UNIT-IFLUID PROPERTIES & FLOW CHARACTERISTICS

Units
and dimensions Properties of fluids mass density, specific weight,
specific volume, specific gravity, viscosity, compressibility, vapor pressure,
surface tension and capillarity

1. UNIT I -FLUID PROPERTIES &
FLOW CHARACTERISTICS

2. UNIT I FLUID PROPERTIES AND FLOW CHARACTERISTICS
Unitsanddimensions-Propertiesoffluids-massdensity,specificweight,
specificvolume,specificgravity,viscosity,compressibility,vaporpressure,
surfacetensionandcapillarity.Flowcharacteristics–conceptofcontrol
volume-applicationofcontinuityequation,energyequationand
momentumequation.

3. Fluid Properties:
1. Density
2. Specific Gravity
3. Viscosity
4. Surface Tension
5. Capillarity
6. Compressibility
7. Bulk modulus
8. Vapour pressure

4. Density:
Mass densityor specific mass: 
Thedensityofafluidisdefinedasitsmassperunitvolumeat
standardtemperatureandpressure.
�=
????????????�����ℎ����??????�
���������ℎ����??????�
Where,
=
�
�
unit -kg/m
3
For water, = 1000 kg/m
3

5. Specific Weight or weight density: w
It is defined as Weight per unit volume at standard temperature and pressure
Where,
w = g ; Unit -N/m
3
;
w
water= 9.81 kN/m
3
�=
��??????�ℎ������??????�
�����������??????�
=
????????????�������??????�
�����������??????�
�
Specific Volume: v
It is defined as Volume per unit mass of the fluid.
v = V/m = 1/
Unit -m
3
/kg;1 m
3 =
1000 litres

6. SpecificGravity:S
Itistheratioofthespecificweightoftheliquidtothe
specificweightofthestandardfluid.
??????���??????�??????�??????�??????�??????��=
specificweightoftheliquid
Specificweightofthepurewater
Where,
WeightDensityofLiquid=SXWeightdensityofwater
=SX1000X9.81N/m
3
DensityofLiquid=SXdensityofwater
=SX1000kg/m
3

7. Problem 1: Calculate the density, specific weight and weight of one litreof
petrol of specific gravity = 0.7
Given:
�=1�??????���=1�1000��
3
=
1000
10
6
m
3
= 0.001 m
3
S = 0.7
Solution:
(i) Density, ρ= S x1000 kg/m
3
= 0.7 x1000 = 700 kg/m
3
(ii) Specifcweight, w= ρxg = 700 x9.81 N/m
3
= 6867 N/m
3
(iii) Weight, W =
��????????????ℎ�
�??????����
w=
�
0.001
or 6867 =
�
0.001
W = 6867 x0.001
W = 6.867 N

8. Kinematic viscosity:
It is defined as the ratio between the dynamic viscosity and mass
density of fluid.
Units: m
2
/s ; 1 Stoke = 10
-4
m
2
/s
Viscosity:
Dynamic viscosity:
Viscosity is defined as the property of fluid which offers resistance to
flow.
Unit -Ns/m
2
; 1 Poise = (1/10) Ns/m
2
; 1 Centipoise= (1/100) Poise
????????????
��
��
,??????=�
��
��
,
�=
�
�

9. Dynamic Viscosity
Force
Area
Shear stress (

)
Shear rate (du/dy)Slopeofline=
�=
??????
ൗ
��
��

10. Problem2:APlate0.025mmdistantfromafixedplate,movesat60cm/sand
requiresaforceof2Nperunitareai.e.,2N/m
2
tomaintainthisspeed.
Determinethefluidviscositybetweentheplates.
Given:
dy= 0.025 mm = 0.000025 m
u = 60 cm/s = 0.6 m/s
du = u –0 = 0.6 m/s
??????= 2 N/m
2
Solution:
µ =
??????
????????????
????????????
=
2
0.6
0.000025
µ = 8.33 x 10
-5
Ns/m
2
(or) 8.33 x 10
-4
poise

11. Problem3:Determinetheintensityofshearofanoilhavingviscosity=1poise.Theoil
isusedforlubricatingtheclearancebetweenashaftofdiameter10cmanditsjournal
bearing.Theclearanceis1.5mmandtheshaftrotatesat150r.p.m.
Given:
µ = 1 poise =
1
10
??????�
�
2
D = 10 cm = 0.1 m
Dy = 1.5 mm = 0.0015 m
N = 150 r.p.m.
Solution:
u =
??????????????????
60
=
??????∗0.1∗150
60
= 0.785 m/s
du = u –0 => u = 0.785 m/s
??????= µ
��
��
=
1
10
x
0.785
1.5�10
−3
??????= 52.33 N/m
2

12. Problem4:Ifthevelocitydistributionoveraplateisgivenbyu=3/4y–y
2
inwhichuis
thevelocityinmertrepersecatadistanceymetreabovetheplate.Determinetheshear
stressaty=0andy=0.15m.Takedynamicviscosityoffluidas8.5poise.
Solution:
u =
3
4
y –y 2
��
��
=
3
4
−2y
At y = 0.15,
��
��
=
3
4
−2∗0.15=0.75−0.30=0.45
Viscosity, µ = 8.5 poise = 0.85 Ns/m
2
We know,
??????= µ
��
��
=0.85∗0.45
??????= 0.3825 N/ m
2

