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Unit - 1
BASIC CONCEPTS AND PROPERTIES
Fluid - definition, distinction between solid and fluid - Units and dimensions - Properties
of fluids - density, specific weight, specific volume, specific gravity, temperature,
viscosity, compressibility, vapour pressure, capillary and surface tension - Fluid statics:
concept of fluid static pressure, absolute and gauge pressures - pressure
measurements by manometers and pressure gauges.
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1.2.4. Specific Gravity. Specific gravity is defined as the ratio of the weight density (or density) of afluid to the weight density (01 density) of a standard fluid. For liquids, the standard fluid is taken water andfor gases, the standard fluid is taken air. Specific gravity is also called rela rive density. It is dimensionless
quantity and is denoted by the symbol S.
¥athematically, Set' u 'd) Weight density (density) of liuui.:or lqUl s = - - - ..I: _
Weight density (density) ofwat::r
S(4' ) _ Weight density (density) of gasJ.orgases - W . h d . (d . '\ f .erg t density ensity J 0 aIT
Thus weight density of a liquid =S x Weight density of water= S x 1000 x 9.81 N/mo,
The density of a liquid = S x Density of water= S x 1000 !'(g)m·i. ...(1.1 A)
Problem 1.1. Calculate the specific weight, density ana soec grav.ty one litre 0;' :1 liquid which
weighs 7N.Sol.Given:
Volume, 1:( ,= 1 litre = ---:- rrr i
JOOO \1 litre = 10\)0 nr3 or l litre = 1000 cm' )
(i) Specific weight (w)
=7N.
= Weight ;:: 7 N = '7( Otl "11 3VI
I 1 \ ,'".101.o ume { )"i 1000 m
Ans.
Weight
(iii) Specific gravity
w 7000 , .' :\= g = 9.81- kg/m = 7U,5Iig!m·.. Ans.
_ Density .9.1' liquid _ 71}.5_- Density ofwater - WOO {. Density of water = 1000 kg/nr'}
= 0.7135. Ans.
(ii) Density (p)
Problem 1.2.Calculate the density, specific .veight ami weiglu oj or ? litre of petrol oJspecific gravity
= 0.7.Sol.Given:Volume ::"1 litre
4 ,,' .• 1 10\ 10 , •:: 1 x lUliU em = =r:«: nr ~ O.i)~)i In
10')
Sp. gravity,(,) Density (p)Using equation (1.1 A),Density (p)(ii) Specific weight (w)Using equation (1.1),
S = 0.7
w=pxg= 700 x 9.81 N/m" :::6867 Nmr". Ans.
(iii) Weight (W)
We know that specific weight_ Weigl'!.- Voiun'.,:
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7 '7
1.3.05.Types of Fluids. The fluids may be classified into the following five types:1. Ideal fluid, 2. Real fluid,3. Newtonian fluid,5. Ideal plastic fluid.
1.Ideal Fluid. A fluid, which is incompressible and is having no viscosity, is known as in ideal fluid.Ideal fluid is only an imaginary fluid as all the fluids, which exist, have scme viscosity.
2. Real Fluid. A fluid, which possesses viscosity, isknown as real fluid. All the fluids, in actual practice, are realfluids.
3. Newtonian Fluid. A real fluid, in which the suearstress is directly, proportional to the rate of shear strait, (orvelocity gradient); is known as a Newtonian fluid.
4. Non-Newtonian Fluid. A real fluid, in which theshear stress is not proportional to the rate of shear strain (orvelocity gradient), is known as a Non-Newtonian fluid.
S. Ideal Plastic Fluid. A fluid, in which shear stressis more than the yield value and shear stress is proportionalto the rate of shear strain (or velocity gradient), is known asideal plastic fluid.
4. Non-Newtonian fluid, and
t{'deo' Fluid
- Velocity gradient (~~ jFig. J .2. Typesof fluids
Problem 1.3.If the velocity distribution over a plate is given by II = ~Y - Y in which II u the velocityin metre per second at a distance y metre above the plate, determine the shear stress at y = 0 and y = 0.15 m.Take dynamic viscosity of fluid as 8.63 poises.
Sol. Given: 2 7U=3"Y-y-
au 2-.:-= -- 2yay 3
(du ) ( du ) 2 2 ,. \ 2 'L-d or d ::"3-- :J):: 3= ,)\)61Y aty-O y y=O
( dU)'Also -dy alyEO.IS (
du ) 2 ') - . 7or d :::3- L. A ..t::l :: .66 - .30 :: 0.367y y - 0.15 .
Value of u = 8.63 poise = 8~~3SI units = 0.863 N s!m2
Now shear stress is given by equation (1.2) as T :: ~ ~; .
(i) Shear stress at y = 0 is given by
~o = !-t ( ~u ) :: 0.863 x 0.667 = 0.5756 N/m2. ADS.y y = ()
(it) Shear stress aty = 0.15 m is given by
(T)y"O.1.5 = !-t ( ~u ) = n 863 x 0.367 = 0.3167 N/m2•Y y=0.15
ADS.
I 1••• Problem 11.,:'-A plate, .0.025 mm distant from a fixed plate, moves at 60 cmls and requires a force• of 2 N per unit area i.e. 2 N/m2 to maintain this speed. Determine the fluid viscosity between the plates.
Sol. Given:• Distance between plates, dv = .025 mm
This is the value of shear stress i.e. 't• Let the fluid viscosity between the plates is u.
• Using the equation (1.2), we have ~ = f.l ~~.
• where du = Change of velocity = u - 0 = U = 0.60 m/sdy = Change of distance :', J25)< 10-3 m
•••
•••••
Velocity of upper plate,
= .025 X 10-3 mII = 60 cm/s = 0.6 m/s
NF=2.02"'m
1 FLa - _.dy=.025mm--=:'-:::-:':_ U=60cm/s
Force on upper plate,T -r=:
FIXED PLATE
Fig. 1.3
F. r 0 )'-1
't = orce per urnt area ::::.'. --;;m""
20- -~~. .. ~ .025x 10-3, 3~ = 2.0 x .025 x 10- = 8.33 x 10-5N s
0.60 m2
;.:8.33 x 10-5 x 10 poise = 8.33 x.10-4 poise.
• Problem 1.5.A flat plate of area 1.5 x 106 mm' is pulled with a speed of 0.4 mls relative to another• plate located at a distance of 0.15 mm from it. Find the force and power required to maintain this speed, if the
fluid separating them is having viscosity as 1poise.
• Sol. Given:• Area of the plate, A = 1.5 X 106mnr' = 1.5 m2
Speed of plate relative to another plate, du = 0.4 m/s• Distance between the plates, dy = 0.15 mm = 0.15 x 10-3 m
•••••••••••
Ans,
Viscosityi' 1 NsI!= j pOIse'" ._--- :to m2
du 0.4 Nr ;:, fA. - = ._--, _. ' = 266 66 -,. dy 10 .15 X 10-3 • m2Using equation (1.2), we have
(z):. Shear force, F = 't X area = 266.66 x 1.5 = 400 N. Ans,(iz) Power" required to move the plate at the speed 0.4 m/sec
= F x u = 4('f) x 0.4 = 160 W. Ans,Problem 1.6. Determine the intensity of shear :.1{an oil having viscosity = 1poise. The oil is used for
lubricating the elearance between a shaft of diameter to em and its Journal bearing. The clearance is 1.5mmand the shaft rotates at 150 r.p.m.
Sol. Given: 1. 1 N s
II = pOIse = -,-r' to m2
*Power :: F x u N mls = F x u W (": N mls = Watt)
___ ~:._~ ~ _~ 4: ~ -: -- - - - ij,
7 '5
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Dia. of shaft, D = 10 em = 0.1 mDistance between shaft and journal bearing,
dy = 1.5,mm :;:1.5 x 10-3 mSpeed of shaft, N = 150 r.p.m.Tangential speed of shaft is given by
u :::1fDN :::~..X 0.1 x 150 = 0.785 e/s60 60
Using equation (1.2), du't=!!-d "
Ywhere du = change of velocity between shaft and bearing ,:;;u ;) e I:
1 (1,'785 . 2= _. X _._ .._.--- :: 52..33N/m Ans.10 1.5 >< 10-3
Problem 1.7•Caleulate the dynamic viscosity of an oil, which is used for lubrication between a squareplate of size 0.8 m x 0.8 m and an inclined plane with angle of inclination 30e as shown in Fig. 1.4.The weightof the square plate is 300 N and it slides down the inclined plane with a uniform velocity 0.( 0.3 mls. Thethickness of oil film is 1.5mm.
Sol. Given:Area of plate,Angle of plane,Weight of plate,
A = 0.8 x 0.8 = 0.64 m2
e = 30°W=300N
Velocity of plate, u = 0.3 m/sThickness of oil film, t = dy
= 1.5 mm = 1.5 x 1(,-3mLet the viscosity of fluid between plate and incl.ned plane is u.
Component of weight W, along the plane
= W cos 60° :::300 cos riO" =: 15i N
Fig.lA
Thus the shear force, F, on the 'bottom surface of the plateF· 150 !and shear stress, 't = -- = -- N/n:!'
Area 0.64 -
Now using equation (1.2), we havedu
't=",,-dy
where du = Change of velocity = u - 0= u = 03 m/sdy = t = 1.5 x 10-3m
;:; 150 N
150 0.30.64 = !! 1.5 x 10-3
150x1.5x1{1-3 l'"7N ~, 1'1017 .!!= 0.64 x 0.3-'- :;;:;1. .I 1 s/m" = 1. ,x 1 = 1 . poise, Ans.
Problem 1.S. Two horizontal plates are placed 1.25 em apart, the space between them being filledwith oil of viscosity 14poises. Calculate the shear stress in oil if upper plate is moved with a velocity of2.5 mls. (AMIE, 1972)
.------------ u.~ _
9
.. Sol. Given:
•..••where•
Distance between plates,
Viscosity,
Velocity of upper plate,
dy= 1.25 em= 0,0"125m14 2
II = 14 poise > _, N s/mr ~ 10
u :: 2.5 m/sec.
Shear stress is given by equation (1.2) as. 1: = !-.I. ~~
du = Change of velocity between plates= II - 0 :::U :: 2.5 m/sec.
dy = 0.0125 m.
••14 2.5 2
't ::: 10 x .0125 = 280 N/m. Ans.
Problem 1.9. The space between two square flat parallel plates is filled with oil. Each side o/the plate.60 em. The thickn,ss of the oil film is 12.5 mm. The upper plate, which moves at 2.5 metre per sec requiresa force of 98.1 N to maintain the speed. Determine:
••(i) the dynamic viscosity of the oil in pOIse, and(ii) the kinematic viscosity o] the oil in stokes if the specific gravity of the oil is 0.95.
(AMIE, Winter 1977)
••••••
Sol. Given ",Each side of ,1 square plate
:. Area,Thickness of oil film,Velocity of upper plate,
=' 60 ern= 0,60 mA 0= 0.6 x 0.6::: (,.36 m2
dy = 12.5 mm= 12.5 x 10-3 mu :: 2.5 m/sec
. . Change of velocity betw~en plates, du :::2.5 n/secForce required on upper plate, F =: 98.1 N
:. Shear stress,Force F 98.1 N,---_ ..-
- Area - .A - 0.36 m2
• (i) Let !-.I. =Dynamic viscosity of oil
••••••••••••
Using equation (1.2),du 98.1 2.5
't = u dy or 0.36:::!l x 12.5 X 10-3
U .=.. 98.1 x 12~~ 10-3 ::: 1.3635 N s0.36 2.5 m2
; 3635 ' 1<)::: 13.635 poise. Ans.
(.: 1:,s = 10poise )
(il) Sp. gr. of oil,S::: 0.95Let v = kinematic viscosity of oilUsing equation (1.1 A),Mass density of oil, P ::::5 x 1000 :c, 0.95 x 1000 :::950 kg/m3
1.3635 ('~ ~.)\ m'"
v ;:.'~--'-"--95G
:= .001435 m2jsec = .001435 x 104 cm2/sUsing the relation, V := ~ , we getp
:::14.35 stokes. Ans.
+
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Problem 1.10.Find the kinematic viscosity of an oil having density 981 kg/m3. The shear stress at apoint in oil is 0.2452 N/m2 and velocity gradient at that point is 0.2 per second.
Sol. Given:·Mass density,Shear stress,
Velocity gradient,
p = 981 kg/nr'"t = 0.2452 N/m2
dudy = 0.2 s
du"t = 11- or 0.2452 = 11x 0.2dy
0.2452 211= 0.200 = 1.226 N s/m
Using the equation (1.2),
Kinematic viscosity v is given by
JJ. 1.226 ... 2v =.c. = -- = .125 x 1O-~m /secp 981= 0.125 x 10-2 x 104 cm2/s = 0.125 x 102 cm2/s
= 12.5 cm2/s = U.S stoke. Ans.
Problem 1.11. Determine the specific gravity of a fluid having viscosity 0.05 poise and kinematicviscosity 0.035 stokes. .
Sol. Given:
Viscosity, '" = 0.05 poise = °i~5N s/m2
Kinematic viscosity, v = 0.035 stokes
= 0.035 cm2/s= 0.035 x 10--<1m2/s
Using the relation v = H-., we get 0.035 x 10--4 =:: 0.05 x 1.P 10 P
0.05 1 3P = -x _ .._._-= 1428.5 kg/rn10 .035 X 10-4
· . Sp. gr. of liquid = Dens~ty 0f.~quid = 1428.5 =. 1.4285::::1.43.Density of water 1000 Ani,
Problem 1.12.Determine the viscosity of a liquid having kinematic viscosity 6 stokes and specificgravity 1.9..
Sol. Given:Kinematic viscositySp. gr. of liquidLet the viscosity of liquid
Now sp, gr. of a liquid
· . Density of liquid
v = 6 stokes = 6 cm2/s = 6 x 10-4 m2,!s= 1.9=11_ Density of the liquid- Density of water
1 9 = Density of hquid. 1000
= 1000 x 1.9....:1900 ~m
or
I........--------------------~------_,Q••••••••
Using the relation v :; ~ c we get. p
6 '( 10-4 ""_~, 1900
!!= 6 x 10-4)1 ;900 = 1.14N s/m2:::L 14 x 10""'11.40 poise. Ans,
Problem 1.13. The velocity distribution for flow over a flat plate is given by u = ~y-I in which u is
•• the velocity in metre per second ai a distance ymetre above the plate. Determine the shear stress at y = 0.15 m.• Take dynamic viscosity of fluid lIS 8.6 poise.
3 ~.u=-y-V'
4
~- == ~-- 2}d,\ 4
Sol. Given:
••••••••••••••••••••••
Viscosity,
-- z;• z;__-2 < 0.15 = 0.75 -0.30 = 0.454
~i := 8.5 poise == 8.5 N~10 m"
(-: Hl poise v I ~~ 1Aty = [) 15,
T :: U ~u == ~!~:~x 0.45 ~ = 0.3825 N . Ans.. dy tlJ m2 m2
Problexu 1.14. The dynamic VIscosity of an oil, used for lubrication between a shaft and sleeve is6 poise. The shaft is of diameter 1.4m and rotates at j90 r.p.m. Calculate the power lost in the bearing for asleeve length 0/90 mm. The thickness oi the oil film L"1.5mm.
Using equation (1.2),
Dia. of shaft,
Speed of shaft,
Sleeve length,Thickness of oil
~l :: 6 poise
_£.Ns_06N~- ? -10 m . m2
D = 0.4 m
N = 190 r.p.rn.
L :: 90 mrn -r-. 90 x W-l TIl
1.!5mm
lc{« « ,,.1, '1"",jSol. Given:Viscosity ~:~:;> >;
90mmSLEEVE
Fig. 1.5
Using the relation
nlJN n x 0.4 x 190.: :: --66-- :;"- 60 = 3.98 m/s
dur=!-l-dy
Tangential velocitv ot suaft,
where du = Change of velocit: :: I, - 0 = II = 3.98 m/sdy = Change of distance == ( := 1.5 l( 10-3 In
1. ,= 10 x 3.98 = 1592 N/m21.', x 10-3
This is shear stress on shaftShear force on the shaft, F = Shear stress x Area
= 1592 x ill xL = 1592 x Jt x .4 x 90 x 10-3 = 180.05 N
-d
Problem 1.17.A 15 em diameter vertical cylinder rotates concentrically 'inside another cylinder ofdiameter 15.10 em.Both cylinders are 25 emhigh. The space between the cylinders isfilled wuh a liquid whoseviscosity is unknown. If a torque of 12.0Nm is required to rotate the inner cylinder at 100 r.p.m., determinethe viscosity of the fluid. (AMIE, Winter 1979)
Sol. Given:Diameter of cylinder = 15 em = I, ..5 mDia. of outer cylinder = 15.10 em :: 0.151 mLength of cylinders, L = 25 em = 0..25mTorque, T= 12.0 NmSpeed', N = 100 r.p.m.Let the viscosity = fA.
T ial I' f lind rrDN n: x 0.15 x 100 07854 m/angenti ve ocity 0 eyer, U = """"6() = 60 =. S
Surface area of cylinder,
Now using relation
A = rrD x L :~:rtX 0.15 x 0.25 :: .1178 m2
du1:=fA.- dy
where du = u -0 = U = .7854m/s
dy _ 0.151 - 0.150 _ 0005... 2 m e . m
_ fA. x .78541: - .0005
!-!)< 785'~F = Shear stress :< Area ::::-~Oi)oj-'-x ,1J 78
T=FxD2
12 0 = fA. x .7854 1178 .15. .0005 x. x 2
12.0 ......0005 x 2 . 2fA.= 0.7854 x .1178 x .15 = 0.864 N s/m
= 0.864 x 10 = 8.64 poise. Ans
Problem 1.18. Two large plane surfaces are 2.4 cm apart. The space between the surfaces is filledwith glycerine. What force is required to drag a very thin plate of surface area 0.5 square metre between thetwo large plane surfaces at a speed of 0.6 mls, if:
(i) the thin plate is in the middle of the two plane surfaces, and(il) the thin plate is at a distance of 0.8 em from one of the plane surfaces? Take the dynamic viscosity
of glycerine = 8.10 x to! N slm2•
. . Shear force,
.. Torque,
Sol. Given:Distance between two large surfacesArea of thin plate,
Velocity of thin plate,
Viscosity of glycerine,
,: 2A em,
A :.:0.5 m2
u '"0.6 m/s!-l ::- 8.10 X 10-1 N 81m2
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Case I. When the thin plate is in the middle of the two plane surfaces [Refer to Fig. 1.7 (a)]Let F1 = Shear force on the upper side of the thin plate
Then
F1 = Shear force on the lower side of the thin plateF = Total force required to drag the plate
F =F1 +F;
'U' 'I""((""T(({ww««((W «{«
1.2 em
2.4 em_.t---- __.Ft 1.2 em
;;;/"",,),1"";»1") ,;""",.,Fig. 1.7 (a)
• where du = Relative velocity between thin plate and upper large plane surface
• = 0.6m/sec
The shear stress ('t1) on the upper side of the thin plate is
• given by equation,
•••••••••••••••••••••
dy = Distance between thin plate and upper large plane surface= 1.2 cm = 0.012 m (plate is a thin one and hence thickness of plate is neglected)
(06 \
.• 't1:: 8.10 x 10-1 x .0'12,1 = 40.5 N/m2
F1= Shear stress x Area= 't1 x A ::.40.5 x 0.5 :::::20.25 N
Similarly shear stress ('t2'\ on the lower side of the thin plate is given by
(' du \ 8' 10 '. _.1 (0.6) 40 5 N/ 21: = ~ . -- .. I.':' X 1\) • X - = . m'2 \ dy /, . 0.012
'\ . ~
Now shea I force,
. . ShuH force,. , Tot.al force,
F2 = t2 ): A ::;;405 x 0.5 7~ 20.25 NF -;Fl + F2 == 20.25 + 2U,25 = 40.5 N. Ans•
Case H. When the thin plate is at a distance of 0.8 em from one of the plane surfaces [Refer to
Fig. 1.7 (b)].Let the thin plate is a dis.ance O.S cm from the lower plane surface.Then distance of the plate from the upper plane surface
= 2.4 -lLS = 1.6 em = .016 m(Neglecting thickness of the plate)
The shear force on the upper side of the thin plate,F1 = Shear stress x Area = 't1 XA
= I'(~~),xA = 8.10x 10-' x (000~6)x 0.5
= 15.18 NThe shear force on the lower side of the thin plate.
1;;' {(du\ Al'~ = 'r~,», /~ cc fA. \ dy r
\ i.""
"" '1''''' '" "'(W ' , """" '"
1.6 em2.4 em 1 .. F
"ill",,: ))~i~~~I)) ~, Ii IiFig. 1.7 (b)
-1 I 0.6 ):; 8.10 x 10 x t.0:8/100 x 0.5 = 30.36 N
., Total force required = P1 + P2 = 15.18 + JU.36 = 45.54 N. Ans.
•••••••••••••
Problem 1.19.A vertical gap 2.2 em wide ofi'lfinite extent contains a fluid of viscosity 2.0N s/m2 andspecific gravity 0.9. A metallic plate 1.2 m x 1.2 m x 0.2 em is to be lifted up with a constant velocity of0.15 mlsec, through thegap. If the plate is in the middle of the gap. find the force required. The weight of theplate is 40N.
Sol. Given:Width of gap = 2.2 cm, viscosity: 1.1. = 2.0 N s/m2Sq. gr. of fluid = 0.9. . Weight density of fluid
= 0.9 x 1000 = 900 kgf/m3 = 900 x 9.81 N/m3
C. 1kgf= 9.81 N)Volume of plate = 1.2 m x 1.2 m x 0.2 em
3 ~= 1.2 x 1.2 x .002 m = .00288 nr= 0.2 em= 0.15 m/sec=40N.
Hg.1.8
When plate is in the middle of the gap, the distance of the plate from vertical surface, of the gap
= (Width of gap _.~ickneSS ofplate_l ' (2.2; 0.2) = I em = .01m.
Now the shear force on the left side of the metallic plate.Fl = Shear stress x Area
(du ) Ar 2 f 0.15 \ N·= '" dy 1x ea ::"~'.0 x \ ill )< 1.2 x 1.2 .
= 43.2 N.
-------- -- . . - . -------------------------------
Thickness of plateVelocity of plateWeight of plate
Similarly, the shear force on the right side of the metallic plate,F2 = Shear stress x Area
= 2.0 x ( :~~ ) x 1.2 x 1.2 = 43.2 N
.. Total shear force = Fl + F2 = 43.2 + 43.2 = 86.4 N.Inthis case the weight of plate (which is acting vertically downward) and upward thrust is also to be
taken into account.The upward thrust = Weight of fluid displaced
= (Weight density of fluid) x Volume of i]uid displaced= 9.81 x 900 x .00288 N
(: Volume of fluid displaced = Volume of plate = .00288)
= 25.43 N.The net force acting in the downward direction due to weight of the plate and upward thrust
= Weight of plate _-Upward thrust= 40 - 25.43 = 14.57 N
. • Total force required to lift the plate up
;::Total shear force + 14.57 = 86.4 + 14..57 = 100.97 N. Ans,
•1IIiiiii --_· ··..__·__ · · - .-----.------
•••• (ii) For Adiabatic Process. Using equation (1.7) for adiabatic process
I !'I(II'. • "t . ;-',~7 "J \'}I J f, ,fI"j! !
, ': ' " f;. "j .~. ..'. ~ r ~' ,"
Pk = Constant or p ''fik = Constantp
Differentiating, we get pd0;ti~ + ';fk(dp) =.0P x k x V'~-1# + Vkdp = 0
pIaN + Vdp = 0 [Cancelling 'fjfk-l to both sides]
kd \.I wd pk __ V dpp v=-vp or - d'rI
•••••••••••••••••••••••••••
or
or
or
Hence from equation (LtD), we haveK = pk ...(1.13)
where K = Bulk modulus and k:: Ratio of specific heats.
Problem 1.23. Determine the bulk modulus of elasticity of a liquid, if the pressure of the liquidisincreased from 70 Nlcm2 to 130 Nlcm". The volume of the liquid decreases by 0.15 per cent.
Sol. Given:Initial pressureFinal pressure.', dp = Increase in pressureDecrease in volume
= 70 N/cm2
= 130 N/cm2
= 130 - 70 = 60 N/cm2
= 0.15%d'V 0.15
--q-= + 100
Bulk modulus, K is given by equation (1.10) as
K = .se..; §ON/cm2 = 60 x 100 = 4 104Nt 2# .15 .15 x cm •- -V 100
Problem 1.24. What is the bulk modulus of elasticity of a liquid which is compressed in a cylinder froma volume of 0.0125 m3 at 80 Nlcm2 pressure to a volume 0/0.0124 m3 at 150 Nlcm2 pressure?
Sol. Given:
Initial volume,Final volume
. . Decrease in volume,
Initial pressureFinal pressure
:. Increase in pressure,
Ans.
't/ = 0.0125 m3
= 0.0124 m3
dV = .0125 - .0124 = .0001 m3
dV .0001v: .0125= 80 N/cm2
= 150 N/cm2
dp = (150 - 80) = 70 N/cm2
Bulk modulus is given by equation (1,10) as
K = ...!!E___::: .._1Q_ = 70 x 125 N/cm2'dV ,0001- ..--rt .0125
= 8.7S x Hr N/em2• Ans,
•••• Problem 1.27. The pressure outside the droplet of water of diameter 0.04 mm is 10.32 N/cm2• (atmospheric pressure). Calculate the pressure within the droplet if surface tensiun is given as 0.0725N/11!-of
••The pressure inside the droplet, inexcess of outside pressure is given by equation (1.14)
= 40 = 4 x O.0~25= 7250 N/m2p d 04' 1'-3• X .L\I
7250N 2= 4 2 = 0.725N/cm10 cm
=p + Pressure outside the droplet= 0.725 + 10.32= 11.045N/cmt. Ans.
1.6.4. Capillarity. Capillarity is defined as a phenomenon of nse or fall of a liquid surface in a smalltube relative to the adjacent general level of liquid when the tube is held vertically in the liquid. The rise ofliquid surface is known as capillary rise while the fall of the liquid surface is known as capillary depression.It is expressed in terms of em or rom of liquid. Its value depends upon the specific weight of the liquid, diameterof the tube and surface tension of the liquid.
water.Sol. Given:
, Dia. of droplet, d = 0.04 rom = .04 x 10-3m= 10.32N/cm2= 10.32 x 104N/m2
a = 0.0725 N/m
Pressure outside the droplet••••••••••••••••••••••••••
Surface tension,
or '
Pressure insidethe droplet
Expression for Capillary Rise. Consider a glass tube of smalldiameter cd' opened at both ends and is inserted in a liquid, say water. Theliquid will rise inthe tube above the level of the liquid.
Let h= height of the liquid inthe tube. Under a state of equilibrium,the weight of liquid of-height h is balanced by the force at the surface ofthe liquid inthe tube. But the fOI'Ce at the surface of the liquid in the tube,is due to surface tension.
Let a = Surface tension of liquide =Angle of contact between liquid and glass tube.
The weight of liquid of height h inthe tube= (Area oftube x h) x p x g
=:tPxh><pxg
where p = Density of liquidVertical component of the surface tensile force
= (0 x Circumference) )( cos e= 0 x xd x ,XlS e
For equilibrium, equating (1.17) and (1.18), we get
1t tP x h x p x g=0 x xd x cos e4
orh = 0 x 3td x cos e = 4 a cos e
1t'_i2 . pxgxd'4a-xp)1g
Fig. 1.13. Capiliary rise.
...(1.17)
...(1.18)
...(1.19)
4ei cos 0It ::: ----------- --
~-;gri...(1.21)
•The value of 8 between water and clean glass tube is approximately equal to zero and hence cos 8 is
• to unity. Then rise of water is given by40It ;: .--- _._-
;_, F-,;': >: d...(1.20)
••••••
:::pg < It X ~d2, 4{ .. p = pgh} Eg.1.14
Equating the two, we get'fl" ,""
o x xd x cos \:.i ::: pgh ,>( ::' d'-s
Value of B fcrmercury and glass tube is 1_28°• Problem 1.28..Calculate the capillary rise in a gla:::srube of2.5 mm diameter when immersed verticallyin (a) water and (b) mercury. Take suriace tensi()n~0 = 0.0725 Nlm for water and (J = ().52 Nlm for mercury• contact with air. The specific grav,:tyfor mercury is given as 13.6 and angle of contact = 1300•
••••
Sol. Given:Dia. of tube,Surface tension, 0 for water
o for mercurySp. gr. of mercury
Density
c1 == 2.5 mm =, 2,5 >: 10-3m;:0' !}.(n25 Nim::= 0.5), N/m::::13.6:::13.6 x 1000 ~~g!m3.
• (a) Capillary rise for water (8:::: 0)
••••••••
Using equation (1.20), we gelIt ::: 40 :::_, 4 x 0.0725
P X g x d lOOO x 9.81 x 2.5 :x 10-3.: 0118 m ""Lj.8 COl. Ans.
(b) For mercuryAngle of cnnstant between mercury and glass tube. 8 = 130"
Using equation (1.21), we gttIt::: 40 cos 8 :::,_, 4 x 0.52 x cos 13(~_
P x g x d 13.6 x 1000 x 9.81 x 2.5 >'. 10-3
:::__.004 m:::--0.4 em. Ans.
The negative sign indicates the capillary depression .
••••••••••••••••••••••••••••••
Problem 1.29. Calculate the capillary effect in millimetres in a glass tube of 4 mm diameter, whenimmersed in (i) water, and (ii) mercury. The temperature of the liquid is 20"C and the values ~fthe surfacetension a/water and mercury at 20°C in contact with air are 0.073575 Nlm and 0.51 Nlm respectively, Theangle of contact for water is zero that for mercury i.30°. Take density ofwater at 200e as equal to 998kglm',
(U.P.S.C. Engg. Exam., 1974)
Sol.Given:Dia of tube, d = 4 mm = 4 x 10-3 mThe capillary effect (i.e., capillary rise or depression) is given by equation (1.20) as
h = 40 cos €_.pxgxa'
where 0 = surface tension in kgf/me = angle of contact, and p = density(i) Capillary effect for water
o = 0.07357S N/m, e = 0°p = 998 kg/m" at 20°CI _ 4 x 0.073575 x cos 0° _,., 51 10-3 - 7 51 .l - 998 9 81 4 10-3 - I. X m -. mm.x ,. x x
Ans.
(ii) CapUlary effect for mercury0= 0.51 Nzm, e = 130° andp = sp. gr. x 1000 = 13,6 x 1000 ::: 13600 kg/m2
It = 4 x 0.5.1 x cos 130°13600 x 9,81 x 4 x 10-3
= - 2.46 X 10-3 m = _. 2.46 mm,The negative sign indicates the capillary depression.
ADS.
