A TextBook of Fluid Mechanics and Hydraulic Machines - Dr. R. K. Bansal.pdf

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Dr. :R.K. Bans t

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CONTENTS Chapter

Pages

Chapter L PrOllerties of Fluids

1-34

1.1. Introduction 1.2. Properties of Fluids 1.2. 1. Density or Mass Density 1.2.2. Spedfic Weight or Weight Density

1.2.3. 1.2.4.

Spocific Volume Spedfic Gravity Solved Problems 1.1- 1.2

2 3

Kinematic Viscosity Newton's Law of Viscosity 1.3.3. 1.3.4. Variation of Viscosity with Temperature 1.3.5. Types of Fluids Solved Problems 1.3-1.19 Thermodynamic Properties 1.4. 1. Dimension of R 1.4.2. Isothermal Process 1.4.3. Adiabatic Process Universal Gas Constant 1.4 .4. Solved Problems 1.20-1.22 Compressibility and Bulk Modulus Solved Problems 1.23-1.24 Surface Tension and Capillarity 1.6. 1. Surface Tension on Liquid Droplet 1.6.2. Surface Tension on a H"llow Bubble 1.6.3. Surface Tension on a Liquid Jet Solved Problems 1.25- 1.27 Capillarity 1.6.4. Solved Problems 1.2&--1.32 Vapour Pre~sure and Cavitation Highlights Exercise

1.3 .2.

1.5. 1.6.

1.7.

1 1

2 2

l.3. Viscosity Units of Viscosity 1.3. 1.

104.

1

Chapter 2. Pressure and Its Measurement

3 5 5 6 6 6 17 18 18 18 19 19 21 22

23 23

"2424 25 26 29 30 30

3....8

2 . 1. Fluid P ressure at a Point 2.2. Pascal"s Law 2.3. Pressure Vllristion in a Fluid at Rest Solved Problems 2.1-2.7

35 35 36 37

(ix)

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IL 2.4. Absolute. Gauge, Atmospheric and Vacuum Pressures Solved Problem 2.8 2.5. Measurement of Pressure 2.5.1. Manomet/'/1 {( shaft IIml" collCl'IIlrir sIN'I'e. Till' s/,,{'\,I' IIIll1illS {/ sP('('(/ of 50 ollis, W/1l'1I II lorn" of 40 N is IIPIJ/inf 10 the S/I'I'I'I' parallel 10 IIII' slmfl. /){'/('flllillf IIIl' SpUI/ if {/ jora of 200 N is (If'l'lin/. Solu t ion. Gi ven: Speed of sleeve, fit = 50 em/s when force. FI =: 40 N.

Let speed of sleeve is "2 when force. F2 = 200 N. Using relation

t =

I.l

till dy

rorce F where t = Shear stress'" - - • Are a A 't'ighl fijlt'd by lilt' h)'llftllllie prt'JJ ",ht'll Iht' foret' applinl (1/ lilt' plllllger is 500 N. Solulion. Give n : Dia. of ram. /) '" 30 em = 0.3m Dia. of plunger. d", 4.5 em = 0.045 m

Force on plunger. Find weight lifted

"'",5OON

"W

Area of ram. Are a of plunger.

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IL 138

Fluid Ml>chanics

Pressure il1! cnsity due to plun ge r '"

Force o n pl un ger A rea of plun ger

F 500 2 = - = - - - N/m .0 0 159

(l

PLU NGER

Due to Pasc al's law . the inte ns it y o f pressu re will be eq ua ll y trans mi u ed in a ll di rcl:1 io ns. He nce the pressure inte ns it y at the ram 500 ::: 3 14465.4 Nfrn 1 .00 159

: -:-_W""";'~'hC'_ '" ~ '"

But pressure intensit y a! ram

Area of ram

w

.07068

A

• Fig. 2.3

_'_V_ N/m 1 .07068

= 3 14465.4

We ight = 3 [4465.4 x.07068 = 22222 N '" 22.222 kt'l. Ans . Problem 2.2 A Il)'liralllir pre SJ bas (/ ralll of 20 011 ,Iimll" fPr (11111 (I pi IIl1ga of 3 Til! (liw/lFler. It is usn i for liftillg (/ wi'iglll of 30 liN. Pilld lilt' fOITe rt' qllirn/ mille p/1I1I8e r.

