First law of thermodynamics

First law of thermodynamics

1. First Law of
Thermodynamics
Dr. Rohit Singh Lather

2. The First Law of Thermodynamics
üThe quantity(Q–W) is the same for all processes
üIt depends only on the initial and final states of the system
üDoes not depend at all on how the system gets from one to the other
üThis is simply conservation of energy
(Qis the heat absorbed and Wis the work done by the system)
The internal energy E of a system tends to increase, if energy is added as heat Q and tends to
decrease if energy is lost as work W done by system
dE= dQ–dW( first law)

3. Internal Energy
•Internal energy (u): that portion of total energy E which is not kinetic or potential
energy. It includes thermal, chemical, electric, magnetic, and other forms of energy
•Change in the specific internal energy
du = CdT
•In case of gases internal energy is given by u = C
vdT
•Specific heat changes with temperature is given by
C = C
o(a+bT), a and b are constants and C
ois specific heat at 0
o
C
•The total energy in the mass m is the sum of internal energy as well as PE and KE in the mass
E = U + PE + KE = m .u + mg. Z +
!
"
mV
2
For unit mass E = em= u +
$
%
+
!
"
V
2
(p = 'gZ)

4. Enthalpy
1Q
2=U
2 − U
1 + P
2V
2 − P
1V
1
= (U
2+ P
2V
2) − (U
1+ P
1V
1)
u = h -pv
The enthalpy is especially valuable for analyzing isobaric processes
Enthalpy h = u + pv
s
•Theheattransferinaconstant-pressurequasi-equilibriumprocessisequaltothechangein
enthalpy,whichincludesboththechangeininternalenergyandtheworkforthisparticular
process
•Thisisbynomeansageneralresult
•Itisvalidforthisspecialcaseonlybecausetheworkdoneduringtheprocessisequaltothe
differenceinthePVproductforthefinalandinitialstates
•Thiswouldnotbetrueifthepressurehadnotremainedconstantduringtheprocess

5. Source: http://www4.ncsu.edu/~kimler/hi322/Rumford-expt.gif
Sir Benjamin Thompson
Count Rumford •Thompson’s theory of heat was demonstrated by a test tube full
of water within wooden paddles
•Water boiled due to friction
•The heat of friction is unlimited

6. The Joules Experiment
Rise in
temperature
One calorie corresponds to the amount
of heat that is needed to get one gram
of water from 14.5 C to 15.5 C
1 Cal = 4.1840 J
( ∮*+= ∮*,
A paddle wheel turns
in liquid water
proportionality constant -mechanical equivalent of heat
W
W
H
Insulating walls
prevent heat transfer from the
enclosed water to the surroundings
As the weight falls at
constant speed, they
turn a paddle wheel,
which does work on
water
If friction in mechanism is negligible, the work done by the paddle wheel on the water equals the
change of potential energy of the weights

7. Joule showed that the same temperature rise could be obtained using an electrical resistor heated
by an electric current
Work can be transformed into heat
Heat and work are of the same nature and constitute different forms of energy

8. First Law of Thermodynamics
“If a system executes a cycle transferring work and heat through its boundaries,
then net work transfer is equal to net heat transfer”
E
2–E
1= Q -W
dE= Q -W
dE= E
2–E
1= ∫.(0−2)
"
!
For a closed system
dU= U
2–U
1= ∫.0−2
"
!

dU= U
2–U
1= Q -W
“During any cycle, the cyclic integral of heat added to a system is proportional to the cyclic
integral of work done by the system”
∮*+= ∮*,

9. Adiabatic processes
Process that occurs so rapidly or occurs in a system that is so well insulated that no transfer of
thermal energy occurs between the system and its environment
Free expansion
Adiabatic processes in which no heat transfer occurs between the system and its environment and no
work is done on or by the system
Cyclical Processes
Processes in which, after certain interchanges of heat and work, the system is restored to its initial
state. No intrinsic property of the system—including its internal energy—can possibly change
Constant-volume processes
If the volume of a system (such as a gas) is held constant,so that system can do no work