Problem 1.30. The capillary rise in the glass tube is not to exceei) 0.2 mm of wa!cr. Determine itsminimum size, given that surface tension for water in contact with air = ().0725 N/m.
Sol. Given:
.'
Capillary rise,Surface tension,Let dia. of tubeThe angle e for waterDensity (p) for water
h = 0.2 mm = 0.2 x 103m0= 0.0725 N/m=d=0= 1000 kl!)m'
Using equation (1.20), we get
h = 40 or 0,2 x W-3 = 4 x 0.0725P x g x d 1000 x 9.81 x d
d = 4 x 0.0725 r = 0.148 m :::14.8 em.1000 x 9.81 x .2 x 10- .
ADS.
Thus minimum diameter of the tube should be 14,8 em
.' Problem 1.31.Find out the minimum size of glass tube that can he used to measure water level if thecapillary rise in the tube is to be restricted to 2.1nm. Consider su.rface tension of water in contact with air as0.073575 Nlm. (Converted to SI Units, A.M.LE., Summer 1985)
_._--_ .._----------
•
••••••••
Sol. Given:Capillary rise,Surface tension,Let dia. of tubeThe angle e for water
h = 2.0 rom = 2.0 x 10-3ma = 0.073575 N/m=d=0
The density for water,Using equation (1.20), we get 4a 4 x 0.073575
h :: 'p X g x d or 2.0 x 10-3 = -1.;...OOO__.;.x..:..;9:....;..~8~1x-d
d._ 4 x 0.073575 - 0015 - III!. ._ . __. m - .., em.
1000 x 9.81 x 2 x 10-.3ADs.
Thus minim:um diameter of the tube should be 1.5 em.problem J32. An oil of viscosity 5 poise is used for lubrication betWeen a shaft and sleeve. The
-diameter of the shaft is 0.5 m and it rotates at 200 r.p.m. Calculate the power lost in oil for a sleeve length of.100 min. The thickness of oil fibn is 1.0 min. [DelhiUnivetSity.DeCember. 1992 (NS)I
Sol. Given:•••••••
Viscosity, ~ = 5 poise_ 2_ - 0 5 N "/m2- 10 -. .1..
D=0.5mN = 200 r.p.m.L = 100 rom = 100 x 10-3m = 0.1 mt = 1.0 rom = 1x 10-3mu = 1tDN = 1t x 0.5 x 200 = 5.235 mls
60 60
du't=IA--dy
Dia. of shaft,Speed of shaft,Sleeve length,.Thickness of oil film,
Tangential velocity of shaft,
Using the relation,
• where, du =Change of velocity = u - 0 = u = 5.235 tal«• dy =Change of distance = t = 1 x 10-3 m
., 1: = 0.5 x 5.2;5 = 2617.5 N/m2• lx1~
•••••••••
This is the shear stress on.the shaft:. Shear force on the shaft, F = Shear stress x Area
Torque on the shaft,
= 2617.5 x :ltD x L (. : Area = '1fl) xL)= 2617.5 x x x 0.5 x 0.1 = 410.95 N
T = Force x ~~= 410.95 x 0.5 = 102.74 Nm:: 2
= T x (I)Watts = T x 21tN W6021t x 200= 102.74 x 60 = 21.50W = 2.15 kW. ADs.
:. Power* lost
.power in case of S.l. Unit = T x (J.) or ~ Watts or ;:,: kW. The angular velocity (J.) = ~ .
•••Sol. Given:
• Dia.ofram,
• :. Area of ram,
D = 20 em = 0.2 mJt_.:it~. 2
A. = - D'" '" -- (.2)" :::0.0314 m4 4
d = 3 em= 0.03 m
a = Jt (.03i = 7.068 x 10-4m24
W = 30 kN = 30 x 1000 N = 30000 N.
Dia. of plunger.
:'. Area of plunger,
Weight lifted,See Fig. 2.3.
P. . dId d 1 Force Fressure tnten~lty eve ope tie to p unger = -Ar :::- .ea a
By Pascal's LaW, this pressur~ is transmitted equally in all directions
•••••••••••
Hence pressure transmitted at the ram = Fa
•••••••••••••M-C:
:. Force acting on ram
But force acting on ram
= Pressure intensity x Area of ram
= F xA = F x .0314 Na 7.068 x 10-4
=Weight lifted ~.30000 N
~OOOO = F x 0.314'7.068 x 10-4
F~ 30000 x 7.0~§ X 10-4 = 675 ~ N
.03Vi ••..Ans •
Height of liquid column, Z = 0.3 m.The pressure at any point in. a liquid is given by equation (2.5) as
p= pgZp = 1000 kglm3p = pgZ = 1000 x 9.81 x 0.3 = 2943 N/m2
2943 N '1 N' l= 104 /cm~ = 0.2943 /em. Ans.
(a) For W:iter,
(b) For air of sp. gr. 0.8,
(c) For mercury, sp. gr.
:. Density,
Po = Sp. gr. x Density of water= 0.8 x P = 0.8 x 1000 = 800 kg/m3
p = Po x g x Z",,800 x 9.81 x 0.3 = 2354.4 N2 = ~5~.4 N 2 .m 10 em
.~.
:= 0.2354 -~"i' Ans.ChI
:::13.6Ps= 13.6 x 1COO= 13600 kg/m3
.~-
•••••••••••••••••••••••••••••••••
p= Ps X gx Z
. 5 N= 13600 x 9.Hl x 0.3 z: 4002 ---::;m"'
40025 N= --4- = 4.002 --2' Ans.10 em
Problem 2.4. Thepressure intensity at apoint in a fluid is given 3.924 N/cm2. Find the correspondingheight offluidwhen the fluid is: (a) water, and (b) oil of sp. gr. 0.9.
Sol. Given:
Pressure intensity, p = 3.924 N 2 = 3.924 X 104 N2 .cm m
The corresponding height, Z, of the fluid is given by equation (2.6) as
Z'= __L_pxg
(a) For water, P = 1000 kg/m''
P 3.924 x 104Z = -- = 10( 't 981 = 4 m of water. Ans.P x g _, x . ~
= 0.9Po = 0.9 x 1000 :~900 kg/m'
Z - -p-- 3.924 x 104 - 444 f oil A- - -. m 0 01. ns.Po x g 900 x 9.81
Problem 2oS.An oil of sp, gr. 0.9 is contained in a vessel. At apoint the height of oil is :tom. Find thecorresponding height of water at the point.
Sol. Given:
(b) For oil, sp. gr.
Sp. gr. of oil,Height of oil,Density of oil,
Intensity of pressure,
So = 0.9Zo=40mPo= 1000 x So = 1000 x 0.9 = 900 kg/nr'
Np = Po x g x Zo :: 900 x 9.81 x 40 -;-m"
D. . Corresponding height of water = D . "f
ensrty 0 water x g900 x 9.81 x 40= 1000 x 9.8J = 0,9 x 40 =:36 [II of water.
Problem 2.6.An open tank contains water upto a depth of2 m and above it an oil of SlJ. gr. 0.9 for adepth of1 m. Find thepressure intensity (i) at the interface of the two liquids, and (ii) at the botto n of the tank:
Sol. Given:
Ans,
Height of water,Height of oil,Sp. gr. of oil,Density of water,Density of oil,
Zl=2mZ2= 1mSo= 0.9PI = 1000 kg/n. 1
P2 = 0.9 x 1000 := 900 kg/m"
•••••••••••••••••••••••••••••••••
2.6.2. If-tube Manometer. It consists of glass tube bent in U-shape, one end of which it; connected toa point at which pressure is to be measured and other end remains open to the atmosphere as shown inFig. 2.9. The tube generally contains mercury 01 any other liquid whose specific gravity is greater than the. specific gravity of the liquid whose pressure is to be measured.
fA
(a) For Gauge pressure. tb) For Vacuum pressure.Fig. 2.9. U-tube Manometer
(a) For Gauge Pressure. LetB is the point at which pressure is to be measured, whose value isp. Thedatum line is A -A.
Let hI = Height oflight liquid above the datum line~ = Height of heavy liquid above the datum lineSI = Sp. gr. of light liquidPI = Density of light liquid = 1000 x S,S2 = Sp, gr. of heavy liquidP2 = Density of heavy liquid = 1000 x S2
As the pressure is the same for the horizontal surface. Hence pressure above the horizontal datum lineA-A in the left column and in the right column of If-tube manometer should be _,arne.
Pressure aboveA-A in the left column = P + PI x g x hIPressure aboveA-A in the right column = 1='2 x g x h2Hence equating the two pressures p + Pighi = P2gh2. P = ({J~~ - PI X g X hI)'
(b) For Vacuum Pressure. For measuring vacuum pressure, the level of the heavy liquid in themanometer will be as shown in Fig. 2.9 (b). Then
Pressure above A-A in the left column = P~h2 + r-llghj + PPressure head in the right column above A-A ;;.;C. . P~~ .+ PIghl + P = 0
...(2.8). Problem 2.9. The right limb of a simple U-tube manometer containing mercury is open to the
atmosphere while the left limb 1..11 connected to apipe in which a fluid of sp. gr 0.9 is flowing. The centre ofthe pipe is 12 em below the level of mercury in the right limb. Find the pressure of fluid in thepipe if thedifference of mercury level in the two limbs is20 em,
Sol. Given:Sp. gr. of fluid, SI = 0.9.'. Density of fluid, PI = SI x 1000 = 0.9 x 1000 = 900 kg/m"
•••• Sp. gr. of mercury, S2 = 13.6
••
: .• Density of mercury,Difference of mercury levelHeight of fluid fromA-A,Let p = Pressure of fluid in pipe
P2 :: 13.6 x 1000 kg/nr'b: = 20 em= 0.2 mhI = 20 - 12 = 8 em = 0.08 m
f T•20em
Equating the pressure above A-A, we get
• p + Plgh1 = p~~.p + 900 x 9.81 x 0.08 = 13.6 x 1000 x () 81 x :p = 13.6 x lOOe x 9,81)' 2. -·900 " Cl,S1x 0.08
.1A
•• Problem' 2.10.A simple Ustu: f manometer containutg mercury is connected to apipe in which a fluidfJ[ sp. gr. 0.8 and having vacuum press ure is [lowing. The other end of the manometer is open to atmosphere."ind the vacuum pr essure in pipe, if tne difference of mercury level in the two limbs is 40 em and the height• fluid in the left from the centre of pipe is 15 em below.
Sol. Given:Sp. gr. of fluid,Sp. gr. of mercury,Density of fluid,
= 26683 -7ut:;: 25977 N/m2::: 2.5:'17 N/cOl2• ADS.Fig. 2.10
• S1 ::0.8,S2 -= 13.6PI = 80u••
A
Density of mercury, P2::: 13.6 x 1000• Difference I)::mercury level, h2 = 40 em = 0.4 m. Height of liquid inleft limb, hl = 15 em = 0.15 m. Let the pressure in pipe =p. Equating pressure
.bove datum line A··A, we getp~/lQ, + P1g::I! +p :::0• .P = _ (P2i:ih~ +- i-'19it11
:;;._ l:; (; /- 1000 x 9,8 i :x 0.4 + 800 x 9.81 x 0.15]c, _ .53Jf)o 4, 177.2 ::.- 54543.6 N/m2 = - 5.454 N/cm2•
Fig. 2.11
• ADS.
• Problem2.H.A Us'Iube manometer 1..\ used to measure the pressure of water in apipe line, which is.n excess of atmospheric pressure. The right limb or the manometer contains mercury and is open toatmosphere. The contact between water and mercury is in the left limb. Determine the pressure of water in the
-main line, if the difference in level of mercury in the limbs of U-tube is 10 em and the free surface of mercury.is in level with the centre of the pipe If the pressure of water in pipe line is reduced to 9810 N/m2, calculate
the new difference in the level of mercury. Sketch the arrangements in both cases. (A.M.LE., Winter 1989)
4t Sol. CHvenDifference of mercury :: 10 em = 0.1 m•• The arrangement is shown m Fig. 2.11 (a)Let PA = pressure of water in pipe line (i.e., at,point A)The points Band C lie on the same horizontal line. Hence pressure atB should be equal to pressure at
• C. But pressure aLB.•• = pressure at A + pressure due to 10 em (or 0.1 m) of water
=PA + P >: g x n
• where p = 1000 kg/m3 and It = (" I m
•". .
.'-- -- ---~ -~-. .'t
••••••••
Pressure above X-X in the right limb = Pg x g x h + P2 x g xv +PI'where PB = Pressure atB.
Equating the two pressure, we have
PIg(h +x) +PA = Pg x g x h + P2RY +PBPA - PB = Pg x g x h -I- p~ - PIg(h + x)
= h x g(Pg - f'l~+ P~ - PIgx.. Difference of pressure atA and B = h x g(Pg ...PI) + p~y - PIgx
• [Fig. 2.18. (b).A and B are at the same level and contains the same liquid of density PI Then
Problem 2.16.A differential manometer is connected at the two points A and B of two pipes as shownin Fig. 2.19. The pipe A contains a liquid of sp. gr. = 1.5 while pipe B contains a liquid of sp. gr. = 0.9. The
• pressures at A and Bare 1kgflcm2 and 1.80 kgflcm' respectively. Find the difference in mercury level in thedifferential manometer.
Sol. Given:Sp. gr. of liquid atA, Sl = 1.5Sp. gr. of liquid atB, S2 = 0.9 .. P2 = 900Pressure atA,PA = 1 kgf/crrr' = 1 x 104 kgf/nr'
= 104 x 9.81 N/m2 ('.: 1 kgf 'c, 9.8 iN)
•••••••••••••••••••••••
Pressure above X-X in right limb = Pg x g x h + PI X g X X +PBPressure above X-X in left limb = PI x g x (h ,..:r) +PAEquating the two pressure
Pg x g x h + PIgx +PB = PI x g x (h + x) +PAPA - PB = Pg x g x h + P1gx- P1g(h + x)
= g x h(pg - Pl}]' ...(2.13)Problem·2.1S. A pipe contains an oil of sp. gr. 0.9. A differential manometer connected at the two
points A and B shows a difference in mercury level as 75 em Find the differenc e of pressure at tile two points.Sol.· Given:Sp. gr. of oil,Difference in mercury level,Sp. gr. of mercury,
S1 = 0.9h = 15 em ::::CI.lS m
Density, Pi = (j,9 x 1000 = 90( sg/nr'
Sg = 13.6 . . Density, Pg = j 3.6 x 1000 kg/m:'
orThe difference of pressure is given by equation (2.13) -
PA - PB = g x h(pg - PI)= 9.81 x 0.15 (13600 - 900) = 18688 N/m2. Ans.
r' :: 1500
Pressure atB,PB = 1.8 kgf/cm2
= 1.8 x 104 kgf/m2
= 1.8 x 104 x 9.81 N/m2 c· j kgf ». 9 81 N)Density of mercury = 13.6 x 1000 kg/m 'Taking X-X as datum line.
Pressure above X-X in the left limb= 13.6 x 1000 x 9.81 x h + 1500 x 9.81 x (2 + 3) +PA= 13.6 x 1000 x 9.81 x h + 7500 x 9.81 + 9.81 x 1()4
Fig. 2.19
...(2.12)
•••above X-X in the right limb
:::900 x 9.81 x (h + 2) +PB:::900 x 9.81 x (I: + 2) + 1.8 x 104 x 9.81
•E.quating the two pressure, we get•. 13.6 x 1000 x 9.81h + 7500 x 9.81 + 9.81 )0: 104• = 900 x 9.8i x (h + 2) + 1.8 X 104 x 9.81
Dividing by 1000 x 9.81, we get• 13.6h + 7.5 + 10::: (h + 2.0) x ,9 + 18• 13.6" + 17.5 = c.st. + 1.8 + 18 = 0.9h;' 19.8
• problem 2:17. A J;fferential manom;ter is connected at the tWOpoints A and B as shown inFig. 2.20.
air pressure is 9.81 N1cm' (abs), fmd tne absolute pressure atA
•
'J
Sol. Air pressure atB :::9.81 N/cro-4 r
PB:: 9.81 x 10 N/m':: 0.9 x 1000:: 900 kg/m3:: 13.6 x 1000 kg/m3
•...(13.6 _ 0.9)h:: 19.8 _-17.5' Of 12.-7h:::2.3
) ':Ih:: _~.:::: O.lSl m= 18.1 em. Ans .2.7
• Density of oil• Density of mercury
• Let the pressure litA is PATaking datum line atX-X
•••••••••
Pressure above X-X in the rigbt iirub:: 1000 x 9.81 x 0.6 +PB:::5886 + 98100::: 103986
Pressure above X-X in the left limb:::13.6 x 1000 x 9.81 x 0.1 + 900 x 9))1x 0.2 +PA
:: 13341.6 + 1765.8 +PA
Equating the iWO pressure bead103986:: 13341.6 -I- 1765.8 + p,\
PA::: 103986 _15107.4:: 88876.8
OIL OFSp.gr.=O.9
X
l'20emi
IDem
T
Fig. 2.20
PA' 88876.8 N/m2 = ~ = 8.887L.10000 cm2 cm2
., Absolute. pressure at A> 8.887 N/cm2, Ans.
2.7.2. Inverted U-tube Differential Manometer. It con.islS of an inverted U-rube, containing a light liquid. The two ends.0f the tube are connected to the poinlS whose difference" f PteSSureis to be measured. It is used for measuring difference of low
• pressures. Fig. 2.21 shows an inverted U-tube differentialmanometer connected to the two points A and B. Let the pressure at
• A is more than t~e pressure atB.• Let"l = Height ofliquid in left limb beloW the datum
line X-X• "2 :: Height of liquid in rigbt limb
•Fig. 2.21
•••• (iI) Pressure by Isothermal Law. Pressure at any height Z by i,othennallaw is given by equation• (2.18) as
•••••
P .-:Po e: gZ/RT
gZxh- ---,-
" 10 143,< 104 e Pc[
Po ' ]... - = RT and Po g = WoPo ,ZPo)< .:
= 10.143 X 104 e- --;;;--
::. 10.143 X 104 e(- 2500 x 1.208 x 9.81)/10.143 x 104
••:::101430 x e" .292 ::: 101430 x 1 :::75743 N/m2
. 1.339175743, .= -'-4- N/cm2:::7.574 N/cOl2. Ans.10
Problem :t23. The barometric pressure at sea level is 760 mm of mercury while that on a mountain• top is 735mm, if the density of air is assumed constant at 12 kglm", what is the elevation of the mountain top.
(A.M.I.E., Summer, 1988)••Sol. Given:Pressure* ill sea,
••••
Po ,.:760 mm ofHg
760. 2 2:::'1000 x 13.6 x 1000 x 9.81 Nzm :::101396 N/m
P :::735 mm of Hg
::: l~~O x 13.6 x 1000 x 9.81 :::98060 N/m2Density of au, f; :: 1.2 kg/nr'Let h = Height of the mountain from sea-level.
Pressure at mountain,•
We know that as the elevation above the sea-level increases, the atmospheric pressure decreases. Heretihed~nsity of air is given constant, hence the pressure at any height 'It' above the sea-level is given by theequanon,
• p :::Po " p x g x It.r it:::: Pl!_=1!.. ::: 101396 - 98060 :::283.33 m. Ans.oJ >,' .>.?' '~ :' x 9.81
• Problem 2.21.1.Calculate thepressure at a height oj -'-SOO m above sea level if the atmospheric pressure,/,I 10.143 N/cm2 and temperature is 15°C at the sea-level, assuming (i) air is incompressible, (ii) pressureTariation follows isothermal law, and (iii) pressure variation follows adiabatic law. Take the density of air at4je sea-level as equal to 1.285 kg/mi. Neglect variation ofg with altitude.
Sol. Given:•• Height above sea-level,Pressure at sea-level,
Z:: 7500 m
Po:= 10.143 N/cm~'= 10.1'43 x 104 N/m:!
• Temperature at sea-level, '1 "'O~lO=1.)\....,
• "Here pressure head (Z) is given as 760 mm ofHg. Hence (P/pg) = 760 mm ofHg. The density (p) formercury3 ,760 2W3.6 x 1000 kg/m . Hence pressure (P) will be equal to p x g x Z i.e. 13.6 X 1000 x 9.81 x iooo N/m .
•
••••••••••••••••••••••••••••••••••
Unit - 2
FLiUD KINEMATICS AND FLUID DYNAMICS
Fluid Kinematics - Flow visualization - lines of flow - types of flow - velocity field and
acceleration - continuity equation (one and three dimensional differential forms)- Equation of
streamline - stream function - velocity potential function - circulation - flow net - fluid dynamics
- equations of motion - Euler's equation along a streamline -
Bernoulli's equation - applications - Venturi meter, Orifice meter, Pitot tube - dimensional
analysis - Buckingham's TT theorem- applications - similarity laws and models.
•••••••••••••••••••••••••••••••••
---. --~-------.-.~--.----------------- -- -
Two-dimensional flow is that type of flow in which the velocity is a function of time and tworectangular space co-ordinates say x and y For a steady two-dimensional flow the velocity is a function of twospace co-ordinates only. The variation of velocity in the third direction is negligible. Thus. mathematically fortwo dimensional flow
U=/I(X,y), v=h(x,. y) and w=O.
Three-dimensional flow is that type of flow in which the velocity is a function of time and threemutually perpendicular directions. But for a steady three-dimeosional flow the-fluid parameters are functionsof three space co-ordinates (x, y and z) only. Thus. mathematically. for three-dimensional flow
U =tv». }',z), v =.f7(x, y, z), w =13(x. y, z).
5.4. RATE OF'FLOW OR DISCHARGE (Q)
It is defined as the quantity of a fluid flowing per second through a section of a pipe or a channel. Foran incompressible fluid (or liquid) the rate of flow or discharge is expressed as the volume of fluid flowingacross the section per second. For compressible fluids. the rate of flow is usually expressed as the weight offluid flowing across the section, Thus
(i) For liquids the units of Q are m3/s or litres/s(ii) For gases the units of Q is kgf/s or NewtonlsConsider a liquid flowing through a pipe in which
A = Cross-sectional area of pipeV :::Average velocity of fluid across the sectionQ=AxVThen discharge
...(5.1)5.5. CONTINVITY EQUATION
The equation based on the principle of conservation of mass is called continuity equation. Thus for afluid flowing through the pipe at all the cross-section, the quantity of fluid per second is constant. Considertwo cross-sections of a pipe as shown in Fig. 5.1.
Let 1'1= Average velocity at cross-section ),-1PI =Density at section 1-1
andAl =Area of pipe at section 1-1
V2• P2. A2 are corresponding valves at section, 2-2.Then rate of flow at section 1-1 = PIA IVIRate of flow at section 2-2 = PzA2 V2According to law of conservation of massRate of now at section i -1 = Rate of flow at section 2-2
PIA 1/1 = P2A2V< ..(5.2)
DIRECTION-OF FI1:1N
<D ®«( r <J c , c c , c' t < c c < I " c < t •
I, I-'to ....,,,1
or Fig. 5.1. Fluid flowing through a pipe.
Equation (5.2) is applicable to the compressible as well as incompressible fluids and is calledContinuity Equation. If the fluid is incompressible, then PI = P2 and continuity equation (5.2) reduces to
A! VI =A2V'1 ... (5.3)Problem 5.1. The diameters of a pipe at the sections J and 2 are 10 cm and 15 em respectively. Find
the discharge througb the pipe if the velocity ofwater flowing through the pipe at section 1is 5mls. Determinealso"the velocity at section 2.
Sol. Given:At section 1. D} ::: 10 em = 0.1 m.
••••••••••••••••••••••••••••••••11
At section 2,
1t 2 1t, 2AI = 4 (DI ) = 4 (.1)- = .007854m
VI =5m/s.D2 = 15 ern = 0.15 m
1t 2 'JA2= 4 (.15) =0.01767 m~
(i)Discharge through pipe is given by equation (5.1)
®!i-.~~---r-.!
,',! . D;:...- ::;·,_:m
'",-,~.t-' ..,
or Q=AI X VI
= .007854x 5 = 0.03927m3/s. Ans.Using equation (5.3), we have AI VI =A2V2
A IVI .007854(ii):. V2=A;- = .01767 x 5.0 = 2.22 mls.
Fig. 5.2
Ans.
Problem S.2. A 30 em diameter pipe, conveying water, branches into two pipes of diameters 20 emand 15 em respectively. If the average,veloeity in the 30 em diameter pipe is 2,5 m/s,find the discharge in thispipe. Also determine the velocity in 15 em pipe if the average velocity in 20 em diameter pipe is 2 m/s.
Sol. Given:
V, =2.5m/Sec--.0, = 30 em
...Fig. 5.3
DI = 30 em = 0.30 m1t 11:
Al = 4 DI2 = 4 X.32 = 0.07068m2
VI =2.5 rn/sD2 = 20 em = 0.20 m
1t 2 1t '.A2= 4 (.2) ='4"x.4 = 0.0314 rrr',
V2= 2m/sD3 = 15ern = 0.15 m
...
... A 1t '15)2 n 0 2 . . r: ')3 = 4 ~. :: 4' x ,.2 .5:; 0.01107 c.:Find (i) Discharge in plpe 1 or QI
(il)Velocity in pipe of dia. 15 em or V3
Let Qb !l2 and Q3are discharges in pipe 1, 2 and :1 respectively.Then according to continuity equation
QI'= Q2+ Q3...(1)
,.~ ..~ ..-.-~--'--- ..---~-,.-~,.--.-.---••••• (i) The discharge Ql in pipe 1 is given by
Qt =At Vt =0.07068 x 2.1 m3/s = 0.1767 m3/s. ADs.
(ii) Value erv,••••••••••••••••••••••••••••....----.-
Substituting the values of Qt and Q2 in equation (1)0.1767 :.:0.0628 + Q~,
Q3 = .1767 _.0.0628 = 0.1139 m3/sQ3 =, A3 x V1::.:.01767 x \"3 or .1139 = .01767 X V3
.1139V:, ::: ~01767 = 6.44 mls. Ans.
But
Problem 5.3. Water flows through a pipe AB 1.2 m diameter at 3 m/s and, then passes through a pipeBC 1.5 m diameter. At C, the pipe branches. Branch CD is 0.8 m in diameter and carries one-third o/the flowin AB. The flow velocity in branch CE is 2.5 m/s. Find the volume rate a/flow InAB, the velocity in BC, thevelocity in CD and the diameter of Cg (AMIE, Winter 1982 ; Osmania University. 1990)
Sol. Given:Diameter of pipe AB,Velocity of flow through AB,Dia. of pipe BC,Dia. of branched pipe CD,Velocity of flow in pipe CE,Let the flow rate in pipeVelocity of flow in pipeVelocky of flow in pipeDiameter of pipeThen flow rate through
and flow rate through
DAB = 1.2 InVAB =3.0 mlsDBc= 1.5 mVcD=0.8 mVeE= 2.5 mlsAB = Q mJ/sBe = VBf.:' ff'JS
CD = VeD misCE=DnCD: Q/j
CE:Q_Q/3=N 3
Fig. 5.4
(i) Now volume flow rate through AB : Q :::;VAB x Area of AB'1'(: ( , 2 3 0 1t (1 2' 2
:= 3.0 »: '4 D..<\B.l = . x 4 .)
: 3.393m3/s. ADS.
Problem 5.4.A 25em diameter pipe carries oil of sp. gr. 0.9 at a velocity of 3m/s. At another section• the diameter is 20 em. Find the velocity at this section and also mass rate offlow of oil.• Sol. Given:. at Section 1,••
•••..•••• or
•••• or••••• or•• or
•
• at Section 2,
••..
or
(it)Applying continuity equation to pipe AB and pipe BC,VABx Area of pipe AB = VBC x Area of pipe Be
x 2 X 23.0 x 4 (DAB) = VBC x 4 (Dsd
3.0 x (1.2)2 = VBC X (1.5)2
3 X 1.22VBC = 2 ~ 1.92 mls. ADS.1.5
[ Divide by : ]or
(iii)The flow rate through pipe
. CD=Q] =~= 3.~3 = LI31 m3/s
QJ = VCDx Area of pipe CD x : (CCD)2...
...X 21.131 = VCDx 4 x.8 = 0.5026 VCD
1.131VCD = 0.5026 = 2.25 mls. ADS•
(iv) Flow rate through CE,
Q2 = Q- QJ = 3.393 -1.131 = 2.262 m3/s...Q2 = VCE x Area of pipe CE = VCE~ (DCE)2
2.263 = 2.5 x ~ x (DCE)2
DCE =.y 2.263 x 4 = -../1.152= 1.0735 m2.5xx
:. Diameter of pipe CE = 1.0735 m. ADS.
D] =25 em =0.25 m
A] = : D]2 = : x .252 = t~049 m3
V] =3m/sD2=20cm =0.2 m
A2 = x (.2)2 = 0.0314 m24Y2=?
Mass rate of flow of oil = ?Applying continuity equation at Sections 1 and 2•
A]V] =A2V2
0.049 x 3.0 = 0.0314 X V2
... t: _ 0.049x3.0 -468 .1t 2 _. -. OilS •.0314 ADS.
...
~~~----------•
Problem 5.S.A jet of water from a 25 mm diameter nozzle is directed vertically upwards. Assuming• that the jet remains circular and neglecting any loss of energy, that will be the dimneterat a point 4.5m above• the nozzle, if the velocity with which the jet leaves the nozzle is 12 m/s. (A.Ml.E., Summer, 1988)
Sol. Given:
• Consider the vertical motion of the jet from the outlet of the nozzle to thepoint A (neglecting any loss of energy).
Initial velocity, u = VI = 12 m1sFinal velocity, V = V2Value of g =- 9.81 m1s2 and h = 4.5 mUsing, \t~ - u2 = 2gh,we get
vl- }22 = 2 x (--9.81) x 4.5
V2 = -..)122 - 2 x-9~1x 4.5' = -../144 - 88.29 = 7.46 m1s
••••••••••••••••••••••••••
Mass rate of flow of oil =Mass density x Q = P XAI X VI
= Density of~,?ilDensity of water
:::Sp. gr. of oil x Density of water3 900kg= 0,9 x 1000 kg/m = 3m
=900 x 0.049 x 3.0 kgls = 131.23 kgls. ADS.
Sp. gr. of oil
:. Density of oil
:. Mass rate of flow
Dia. of nozzle,Velocity of jet at nozzle,Height of point A.
D! = 25 mm ,:.0.025 mV,=12m1sI:! :::4.5 m
A-4--ITII 4.5m
'1JET OFWATERLet the velocity of the jet at a height 4.5 m :::V;
Now applying continuity equation to the outlet of nozzle and at point A,
we getFig. 5.5
or
1t 2A V -Dj X VI 2A2 ::: :_, t_ = ~~_._. = 1t x (0.025) x 12 = 0.0007896Vi v~ 4 x 7.46
u). = Diameter of jet at point A."It . 1t . 2-
A2::: 2:D2' or 0.0007896 = '4 x D2
D2 = ~9.0007!9~X 4 = 0.0317 m =31.7 mm. Ans.