Solution. Give n : /) '" 20 em = 0.2

Di a. of ram.

Area of ram. Dia. of plunger Area of plun ge r. Weig lll lifted.

A '"

III

~ /)2 '" ~ ( . 2) 2 '" 0.03 14 111 2

4 4 ll=3cm =O.03 m

"'" ~(.03) 2 = 7.068 X 10- 4

rn !

4 W =30 kN = 30x IOOON =30000 N.

See Fig . 2.3. Fo rce F Pressure int ensi ty develo ped due 10 p lun ge r ", - - == - . Area " By Pasca l" s La w. th is pressure is tralls mi ltcd equa ll y in all d irect io ns Hence press ure tran sm itted at th e ram == Fo rce actin g o n ram

!...

" inte nsity X Area o f ram = Pressure F

= - x A ", But force al:tin g o n ram

"

F x .03 14 N 7.l168 x 10-4

= Weig ht li fK"d '" 30000 N

30000 :

Fx .03 14 7.068 x 10-4

F: 30000 x 7.068 X 10-4

.03 14

'" 675.2 N. Ans .

Problem 2 .3 Ca/n'/(/((' f/If prnsllr(, (/II(' to a CO/UIIIII of0.3 of(n ) w(/(a. (b) (III oil ofSI'. (c) Ilwrcur )' of sp. gr. 13.6. Til/.;(' i/('Il$ify ofw(lfa. p '" /000 kg/IIIJ. Solution. Give n : Height o f liq uid co lumn. Z=O.3 m.

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,~ r.

0.8. (II ld

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IL Pressure ilnd its Measurement

39 1

The press ure al an y point in a liquid is given by equation (2.5) as p'" pgZ «(I) For

wal~r.

p

= 1000 kg/Ill)

I' = pgZ '" 1000 x 9.&1 x 0.3 '" 294) Nlm 2 -_ ['(j'T 294] Ntem !-_ O"94\NI ! A liS. . ._ . {fll. (b) For oil of sp. gr. 0.8.

From eq uation ( 1.IA). we know lhat Ihe densit y of a fluid is eq ual to specific grav ity of Iluid multiplied hy density of water.

Density of oil. Now pressu re,

Po = Sp. gr. of oi l x Dens it y of wakr = 0.8 x P '" 0.8 x HXXl '" 800 I;g/m3

(Po= [k nsi ly of oil)

1'=PoxgxZ = 800 x 9.8 1 x 0.3 = 2354.4

~

= 23s:.4

10

Ill '

= 0.2.\54

~. em'

~. AilS.

,m

(c) For mercury. sp. gr.

'" 13 .6 Fro m equa tion (l. IA) we know that tile den s ity of

J

fluid is eq ua l to specific g ravi ty of fluid

multiplied by densit y o f water

Densi1y of mercury.

p. = Spcl:ific grav it y of mercury x Density of water = 13.6 x 1000 = 13600 kg/m l P=P.xgxZ

'" 13600 x 9.81 x 0. ] " 40025 ~

40025 - - , - " 4.002

10

N

--!'

em

N

~ , 111 -

Ans.

Problem 2.4 Till'" prnSlln illlnlsily (1/ (I poim ill (lfluid is gil"nl 3.924 Nlnl/. Find Iltf ("orrnpolll/illg Iti'ighl offluid ...hi'llilti' fluid i~ ; ( II) W(I(Fr. lind (b) oil OfS!" gr. 0.9. Solullon. Given: N , N Pressure imensity. 1''' ].924 - , "].924 x 10 ~,. cm -

m-

The cU lTcsponding heigh1. Z. of 1hc fluid is givcn by eq ua1i u ll (2 .6) as

z== - " -

I'X ,

(u ) For watcr.

p " 1000 kghn·l

z== ~"

I'X,

(b) For uil. s p. gr.

:. Density of oil

lOOOx 9 .81

- 4 III uf wa ter. An s.

