | WINCH | samc | sich | ranch
Free | 80 125] 80 125| 90 128| 90 125
‘air | PS PSI] psı PSI] PSI PSI| psı PSI
410] 175 120
515| 225 150| 5
mo] as alu 08] 04 02
20| 176 115| 40 28/15 08
30| 385 255| 90 60/ 30 20
40] 695 455] 155 105] 45 30
so] 105 700| 240 160| 75 50
© 235| 100 70
70 315| 135 90
E
El
100 635 | 270 180| .65
125 9so| 420 280| 105
150 145| 575 400| 145
5as | 200
7,0 | 260 115 0
180 120
255 170
790 355 23
103 455 305
--- 580 380
= 710 470
Air Flow Loss Through Pipes
Table of Factors - See Instructions for use on page 46.
ene nego ara
2a Ssase|
azgag serge ganas
SCFM
5
0
13
2
BJ
20
3S
20
RER Ina aasıa NESZIELERR SSE IL Sadia SITES ZEN]
alii | isczena suns? 2208880808 40884) | leas sana 2020 anges 29888
; sense SSIEN Roses Bune SEHE El lens sance sono cease SEES 244)
E 23988 SEE STE 1478 press eee 2200 2885 Beas 23558
Lives 28393 282) eset 22582 Saaae 82829 Sages Seek
is oehS3 28838 883) seeds 33808 aces)
DEREN (¿8863 SEES pass grasses
NS BES EIRSBSESSSERERGBS GEARS 88858 58889
45
Table of Factors - See Instructions for use on page 46.
ene nego ara
2a Ssase|
azgag serge ganas
SCFM
5
0
13
2
BJ
20
3S
20
RER Ina aasıa NESZIELERR SSE IL Sadia SITES ZEN]
alii | isczena suns? 2208880808 40884) | leas sana 2020 anges 29888
; sense SSIEN Roses Bune SEHE El lens sance sono cease SEES 244)
E 23988 SEE STE 1478 press eee 2200 2885 Beas 23558
Lives 28393 282) eset 22582 Saaae 82829 Sages Seek
is oehS3 28838 883) seeds 33808 aces)
DEREN (¿8863 SEES pass grasses
NS BES EIRSBSESSSERERGBS GEARS 88858 58889
45
Instructions
For Estimating Air Flow Loss Through Pipes.
To estimate the in pressure loss through pipe And the Factor frm the
chart on page 43 according to the pipe size and SCFM flow. Take the factor and
Sivide it by the ratio of compression (calculated in absolute pressure values)
Compression ratio will be [gauge pressure + 18.5] + 14.5 PSI. Then multiply
th umher y acta eng of pps fs, hen ide by 1000, Thi il ne
Pressure loss in
Pressure Loss Through Fittings
Figure in he body of this chart ae ai pease lo loses Uhr screw
Aktings expressed inequivalent lengths of straight pipe of che same diameter.
ics on a gate valve the How resistance would be the same as 0.57
foot of straight 1 pipe
Pipe Long Medium Close Too
Size | Gate Radius Radius Standard Angle Return Thru Globe
NPT | Vaive _Ellor® Ellor” Ellor™ Valve Bend Side Valve
TT 04 08 OB 11 19 17 25
3% | 032 08 073 12 16 18 28 33
108 07 09 16 21 23 31 47
mej og CT Ya 22 29 33 44 68
vá | 098 13 18 26 3s 33 52 ?8
2| 13 17 22 38 48 53 71 108
EU 18 22 38 38 59 88 87 191
3| 21 38 38 87 77 88 114 11
al 39 50 79 107 18 188 27
sl 39 51 85 «04 139 188 207 9
“Oy on run of standard tee
Gran tun oto reduced in size 25%
“Gen run of standard tee reduced in size 50%
Friction of Air in Hose
Pressure drop per 25 feet. In proportion for longer or shorter lengths
Size | SCFM TOPS! BOPSI_9OPSI_100PSI_110P8I
veo] 20 10 09 08 07 0
Ed 32 28 24 23 20
Ed 79 60 53 48 43
El Ra 18 $5 84 78
Ed 200 174 138 133 120
7 284 202 220 193 178
E Ble 308 272 28
saw] 20] 04 03 02 02 02 02 01
So | 08 08 05 08 04 0a 03
30 | 5 12 0s 08 07 08 05
S| 28 19 33 13 11 10 08
so | 35 28 23 18 18 14 13
do | 33 33 32 28 23 20 18
a| 65 32 42 38 3: 37 24
S| 83 88 53 37 40 8% $1
so! 118 86 70 58 §0 44 39
MO | 12 112 a8 72 62 54 49
rio | ale 02 of of of 01 01
do | 03 03 02 02 02 02 02
S| 03 Oa 04 03 03 02 9
Bo | 08 08 05 05 04 04 03
Ol 11 08 67 07 08 05 04
ml 18 12 10 08 07 08 0
%| 20 10 13 11 09 08 oF
wo] 38 20 18 14 12 10 0
mo] 35 28 20 17 14 12 1
Data on pages 45 and 46 were adapted from the Trade Standards adopted by the
‘Compressed Air and Gas Institute.
46
For Estimating Air Flow Loss Through Pipes.
To estimate the in pressure loss through pipe And the Factor frm the
chart on page 43 according to the pipe size and SCFM flow. Take the factor and
Sivide it by the ratio of compression (calculated in absolute pressure values)
Compression ratio will be [gauge pressure + 18.5] + 14.5 PSI. Then multiply
th umher y acta eng of pps fs, hen ide by 1000, Thi il ne
Pressure loss in
Pressure Loss Through Fittings
Figure in he body of this chart ae ai pease lo loses Uhr screw
Aktings expressed inequivalent lengths of straight pipe of che same diameter.
ics on a gate valve the How resistance would be the same as 0.57
foot of straight 1 pipe
Pipe Long Medium Close Too
Size | Gate Radius Radius Standard Angle Return Thru Globe
NPT | Vaive _Ellor® Ellor” Ellor™ Valve Bend Side Valve
TT 04 08 OB 11 19 17 25
3% | 032 08 073 12 16 18 28 33
108 07 09 16 21 23 31 47
mej og CT Ya 22 29 33 44 68
vá | 098 13 18 26 3s 33 52 ?8
2| 13 17 22 38 48 53 71 108
EU 18 22 38 38 59 88 87 191
3| 21 38 38 87 77 88 114 11
al 39 50 79 107 18 188 27
sl 39 51 85 «04 139 188 207 9
“Oy on run of standard tee
Gran tun oto reduced in size 25%
“Gen run of standard tee reduced in size 50%
Friction of Air in Hose
Pressure drop per 25 feet. In proportion for longer or shorter lengths
Size | SCFM TOPS! BOPSI_9OPSI_100PSI_110P8I
veo] 20 10 09 08 07 0
Ed 32 28 24 23 20
Ed 79 60 53 48 43
El Ra 18 $5 84 78
Ed 200 174 138 133 120
7 284 202 220 193 178
E Ble 308 272 28
saw] 20] 04 03 02 02 02 02 01
So | 08 08 05 08 04 0a 03
30 | 5 12 0s 08 07 08 05
S| 28 19 33 13 11 10 08
so | 35 28 23 18 18 14 13
do | 33 33 32 28 23 20 18
a| 65 32 42 38 3: 37 24
S| 83 88 53 37 40 8% $1
so! 118 86 70 58 §0 44 39
MO | 12 112 a8 72 62 54 49
rio | ale 02 of of of 01 01
do | 03 03 02 02 02 02 02
S| 03 Oa 04 03 03 02 9
Bo | 08 08 05 05 04 04 03
Ol 11 08 67 07 08 05 04
ml 18 12 10 08 07 08 0
%| 20 10 13 11 09 08 oF
wo] 38 20 18 14 12 10 0
mo] 35 28 20 17 14 12 1
Data on pages 45 and 46 were adapted from the Trade Standards adopted by the
‘Compressed Air and Gas Institute.
46
Air Flow Through Orifices
Figures in this chart show theoretical SCFM air flow through sharp edged
ries practice, nly about 2rd of tie flow i obtained. The char may be
lui or rouehly'estimating travel speed of a loaded air elinder Asume
about 75% of the line PSI is actually working on the load, with the remaining
7 consumed in flow losses in the 4 way valve and connecting lines. Caleulate
of your Incoming line PSL and use this figure to enter the first column in
this chart. Move across the table to the column headed by the actual port size
of the day valve in the circuit. Use about half the flow shown, because a
‘Sway valve is not a sharp edged orifice, and will usually pass only about half
as much air as a sharp edged orifice
‘After finding the SCFM free air low, convert this to CEM (compressed air
flow) at the pressure required to move the load. From this the speed of travel of
the air eylinder can be estimated
Chart shows approximate SCFM (tre air) flow through sharp edged orifices.
Across rie Diameter, in inches
Once [1764152 16 m a a 12 58 Ji 76 1
06? 249 983 397 159 357 G35 003 149 195 254
OS 272 109 434 174 391 695 109 188 213 278
073 289 117 488 187 422 790 117 188 230 300
083 331 132 590 212 477 847 132 191 280 309
(095 278 152 Gor 243 546 970 152 218 207 38
105 158 672 269 605 108 168 242 329 400
se
196 786 314 707 126 106 283 385 503
592 237 988 213 379 802 653 161 1816
849 260 104 20% 218 649 93% 1272 1061
705 202 113 254 452 703 1016 1383 1805
792 305 122 27a sen 702 1007 1404 1081
790 318 128 204 1938 1549 2029
Vacuum Flow Through “Orifices
‚This chart approximates the flow that might be expected through a practical
orifice: Flows are about 2/rds the theoretical low obtained through a sharp
‘aged orifice, A best, these figures are only approximate because the How
‘Seleristic of your orifice ean only be determined by actual measureme
Specified conditions
DESIGN NOTE: This chart shows that multiple-hole grippers work more ef
ficiently at reasonably high vacuums. For example, looking at the chart for a
YA” diameter hole, the first 6° Hg of vacuum flows 8.25 SCEM, while the in
{crease inflow over the last 6, from 18" to 24 is only 2.2 SCFM. The more eff.