Let
Then
5.6. CONTINUITY EQUATION IN THREE-Dli\-'IENSIONSConsider a fluid element of lengths dx, dy and dz in the direction of x, y and z, Let u, v and w are the
inlet velocity components in x. y and z directions respectively. Mass of fluid entering the faceABCD per second:::p x Velocity in --direction x Area of ABCD= P x u x (dy x dz)
Then mass of fluid leaving the face EFGH per second = pu dydz + :x (pu dydz) dx
----~--.-
••••••••••••••••••••••••••••••••
...dx=u f!!=vanddt,:wdt ' dt dt
du au au au auClz= dt = u ax + v cry +w az + at
dv av av av avay= dt = u ax+ v ay +w az+ at
dw aw dw aw awaz= dt =u ax +v ~ +w az +-at
For steady flow, ~~ = 0, where Vis resultant velocity
But
...Similarly,
or au av aw-=0 -=Oand-=Oat ' at atHence acceleration in x, y and z directions becomes
du au au aua =-=U-+V-' +w-Xdt ax ay azdv av av avay=-=u-+v-+w-dt ax dy azdw aw aw awa =-=u-+v-;--+w-
z dt ax 0)' azA=axi+ayi+alc 'j=....Jax2+a/+a/" .
Acceleration vector
i...(5.6)
.l
...(5.7)
...(5.8)
5.7.1. LocalAcc:eleratlon and Convective Ac(!eleration
LocaI acceleration is defined as the rate of increase of velocity with respect to time ,:it a given point
in a flow field. In the equation given by (5.6), the expression ~~ , ~; or ·t~is known as local acceleration.
Convective acceleration is defined as the rate of change of velocity due to the change of position of
fluid particles in a fluid flow. The expressions other than ~~ , ~; and ~~ in the equations (5.6) are known as
convective acceleration ..
Problem 5.6. The velocity vector in a fluid flow is givenV= 4ri -10ryj + 2tk.
Find the velocity and acceleration of a fluid particle at (2, 1, 3) at time t = 1.
(Delhi University, 1987)Sol. The velocity components u, v and w are u = 4i3, V =- lOr y, w =2tFor the point (2, 1, 3). we have x = 2, y = 1 and z = 3 at time t = 'I.
Hence velocity components at (2, 1, 3) are
u = 4 X (2)3 = 32 units
V = - 10(2)2(1) = - 40 units
w = 2 x 1 = 2 units
••
.1----------·----·---••. . Velocity vector Vat (2, 1, 3) = 32i- 40j + 2k
Resultant velocity =~"2+ ;. + w!
=~322+(-40i·+22 ="1024+1600+4 =51.26. uaUs. ADs.
Acceleration is given by equation (5.6)au au au au
a)::: u~+ v~+w~+-;-ox oy oz otdV av av avav== u-+v-+ w-+-. ax ay az atdW dW aw awa.>u- +v- +w- +-~ ax ay az at
Now from velocity components, we have
au ::.:lW, au :: 0, au == 0 and au = 0ax ay az atav dv 2dv avax ==, - 20xy, 3y = - lOx-, dz = 0 at = 0
ow dw dW Owax = 0, -ay = 0, az = 0 and at = 2.0.
••••••••••••••••• or•
Substituting the values, the acceleration components at (2, 1,3) at time t = 1 areax = 4~(12X2)+ (- 10~y)(0) + 2t x (0) + 0
=,48.0 == 48 x (2)5 = ...~)x 32 = 1536 unitsay =:4~(- 20xy) + (- 10ry)(- lor) + 2t(0)+ 0
:::- 80x4y + lOOx4y:::_ 80(2)4(1) + 100(2)4 X 1 = - 1280 + 1600 = 320 units.
a, == 4~(0) + (-10~Y)(O)+ (2t)(0) + 2.0 =2.0 unitsA :::a) + aj + ill' = 1536i + 320j + 2k. Ans.
:: ~(1536)2 + (320)2+ (2)2 units
~ "2359296 + 102400 + 4 = 1568.9 units. Ans.
Acceleration is
Resultant A
• Problem 5.7. The following cases represent the two velocity components, determine the third com-ponent of velocity such that they satisfv the continuity equation:
• (i) u = ).2 +I + t :::::.~l " ...C'• (ii) v = 2}2, W = 2xyz.
••••••••.._.-
Sol. The continuity equation for incompressible fluid is given by equation (5.4) as
au + av + dW =0ax ay az
Case I.au =2x.. axav.. ay=2xy-ri-x./
. he values of au d dv . ., .Substituting t e va ues ot ax an ay Incontinuity equation.
••••••••••••••••••••••••••••••••
_2 aW2x+2xy-r+x+-az=O
~ 2 2or az =-3x-2xy +z or aw = (- 3x-2xy + z )az
Integration of both sides gives J dw = J (- 3x - 2xy + z2)dz
or w =( - 3xz - 2tyz + ; ) +Constant of integration,
where constant of integration cannot be a function of z, But it can be a function of x and y that isj{x. y).
• w = ( -3xz-2xyz + ~ )+f(1<, y). ADs.
avCasen. v = 2y. :. ()y = 4y
~w =2xyz . . az = 2xy
Substi . th al fav d~' .. .u stitutmg e v ues 0 ay an az m contmuity equation, we get
auax +4y + 2.xy = 0auax = - 4y - 2xy or du = (- 4y - 2xy)tU
~u = -;-4xy - 2y 2" + fey, Z) = - 4xy - xV + f(y, z), ADS.
or
Integrating, we get
Problem 5.8. A fluid flow field is given byV = ryi +1'lj - (2xyz + yi) k.
Prove that it is a ctlfe of possible steady incompressible fluid flow. Calculate the velocity andacceleration at the point (2, 1. 3).
SoLFor the given fluid flow field u =ry
v=y.z
au =2xy.. ax
av()y = 2yz
Ow.. - =--2xv··-2yz.az .c
i.e.,For a case of possible steady incompressible fluid flow, the continuity equation (5.4) should besatisfied.
au+av+aw_oax ay az-'
S bsti . th I f au av d ~u sntunng eva ues 0 ax' ayan az' we get
au av awax + ay + az = 2xy + 2yz - 2xy - 2yz = 0
Hence the velocity field V = ryi +yl'lj - (2xyz + yz2)k is a possible case of fluid flow. ADS.
••Velocity at (2, 1, 3)
• Substituting the values x = 2,y = 1 and z = 3 in velocity field, we get• V= x?yi + l{j - (2xyz +yi)k
= 22 xli + )2 X 3j -. (2 x 2 x 1 x 3 + 1X 32)k• =4i+3j-21k. Am.
• md Resultant velocity == '/42 -i~-};_-:--21)2 '" '1/16+9 + 441 = ...)466= 21.587 units. ADS.
.Acceleration at (2, 1,3)The acceleration components ax, Qy and az for steady flow are
au au aua :::u-+v-+w-x ax dY dZ
av dv (Iva . :::U - + v - + w --., ax ay dz
Ow aw d14'at. :::U ax + v dy + w -a;.'
2 au au 2 auu =x-y, dX = 2xy, ay =.r and az =0
2 av av aV.2v = y z, ax = 0, ay = 2yz. dz = y
law ,aw _laww .:C - 2xyz ._YZ '-a = -- 2yz. -a = - 2xz - Z-, -a- = - 2xy - 2yz.. x y Z
••••••••Substituting these values in ecceleration components, we get acceleration at (2, I, 3)
ax = ,t2y(2xy) + y2Z(~) - (2xyz +YZ2)(0)::.\ '),2 .j. 2.,,2~"r . x~,,/ ..._= 2(2)312 + 22 X 12 X 3~:::2 x 8 + 12= 16 + 12 ::;28 units
ay = x2y(0) + Iz(2yz) - (2xyz +YZ2)(I)=2y3Z2 _ 2xy3z _ y3Z2
=2xI3x32-2x2x13x3_ 13x32=18-12-9= -3 unitsaz = x2y(_ 2yz) + y2Z(_ 2xz - i) - (2xyz +yi)(- 2xy - 2yz)
= _.ulz -2x1i - 1£'+ [4~-;z + 'lxIt+4x1z2 + 21z3]= - 2 X 22 X (2 X 3·_ 2 x 2 X 12X 32 _12 X 33 + [4 X 22
x 12X 3 + 2 x 2 X 12 X 42 + 4 x 2 X 12X 32 + 2 X 12 X 33]= - 24 - 36 - 9 + [48 + 36 + 72 + 18]= - 24 - 36 - 9 +48 + 36 + 72 + 18 = 105
. Acceleration = ati + ayi + axk = 28i _.•lj + tOSk. Ans.
Resultant acceleration = ~282 + (--3)2 + 1052-:= .,f784 + 9 + 11025
=" 11818 = 108.71 units. Ans.• Problem 5.9. Find the convective acceleration at the middle of a pipe which converges uniformly from0.4 m diameter to 0.2 m diameter over 2 m length. The rate offlow is 20 litis. If the rate offlow changes
.niformly from 20 litis to 40 litis in. 30 seconds, find the total acceleration at the middle of the pipe at 15thsecond.••
••••••••••• ./or•
For a line of constant stream function
• = d'I'= 0 or vdx- udy = 0!ll. _ y_dx- u
••••••••• 'or••
But~ = Slope of equipotential line.
5.8.4. Line of Constant Stream Function
\jf = Constantd\jf = 0
d\jf = ~ dx + ~ dy = + vdx _ udyox oy .
...But
{".. ~=V'~=-u}ox ' oy
But ~ is slope of stream line....(5.14)
From equations (5.13) and (5.14) it is clear that the product of the slope of the equipotential line and• the slope of the stream line at the point of intersection is equal to - 1.Thus the equipotential lines are orthogonalto the stream lines at all points of intersection.•
5.8.5. Flow Net. A grid obtained by drawing a series of equipotential lines and stream lilies is called• a flow net. The flow net is an important tool in analysing two-dimensional irrotational flow problcms.
5.8.6. Relation between Stream Function and \, "'''City i',,'em,.l FMellO', From eq""'ion (5.9).• wehave•••••-and
•••••••td•••
U -_E.!- oxu=-~U -_E.!_ ~- ox - - cry
E.!_~ }ox - cry~--~i)y - ox "
Problem S.10. The v~ltx.ity potentialjunction (q,) is given by an expressto«-v-+9+y2.
and ~v =._..,oy
1I=~ oxo ~
V --!!!-~- i)y - ox
From equation (5.12), we haveand
Thus, we haveand
Hence
...(5.15)
(i) Find the velocity components in x and y direction.(ii) Show that q, represents a possible case offlow.
S J "".3 _2 x\ "'o • Given: ""= .,..~ _ .r +_...:....Y'~'f' 3 ::i
The partial derivatives of q,w.r.t, to x and y are
E.!.i 3t2):_ox = - 3 - 2x + --3"~ 3 2 xl~=-T+3+2Y
...(1)
...(2)
•••••••••••••• and
•••••••••..•••••• point.
•
(i) The velocity component~ II and v are given by equation (5.9). r .,3 "Xl] 3u=_~=_lL_2x+:".:£l =L+2x-,n,ax 3 3 3
u ~';i + 2x - x2y. Ans .3
v c-: _ ~'. = _. 3xl.+~~+ 2Y] = ~ - ~ - 2y =,xy2 -~ - ly.dy 3.J 3 3 3
...ADs.
(ii) The given value of e!>, wil! represent a possible case of flow if it satisfies the Laplace equation. i.e.•
a2e!> a2<\>-+-=0aXl alFrom equations (1) and (2), we have
Now ~ = -l13 - 2x + ,Xly
iP<\>aXl=-2 +2xy
...
ADS.. • Laplace equation is satisfied and hence $ represent a possible case of flow.
Problem 5.11. The veloclty potential function is given by <\> = 5(r - y).Calculate the velocity components at the point (4,5).Sol. <\> = 5(xl - ;)
:. ~ = lOx
~=-IOy.
But velocity components u and v are given by equation (5.9) as
u=-~=""IOx
v = - ~ = - (- lOy) = lOy
The velocity components at the point (4.5), i.e., at x =4, Y = 5u = - lOx 4 = - 40 unitsv = 10x 5 = 50 units. ADs.
Problem 5.11. A stream function is given by 'IV = 5x - 6y.Calculate the velocity components and also magnitude and direction of the resultant velocity at any
.----~----- ----.--~..-
•••••••••••••'.•••••••••••••
1
...'II = 5x- 6y
~ = 5 and ~ = - 6.ax dyBut the velocity components u and v in terms of stream function are given by equation (5.12) as
u =- ~ = - (- 6) = 6 units/sec. Ans.
v = ~ =S units/sec. Ans.
= -vu2 + v2 = -v62+ 52 = 36 + 25 = -v6f = 7.81 unit/secResultant velocity
Direction is given by tan e = ~ = .1= 0.833u 6e = tan-I .833= 39° 48'. Ans.
...Problem S.13·If/or a two-dimensional potential flow, the velocity potential is given by
cj) = x(2y-J)
determine the velocity at the point P(4, 5). Determine also the value of stream function 'I' at the point P.
(AM IE.,Winter, 1977)Sol. Given: cj) = x(2y - 1)(I)The velocity components in the direction of x and yare
u = -~ =_1.. [x(2y -- 1)] = - [2y - 1] = 1 - 2yax ax
v = - ~ =_1.. [x(2v - 1)] =- [2x] =--2xay ay -At the point P(4, 5), i.e. at x = 4, y = 5
u = 1- 2 x 5 = - 9 units/secv =- 2 x 4 = - 8 units/sec
:. Velocity at P = - 9i - 8}
or Resultant velocity at P = -V92+ 82 = "81 +-64, = 12.04units/sec ;::,12.04 unltsrsec, Ans.(ii) Value of Stream Function at P
We know that Wi = - u =- (1- 2y):; 2y -1 (i)
and ~; =v =- 2x (ii)
Integrating equation (i) w.r.t. 'y', we get
Jd r(2 l)d 2y' C f.. .. 'II = JI Y - y or 'V~" '1 ._y" .onstani : rntegration.
The constant of integration is not a function of y but it can be a function ofx. Let the val ue of constant• of integration is k. Then
•••••
'I'=y2_y+k.Differentiating the above equation w.r.t. 'x', we get
~_akax-ax'
...(iil)
•••••••• and
•••••••••••••••••••••••••
5.9.5.Vorticity. It is defined as the value twice of the rotation and hence it is given as = 200.Problem S.lS.A fluid flow is given by V = 8:!i - JOryj.Find the shear strain rate and state whether the flow is rotational or irrotational.SoLGiven :. V = 8ri - 10Dj
:. u=8r, aaU'=24~,~:
v =-lory, av-a"=-20xy,~
(i) Shear strain rate is given by equation (5.16) as
-1 (av au)' 1=- -a +~ = :;(--20X)'+ 0) = -lOxy.2 x oy "
(il) Rotation in x - y plane is given by equation (5.17) or
(I) =! (av _ au )= .!.l__20X)' - 0) = -- lO!X)'z 2 ax iJy i:
As rotation Wz:F- O.Hence flow is rotational. Ans,
ADS.
Problem 5.19. The velocity components in a two-dimensional flow areu = 1/3 + 2x - ry and v:: xI- 2y - x1/3.
Show that these junctions represent a possible case of an irrotational flow.Sol. Given: u = yl/3 + 2x -~y
:. ~: =2-2xy
au =li.__2 = y2 _ .;2dy 3 x- .•
v =xy2 - 2y -r/3dv'-=2xy-2dy
Also
...dv 3rax =l-3=i-)l2
F t -di . al fl .. "au dVor a wo nnension ow, continuity equation IS -a +.:;-:::: (lX oy
Substituting the value of ~: and ~~ , we get
au au ,ax + ay = 2 - 2X)' + ,~ - 2 = 0
:. It is a possible case of fluid flow.
1 (av au 1COz = 2 ax - ay )=~[(i--r)_..(i --~)]=0
(ii) Rotation, OOz is given by
:. Rotation is zero, which means it is case of irrotational flow. Ans,
••••• or
• or••••
p . yz- + z + ~- = constantpg 2gi.1 !T'~- + ~- i ;.;;:constantpg Zg
Equation (6.4) is a Bernoulli's equation in which.0_:'_::::pressure energy per unit weight of fluid or pressure Head.pg
v'~/2g.::kinetic energy per unit weight or kinetic Head.z :: norential energy per unit weight or potential Head.
...(6.4)
• 6.5. ASSUMPTIONS
• The following are the assumptions made in the derivation of Bernoulli's equation:(z) The fluid is ideal, i.e. viscosity is zero (if) The flow is steady
(iiI) The flow is incompressible (iv) The flow is irrotational.
• Problem 6.1. Water isflowing through apipe of5 emdiameter under apressure 0/29.43 Nlcm2 (gauge)and with mean velocity of 2.0 mls. Find the total head or total energy per unit weight of the water at a
• cross-section, which is5m above the datum line.Sol. Given:
• Problem 6.2. A pipe, through which water is flowing, is having diameters, 20 em and 10 em at the• cross-sections 1and 2 respectively. The velocity atwater at section 1 is given 4.0 mls. Find t~ velocity head• at sections 1and 2 and also rate of discharge.
Sol. Given:
•
•••••••
••••••••
Diameter of pipePressure,Velocity,Datum head,Total head
"..') em = 0.5 m;,,) =, 29.43 N/cm2 = 29.43 x 104N/m2'" 2.0 m/s
oressure head + kinetic head + datum head
: L._ = 29.43 X 104 = 30 m.:)g 1000 x 9.81;,:' 2" '2" ._. = - ..-- = 0.204 m2g 2 x 9.81fl v2 •
~.. L.. + - + z = 30+ 0.204 + 5 = 35.204m.og 2g ADs.
Pressure head
Kinetic head
:. Total head
.. Area,
D1 = 20 em= 0.2 m1C?:Tt 2 2A1 = 4 D1- = 4 (.2) = 0.0314 m
\/1 = 4.0 m/sD2 = 0.1 m
®
(l) Velocity head at section 'f
= VI" = 4.0 x 4.U = O.8J.Sm.2g 2 x 9.81 Ans.
Fig. 6.2
, .• -r.' __-::-__.....:::....::.__ --- ~_____ __
•••••••••••••••••••
(il) Velocity heat at section 2 = Vl/2gTo find V2J apply continuity equation at 1and 2... or A1Vj .0314
'V := --~ ::-: --- X 4.0 = 16.0 m/s2 4" ,00785''-
:. Velocity head at section 2
(iii) Rate of discharge
= vl = 16.0 x 16.0 = 83.047 m. ADs.2g 2 x 9.81
=A1V1orA2l'2
= 0.0314 x 4.0 = 0.1256 m /.
= 125.6Iitresi:i. ADs.
Problem 6.3. State Bernoulli's theorem for steady flow of an incompressible fluid. Derive anexpressionlor Bernoulli's equationfrom first principle and state the assumptionsmadefor such a derivation.
(AM'IE, May 1974)Sol. Statement of Bernoulli's Theorem. It states that in c steady, ideal flow of an incompressible
fluid, the total energy at any point ofthe fluid is constant 'The total uicrgy consists of pressure energy, kineticenergy and potential energy or datum energy. These energies ner II: rv > nght (It the fluid are: ' '
Pressure energy = E:pgy2
Kinetic energy =-2g
Datum energy = zThus mathematically, Bernoulli's theorem is written as
p V- + -2 + z = Constant.w g
Derivation of BernoulU's theorem. For derivation of BernGulU's theorem, the Articles 6.3 and 6.4• should be written.• Assumptions are given in Article 6.5.
Problem 6.4. Thewater isflowing throughapipe havingdiameters20 emand 10em at sections 1 and• 2 respectively. The rate offlow throughpipe is 35 litresls. The sectio» f " 6m above datum and section 2 is• 4 m above datum. If thepressure at section 1 is 39 24 Nlcm/, find the »aensity ofpressure at section 2.
Sol.Given:•••••"••••
At Section 1, DI = 20 em = 0.2 m
Al = : (.2)2 = 0.0314 In
PI = 39.24 N/cm2
= 39.24 x 104 N/m2ZI = 6.0m
At Section 2, D2 = 0.10 m
A2 = : (O.li = .00785 m2
z2=4mP2= ?
••••••...ld
Q:-:: 35lit/s = 1~0 = 0.035 m3/s
Q = A1V1 =A2V2
Q 0.035VI = Ai = .0314 = 1.114 m/s
V. ;c: Q,:: 0.035 = 4.456 m/s, Ac .00785
Pi VI P V2~- + - .+ 1 ::c- _.- + - + z"pg 2!' pg - 2g ~
42"39.24 x 10 + (1.114) + 6.0 ::; P2 + (4.456)", + 4.01000 x 9.81 2 x 9.81 1000 x 9.81 2 x 9.81
40 + 0.063 + 6.0 :::9~~O+ 1.012 + 4.0
P246.063 :::9810 + 5.012
9~;0- '" 46.063 - 5.012 = 41.051
P2 = 41.051 x 9810 N/m2
- 41.051 x 9810 N/ 2 _ 40 "7 N/ Z- 104 em - .1iI em •
Rate of flow,
Now
••••
Applying Bernoulli's equation at secnons and 2. we get
•i••••
Ans.
Problem 6.5. Water is flowing through apipe having diameter 300 mm and 200 mm at the bottom and.er end respectively. The intensity of pressure at the bottom end is 24.525 N/cm2 and the pressure at the.per end is 9.81Nlcnt'. Determine the difference in datum head if the rate of flow through pipe is 40 luls.
Sol. Given:••••••••••••
Section 1, DI = 300 mm :;,;0.3 '1
PI = 24.525 Nlcm2 ,= 24525 x 104N/m2D2 = 200 mm ::;0.2 In
2 __- 4 2P2 = 9.81 N/cm = 9.81 " 10 N/m= 40 litis
40 3Q = 1000 = 0.04 m Is
Al VI =A2V2 = rate of flow e 0.04
VI = .04 = ~~ = 0.04_ = 0.5658 m/s.AI iD12 ~ (0.3)2
:0.566 m/s
V2 = .04 = _~ = _Q.04 = 1.274 m/sA2 ~ (D..,-)2 ~- ( ?,24 "". a "-i
Section 2,
Rate of flow
Now
DATUM LINE
Fig. 6.4
••••••••••••••••••••••••••••••••111..
or
ApplyiDg Bemoulli's equa~on at (1) and (2), we get
PI Vl2 P2 vl-+-+Zl=-+-+qpg 2g pg 2g24.52S X 104 .566 x .566 9.81 x 104 (1.274)"-1000 x 9.81 + 2 x 9.81 +Zl= 1000 x 9,81 + 2 x 9,81 +q
25 + .32 +Zl= 10 + 1.623+ q25.32 + Zl= 11.623 +q
q-Zl = 25.32 -If.623 = 13.697 = 13,70 m
oror ...
:. Difference indatum head =q- zl = 13.70 m. AnsProblem 6.6. Thewater is flowing through a taper-pipe . ,to length 100 m having diameters 600 mm at
the upper end and 300 mmat the lower end, at the rate 450 litrests. Thepipe has a slope of 1in 30. Find theJ!Tesmre at the lower end if the pressure at the higher level is lO 62 N Iem2
SoL Given':Length of pipe, L = 100 mDia. at the upper end; DI = 600 mm = 0.6 m
:. Area, At = :'Dt2=: x (.6)2 = 0.2827 m2
PI = pressure at upper end = 19.62 N/cm2= 19.62x 104N/m2
Dia. at lower end, D2 = 300 mm = 0.3 m Fig. 6.5
A2 = =D22 = = (.3'? = 0.07068 m
Q = rate of flow = 50 litres/s »T~~o= ('.05 m3;s
Let the datum line is passing through the centre of the k,we." end.Then 2'2;: 0
:. Area,
As slope is 1in30 means
Also we know
... TT Q 0,05 0 1'768mJ 7q = A = ,2827 = . sec:: 0.17 mls
aDd v: Q 0.5 02 = A2 = .07068 = .7074 m/sec = 0.707 mls
ApplyiDg Bernoulli's equation at sections (1) and (2), WI"; ge,;
Pt V12 P2 V2"-+-+zl=-+ '--TV;pg 2g pg 19 4-
19.62 x 104 .1772 10 P2 .70721000 x 9.81 + 2 x 9.81 +"3 = pg + 2 ~ 9.81 'V '"
20 + .001596 + 3.334 :: P2 + {)i.()254. pg
or
or
••"
or•fir23.335 - 0.0254:: -. oF"- .-
1000 x 9.81P2:: 233,< 9810 N/ml = 228573 N/m2 = ZZ.857 N/cm'1..
fI.6•BERNOULLI'S EQUATION lOR REAL FLUIDThe Bernoulli's equation was derived on tb~ assumption that fluid is inviscid (non-viscous) and
.herefore frictionless. But all the rea! fluids are:viscous and hence offer resistance to flow. ,Thus there arealways 'some losses in fluid flows and hence in the application of Bernoulli's equation; these losses have to be
_aken into consideration. Thus the Bernoulli's equation for real fluids between point 1 and 2 isgive,n asPl v/ P2 vl-.+--+Z; ;,::-+-+.q+hLpg 2g , pg 2g
• where hL is loss of energy between point 1 and 2• Problem 6.7. A pipe of diameter 400 mm carries water at a velocity of 25 mls. Thepressures at the
points A and B are given as 29.43 N/cm2 and 2.Z.563 N/cm'2respectively while the datum head alA andB are.28 m and 30 m. Find the loss of head between and B
Sol. Given :
•
•••••••••••••••
•..(6.S)
D=400mm::04m.V=25mhPA = 29.43 N/cm~::, 29.43 x 10" N/m2
ZA = 28mVA = V = 25 in/s
:. Total Energy atA,
Dia. of pipe,Velocity,At point A
AtpointB,
PA viEA :: - + ~ -r z,pg .ig44-2= 29. ~ x 10 -+._ 25 + 28
1000 x 9.81 2 x 9.81= 30 + 31.85 + 28:; 89.85 m
PB = 22.563 N/em2 "" 22.563 x 104 N/m2
Fig. 6.6
Z13 = 30 m
. . Total Energy at B.
. . Loss of Energy
22563 x Hr! 25~~ ··woo x 9.8i + 2 x 9.81 + 30 = 23 + 31.85 + 30 = 84.85 m
:= fA - E13 :: 89.85 - 84.85 = 5.0 m. ADs.
Problem 6.8. A conical tube of length 2.0 m is fixed vertically with its smaller end upwards. The• velocity of flow at the smaller end is 5 mls while at the lower end it is 2 mls. The pressure head at the
. . . . .. ,0.35 (Vl - v2i . .• smaller end IS 2.5 m of liquid. The loss of head tn the tube 1..'1 2g , where Vl IS the velocity at the
• smaller end and V2 at the lower end respectively. Determine thepressure IU!4d at the lower end. Flow takesplace in the downward direction.
•
I••• . . The discharge
•••
Q :.:::11" x (l'J ;; V'2 x anCt
<lo t c ffih
= \f:--( ~ f c,:•.•(iv)
The above expression is simplified by using•••••••••••••••••
Substituting this value 01C, in equation (zv),we get
Q :.:::an " (
: !i~y,_v'_2g_7_1. _. __ Cd aoalv'2ihVa12 - ao2
I
...(6.13)
lifere Cd = Co-efficient of discharge for orifice meter.The co-efficient of discharge for orifice meter is much smaller than that for a venturimeter.
Problem 6.22.An orifice mete,' with orifice diameter 10 em is inserted in apipe of 20 cm diameter.~ pressure gauges fitted upstream and downstream of the orifice meter given readings of 19.62Nlcnf and9.81 N/cm2 respectively. Co-efficient of discharge for the meter is given as 0.6. Find the discharge of water.rough pipe.
•• Sol.Given:
•••Dia. of orifice, do = 10 em
:. Area,:it . ". 'ao:: "4 (10)" '" '78.54 ~~
•••••••••••••••••••••••••••••••••
Dia. of pipe, d1= 20 em
al = ~ (20)2:: 314.16 cnr'
Pi :: 19.62 N/cm2:: 19.62 x 104N/m2
PI 19.62 x 104 20 f te----:: mo wa rpg 1000 x 9.81
P2 = 9.81 X 104 .pg 1000 x 9.81 = 10 m of water
:. Area,
Similarly
. Pl P2 '1("• h:: .--~,'-" '" L. 1.Upg pg
c,: 0.6The discharge, Q is given by equation (6.13)
Q =c, aOal x {2gh'Va/ _.ao2
= 0.6 x 78.54 x 314.10 x >/2x 981 x toOOV(314.16)2 ~ (78.54)2
'I.; rn of water = 1000 em of water
207368.38.09 68" 2R 3 68 21 Ii ' An:: 304 :: d3..<. em I~i = . tres/s, s.
Problem 6.23.An orifice meter with orifice diameter 15 em is inserted in a pipe of 30 cm diameter.Thepressure difference measured by amercury oil differential manometer on the two sides of the orifice metergives a reading of 50 em of mercury. Find the rate ojjtow of oil otsp gr. O.Q when the co-efficient of dischargeof the meter = 0.64.
Sol. Given:Dia. of orifice, do =; 15 em
:. Area,
Dia.ofpipe, .11 "'·30 cmIT ~
111 0.' 4' (30y:= 706,8:- em:. Area,
Sp. gr. of oil,Reading of Diff. manometer,
So::; 0.9x :: 50 em of mercury
Differential head, !S 1 113:' 1h ::x I zs. - 1 = 50 I ~ :: - 1 'em of oil, So' I ! P,,:! It... .: t., .I
.5 em or, ·il
The rate of the flow, Q is given equatio.. i ~3)
••••••••••••••••••••••••••••••••
3. Pitot-tUbe and vertical piezometer tube connected with a differential U-tube manometer as showninFig. 6.16.
PIEZO METERTUBE
1 t1c~..1.. d
" ~
'=- -.'jr- ''-[-ill]- -....... -~.~---:. _ '. R =,: :::_-
- ~.". ,.- ._ -._ '.~ - - -.-, ~......-...._.. '~"' __-- ---.~ ._.=- ",'~ - - -
",- -'-'-'-~ _. --._. - -.~ ._ - - -,~-~-~--.~.--
--Fig. 6.14
- - - ---.-:---- - --Fig. 6.16 '~ig.6.17
4. Pitot~static tube, which consists of two circular concentric tubes one inside the other with someannular space iIi.between as shown inFig. 6.17. The outlet of these two tubes are connected to the differentialmanom~ter where the difference of pressure head 'h' is measured by know mg the difference of the levels of
the manometer liquid say x. Then h =x [~ - 11 .o I
,)
Problem 6.2A.Apitot-static tube placed in the centre of ,J 300 mm pipe line has one orifice pointingupstream and other perpendicular to it. The mean velocity in the pipe is a.8Do/the central velocity. Find thedischarge through thepipe if the pressure difference between tne two oriiices '8 60 mm of water. Take theco-efficient ofpitot tube as C; = 0.98.