10. First Law Corollaries
Corollary 3
“The perpetual motion machine (PMM) of first Kind is impossible”
Corollary 2
“The internal energy of a closed system remains unchanged if the system is isolated from its
surroundings”
Corollary 1
“There exists a property of a closed system such that the change in its value is equal to the
difference between heat supplied and work done during the change of state”

11. Corollary 1
“There exists a property of a closed system such that the change in its value is equal to the
difference between heat supplied and work done during the change of state”
∮40= ∮42
∫40
"
!
A +∫40
"
!
B =∫42
"
!
A + ∫42
"
!
B
∫40
"
!
C +∫40
"
!
B =∫42
"
!
C + ∫42
"
!
B
∫40
"
!
A -∫40
"
!
C =∫42
"
!
A -∫42
"
!
C
Subtracting and rearranging
∫(40
"
!
-42)
6= ∫(40
"
!
-42)
7
Volume
Pressure
1
2
C
B
A
A, B, & C are arbitrary processes
between state 1 and state 2

12. *+ −*, 89 9:;< =>? :@@ AB< C?>D<99<9 E<AF<<G 9A:A< H :GI 9A:A< J
Therefore *+ −*, depends only on the initial and final states and not the path followed
between the two states
dE= 40 −42
E
2–E
1=
1Q
2–
1W
2
•The physical significance of the property E is that it represents all the energy of the system in the given
state
•In thermodynamics, it is convenient to consider the bulk kinetic and potential energy separately and then to
consider all the other energy of the control mass in a single property that we call the internal energy and to
which we give the symbol U
E = Internal energy + kinetic energy + potential energy
E = U + KE + PE

13. The perpetual motion machine (PMM)
HYPOTHETICAL SYSTEM
produce useful work
indefinitely
produce more work or
energy than they
consume
There is undisputed scientific consensus that “Perpetual motion would violate the Laws of Thermodynamics”

14. What is “Perpetual Motion”?
•Describesatheoreticalmachinethat,withoutanylossesduetofrictionorotherformsof
dissipationofenergy,wouldcontinuetooperateindefinitelyatthesameratewithoutanyexternal
energybeingappliedtoit
•Machineswhichcomplywithboththe1
st
&2
nd
LawsofThermodynamicsbutaccessenergyfrom
obscuresourcesaresometimesreferredtoas“PerpetualMotion”machines

15. Perpetual Machine of First Kind (PMFK)
•A “perpetual motion” machine of the first kind produces work without the input of energy
It thus violates the 1
st
Law of Thermodynamics: the Law of conservation of energy
•First law states “that the total amount of energy in an isolated system remains constant over”
•A consequence of this law is that energy can neither be created nor destroyed: it can only be
transformed from one state to another
•So, It is clearly impossible for a machine to do the work infinitely without consuming energy
Examples of the 1
st
Kind of “Perpetual Motion” Machine
The Overbalanced Wheel The Float Belt The Capillary Bowl

16. Laws of Conservation
Law of Conservation of Mass: Mass can neither be created nor be destroyed, but may be converted
from one form to another form
A
1V
1'
1= A
2V
2'
2
(One dimensional continuity equation)
Law of Conservation of Momentum: If no external force acts on a system, linear momentum is
conserved in both direction and magnitude
m
1V
1+ m
2V
2= (m
1+m
2)V
Law of Conservation of Energy: Energy can neither be created nor be destroyed

17. Energy and System
Change in Total Energy
of the System
Total Energy Entering
the System
Total Energy Leaving
the System
= -
∆E = E
in-E
out
∆E
system= E
finalstate–E
initialstate
Mechanism of Energy Transfer
Work TransferMass TransferHeat Transfer
Energy balance in rate form, E
in–E
out=
IK
9L9A<;
IA
0 for steady state
For a closed system undergoing a cycle , ∆ E
system= E
2 –E
1 = 0 -> E
in= E
out