‘lent design would be to use more smaller holes working at a higher vacuum.
Figures in body of chart are air flows in SCFM (standard cubic feetiminute)
Ori
476
98
5
8
7
3
12
15
2
3
50
35
40
as
50 | 229
El
”
80
©
100
0
120
130,
Degree of Vacuum Across Orifice, Inches Mercury (Hg)
PANA A
‘018 025 0 07 MT US US 0 0
074 (100 128 18 165 180 19 220 20
300 420 $17 595 560 725 700 280 100
120 168 206 237 264 289 312 353 404
A7 674 825 982 106 116 124 140 102
108 152 185 214 238 260 280 318 364
191 270 330 385 423 463 500 565 546
0 422 517 503 662 728 780 880 101
330 606 740 853 952 104 112 127 145
8
5
828 ‘tor 116 190 12 183 173 198
108 191 12 169 185 200 225 288
47
Figures in this chart show theoretical SCFM air flow through sharp edged
ries practice, nly about 2rd of tie flow i obtained. The char may be
lui or rouehly'estimating travel speed of a loaded air elinder Asume
about 75% of the line PSI is actually working on the load, with the remaining
7 consumed in flow losses in the 4 way valve and connecting lines. Caleulate
of your Incoming line PSL and use this figure to enter the first column in
this chart. Move across the table to the column headed by the actual port size
of the day valve in the circuit. Use about half the flow shown, because a
‘Sway valve is not a sharp edged orifice, and will usually pass only about half
as much air as a sharp edged orifice
‘After finding the SCFM free air low, convert this to CEM (compressed air
flow) at the pressure required to move the load. From this the speed of travel of
the air eylinder can be estimated
Chart shows approximate SCFM (tre air) flow through sharp edged orifices.
Across rie Diameter, in inches
Once [1764152 16 m a a 12 58 Ji 76 1
06? 249 983 397 159 357 G35 003 149 195 254
OS 272 109 434 174 391 695 109 188 213 278
073 289 117 488 187 422 790 117 188 230 300
083 331 132 590 212 477 847 132 191 280 309
(095 278 152 Gor 243 546 970 152 218 207 38
105 158 672 269 605 108 168 242 329 400
se
196 786 314 707 126 106 283 385 503
592 237 988 213 379 802 653 161 1816
849 260 104 20% 218 649 93% 1272 1061
705 202 113 254 452 703 1016 1383 1805
792 305 122 27a sen 702 1007 1404 1081
790 318 128 204 1938 1549 2029
Vacuum Flow Through “Orifices
‚This chart approximates the flow that might be expected through a practical
orifice: Flows are about 2/rds the theoretical low obtained through a sharp
‘aged orifice, A best, these figures are only approximate because the How
‘Seleristic of your orifice ean only be determined by actual measureme
Specified conditions
DESIGN NOTE: This chart shows that multiple-hole grippers work more ef
ficiently at reasonably high vacuums. For example, looking at the chart for a
YA” diameter hole, the first 6° Hg of vacuum flows 8.25 SCEM, while the in
{crease inflow over the last 6, from 18" to 24 is only 2.2 SCFM. The more eff.
‘lent design would be to use more smaller holes working at a higher vacuum.
Figures in body of chart are air flows in SCFM (standard cubic feetiminute)
Ori
476
98
5
8
7
3
12
15
2
3
50
35
40
as
50 | 229
El
”
80
©
100
0
120
130,
Degree of Vacuum Across Orifice, Inches Mercury (Hg)
PANA A
‘018 025 0 07 MT US US 0 0
074 (100 128 18 165 180 19 220 20
300 420 $17 595 560 725 700 280 100
120 168 206 237 264 289 312 353 404
A7 674 825 982 106 116 124 140 102
108 152 185 214 238 260 280 318 364
191 270 330 385 423 463 500 565 546
0 422 517 503 662 728 780 880 101
330 606 740 853 952 104 112 127 145
8
5
828 ‘tor 116 190 12 183 173 198
108 191 12 169 185 200 225 288
47
Oil Flow Through Orifices
‘These charts show PSI pressure drops to be expected in hydraulic oil when
flowing through sharp edged once, Cautiont Calculated pressure drops are
‘only approximate because factors such as specific gravity, viscosity, shape of oF
{ice and plumbing ahead of and following the orice may cause variations. IL
is beat to make the orifice slightly undersize to start. then to gradually enlarge
At wile measuring actual pressure drop
‘hy making the orice al sharp waged as possible it becomes less sensitive to
oil temperature changes (which affect oil viscoso)
Specie gravity of the fluid significantly influences the pressure drop, which
increases approximately as the square of the increase of specific gravity. The
‘charts were calculated far où with a gravity of 0.9, close approximation for hy.
Äraulicoil Using other Auide, a multiplying factor must be applied to chart val
es For example to find the pressure drop of water, which has a gravity of 1.00,
find the multiplier as follows:
(100)? (09)? = 100 + 081 = 1.23 Multiplying Factor
‘Therefore, multiply all chart values by 1.23 when caleulating for water flow.
‘These charts were calculated from information supplied by Double A Prod:
ets Co, The constant, 23.5, shown in the formula below was developed exper
‘mentally by measuring pressure drops across average orifices, Values not shown.
may be calculated from the same basic formula usud in enleulating the chart
Presure Drop (AP) = (GPM «(28S A)
Pressure Drop Across Orifices from 3/64” to 3/16"
Figures in the body of these charts are PSI pressure drops to be expected in
a flow of hydraulic oil across sharp edged orifices of various diameters
Orifice Diameters in Inches
GPM| aes 116 5/64 3/32 7/64 WB 9/64 G2 11160 96
3| 5445 1730 70 HO 185 NO 68 34 30 2
al 1260 608 328 192 120 79 54 38
222-4803 1070 950 613 300 188 123 84 59
FAT == 4490 2140 1185 677 42 277 189 19
10 3600 2050 1205 750 493 336 238
12% 3205 1880 1175 770 528 31
15 4615 2705 1890 1110 767 5%
3810 3005 1970 1325 050
1705 1205
2100 1485
2648 1798
2025 2140
4120 2910
=== 3800
Pressure Drop Across Orifices from 13/64” to 1/2"
Orifice Diameters in Inches
Gem | 13/58 7192 15/64 wa 9/32 5/16 11/32 Ye 76 12
3] 16 12 = =
|» 2 ‘Chart Values are in PSI === "|
| 43 2 2 1 12 --
Te] 97 Rn 8 © % 17 72
To] 172 128 8 75 47 1 2
nan 270 200 153 117 73 48 33
15] 388 296 220 160 106 6 47 3 18 M1
17% | 528 393 300 20 148 94 68 45 25 14
20| 690 515 392 301 168 123 Bi 59 32 19
ZA] 873 649 46 380 297 158 106 7 4 24
25 | 1075 800 612 470 293 192 131 93 50 29
an [1308 970 Tai 568 35 233 189 112 81 36
30] 1850 1185 0 675 420 27 189 194 72 42
35 | 2115 1570 1200 920 575 377 258 182 98 sa
3012760 2050 1570 1200 751 492 306 237 128 75
‘These charts show PSI pressure drops to be expected in hydraulic oil when
flowing through sharp edged once, Cautiont Calculated pressure drops are
‘only approximate because factors such as specific gravity, viscosity, shape of oF
{ice and plumbing ahead of and following the orice may cause variations. IL
is beat to make the orifice slightly undersize to start. then to gradually enlarge
At wile measuring actual pressure drop
‘hy making the orice al sharp waged as possible it becomes less sensitive to
oil temperature changes (which affect oil viscoso)
Specie gravity of the fluid significantly influences the pressure drop, which
increases approximately as the square of the increase of specific gravity. The
‘charts were calculated far où with a gravity of 0.9, close approximation for hy.
Äraulicoil Using other Auide, a multiplying factor must be applied to chart val
es For example to find the pressure drop of water, which has a gravity of 1.00,
find the multiplier as follows:
(100)? (09)? = 100 + 081 = 1.23 Multiplying Factor
‘Therefore, multiply all chart values by 1.23 when caleulating for water flow.