Sol. Given:Oia. of pipe,Diff. of pressure head,
d = 300 mm :..0.3U
h = 60 mm ot wale! .u6 Inot water
eli = 0.98
Mean velocity, V = 0.80 x Central velocity
Central velocity is given by equation (6.14):: C,/ ..,j2gh = 0,98 x \ ~r,<9.81 :.-.))6 ""1.063 m/s
... v,,~0.80 x 1.063 = 0.85,}4 m.s
Discharge, Q::: Area of pipe x V
1t '.- 1t ,.:::-- d2 xV;: - (.30)" < 0.8504::: 0.06 m3/s. Ans,44'
•
••• Problem 6.25.Find the velocity of the (low of an oil through a pipe, when the difference of mercuryi in.a differ""tial U-tube manometer connected 10 the ,....0 tappings of the pitot-tube is 100 mm. Take~c!fficlentofpaot-tube 0.98 and sp. gl. oj OIi c; t) i_
• Sol. Given:
•••
Diff. of-mercury level,Sp. gr. of oil,Sp. gr. of mercury,
x .CO. lUOmm ""0,1 mSo"" 1),8
jir' __ • "(I 13,6 1) 16 f oil!. \! O,g '-. = . m 0 01
• Diff. of pressure head,
• ",",r 12-'9.81 x 1.6 = 5.49 m/s. ADS.. . Velocity of flow
• Problem 6.26. A pitot-stanG tube is !ISH' '(i measul't' the velocity of water in a pipe. The stagnationiressure head is 6 m and static pressure head I,;, 11', Calculate the velocity offlow assuming the co-efficient~be equal to 0.98. (A.M.l.E., WinteI, 1979)
Sol. Given:• h, ~;.,') U!
II, ~.:; c:1Stagnation pressure head,
Static pressure head,••• Velocity of flow, \' , virgh' ""0:98 v'Ix9:81 x 1 = 4.34 m/s. ADS.
problem 6.27 .A sub-marine moves horizontally in .sea and has its axis 15m below the surface of water .• pitot -tube properly placed just in front of the sub-marine and along its axis is connected to tire two limbs of11.. II-tube containing mercury. Tire difference of mercury level is found to be 170 mm. Find the speed of tire.ab-marine knowing that the sp. gr, ofmercurv is 13.6 and that of sea-water is 1.026 with respect offresh.ater. (A.M.l.E., WinteI, 1975)
•••••
Sol. Given:Diff. of mercury levelSp. gr. ofme,rcury,Sp. gr. of sea-water,
.. n\' 13.6 1) 20834I !".. 0 i ; " -.-.'---- =. m, ~ 1.026"
{, mm " n 17 m')' j;
"' v).,15" :: v2 ;(9.sf>(2Ii834 = 6.393mls6.393 x 60 x 60 "h~ ,__ ._ km/hr = 23.01knll r. ADS.
1000•• Problem 6.ZS. A pitot -tube is insertec tn a pipe 0[300 mm diameter. The static pressure inpipe is100 mm of mercury (vacuum). The stagnation pressure at the centre of the pipe, recorded by the pitat-tube is
.0.981 N/cm2. Calculate the rate off1ow of wc!pr through pipe, iftlte mean velocity offlow is 0.85 times the• central veloc"!. Take C, = 0.98. (Converted to S.1. Units, A.M.I.E., SumrneI, 1987)
Sol. Given:
• Dia.ofpipe,d :~300 mm := 0.30 1'n
•••.. Area,
:rt ,2 1t! 3)2 0 07068 2~,.: -,.'"a ::: """'\.," =. mJ 4
••••••••••••••••••••••••••••••••
Static pressure head = 100 mm of mercury (vacuum)
= - 1~~0 x 13.6= -1.36 m of water
= .981 N/cm2 = .981 )< 104N/m24 ..'= .981 x 10 = .981~. 10"_::: 1 m
pg 1000 )< 9.81It = Stagnation pressure bead Static pressure head= 1.0 - (- 1.36):= l.n '.~6::: 236 m of water
= c,V2gh= 0.98 x V2x 9.81 x 236 ::;.:6,668 m/s
V = 0.85 x 6.668:= 5.667R m..
Stagnation pressure
. . Stagnation pressure head
.. Velocity at centre
Mean velocity,
. . Rate of flow of water '= V x area of pipe= 5.6678 x .07068 m3/s = 0.4006 013/S. Ans,
6.8. THE MOMENTUM EQUATIONIt is based on the law of conservation of momentum or on the momentum principle, which states that
the net force acting on a fluid mass is equal to the change in momentum of flow per unit time in that direction.The force acting on a fluid mass 'm' is given by the Newton's second law of motion,
F=mxawhere a is the acceleration acting in the same direction as force F
dvBut a=-dtdvF=mdt
= d(mv)dt me can be taken inside the differential}
F=~V)dt ...(6.15)
Equation (6.15) is known as the momentum principle.Equation (6.15) can be written as F.dt = d(mv) ...(6.16)
which is known as the impulse-momentum equation and states that the impulse of a force F acting on a fluidmass m in a short interval of time dt is equal to the change of momentum d(mv) inthe direction of force.
Force exerted by a flowing fluid on a Pipe- Bend
The impulse-momentum equation (6.16) is used to determme the resultant force exerted by a flowingfluid on a pipe bend.
Consider two sections (1) and (2), as shown in Fig. 6.18.'Let vl = velocity of flow at section Ift,
PI = pressure intensity at section (1).Al = area of cross-section of PII'" ,\ section (1) and
v2,P2, A2 = corresponding values of velocity, pressure and area at section (2).Let Fx and Fy be the components of Theforces exerted .''', tne flowing fluid on the bend in x-and
y-directions respectively. Then the force exerted by the bend '1' no' j I the :iirections of x and y will be
~-------------------------------------------•
••••••••••••••••••••••••••••••••••
Unit - 3
INCOMPRESSIBLE FLUID FLOW
Viscous flow - Navier - Stoke's equation (Statement only) - Shear stress, pressure
gradient relationship - laminar flow between parallel plates - Laminar flow throughcircular tubes (Hagen poiseulle's)- Hydraulic and energy gradient - flow through pipes -
Darcy - weisback's equation - pipe roughness -friction factor - Moody's diagram-minor
losses - flow through pipes in series and in parallel - power transmission - Boundary
layer flows, boundary layer thickness, boundary layer separation-drag and liftcoefficients.
:. Average velocity,
or
_ 1! - op \u=---I ..-- !S~l \ ax i
...(9.5
Dividing equation (9.4) by equation (9,5)_, .3]} '(
Umax 4"" ax--~ :': --- ..~-'---Ii t; dj
811 l_--ay iR .\
:. Ratio of maximum velocity to average velocity := 2.0.
(iii) Drop of pressure for a given Length (L) of a pipeFrom equation (9.5), we bave
.- 1 ~I - an \ R'u=>:= ~ I8f.l ax \
, "Integrating tbe above equation w.r.t. z, we get
II f18-_ dp » ~dx2 2 R
_ [PI - P2] =8; [Xl ---X2i or
=8;L- {
or
_ 8t-WL- (DI2)2
32t-WL(PI -pz) = D2 '
Fig. 9.3
where Pl - P2 is (he drop of pressure.
:. Loss of pressure bead:: £~_::_P2
pgPI - Pz . 32f.!UL____ ~. /1. _. ------.
pg -" - pf[D2
Equation (9.6) is called Hagen Poiseuille Formula.Problem '.l.A crude oil of viscosity 0.97POlSt' and relative density O.9 is flowing through a horiz
circular pipe of diameter 100 mm and of length 10m. Calculate the difference of pressure at the two eltkepipe, 'if ico kg of the oil is collected in a tank in 3(Jseconds.
. 0.97 2!-A ~: ~).97poise '"'_.-0'- ""0.097Ns/mSol. Given:
II
Relative Df-·t\Sity:. Po. or Density,Dia. of pipe,
__ ).9 x J '), oilD :;,;100 miu
L '" iU In
•••••••••••••~'•
Mass of oil collected,in time,
M:; 100 kgi t: 30 seconds
Calculate difference of pressuThe difference of pressure ~7)
1),;L/'
Now, mass of oil/sec
I viscous or laminar flow is given by
~'*2uL_ where u= average velocity = _£_n ka
HIDkg!:::. --- S30
.,= :;'10 X Q :: 900 x Q (.,' Po= 900)lUG-- :; ;)O(} x Q30 r
~ For laminar or viscous flow, the Reynolds number (Rt:)is less than 2000. Let us calculate the-Reynolds~ number for this problem,
'00 J 3,.) "" '_ X _. - = 0.0037 m Is-_ iO 900
Ii ::-L= .0037 = .0037 = 0.471 mls.Area : D2 ~ (.1)2
•~• where P = Po= 900, V = U::;;.0.471~,
Reynolds number, 8,
Ii
In, p:::: Il097.471 x 0.,1
x ------: 436.910.097As Reynolds number is less than 2000 the flow is laminar.
__p_;~ ~:2",iiL= ~?x 0.097 x;471 x 10 N/m2" p "D2 (.1)
:: 1462.28 N/m2 = 1462.28 ><.1'0-4N/cm2 = 0.146% N/cm1.. Ans.~bl.m 9.l.An oil of viscosity 0.1 Nslm' and relative density 0.'9 is flowing through a circular pipe
'Of diameter 50 mm and of length 300 1'1 Th..::rate of flow of fluid through the pipe is 3.5 lares/so Find thepressure drop in a length of 300 m and also the shear stress at the pipe wall. (Delhi University, 1987)
Sol. Given: Viscosity, r-t :::: 0 1 Ns/nr'Relative Density :: O. <;I
:. Po or density of oil:; 0,9 x 1000::: 900 kg/nr' ('.: Density of water = 1000 kglm3)
D := 50 mm = .05 m
"For derivation, please refer to jut. 12.8 i
~-~tres/s
Find (i) Pressure drop, PI .P:
(it) Shear stress at pipe wall. TO
•••••••••••••fI:.:.•••
•
(i) Pressure Drop (Pt - pz)where u = .~ = .0035 = .0035 = 1.782mls
Area : D2 ~. (.05)2
where
The Reynolds number (Re) is given by, R; =P:p = 900 kglm3, V = average velocity =u= 1.782m/s. R = 900 1.782 x .05 = 801 .~
e . X 0.1 .
As Reynold number is less than 2000, the flow is viscous or laminar32 x 0.1 x 1.782x 3000
Pi - P2 = (.05)2-= 684288 N/m2 :; 68428 x 10·-4N/cm2 = 68.43 N/cm1.. Ans.
i.e.,
(ii) Shear Stress at the pipe wall ('to)The shear stress at any radius r is given by the equation (9.1)
op r1· -::--- ax 2
. . Shear stress at pipe wall, where r = R is given by- op R
't- - _._-,-'j- ax 2.
Now
and
= 684288 N/m" :: 2280.96 N/m3300 m
R =D = .05 = .025m2 2
.025 N 1.'to = 2280.96 x2 m2 = 28.512 N/m. Ans.
Problem 9.3.A laminar flow is taking place in a plpe of diameter of 200 mm. The maximum velociiis 1.5mls. Find the mean velocity and the radius at which this occurs. Also calculate the velocity at 4 em fro.the wall of the pipe. (Delhi University, April 1981
Sol. Given: Dia.ofpipe, D = 200 mm = 0,20 m
Find (i)Mean velocity, Ii(ii)Radius at which u occurs
(iii) Velocity at 4 em from the wall.
(I)Mean Velocity, uU j
Ratio ot :ax =: 2.0 ,V :'_:;'.o 2.0u u_ 1.5u:; - = 0.75 tel». Ans.
2.0
-------~--- - - ..
••(ii) Radius at which u occursThe velocity, u, at any radius 'r' is given by (9.3)
, ap "II .. : --[R"
4fJi- axBut from equation (9.4) Uma;: IS given t-
•••••••••••••••••••
Now the radius r at whicu
;).75J.se ..
" \_._ !D.l J .-
2 1 ap 2 [ r'l]r 1 ::: _. 4!! ax R 1 - R2
...(1)
j r \." 1! -I --.-. I !
t, '!.! r \::i r (r)2]1 -l (~ii2) J = 1.5l1 - 0.1
.i: )0.1
751.50 -
1 1-2 2
(iii) Velocity at 4 em from tne 1/Hi j
.'. The velocity at i'
4 em from pipe wall is uiven ,;:,;.1' 'I:
= 1.5[ .0---::,,;
v'3::- OJ x .707 c:: .0707 m = 70,' mm. Ans,
4 ., 6.0 em :; 0.06 m
.5 f 1L
/ .06 \2 -,(-I,.1 I I\ I J
R--R = IOem
64 :::0..96 01/ s. Ans. Fig. 9.4
Problem 9.4. Crude oil 0 ~ " r'C":,:_ and relative density 0.9• flows through a 20 mm diameter \ ertical P'P'' The pressure gauges fixed 20 m apart read 58.86 N/cm2 and• 19.62N/crrf as shown in Fig. 9.5.Find the direction and rate of flow through thepipe. (AMIE, Winter 1976)
G. ~ . 1.5 0 15 N / 2Sol. iven: ;l::C. 1 :: poise :::-in=. s m•••••••l~;•
Relative Density:. Density of oilDia. of pipe,
o 1000 :!, 900 kg/m3
_ :':, mm ~,0.02 m~.::-:20 m
8.86 N/cm2 "= 58.86 x 104 N/m~(, 62 Nff:.m:;:: IQ.62 x 104 N/m2.
,J - --- -- -= -----~- ._- -
-~-.
I,I,,,I•f•if•~~•,•,~••~
L,I•l:•,•,L,,~
L~
~~. Problem 9.6.What power is requiredper kilometre of a line to overcome the viscous resistance to the• flow of glycerine through a horizontal (J/{}(:' oidiameter 100 mm at the rate of 10 litresls ? Take I-l = 8poise
and kinematic viscosity (v) = 6.0 stoke (Delhi University, 1982)
. . Rate of flow ::;average velocity x area
-' '" O.2827Iitres/s. Ans. ('.: 10-3m3 = 1litres)
'''/Problem 9.S.Afluid of viscosity O.7Ns/m2 and specific gravity 1.3 isflowing through a circular pipeof diameter 100 mm. The maximum shear stress at the pipe wall is given as 196.2 N/m2, find (a) the pressuregradient (b) the average velocity and (e) Revnold number of the flow.
Sol. Given: ,:- \ 7 Nsm"
.. Density ~J x 1000 = 1300 kg/nr'Dia. of pipe, D :" 100 mm e 0.1 mShear stress, '1;(; -::-. 196.2 N/m2
Find (l) Pressure gradient, 1;:(il)Average velocity, U(iil) Reynold number, R"
(l) Pressure gradient, *The maximum shear stress (to) is given by
ap R
dp
ap D ap 0.1or 196.2=--x-=--x-ax2 ax 4 ax 4
196.2 x 4 _ 7848 N/ 2- - m perm0.1. , Pressure Gradient 7848 N/m2 per m. Ans.(iz) Average velocity. i;
1 I 1 rJp "14~axR'" I
j
-: 50 m/s(iil) Reynold number, R,
R; :: U x D = ~ x D = P x u x D. V clip I-l
= 1300 x ~.5I) x_9_l_ = 650.00.0 '7.1
Ans,
••,~-
. "
. or~.•~•..: ; _.PI =.E~. ,~'ti1.:i.. .( ..' 1"'.,... 11.. _ . - - - - \) ."))I J pg pgl-' . '-
.,; (iv) Shear Stress Distribution. It IS obtained by substitutingthevalue of u from equation (9.9) into
~.•l'•~•~I a• In equation (9.14), a~and t are constant. Hence r varies linearly with y. The shear stress distribution
'is shown in Fig. 9.7 (b). Shear stress ismaximum wheny = °or t that at the walls of the plates. Shear stress isI zerowheny = tl2 that is at the centre line between the two plates. Max. shear stress (to) is given by
• 1 upi 1:'1; '7 -:: -. t. ...(9.15)~ .~
Problem 9.7. Calculate ((.')the ft' r, gradient along flow, (b) the average velocity, and (c) the.discharge for an oil of viscosity O.02 A slm: 'lowing between two stationary parallel plates 1 m wide maintained.. 10mm apart. ~he velocity midway between the olates is 2 mts (Delhi University, 1982)~ Sol. Given:.. Viscosity, 1.1. = .02 N s/nr'
Width, b = ] mDistance between plates, t = 10 mm = ,01 m
..••••••
(iiI)Drop oCPressure head for a given Length. From equation (9.11), we have
1 ap 2 ap 12I-'Uu;;:: --'--1 or -=---121.1. ax ax j2
Integrating this equation w.r.t. x, we gel
CD ®« <' , I {, , , , , , . (. , ( ~,"" " .
If hfis the drop of pressure head, Iher,xz--..,
Fig. 9.8
a a [ 1 iJp 1 [ 1 ap 1t ~.:Il· - u. :: U - - - -. (ty - y2) = I-' - - - (t - 2y)d.Y ay 21-' ax 21-' ax1 dp _
'T;-::.- - - [t-· 2v;2. ax '.; } ...(9.14)
Velocity midway between the plates,
(I) Pressure gradient ( : )
lJmax = 2 m/s.
Using equation (9.10), 1 dp 2 1 (dP) 2Uma:( = - 8J.ldx t or '2.0 = - 8 x .02 dx (.01)
dp 2.0 x 8 x .02 3200 NI 2 A- = - = - m per m. os.dx .01 >< .01
~ ~~,_,,_;;..- - -- T -- --- - - <- -~~'"~~"''''-' __ '''''~~''''''_- .
•••••••••••••••••••••••••••••••••
(il)Average velocity (u )Using equation (9.12),
Umax _ ~U - 2
Ii~.::3_,~~,~ ~-~.-:.?_"..~ ;3 ru/s. '\us.
(iii) Discharge (Q) C" Area of flow •...:U ,,-.:, . U = 1 x .\)1 x 1.33 = .0133 m3/s. Ans.Problem 9.8.Determine (a) the pressure gradient, (b) the shear stress at the two horizontal parallel
plates and (c) the discharge per metre width for the laminar flow of oil with a maximum velocity of 2 mlsbetween two horizontal parallel fixed plates which are 100 mm apart. Given I-! = 2.4525N slTrf.
Sol. Given:Umax ::::2 m/s, . :~100 mm :: i) lm, !!= 2.4525 N s/m2dp
(z) Pressure gradient, dx
(iz) Shear stress at the wall, 'to.(iii) Discharge per metre width, Q.
dp(z) Pressure gradient, dx
Find
Maximum Velocity, UmQX) is given by equation (9.10,'fJ _ .. J:.. dp ,2, max _ - 8"" a.x '
Substituting the values
or 2.0 == ._ ._. __ 1_....8)< 2.4525
ap 0' 0 x S'x_.::: _'::':_ •.._-' 0:: -3924 N/m2 perm. Ans.ax 1x .1
(ii) Shear stress at the wall, 'to
'to is given equation (9.15) as1 iJp
'to = -"2 ax x t = - t (:""'"3924»)( 0.1 = 196.2 N/ml. Ans.
(iiz)Discharge per metre width, Q= Mean velocity x Area
= ~ Uma."t X (t X .n = *)<; 2.0 x 0.1 x 1 = 0.133 m3/s. Ans.
Problem 9.9.An oil of viscosity 10 poise flows between two parallel fixed plates which are kept at adistance of 50 mm apart. Find the rate of flow of oil between the plates if the drop of pressure in a length of1.2m be 0.3 Nlcm', Thewidth of the plates is 200 mm.
;Sol.Given:1J..::::: 10 poise
10.. 2 . .z: 1C N s/m '" 1 )~ ",;nt"
""5(,mm :::::OJb ]1·
( 1 . 1 N&).: pOISe - 10 m2
0 '1 'N" '2PI --P2:: .."] 1m::
Width,
L:: 1.20 mB ,,;200 mm= 0.2(i tu
•••I'•••••••••
Find Q, rate of flowThe difference of pressure is giv.
Substituting the values <;lie; E "
U.3 ,>. l'
. . Rate of flow
l l"i ," 0 U5 05; 2____l::I_. J(. 1.. x. x. _ 0 52 m/s12 x 1.20 - .
it Area = 0.52 x (B x t)- C. ':;2 x 9.20 x .05 m3/s = .0052 m3/so 0052 x 103 litre/s = 5.2litrels. Ans.
• ~roblem 9.10..Water at l~oC /101,vs between two large p~rallel plates at a distance of 1.6mm apart.; Determine (a) the maxunum velocity (b) tne pressure drop per unu length and (c) the shear stress at the walls• of theplates if the average velocity is 0.2 mis. The viscosity of water at 15°C is given as 0.01 poise.~ Sol.Given: ':: 1.6mm:: 1,6 X 10-3 m• 0.11016m
,. fJ m/sec;;t::, .0] poise = .~~ = 0.001 N s/m2• (z) Maximum Velocity~Um!]," IS !21\'C"n equation (9,12\
.. i.e., j.5 ' 0.2 '" 0.3m/s, Ans.!;.'•~,•~, or
•ap 01 02
" or 1:.' r. -=-1-0x .._-' -2 = 937.44 N/m2 per m.;,; (.0016)L (iiz) Shear stress at the walls is given equation (9.15)
"
.. 1 ap . 2T(J ;; -,,)' -. x t = ~ x 937,44 x .0016 = 0.749N/m. Ans., -~
. Problem 9.11.There is a horizontal crack 40mm wide and 2.5 mm deep in a wall of thickness 100 mm., .Waterleaks through the crack. Find the rate of leakage of water through the crack if the difference ofpressure
between the two ends of the crack is ·],02043 N!cm2. Take the viscosity of water equal to 0.01 poise., Sol.Given:, ~~;::::;:~~:
Length of crack,
(iz)Thepressure drop, (p- -P;;) given by equation (9.13)
pressure drop per unit length
.rm ;,. ,J025 Inl!'\) mm = J.t m ,
PI ---P2 •.::0.02943 N/cm2 = 0.02943 x 104 N/m2 = 294.3 N/m2. .01 N s
k r-: .01 poise = 10 m2
••••••••••••••I"i•~i'\!••H••
•
Find rate of leakage (Q)(PI - P2) is given by equation (9.13) as
12~ .01 U x 0.1PI -P2 = P or 294.3 = 12 x 10 x (.0025 x .0025)
U = 294.3 x 10 x .0025 x .0025 = 1.5328 m/s12x.OlxO.l
=u x area of cross-section of crack
= 1.538 x (b x t)= 1.538 x .04 x .0025 m3/s = 1.538 x 10-4m3/s= 1.538 x 10-4x 103litre/s = 0.1538 litre/so Ans.
problem 9.12. The radial clearance between a hydraulic plunger and the cylinder walls is 0.1 mm ;the length of the plunger is 300 mm and diameter 100 mm. Find the velocity of leakage and rate of leakagepast the plunger at an instant when the difference of the pressure between the two ends of the plunger is 9m
of water. Take lA. = 0.0127 poise.Sol. Given:
Rate of leakage
The flow through the clearance area will be the same as the flow between two parallel surfaces,t = 0.1 rom :: 0.0001 mL = 300 rom = 0.3 m
Diameter, D = 100 rom = 0.1 m
Difference of pressurePi -P2= :c 9mof water
pgPI - P2 = 9 x 1000 x 9.81 N/m2 = 88290 N/m2
.0127 N s= ---ul m2Viscosity, JA. = ~0127poise
Find (i)Velocity ofleakage, i.e., mean velocity Ii(ii) 'Rate of leakage, Q
(i)Velocity ofleakage (uj:-The average velocity (Ii) is given by equation (9.11)
_. 1 ap_"u = ---- <t"12~ ax1 P - n-
= -'---.61'2.7 X_l T~_!' .( (OOOl) x (.0001)12 x ----
10
10 88290. 0001' (00 )= 12 x .0127 x 0.3 "x (. ) x . 01
= .193 m/s = 19.3 cm/s. Ans.
(ii) Rate ofleakage, QQ = u x area of flow
= 0.193 x rrD x t m3;s = 0.193 x 3t x .1 x .0001 m3/s= 6.06 x 10-6m3;s = 6.06 x 1O~ x l03litre/s
:: 6.06x 1o-3lltre/s. Ans.
_.------------------------ ------- - - - ------
•••
L = length of pipe,V = mean velocity of flow,d = diameter of pipe.
•where hI = loss of head due to friction,•
• (b) Chezy's Formula for loss of head due to friction in pipes. Refer to chapter 10 article 10.3.1 inwhich expression for loss of heao due o frrction in pipes 1& derived. The equation (iil) of article 10.3.1, is• ' t:ht' , ~'x L »: ~,,: ... (11.2)
pg A.
P :::wetted perimeter of pipe,L ::;:Length of pipe,A = area of cross-section of pipe,
V = mean velocity of flow.
• N th . fA ( Area of flOW')' II h d I' d th h d l' di d .ow e ratio 0 p = Peri . ( d)- .. IS ca eo v rau ICmean ep or y rau IC ra IUS an ISenmeter wette ., .
• noted by m.
:. Hydraulic mean depth,
:It ,.
A 4d-< dm=-=--':-p :ltd 4••••••
Substituting ~ = m or ~ =!in equation (11.2), we get
h ::::L~x .. x F' or V'2 - II x pg x m x !-pg x m x !!t.f pg - f t' L - f' L
V = - IPf o~· '~~ .- mY/m '!ty~ L V~ ,. L ...(11.3)
• Let yPf = C, where C is a constant known as Chezys constant and t = i, where i is loss of head.r unit length of pipe.
• Substituting the values OfyPf andV¥ in equation (11.3), we get
• y"" C Yin[ ...(11.4)• Equation (11.4) is known as Chezy s formula. Thus the loss of head due to friction in pipe from Chezy' sformula can be obtained if the velocity of flow through pipe and also the value of C is known. The value of m.r pipe is always equal to d/4.• Problem 11.1.Find the head lost due tofriction in apipe of diameter 300 mm and length 50 m, throughwhich water is flowing at a velocity of 3 mls using (i) Darcy formula, (ii) Chezy's formula for which C = 60.• Take v for water = 0.01 stoke.
• Sol. Given:
•••••
Dia.ofpipe,Length of pipe,Velocity of flow,Chezy's constant,Kinematic viscosity,
t 300 mm ',,,0.30 mSO In
V:;:::3 m/sC = 60y :: 0.01 stoke = 0.01 cm2/s.".0.01 x 10" 4 m2;s.
~ - " ---~. ~ ~"' """.""'_~""'~=;o_~- -Jr.-iF-jlP1'". - -,. " .~ .
•••••••••••••••••••••••••••••••••
(I)Darcy Formula is given by equation (11.1) as
4. f. L . V2h :::..._:_L ..__ ._.
1 d x 2g
where 'f' = co-etfictent of friction is a function of Reynold nun» 't1. e'
But R, is given by
:. Value of
.. Head lost,
R ill< d 3:J x: 1).30 . ....J:;e ~ . ---~:: . ::::y >< } ~l' .01 x 10-4
f = 0.079 = .079 :::.00256s,114 (9 x 105)114
II __4 X .00256)( 50 x 32{'I 0.3 x 2Jl x 9.81 .. -7828 m. ADs.
(it) Chezy's Formula. Using equation (11.4)
v= CYmld 0.30where C = 60, m = - = - = 0.075 m4 4
But
.} \ 1
'" :)0 1 ,. (PS :::0.0333or
Equating the two values of i,we have it5._::, 03T"50 . -.
ht~: SC!i< 0333·· 1.665 Hit. Ans.
Problem 11.2. Find the diameter of a pipe of lengtn 2000 m when the rate of flow of water throughthe pipe is 200 litresls and the head lost due to friction ;s4 m. Take the value GfC = 50 in Chez"I"sformulae.
Sol. Given:
Length of pipe, L = 2000 m
Chezy's formula is given by equation (11.4) as v -::(' vm[Substituting the values of V,m, i and C, we get
0.2 ~ 4 = 50 ~2 or Vfj_4x .(j02xd 4 .
Discharge,Head lost due to friction,Value of Chezy's constant,Let the diameter of pipe
Velocity of flow,
Hydraulic mean depth,
Loss of head per unit length,
Q = 200 litre/s = 'J 2 m3/shf=4mC=50
::::.dDl'scharge Q' un .~. (", aV = ( ::: __ . :::__ =_ ~ '!_':'~ :_.
Area :. J2 ~ d' xd:4- 4
dm:::-4
,h( c)I = --L = '"L 2000
0.2 x 4 .00509'1UPx50=~-
7'•• . .1.,_J • . . •
• ' 1'1;
•• Squaring both sides,~ 002 _ .005092 - .0000259 .5 _ 4 x .0000259 - 005184 x . - cf - cf or a- - .002 -'.
d :; ~0.0518 :: (.0518)1/5 = 0.553 m = 553mm. Ans.
Problem 11.3.A crude oil of kinematic viscosity 0.4 stoke is flowing through apipe of diameter 300 mm.. the rate of 300 litres per sec. Find the head lost due to friction for a length of 50 m of the pipe.
Sol. Given:
•
. . Reynold number,
" _ 0.4 stoke = 0.4 cm2/s = .4 x 10- 4m2/sd::: 300 mm = 0.30 mQ :::300 litres/s = 0.3 m3/sL == 50 m
V :: ~ = -.-~~ = 4.24 m/sArea :rt:O 3)24 \ .
R :: V x d = 4.24 x 0.30 = 3.18 x 104e v 0.4 x 10-4
••••••••••
Kinematic vjscosity,Dia.ofpipe,
.' Discharge,Length of pipe,
Velocity of flow,
As Re lies between 4000 and 100,000, the value ofjis given byf - .079 _ .079 - 00591- (Re)1I4 - (3.18 x 104)1/4 - .
. . Head lost due to friction,h, c::. 4 .f.L ' y2 = 4 x .00591 x 50 x 4.242 = 3.61 m. .t d x 2g 0.3 x 2 x 9.81 Ans.