" 0.9 l Po" 0.9 x 1000 " 900 kg/m·

Z

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3.924 x 104

__ _ ,_, _ __ 3 o~ 92~4~X~'~O~' -:; == 4.44

Po xg

900 x9.81

11\

uf uil. Ans.

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Fluid Ml>chanics

Problem 2.5

All oil of sp. gr. 0.9 is

("OIlWillt'd

ill

(I

"rcssure at J)* + Pressore due to

(10 - 2x) ern of mercury

1910 + 1000 x 9.8 1 x

i'OO ("""'10-")

..--1+"' ~

to em

(10-2.,)

",0 + (13.6 x I OOO) x 9.8 1 x

or or

--------

-- -- - -

100

J,j ,-r

( 10 _ 2.)

11,. __ ,,- _L

~B

- -1:0,- -

Di vid illg by 9.81. we get 1000 + 100 - lOx = 1360 - 272x

272x - l0-,= 1360-1100 262x'" 260

or

2 6() ,= -'" 0.992 em 262

New diffcfCIlCC of mercury '" 10 - 2 l em =10 - 2 x 0.992

Fig. 2.11 (b)

'" IWI6 em. Ans.

U-t"""

Problem 2.12 Fig. 2.12 .\110;"$ "colliC(IIl'l'Sse/ /I(I\'illg iu 011//1'1 til A /0 wllirh fI mlUlOllli'll'r i.1 rOlllll'("/I'l/. H", rNII/ill8 oflhl' /II"HomNer g;,'''" ill IIIl' figllre S/Wlt'" wlwllll", ""$.W'I is ('/"1'1)'. Filld 1/'" rl'{I{lillg "1'/'" ""IIl"III('/l'r Wlt,," Ih" w ssd i. (""",[,Indy filled ,..ith .m/('T.

Solu tion. Vcssd is elllll' )'. Given: Difference of mercury level Le t ill "" He igh t of water abo ve X·X

Sr . gL of mercury, Sr . gr. of watu.

Sl = 13,6 SI

= 1.0

Density of nU'rcury. P2 '" 13.6 x lOoo Dcnsity o f wmc r. PI '" 1000 Eq uatin g the pressure above datum line X-X. we have

P2 xgxil 2 ""Plxgxil l 13 .6x IOOOx9.8 1 x 0.2"" lOoox9.8 1 xiii

or V~ssel

,

,M I

1 I

"

1

III "" 2.72 111 of water. is full uf wa te r. When ves.w l is full of water. the

Fi g . 2. 12 pressure in the right limb will increase and mercury leve l in the rig ht limb will gu down. Let the distance throu gh which mercury goes down in the right limb be . ), cm as shown in Fig. 2. 13. The mercury wi ll rise in the left hy a d ist an ce of)' Clll. Now the datum line is Z-z. Equat ing the pressure aoove th e datum line Z·z. Pressure in left lim b"" Pressure in righ t limb 13.6 x WOO x 9.81 x (0.2 + 2)"1100) '" lOOOx9.8 1 x(3+II I +yI I 00)

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IL Pressure ilnd its Measurement

13.6 x (0.2 + 2y1lOO) '" (3 + 2.72 + yllOO) 2.72 + 27.2yf1OO '" 3 + 2.72 + )'/ 100 (27.2)' - y)l100 '" 3.0 26.21' = 3 x 100 = 300 300 y ~ - - = 11.45clIl 26.2 The d ifference of mc n; ury le\'e l in twO li mbs = (20 + 2y) e m of men:ury

47 1

hi = 2.72 e rn )

-- -1 3m

1 1,

= 20 + 2 x 11.45 = 20+22.90

.J

1- -

- T

1x (2{l + 2y)cm i - -'-

T

1",

F T'

"',

= 42.90 em of mercury

Re ading of manometer'" 42.90 em. AilS.