‘These charts were calculated from information supplied by Double A Prod:
ets Co, The constant, 23.5, shown in the formula below was developed exper
‘mentally by measuring pressure drops across average orifices, Values not shown.
may be calculated from the same basic formula usud in enleulating the chart
Presure Drop (AP) = (GPM «(28S A)
Pressure Drop Across Orifices from 3/64” to 3/16"
Figures in the body of these charts are PSI pressure drops to be expected in
a flow of hydraulic oil across sharp edged orifices of various diameters
Orifice Diameters in Inches
GPM| aes 116 5/64 3/32 7/64 WB 9/64 G2 11160 96
3| 5445 1730 70 HO 185 NO 68 34 30 2
al 1260 608 328 192 120 79 54 38
222-4803 1070 950 613 300 188 123 84 59
FAT == 4490 2140 1185 677 42 277 189 19
10 3600 2050 1205 750 493 336 238
12% 3205 1880 1175 770 528 31
15 4615 2705 1890 1110 767 5%
3810 3005 1970 1325 050
1705 1205
2100 1485
2648 1798
2025 2140
4120 2910
=== 3800
Pressure Drop Across Orifices from 13/64” to 1/2"
Orifice Diameters in Inches
Gem | 13/58 7192 15/64 wa 9/32 5/16 11/32 Ye 76 12
3] 16 12 = =
|» 2 ‘Chart Values are in PSI === "|
| 43 2 2 1 12 --
Te] 97 Rn 8 © % 17 72
To] 172 128 8 75 47 1 2
nan 270 200 153 117 73 48 33
15] 388 296 220 160 106 6 47 3 18 M1
17% | 528 393 300 20 148 94 68 45 25 14
20| 690 515 392 301 168 123 Bi 59 32 19
ZA] 873 649 46 380 297 158 106 7 4 24
25 | 1075 800 612 470 293 192 131 93 50 29
an [1308 970 Tai 568 35 233 189 112 81 36
30] 1850 1185 0 675 420 27 189 194 72 42
35 | 2115 1570 1200 920 575 377 258 182 98 sa
3012760 2050 1570 1200 751 492 306 237 128 75
A
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Straight Thread Fitting Sizes
Is chart gives thread size and O-ring size us used on straight thread con
rector, stright thread tube Rings; ce These sls are applicable for SAB,
AS sii MS tnneetiony Orig ed for tra connectors o ot conf 1
called “standard size: They should be purchased specially for Ihe service
nd should conform te dimensions shown.
ARP 568
Fit Tubing Thread Uniform O-Ring _ O-Ring
DashNe. | "O0. Size DashNo. 1D.” Thickness
2 ve BEA M2 02 004
3 ane 38-24 “203 0301 0.064
3 vs" TI 908 0351 0072
3 516 12-20 “gs 0414 0072
3 38. anse 906 0468 0078
= m 3-16 “90s 064 0.087
0 EN Tea, “10 0755 0097
2 En Metz 2 084 on
en mer 1 ne
6 Wet? 6 m 1
20 men 0 1475 ons
As v2 4 1720 018
= 242 2 2337 0m
Equivalent Pipe and Tubing Sizes
‘This table suggests a comparable size when going from pipe into tubing and
vice versa, These sizes have approximately equal How capacity. For sizes over
Une inch, use pipe and tubing of the same size rating.
Tubing OD. Inches| a 56 8 12 Se au 70 1
Pipe Size NPT | 18 V6 14 36 12 A 34
ISO Standardization Effort
‘The International Standards Organization (ISO) is attempting to establish
a set of port and tubefhose connection standards for worldwide use They intend
to recognae 10 standards as outlined inthe table bel: They endorse strongly
‘Matos Connection
es mer [Has
PP ESA
roraseame| ie [io ee [so sc sone] so mora
(RE,
Ta
Rue, | SE [soma sore] ——— [some
Gear data
Par Metric.
pl, [osea] — | ——— | sos
are
pee
| O |
ER
50
Is chart gives thread size and O-ring size us used on straight thread con
rector, stright thread tube Rings; ce These sls are applicable for SAB,
AS sii MS tnneetiony Orig ed for tra connectors o ot conf 1
called “standard size: They should be purchased specially for Ihe service
nd should conform te dimensions shown.
ARP 568
Fit Tubing Thread Uniform O-Ring _ O-Ring
DashNe. | "O0. Size DashNo. 1D.” Thickness
2 ve BEA M2 02 004
3 ane 38-24 “203 0301 0.064
3 vs" TI 908 0351 0072
3 516 12-20 “gs 0414 0072
3 38. anse 906 0468 0078
= m 3-16 “90s 064 0.087
0 EN Tea, “10 0755 0097
2 En Metz 2 084 on
en mer 1 ne
6 Wet? 6 m 1
20 men 0 1475 ons
As v2 4 1720 018
= 242 2 2337 0m
Equivalent Pipe and Tubing Sizes
‘This table suggests a comparable size when going from pipe into tubing and
vice versa, These sizes have approximately equal How capacity. For sizes over
Une inch, use pipe and tubing of the same size rating.
Tubing OD. Inches| a 56 8 12 Se au 70 1
Pipe Size NPT | 18 V6 14 36 12 A 34
ISO Standardization Effort
‘The International Standards Organization (ISO) is attempting to establish
a set of port and tubefhose connection standards for worldwide use They intend
to recognae 10 standards as outlined inthe table bel: They endorse strongly
‘Matos Connection
es mer [Has
PP ESA
roraseame| ie [io ee [so sc sone] so mora
(RE,
Ta
Rue, | SE [soma sore] ——— [some
Gear data
Par Metric.
pl, [osea] — | ——— | sos
are
pee
| O |
ER
50
Thread Forms of Fluid Connectors
National Pipe Thread Fuel (NPTF)
e Thread conforms to ANSI B1 203 CE
+ Physically interchangeable with NPT but has modified [~~]
threads for better pressure tight sealing id
(© Tapered thread profile seals by metal to metal tt
erierence tually requires Sealing compound
dor pressure tight connections
+ Pitch and diameter are measured in
fe Taper angle is 0.75" per foot or 1°47"
Thread angle is 60
Straight Thread O-Ring (SAE)
“e Thread conforms to 180 263 and ANSI B1.1 Unified
+ Port conforms to ISO 11926 and SAE J1926
‘+ Commonly called straight thread O-ring fittings
+ Pitch and diameter are measured in inches,
Del UND
+ Threads are parallel and requires O-ring for pressure
ight connection
e Thread angle is 60
ches
British Standard Pipe Tapered (BSPT)
Thread conforms to 180 7
Pitch and diameter are measured in inches,
ee Gund
+ Tapered thread profile seals by metal to metal
iterterence usual requires sealing compound
for pressure tight conrecton
+ Taper angle is 1" 47, the same as NPTIF)
Thread angle is 55
© Not interchangeable with NPT)
British Standard Pipe Parallel (BSPP)
Thread conforms to 150 228-1
Port conforms to ISO 1179
Pitch and diameter measured in inches,
5004.10
‘© Parallel threads require O-ring, rush
washer, gasket or metal to metal seal
between connections for pressure ight
connection
e Thread angle is 55
© Not interchangeable with SAE or NPTIF)
Metric Straight Thread O-Ring
(© Thread conforms to ISO 261
© Port conforms to ISO 6149 and SAE J2244
Pitch and diameter
eg. M22 13
+ Parallel threads require O-ring for pressure tight
e Thread angle is 60
+ Easily identified by raised ridge on female port
‘counterbor
+ Not interchangeable with SAE or BSPP
51
National Pipe Thread Fuel (NPTF)
e Thread conforms to ANSI B1 203 CE
+ Physically interchangeable with NPT but has modified [~~]
threads for better pressure tight sealing id
(© Tapered thread profile seals by metal to metal tt
erierence tually requires Sealing compound
dor pressure tight connections
+ Pitch and diameter are measured in
fe Taper angle is 0.75" per foot or 1°47"
Thread angle is 60
Straight Thread O-Ring (SAE)
“e Thread conforms to 180 263 and ANSI B1.1 Unified
+ Port conforms to ISO 11926 and SAE J1926
‘+ Commonly called straight thread O-ring fittings
+ Pitch and diameter are measured in inches,
Del UND
+ Threads are parallel and requires O-ring for pressure
ight connection
e Thread angle is 60
ches
British Standard Pipe Tapered (BSPT)
Thread conforms to 180 7
Pitch and diameter are measured in inches,
ee Gund
+ Tapered thread profile seals by metal to metal
iterterence usual requires sealing compound
for pressure tight conrecton
+ Taper angle is 1" 47, the same as NPTIF)
Thread angle is 55
© Not interchangeable with NPT)
British Standard Pipe Parallel (BSPP)
Thread conforms to 150 228-1
Port conforms to ISO 1179
Pitch and diameter measured in inches,
5004.10
‘© Parallel threads require O-ring, rush
washer, gasket or metal to metal seal
between connections for pressure ight
connection
e Thread angle is 55
© Not interchangeable with SAE or NPTIF)
Metric Straight Thread O-Ring
(© Thread conforms to ISO 261
© Port conforms to ISO 6149 and SAE J2244
Pitch and diameter
eg. M22 13
+ Parallel threads require O-ring for pressure tight
e Thread angle is 60
+ Easily identified by raised ridge on female port
‘counterbor
+ Not interchangeable with SAE or BSPP
51
Three-Phase Motor Data
Frame Assignments and Dimensions for Squirrel-Cage
Induction Motors — 3-Phase, 60 Hz, Design B
Mel à v
a | ut
|
Drip-proof (Open Type) Enclosures
Speed NEMA Sha
Hp “APM Frame UC A D 0 v key
vast Je Hu ET
7200 EE
3800 Wer 78 10% 8% 3% & 24 3
3
e
E ek &
EEE
19360 A la 15% 1 Ba
O
MU MA ES
RE:
SEHEN
EEE BEE
a du on
OE
ENE MEE A E
| u u
Ei.) N
[503600 Seats m 244 18e 8 18% 3% 12
oe ee Rm Pe ee
[60 3600 dors Ve 26 18% 8 16% Mm 1/2
EE SEE TMER
TE MA EEE
EE E ee
100 3600 Sests i 27% tem 9 tem 54 12
(This table is continued on the next page)
52
Frame Assignments and Dimensions for Squirrel-Cage
Induction Motors — 3-Phase, 60 Hz, Design B
Mel à v
a | ut
|
Drip-proof (Open Type) Enclosures
Speed NEMA Sha
Hp “APM Frame UC A D 0 v key
vast Je Hu ET
7200 EE
3800 Wer 78 10% 8% 3% & 24 3
3
e
E ek &
EEE
19360 A la 15% 1 Ba
O
MU MA ES
RE:
SEHEN
EEE BEE
a du on
OE
ENE MEE A E
| u u
Ei.) N
[503600 Seats m 244 18e 8 18% 3% 12
oe ee Rm Pe ee
[60 3600 dors Ve 26 18% 8 16% Mm 1/2
EE SEE TMER
TE MA EEE
EE E ee
100 3600 Sests i 27% tem 9 tem 54 12
(This table is continued on the next page)
52
(This table is continued from the previous page)
wea gun
we ‘Wit ME y ao o y À
BRE SEE RP 2 RES
Be, A Bi
a
Se a ey À à à
EE 150 E
ESS TE |
CEE EEE En coe
ER MEET ENG
Totally Enclosed, Fan Cooled (TEFC)
Pa vi
Tie tt 3
EE.