18. •Theconservationofmassandtheconservationofenergyprinciplesforopensystemsorcontrol
volumesapplytosystemshavingmasscrossingthesystemboundaryorcontrolsurface
•Thermodynamicprocessesinvolvingcontrolvolumescanbeconsideredintwogroups:steady-flow
processesandunsteady-flowprocesses
•DefiningaSteadyFlowProcess:Aprocessduringwhichthefluidflowssteadilythroughthe
controlvolume(CV)
-FlowprocessàfluidflowsthroughCV
-Steadyànotchangingwithtime
-Duringasteadyflowprocess:
-Conditions(fluidproperties,flowvelocity,elevation)atanyfixedpointwithintheCVareunchangingwithtime
-Properties,flowvelocityorelevationmaychangefrompointtopointwithinCV
-Size,shape,massandenergycontentoftheCVdonotchangewithtime
-Rateatwhichheatandworkinteractionstakeplacewithsurroundingsdonotchangewithtime
-Devices/systemswhichundergosteadyflowprocess:
Conservation of Energy for Control Volumes
compressors, pumps, turbines, water supply pipes, nozzles, heat exchangers, power plants, aircraft engines etc.

19. •Duringasteady-flowprocess,thefluidflowsthroughthecontrolvolumesteadily,
experiencingnochangewithtimeatafixedposition
V
cm
m
iV
i
m
eV
e
W
net
Q
net
Control Surface
Z
i
Z
cm
Z
e
The mass and energy content of the open system may change when mass enters or leaves the control volume
e –exit
i-inlet
Z –height
Q –Heat
W –Work
m –Mass
V –Velocity
For Steady-State, Steady-Flow Processes
•Mostenergyconversiondevicesoperatesteadilyoverlongperiodsoftime
•Theratesofheattransferandworkcrossingthecontrolsurfaceareconstantwithtime
•Thestatesofthemassstreamscrossingthecontrolsurfaceorboundaryareconstantwithtime
•Undertheseconditionsthemassandenergycontentofthecontrolvolumeareconstantwithtime
MN
OP
MQ
= ∆ṁ
cv= 0
MR
OP
MQ
= ∆Ė= 0

20. Steady-state, Steady-Flow Conservation of Mass: ! ! (/)m m kgs
in out∑∑=
Steady-state,steady-flowconservationofenergy:Theenergyofthecontrolvolumeisconstant
withtimeduringthesteady-state,steady-flowprocess
! ! !E E E kW
in out system− =
Rate of net energy transfer
by heat, work, and mass
Rate change in internal, kinetic,
potential, etc., energies
()
"#$%$ "#$%$
Δ
0

21. •Theconservationofenergyprincipleforthecontrolvolumeoropensystemhasthesameword
definitionasthefirstlawfortheclosedsystem
•Expressingtheenergytransfersonaratebasis,thecontrolvolumefirstlawis
! ! !E E E kW
in out system− =
Rate of net energy transfer
by heat, work, and mass
Rate change in internal, kinetic,
potential, etc., energies
()
"#$%$ "#$%$
Δ
Consideringthatenergyflowsintoandfromthecontrolvolumewiththemass,energyenters
becausenetheatistransferredtothecontrolvolume,andenergyleavesbecausethecontrolvolume
doesnetworkonitssurroundings,theopensystem,orcontrolvolume,applyingthefirstlawof
thermodynamics
Conservation of Energy for General Control Volume
Energy balance in differential form dE= 4Q -4W
Time rate form of energy balance
MR
MQ
= Q –W
. .

22. Time rate change of energy is given by
MR
MQ
=
MSR
MQ
+
MTR
MQ
+
MU
MQ
Time rate change of energy is given by Q –W =
MSR
MQ
+
MTR
MQ
+
MU
MQ
Where the time rate change of the energy of the control volume has been written asΔ
!
E
CV
Consideringthatenergyflowsintoandfromthecontrolvolumewiththemass,energyenters
becauseheatistransferredtothecontrolvolume,andenergyleavesbecausethecontrolvolume
doesworkonitssurroundings,thesteady-state,steady-flowfirstlawbecomes

23. Total energy crossing
boundary per unit time
Total energy of mass
leaving CV per unit time
Total energy of mass
entering CV per unit time
! ! !
! ! !
Q Q Q
W W W
net in out
net out in
= −
= −
∑∑
∑∑
Mass flow rate (kg/s) =
V
!
W
!
P
!
=
V
"
W
"
P
"
Steady Flow process Involving one fluid stream at the inlet and exit of the control volume
Where, V = Velocity (m/s)
v = specific volume (m
3
/kg)