Problem 1l.4.An oil of sp gr 7!0' ilowing through a pipe of diameter 300 mm at the rate ofttoo litresls.Find the head lost due If) Iriaion andpower required tomaintain theflow for a length ofwoom. Take
• = .29 stokes.Sol. Given:
. . Head lost due to friction,
S::: 0.7d :: 300 mm= 0.3 mQ = 500 litres/s = 0.5 m3/sL:: 1000 illV == _lL_ = 0.5 = 0.5 x 4 = 7.073 m/s
Area . ~ d2 :rt x 0.324
R = Vx d = 7.073 x 0.3 = 7.316 x (10)4e v 0.29 x 10-4
f= .079 = 0.79. = 0048R/14 (7.316 x 104)114 .
h,= 4 x f~ Lx V2 4 x .0048 x 1000 x 7.0732 = 163.18 illd x 2g 0.3 x 2 x 9.81
pg.Q.hf,-;: 1000 KW
••••••••••
Sp. gr. of oil,Dia. of pipe,Discharge,Length of pipe,
Velocity,
.. Reynold number,
. . Co-efficient of friction,
Power required
.where p = density of oil = 0.7 x 1000 == 700 kg/m3
• :. Power required " 700 x Y.8\~g~x 163.18 = 560.28 kW. Ans,
•
•,.••••..••..,.•••..•..••.,•••••••••..•••
Problem 11.5. Calculate the discharge through a pipe of diameter 200 mm when the difference ofpressure head between the two ends of apipe 500 m apart is 4m of water. Take the value of T = 0.009 in theF. I Ir. 4. f· L . V2
. jormu a '7 = d 2x gSol. Given:Dia. of pipe,Length of pipe,
ht= 4 m of waterf= .009
h = 4 xfxL X V2f dx 2g
4 0 = 4 x .009 x 50q2___,. or V2 = 4.0 x 0.2 x 2 x 9.81 = 0.872. 0.2 x 2 x 9.81 4.0 x .009 x 500V = YO.872 = 0.9338:::. 0.934 m/sQ = velocity x area
= 0.934 x :Tt d2 = 0.934 x :'!. (0.2)24 4
= 0.0293 m3/s = 29.3 litre sis. Ans.
Problem 11.6.Water is flowing through apipe of diameter 200 mm with a velocity of3 mls. Find the.
head lost due to friction for a length of 5m if the co-efficient of.friction is given by f = .002 +~0:'3' where R,. e
Difference of pressure head,
Using equation (11.1), we have
or'
. . Discharge,
d = 200 mm = 0.20 mL = SOOm
is Reynold number. The kinematic viscosity of water = .01 stoke.Sol. Given:Dia. of pipe,Velocity,Length,Kinematic Viscosity,
:. Reynold number,
Value of
. . Head lost due to friction,
d = 200 mm = 0.20 rnV = 3 m/sL=5mv = 0.01 stoke = .01 x 10-4m2js
R, = V x ~ = 3 x 0 }O_ ~ 6 x 105'i 01" n'"
,_ n2' 0.9 _ I ;'J . __ .O~__ 02 0.9 _J -.U T 03 - . ~ +': . 5 3 -. + .Re " (6 x to r 54.13= .02 + .00166 = 0.021-66
It = 4 x fx Lx V2 = 4.0 x .02166 x 5.0 x 32f d x 2g 0.20 x 2.0 x 9.81= 0.993 m of water. ADS.
Probleml1.7.An oil of sp. gr. 0.9 and viscosity 0.06 poise is flowing through apipe of diameter200 mm at tht!f.ate of 60 litresls. Find the head lost due to friction for a 500 m length of pipe. Find the powerrequired to maintain this flow.
Sol. Given:Sp. gr. of oil = 0.9
Viscosity, O06 ' 0.06 N / ?fA. =. pOIse = 1c· s.m .
•••. I
of pipe,~ischarge,Length,
·Density,
•••Reynold number,
Q = 60 litres/s = 0.06 m3;sL::: 500 mp =,0.9 x 1000 = 90Gkg/m"
R = pV d = 900 V x 0.2e f.t x 0.06
10
. . Power required•= 9.48 m of water. Ans.
:;: pg.Q.h[ = 900 x 9.81 x 0.06 x 9.48 = II. 0'" kW1000 1000 ~. *' • Ans.
Re " 900 ,. 1.91 x O.~x 10 = 573000.06
• As R, lies between 4bOO and 105, the value 01 co-efficient of friction, f is given by
• ;' _ 0.079 _ 0.079 _ 0051J - R/.25 - (57300)0.25- .
•
.:.h :: 4 xix Lx V2 = 4 x .0051 x.500 x 1.912f d x 2g 0.2 x 2 x 9.81. . Head lost due to friction,••
WIINOR ENERGY (HEAD) LOSS ESThe loss of head or energy due to friction in a pipe is known as major loss while the loss of energy due
:~e of velocity of the following fluid in magnitude or direction is called minor loss of energy. The minorjf energy (or head) includes the following cases: .
1.Loss of head due to sudden enlargement,• 2. Loss of head due to sudden contraction,• 3. Loss of head at the entrance to a pipe
4. Loss of head at the exit of .1 pit»:
• 5. Loss of head due to an obstruction j;
• 6. Loss of head due to bend in the pipe7. Loss of head in various pipe fittings
) ·,t
• In case of long pipe the above losses are small as compared with the loss of head due to friction andIe they are called minor losses and even may be neglected without serious error. But in case of a short pipe,ese losses are comparable with the loss of head due to friction.• 11.4.1. Loss of head due to sudden enlargement. Consider a liquid flowing through a pipe which hasijen enlargement as shown in Fig. 11.1. Consider-two sections (1)-(1) and (2)-(2) before and after the
llargement.
• Let•PI = pressure intensity at section 1-1,
Vl = velocity of flow at section 1-1,
Al = area of pipe at section 1-1,
P2, V2 andA; = corresponding values at section 2-2.
,
Then he becomes askvl vlh=-=0.375-
c 2g 2gIf the value of C,is not given then the head Joss due to contraction is taken as
y,2 y,2= 0.5 -2._ or he = 0.5 22-. ...(11.7)
2g gProblem 11.8.Find the loss of head when apipe of diameter 200 mm is suddenly enlarged to a diameter
of 400 mm. The rate offlow ofwater through the pipe is 250 litresls,Sol.Given:Dia. of smaller pipe, Dr = 200 mm ::::n.20 m
:. Area, :ur:;,141 m2
Dia. of large pipe, D2 = 400 mm = C,A HI
:. Area,....... .-
A2 = ~ x (0:4f :: 0.12564 mol
Q = 250 litres/s = 0.25 m3/s
V Q 0.25 '7 9' /1 = At = .03141 = . 6m S
lT2 = g_ = 0.25 9Q IY' Al .12564 = 1. ,m s
Discharge,
Velocity,
Velocity,
Loss of head due to enlargement is given by equation (11.5) as
h - (VI - V2)2 _ (7.96 -:._!:_?~~-1816 f te Ae - 2g - , 2g -. m 0 wa r. ns.
Problem 11.9. At a sudden enlargement of a water main from 240 mm to 480 mm diameter, thehydraulic gradient rises by 10 mm. Estimate the rate offlow. (A.M.LE., Summer, 1977)
Sol. Given:Dia. of smaller pipe, D, = 240 nun ;;0:: 0.24 m
it, " J! ..Ai;:: ;,D/= '4 \24(D2 = 480 mm » 0.481'll
A2 = 1(0.48)2
:. Area,
Dia. of large pipe,
:. Area,
Rise of hydraulic gradient', l.e., (Z2 + ::; ) - (=~+z, j= 10 rum = l~O = 1~ m
Let the rate of flow =QApplying Bernoulli's equation to both sections, i.e. smaller pipe section, and large pipe section.
2 2'Pl VI P2 V2- + -2 + z, = -. + -2 + z" + Head loss due to enlargement ...(i'pg g ., pg g .. "
But head loss due to enlargement,
...(it)
----------------- ...._---,,---_._-_._._----,-------·Please referArt. 11.5.1.
•••••••••••••••••••••••;.••••••If'.~,•~
Substituting this value of Vc in equation (iii), we get2
( AxV _il(V _ V2) \ Cc(A - a) .
Head loss due to obstruction =_c .:._:::_' -----.2g 2g
...(11.10)
11.4.6. Loss of Head due to Bend in pipe. When there ,8 a ny bend in a pipe, the velocity of flowchanges, due to which the separation of the flow from the boundary and also formation of eddies takes place.Thus the energy is lost. Loss of head in pipe due to bend is expressed as
kV2h ---b - 2g
where hb = loss of head due to bend, V = velocity of flow, k = co-efficient of bendThe value of k depends on,(I) Angle of bend, (it) Radius of curvature of bend, (iii) Diameter of pipe.
11.4.7. Loss of Head in Various Pipe Fittings. The loss of head in the various pipe fittings such asvalves, couplings etc. is expressed as
ky2- 2g ...(11.11)
where V = velocity of flow, k = co-efficient of pipe fitting.
Problem 11.1S. Water isflowing through a horizontal pipe of diameter 200 mm at a velocity of 3 mls.A circular solid plate of diameter 150 mm isplaced in the pipe to obstruct the flow. Find the loss of head dueto obstruction in the pipe ifC, = 0.62.
Sol. Given:Dia. of pipe,Velocity,
Area of pipe,
Dia. of obstruction,
D = 20(} mm > O,lL n'V :::3.0 m/s
A = ~ D2 = : (0.2)2::; 0.03141 m2
d = 150 mm = 0,15 m
a = ~ (.15)2 = 0.01767m2
c,= 0.62The head lost due to obstruction is given by equation (11.10) as
V2 (A ,\~~= 2g Cc(A_ a) ,~1.0 )
3 x 3 [.03141 ]2:::2 x 9.81 0.62[.03141_.. fl1767] - 1.0
= 2;~.81[3.687-- 1.0]2= ~l311m. Ans.
:. Area of obstruction,
Problem 11.16. Determine the rate of flow of water through a pipe of diameter 20 em and length 50 mwhen one end of the pipe is connected to a tank and other end of the pipe is open to the atmosphere. The pipeis horizontal and the height ofwater in the tank is 4 ;11 above the centre of the pipe. Consider all minor losses
4 f L '.12and take f = .009 in the formula ht = .d ~ 2~
••• > I
. -v -• Sol.Dia. of pipe,• Length of pipe,
Height of water,• Co-efficient of friction,
d = 20 em = 0.20 mL "'"50 mf-I :::.4m
• Let th~ velocity of~ater in :ipe. V ~u/s', . .Applying Bernoullis equation at the top of the water surface IIIthe tank and at the outlet of prpe, we
,.Taking point 1 on the top and point 2 at the outlet of pipe].
•P1 V12 P2 vl -- + - + z1 = - + - + Z2 + all lossespg 2g pg 2g
• Considering datum line passing through theIiof pipe• 0 + 0 + 4.0 = 0 + ~~ + 0 + (hi + hi)
vl4.0 = 2g + hi + hf
rWATER SURFACE--:::.=--==-F -
•• But the velocity in pipe = V, . . V = 'V:;
V2 .4.0 = 2g + hi + It,
4m
L._._.__----.E..r--~_:_:_----..!?~~,los 50m ~'...
d-2Qcm
Fig. 11.4
• ...(i)
• l~From equation (11.8), hi ,= D,5 <. and "i'll, ,n equation (11.1) is given as
• -s•••••••
Substituting these values, we have\/2 0.5 V2. 4 x f. L . V24.U::: -- + .-....---- + ---',---g 2g u x 2g
V2 r " ( 4 x .009 x 50 1 V2 . V2=: 2g t 1.0 + 0.5 + - 0.2 = 2g [1.0 + 0.5 + 9.0] = 10.5 x 2g
V ='" /"4 x 2 x 9.81 .= 2 '734 /V 10.5 '. m sec
.. Rate of flow, Q : A x V = ~ x (0.2i x 2.734 = 0.08589 m3/s = 85.89Iitres/s. Ans,
• Problem 11.17.A horizontal ptpe line 40 m long is connected to a water tank at one end and dischargesf.ly into the atmosphere at the other end. For the first 25 m of its length from the tank, the pipe is 150 mmrilifneter and its diameter is suddenly enlarged to 300 mm. The height of water level in the tank is sm abovertf centre of the pipe. Considering all losses of head which occur, determine the rate offlow. Take f = .01 for.h sections of the pipe. (Osmania University, 1992; A.M.LE., Summer, 1978)
Sol. Given:Total length of pipe,Length of 1st pipe,Dia. of 1st pipe,Length of 2nd pipe,
• L = 40mL1::: 25 m••d) = 150 mm = 0.15 mL:~:: 40 - 25 = 15 m
•
--
i'.... . "':i ",'• ,'j> 1 ' .
Dia. of 2nd pipe,Height of water,Co-efficient of friction,
d2 = 300 nun = 0.3 mH=8mi= 0.01
Applying Bernoulli's theorem to the free surface of water in the tank and outlet of pipe as shown inFig. 11.S and taking reference line passing through the centre of pipe.
am
l_._-~~!-LL,:25m
d,aO.15m
or
Fig. 11.5V 2
(' 0 8 P2. 2 " t} 11 + +. = - + - ...'J +;}J ossespg 2g
V280 0 2. ' 1. ' •. ',' + -2 . .,. tti + rl-f + i'le + Ittg .1 2
...(i)
wherev2.
hi = loss of head at entrance = 0.5 _12g... 4xfxL1XV12
h/1 = head lost due to friction Inpipe 1 = . -d1 x 2g. . 2.(V1 - v,.,\
he = loss head due to sudden enlargement = "',2g. 4xfx12xV,2
hh =Head lost due to fnetion in pipe 2 ::;.:; 2d2x 2g
But from continuity equation, we haveA1V1 =A2V2
~ : .A 'T - d2« ).. V" 'd 2. .,V1 =2~.::_4__ ._~;: I. _l \ X V2 = (0.1.35)'"x V2 = 4V2
AI ~ , I d I~-a, '4
Substituting the value of V1in different head losses; we haveO.S V12. 0.5 x (4V2)2 8vl
hi=---= -=--2g 2g 2g4 x 0.01 x 25.x (4Vl)
hf -1- 0.15x2xg
= 4 x .01 x 25 x~x vl = 106.67 V22.0.15 2g 2g
•. 2 2 y.2
4 x .01 x 15 x V2 4 x .01 x 15 V2 2!tt2 = 0.3 x 2g . = . 0.3 x 2g = 2.0 x 2g
t.bstituting the values of these losses in equation (i),we get• vi 8vi vi 9vl vi8.0 =-+ -+ 106.67 -+ -+ 2 x-• 4 ~ ~ ~ ~
V 2 \1.2::;: -;;,?:_ [t + 8 + 106.67 + 9 + 2] = 126.67 22
....g
~'f~""EI = V'§>·O x 2 x 9.S[ =v'i.2391 = 1.113mi.126.67 126.67
••••• Rate of flow,Q ~-:;.A2X V:l:: .•~ (0.3)2 x Ll13 :: 0.07867 m3/s::: 78.67Utres/s. Ans.
r/:"blem 11.18.Determine the difference in the elevatio,,, between the water surfaces in the two tanks~ ure connected by a horizontal pipe of diameter 300 mm and length 400 m. The rate of flow of water1ithe pipe is300 litres/s. Consider all losses and take the value off = .OOB.
Sol. Given:·Dia.ofpipe,• Length,
d:: 30e mm= 0.30 m
Discharge,• Co-efficient of friction;
• Velocity,
•
L lC 400 mQ :: 300 litis = 0.3 m3/st=0.008
v -::_Q__- ::: 0.3 = 4.244 m/sArea ~ x (.3)2
• Let the two tanks are connected by a prpe as shown in Fig. 11.6.
•••••
! iH. I L=4OCm -I L._. ~ d:=O.3C'" I HzL--===~__ ._"V . _._- _- ~
• H2 = height of water in 2nd tank above the centre of pipe
Fig. 11.6
H1= height of water in 18t tank above the centre of pipeLet
•••Then difference in elevations between water surfaces= H1 -H2Applying Bernoulli's equation to the free surface of water in the two tanks, we have
H1 =H2 + lasses=H2 +hi+Hfl +b«
V2 OS x 42442hi = Loss of head at entrance= 0.5 2g = .2 x 9'.81 = 0.459 m
_ . . ,_~_~:1x L xr._4 x .008 x 400 x 4.244~-hIt _ Loss of head due to fnctlO.l _. d.' .., ~ 03 2 981 - 39.16 m:>.... g . X X •
...(i)
~re
•••
Ja~---------------------••••
...(11.14)
Problem 11.30. The difference in water surface levels in two tanks, which are connected by three pipes,.ries of lengths 300 m, 170 m and 210m and of diameters 300 mm, 200 mm and 400 mm respectively, is1r'Determine the rate of flow of water if co-efficient of friction are .005, .0052 and .0048 respectively,ansidering: (i) minor losses also (ii) neglecting minor losses (Delhi University, 1987)
• Sol. Given:.Difference of water level,Length of pipe 1,Length of pipe 2, L2 == 170 m and dia., d2 = 200 mm = 0.2 mLength of pipe 3, L3 = 210 m and dia., d) = 409 mm = 0.4 mAlso, f1 = .005,12 = .0052 andh = .0048
• (i) Considering Minor Losses, Let V}, V2 and V~ are the velocities in the Ist, 2nd and 3rd pipe
'wectively.From continuity, we have AiVI = A2 V2 = A3 V3•
•••L, == 300 m and dia., d, =300 mm = 0.3 m
••..••••••••••••••
Now using equation (1L 12), "';~haveO - V 2 4(1 L,2 0 5V' ? 4fL V 2 (V 1/ )2 4f'-1 V 2 2.J l _1 'j !' • 2 n 2 2 2 - V 3 3'-'3 3 V3H = ----- + --- --_.,_,+ --,- + ---- + + + -
2g !J. '" -: 2g d2 x 2g 2g d3 2g 2g
Substituting V2 and V3,~ U 5V,~ 4 x .Oe5 x 300 x V12 0.5 x (2.25 V12)
1.L.0::: --~---- + --- + '2g 0.3 x 2g 2g
4'O(}52 7 (2.25 Vl)2 (2.25 Vl - .562 Vl)2+ )( > x lOx + -----=-----=,;,_
0.2 x 2g 2g4 x .0048 x 210 x (.5625 vli (.5625 Vl)2+ -_- + -,- __ ;;.;.__
0.4 x 2g 2g
.. Rate of now,
V1'"12.0 = 2g [0.5 + 20.0 + 2.53 + 89.505 + 2.847 + 3.189 + 0.316]
v2=+ [118.8871L-g
V :::'" /12 x 2 ~ 9.81 = 14(n /1 V 118.887 . m sQ :;;:Area x Velocity =Al x V,
It , ,2 T Jt ·2 3:: -i' lOl) X" 1= ~f(.3) x 1.407 = 0.09945 m /s
:=- 99.45lires/s. Ans.
or
(ii) Neglecting MInor LOsses. Using equation (11.13), we have
4fJ!..lVl2 4fiL2 V22 4/3L3vlH = + + ~_;;__;;_dlx2g d-zx2g d3x2g
12.0 = Vl2 [ 4 x .005 x 300 + 4 x .0052 x 170 x (2.25l + 4 x .0048'x210 x (.5625f].2g 0.3 0.2 0.4
V? ~2= 2g (20.0 + 89.505 + 3.189] = 2g x 112.694
V = ... /2 x 9.81 x 12.0 = 1 445 m/1 V 112.694 . s
. . Discharge, Q = Vl XAl = 1.445 x : (.3)2 = 0.1021 m3/s = 102.1 Htres/s. ADs.
Problem 11.30.A. Threepipes of 400mm, 200 mm and 300mmdiametershave lengths of 400 m; 200 m;and 300 m respectively. They are connected in series to make a compound pipe. The ends of this compoundpipe are connected with two tanh whose difference of water levels is 16m. IIco-efficient of friction for thesepipes is same and equal to 0.005, determine the discharge through the compound pipe neglecting first theminor losses and then including them. (A.M.I.B.,Summer, 1990)
Sol. Given:Difference of water levels, H =16 mLength and dia. of pipe 1, L1 = 400 m and dl = 400 rom = 0.4 mLength and dia. of pipe 2, L2 ::i: 200 m and d-z= 200 mm = 0.2 mLength and dia. of pipe 3, L3 = 300 m and d3 = 300 mm = 0.3 mAlso h =12=/: :,.-0.005(i) Discharge through the compound pipe first neglecting minor losses.Let Vl, Vz and V3are the velocities in the tst, 2nd and 3rd pipe respectively.
From continuity, we have AlVl =AZV2 =A3V3
3t . 2A V 4 cil d 2 ( 0 4, )2 0
V2= Ai 1=_·-x Vi = .2....2 V1 = -7" Vi = 4Vi2 3t i 2 d') 0.2
-12 . "4 .
and
n: 2AiV1 4 d, U12.. l! U.4)2 .V3= -2- = _ .....- x Vl =2 V1:: .- Vi = 1.77V,A3 :It 0,32 d3 . 0.2 -
4 " . .
Now using equation (11.13), we have
H = 4flLlVl2 + 4hJ..2V22+ 4/~3V32dl x 2g d-zx 2g d3 x 2g
14 x 0.005 x 400 x V12 4 x 0.005 x 200 x (4'Vi)2 4 x 0.005 x 300 26= . . + ._ + . . . x (1.77 Vi)
0.4 x 2 x 9.81 0.2 x 2 x 9.81. 0.3 x 2 x 9.-81. zVl (-4 x 0.005 x 400 4 x 0.005 x 200 x 16 4 x 0.005 x 300 x 3.157 )
= + +2 x 9.$1 0.4 0.2 0.3
or
•...---------- ..--~------~,-..•• j : I,' /, ,,~ ,,' ~ wu , "~'; I '
'. ~'if I ,(, f ~=~ ~'if" 'ii, I ~'"I"~ , 't;, ~ ,.. ~ "-, ,~
• Equation (11.17) is known as Dupuit's equation. In this equationL =L1 +L2 +L3 and dhd2 and d3 areknown. Hence the equivalent size of the pipe, i.e. value of d can be obtained.• Problem 11.31. Threepipes of lengths 800 m, 500 m and 400 m and o/diameters 500 mm, 400 mm
.nd 300 mm respectively are connected in series. Thesepipes are to be replaced by a single pipe.of length1700m. Find the diameter of the singlepipe.•
.11.9. FLOW THROUGH PARAI LEI PIPESConsider a main pipe which divides into two or more branches as shown inFig. 11.17 and again join
.together downstream to form a single pipe, then the branch pipes are said to be connected in parallel. Thedischarge through the main is increased by connecting pipes in parallel.• • BRANOi PIPE2
• Lz,dz,Vz
••
••••••or•••
••
Sol. Given:Length of pipe 1, L1 = 800 m and dia., d1 = 500 rom = 0.5 mLength of pipe 2, L2 = 500 m and dia., d2 = 400 rom = 0.4 m
Length of pipe 3, L3 = 400 m and dia., d3 = 300 mm = 0.3 m
Length of single pipe, L '::1700 m
Let the diameter of equivalent single pipe = d;. z., L2 L3
Applying equation (11 :<: "~ ;S + ------:f" + ~:,:- d1 d( d3
:700 = 800 + 500 + 400 = 25600 + 48828.125 + 164609= 239037d"~ .55 .45 0.35
'i 1700d :::239037 ::: ,007118
d > (.007188)(12= 0.3718 = 371.8 mm. Ans.
BRANCHPIPE1
Fig. 11.17
• The rate of flow in the main pipe IS equal to the sum of rate of flow through branch pipes. Hence from
• Fig. 11.17, we have
••.' or•••
Q=Ql+Q2In this, arrangement, the loss of head for each branch pipe is same.. . Loss of head for tJranch pipe 1 = Loss of head for branch pipe 2
2 24f1L1V1 4f~2V2=d1 X ~g d2 X 2g
...(11.18)
..•(11.19)
If f1 =12, then ...(11.20)
Problem 11.32. A main pipe divides into two parallel pipes which again forms one pipe as shown inFig. 11.17. nre length and diameter for the first parallel pipe are 2000 m and 1.0 m respectively, while thelength and diameter of 2nd parallel pipe are 2000 m and 0.8 m. Find the rate of flow in each parallel pipe, iftotal flow in the main is 3.0 m3js. The co-efficient of friction for each parallel pipe is same and equal to .005.
Sol. Given :Length of pipe 1,Dia. of pipe 1,Length of pipe 2,Dia. of pipe 2,
L1 = 2000 md, = LO m
Total flow,
L2 = 2000 md2 = 0.8 m
3Q = 3.0 m /s11=:h ::I:; .005
Let Q1 = discharge in pipe 1Q2 = discharge in pipe 2Q = Ql + Q3 = 3.0 ...(l)From equation (11.18),
Using equation (11.19), we have4flL 1V12 4[-).2V2
2=d1 X 2g il2 X 2g
4 x .005 x 2000 x VI 4 x .005 x 2000 x Vl
or
1.0 x 2 x 9.81 0.8 x 2 x 9.81V 2 lj 2 v:;_1_ = .:.L or V.2 - _?--1.0 0.8 1 - 0,8
...(ii)
[.: Vl-.~41and
jt 2 - _ it ~,'I _ :/2Ql = '4d1 X \II ,: ~1(1(x ~:{94
Q2 = : dl x V2=.= (.8)2 x )12= = x .64 x "'2
Substituting th~ value of Ql and Q2 in equation (l), we get:7t V2:7t _'4x 0.894 + '4x .64 V2 = 3.0 or
Now
0.8785 V2 + 0.5026 V2 = 3.0
or V:i[.8785+ .5026] = 3.0 V=~-=217m/s.1.3811 .01
Substituting this value in equation (it),
V V2 2.17 2 4 7 '1= .894 " 0.894 = . 2 m/s
HenceJt~. 1t.. 3Ql = '4 dl~ x VI = 2 x 1'"x 2.427 = 1.906 01 Is. Ans,
Ql = Q - Ql = 3.0 -1.906:; 1.094013/S. Ans.
Problem 11.33. A pipe line of 0.6 m diameter is 1.5 km long. T( increase the discharge, another lineof the same diameter is introduced parallel to the [irs: in the second ha/fofthe length. Neglecting minor losses,find the increase in discharge. if 4f = 0.04. The head at inle. I. 300 mm (A.M.I.E., December, 1975)
•ti0l• Given:Dia. of pipe line,
~ngth of pipe line,
•Head at inlet,• Head at outlet,.:. Head lost,
• Length of another parallel pipe,
D=O.6mL = 1.5 km = 1.5 x 1000 = 1500 m4f:: 0.04 orf= .01It = 300 rom = 0.3 m= atmospheric head = 0
ht= 0.3 m1500
L1=2=750m
• Dia. of another parallel pipe, d, = 0.6 mFig. 11.18 shows the arrangement of pipe system.•••••
Q ---
LI'" 750m, d.cO.6m
B--0, C--O-=-O-.6-m---'''~02--------~-~ 0:L=I500mO.3m
~--.-
..~
Fig. 11.18
• 1st Case.Discharge for a single pipe of length 1500 m and dia. = 0.6 m.
• Thi h did f···· 1 . . I 4JLV*2IS ea ost ue to riction Insing e pipe IS It = d 2• x gbere V* = velocity of flow for single pipe
• y*20.3 = 4 x .01 x 1500 x• 0.6x2g
• •
•V* = 0.3 x 0.6 x 2 x 9.81 2426 mI
4 x .01 x 1500 = O. s
: .. Discharge,Q* = V* x Area = 0.2426 x : (.6i = 0.0685 m3/s
...(i)
• 2nd Case. When an addition pipe of length 750 m and diameter 0.6 m is connected in parallel with the
• half length of the pipe.Let••
Ql :: discharge in Ist parallel pipeQ2 = discharge in 2nd parallel pipe
Q:: Ql + Q2Ilere Q = discharge in main pipe when pipes are parallel.
But as the length and dia~eters of each parallel pipe is same
• . Ql = Q2=Q/2
•••Consider the flow through pipe ABC orABDHead loss through ABC = Head lost through AB +.head lost through Be
...(ii)
But head lost due to friction through ABC = 0.3 m given. . 4 x f x 750 x V2Head lost due to frictIon throughAB = 0.6 x 2 x 9.81 where V = velocity of flow throughAB
__ " CC" ..
••••••
=~= Q =Area ~ (0.6)2 3t x .36
.. Head lost due to friction throughAB
= 4 x .01 x 750 x ( 4Q Y ""i X7/!'l.0.6 x 2 x 9.81 \ jt x .36 ! -= c'l s ", ~'.'
\ '
Head lost due to friction through Be
ri , .
lV1 = Discharge = Q 1
jt 2 3t 24 (.6) 2 x 4 x (.6)
= 4x.01x750 x 16 Q2=7.969Q20.6 x 2 x 9.81 4 x 3t2 x .362
Substituting these values in equation (ii), we get'. 2 2 . 20.3 = 31.87Q + 7.969Q = 39.839Q
Q=~ =0.0867m3is
. . Increase in discharge = Q - Q* = 0.0867 - .0685 = 0.0182 m:i/s. Ans.problem 11.34.A pumping plant forces water through a 600 mm diameter main, the friction head
being 27 m. In order to reduce the power consumption, it is proposed to lay another main of appropriatediameter along the side of the existing one, so that two pipes may work in parallel for the entire length andreduce the friction head to 9.6 m only. Find the diameter of the nel1'main if: with the exception of diameter, itis similar to the existing one in every respect. (A.M.LE.,Winter, 1974)
Sol. Given:Dia. of single main pipe, d = 600 mm= 0.0 m
It,= 27 m=9.6m
Friction head,Friction head for two parallel pipes
1st Case.For a single main [Fig. 11.19 (a)]
hf= 4.fL.V2 or 27.0 = ~xL x V2. d x 2g 0.6 x 2. x 9.81
JLV2 = 27.0 ~.2 x 9.81 :::317.844 = 79.461,4 5
Q2f.L. 2 = 79.461
A
whcreV=Q.A
...(i)...2nd Case. Two pipes are in parallel (Fig. 11.19 (b)lLoss of bead in anyone pipe :::9.6m
:. For 1st pipe,
J;;..._ _
••••3. The repeating variables selected should not form a dimensionless group.4. The repeating variables togerher must have the same number of fundamental dimensions.5. No two repeating' arial.les ~hO'Jld have the same dimensions.In most of fluid mechamcs prcruems, the choke of repeating variables may be (I)d, v, P (il) 1,V,P or
• (iii) 1, v, IAor (i.,) d, v, IA.
•
•••••
Variables with Geometric Property are(i) Length, 1 (ii) dVariables with flow property are(I) Velocity, V
(iiI) Height, H etc.
(it) Acceleration etc.Variables with fluid property' r(\ (n) p, (iil) wetc.