13. Problem 5: A 400 mm diameter shaft is rotating at 200 r.p.m. in a bearing of length 120 mm. If
the thickness of oil film is 1.5 mm and the dynamic viscosity of the oil is 0.7 N.s/m
2
determine:
(i) Torque required to overcome friction in bearing (ii) Power utilised in overcoming viscous
resistance.
Given: D = 400 mm = 0.4 m
N = 200 rpm
L = 120 mm = 0.12 m
dy= 1.5 mm
µ = 0.7 Ns/m
2
To find:
(i) Torque
(ii) Power
Solution:
u
1 = 0; u
2=
??????????????????
60
=
??????∗0.4∗120
60
= 2.5 m/s
du = u
2–u
1= 2.5 m/s

14. µ =
??????
????????????
????????????
=> 0.7 =
??????
2.5
0.0015
=> ??????= 1172.26 N/m
2
F
shear= ??????* A = ??????* �* D * L
= 1172.26 * �* 0.4 * 0.12
F
shear= 176.77 N
T = F
Shear* r
= 176.77 * 0.2
T = 35.2 N m
P =
2??????????????????
60
=
2??????120∗35.2
60
P = 737.2 W

15. Problem6:Calculatethedynamicviscosityofanoil,whichisusedforlubricationbetweena
squareplateofslice0.8mx0.8mandaninclinedplanewithangleofinclination30
o
asshown
infig.Theweightofthesquareplateis300Nanditslidesdowntheinclinedplanewitha
uniformvelocityof0.3m/s.Thethicknessofoilfilmis1.5mm.
Given:
A = 0.8 x 0.8 = 0.64 m
2
θ= 30
o
W = 300 N
u = 0.3 m/s
du = u –0 = 0.3 m/s
t = dy= 1.5 mm = 1.5 x 10
-3
m
To find:
µ

16. Solution:
Component of weight W, along the plane = W cos 60
o
= 300 cos 60
o
= 150 N
Shear force F, on the bottom surface of the plate = 150 N
??????=
??????
??????��??????
=
150
0.64
=234.375
Using Newton’s law,
µ =
??????
????????????
????????????
µ =
234.375
0.3
0.0015
= 1.17 Ns/m
2
µ = 11.7 poise.
30
o
60
o
W

17. Homework: Two large plane surfaces are 2.4 cm apart. The space between the surfaces is filled
with glycerine. What force is required to drag a very thin plate of surface area 0.5 square metre
between the two large plane surfaces at a speed of 0.6 m/s, if (i) the thin plate is in the middle of
the two plane surfaces, and (ii) the thin plate is at a distance of 0.8 cm from one of the plane
surfaces? Take the dynamic viscosity of glycerine= 8.10 x 10
-1
Ns/m
2

18. Types of Fluids:
Shear stress (
??????
)
Shear rate(du/dy)
Ideal Solid
Ideal Fluid

19. IdealFluid:
Afluid,whichisincompressibleandishavingnoviscosity,isknownasIdeal
fluid.ItisonlyimaginaryFluid
RealFluid:
Afluid,whichishavingviscosity,isknownasRealfluid.AllfluidsareReal
fluids.
Newtonianfluids:
ThesefluidsfollowNewton’sviscosityequation.Forsuchfluidsdoesnot
changewithrateofdeformation.Exp.Water,Keroseneetc.,
NonNewtonianfluids:
Therefluidswhichdonotfollowthelinearrelationshipbetweentheshear
stressandtherateofdeformation.Exp.Polymersolutions,Bloodetc.,

20. SurfaceTensionofLiquids
Thephenomenonofsurfacetensionarisesduetothetwokinds
ofintermolecularforces.SurfaceTensioniscausedbytheforceof
cohesionatthefreesurface.
(i)Cohesion:Theforceofattractionbetweenthemoleculesofthe
sameliquid.Surfacetensionisduetocohesionbetweenparticlesat
thefreesurface.
(ii)Adhesion:Theforceofattractionbetweenunlikemoleculesofa
solidboundarysurfaceincontactwiththeliquid

21. Case I -Water droplet:
Let, P = Pressure inside the droplet above outside pressure
D = Diameter of the droplet
= Surface tension of the liquid
i. Pressure force = ??????×
??????
4
�
2
and
ii. Surface tension force acting around the circumference = d
Under equilibrium conditions these two forces will be equal and opposite,
??????×
??????
4
�
2
=??????��
�=
4??????
�

22. Case II -Soap bubble:
i.Pressure force =??????×
??????
4
�
2
, and
ii.Surface tension force acting around the circumference =2(d)
??????×
�
4
�
2
=2??????��
�=
8??????
�
Case III-A liquid Jet:
i.Pressure force = p ld , and
ii.Surface tension force acting around the circumference =2 l
??????×�×�=??????×2�
�=
2??????
�

23. Surface Tension Formulae:
S. No. Pressure forceSurface TensionPressure
Water Jet p ld 2 l �=
2??????
�
Water Droplet??????×
�
4
�
2
d �=
4??????
�
Soap Bubble ??????×
�
4
�
2
2 (d) �=
8??????
�

24. Problem 7: Find the surface tension in a soap bubble of 40 mm diameter when
the inside pressure is 2.5 N/m
2
above atmospheric pressure.
Given:
d = 40 mm = 0.04m
p = 2.5 N/m
2
To find :
Surface Tension (σ)
Solution:
p =
8??????
�
σ=
p∗d
8
=>
2.5∗0.04
8
σ= 0.0125 N/m

25. Capillarity:
➢Capillarityisaphenomenonbywhichaliquidrisesintoathinglasstube
aboveorbelowitsgenerallevel.surfacerelativetotheadjacentlevelof
thefluidiscalledcapillarity.
➢Thisphenomenonisduetothecombinedeffectofcohesionandadhesion
ofliquidparticles.
For water = 0
For mercury = 140
0

27. Expression for Capillary Rise:
= Surface tension of the liquid
= Angle of contact between the liquid and glass tube
= Density of the liquid
The weight of the liquid of height h in the tube = (Area of the tube x h) x x g
=
�
4
�
2
×ℎ×�×�
Vertical component of the surface tensile force=(??????×????????????�����������)×cos??????
=??????×��×cos??????
Equating these two eqn.,
�
4
�
2
×ℎ×�×�=??????×��×cos??????
ℎ=
??????×��×cos??????
�
4
�
2
��
=
4??????cos??????
���
ℎ=
4??????
���
If is equal to zero

28. Expression for Capillary Fall:
= Surface tension of the liquid
= Angle of contact between the liquid and glass tube
= Density of the liquid
Intensity of the pressure at the depth = (Area of the tube x h) x x g
=
�
4
�
2
×ℎ×�×�
Vertical component of the surface tensile force=(??????×????????????�����������)×cos??????
=??????×��×cos??????
Equating these two eqn.,
�
4
�
2
×ℎ×�×�=??????×��×cos??????
ℎ=
??????×��×cos??????
�
4
�
2
��
ℎ=
4??????cos??????
���
Note: For mercury
and glass tube is 138
o

29. Problem8:Calculatethecapillaryriseinaglasstubeof4mmdiameter,when
immersedin(i)water,and(ii)mercury.thetemperatureoftheliquidis20
o
Candthe
valuesofthesurfacetensionofwaterandmercuryat20
o
Cincontactwithairare
0.073575N/m&0.51N/mrespectively.Theangleofcontactforwateriszerothatfor
mercury130
o
.Takedensityofwaterat20
o
Casequalto998kg/m
3
.
Given:
d = 4 mm = 0.004 m
Solution:
(i)Capillary effect for water
σ = 0.073575 N/m, θ= 0
o
ρ= 998 kg/m
3
@ 20
o
C
h =
4σcosθ
ρ??????�
=
4∗0.073575∗cos0
998∗9.81∗0.004
h = 0.00751 m or 7.51 mm

30. Contd…
(i)Capillary effect for mercury
σ = 0.51 N/m, θ= 130
o
ρ= 13600 kg/m
3
(ρ= S * 1000 => 13.6 * 1000)
h =
4σcosθ
ρ??????�
=
4∗0.51∗cos130
13600∗9.81∗0.004
h = -0.00246 m or –2.46 mm

31. Bulkmodulus:
Itisdefinedastheratioofcompressivestresstovolumetric
strain.
Unit-Τ
??????
�
2
Thenegativesignindicatesthevolumedecreasesaspressure
increases.
Compressibility:
Whenafluidissubjectedtoapressureincreasethe
volumeofthefluiddecreases.Itisknownascompressibility.It
isreciprocalofbulkmodulus.
Compressibility=1/k
�=−
��
���

32. Vapourpressure:
Allliquidshaveatendencytoevaporatewhenexposed
toagaseousatmosphere.Thevapourmoleculesexertapartial
pressureinthespaceabovetheliquid,knownasvapour
pressure.

33. Gas Laws:
Boyle’s Law
Atagiventemperatureforagivenquantityofgas,theproductof
thepressureandthevolumeisaconstant.
P
1V
1= P
2V
2
Ideal Gas Law:
Ideal Gas Law relates pressure to Temp for a gas
P = RT
Where, T = Absolute temperature in
o
K units
R= Gas constant = 287 Joule / Kg-
o
K

34. Universal Gas constant:
PV = mRT
Where,
m = Mass of gas in kg;
v = Specific Volume.
p = Absolute pressure ;
T = Absolute temperature

35. Pressure in a Fluid
�=��ℎ

36. Pressure:
Forceperunitarea:1N/m
2
=Pascal=Pa
StandardAtmosphereP=101.33kPa=101.3kN/m
2
=10.3mofmg

37. Thank you