24. •Anumberofthermodynamicdevicessuchaspumps,fans,compressors,turbines,nozzles,
diffusers,andheatersoperatewithoneentranceandoneexit
•Thesteady-state,steady-flowconservationofmassandfirstlawofthermodynamicsforthese
systemsreduceto
mmm
ie
!!! ==∑∑since
)
2
()
2
(
22
i
i
ie
e
e gzhmgzhmWQ ++−++=−
VV
!!!!
Using subscript 1 and subscript 2 for denoting
inlet and exit states
)](
2
[
12
2
1
2
2
12 zzghhmWQ −+
−
+−=−
VV
!!!
)(
2
12
2
1
2
2
12 zzghhwq −+
−
+−=−
VV
Dividing the equation bym!
yields
wzghq +Δ+
Δ
+Δ=
2
or
2
V
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+Δ+
Δ
+Δ+Δ= wzgpvuq
2
2
V
Steady-State, Steady-Flow for One Entrance and One Exit

25. ;
H(
X
H
Y
HZZZ
+
\
H
J
JZZZ
+B
H)++
G<A= ;
J

X
J
Y
HZZZ
+
\
J
J
JZZZ
+B
J + ,
G<A
Solving Steady Flow Energy Equation
[h, W, Q should be in kJ/kg and V in m/s and g in m/s
2
]
;
H(X
HY+
\
H
J
J
+B
H)++
G<A= ;
J

X
JY+
\
J
J
J
+B
J + ,
G<A
[h, W, Q should be in J/kg and V in m/s and g in m/s
2
]
Our aim is to give heat to the system and gain work output from it.
So heat input →+ive(positive) Work output →+ive(positive)
+ive(positive) +ive(positive)
-ive(negative) -ive(negative)
Work Work Heat
Heat

26. Applications of SFEE
•Nozzles and diffusers (e.g. jet propulsion)
•Turbines (e.g. power plant, turbofan/turbojet aircraft engine), compressors and pumps (power
plant)
•Heat exchangers (e.g. boilers and condensers in power plants, evaporator and condenser in
refrigeration, food and chemical processing)
•Mixing chambers (power plants)
•Throttling devices (e.g. refrigeration, steam quality measurement in power plants)
All elements of a simple power plant/ refrigeration cycle and more! In principle, you can take the
elements together to calculate power generated/required, heat removed/supplied.
Source: www.google.com
Throttling Devices
Heat ExchangersGas Turbines

27. Water Turbine
X
HY+
\
H
J
J
+]
H+C
H^
9H−F+_=X
JY+
\
J
J
J
+]
J+C
J^
9J

X
HY+
\
H
J
J
+]
H+C
H^
9H+_=X
JY+
\
J
J
J
+]
J+C
J^
9J + F
^
9H= ^
9J= ^
9
X
HY+
\
H
J
J
−F=X
JY+
\
J
J
J
+ ^
9(p
2–p
1)
X
H
−X
J
,
Datum
Water Turbine

28. Steam / Gas Turbine
X
HY+
\
H
J
J
+B
H−_=X
JY+
\
J
J
J
+B
J

+F
X
HY+
\
H
J
J
+]
H+C
H^
9H+_=X
JY+
\
J
J
J
+]
J+C
J^
9J +F
IX=Z
,
Steam/Gas
Turbine
\
H
J
J
+B
H=
\
J
J
J
+B
J

+ F
Steam/Gas In
Steam/Gas Out

29. Steam Nozzle
X
HY+
\
H
J
J
+B
H+_=X
JY+
\
J
J
J
+B
J + F
X
HY+
\
H
J
J
+]
H+C
H^
9H+_=X
JY+
\
J
J
J
+]
J+C
J^
9J

+F
\
H
J
J
+B
H=
\
J
J
J
+B
J

Steam In
Steam Out
\
J=\
H
J
+J(B
H−B
J)
\
J=J(B
H−B
J)