12.4.4. Procedure for Solving Problems by Buckingham's n-theorem. The procedure for solvingproblems by Buckingham's x-tbeorem is explained by considering the problem 12.6 which is also solved by
• the Rayleigh's method. The problem isThe resisting force R of a supersonic plane during flight can be considered as dependent upon the length
• of the aircraft 1, velocity V, air viscosity ~. air density p and bulk modulus of air K. Express the functional• relationship between these variables and the resisting force (A.M.I.E.,Summer, 1977)
Sol. Step 1.The resisting force R depends upon (1) I, (it)V, (iii) lA,(iv) p and (v)K. HenceR is a function• of 1, V, lA,P and K. Mathematicall y,
• R = f(/, V, u, p, K) ...(i)...(it)or it can be written as 11 (R, I, V, p, KJ :;::0•• [m is obtained by writing dimensions of each variables asR=MLT-2, V=LT-\ IA=MT-1t+,P =ML-3,
• K =MLT -2. Thus as fundamental dimensions in the pioblem areM, L, T and hence m = 3.]Number of dimensionless n-terms n ---m :: 6 _..3 = 3.Thus three n-terms say ;ttl, Jt2 and Th are formed. Hence equation (il) is written as
fI (nt, Jt2, Jt3) =~O. ...(iil)Step 2. Each Jt term = m -s- 1 variables, where m is equal to 3 and also called repeating variables. Out
of six variables R, 1,V, lA,P and K, three variables are to be selected as repeating variable. R is a dependent• variable and should not be selected as a repeating variable. Out of the five remainingvariables, one variable• should have geometric property, the second variable should have flow property and third one fluid property.
These requirements are fulfilled by selecting 1, V and p as repeating variables. The repeating variables• themselves should not form a dimensionless term and should have themselves fundamental diamensions equal
to m, i.e., 3 here. Dimensions of 1,Vand pare L, LT -1,ML-3 and hence the three fundamental dimensions• exist in 1,Vand p and they themselves do not form dimensionless group.
Step 3. Each n-term is written as according to equation (12.4)
Jtl. = fl . V'! . pCl . R }Jt2 = f2 . V'2 . pC2 • IAJt!- :: f3 . V'3 . pC3 • K
•••
••••••have
:. Total number of variables; 1'1 :::: 6.Number offundamenta1 dimensions. m = 3.
...(iv)
Step 4. Each n-term is solved bv the principle of dimensional homogeneity. For the-first n;..term,we
••••••'.or
Equating the powers ofM,.L, T on both sides, we getPowerofM, 0:; Cl + 1 c1::;-1Power ofL, 0 = a1 + b1 - 3Cl + 1,:. ai = - b1 + 3C1 - 1 '"-"2 - :3 - 1 = - 2
power ofT, 0 = -bl-2 ., bi = - 2
Substitlitingthe values of ab b1 and CI in equation (iv),11:1= 1-2 .V -2 . p-l _R
R R:Tt ----- ...•, 1 - 12V2p - p[2V2
Similarly for the 2nd jt-tenn, we get ~ = AfL oro := L (ie) • (LT -1l, . (ML -3)bz . ML·-1T -1 .
Equating the powers ofM, L, T on both sidesPowerofM, 0 = c,: + 1, C2 =-1
...(v)
o :; a2 + b1 - 3("2 .
az = - b2 + 3C2 of ;:
Power ofT, 0 = _.(;2 i,
Substituting the values of a2, b2 and '2 in n? of (iv)
PowerofL, 3+1=-·t::::-1
_£_iVr
3rd ""'term.
or
11:3= la3 . V'3 , pC'; . KItfL 0To = La} . (LT-3)b3 • (MI,-3)u3 •Ml:.-IT-2
Equating the powers of M, L, T on both sides, we have
PowetofM,PowerofL,
o = ..::? + L C3::::: - 1o := fl;l + b3 - 3(, . . :13:::: - b3 + 3C3 + 1= 2 - 3 + 1 = 0
Power of T, 0 :: - b3 -- 2, ~::- !Substituting the values of a3,.b3 and C3 in ·11:3 terms
oj V-2 -, K K11:3= t '. . p . ..""~ - .VI'
Step 5. Substituting the values of :1tl, :rtl and 1t3 in equation (il), we get
f( R u K \ -)1 pPV2' Wp' " \fip ) :::{ or
or11.. IV:;:f i Ans.
Problem 12.8. (a) State Buckingham's 1(- theorem,:/ (b) The efficiency 'YI of a fan depends on ttensity p, dynamu.:viscosity ~ of the flUId, angular velocity 00,
diameter Do/the rotor and the discharge Q. Express II in terms o{divr,ensionless parameters.(A.'M,.I.E.,Winter, 1976)
Sol. (a) Statement of Buckingham' s 11:-th~.oremis given in Article 12.4.2.
(b) Given: 'YI is Iifunction of p, ~, 00,D and Q ...(~'11 :: f(p, ;l, co, D; Q) ".rf· - (·n. 0 II 00 D Q) = 0.. 1. . IT. !' t' r" , , .
•••
Number of n-terms ""n -"m 0: 6 - 3 = 3Equation (l) is written as f (it 1 '(2: n3) :::0 ...(ii)Each 1t-term contains m + 1 variables, where m is equal to three and i,salso repeating variable.ChoosingD, co and p as repeating variables, we have
:It1 :::D": O}l. pC, , II:It:: :::DU') robl, pC2 , J.1
Jt;; '" noo rob" pC3 , QFirst 1t-term (1tl)' nl ::;D(J, , rob1 . pCI . IISubstituting dimensions on both sides of nh
AfJL0'fJ =L01. (T-1)b1 (ML-3)Cl.J.ilLoTo
Equating the powers of M, L, Ton both sides
Hence total number of variables, T; =, 6• The value of m, i.e., number of fundamental dimensions for the problem is obtained by writingdimensions of each variable. Dimensions of each variable are• II = Dimensionless, ML'1 f..t;;; Mr1r-1 ill::: T-l,D =L andQ =L3T-1... m=3
•••••••••••
0:;: c: + 0,PowerofM,PowerofL,Power ofT,Substituting the values of a , b- iL:\ " in It"we get
• [If a variable is dimensionless, it itself is a x-term. Here the variable II is a dimensionless and hence T)is a n-term. As it exists in first n-term and hence :Ttl = ll. Then there is no need of equating the powers. Directly
.the value can be obtained.]~ d terr DOl b- c-~n 1t-erm. ~ = . ill ._ • p" u.••••••••••••
Substituting the dimensions on both sidesMlLoro =LOl. (T-1)bl• (ML -3)C2. ML -IT-1
Equating the powers of M, L, T on both sidesPowerofM,PowerofL,Power «r,
0::: C2 + 1,o = a2 - 3C2 - 1,o =-b2-1,
•• C2 = -·1• • a2 = 3c2 + 1~-3 + 1= _;2.. b2=-1
Substituting the values of a2l b2 and C2 in ~,__ D-2 -1 -1 _ ___I:!:._
lit,. - ,u). P . II - 2,. .'" Drop
3rd rc-term. fL, :::DQ3 , jl)b, ,pc3 QSubstituting the dimensions on both sides
,\1'r()'T ::::L03" (T-1)b3, (ML-3)C3 .L3T-1
Equating the powers ofM, Land T on both sidesPowerofM, 0= C3, .. C3=0
PowerofL, (I = a3 - 3C3 + 3, .. a3 = 3C3- 3= - 3..Power ofT, I) = -b3-h
. b3=-1..
••••••••••••••••I.
Substituting the values of a3, b3 and C3 in :lt3,
.,.._D-3 ,.~-1 pO Q = _Q--J~3 - • UJ ., 2D co
Substituting the values of :It}>~ and :lt3 in equation (ii)
It (TI' ~ , _Q_) = 0 or TI= CPl' -f--, ~ 1.D2(Op D2(O, D (UP n (0 jProblem 11.9. Using Buckingham's :It-theorem, show that the velocity through a circular orifice is
given by V = "2gB 01> ( ~', p~lwhere H is the head causing flow, D " the diameter of the orifice, I' is
co-efficient of viscosity, p is the mass density and g is the acce/Fratre,.; to gravit»,(A.M.LE., Winter, 1977)
Ans.
Sol. Given:V is a function of H, D, ~, p and g.. V = f(H, D, u, p, g):. Total number of variable, n = 6Writing dimension of each variable, we navi
V::;Ll ....1-1::; J)
Thus number of fundamental dimensions, m :: 3:. Number of rt-terms = n - m = 6 - 3 = 3.Equation (i) can be written as 11 (:lt1' :lt2, :lt3) = 0
(!'- LJ D: p, g) = 0... (z)
...(it)
Each :It-term contains m + 1 variables, where m = :3 and is also equal to repeating variables. Here Visa dependent variable and hence should not be selected as repeating variable. Choosing H, g, P as repeatingvariable, we get three n-terms as
:ltl = H" ,ghl . pCl " ".
:lt2 = Jr2 .l2 .PC2 • D
:lt3 =Ba3 • g/'> . f/3 . ~t
First n-term. :ltl = J{l1 . i1 pCl. vSubstituting dimensions on both sides
AfL °T i) = L al (LT ":"Y)l eM [ .
Equating the powers of M, L, T on both sides.
, ,
:0
: - b] + :::'C1 - 1= t - 1= - t=-2.
2
PowerofM,PowerofL,
0=(
Power ofT,
Substituting the values of al, b1and C1 in It;,1 1_- -- _ V
:rtl = B 2 .g 2. pO . v - .--1gB'
Second n-term. :rt2 = J{l2 . gb2 . pC2 . DSubstituting the dimensions on both sides,
fiJL or 0", La2 . \LT' 2)h2 (MI, 3"
------- -
Multiplying by a constan: does not change the cbaracter of 3t-tenns.
Problem 12.10. The pressure difference IIp tn apipe of diameter D and length I due to turbulent flow.. depends on the velocity V, viscosity !-" density p and roughness k. Using Buckingham's x-theorem; obtain an
expression for I1p. (DelhiUniversity, Nov., 1982)" Sol. Given:l1p is a function ofD, I, V; u, p, k:. l1p = ltD, I, V, J.l, p, k):. Total number of variables, n ~ '7'
Equating the powers ofM, L, T,PowerofM,PowerofL,Power ofT,
o = a2 + b2 - 3c2 + 1,a2 = - b2 + 3C2-1 = -1o = - 2b2, :. b2 = Q
Substituting the values of az, b2, (;2 in 3tz,
Third 3t-tenn.
-1 6 ° D3tz=H .gp .D=H'
3t3=1fl3 •~ • pe3 • J.lSubstituting the dimensions on both sides
M1LoTo = La3 • (LT-2)a3 • (ML-3)C3• ML-1T-1Equating the powers of M, t.. T on both sidesPowerofM, 0 == c~+ 1,PowerofL,
Power ofT,
C3=-1
:. a3=-~+3c3+1=~-3+1=_~
SubstitUting the values of a3, b3 and c~in 3t3,
ts--3/2 - 112 -1 J.l1t3= II ,g . p . fA..:: n3/2 _r~
rt: vgp_~__ JA.V- HpvgH - HpVVgH [Multiply and Divide by V]
or
-.1_ ...- HpV' '''1
Substituting the values of 3th 3tz and 3t3in equation (il),
fi ( ~ D 1t _H._) - 0 or _y_ - '"[D 3t _j!_ J1 VgH' H' 1HpV - VgH - 'f' H' 1HpV
V = YZgH ~ [ : ' ~ J: .Am.
or ... (1)
Writing dimensions of each variable,
Dimension of l1p = Dimension of pressure = AlL-IT-2
D =L, 1= L, V =LT-1, JA. =ML-1T-1, P =ML-3,k=L.'. Number of fundamental dimensions, m = 3Number of x-terms = n - m = 7 - 3 = 4.
-
~ . .
•••••..•••....•••••..•"••"..flit
"•....,.•""
Now equation (I) can be grouped in 4 x-terms as
11 (;Ttl>~, ;Tt3,tt4) = 0 ...(iz)Each n-tenn contains m + 1or 3 + 1= 4 variables. Out of four variables, three are repeating variables.
ChoosingD, V,p as the repeating variables, we have the four n-terms as
ttl = D"» . V'1 . pel. I1p
~ =Da2 • V'2 . pCl ,
tt3 :::Da3 • "03 . ~{3. !-t,
tt4 = Da4 • v"l ,.pC4 . kttl =Dal . ybl . pCl . IIpFirst tt-tenn
Substituting dimensions on both sides,
MlL°T' =La1 • (LT-1)b1 • (ML .-3)cJ • ML-'.T-2Equating the powers ofM, L, T on both sides,PowerofM,PowerofL,Power ofT,
0= Cl + 10= a1 + b, -, 3Cl - 1,0= -bi -2,
(, :::-1u, ~o - b, + 3el + 1= 2 --3 + 1= 0b ""- 2
Second n-term ~ =Da2 • yb2 . pC2 . ISubstituting dimensions on both sides,
MlLoTo =La2. (Ll,,-1)b2• (Ml 'Equating powers of M, L, T on both sides,PowerofM,
Power ofL, 0 = a2 D2 - 3,: i,Power of T, 0 = - b2,
Substituting the values of a2, b2, c2 in tt2,
D-1 nO 0 1 I~= ,V,P'=D
:1.2 := b2 + 3C2 - 1 = - 1b2:::: 0
Third n-tenn
Substituting dimensions on both sides,
MlLoTo = La3 • (LT ..1)b3 . (ML-'t3 ML -i1'-1
Equating the powers of M, L, T on both sides,PowerofM,
PowerofL,
Power ofT,
0= C3 + 1,
0= a3 + b3 - Jc., ... 1,(3""--1
u3 '" ,--[..3 + 3C3 + 1= 1- 3 + 1= - 10= ....b, - !
Substituting the values of a3, b3 and c, in :lt3,
;Tt3=D-' V··j,
0'" 1t:::: -.~-r- • 1>\/(:
,-~-..••••
r"~ /,1 '., f" "'~ I 'i" ~", 11/"/ ' in .-
• ~ "Ii! f$ •
• or
••••••••
'Itt! "" Da4 , yb4 , pC4 . kAfL O'f1 == Lad. (LT-1)b4 • (ML-3)c4 • L
Equating the power of M. L) T on both sides,
Fourth n-term{Dimension of k =L}
PowerofM,
PowerofL,
Power ofT,
.• C4 =0• . a4 = b4 + 3C4 - 1 = - 1.. b4=0
I) ;:::.il4 .- b4 - 3C4 + 1,
0= -b4,
Substituting the values of a4, h4' C4 in Jt4,
tr' 00 0 k kTC;=,. .V'.p, =D
Substituting the values of;:J, IT, it~and Jt4 in (il), we get
f ( 6.p. L _...E. ',f:.. i ~ 01 pV." D n t ; or
Expression for hr (Difference .A pressure-head). From experiments, it was observed that pressure
• difference, lip is a linear function of ~ and hence it is taken out of function
••.'•••
Sp .-l. [_L .!.lpy2 - D <I> DVp' D ~
!:~~V2 • ~ ¢ rnt,~]Dividing by g to both sides, we have ~ = ;~ ~ $ [nt,~].
• Now <I> [D~T 'Dk1contai ns two terms. Find one isD+ which is R 1 or _!_ and secondY PI' Y P eynold number Re
• Jone is Dk which is called roughue S''; h" 'Ie! :\,lowQ:l f .L, k 1 is put equal to f, where f is the co-efficient of•
! 11.,. D! 'I. ,.
friction which is a function (>1' Rr Tl101d nun 1)('1and roughness factor.••••
~ip_,~ V2J-- ? .pg __ gD
Multiplying or dividing by any constant does not change the character of n-terms.
se. 4f.LV2.. "";ht= D 2 . Ans.Pg x g
• Problem 12.11. The pressure difference sp in a pipe of diameter D and length I due to viscous flowdepends on the velocity V,viscosity !-.t and density p. Using Buckingham's x-theorem, obtain an expression for• Sp.
• Sol. This problem is similar to problem 12.10. The only difference is that lip is to be calculated for• viscous flow. Then in the repeating variable instead of P, the fluid property IA- is to be chosen.
•---~-----
•
• Problem 11.13. The frictional torque T of a disc of diameter D rotating at a speed N in a fluid ef
• viscosity i!and density jl.;,. a turbulent flow is given by T = [hi' pq, [ ~ 1·• Prove this by the method of dimensions.
Sol. Given: T = f(D, N, JA., p)••••••••••••••••••••••
Substituting the values of 1I:J, :rt2' :rt3and :rt4in equation (ii),
f (p Dw .u, C ';...0 or p - 1.2 (DW __I:!_ C)
. 1 nlv2p' V ' DVp , v ! _, , D2V2p - J: V' DVp , V
Z Z (Dw _lL C \P=DVpfz\V'DVP'V-j-': ADs.
or ft(T, D,N, JA., p) = 0
(A.M.I.E., May, 1975)...(i)
:. Total number of variable, n = 5Dimensions of each variable are expressed as
T = ML2T-2, D =L, N = T-1, JA. =ML-lT-l, P =ML-3
:. Number of fundamental dimensions, m = 3Number of n-terms = n - n. = 5 ':l.,-?Hence equation (z) can be written as /i(:rtb~)::: 0 ...(iz)Each 1t-term contains m + ::variable, i.e. 3 + 1 = 4 variables. Three variables are repeating variables.
Choosing D, N, P as repeating variables, the n-terms are:rtl = tr: .Jtl . pCl , T
1t2 = Daz .Jtz . pez. "'"
Equating the powers ofM, L, T on both sides,PowerofM,PowerofL,Power ofT,
0= <i + 1,0= al -3Cl + 2,0= -bl -2,
.. cl =-1
.. a1 =3Cl-2 =-3 -2 =-5
.. b1=-2
Substituting the values of at> b1, c) in n,
0 ,-5 11.1""2 -1 T TJl \ :; .1Y. P , = D2N2p .
Dimensional Analysis of Tt2
Tt2:-; Da2 .Jtz, pez. JA.
Substituting dimension.on both sides,ftlJL°To = La2 • (T -1)b2 • (ML -3t2 .ML-1T-1•
Equating the powers ofM, L, T on both sides,PowerofM,PowerofL,Power ofT,
0= C2 + 1,0= a2 -3C2 -1,O=-b2-1,
•• C2 =-1• • a2 = 3C2 + 1=- 3+ 1=- 2.. b2=-1
, "' ~ r f I ~
, <' - J, "I'if
Substituting the values of a2, b2 and C2 in ~,
D-2N-1 -1 11:lt2= P .!-t = _...c:_,D2Nr'
Substituting the values Of:ltl and :lt2in equation (ii),
Ii ( T .:.....H:.._) - 0 0"1 '2·2' 2 - •D'N p DNp
r= DSN2 P q,[+]. Ans.DNp
Problem 12.14. Using Buckingham' s :It- theorem, shown that the discharge Q consumed by an oil ringis given by
or
Q -Ntftt-.[_l!.- _L ~]- 'I' pNcP' pN2d3 J pN2d
where d is the internal diameter of the ring, N is rotational speed, p is density, IJ. is viscosity, p is surfacetension and w is the specific weight of oil.
Sol. Given: Q = fed, N, p, !-t, 0, w) or fl(Q, d; N, p, f.,l, 0, w) = 0 ...(t):. Totalnumber of variables, n = 7Dimensions of each variables are
Q =L3T-J., d = L, N:;' T'[ ;.; "" ,~t[,-3 U = ML-1T "1, c = MT-2w=ML-2r,··2and
:. Total number of fundamental dimensions, m = :.\.: Total number of n-terms = n - m = 7 - 3 = 4
.. Equation (l) becomes as fl(:ltt> :lt2,1£3, :lt4)= 0Choosing d,N, p as repeating variables, the n-terms are
:lt1= cfl .1';'1 . pCl •Q~ = rflz . Xbz . pCz . j.A
:lt3= rfl3 . ,~3 • pC3 , (}
:lt4= cf4 •~4 • pC4 W.
:lt1= rfll .N"1 . pCl .Q.
...(it)
First :It-term
Substituting dimensions on both sides,kfLOrO=Lat. (T-1)b1• (ML
Equating the powers of M, L, T on both sides.
PowerofM,
Powerof Z,
Power ofT,0= al - 3cl + 3, al = 3Cl - 3 = 0 - 3 = - 3
0= - hI - 1, hI:: -1
d -3 N-1 ° Q _Q_:ltl = . . . p. = d3N
PowerofM,PowerofL,Power ofT,
0= C4+ 1,o == a4 - 3C4- 2,o =-b4-2,
c4=-1:. a4 = 3C4+2= - 3+ 2=-1.. b4=-2
Second,,·term•••"••A.."••••..'It
••....•".."•
Substituting the dimensions on both sides,},fLoTo =L02. rJ-1)~.(ML-3t2 .ML-1 T-1
Equating the powers ofM, L..Ton both sides,
Power ofT,' .
o :;::(~2+ I,o :: a2 - 3C2- 1,a2 -,:·3(;2 + 1 :=I - 3 ...1= - 2I) ~ - b,;: ". 1,
.. c2=-1
.. c2=-1
.. b2=-1
PowerofM,PowerofL,
Substituting the values of a2, b2, '2 in 1t:z,
'1t7. = a" .N-1 • p•.l .~=kor p~cp'
Third ,,·termSubstituting dimensions on both sides,
AfLoTo = LaO,. (T-1)b3 • (ML-3)C3 •MT-2•Equating the powers ofM, L, T on the sides,PowerofM, 0 = C3+ 1, :. C3 =-1
Powe,:'ofL,power ofT,
0= a3-3c3,O=-b3-2,
:. a3= 3C3=-3:. b3=-2
Substituting the values of a3, b3, C3 in 11:3,3 2 -1 0
Jt3 =a .N" . p ,0 = ~ N2p .
Fourth ".t~rm )1;4 ~ cf4 ~4, pC&. • W
Substituting dimensions on both sides,MOLofl = ir-, (T-1)b4. (ML-3)C4.ML-2T-2•
Equating the powers ofM, L, T on both sides,
. Substituting the values of a4, b4 andC4in :71:4,
d -1 N-2 -1 W11:4 = . . p . w = cJ.N2p .
Now substituting the values of 11:1, 11:2, 11:3, :71:4in (ii),
f( jN'~' d';2p'';p) =0 or ~=ft [~. d';'p.';p]" or.. 3 [_1_ 0 w]Q :: d Ncj> 2.' 3 2. ,--z'. Ans.pNd d N f dN P .
12.5.MODEL ANALYSISFor predicting the performance of the hydraulic srructures(such as dams, spill ways etc.) or hydraulic
• machines (such as turbines, pumps etc.), before actually constructing or'manufacturing, models of thestructures or machines are made and tests are perfonned on themto obtain the desired information.
•
•. ~.~ ~f . ' <
,
• ;'(iii)'Co"efticient of Drag (CD)
From equation (13.14), we haveF
r> D. ") - .,------1 - 1pA U22 1'1. ,<
" whereFD is given by equation (13.12) as
f l fL HQFD = to X b x dx = 0.365 V1f; x b x dxo 0 x r
z: 0.365 fL HY._ - IPJJI x b x dxJo x V~-
.. ;., c: fL ,,[I- ~)( ..1_ x b x dx'--Jo'-' V'--,;'..fX
. ...rsr fL ...rsr [xV2 ]L'" 0.365 !-tUV 1!i~- X b . xI!2 dx = 0.3651lU V ~ x b x -1-I.! (I Il'-, " 2 0
== 0.365 x 21lU.y-;Q x b x ..;rIl
"" 0,73 b~lU~Il
0.73bllU\{P!!!Il
CD=
...(13.18)
.. whereA = Area of plate = Length of plate x width = Lx b
. . C, ""__,'0.73 bill) - tpUL = 1.46 Il ... tpUL, ~ i' x L x b x U2 V ~ pLU V7
<
.. 46 ViA _ _ ~ _ 1.46 { .. ---'!..... __ 1_}- \ipL -l.46Vr.:ur'-VReL ...(13.19) . pUL vReL
Problem 13.4. For the velocity profi le given in problem 13.3, find the thickness of boundary layer at,. the end of the plate and the drag force on one side of a plate 1m long and 0.8 m wide when placed in water" flowing with a velocity of 150 mm per second. Calculate the value of co-efficient of drag also. Take Ilfor water
= 0.01 poise.
•
... Sol. Given:Length of plate,Width of plate,Velocity of fluid (water),
L = 1mb = 0.8mU = 150 mm/s :.:0.15 sel»
. 0.01 N-s . N-sJA. for water = 0.01 pOise =10m2 = 0.001 m2
Reynold number at the end of the plate i.e. at a distance of 1m from leading edge is given byR = pUL = 1000 x 0.15 x 1.0 (.,' P = 1000)
eL Il .001::: 1000 x .15x 1.0 = 150000
0.001
•••••••••••••••••••••••••••..•••"It
As laminar boundary layer exists upto Reynold number = 2 x lOS. Hence this is the case oflaminarboundary layer. Thickness of boundary layer at x = 1.0 m is given by equation (13.17) as
0= 5.48 v~ =~:~~~oo= 0.01415 m = 14.15mm. Ans.ex
(iz) Drag force on one side of the plate is given by equation (13.18)
FD=0.73b~U~~.t
= 0.73 x 0.8 x 0.001 x 0.15 x v'f50000 { .., PC: ~R'L }
= 0.0338 N. Ans.(iiz) Co-efficient of drag, CD is given by equation (13.19) as
1.46 1.46CD = vR = v150000 = .00376. Ans.
eL
2
Problem 13.5.For the velocity profile for laminar boundary layer ~ = ~ ( f )- ~(f )Determine the boundary layer thickness, shear stress, drag force and co-efficient of drag in terms of
Reynold number.Sol. Given:
Velocity distribution,
p~2 =1;[1: ~(1- ~ )dy 1Substituting the value of ~ = ~ ( f )-~(t )l in the above equation
Using equation (13.10), we have
if,=! [1:[~(f)-~(f)3][ l-{~(f)-~(f)'}]dyl=! [1: (~-~)( 1-~+~ )dY 1=1-[r(~_~+~_L+~_.L)d l'ax J0 20 462 204 203 464 466, Y
/
_1- [31 913 3l l 3l . 1 ]6- ax 2 x 20 - 3 x 462 + 5 x 464 - 4 x 263 + 5 x 464 - 7 x'4l,6 0
= _!_ [302 _ 303 + 1._~=_.!. 04 + ~ 05 _ .1..071ax 46 402 20 f.l4 8 03 20 (')4 2F 66 "'
d[3 3 3. I ~',... i= 'ax 4' 0 - 4'6 + 20 b •. 8" s + 26 \) . Zg ,~
(ii)Shear stress at any distance from leading edge is given by 'to = 0.327 .I:!!£..fK:x ex
U .. .x 4 x 0.2Atx = 200 nun = 0.2 m,Re '" -- = 4 = 53333_,' ',' D.15 x 10-
~... _ 0.327 x IIx 4 X "'53333to - 0.1
But ll=VXP
== 0.15 X 10-4 x·1.24 = 0.186 X 10-4
... 'to = 0.327 X 0.186 X 10-4 X 4 X "'53333 = 0.02805 N/m1.. Ans •0.2
(iii) Drag force on one side of the plate is given by equation (13.33)
FD = 0.655 X llU X b x ~
~= 0,655 x 0,186 x 10-4 x 4,0 x 0,6 xVff!••
::c 0.29234 X 10-4 x'" J 4 X 0.5-4 = 0.01086 N. Ans,V.15x10
• Problem 13.11.A thinplate is moving in still atmosphericair at a velocity 0/5 mls. The length of the.' plate is 0.6m and width 0.5 m. Calculate (i) the thickness of the boundary layer at the end oftbe plate, and• (ii) dragforce on one side 0/ theplate. Takedensity of air as 1.24 kglm3and kinematic viscosity 0.15 stokes.
••••••••••••• where CD from Blasius's solution,
Sol. Given:Velocity of plate,Length of plate,Width of plate,Density of air,Kinematic viscosity,
U = 5 m/sL = O.6mb = 0.5 mP = 1.24 kglm3
V = 0.15 stokes = 0.15 x 10-4m2js
R = UL = 5 X 0,6 = 200000.e V 0.15 X 10-4Reynold number,
~s Re is less than 5 x 1as, hence boundary layer is laminar over the entire length of the plate.(i) Thickness of boundary layer at the end of the plate by Blasius's solution is
" _ 4,9lx _ 4.91 L _ 4.91 X 0.6 _ 00658 - 6':8 An'J - VRe", - "'Ret - "'200000 -. m - .., mm. s.
(ii) Drag force on one side of the plate is given by equation (13.14) as~ FD(;D=---
_!_ PAU22
FD = t p AU2 x CD
1.328 1.328CD = "'R = v'''00000 = 0.002969 = .00297
eL ..
••••••••••••••••••..••••••••••
FD = ~x 1.24 x 0.6 x 0.5 x 52 x .002970
= 0.01373 N. Ans.Note. Hno velocity profile is given in the numerical problem but boundary layer is laminar, then Blasius's solution
is used.
... {.: A=Lxb=0.6xO.5}
Problem 13.1Z.A plate of 600 mm length and 400 mm wide is immersed in a fluid of sp. gr. 0.9 andkinematic viscosity (v =) to" m2/s. The fluid is moving with a velocity of 6mls. Determine (i) boundary layerthickness, (ii) shear stress at the end of the plate, and (iii) drag force on one side of the plate.
Sol• .As no velocity profile is given in the above problem, hence Blasius's solution will be used.Given: Length of plate, L = 600 mm = 0.60 mWidth of plate, b = 400 mm = 0.40 mSp. gr. offluid, S = 0.9:. Density,Velocity of fluid,Kinematic viscosity,
p = 0.9 x 1000 = 900 kg/m3
U = 6m/sV': 10-4m2/s
Reynold number, R .. U x L _ 6 x 0.6 _ ';. 104e - - '----::a-- - _"U x. .L V 10 .
.AsRq. is less than 5 x lOS, hence boundary layer IS laminar over the entire length of the plate.
(i)Thickness of boundary layer at the end of the plate from Blasius's solution is
s, _ :!_.91x h 06 d R 3 6 104u - vRe ' w ere x = . m an ex= . Xx
= 4.91 X 0.6 = 0.0155 m = 15.5 mm. ADs.V3.6 X 104
(ii) Shear stress at the end of the plate is_ 0332 .E!!:_ _ 0.332 X 900 x 62 _ IU LNt 2 An"to-. ...flf:: - _I - ;:IU.U m. s.
ReL V 3.6 X 104(iii)Drag force (FD) on one side of the plate is given by
FD='~pAU2xCD
b C f BI " 1" CD -_ 1.328 -_ 1.328 ·_'·0.00699W, ere D rom asius s so ution IS _~vReL V3.6 x 104
FD = 1pAU2 X Cn
= :~x 900 x 0 6 y <14 x 62 x .006992
= 26.78 N. Ans.
13.4. TURBULENT BOUNDARY LAYER ON A .FLA r PLATl
...{'.: A=Lxb=O.6x.-'
The thickness of the boundary layer, drag force on one side of the plate and co-efficient of drag due t.turbulent boundary layer on a smooth plate at zero pressure gradient are determined as in case of laminaboundary layer provided the velocity profile is known. Blasius on the basis of experiments give the followimvelocity profile for turbulent boundary layer
~=(*r1 7 jwhere n = "7 for Re < 10 but more than 5 x la'
...(13.35
-~-----------
••••••••••••••••••••••••••••••••••
Unit - 4
HYDRAULIC TURBINES
Fluid machines: definition and classification - exchange of energy - Euler's equation for
turbo machines - Construction of velocity vector diagram's - head and specific work -components of energy transfer - degree of reaction.