NE:
nn
wu
ME
|;
¡EE
HE!
¡EE
a
(BE:
HE:
i
TR E
NE:
DE:
RR
ar
E ae,
E a.
CNE
wi
RE:
eer
iS ae
EME Bis
er
BE:
a
.-—.;
a à
RR Si.
weg E
DR de de à
ee E
oe
ENE
NOTES: Dimensions ane cen A the nearest sntenth fa inch to stand
[A specification. “Dim: C will vary with motor brand as this is not specified
by NEMA ” ”
53
wea gun
we ‘Wit ME y ao o y À
BRE SEE RP 2 RES
Be, A Bi
a
Se a ey À à à
EE 150 E
ESS TE |
CEE EEE En coe
ER MEET ENG
Totally Enclosed, Fan Cooled (TEFC)
Pa vi
Tie tt 3
EE.
NE:
nn
wu
ME
|;
¡EE
HE!
¡EE
a
(BE:
HE:
i
TR E
NE:
DE:
RR
ar
E ae,
E a.
CNE
wi
RE:
eer
iS ae
EME Bis
er
BE:
a
.-—.;
a à
RR Si.
weg E
DR de de à
ee E
oe
ENE
NOTES: Dimensions ane cen A the nearest sntenth fa inch to stand
[A specification. “Dim: C will vary with motor brand as this is not specified
by NEMA ” ”
53
OL AH COW HS GLS
Bisssse==]
MHA NHL HH — 9215 om ui
fRagse….
ion
(06 ‘dH JOIO AH O02 OL Sz
AH WHR HN — ong a
54
slouvis Jon oseud-e
DIDA UOD]]DJSU] 1070
06 aH soion aH OZ OLE
Bisssse==]
MHA NHL HH — 9215 om ui
fRagse….
ion
(06 ‘dH JOIO AH O02 OL Sz
AH WHR HN — ong a
54
slouvis Jon oseud-e
DIDA UOD]]DJSU] 1070
06 aH soion aH OZ OLE
Wire Selection Guide
“Two important considerations in choosing the conductor sie for electric wir.
ing are: (Ih the sale current carrying capacity, and (2) the voltage loss due to
wire resistance. On short runs. up to 20 feet, voltage los is very low and need
idee Wire sio shu be sli ors cree expat ax shown
‘On longer runs, several hundred feet or more, the voltage loss may be too
hgh we ie plc solely on the banc a arent capac Tanger ire
Size should be used to keep voltage lose toa selected minimum. Chart 3 may be
ted for thi oe sf
Permissible Voltage Loss ...
‘There is always a voltage loss on any wiring run. The designer must dec
on how much loss can be tolerated without seriously affecting performance, and
‘must select a wire size in which this loss will not be exceeded. A rule-of-thumb
Suggests that electric motors should not be run on a voltage les than about 10%
of their nameplate rating. In deciding on allowable voltage losa in the wiring a
designer must consider the minimun available power line voltage which may
Sceur at a certain time of the day. For example, a 200.vol rated motor should
fot opens than 209 volt which Oe les than namepate voltage Rat
inp) If the lowest power line voltage is 220 volta, then the wiring should not
hatte more than 12 vols loss à
“A rule that works in most cases is to choose a wire size which does not give
more than a 5% lows of input voltage
Chart 1 - Wire Ampacity For Short Wiring Runs
“Ampacity” is an abbreviation for ampere capacity. This chart is for short
wiring runs of less than 20 feet, Ampere capacity is taken from the NEC (Na-
Sonal Electrical Code) on wire sizes of No. 14 and larger It is for insulated cop-
per wire ofthe kind that is widely used for house and building wiring. A larger
pare capacity allowed on wre th certain types of insulin when used
‘under certain conditions, but the NEC handbook should be consulted
‘Amperage Rating for Copper Wires
Mire sizo, AWG 18161412108 6 4 3 2 1 0 00 om
Tnraceway orcable” $3 182030 49 TD 125 ius 165
In open a 8 12 35 28 38 28 88 108 120 180 tes 188 208 280
Chart 2 — Voltage Loss on Long Wiring Runs
"This charts for long wiring runs of several hundred fet or more, To use the
chart, several facts must be established: (1) the current draw ofthe device to be
‘perated must be determined: (2) the amount of voltage fos that can be tler
led must be decided on: and (3) the length of wire must be estimated or mea
Sed, sing the sum of outgoing and retuen wie length
‘On S-phase devices such as electric motors, cach of the three wires must car-
re cren shi onthe motor nameplate Wi log is he sum afoot
Read across the top of the chart to find the column which matches the a
perage rating of the device. Figures in this column show voltage fosses for 1000
Reet of wire, outgoing plus return, Y for example, our total wire length was 250
fet, voltage losses would be Ith that shown in Ihe chart
Figures in Chart Show Voltage Loss Per 1000 Feet of Wire
sre Current Flow, Amperes.
|
© [RE wa a:
PURE de Be eE
MEA NE
ln ces le bass oon re ae
3 EEE
i 1B me He IE IE IE a
i ELE
‘ ee ee o
MS ee je Ju ee
MES BR HS 18 18 1
EN
a
“Two important considerations in choosing the conductor sie for electric wir.
ing are: (Ih the sale current carrying capacity, and (2) the voltage loss due to
wire resistance. On short runs. up to 20 feet, voltage los is very low and need
idee Wire sio shu be sli ors cree expat ax shown
‘On longer runs, several hundred feet or more, the voltage loss may be too
hgh we ie plc solely on the banc a arent capac Tanger ire
Size should be used to keep voltage lose toa selected minimum. Chart 3 may be
ted for thi oe sf
Permissible Voltage Loss ...
‘There is always a voltage loss on any wiring run. The designer must dec
on how much loss can be tolerated without seriously affecting performance, and
‘must select a wire size in which this loss will not be exceeded. A rule-of-thumb
Suggests that electric motors should not be run on a voltage les than about 10%
of their nameplate rating. In deciding on allowable voltage losa in the wiring a
designer must consider the minimun available power line voltage which may
Sceur at a certain time of the day. For example, a 200.vol rated motor should
fot opens than 209 volt which Oe les than namepate voltage Rat
inp) If the lowest power line voltage is 220 volta, then the wiring should not
hatte more than 12 vols loss à
“A rule that works in most cases is to choose a wire size which does not give
more than a 5% lows of input voltage
Chart 1 - Wire Ampacity For Short Wiring Runs
“Ampacity” is an abbreviation for ampere capacity. This chart is for short
wiring runs of less than 20 feet, Ampere capacity is taken from the NEC (Na-
Sonal Electrical Code) on wire sizes of No. 14 and larger It is for insulated cop-
per wire ofthe kind that is widely used for house and building wiring. A larger
pare capacity allowed on wre th certain types of insulin when used
‘under certain conditions, but the NEC handbook should be consulted
‘Amperage Rating for Copper Wires
Mire sizo, AWG 18161412108 6 4 3 2 1 0 00 om
Tnraceway orcable” $3 182030 49 TD 125 ius 165
In open a 8 12 35 28 38 28 88 108 120 180 tes 188 208 280
Chart 2 — Voltage Loss on Long Wiring Runs
"This charts for long wiring runs of several hundred fet or more, To use the
chart, several facts must be established: (1) the current draw ofthe device to be
‘perated must be determined: (2) the amount of voltage fos that can be tler
led must be decided on: and (3) the length of wire must be estimated or mea
Sed, sing the sum of outgoing and retuen wie length
‘On S-phase devices such as electric motors, cach of the three wires must car-
re cren shi onthe motor nameplate Wi log is he sum afoot
Read across the top of the chart to find the column which matches the a
perage rating of the device. Figures in this column show voltage fosses for 1000
Reet of wire, outgoing plus return, Y for example, our total wire length was 250
fet, voltage losses would be Ith that shown in Ihe chart
Figures in Chart Show Voltage Loss Per 1000 Feet of Wire
sre Current Flow, Amperes.
|
© [RE wa a:
PURE de Be eE
MEA NE
ln ces le bass oon re ae
3 EEE
i 1B me He IE IE IE a
i ELE
‘ ee ee o
MS ee je Ju ee
MES BR HS 18 18 1
EN
a
Table of Equivalents
‘To convert units appearing in Colum 1 (left column) into equivalent values
in Column 2 (center column), multiply by factor in Column 3. Exam
Vert gallons into cubie inches multiply 7 x 231 = 1617
"To convert units appearing in Cofumn 2 (center) into equi
units in Column I (eh). divido by factor in Column 3. EXampl
Horsepower into BTU per minute, divide 25 by 0.02356 = 1061
ToComert.. — Into... 2» Multiply By.