30. Boiler
X
HY+
\
H
J
J
+]
H+C
H^
9H+_=X
JY+
\
J
J
J
+]
J+C
J^
9J

+ F
X
HY+
\
H
J
J
+]
H+C
H^
9H+_=X
JY+
\
J
J
J
+]
J+C
J^
9J

+F
h
1+ q = h
J
aA<:;
+
,:A<?
Control Surface
Boiler
q = h
J-h
H

31. Heat Exchanger
X
HY+
\
H
J
J
+]
H+C
H^
9H−_=X
JY+
\
J
J
J
+]
J+C
J^
9J + F
X
HY+
\
H
J
J
+]
H+C
H^
9H+_=X
JY+
\
J
J
J
+]
J+C
J^
9J

+ F
h
H-q = h
J
aA<:;
+
,:A<?
Control Surface
1 kg steam
q= (h
w2–h
w1) m
w= m
wC
w(T
w2–T
w1)
h
H-(h
w2–h
w1) m
w= h
J
h
H-h
J= (h
w2–h
w1) m
w=m
wC
w(T
w2–T
w1)
q-heat gained by water by passing through condenser
T
w1
T
w2
m
w–flow of water per kg steam
q-heat lost by 1 kg steam to water passing through condenser

32. Reciprocating Compressor
X
HY+
\
H
J
J
+]
H+C
H^
9H−_=X
JY+
\
J
J
J
+]
J+C
H^
9J

− F
X
HY+
\
H
J
J
+]
H+C
H^
9H+_=X
JY+
\
J
J
J
+]
J+C
J^
9J

+ F
b<D8<^<?
+
c8? dG
Control Surface
,
B
H+F−_=B
J

If Velocity changes are neglected and flow process is
treated as adiabatic
Due to large area in contact and low flow rates appreciable heat transfer can take place between the system
and the surroundings. Therefore water cooling is required

33. Rotary Compressor
X
HY+
\
H
J
J
+]
H+C
H^
9H−_=X
JY+
\
J
J
J
+]
J+C
J^
9J − F
X
HY+
\
H
J
J
+]
H+C
H^
9H+_=X
JY+
\
J
J
J
+]
J+C
J^
9J

+ F
c8? e]A
+c8? dG
Control Surface
,

34. Centrifugal Pump
X
HY+
\
H
J
J
+]
H+C
H^
9H+_=X
JY+
\
J
J
J
+]
J+C
J^
9J

+ F
,
Water Sump
\
H
J
J
+X
HY+F=
\
J
J
J
+X
JY+ fg(C
J −C
H)
Control Surface
Steam/Gas Out
X
H
X
J
X
HY+
\
H
J
J
+]
H+C
H^
9H+_=X
JY+
\
J
J
J
+]
J+C
J^
9J

+F

35. Ablowerhandles1kg/sofairat20°C.Findtheexitairtemperature,assumingadiabatic
conditions.Takecpofairis1.005kJ/kg-K.
+=Z
100 m/s 150 m/s
W = 15 kW
X
HY+
\
H
J
J
+]
H+C
H^
9H+_=X
JY+
\
J
J
J
+]
J+C
J^
9J

+ F
X
HY+
\
H
J
J
+B
H+_=X
JY+
\
J
J
J
+B
J

-F
m
1= m
2= kg /s

36. Throttling process
•Aflowisthrottledwhen,forexample,itflowsthroughapartiallyopenvalve
•Whenitdoesso,wenoticethattherecanbeasignificantpressurelossfromonesideofthe
partiallyopenvalvetotheother
So, we can say that such a throttling device is one in which pressure drops and enthalpy remains constant
In throttling devices there may be a change in velocity due to compressibility effects, but it is
observed to be small when the flow velocity is much less than the speed of sound.
We shall assume here the velocity is small relative to the speed of sound so as to recover v
1∼v
2
Thus,h
1= h
2
•Athrottlingprocessismodeledassteadydevicewithoneentranceandexit,withnocontrolvolumeworkor
heattransfer
•Changesinareaaswellaspotentialenergyareneglected

37. This shows that enthalpy remains constant during adiabatic throttling process.
The throttling process is commonly used for the following purposes :
1.For determining the condition of steam (dryness fraction)
2.For controlling the speed of the turbine
3.Used in refrigeration plants
4.Liquefaction of gases