Hydro turbines: definition and classifications - Pelton turbine - Francis turbine - propeller
turbine - Kaplan turbine - working principles - velocity triangles - work done - specific
speed - efficiencies -performance curve for turbines.
" (vii) NUDlberof Jets. It is obtained by dividing the total rate of now through the turbine by the rate of
now of water through a single jet.
" Problem 18.1.A Pelton wheelhas a mean bucket speed of 10 metresper second with a jet of water""wing at.the rate of 700 litresls under a head of 30 metres. The buckets deflect the jet through an angle of1600; Calculate the power given by water to the runner and the hydraulic effICiencyof the turbine.Assume,.-efficient afvelocity as 0.98. (AMIE, Summer, 1980)
",.""•"..••
Sol. Given:Speed ofbucket,Discharge,Angle of deflection:. Angle,Co-efficient of velocity,The velocity of jet,
'I ::!. ii, :::U') ::: 10 m/sQ '" 700 !irrrs/s :::0.7 m3/s, Head of water, H = 30 m
. 'i
.:p := 180· j 60 = 20°C, = fl,98.V1 = C,V2gH = 0.98 ';2 x 9.81 x 30 = 23.77 m/sV..
1::VI - III :: 23.77 --10 = 13.77 m/s
1/VJ :: V! ~ 23 77 m/sl' "
From outlet velocity triangle,Vr2 :: Vr1 = 13.77 mil.
VW2:: Vr2 cos 4> - U2
= 13.77 cos 20 -10.0 = 2.94 m/s" Work done by the jet per second on the runner isfl#iven by equation (18.9) as. :: paY! (VwJ + VW2])( U
• ~ 1000 x 0.7 x (23.77 !. 2.941 x 10
c: aV ""Q ,",0.71113/S)
= 186970 Nm/s'It ~
...•,.It
:. Power given to turbine
Fig. 18.6
oilI ,.,0 '. if ' f186970"" 1000 = 186.97 kW. Ans.
•
probleDl18.2.A Pelton wheel is to be designedfor thefollowing specifICations:Shaft power = 11,772 kW,' Head:: 380 metres ..Speed = 750 r.p.m. ,.Overall efficiency = 86% ,.Jet
• diameter is not to exceed one-sixth of the wheel diameter.Determine:" (l) The wheel diameter, (il)The number ofjets required, and
(iii)Diameter q/the jeJ.TakeKvl = 0.985 andKs: ::.0.45.
"
. ,fThe hydraulic efficiency of the turbine is given by equation (18.12) as ) i '
= 21VWl+ Vw21xu = 2 123.77+ 2.941 x 10 =09454 941t411l Anllh V12 23.77 x 23.77 .. or.., -/0. s.
Sol..Given:Shaft power,Head,Speed,
s.P. =- 11,772 kWH:: 380 mN ..,.150 I.m.p
(AMIE, Winter, 1980)
•••,. Overall efficiency, T)o = 86% or 0.86
Ratio of jet dia. to wheel dia. = ~ =i
Dischargeof onejet, q =Area of jet x Velocity of jet1t 2 1t= - d x Vl = - (.165)4 4
_ S.P. _ 117,72T)o - W.P. - pgx Q xH
1000
Co.efficient of velocity,Speed ratio,Velocity of jet,The velocity of wheel,
But
...or
But
:. Oia.ofjet,
Now
:. Total discharge,
:. Number~Ue~__.~-------------_.
KVl = c; = 0.985KUl =0.45Vl = c,"';2gH= 0.985"';2x 9.81 x 380 = 85.05m/su = U1 = ~= Speed ratio xV2iH = 0.45 x "';2x 9.81 x 380 = 38.85m/s
1rDNu=--60
38.85 = 1rDN60
D = 60 x 38.85 = 60 x 38.85 = 0.888m, An3t xN 3t x 750 . :7 :7 S.
d 1D="6d = *xD = 0.~89 = 0.165m. Ans.
x 85.05m3/s = 1.818m~is ... (1
0.86 = 11772x 1000 whereQ = Total discharge1000x 9.81 x Q x 380
11772x 1000 3Q = 1000x 9.81 x 380 x 0.86 = 3.672m Is
= Total discharge = Q= 3.672 =1jets. Ans.Dischargeof one jet q 1.818
Problem 18.3. Thepenstock supplies water from a reservoir to the Pelton wheel with a gross head c500 m. One-third of the gross head is lost in f. iction in the penstock The rate of flow of water through th·.nozzle fitted at the endof the penstock is 2.0 m'Is. The angle of deflation of the jet is 165°.Determine thpower given by the .water to "the runner and also hydraulic efficiency of the Pelton wheel. Take speed rati= 0.45 and C; = i .0. (AMIE, Fluid Power Engg., I 988 ; OsmaniaUniversity, 1992
Sol. Given:Gross head;
Head lost infriction,
:. Net head,Discharge,Angle of deflection..:. Angle,Speed ratio
Hg=500mn, 500
h,/= - = -- = 166.7m3 3H=n,-'1,= 500 -166.7 = 333.30mQ = 2.0 m /s
= 165°4> = 180-165 = 15°= 0.45
•. ,;$3 jJ•••••••
•
•••••••••••••••••••••••• or
•
.Co-efficient of velocity,Ve.l~ity of jet,Vdocity of wheel,
c; = 1.0VI = c,v2gH = 1.0 x v2 x 9.81 x 3333 = 8O.S6m/su = Speed ratio x v2gH
u = U1 = U2 = 0.45 x v2 x 9.81 x,333.3 ~ 36.387 m/sVr, = V1 - U1 = 8O.8~- 36.387 = 44.473 m/sVw, := v, = 80.86 mls
or.'Also
From outlet velocity triangle, we have
Vrz = V't :::44.47:
Vrz cos q, = u2 + Vw~or 44.473 cos 15 = 36.387 4- VWz
or VW:z = 44.473 cos 15 -- 36.387 = 6.57 m/s. A~--~~--~~~~~~~u,1011--- Vw,-~
Work done by the jet on the runner per secondis given.by equation (lS.9) as
= paV1[VW1 + Vwz] x u = pQ[VW1 + Vwz]x U
;,.';' aV1 = Q)= 1000 x 2.0 x [SO.86+ 6.57] x 36.387 = 6362630 Nm/s. . Power given by the water to the runner in kW
= Work dO~~JDr second = 6316~~0 = 636Z.63 kW. Ans.
Fig.lS.7
Hydraulic efficiency of the turbine is given by equation (18.12) as2[-Vwj + VW2] x U 2 [SO.80+ 6.57] x 36.387
'fIlI;,o: -.....:-1,;:--,--- = 80.86 x SO.86 = 0.9731 or 97.31%. Ans.
Problem lS.4.A Pelton wheel is having a mean bucket diameter ~I1m and is running at 1000 r.p.m.The net head on the Pelton wheel is 700m. If the side clearance angle is150and discharge through nozzle is0.1 m3/s, find:
(it) Hydraulic effICiency olthe turbine.(t) Power available at the nozzle, andSol. Given:Diameter of wheel, D = 1.0mSpeed of wheel, N = 1000 r.p.m.
.. 1fDN ~ x 1.0 x 1000 .. . Tangential velocity of the wheel, u = """'6(} = 60 = 52.36 m/s
Net head on turbine,Side clearance angle,Discharge,
Velocity of jet at inlet,
H=700mcj> = 150Q = 0.1 m3/sVI = c;v2gH = 1 x v2 x 9.81 x 700
(.: Value of Cv is not given. Takeit= 1.0)VI = 117.19 m/s,
. -- -- ..... _
,.• (i) Power available at the nozzle is given by equation (18.3) as
wp = WxH = pxgxQxH:: 1000x.9.81xO.1x.700 = 6867kW.. 1000 1000 1000 ••
(il) Hydraulic efficiency is given by equation (18.13) as2(Vl - u)(1 + cos cj» U
llh= V12
2(117.19 - 52.36) (1 + cos 15) x 52.36= 117.19 x 117.19
Ans. I
= 2 x 64.83 x 1.966 x 52:~~ ,;;:0.9718:: 97.18%. Ans.117.19 x 117.19
Problem 18.5.A Pelton wheel is working under a head of400 m. The water is supplied througpenstock of diameter 1m and length 4 km from reservoir to the Pelton wheel. The co-efficient of friction fe.the penstock is given as .008. The jet of water of diameter 150 mm strikes the buckets of the wheel and ge.deflected through an angle of 1650• The relative velocity of water at outlet is reduced by 15% due to fricti(Jbetween inside surface of the bucket and water. If the velocity of the buckets is 0.45 times the jet velocity I
. inlet and mechanical efficiency as 85%, determine:(i) Power given to the runner, (it) Shaft power,
(iii) Hydraulic efficiency and overall efficiency.Sol. Given:Gross head,Diameter of penstock,Length of penstock,Co-efficient of friction,Diameter of jet,Angle of deflection:. Angle,
Hg = 4UO mD:: 1.0mL = 4 km = 4 x 1000 :: 4000 TO
i= .008d = 150 mm= 0.15 m= 1650
cj> = 180 ._ 165 ::: is'
,..(i)Fig. 18.8
or
Relative velocity at outlet, Vr2 = 0.85 Vr1
Velocity of bucket, U = OA5 x Jet velocityMechanical efficiency, llm:: 85% :: 0.85Let V* =Velocity of water in penstock, and
V1 =Velocity of jet of water.Using continuity equation, we haveArea of penstock x V* = Area of jet x Vl
1tD2xV*=1td2xV4 4 1
d2. 0.151V* = - x V l = ---- x V1:::, 0225 VID2 . 1.02
get
Applying Bernoulli's equation to the free surface of water in thereservoir and outlet of the. nozzle, .
V1Hg = Head lost due to friction -\-.._" .
•"•
u
4. Depth of the buckets which is = 1.2 x a,5.Number of buckets on the wheel.Size of buckets means the width and depth of the buckets,Prob~em 18.11.A Pelton wheel is to be designed for a head of 60 m when running at 200 r.p.m. T~
Pelton wheel develops 95.6475 kW shaft power. The velocity of the buckets = 0.45 times the velocity of the jetoverall efficiency = 0.85 and co-efficient of velocity is equal to 0.98.
Sol. Given:Head; H=60mSpeed N = 200 r.p.m,Shaft power, S.P. = 95.6475 kWVelocity of bucket, u = 0.45 x Velocity of jetOverall efficiency, flo = 0.85Co-efficient of velocity, C; = 0.98.Design of Pelton wheel means to find diameter of jet (d), diameter of wheel (D),Width and depth 0
buckets and number of buckets on the wheel.(i) Velocity of jet, .V1 = C; x..ffiH = 0.98 x ";2 x 9,81 x 60 = 33.62 mls:. Bucket velocity, u = U1 = U2 = 0.45 x V1 = 0.45><33.62 = 15.13 mls
But u - rrDN wbere.D = Diameter of wheel- 60 '
:. 15.13 = 1£ x ~; 200
(ii) Diameter 01 the jet (d)
or D:=. 60 )( 15.13 = 1.44m. Ans •.n: x 200
Overall efficiency
But
flo = 0.85S.P. 95.6475 95.6475 ;'( 1000
fl - ______.- - .o-W.P. - (W.P.)- pxgxQxH
100095.6475 x 1000= -.;..;;.;.;....;..~_.;;,..;..;;..;-
1000 x 9.81 x Q x 6095.6475 x 1000 95.6475 x 1000 .' 3
Q = flo x 1000 x 9.81 x 60 = 0.85 x 1000 x 9.81 x 60 = 0.1912 m Is .
(.: W.P. = pgQ/:
...But the discharge, Q =Area of jet x Velocity of jet
1£..2 1£2:. 0.1912 = 4 a- x Vi = 4"d x 33.62 .
.. /4 x 0.1912d = V ... . = 0.085 m :: 8Smm. Ans •et )( 33.62
...(iiz)SizeofbucketsWidth of bucketsDepth of-buckets
= 5 x d = 5 x 85 = 425 mm= 1.2 i< d = 1.2 x 85 = 102mm. Ans,
(iv) Number of buckets on the wbeel is given by equation (ULl7) asD 1.44Z = 15 +'2d = 15 + 2 x .085 :;;15 + 8.5 = 13.S say 24. Ans.
••..Problem 18.12.Determine the power given by the jet of water to the runner of a Pelton wheel which
.having tangential velocity as 20 mls. The net head on the turbine is50 m and discharge through the jet waterts 0.03 m3/s. Theside clearance angle is 15° and take C.V = 0.975.
• Sol. Given:
• Tangential velocity of wheel, U:: Ul ; /1,2 = 20 mlsNet head, H =< 50 m
tilt
••••••••••,.•
Discharge,Side clearance angle,Co-efficient of velocity,Velocity of the jet,
From inlet triangle,
Q:::: 0.03 m3/scp:: 15°
Cy= 0.975"1= c,x "';2 gH= 0.975 x "=2-x""""9:-::.8=1-x--=S=0.:::30.54m1s.
VWl = Vi = 30.54 mlsVrt :: VWi - U1 = 30.54 - 20.0 = 10.54 mls.
Fig. 18.9
From outlet velocity triangle, we havev,= Vrt = 10.54 mls
Vrzcos cP = 10.54 cos 15 = 10.18 mls.As V,z cos cp less than U2, the velocity triangle at outlet will be as shown ~nFig. 18.9.. VW2:: U2 - Vr2 cos cp = 20 - 10.18 = 9.82 mls.Also as f\ is an obtuse angle, the work done per second on the runner,
z: pav':[VW1 - Vwz] x u = pQ [VW1 - Vwz] x u;;: WOO >'.• 03 x [30.54 - 9.82] x 20 = 12432 Nm/s
. . . Work done ~r second 12432 . ADs.Power given to the runner IIIkW ::: 1000 = 1000 = 12.432 kW.
• Problem 18.13. The three-jet Pelton turbine is required to generate 10,000 kW under a net head of400 m. The blade angle at outlet is 15° and the reduction in the relative velocity while passing over the blade
.is 5%. If the overall effICiency of the wheel is 80%, C; = 0.98 and speed ratio = 0.46, then find: (i) the diameter• of the jet, (ii) total flow in m3/s and (iii) the force exerted by a jet on the buckets.
If the jet ratio is not to be less than 10-,find the speed of the wheel for a frequency of 50 hertz/sec and.the corresponding wheel diameter.
Sol. Given:No. of jetsTotal power,Net head,
•,.••or••••
Blade angle at outlet,Relative velocity at outlet
Overall efficiency,Value ofSpeed ratioFrequency,
=3p= 10,000kWH=400mcp = 15°= 0.95 of relative velocity at inlet
Vrz = 0.95 V"1l1c = 0.80Cy=0.98
:: 0.46f = 50 hertz/s
•t~ ~ - . " if'
ft
"•",.,.
(i) Vane angles at inlet and OI•sleiVII 3.183
From inlet velocity triangle, tan 8 = Vw _ Ul = 56.1 _ 26.18 = 0.10641
e = tan··1 .1064:: 6.072° or 6° 4.32'. Ans.
V"i 2.315From outlet velocity triangle tan <p ::: .... -:: _-:: 0.0643
. '!~ 36.0., j.;\' ".0643 :::3.68° or 3° 40.8'. Ans.
(ii) Velocity of flow at inlet and outteiVT, : :: l83 m/s and "[2 = 2.315 m/s. Ans.
'8.8. FRANCIS TURBINE.. The inward flow reaction turbine having radial discharge at outlet is known as Francis Turbine, after.J..hename of J.B.Francis an American engineer who in the beginning designed inward radial flow reaction type"'!If turbine. In the modem Francis turbine, the water enters the runner of the turbine in the radial direction attIIutlet and leaves in the axial direction at the inlet of the runner. Thus the modem Francis Turbine is a mixed
flow type turbine.• The velocity triangle at inlet and outlet of the Francis turbine are drawn in the same way as in case of...inward flow reaction turbine. As in case of Francis turbine, the discharge is radial at outlet, the velocity of~hirl at outlet (i.e., VW2) will be zero. Hence the work done by water on the runner per second will be
,. = pQ [VWl ul]
•"•
And work done per second per unit wught of water striking/s =i[VWl ullVWtUl
qh = _'-,-' .gFiHydraulic efficiency will be gIven by
• 18.8.1. Important Relations for Francis Turbines. The following are the important relations for
Francis Turbines:
••to .40.••'It
"
B1.The ratio of width of the wheel to its diameter 1S given as n = D~.The value of n varies from 0.10
2. The flow ratio is given as,
Flow ratio = ..,f~;H and varies from 0 15 to O.3().
3. The speed ratio = ..,f;~H varies from 0.6 to 0.9.
.. Pro blem 18.23. A Franc is turbi ne Wit" an overall efficiency of 75% is required toproduce 148.25 kW.. power. It is working under a head of 7.62 m. The peripheral velocity = 0.26 12gH and the radial velocity of
• flow at inlet is 0.96 ..,f2gH . The wheel runs at 150 r.p.m. and the hydraulic losses in the turbine are 22% ofthe available energy. Assuming radial discharge, determine :
A (i) The guide blade angie. (iz) The wheel vane angle at inlet,(iii) Diameter of the wheel at inlet, ana (iv) Width of the wheel at inlet.
(AMIE, Fluid Power-Winter, 1975)•
•••.1: I
Sol. Given :Overall efficiency,power produced,Head,Peripheral velocity,
'flo = 75% = 0.75S.P. = 148.25 kW
H= 7.62mUl = 0.26 v2gH = 0.26 x..fi x 9.81 x 7.62 = 3.179 mls
Velocity of flow at inlet, Vlt :: 0.96 v2gH = 0.96 x ";2 x 9.81 x 7.62 = 11.738 mls.
Speed, N = 150 r.p.m.Hydraulic losses = 22% of available energy
Discharge at outlet = RadialVW2 = 0 and Viz= V2
Hydraulic efficiency is given asT}h = Total head ~::!e~t~:;tdraUl!~}oS~~
=H - .22 H = 0.78 H :::;0H H "
VWt Ul
T}h = gHBut
VW1UlgH :0.78
v: _ 0.78 x g x HW1 - Ul
t:= .u, .__' Ivw,~
0.78 x 9.81 x 7.62 1834 .= -= filS3.179 . Fig. 18.21
(i) The guide blade angle, i.e., a. From inlet velocity triangle,
VII 11.738 0 '-4tan a ::V = 18 34 ::: .0
Wi .'
. a:: tan-1 0.64:: 32.619= or 32° 37'. Ans.
(ii) The wheel vane angle at inlet, I.e., £)Vr .i . >:18
tan H :=: ..._-_. '. -.~ -_··--·--",0774VW1 •.. U ;",34··· ,).179 .
e :::tan-1 .774::;:37.74 or 37° 44.4'. Ails.
(iii) Diameter of wheel at inlet (D1)·
Using the relation,ill1N
Ul=- 60
D1~ 60 x Ul = 60 x 3.rr?. = 0.4047 m. Ans.1t: x N 1t ): 50
,(iv) Width afthe wheel at inlet (B1)
S.P. 148.251)0 :: W-p = -W p.n.. . .'"
•••••••• or
••••
w p, == _WH = P x S x Q x H = 1000 x 9.81 x Q x 7.62.. 1000 1000 1000
, 148.25 148.25 x 1000flo == 1000 x 9.81 x Q x 7.62 = 1000 x 9.81 x Q x 7.62
1000
But
148.25 x 1000 148.25 x 1000 3Q = 1000 x 9.81 x 7.62 x '10 = 1000 x 9.81 x 7.62 x 0.75 = 2.644 m Is
Using equation (18.21), Q == rrJ)1 x Bl X "1,2.644:= :rt x .4047 x Bl x 11.738
B z: ,~'; AO~~11.738 = 0.177 m. ADs.
Problem 18.24. The following dau is given for a Francis Turbine. Net head H = 60 m j Speed• N = 700 r.p.m ; shaft power == 294.3 kW tlo;: 84% ; l1h ::::93% ; flow ratio = 0.20; breadth ratio n = 0.1 ;
Outer diameter of the runner :; 2 inner diameter of runner. The' thickness ot vanes occupy 5% of• circumferential area of the runner, velocity offlow is constant at inlet and outlet and diScharge is radial at
outlet. Determine:•••••••••••••••••••
(I) Guide blade angle, (it) Runner vane angles at inlet and outlet,(iiI) Diameters of runner at inlet and outlet, and (VI)Width of wheel at inlet.Sol. Given:Net head,Speed,Shaft power
Hc60mN::.: 700 r.p.m.
= 294.3kW
Fig. 18.22
Overall efficiency, '10:: 84% = 0.84Hydraulic efficiency, '1l: = 93% = 0.93
VIlFlow ratio, Y2gH = 0.2,0
V4' ::::0.20 x ;!2gH/1
7. '/2 x 9.81 x-60 = 6.862 mls
BID;':(U
Outer diameter, Dl = 2 x Inner diameter = 2 x D2
Breadth ratio,
Velocity of flow, VIl :. Vh = 6.862 m/s.
Thickness of vanes :: 5% of circumferential area of runner
:. Actual area of flow;.: 0.95 reD1 x B1
Discharge at outlet :,.:RadialVW2 = 0 and Vh ::V2
S.P.110=W.P.
0.84:: 2~4.3W.P.
Using relation,
""•" ... W.P. = 26-S: = 350.357 kW.
W P_ Jf!l_ _ p x g x Q x H _ 1000 x 9.81 x Q x 60
. . - 1000 - 1000 - 1000
1000 x 9i~; Q x 60 = 350.357
Q= 350.3~7 x 1000 = 0 595'" 3/.60 X 1000 X 9.81 . t.. m s.
But
...
Using equation (18.21),Q =Actual area of flow X Velocity of flow
= 0.95n:D1 x B1 x VI1
= 0.9S X 1tXDl x O.lDl x Vh
0.5952 = 0.95 x 1t x DI x 0,1 x D) x 6.862 = 2.048D12
D1=~=O.54m
BlDl =0.1
or
But:. B1 = 0.1 XDi = 0.1 x .54= .054m= 54mm
TqeJdial speed of the runner at inlet,U1 = 'JfD# =:It x 0.54 x 700_= 19.79 m/s.
60 60
Using :relation for hydraulic efficiency, Vw, x 19.790.93 ;::9~81x 60
ToT _ 0.93 x 9.81 x 60 - 27 -, mIYWl - 19.79 - .ee s.
(i) Guide blade angle (e)VI1 6.862
From i.i,letvelocity triangle, tan o. :::_. = - :: 0.248Vw• 27,66,•• (1,:: tan-i 0,248 "'.13.92.8°or 13° 55.7'. Ans.
(ii)Runner vane angles.at inlet and outlet (8 and $)Vi1 6.862
tall e = VWl_ Ul = 27 .66 '~_~19-.7-9= 0.872
e = tan-i 0.872 = 4L09° or 41° SA'. ADs•...F
...1 e1' .ang1 .... ':'h Vl1 6.862. 10m ouuet v ocny tn e, tan 'I' = - = - = -u2 u2 U'"
But'1fD7f'/ Jt x D} N
U2=-= x-60 2 60.54 700= :ItX 2 x :60 '::::9.896 mls.
••••••••••
Substituting the value of ll2 in equation (i),6.862
tan 4> :: 9.896 = 0.6934
;.4> = tan-1 .6934° :: 34.74 or 34° 44.4'. ADs.
(iii) Diameters of runner at inlet and outletD, = 0.54 m,D2 = 0.27 m. ADs.
(iv) Width of wheel at inletBl = 54mm. Ans.
Problem 18.25. (a) Show that the hydraulic efficiency for a Francis turbine having velocity offlow• through runner as constant, is given by the relation.
•••• where
••••••••••• or
•••••••
1
i tar? a+-_._--_ ..'11-~\1
tan e I. f
ex. = Guide blade angle and e::: Runner vane angle at inlet.The turbine is having radial discharge at outlet.
(b) If vanes are radial at inlet, then show llh = 2,;.2 +ta a
Sol. Given:Velocity of flow = Constant.. V/1 = Vf2Discbarge is radial at outlet.
From the inlet velocity triangle,V/l
tan a. :;:--v;!
...(i)
Also Fig. 18.23
(": V't = VWt tan a)
...(u)
(': V.." =0)
•j • ,. ~,I c .; '" • !
•,."•••,.••••••
SCROLL CASING
GUIDEVANES
RUNNERVANES
\
\\
•••Fig. 18.26. Main components of Kaplan turbine.
18.9.1. Some Important Points for Propeller (Kaplan Turbine). The following are the importantpoints for propeller or Kaplan turbine .••••
"..•••••••"
1. The peripheral velocity at inlet and outlet are equalrrDcJl
;;.= ....._.60
whereDo = Outer dia. of runner.
2. Velocity of flow at inlet and outlet are equal
::.Area of flow at outlet
rt 2 2="4(Do-Db) •
Problem 18.27.A Kaplan turbine working under a head of 20 m develops 11772 kW shaft power. Theouter diameter of the runner is .3.5 m and hub diameter 1.75m. The guide blade angle at the extreme edge ofthe runner is 35°. The hydraulic: and overall efficiencies of the turbines are 88% and 84% respectively. If thevelocity of whirl is zero at outlet, determine:
(i) Runner vane angles at inlet and outlet at the extreme edge of the runner, and
(ii) Speed of the turbine.Sol. Given:
3. Area of flow at inlet
Head,Shaft power,Outer dia. of runner,
H=20mS.P :' 11772 kWDr> '''' 3.5 m
--------------- - - - - -------------
·suy '/Z 06£ 10 S£O'6£ = [18 I__UR1= <IIlZ'Zl '{,n
118'0 = -"6-6" = tjA = <II Uf.l
...
u ~-sfUI 6'6 = :fA := 'ill pUR S/UI lZ'Zl = -en = In
·SUV '/SS "l8L10 oL6'SL = En; I_uet.= e(IZ'ZI - .,r.,l) l.n _ l~
£1"S = = =eU816'6 tji\: SBu~Api ~lI! i~tu ~ql}O ~~p~ ~UI~llX~ ~1[llBl~llno pUB1~'(Ul18S;l{iue attA ~nlI (])
'SjUI['7''7T _ Pl"Pl - l.n .:'" "'~ OZx 18°6 x 88°0
OZx 18°6 = 88°0In x v1"17l
nS___ =lillIn l'itA.&)U~P!JJ~:l!lnlUpAq10JUOll8J~ ~1[lS1qsn
i., ~£ UBl D UBl 1M
sfUI Vrv ( == 6~-= 6"6= -VA = 1i
~M.:::~ = '0 uet. '~pmeJllApoI~A l~JUJmOld':'A
'sfUl6'6 = 9r;~L= lJi\8Zv'rL
lJA 91zoL = 1jA (SZ90'£ - SZ·Zl).!.:= ;jA x (SL"I- S·t:) v = 8Zv'ILlC Z z lC10
(0001 = d- :.)
zzst 'S!tI
lJi\ x (tqa- tOa)~ = e0s/ UI 8Zv'IL := _OZ x 18'6 x 0001 x vsoO= ~
E 0001 x ZLLll
oz. x 0 x 18'6 x 0001 :=0001 x ZLLH
0001
° °.
yx()xaxd: ZLLH' = vsoO
'1~~~.M. 0001 = 0001 = °do/A~1(lAHxljx8xd elM
.~~ := °ll 'uo~BI~l ~ql~sn °0 = l~llno 18J1N.M.JOAPOI~A
%V8 =Oll
%88 = 'Ill
oS£='OUI ~L"1 = qa
'k>u~pm~ n81~i\()
'A:lU~pm~ ann81pAH'~pme~PBlq~P!BD
'l~~me,p qnHIn1, '(
_ .., I' ; ~ c ",;;J. \ _ ~~ ,,7' i!' h "''''IJ!o:!,C: *.".,_>;:;~ - ,~ "- .- ''tr .,_ "" t (i
••..
•It
"•""'"•••......•••••..••..•..".."••••••
rrDjV(ii\ Speed of turbine i;;. u, 11..-, =: ----.-') .: 6l)
i2.21 = ~_.)(3.5 ~._N60
60 x 12.21. . N = ':Y 3 s.O - = 66.63 r.p.m. Ans.·
r». x ..
Problem 18.28. A Kaplan turbine develops 24647.6 kW power at an average head of 39 metres.Assuming a speed ratio of 2, flow ratio of 0.6, diameter of the boss equal to 0.35 times the diameter of therunner and an overall efficiency of 90%, calculate the diameter, speed and specific speed of the turbine.
(AMIE, Summer, 1981)Sol. Given:Shaft power,Head,Speed ratio,
S.P. = 24647..6 kWH=39m
u1N2gH = 2.0III = 2.0 x V2gH = 2.0 x v2 x 9.81 x 39 = 55.32 mls
Overall efficiency
v;]oJ lIH=t = 0.6...-g.
I, ~.::0.6 < I2gH '"0.6 x v2 x 9.81 x 39 = 16.59 m/s
.:iL,S .><: Diameter of runner:;:;0.35 x Do
lio = 90% :; 0.90
Flow ratio,
Diameter of bos.
Using the relation, -flo = ~~~. where W.P. = P x ~~o~x H190 = 24647.6 = 24647.6 x 1000
p x g x Q xH 1000 x 9.81 x Q x 391000
Q = 24647.6 ~ 1000 = 71.58 m3/s.0.9 x 1000 x 9.81 x 39
But from equation (18.25), we have
Q = ~ (Do2 -Db2) x VIl
71.58 = ~ [Do2._ (.35 Doi] x 16.59
(l) .,
:::~ [D.:? .-. 1225 Do2] x 16.59
~ x .8775 Do"~x 16.59 = 11.433 D/
Do = -v'1fl!i = 2.5 01. Ans.
.. Do = 0.35 x Do = 0.35 x 2.5 = 0.875 DI. Ans.
(iz) Speed of the turbine is given by III =~t'
55.32 = :n: x ~~ x A~
60 x 55.32N = .- 2 " := 422.61 r.p.m. Ans.:n: x ._.
NIPN, = '---s/4 where P = Shaft power in kW
H
N - 422.61 x v'24647.6 _ .122.61 x 156.99 _ 68076 Ars - (39)5/4 - 97.461 - • r.p.m.
(iil) Specific speed* is given by
Problem 18.29.A Kaplan turbine runner is to be designed tp develop 9100kW. Thenet available he"is 5.6 m. If the speed ratio = 2.09, flow ratio = 0.68, overall efficiency 86% and the.diameter of the boss113 the diameter of the runner. Find the diameter of the runner, its speed and the specific speed of the turbin.