ent values of
"To convert 25
AAA
Bu. Re
Mees Barker, 2e
BTU Fool Pounds Yo 7783
ere i
Meta, on ¿Ela
ute, — re
a da
A Lee
Fr =
=e Gene 5
Er eee un Ve
a en
es CECE say ORE
ee = 3
E ga =
Bee vs a sue
= de os
ur A u
Si eae, at
Ei Senken, 188
ÉRIC Des one,
See eb 8
RE a ale
ee ar ques
Fool-Pounds per Second Horsepower 0.001818
Soie Le se
SES der a
A Eu
aaa Seat hs
ee ee ey — Sits
Basar Er 5
Be = e
jr we Be
Be oes, Sele,
ee ee. 38
eg aa a
MURS ne 3
Ed a Le
Be fesse Soin ii
(ee li E
de Sie Be
= = u
Eto E ES
DS Be ou arinse E
wen Beim, a
MR 2e te,
ier ge Bin =
mu Sas bn eo
E 3 e
ine Sen, 2
A Bas
bee seine Am 8
a He —
Fra a;
A A ¡a
EN a a
Ee Seer Ben
56
‘To convert units appearing in Colum 1 (left column) into equivalent values
in Column 2 (center column), multiply by factor in Column 3. Exam
Vert gallons into cubie inches multiply 7 x 231 = 1617
"To convert units appearing in Cofumn 2 (center) into equi
units in Column I (eh). divido by factor in Column 3. EXampl
Horsepower into BTU per minute, divide 25 by 0.02356 = 1061
ToComert.. — Into... 2» Multiply By.
ent values of
"To convert 25
AAA
Bu. Re
Mees Barker, 2e
BTU Fool Pounds Yo 7783
ere i
Meta, on ¿Ela
ute, — re
a da
A Lee
Fr =
=e Gene 5
Er eee un Ve
a en
es CECE say ORE
ee = 3
E ga =
Bee vs a sue
= de os
ur A u
Si eae, at
Ei Senken, 188
ÉRIC Des one,
See eb 8
RE a ale
ee ar ques
Fool-Pounds per Second Horsepower 0.001818
Soie Le se
SES der a
A Eu
aaa Seat hs
ee ee ey — Sits
Basar Er 5
Be = e
jr we Be
Be oes, Sele,
ee ee. 38
eg aa a
MURS ne 3
Ed a Le
Be fesse Soin ii
(ee li E
de Sie Be
= = u
Eto E ES
DS Be ou arinse E
wen Beim, a
MR 2e te,
ier ge Bin =
mu Sas bn eo
E 3 e
ine Sen, 2
A Bas
bee seine Am 8
a He —
Fra a;
A A ¡a
EN a a
Ee Seer Ben
56
Decimal and Metric
Equivalents of Common Fractions of an Inch
eee ee ns
E
E a EE EP ER EE
|
Sen
E
From English to Metric Units
i
A Al y e 8 3 A A
& E a El E S 3 E 1528
223
fo
Sans
Tea
3164
EC]
7164
ver
1/64
196
15164
TE
19164
EL]
23464
EC]
27164
2067
31164
Er
35164
Eu
004
amé
43164
567
ara
LL
ses
Sar
ss
EL
5064
Bee
ques
57
Equivalents of Common Fractions of an Inch
eee ee ns
E
E a EE EP ER EE
|
Sen
E
From English to Metric Units
i
A Al y e 8 3 A A
& E a El E S 3 E 1528
223
fo
Sans
Tea
3164
EC]
7164
ver
1/64
196
15164
TE
19164
EL]
23464
EC]
27164
2067
31164
Er
35164
Eu
004
amé
43164
567
ara
LL
ses
Sar
ss
EL
5064
Bee
ques
57
no mn nu ou u sg a
ose 00 ose
walten
pos
un Suto 01 wsi6u3 ion posan mols MY MOS
(ce 4.) 6S = 9.
wo
wily ge = uni
(ayo
ope uojsianuoo ars Aeworsno
"uoneinau une 40) Posedoud siuownsop uo pasn aq pinoys Sıjun (spIePURYS jeuonewoy
spup PADPUDIS (IS) PUD yssug uaanjog SUOISLIAUO)
ose 00 ose
walten
pos
un Suto 01 wsi6u3 ion posan mols MY MOS
(ce 4.) 6S = 9.
wo
wily ge = uni
(ayo
ope uojsianuoo ars Aeworsno
"uoneinau une 40) Posedoud siuownsop uo pasn aq pinoys Sıjun (spIePURYS jeuonewoy
spup PADPUDIS (IS) PUD yssug uaanjog SUOISLIAUO)
Interchange Between Units
International Metric - Old Metric - U.S. Customary Units
“These charts wil interchange values between al Standard, the
Uster Eni sytem. and the older meri tomo. Te eh column ofeach chart.
hoa ane ti he te an cane me
the charts on dow the athe nit whic mer and nd the
one on which the igure 1" appears Then move tothe le or right on the same one lo
the glum ofthe new unit the value shown ina multiplier to convert tothe new un
Eonversions can be easily made with a packet calculator whieh has an expo
hey or ean be made manually
or manual caleulaions remember thatthe + or - sign in font ofan exponent tells
whether to move the decimal part to the right (ora saga) or tothe let for sign)
$l fro me Examples: DO 10500004, ad 048% 102308
Convert 627 inches into centimeters. In the LENGTH chart, look down the Inch
on lt fare sI The aye ll où thie ine tothe Cine can, Uso
plier 2 540! 821 x 2.840 = 1592.58 centimeters = 15.03 meter
ounfig, inch) ing Bar Use the UNIT PRESSURE chart on
page, Look doen the Pbundainch alum athe gure
Cline tothe Bar column. The figure 0.06897 i 3 ul
SO OGSST = 344 6 bar.
Torque Gravity Acceleration
Meer Aiepond Footibe ines | acceleration due to Era
eer regia 2.2 fect per second
nl 1020.107 TEXTO” es | per second. In Ihe metre
[9507 : 7233 80 | System ze W081 meters per
6 Nano À = SEO per econ
0x0" 182210? ame? 1
Length (Linear Measurement)
Meter Cemimeter Kilometer Mie Inem Foot
q 100 TEE EE
oo 12109 égale Sarat! 2281x10%
New? 010 12100 62102107 —Soareto? 3201x102
Teo mie 1 SiO IIA NT
600% 10%, 1500x105 1.000 ‘ Smet Seo
j2stoxto2 250 25402108 sexos 1 ax io
texto: mar Gotexto! 1ammtot le
mater = 0001 mear = 010 center = 000001 RIT» OORT WER = OIGA
Area (Square Measurement)
Sauar Meter Sq Centimeter Sa Klometer Sauareineh Square Foot Square Mie
A Tei IHM, SOI 1076 861x107
Boa Teo" 1880410 Norexto2 2281110"
OOO 160x109 10701105 20012107
en Fox NOOO, 36110"
Básexio! Gas gas 1 Saxo 21x10
Gants 20x10, 9200" 108 1m : Ser 10%
250010" _2590% 100 _ 2.550, AO? 2880 À
are mireia = 0000007 square malt = 0.0188 saure mc » 0100007078 Sia FT
Volume (Cubic)
Cube Meter Cu Decimeter CuCentimeter US.Gullen _Cubieineh Cubie Foot
+ TID NU ET TI er TI ern
few? 1 D Zero! mim Astro
trie tx? Zero anzu 35a xto®
PRE Asisxio) 1200 274x102 1605x107
Brio 378 EA! ÉCRAN
10“ 10° 180x008 1007 40x10 À Sato
282210? 217 Zenit ra Irene 1
‘eral gafa
ZUS Galo = 0004548 be mein 15T Wor = 4546 cube comite
59
International Metric - Old Metric - U.S. Customary Units
“These charts wil interchange values between al Standard, the
Uster Eni sytem. and the older meri tomo. Te eh column ofeach chart.
hoa ane ti he te an cane me
the charts on dow the athe nit whic mer and nd the
one on which the igure 1" appears Then move tothe le or right on the same one lo
the glum ofthe new unit the value shown ina multiplier to convert tothe new un
Eonversions can be easily made with a packet calculator whieh has an expo
hey or ean be made manually
or manual caleulaions remember thatthe + or - sign in font ofan exponent tells
whether to move the decimal part to the right (ora saga) or tothe let for sign)
$l fro me Examples: DO 10500004, ad 048% 102308
Convert 627 inches into centimeters. In the LENGTH chart, look down the Inch
on lt fare sI The aye ll où thie ine tothe Cine can, Uso
plier 2 540! 821 x 2.840 = 1592.58 centimeters = 15.03 meter
ounfig, inch) ing Bar Use the UNIT PRESSURE chart on
page, Look doen the Pbundainch alum athe gure
Cline tothe Bar column. The figure 0.06897 i 3 ul
SO OGSST = 344 6 bar.