38. In this experiment gas is forced through a porous plug and is called a throttlingprocess
iiiTvP
fffTvP
piston
porous plug
adiabatic walls
The Joule-Thomson Experiment
•In an actual experiment, there are no pistons and there is a continuous flow of gas
•A pump is used to maintain the pressure difference between the two sides of the porous plug
In this experiment, as pressures are kept constant work is done
iiff
v
0
f
0
v
i vPvPvdPvdPw
f
i
−=+= ∫∫
h_=I]+ hF
iiifffiiffif vPuvPuor)vPvP()uu(0 +=+−+−=
From the definition of enthalpy
ifhh=
Hence, in a throttling process, enthalpy is conserved

39. 39
•Intheregionwheretheatomsormoleculesareveryclosetogether,thenrepulsiveforces
dominateandasthevolumeexpands,theenergygoesdown.Thus,fortheseconditions,p
Tis
negative
•Intheregionwheretheatomsormoleculesarecloseenoughthatattractiveforcesdominate,
thenasthevolumeexpands,theenergygoesup.Thus,fortheseconditions,p
Tispositive
•Formostgasesatnottoolargepressures,themoleculesdon'tinteractverymuchandsothereis
littledependenceofenergyonvolumesop
Tisverysmall.Intheextremeofzero
interaction,p
Tiszero.Thisisthedefiningconditionforanideal(perfect)gas.

40. 40
Pump
Porous plug
P
i T
i P
fT
f
Throttling Process
(Joule-Thomson or Joule-Kelvin expansion widely used in refrigerators)
•ThepumpmaintainsthepressuresP
iandP
f
•IntheexperimentP
i,T
iandP
faresetandT
fis
measured
•Consider a series of experiments in which P
iand T
iare
constant (h
i constant) and the pumping speed is changed to
change P
fand hence T
f
•Since the final enthalpy does not change, we get points
of constant enthalpy
The enthalpy is the same on the
two sides of the porous plug i.e.,
h
f= h
i.
We plot T
fas a function of P
f
P
i> P
f

41. 41
Pressure
f
Temperature
f
-A smooth curve is placed through the points yielding an isenthalpic curve
-Note that this is not a graph of the throttling process as it passes through irreversiblestates
•
•
••
•
•
•
•
P
f , T
f
P
i, , T
i
Isenthalpic Curve

42. 42
WenowchangeP
iandT
iandobtainanotherisenthalpiccurve
Pressure
f
Temperature
f
Maximum Inversion T
Cooling
•
d
•
c
•
b
•
a
Heating
Inversion Curve
Inversion Curve
Ideal Gas
Weareinterestedinthetemperaturechange
duetothepressurechange,thereforeitis
usefultodefinetheJoule-Thomsoncoefficient
i
This is the slope of an isenthalpic curve and
hence varies from point to point on the graph
()
hP
T
∂
∂
=µ
•Apointatwhichi=0iscalledaninversionpoint
•Connectingallofthesepointsproducestheinversioncurve

43. 43
Pressure
f
Temperature
f
Maximum Inversion T
Cooling
•
d
•
c
•
b
•
a
Heating
Inversion Curve
Inversion Curve
Ideal Gas
If point aon the diagram
(i< 0) is a starting point
and point bis the final
point, then the T of the
gas will rise, i.e. we have
heatingIf we start at point c
(i> 0) and go to point
d, then the T of the
gas will drop, i.e. we
have cooling These curves are horizontal
lines for an ideal gas
As higher initial starting temperatures are used, the isenthalpic curves become flatter and more
closely horizontal
Maximum inversion T, the value of which depends on the gas.
For cooling to occur, the initial T must be less than the maximum inversion T,for such a T the
optimum initial P is on the inversion curve

44. •Thisalsotellsusthatwecannotjustuseanygasatanysetofpressurestomakearefrigerator,
forexample
-Atagivenpressure,somegasesmaybecooling(m>0)butothersmaybeheating(m<0)
•Theproperchoiceofrefrigerantwilldependonboththephysicalproperties,esp.theJoule-
Thompsoncoefficientaswellasthemechanicalcapacityoftheequipmentbeingused.
•Thus,wecannotjustexchangeourozone-depletingfreoninourcar'sairconditionerwithany
othercoolantunlessthetwogasesbehavesimilarlyinthepressure-temperaturerangesofthe
mechanicaldevice,i.e.,theymusthavethesamesignofmatthepressurestheequipmentis
capableofproducing.
•Generally,touseamoreenvironmentallyfriendlycoolant,weneedtoreplacetheoldequipment
withnewequipmentthatwilloperateinthetemperaturerangeneededtomakempositive.