Sol. Given:Power,Net head,Speed ratioFlow ratioOverall efficiency,
Diameter of boss
P= 9100kWH:-::5.6m= 2.09= 0.68
Y10 :: 86% = 0.86
=: t of diameter 01runn:1
or
Now speed ratioUl
= v'2gHul = 2.09 x v'2 x 9.81 x 5.6 = 21.95 mls
VIi= v'2gH
VIl = 0.68 x v'2 x 9.81 x 5.6 = 7.12 mls
Flow ratio
or
The overall efficiency is given by, Y10 c:;: I PQH \I P xg. It 1000 -)
P x 1000 9100 x 1000Q = 'p x g x H x Y10 = 1000 x 9.81 x 5.6 x 0.86
(.,' pg = 1000 x 9.81 N/m:.: 192.5 m3Is,
The discharge through a Kaplan turbine is given bv
Q:: ~ [Do2 -L;,~l )< Vf1
or :n: [ 2 Ii Do ,)1. '1192.5 = 4 ~o - \3' j x 7.12
"For the definition and derivation, please refer to page 783 Arts, 18.11 and 18.11.1.
•..A
••••••
= : [ 1 - i1Do' x 7.12
Do =... /4 x 192.5 x 9V :1t x 8 x 7.12 = 6.21 01. ADS.
The speed of turbine is given byreDN
Ul=~
" 60 x ul 60 x 21.95i\j :;;:. = -.- = 67.S r.p.01. AnS'~
nxD 1tx6.21
Th.~ d i b N NVP 67.5xv'9100 7-4£ A. e specitic spee IS given y, s = ---s74 = 5/4 = ....,. OS.. H 5.6 .
Problem 18.30. The hub diameter of a Kaplan turbine, working under a head of 12 m; is 0.35 times• the diameter of the runner. The turbine is running at 100 r.p.m. If the vane angle of the extreme edge of the
runner at outlet is 15° and flow ratio 0.6, find:
••••••..•••••..••••••••
(l)Diameter of the runner, (if) Diameter of the boss, and
(iii)Discharge through the runner.The velocity ofwhiri at outlet is given as zero.
Sol. Given:H= 12m
Db = 0.35 x Do where Do = Dia. of runnerN == 100 r.p.m.<p :::: 15°
VII'" .,12gB = 0.6
Ffl = 0.6 x v2gH = 0.6 x .[2 x 9.81 x 12 •
= 9.2 m/s.From the outlet velocity triangle, VW2 = 0
Viz VI1tanq,=-= - (.: Vtz=Vft =9.2)
U2 U2
Head,Hub diameter,Speed,Vane angle at outlet,
Flow ratio
Fig. 18.28
159.2
tan =Ul
U2 = 9.215= 34.33 m/s.tan
But for Kaplan turbine, U1 = U2 = 34.33
...
reDo x NNow using the relation, U1 = 60
:1t x Do x 100or 34.33 = 60
60 x 34.33 £,.~ ADo = 1 0 = U.;..I"01. OS.:1t x 0 -
Db = 0.35 x Do = 0.35 x 6.35 = 2.3 01. ADS.
Discharge through turbine is given by equation (18.25) as
1.: 2 2 :'1: 2 2Q = 4 [Do -Db] x Vl1 = 4"[6.55 -2.3 ] x 9.2
= ~ (42.9026 - 5.29) x 9.2 = 271.77 013/S. Ans.
Proble0118.31.A Kaplan turbine runner is to be designed to develop 7357.5 kW shaft power. The navailable head is 5.50 m. Assume that the speed ratio is 2.09 and flow ratio is 0.68, and the overall efficien:is 60%. Thediameter of the boss is trd of the diameter of the runner. Find the diameter of the runner, its spesand its specific speed. (AMIE, Fluid Power-Summer, 197.
Sol~Given:Shaft power,Head,
Speed ratio
·..Flow ratio
·..Overall efficiency,
Diameter of boss,
Using relation,
or
p= 7357.5 kWH=5.50m
Ul= ";2gH = 2.09
Ul = 2.09 x ";2 x 9.81 y 5 50' ""21.71 m/s
::v:;~i";0.68'V" :::2.68x v2 x 9.8fxs30- ;: 7.064 m/s.1
flo = 60% =: 0.60
Db =txDoShaft power 7357.5
flo = Wate~power = p x g x Q x H1000
0.60 = 7357.,? x 1000 = 7357.5 x 1000P x g x Q x H 1000 x 9.81 x Q x 5.5
7357.5 x 1000 . 3Q = 1000 x 9.81 x 5.5 x 0.60 = 227.27 m Is.
Using equation (18.25) for discharge,
Q = ~(D/ -Db2»( 'V.
or
·..
And
Using the relation,
~ / D '.227.27 = ~ [D} - tt j
= ~ x ~ Do2 x 7.064:: 4.931<5D/
). !J)64
Do = ~ = 6.788 01. Ails.
1Db = '3 x 6.788 = 2.262 01. Ans.
l· '"
••••••••••••••••••••••••••••••••••
Unit - 5
HYDRAULIC PUMPS
Pumps: definition and classifications - Centrifugal pump: classifications,
working principles, velocity triangles, specific speed, efficiency and performance curves- Reciprocating pump: classification, working principles, indicator diagram,
work saved by air vessels and performance curves - cavitations in pumps, Rotary
pumps:working principles of gear and vane pumps
••••••••••••••••••••••••••••••••••
(b)Mechanical Efficiency Cllm).The power at the shaft of the centrifugal pump is more than the power• available at the impeller of the pump. The ratio of the power available at the impeller to the power at the shaft
of the centrifugal pump is known as mechanical efficiency. It is written as
••••••••••••
:rh . WHm~'.'.e:power given to water at outlet of the pump = 1000 kW
Th~,power at the impellerWork done by impeller per second '
= 1000 kW
• whereS.P. = Shaft power.(c) Overall Efficiency (llo). It is defined as the ratio of power outpur'ofthe pump to the power input
• to tlle pump ..The power output of the pump in kWWeight of water lifted x H", WH",
= 1000 =,1000••••••••••
W VW2XU2::-x kW
g 1000WxH",1000
The power at the impeller in kW
Power at the impellerYl -m - power at the shaft
Work done by impeller per second= 1000
W Vw21l2=-x-
g 1000
W(VW2U2 )'g 1000,
S.P.
[Using equation (19.2)]
...{19.9)ll", =
Power input to the pump ::::Power supplied by the electric motor,; S.P. of the pump.
(~)llo= S.P.
•..{19.10)
Also llq.,= ll""lIt X T\",. ...{19.11)
problem 19.1. The internal and external diameters olthe impeller 01a centriJUgalpump are 200 min
and 400 mm respectively. The pump is running at 1200 r.p.m. The vane angles 01 the impeller at inlet andoutlet are 200 and 300 respectively. The water enters the impeller radially and velocity of flow is constant.'Determine the work done by the impeller per unit weight of water.
Sol. Given:Internal diameter of impeller, Dl = 200 rom = 0.20 mExternal diameter of impeller, D2 :::..400 rom = 0.40 m
l: ..__
••••••••••••••••••••••••••••••••••
Tangential velocity of impeller at inlet and. outlet art,
Ul = ~~N = Jt x 0.2~ x 1200 = 12.56 m/s
rrD-JV Jt x 0.4 x 1200u2 = tic) = 60 = 25.13 m/s.
VA VA UItan9=-=--
Ul 12.56 Fig. 19.4
Vfl = 12.56 tan a = 12.56 x tan 20 = 4.57 mls
Speed,Vane angle at inlet,Vane angle at outlet,Water enters radially * means,
Velocity of flow,
N = 1200 r.p.m.a = 20°cp = 30°a = 90° and Vw,
V. =V,'11 12
and
From inlet velocity triangle,
n
From outlet velocity triangle,
25.13 - Vw~= 4.57 = 4.57 - = 7.915. tan cp tan 30
VW2 = 25.13 - 7.915 :::17215 m/s.
or
The work done by impeller per kg of water per second i~,given by equation (19,1) as1 7215>; 25.13= - V u' = ..------- = 44.1 NOlIN. Ans.g W2 " 9,81
Problem 19.Z.A centrifugal pump is to discharge 0.118 m'Is at a speed of 1450 r.p.m. against a healof 25 m. The impeller diameter is 250 mm, its width at outlet is 50 mm and manometric efficiency is 75%Determine the vane angle at the outer periphery of the impeller. (AMIE, Winter, 1982
Sol. Given:
,"'J'I.I~it'iT:...•..It,_.,.I
\~".~\h~
Q = 0.118 m3/sN = 1450 r.p.m.
Rm: 25 mD2 = 250 mm :: 0.25 m
B2 = 50 mm :: 0.05 m
Discharge,
Speed,
Head,
Diameter at outlet,
Width at outlet,
Manometric efficiency, llman = 75%:: 0.75,
Let vane angle at outlet = cpTangential velocity of impeller at outlet,
u2 = rrD:' = Jt x O.2~x 145(~:: 18.98 m/»
--------------------~.-...----".--------------Fig. 19.5
*lfin the problem, this condition is not given even then the water is assumed to be enteringradially unless statedotherwise .
••••••••••••••••••••••••••••••••• ~
••Q ",.JrD2f32 x VI., Q 0.118yh ~ it-D-J32 = :n:x 0.25 x .05 = 3.0 mls.
gHm 9.81 x 25llman = VwP2 = VW1 X 18.98
Vw::;; 9.81 x 25 = 9.81 x 25 = 17.23.7. ilman X 18.98. 0.75 x 18.98
••••••••
Discharge is given by
Using equation (19.8),
...From outlet velocity triangle, we haw:
Vh 30tan cp = (U2 _ Vwz) = (18.98 ~ 17.23) = 1.7143
• cp = tan-11.7143 = 59.74° or 59° 44'. ADs.
• Problem 19.3.A centrifugal pump delivers water against a net headof 14.5metres and a (lp.signspeedof1000 r.p.m: The vanes are curved back to an angle of 300 with theperiphery. The impeller diameter is 300 mm
• and outlet width 50 mm. Determine the discharge of the pump ifmanometric effICiency is 95%.• (AMIE,Winter, 1983; Osmania University, 1992)
••••••••••••
Sol. Given:Net head,Speed,Vane angle at outlet,
h", :c. 14',10
N..,: 1000 r.p.m,Cfl ;-:: 30'"
Impeller diameter means the diameter of the impeller at outlet:. Diameter, D2 = 300 mm = 0.30 mOutlet width, B2 := 50 mm = 0.05 mManometric efficiency, l1man= 95% = 0.95Tangential velocity of impeller at outlet,
_ rrlhN _ :n:x 0.30 x 1000 _ 1570 /U2 - 60 - 60 -. m s.
NoW'using equation (19.8), llman = V gHmWz X U2
. . Discharge,
0.95 = 9:81 x 14.5·V;J2 X 15.70
" 0.95 x 14.5 954 /\I . -,- m s",' 0.95 x 15.70 -. .
Refer to Fig. 19.5. From outlet velocity triangle, we have
• ~ 0 % %• tan ct>:::: (~.;_ Vwz) or tan 30 = (15.70 _ 9.54) = 6.16
••••
Vf2 = 6.16 x tan 30° = 3.556 mls
Q = :n:D-J32x Viz=:n: x 0.30 x 0.05 x 3.556 m3/s = 0.1675 m3/s. Ans.
••••.,.'••••.'•••••••••••••••••••••••
\ ' -- --------
"!..•
Prohlem 19A.A ce!Jtrifugal pump having outer diameter equal to two times the inner diameter tnrunning at 1000 r.p.m. work's against a total head of 40 m. The velocity of flow through the impeller is constaand equal to 2.5 in/soThevanes are set back at an angle of 400 at outlet. If the outer diameter of the impellis 500 mm and width at outlet is 50 mm, determine:
(i) Vane angle at inlet,(iil) Manometric efficiency.Sol. Given:Speed,
(iii Work done by impeller on water per second-ana(Delhi University, May 199
N:: 1000 r.p.m.
Head,Velocity of flow,
Vane angle at-outlet,Outer dia. of impeller,
Hm=40mV'I:: Vh = 2.5 migq, = 400
D2 = 500 mm = 0.50 mD2 0.50o,=2=2= 0.25 m
B2 = 50 mm = 0.05.m
Innerdia.ofimpeller,
W!dth at outlet,
and
Tangential velocity of impeller at inlet and outlet are_ 1EDtl{ _ 3t x 0.25 x 1000 _ 13 09 I
U1 - 60 - 60 -. II1;S
U2= 1rD.Jtl = 3t x 0.50 x 10QQ.= 26.18 m/s.60 60
Fig. 19.6.
Discharge is given by,
(l)Vane angle atinlet (8).
From inlet velocity triangle tan 8:: ~1 = 1~:~9:: 0.191
. 8:: tan? .191 :: 10.810 or 10048'. Ans.(ii)Work done by impeller on water per second is given by equation (19.2) as
w pxgxQ~-Vw U2= -xVw xU2g 2 g 2
:: 1000 x 9.81 x 0.1963 t r 2618.9.81 x YW2 X "
But from outlet velocity triangle, we haveVI' 2.5
tan q,:: J2 =U2- VW2 (26.18 "' V";2)
... 26.18 - V. :: 2.5 = ~:: 2.979W2 tan q, tan·40
Vw2::: 26.18..."2.979 = 23.2.m/s.
Substituting this value of VW2 in equation (z),.we get-theework done by impeller as
~ 1000 x 9~~:lx}.196;:x 23.2 x 26.18
:::1192.2.7.9Nm/s. ADs.
.................................. '~\
.:1
"(iii) MaDometric emciency (TIman)'Using equation (19.8), we have
gHm 9.81 x 40TIman= V = 23 2 26 18 = 0.646 = 64.6%. Ans.
" W2U2 .x.Problem 19.5.A centrifugal pump discharges 0.15 m3/s of water against ahead 0/12.5 m, the speed
"tthe impeller being (JOOr.p.m. The outer and inner diameters ofimpeller are 500 nun and250 mm respectively«:the vanes are bent back at 35° to the tangent at exit. If the area offlow remains O.07m2from inlet to outlet,
'"i.:alculate:.. (z) Manometric efficiency of pump, (it) Vane angle at inlet, and
(iii) Loss of head at inlet to impeller when the discharge is reduced by.40% without changing the speed(AMIE, Summer, 1988)
~~2 ~IVft4'
Vr2.
As area of flow is constant from inlet to outlet, then velocity of"flow will be constant from inlet to outlet
••
"or.."•,...
and
"""".."""•..
Sol. Given:Discharge,Head, n;» 12.5m
Speed, N = 600 r.p.m,
Outer dia.,-.lInnerdia.,Vane angle at outlet,Area of flow,
D2 = 500 mm = 050mD, = 250 mm = 0.25 m
cp = 35°= 0.07 m2
Discharge ::.Area of flow x Velocity of flowO. i5 = 0.07 x Velocity of flow
V f fl U.i5 -l 14 /. . elocity 0 ow::: ).07 '0:: i.. m s
Vii :::Vh = 2.14 m/s
Fig. 19.6 (a)
Tangential velocity of impeller at inlet and outlet are_ Jt[)lN _ jt x 0.25 x 600' ~ 7 8Sm/
U1 - 60 - 60 -. s
_ Jt[)-JV _ jt x 0.50 x 600 _ 15 70 m/U'2 - 60 - 60 -. s
Vh 2.14VW = U2 - _- = 15.70 - 350 = 12.64 m/s
2 tan cp tanFrom outlet velocity triangle,
(i)Manometric efficiency of the pumpg x n; 9.81 x 12.5
Using equation (19.8), we have TImon= V. = 1264' .157 = 0.618 or 61.8%.W2XU2 • x .
Ans.
(it) Vane angle at inlet (8)
tan e = Vii = 2.!~ = 0.272Ul 7.85
e = tan-1 0.272 = 15° 12.'. Ans,
From inlet velocity triangle,
·...........'.....................
Problem 20.1.A single acting reciprocating pump, running at 50 r.p.m; delivers 0.01 m3/s of water.~ diameter of.the piston is 200 mm and stroke length 400 mm. Determine:
,(i) The theoretical discharge of the pump, (il) Co-efficient of discharge, and
• (iiI) Slip and the percentage slip o{ the pump.Sol. Given:Speed of the pump,Actual discharge,
• N= 50r.p.m.Qact= .01 m3;sD = 200 rom = .20 m•• Dia.ofpiston,
•••••••
:. Area,
Stroke, L :: 400 mm = 0.40 m.(l) Theoretical discharge for smgle-acting reciprocating pump is given by equation (20.1) as
Qlh =. '~)(~oxN_:: .031416~.40x50=0.01047m3/s. Ans.
(ii) Co-efficient of discharge s given by
C = Qaef = _Q:~_!_ = 0855-a Qfh .01C47 .. 7 • Ans.
(iii) Using equation (20.8), we getSlip = Qth - Qae(:': .01047 - .01 = 0.00047 m3/s.
:: (Qth ,- Qaet), x 100 = (.01047 - .01) x 100a; .01047
Ans.
••• Problem 20.2. A double-acting reciprocating pump, running at 40 r.p.m., is discharging 1.0 m3 of.water per minute. The pump has a stroke of 400 mm. The diameter of thepiston is 200 mm. Thedelivery and.'mction heads are '20m and 5m respectively. Find the slip of the pump and power required to drive the pump.
Sol. Given:
And percentage slip
••••••••••••
tV 40 r.p.m.Speed of pump,. . . 3, 1.0 3/ 001666 3/(jact::;' }.O m fmm = 60 m s =. m sActual discharge,
Stroke,Diameter of piston,
L :::::400 mm :: 0.40 m
D :: 200 rom :::0.20 m
A:: Jt D2 = Jt (.2)2 = 0.031416 m24 4
hs= 5 mhd= 20 m.
:. Area,
Suction head,Delivery head,Theoretical discharge for double acting pump is given by equation (20.5) as,
Q_ 2ALN _ 2 x .031416 x 0.4 x 40 _ 01675 3/th _ _ -. m s.
60 60 '
Slip = Qth - ~et = .01675 - .01666 = .00009 m3/s. Ans.Using equat~-(W:8),
.>:
• •••••••••••••••••••••••••••••••••
i_) • __ _~ _ .
Power required to drive the double acting pump is given by equation (20.7) as,p =2 x pg xALNx (hs + hd) = 2 x 1000 x 9.81 x .031416 x .4x 40 x (5+ 20,
60,000 60,000 .= 4.109 kW. Ans.
20.6. VARIATION OF VELOCITY ANDACCELERATION IN THE SUCTION ANDDELIVEJPIPES DUE TO ACCELERATION OF THE PISTONIt is mentioned in Art. 20.3 that when crank starts rotating, the piston moves forwards and backwa
in the cylinder. At the extreme left position and right position of the piston in the cylinder, the velocity ofpiston is zero. The velocity of the piston is maximum at the centre of the cylinder. This means that at the stof a stroke (may be suction or delivery stroke), the velocity of the piston is zero and this velocity beconmaximum at the centre of each stroke and again becomes zero at the end of each stroke, Thus at the beginniof each stroke, the piston will be having an acceleration and at the end of each stroke, the piston will be havia retardation. The water in the cylinder is in contact with the piston and hence the water, flowing from 1
suction pipe or to the delivery pipe will have an acceleration at the beginning of each stroke and a retardatiat the end of each stroke. This means the velocity of flow of water in the suction and delivery pipe will notuniform. Hence an accelerative or retarding head will be acting on.the water flowing through the suctiondelivery pipe. This accelerative or retarding head will change the pressure inside the cylinder.
If the ratio of length of connecting rod to the radius of crank (i.e.,Llr) is very large, then the motiof the piston can be assumed as simple harmonic in nature. Fig. 20.3 shows the cylinder of a reciprocatisingle acting pump, fitted with a piston which is connected to the crank. Let the crank is rotating at a consuangular speed.
Fig. 20.3. Velocity and acceleration r.f oiston.
(l) =Angular speed of the crank in rad./s,A = Area of the cylinder,a = Area of the pipe (suction or delivery),1= Length of the pipe (suction or delivery), andr = Radius of the crank.
In the beginning, the crank is atA (which is called inner dead centre) and the piston in the cylinderat a position shown by dotted lines. The crank is rotating with an angular velocity (l) and let in time 't' seconcthe crank turns through an angle e (in radians) fromA (i.e., inner dead centre). The displacement of the pisnin time 't' is 'x' as shown in Fig. 20.3.
Now e = Angle turned by crank in radians in time 't'
Let
= wtThe distance x travelled by the piston is given as
x = DistanceAF = AD -_.FO
... 1
· :~"
••••••
tin (A ..]2hf= _,~x """'00".dx 2g a
3. When e = 180°, sin 180" :::.:) h.> 0. . Maximum value of loss of head due to friction;
2. When e = 90°,
4[1 (A ']""(hf)max = d x 2g x -;; mr...(20.20)
Problem 20.3. The cylinda bore diameter of a single acting reciproc.ating pump is 150 mm and its.stroke is 300 mm. The pump runs at 50 r.p.ln. and lifts water through a heighr of25 m. The delivery pipe is
22 m long and 100mm in diameter. Find the heorettcu] }isclwrge and the theoretical power required to run• the pump. If the actual discharge is 4.2 iitres find the percentage slip. Also determine the acceleration head• at the beginning and middle of the .ieIi>2' .', (Fluid Power Engg., AMIE, Summer, 1987)
Sol. Given :
•
•••••••••••••••••••••
Dia. of cylinder, D -r: 150 rum ,: (US m
A = (:)( 0.152:= 0.01767 m2
L :::300 rum := (L3 m
N -= 50 r.p.rn
Ans.
:. Area,
Stroke,Speed of pump,Total height through which water is lifted,
H·;;25m
Length of delivery pipe,Diameter of delivery pipe,
ld'~ 22 mdd = 100 mm = 0.1 m
Qaet:::: 4,2 litres/s = 1~'~0ru3/s = 0.0042 m3/s.Actual discharge,
(l) Theoretical discharge (QthjTheoretical discharge for :J single~acting reciprocating pump is given by equation (20.1), as
r: A._.. ><._b..:-}'!_ z: (l~01767x 0.3 x 50 := 0.0044175 m3/s,,' 60 60
! i)'l.~' I 7"-. , idOn litres/s = 4.41 '15Htres/s. Ans.
(ii) Theoretical power (I'I
Theoretical power is given by'oJ _ crJleort'tiC(~~eight of water lifted/s) x Total height: I _" 1000
P x g x Qth x H- --,------_.._--- 1000
1000 x 9.81 x 0.0044175 x 251000
::::1.0833kW. ADS.
(iil) The percentage slipThe percentage slip is given by,
(o;-a: ) (4.4175 - 4.2) 100 492111% slip = ~,-., - x 100 = 4.4175 x =. 70.
':
•e·•••••.'••••••.;•••••.'•••••e·
••••••••
--- -
(iv) Acceleration head at the beginning of delivery stroke.The acceleration head in the delivery pipe is given by equation (20.15) as :
Id A ~hod = - x - (J)~r x cos eg ad
where ad =Area of delivery pipe = ~ x (o.li = 0.007854
21fN 2" x 50(J) =Angular speed = 60 = 60 = 5.236
C nk di L 0.3r = ra ra IUS = 2 =2= 0.15 m
_ 22 0.01767 :i;had - 9.81 x 0.007854 x 5.236 x 0.15 x cos e = 20.75 x cos e
At the beginning of delivery stroke, e = 0 and hence cos e :::1• • had = 20.75 m. Ans.(v) Acceleration head at themiddle of delivery stroke.At the middle of deliver}' stroke, e = 90° and bence cos 6 = U,
hod = 20.75 x 0 = O. Am,
20.8.INDICATORDIAGAAM
= 10.3 m of water,L = Length of the stroke,hs :: Suction bead, andb« = Delivery bead.
During suction stroke, the pressure head in thecylinder is constant and equal to suction bead (hs), wbicb isbelow the atmospberic pressure bead (Harm)by a beigbt of hs·The pressure bead during suction stroke is represented by aborizontalline AB wbich is below tbe line EF by a height of
•1~lw ,~:1:1 IWi0:1::>if/\lIniI.IJ1%a.
+
DEUVERY STROKE
The indicator diagram for a reciprocating pump is defined as tbe grapb between the pressure beadthe cylinder and tbe distance travelled by piston from inner dead centre for one complete revolution of 1
crank. As the maximum distance travelled by tbe piston is equal to the stroke length and bence the indica'diagram is a graph between pressure head and stroke length of tbe piston for one complete revolution. 1pressure bead is taken as ordinate and stroke length as abscissa.
20.8.1. Ideal Indieator Diagram. The grapb between pressure bead in the cylinder and stroke lenlof the piston for one complete revolution of the crank under ideal conditions is known as ideal indicadiagram. Fig. 20.4 shows the ideal indicator diagram, inwhich lineEF represents tbe atmospheric pressure head equalto 10.3 m of water.
Let Harm = Atmospberic pressure headcto
hd
1 E ----r~A B S
Hotm SUCTION STROKE t la-3m
l * ,. STROKE LENGTH.-I ~ •
_ STROItE LENGTH
Fig. 20.4. Ideal indicator diagram.
'h's' .During delivery stroke, the pressure bead in the cylinder is constant and equal ~ deh~ery bead (which is above tbe atmospheric bead bya height of (hd)' Thus, the pressure bead dunng delivery strokrepresented by a horizontal line CD which is above the line EF by a height of h~.,Tbus, for oiJ,CQmIrevolution of tbe crank, tbe pressure head in the cylinder is represented by the diagram A-B-C-D-A. 1
diagram is known as ideal [ndicator diagram.
Problem 20.4. The length and diameter of a suction pipe of a single acting reciprocating pump a5 m and 10 em respectively. The pump has a plunger of diameter 15 em and a stroke length of 35 em. 1centre of the pump is 3m above the water surface in the pump. The atmospheric pressure head is 10.3 mwater and pump is running at 35 r.p.m. Determine:
(l) Pressure head due to acceleration at the beginning of the suction stroke,(il) Maximum pressure head due to acceleration; and(iil) Pressure head in the cylinder at the beginning and at the end of the stroke.Sol. Given:Length of suction pipe,Diameter of suction pipe,
:. Area,
(,::5md; ,;,10 em ~ ;i. ,;'
a ..~ :,~.d L:::c. - .; I ' -'- I)C7854m2J 4 ,. 4
Diameter of plunger,
:. Area of plunger,
Stroke length,
D:;; 15 cm e (1.15n,
Crank radius,
1tD2 :it 1-2 01'767 2.4 :' - :: - X , , ::. m4 4
L c::: 35 em = 0.35 mL 0.35 .r ;;:;;- :: -- :: 0,175 m2 2
Suction head, hs :: 3 mAtmospheric pressure head, Hatm:.: 10.3 m of waterSpeed of pump, N =' 35 r.p.m.Angular speed of the crank is given by,
(I) _. 2JTN :: 2:itx 3S :: 3.665 rad/s._. 60 60, he : . , . iven by equation (20 14) as(z) The pressure head due to acceleration in t e suction pIpe IS gr .
h :: ~'-X :i_ x (1)« cos 8as 0 a .
b S
At the beginning of stroke, B :: Oand lienee cos 8/., A
It ...;:;; -- '" .-. )(•• a._, g as
(ii) Maximum pressure head due to acceleration in suction pipe is given by equation (20.16), as
(h) ::!!. x A x o}r:: 2.695 m. Ans.as max g as .
(iil) Pressure head in the cylinder at the beginning of the sucti~n stroke (Refer to FIg. 20.S):: hs + has:: 3.0 + 2.695 == ).695.
This pressure head 'in the cylinder is below the atmos~he~c pr~ssur~ hood.:. Absolute pressure head in the cylinder at the beginning or suction stroke
= Atmospheric pressure head - 5.695= 10.3 - 5.695 = 4.605 m of water (abs.) Ans,
di h li de r at the end 01' suction strokeSimilarly, pressure hea m t e cy Ill. f' _ 0 2 "'-q'" :: 0.305 m below atmospherictlr.essure)l:::h ·-has - 3, .u,:J
':':1·~i.3_·0.305 " 995m .)fwater (abs.) Ans.
;:_--x -,01767 x 3.6652 x .175 = 2.695 m. A- 9.81 :)07854
• ••••••••••• I ••••••••••••••• I I I
••
..
respect~velyand wate;'is d:~rro~Ze"~20.4 the lengthanddiametero· . .: . .' Ndetermine : vered I: y fhe Dump to a tank hi h . {the,dellverypipe are 30 m and 10. W tc 1S 20 m above th em(I) Pressure head due to. " I . e centre of the pump,
('~ .zcce eratton at the b . .ll) Pressure head in the ..vli d eginntngof delivery stroke
(
•• ....J In er at the beg' . .f 'lll) Pressure head in the cyli J ' tnntng OJ the delivery stroke and, .naer at the end ft' d li 'Sol. Given : ~. ne e Iverystroke.
~ngth of delivery pipe, id = 30 mDIameter of delivery pipe,
.', Area of delivery pipe,
de' = 10cm = 0.1 m_][ 2 "l,
ad - 4 dd = 4 (.ll= .007854m2
Diameter of plunger,
:. Area of plunger,
Stroke length,
D = 15em == 0.15 m" ][ 2 rr" ::.:'4D ,: 4 (.15)2= .01767m2
35 cm « 'US m
Crank radius ,} r -v r1... d.:.U.~- :: --;;-' =: 0.175m:.~ '-,
Delivery head,Speed of pump,
Angular speed,
N "" 35 f.p.m.. .. 2rrN 2][ x 35U! :;:;60 = 60 = 3.665 rad/s.
(1) Using equation (20.15) we get the pressure head due to acceleration in delivery pipe as
I fd A 2"ad = - X - (l) r COS e
g ad
At the beginning of delivery stroke, 8 = 0 and hence cos e = 1.:. Pressure head due to acceleration at the beginning of delivery stroke becomes as
. I fd A 2lad = - x --- (I) rg ad
.~,i2_ x .:01767 2 -9.81 '.007854 x (3.665) x 0.175 -16.17 m. Ans.
(il) From Fig. 20.5, the pressure head in the cyHnder at the beginning of the delivery strokez: Fe' .: FC + Ce' = (hd+ hod)m of water above atmospheric head-,;,20 + tlt.17::: 36.17m of water above atms.:: 36.11 -to Atmospheric pressure head.:.::36.17 + 10.3:::46.47m (abs.) Ans.
(iii) The pressure head in the cyHnder at the end of delivery stroke= ED' above atmospheric pressure head
::: (ED - DD') ::: (ftd - hod).: 20 -16.17 = 3.83m of water above atms.= 3.83 -+ 10.3= 14.13 in (abs.) Ans.
problem 20.6.A single acting reciprocating pump has piston diameter 12.5em and stro1relength30 em. The centre of the pump i.; 4 m above the water level in the sump. The diamerer. and length of suctionpipe are 7.5em and 7m respectively. The separation occurs if the absolute pressure head in the cylinder tlMring