Torque Gravity Acceleration
Meer Aiepond Footibe ines | acceleration due to Era
eer regia 2.2 fect per second
nl 1020.107 TEXTO” es | per second. In Ihe metre
[9507 : 7233 80 | System ze W081 meters per
6 Nano À = SEO per econ
0x0" 182210? ame? 1
Length (Linear Measurement)
Meter Cemimeter Kilometer Mie Inem Foot
q 100 TEE EE
oo 12109 égale Sarat! 2281x10%
New? 010 12100 62102107 —Soareto? 3201x102
Teo mie 1 SiO IIA NT
600% 10%, 1500x105 1.000 ‘ Smet Seo
j2stoxto2 250 25402108 sexos 1 ax io
texto: mar Gotexto! 1ammtot le
mater = 0001 mear = 010 center = 000001 RIT» OORT WER = OIGA
Area (Square Measurement)
Sauar Meter Sq Centimeter Sa Klometer Sauareineh Square Foot Square Mie
A Tei IHM, SOI 1076 861x107
Boa Teo" 1880410 Norexto2 2281110"
OOO 160x109 10701105 20012107
en Fox NOOO, 36110"
Básexio! Gas gas 1 Saxo 21x10
Gants 20x10, 9200" 108 1m : Ser 10%
250010" _2590% 100 _ 2.550, AO? 2880 À
are mireia = 0000007 square malt = 0.0188 saure mc » 0100007078 Sia FT
Volume (Cubic)
Cube Meter Cu Decimeter CuCentimeter US.Gullen _Cubieineh Cubie Foot
+ TID NU ET TI er TI ern
few? 1 D Zero! mim Astro
trie tx? Zero anzu 35a xto®
PRE Asisxio) 1200 274x102 1605x107
Brio 378 EA! ÉCRAN
10“ 10° 180x008 1007 40x10 À Sato
282210? 217 Zenit ra Irene 1
‘eral gafa
ZUS Galo = 0004548 be mein 15T Wor = 4546 cube comite
59
Force (Including Force Due to Weight)
Newton Dime Klopond _MerieTon USTON Pound
1 Txi0® 1020x107 1020x107 114x107 22482705
ti Fonte 1020x10° 1ı2axo? 22489 10%
aor Sexo 1 Do ES
807x102 9807x108 10 ; Sige" 3208x108
Brie SX 100x103 1016 10 oo
mr Sex Sorento Sorex" 1 2000
(da Aare detesto! 480610 Sun: 1
Thon = 900 Newons= TOTS Koponds = TOTS mations = 1.120 US Tons = 2340 pounds
Mass (Not Weight)
Kiogram rom MetieTon_ Newton Pound US.Ton
5 000 IT 2285 +100 10"
ix? 110% Sang 2205103 Lien to
Be tayo À E 110
Voss! 1a hoaxes 1 2208210" Vago
10210" dx Asa: 4448 i 320%
Mama fun FASO IO? VAN dam oO
Boni Sox darznto? _Ba06x10* 2000 ï
Velocity
IMotersec _Kilometerafr Meur Feeuin._ FeetSee inches.
nl 36 227 968 1973281 FESTE
eto’ Tato! éélexiot Serie S1rdx10% 6560x102
lerexto' + Enano? Sama Saxo Gex tae
0210 1000 ‘ e ter 05610
Éœnrio 29x10? 16x10 1 Keerx1o? 12
E CIN] : Taro?
Henriot 1x1 DA7OXIOS exe? tomxto? 1
Y dicielarsocond = 0.1 maarTcend = 0005968 Win 00656 win
Unit Pressure (Either Fluid or Mechanical)
E Newton? Kiopondimt Atmosphere PoundsiFt? _Poundsinch?
Ta PETITE HET TE TU
: RC SO
ori: a Serato? esto 1422410?
Brio" Bora um | Sorento ZOO 14220
tor) de dore 1 Re}
ae E Pan)
[6807x10° Speo Tomxıd: 6806x10° 1a 1
"lope em = 0.007 bar = 06070 Pascal 0.9678 ainos = 2048 LRQ ia Te 22 LIRE
Power (Fluid, Electrical, or Mechanical)
Iowa Yatsoulls Footpounde Foot m
tee pechinas” perSocona’ TUN Brun
y 1000 EAT Sain TO SOD
hero t Me ani! da? Senex 10?
cor 16 ET]
B2c0%10% 2x0: 3 Hansa? Front? texto
Er : u pronos
Bains agrio? Men tarea? 1 HO
‘oro? +7 _preouto?_i2are do i
US 1 UR Worsoponer= 07168 A = DR 25 GTN AS BTN
Energy or Work
Kowttiour Watsecand, Horsepower: Foot Sound inchPound BTU
7 E UE TEE ES
Error, À Au, PEO Bast 807x104
Borer nso? | IO eue Bastx0? daran
Tas‘, base ioe Tas EI 280.0
Error 16 Stunt 1 2 128s. 1?
Bios Ameio) Anime Smet 1 Serie
Banixios AECI aout PR amd 1
60
Newton Dime Klopond _MerieTon USTON Pound
1 Txi0® 1020x107 1020x107 114x107 22482705
ti Fonte 1020x10° 1ı2axo? 22489 10%
aor Sexo 1 Do ES
807x102 9807x108 10 ; Sige" 3208x108
Brie SX 100x103 1016 10 oo
mr Sex Sorento Sorex" 1 2000
(da Aare detesto! 480610 Sun: 1
Thon = 900 Newons= TOTS Koponds = TOTS mations = 1.120 US Tons = 2340 pounds
Mass (Not Weight)
Kiogram rom MetieTon_ Newton Pound US.Ton
5 000 IT 2285 +100 10"
ix? 110% Sang 2205103 Lien to
Be tayo À E 110
Voss! 1a hoaxes 1 2208210" Vago
10210" dx Asa: 4448 i 320%
Mama fun FASO IO? VAN dam oO
Boni Sox darznto? _Ba06x10* 2000 ï
Velocity
IMotersec _Kilometerafr Meur Feeuin._ FeetSee inches.
nl 36 227 968 1973281 FESTE
eto’ Tato! éélexiot Serie S1rdx10% 6560x102
lerexto' + Enano? Sama Saxo Gex tae
0210 1000 ‘ e ter 05610
Éœnrio 29x10? 16x10 1 Keerx1o? 12
E CIN] : Taro?
Henriot 1x1 DA7OXIOS exe? tomxto? 1
Y dicielarsocond = 0.1 maarTcend = 0005968 Win 00656 win
Unit Pressure (Either Fluid or Mechanical)
E Newton? Kiopondimt Atmosphere PoundsiFt? _Poundsinch?
Ta PETITE HET TE TU
: RC SO
ori: a Serato? esto 1422410?
Brio" Bora um | Sorento ZOO 14220
tor) de dore 1 Re}
ae E Pan)
[6807x10° Speo Tomxıd: 6806x10° 1a 1
"lope em = 0.007 bar = 06070 Pascal 0.9678 ainos = 2048 LRQ ia Te 22 LIRE
Power (Fluid, Electrical, or Mechanical)
Iowa Yatsoulls Footpounde Foot m
tee pechinas” perSocona’ TUN Brun
y 1000 EAT Sain TO SOD
hero t Me ani! da? Senex 10?
cor 16 ET]
B2c0%10% 2x0: 3 Hansa? Front? texto
Er : u pronos
Bains agrio? Men tarea? 1 HO
‘oro? +7 _preouto?_i2are do i
US 1 UR Worsoponer= 07168 A = DR 25 GTN AS BTN
Energy or Work
Kowttiour Watsecand, Horsepower: Foot Sound inchPound BTU
7 E UE TEE ES
Error, À Au, PEO Bast 807x104
Borer nso? | IO eue Bastx0? daran
Tas‘, base ioe Tas EI 280.0
Error 16 Stunt 1 2 128s. 1?