45. http://faculty.chem.queensu.ca/people/faculty/mombourquette/Chem221/3_FirstLaw/ChangeFunctions.asp
•The sign of the Joule–Thomson coefficient, µ, depends on the conditions
PositiveNegative
•The temperature corresponding to the boundary at a given pressure is the ‘inversion temperature’
of the gas at that pressure
The maximum inversion temperatures of
some gases are given below :
(i) He=24K
(ii) H
2=195K
(iii) Air=603K
(iv) N
2=261K
(v) A=732K
(vi) CO
2=1500K

46. 46
To make the discussion clear, we have exaggerated the slopes in the above T-P diagram. In fact, for
most gases at reasonable T’s and P’s the isenthalpic curves are approximately flat and so0≈µ
It can be shown that
µ
P
h
P
T
c
P
T
c
P
h
−=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
−=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
)(00 Thhsoand
P
h
then
T
==
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
=µ
We now have
)1.5oblem(Pr0
P
h
v
u
TT
=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
Constant temp. coefficient and can be
determined by Joules Thompson experiment
for an ideal gas
•Foragivenpressure,thetemperaturemustbebelowacertainvalueifcoolingisrequired
but,ifitbecomestoolow,theboundaryiscrossedagainandheatingoccurs
•Reductionofpressureunderadiabaticconditionsmovesthesystemalongoneoftheisenthalps,or
curvesofconstantenthalpy

47. 47
•Somegasescanbeliquefiedinasimpleprocess
-Forexample,carbondioxidecanbeliquefiedatroomtemperaturebyasimpleisothermal
compressiontoabout60bar
•Toliquefynitrogenorairisnotsosimple.
-Atroomtemperature,regardlessofanyincreaseinpressure,thesegaseswillnotundergoa
phasetransformationtotheliquidstate
•Amethodforthesegases,usingthethrottlingprocess,wasinventedin1895andiscalledthe
Hampson-LindeProcess
-Thebasisideaistousethegascooledinthethrottlingprocesstoprecoolthegasgoingtowards
thethrottle
untiltheTisbelowthemaximuminversionT
-Startingfromroomtemperature,thiscyclecanbeusedtoliquefyallgasesexcepthydrogenand
helium
-ToliquefyHbythisprocess,itmustfirstbecooledbelow200KandtoaccomplishthisliquidN
at77Kisused
-ToliquefyHebythisprocess,itmustfirstbecooledbelow43KandtoaccomplishthisliquidH
canbeused.(AdevicecalledtheCollinsheliumliquifierisusedtoliquefyHe.
Liquefaction of Gases

48. Hampson-LindeProcess

49. 49
Ifathrottlingprocessisusedtoliquefyagas,thecooledgasisrecycledthroughaheatexchangerto
precoolthegasmovingtowardsthethrottle.Thegascontinuestocoolandwhenasteadystateis
reachedacertainfraction,y,isliquefiedandafraction(1-y)isreturnedbythepump.
Usingthenotation:
-h
i=molarenthalpyofenteringgas
-h
f=molarenthalpyofemerginggas
-h
L=molarenthalpyofemergingliquid
Sincetheenthalpyisconstantwehave
h
i=yh
L+(1-y)h
f
Ofcourse,assomeofthegasliquefies,additionalgasmustbeaddedtothesystem.
It should be mentioned that Joule-Thomson liquefaction of gases has these advantages:
•No moving parts that would be difficult to lubricate at low T.
•The lower the T , the greater the T drop for a given pressure drop.

50. Controlling the speed of the steam turbine by Throttling