Bios Ameio) Anime Smet 1 Serie
Banixios AECI aout PR amd 1
60
Temperature Conversion Chart
Enter the table in the column marked “Temp” with the temperature either
Fahrenheit or Celsius (Centigrade) that you wish to convert. Hf converting into
Celsius, read the equivalent value in the column to the left IE converting into
Fahrenheit, read the equivalent value in the column to the right,
Cc Temp E C Temp E Cc Temp E
7% E 185 60 100 143 290 5%
72 1 338 Wer 6 i418 149 300 972
Nee 2 Be 166 62 1436 16% 310 500
161 3 7 i 6 454 160 320 608
155 4 392 177 64 172 165 30 6%
130 $ 418 162 68 1490 171 Sada
a6 423 188 6 1808 177 380 602
39 7 446 193 67 1820 182 360 080
193 8 464 199 68 1564 188 37 6%
127 9 482 204 88 562 19% 380 70
333 do 500 510 70 1580 1099 380 70%
E DIS 71 198 204 400 782
ii 12 586 22 72 1616 20 410 70
108 13 84 227 73 103% 28 420 7
100 18 872 233 7a 1082 22 430 806
94 18 590 238 7 1070 226 40 824
88 16 508 244 76 1688 232 450 842
#5 7 ge 260 TR 1708 2% 480 86
77 18 Baa 23 ma 2 470 878
72 19 862 262 m2 24 480 8%
86 20 680 208 1760 254 490 01
51 21 098 23 178 260 $00 92
3 2 Te 277 m 2 510 9%
50 2 734 262 83 tera u 98
44 2 762 288 94 1832 276 930 90
as 3» 70 203 06 1880 282 SD 100
33 2% 788 299 88 1808 288 880 1022
28 à 6e 30% 87 180 203 580 10%0
22 2% 824 310 88 1904 299 570 1058
F8 2% Bea 518 89 192 So 1078
11 30 860 321 90 10 310 500 108%
oe Bre Bo 8 168 1 60 1112
9. 3 es 353 92 1976 31 610 113
os 3 94 HS OS 192 26 620 sa
Ma 932 a Si 202 32 60 16
15 35 950 349 98 2050 33 O0 181
22 36 a 355 96 2088 363 650 1202
57 37 86 Sei 37 2008 9 60 120
$3 38 100% 360 98 208% 3% 670 1238
38 39 102 371 98 202 30 800 1250
44 40 1040 38 100 212 ges Go 1274
49 41 108 43 110 20 Si Yoo 1%
55 4 106 4 120 28 Se 70 1310
80 43 104 54 130 260 2 70 1328
66 44 1112 60 140 284 27 70 146
F1 4 180 10 302 3o3 Tao 14
77 46 M8 A 160 20 Seo 70 1382
82 47 166 76 170 38 308 760 1400
BB 48 118% 89 180 356 310 770 1418
93 43 1202 88 190 37% a oo 1436
39 50 120 9% 20 3% Fre 14
fos 81 1238 9 210 410 26 600 1872
ni 1255 10 212 413 332 810 1300
M5 83 1874 103 20 428 6 820 1508
127 M 1202 NO 20 46 43 60 18
126 55 1810 118 20 36% 9 bao 10
132 86 1328 121 250 482 ass 950 1562
187 Br 138 127 20 500 de 860 1560
133 58 1804 12 20 518 des 870 18
1 280 5% a
61
Enter the table in the column marked “Temp” with the temperature either
Fahrenheit or Celsius (Centigrade) that you wish to convert. Hf converting into
Celsius, read the equivalent value in the column to the left IE converting into
Fahrenheit, read the equivalent value in the column to the right,
Cc Temp E C Temp E Cc Temp E
7% E 185 60 100 143 290 5%
72 1 338 Wer 6 i418 149 300 972
Nee 2 Be 166 62 1436 16% 310 500
161 3 7 i 6 454 160 320 608
155 4 392 177 64 172 165 30 6%
130 $ 418 162 68 1490 171 Sada
a6 423 188 6 1808 177 380 602
39 7 446 193 67 1820 182 360 080
193 8 464 199 68 1564 188 37 6%
127 9 482 204 88 562 19% 380 70
333 do 500 510 70 1580 1099 380 70%
E DIS 71 198 204 400 782
ii 12 586 22 72 1616 20 410 70
108 13 84 227 73 103% 28 420 7
100 18 872 233 7a 1082 22 430 806
94 18 590 238 7 1070 226 40 824
88 16 508 244 76 1688 232 450 842
#5 7 ge 260 TR 1708 2% 480 86
77 18 Baa 23 ma 2 470 878
72 19 862 262 m2 24 480 8%
86 20 680 208 1760 254 490 01
51 21 098 23 178 260 $00 92
3 2 Te 277 m 2 510 9%
50 2 734 262 83 tera u 98
44 2 762 288 94 1832 276 930 90
as 3» 70 203 06 1880 282 SD 100
33 2% 788 299 88 1808 288 880 1022
28 à 6e 30% 87 180 203 580 10%0
22 2% 824 310 88 1904 299 570 1058
F8 2% Bea 518 89 192 So 1078
11 30 860 321 90 10 310 500 108%
oe Bre Bo 8 168 1 60 1112
9. 3 es 353 92 1976 31 610 113
os 3 94 HS OS 192 26 620 sa
Ma 932 a Si 202 32 60 16
15 35 950 349 98 2050 33 O0 181
22 36 a 355 96 2088 363 650 1202
57 37 86 Sei 37 2008 9 60 120
$3 38 100% 360 98 208% 3% 670 1238
38 39 102 371 98 202 30 800 1250
44 40 1040 38 100 212 ges Go 1274
49 41 108 43 110 20 Si Yoo 1%
55 4 106 4 120 28 Se 70 1310
80 43 104 54 130 260 2 70 1328
66 44 1112 60 140 284 27 70 146
F1 4 180 10 302 3o3 Tao 14
77 46 M8 A 160 20 Seo 70 1382
82 47 166 76 170 38 308 760 1400
BB 48 118% 89 180 356 310 770 1418
93 43 1202 88 190 37% a oo 1436
39 50 120 9% 20 3% Fre 14
fos 81 1238 9 210 410 26 600 1872
ni 1255 10 212 413 332 810 1300
M5 83 1874 103 20 428 6 820 1508
127 M 1202 NO 20 46 43 60 18
126 55 1810 118 20 36% 9 bao 10
132 86 1328 121 250 482 ass 950 1562
187 Br 138 127 20 500 de 860 1560
133 58 1804 12 20 518 des 870 18
1 280 5% a
61
Table of Standard Wire Gauges
DE ce
Su L LONG Cop- La PP ohms Ohms
lues ses! moe ou | Fe cu i io
soins qe teres | Bs gun Be,
A E O
HEEE
EE LEE EEE
E EE E EE
ia ae
i a | Re ae ae
i BE D où à | RR 28 Be Is
; ie me as ia | ae aoe ae Le
EEE E
LES Be | ee Ae
EE AREA E
ee de ise ie | ome de le du
EE AREA
E eee |e ee ie
EA eee E:
da de ie i ss
MENE EE
E HS ie
escasas
E EA Te |
2 oe du we | Be ds an
EE eR | MEE:
¡SONES Sd
aaa ass
aim a oo is | ae de ips
1000231000 pu
FE on LE as
SHG D BB | à & a
Soop oe op fe | dE Be de ai
GE we | Bee ge
au a dé
do non | do de
LEER ET
EEE tek #
(1) Manufacturers standard gauge for hot and cold roled sheets, based on a
Ja of 41.82 bs per inch of thickness per square font
2) US, Steel Wire Gauge. Also known as Washburn and Moen, American
‘Steet and Wire, and Roebling gauges. Used by most of the steel producers in che
US and has replaced the Birmingham (Stubs) gauge in most instances
(8) Special gauge for piano and music wire for mechanical springs.
(9) Stubs Stet Wire Gauge, Used sometimes or cacon se dil ad. Dil
May also be called out la other gauges such as Morse Twist Drill Ga
‘(Manufacturers Standard Gauge for twist dils) or by the American Standard
‘Straight Shank Twist Drill standard dimensions
(6) Copper wire is measured by the B & $ gauge (als called AWG). Res
“values are at 77°F or 25°C. Convert between “C and *F with chart on page
"To convert chart resistance values toa new °C temperature:
AT = R «(1 + 0.00585 (1-25)
Tis resistance at new °C temperature; is new temperature in *C; and A
dore ee
‘Acireular mils a unit ofarca that is applied to electrical wires and is equal
to the area ofa circle 1 mul (0.01 inch) de diameter The area of any cree i
‘equal to the square ofits diameter in mil
62
DE ce
Su L LONG Cop- La PP ohms Ohms
lues ses! moe ou | Fe cu i io
soins qe teres | Bs gun Be,
A E O
HEEE
EE LEE EEE
E EE E EE
ia ae
i a | Re ae ae
i BE D où à | RR 28 Be Is
; ie me as ia | ae aoe ae Le
EEE E
LES Be | ee Ae
EE AREA E
ee de ise ie | ome de le du
EE AREA
E eee |e ee ie
EA eee E:
da de ie i ss
MENE EE
E HS ie
escasas
E EA Te |
2 oe du we | Be ds an
EE eR | MEE:
¡SONES Sd
aaa ass
aim a oo is | ae de ips
1000231000 pu
FE on LE as
SHG D BB | à & a
Soop oe op fe | dE Be de ai
GE we | Bee ge
au a dé
do non | do de
LEER ET
EEE tek #
(1) Manufacturers standard gauge for hot and cold roled sheets, based on a
Ja of 41.82 bs per inch of thickness per square font
2) US, Steel Wire Gauge. Also known as Washburn and Moen, American
‘Steet and Wire, and Roebling gauges. Used by most of the steel producers in che
US and has replaced the Birmingham (Stubs) gauge in most instances
(8) Special gauge for piano and music wire for mechanical springs.
(9) Stubs Stet Wire Gauge, Used sometimes or cacon se dil ad. Dil
May also be called out la other gauges such as Morse Twist Drill Ga
‘(Manufacturers Standard Gauge for twist dils) or by the American Standard
‘Straight Shank Twist Drill standard dimensions
(6) Copper wire is measured by the B & $ gauge (als called AWG). Res
“values are at 77°F or 25°C. Convert between “C and *F with chart on page
"To convert chart resistance values toa new °C temperature:
AT = R «(1 + 0.00585 (1-25)
Tis resistance at new °C temperature; is new temperature in *C; and A
dore ee
‘Acireular mils a unit ofarca that is applied to electrical wires and is equal
to the area ofa circle 1 mul (0.01 inch) de diameter The area of any cree i
‘equal to the square ofits diameter in mil
62
Densities and Specific Gravities
of Common Materials
Specific gravity is a number indicating how many times a certain volume of
‘a material is heavier than an equal volume of water. As the density of water dif
fers slighty at diferent temperatures, the temperature of 2. Fs une ar com
parison. The weight of one cubic inch of pure water at 62° F is 0.0361 pou
ithe medie gravity Samy material Known’ tne weight ofa cub inch af
‘the material ean be found by multiplying its specific gravity by 0.0361
"To find the weight per eubie foot of a materia, the specific gravity of which
is known, multiply the specific gravity by 62.335
Tf the weight of a cubie inch of material a known, the specific gravity is
by dividing the weight per cubie inch by 0.0361
If the weight of a cubic foot of a material 12 known, the specific gravity is
found by multiplying the weight por eubie foot by 0.01604,
is found
fie | Weight Per Cubic | Weight Per Cubic
Material ravity | "Ingh (Pounds). |" Foot (Pounds)
Aluminum 270 00075 16850
Brass: 800. 202 860 0.3105 536.60
Brass: 706. 302 844 0.3048 526.70
Brass: 606. 402 836 03018 52170
‘Brass: 500. 502 820 02961 51170
Bronze: 966.107 | 878 am 54790
‘Chromium 693 02502 432.40
Concrete 231 0088 144.00
Copper. 880 03210 55470
God 1925 0695 1200.00
Iron, Cast 703779 | 02540270 438 70-482.40
Iron. Wrought 730790 | 02820288 186.70-406 00
Lead 1137 0470 70900
Magnesium 175 0062 109.00
Steel, Carbon 733-787 | 02830284 489.00-400.80
Steel, Stainless. 770 0278 480.00
Tin 729 02683 454.90
Mechanical Properties
of Common Materials
timate JE ete
eitisiehy of
Compas Tension o! ein
tai Tension [Cepa or |campeesen 0] 3) |)
Ties, frgedoled
diode | somo | som gascon | 028
02000 || 43000 Pom | 035
Sse | mon| dm Pom | 035
omomc |1som| com como | 02
cast ron
Sols | 20000 | so sovome | 028
Gray (ASTM | St | 123000 as
Sry ase) | 60000 | nam som | 02
Naleade 000 | 120000 52000 | 02
Wrougiton | 4800 | "25000 | a
Ste cast,
tono 0000 02
Medic ‘ana 0
tne 00 == 08
Brass
Cat 0000 | -—— ---| 09
Fines 56000 | 1800 ===} om
Coes E | 22 saone | 030
Bronze
Cast 2200 == | où
Coran | 86400 scan | 031
of Common Materials
Specific gravity is a number indicating how many times a certain volume of
‘a material is heavier than an equal volume of water. As the density of water dif
fers slighty at diferent temperatures, the temperature of 2. Fs une ar com
parison. The weight of one cubic inch of pure water at 62° F is 0.0361 pou
ithe medie gravity Samy material Known’ tne weight ofa cub inch af
‘the material ean be found by multiplying its specific gravity by 0.0361
"To find the weight per eubie foot of a materia, the specific gravity of which
is known, multiply the specific gravity by 62.335
Tf the weight of a cubie inch of material a known, the specific gravity is
by dividing the weight per cubie inch by 0.0361
If the weight of a cubic foot of a material 12 known, the specific gravity is
found by multiplying the weight por eubie foot by 0.01604,
is found
fie | Weight Per Cubic | Weight Per Cubic
Material ravity | "Ingh (Pounds). |" Foot (Pounds)
Aluminum 270 00075 16850
Brass: 800. 202 860 0.3105 536.60
Brass: 706. 302 844 0.3048 526.70
Brass: 606. 402 836 03018 52170
‘Brass: 500. 502 820 02961 51170
Bronze: 966.107 | 878 am 54790
‘Chromium 693 02502 432.40
Concrete 231 0088 144.00
Copper. 880 03210 55470
God 1925 0695 1200.00
Iron, Cast 703779 | 02540270 438 70-482.40
Iron. Wrought 730790 | 02820288 186.70-406 00
Lead 1137 0470 70900
Magnesium 175 0062 109.00
Steel, Carbon 733-787 | 02830284 489.00-400.80
Steel, Stainless. 770 0278 480.00
Tin 729 02683 454.90
Mechanical Properties
of Common Materials
timate JE ete
eitisiehy of
Compas Tension o! ein
tai Tension [Cepa or |campeesen 0] 3) |)
Ties, frgedoled
diode | somo | som gascon | 028
02000 || 43000 Pom | 035
Sse | mon| dm Pom | 035
omomc |1som| com como | 02
cast ron
Sols | 20000 | so sovome | 028
Gray (ASTM | St | 123000 as
Sry ase) | 60000 | nam som | 02
Naleade 000 | 120000 52000 | 02
Wrougiton | 4800 | "25000 | a
Ste cast,
tono 0000 02
Medic ‘ana 0
tne 00 == 08
Brass
Cat 0000 | -—— ---| 09
Fines 56000 | 1800 ===} om
Coes E | 22 saone | 030
Bronze
Cast 2200 == | où
Coran | 86400 scan | 031
Womack Fluid Power Textbooks
information un this booklet
sek gat the way fd
Ich PSN Bower
vente Womack
textbooks are writen by the
A
almost every topic on both
ina Buts nd do
Practical wis. it can “be
Undersea by anyone who à
SMO ie elemenar
and theory ie sold wo the A Really Practical Treatment of Fluid Power!
Praca ops
INDUSTRIAL FLUID POWER — Volume 1
“Thi na complete baie cour on air and hydraulic Guid power Used qa a textbook in
sone hi needy ont od pe ci y
‘Specifications: 238 pages. 87 x 11, Over 330 illustrations; Saft bound
ee ‘Now availabe in Spanish:
INDUSTRIAL FLUID POWER — Volume 2
A or advance nk wich coves deta and peur con finders ta greater
depth nore mor design Information and ern tha in Glue 1
‘Specittations: 272 pages SL 11“ Over 30 ilastations, Sa bound
INDUSTRIAL FLUID POWER — Volume 3
“Tic bok covers applications with rotary power output from Hydraulic and air motors
arta trast ary aun PO Bow der Other tps bs apn
‘Specifications: 242 pages 8¥'x 11°; Over 250 illustrations: So bound
FLUID POWER IN PLANT AND FIELD.
Written especially far people working i the Sel. Many suggestions on installation
maintenance and troubleshooting, but thor Is much supplementary Information for readers of
Specifications: 232 pages, SH x 11: Over 50 illustrations: Soft bound.
ELECTRICAL CONTROL OF FLUID POWER
“This bak explains components ved in letra control and includes chapter n Inder
iagrame, hard Wird Gres grammaticales serv valves prepara! soon
{ales mir starters ander leet topics
‘Specitcations! 228 pages BE 11 Over 200 ustration; Soft hound
SELECTED TOOLING TREASURES
‘hie ok ie writen in loymars language and can bel
business owners SANS Puri, agents and
maintenance personnel e understand the alo spe
teense gg Te author RE Wonach. hs vu 6
‘Torusace, rer, electronic, and the el istry which
‘an bp jou avd many tthe problems and pe that
fre crient found in ling
Specliatens’ 238 fare, su x
Rens bound,
FUNDAMENTALS OF INDUSTRIAL
CONTROLS & AUTOMATION
focused on a beginners study of Sicario, electronics
Senta component pd automaton rested dol
Emi to maximise pd miaimise downtime
Simply maintenance, Imprute ca and provide
Specitestions! 281 Yagi Ser" TES Over 24
ne Sof wound. 0 0
Write, phone or fax for latest price list; or to place an order.
‘PHONE: (800) 569-9801 PAX: (214) 680-5314
information un this booklet
sek gat the way fd
Ich PSN Bower
vente Womack
textbooks are writen by the
A
almost every topic on both
ina Buts nd do
Practical wis. it can “be
Undersea by anyone who à
SMO ie elemenar
and theory ie sold wo the A Really Practical Treatment of Fluid Power!
Praca ops
INDUSTRIAL FLUID POWER — Volume 1
“Thi na complete baie cour on air and hydraulic Guid power Used qa a textbook in
sone hi needy ont od pe ci y
‘Specifications: 238 pages. 87 x 11, Over 330 illustrations; Saft bound
ee ‘Now availabe in Spanish:
INDUSTRIAL FLUID POWER — Volume 2
A or advance nk wich coves deta and peur con finders ta greater
depth nore mor design Information and ern tha in Glue 1
‘Specittations: 272 pages SL 11“ Over 30 ilastations, Sa bound
INDUSTRIAL FLUID POWER — Volume 3
“Tic bok covers applications with rotary power output from Hydraulic and air motors
arta trast ary aun PO Bow der Other tps bs apn
‘Specifications: 242 pages 8¥'x 11°; Over 250 illustrations: So bound
FLUID POWER IN PLANT AND FIELD.
Written especially far people working i the Sel. Many suggestions on installation
maintenance and troubleshooting, but thor Is much supplementary Information for readers of
Specifications: 232 pages, SH x 11: Over 50 illustrations: Soft bound.
ELECTRICAL CONTROL OF FLUID POWER
“This bak explains components ved in letra control and includes chapter n Inder
iagrame, hard Wird Gres grammaticales serv valves prepara! soon
{ales mir starters ander leet topics
‘Specitcations! 228 pages BE 11 Over 200 ustration; Soft hound
SELECTED TOOLING TREASURES
‘hie ok ie writen in loymars language and can bel
business owners SANS Puri, agents and
maintenance personnel e understand the alo spe
teense gg Te author RE Wonach. hs vu 6
‘Torusace, rer, electronic, and the el istry which
‘an bp jou avd many tthe problems and pe that
fre crient found in ling
Specliatens’ 238 fare, su x
Rens bound,
FUNDAMENTALS OF INDUSTRIAL
CONTROLS & AUTOMATION
focused on a beginners study of Sicario, electronics
Senta component pd automaton rested dol
Emi to maximise pd miaimise downtime
Simply maintenance, Imprute ca and provide
Specitestions! 281 Yagi Ser" TES Over 24
ne Sof wound. 0 0
Write, phone or fax for latest price list; or to place an order.
‘PHONE: (800) 569-9801 PAX: (214) 680-5314
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