Problem 2.13 A pressure gmlge cOll sis /s of ","0 c)'lilUlrim/ bulbs IJ /1111/ C r(/cl, of /0 sq. em crossseclio/l11/ oretl, which ""- cOIIIIl'Cled b y /I U-lUhl' Willi va/jell/limbs ('(wl, of 0.25 sq. em cross-sUliO/wl /lrn l. A rnl/iqllid of sl'l'cifk gral'il), 0.9 is filled ill/a C 1/1111 c/('(l f Will'" is jilin/ imo B, IIII' S II'ftirF of Jf'/Ulrtlliu/l bring ill Ihl' limb '1I/Ilci1n{ 10 C. Filii/II/(' ili SplaCf'1II1'I1I of 1/11' sur/art' of sepllrm ;011 will'll IIII' pressure 011 lilt' sur[net' ill C i s gr,.fllu 1IIfIllilru i illlJ by wr WlIOIIIIII'qUlliio / olr Irpm/ o[II'/II,.r. Solution. Give n : Ar~ a o f eac h bulb Band C, Ar~ a

of eac h v.:rtical li mb ,

1\= IOc rn 1 (I

= 0.25 ern 1 Its d~n si l y = 900 kg/m3

= 0.9

Sp. gr. of red liquid

X·X = In ilial se parali on leve l

Lo<

11(." = Heighl of red liquid above X-X

11[1 = He ighl of watc r abo l'c X-X Press ure above X-X in

lh~

Idllimb = 1000 x 9.81

X "B

Press ure above X-X in the ri g ht limb = 900 x 9.81 x Ire Eq uatin g the lWO prcssure . we get IOOO x 9 .81 xll ll ",900x9.81 Xll c

1111 = 0.9

.. ·(0

II C

lJ40

Wh en the pressure head o\'er lh e surface in C is inc reased by I cm of wale r. let the separat ion leve l falls by an amo unt equa l 10 Z. Th en Y- Y b11 '" Fall of heavy liquid in reservoir

r

I,! '" Rise of heavy liquid in right limb h,

= Heighl of centre of pipe above

X-X

T

/I" = Pressure at A. w hich is \0 bll measured

",

A = Cross-section!11 area of the reservoir (I '"

1

Cross-sc(;t;onal area of Ihe right limb

T

S, = Sp. gr. of liquid in pipe $1

_,r!;~'::'~_:'1R~Yo~:_=~"'Qj y

y

'"

= Sp. gr. of heavy liquid in reservoir and right limb

Fig. 2.15 V"lical si'lgie coll/mn

PI = Density of liquid in pipe

malwmeur. Pl = Density of liqu id in reservoir Fall of heavy liquid in rc!,crvoir will c ause a rise of heavy liquid kvcl in lh", right limb. A

X ~,

= (/

llh =

X

(1)(

hl

112

... ( i)

A

Now consider the datum line Y-Y as shown in ri g. 2.[5. Then pressu re in the right limb above Y-Y.

Pz x g x (llh + /'2) Pressure in lhe [eft limb above Y-Y = p, x g x (1'111 + Il,) + IJ" =

Equaling these pressures. we ha ve Pl xg x(l'1h + " 1) = p, Xg x {l'1h +/i,)+p"

1'" = P!8 (t'1I! + "0

"'

-

p,g(tlh + ",)

= I'1h lPlg - p,g] + "ZPlg - ",p,g

But from eq uation (i),

I'1h = a x "1 A

.. .(2 .9)

As the area A is very large as compared 10

fl.

hence ratio!!... becomes very s mall and can be A

neglecled. Thcnp,,="lP!8 - ",P,g ... ( 2. 10) From equation (2. 10), it is c lear lhal as ", is known and hence by knowing /'1 or rise of heavy liq uid in the right limb. lhe pressure at A call be ca lcu laled . 2. Inclined Single Column Manometer

Fig. 2.16 shows the incli l1ed sin gle (o lumn mal101l\e\Cr. This manometer is more sensitive. Due to in(lination th r dis\ance moved by the heavy liquid in the rig ht limb will be more.

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Fig. 2.16

Inclined single coll/mn manometer.

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Fluid Ml>chanics

Let

L", Length o f licavy liquid moved in right limb from X -X

e = Inclination ofrighllirnh Wilh horizontal 11 2 '" Vertical rise of heavy liquid in ri gh t limb from X-X ", L x sin

e

From equ ation (2.10), Ihe pressure at A is II... = "~P2g - "IPlg· Substituti ng the val ue of 11 2, we gCI

e

... ( 2 . 1 I ) /I,., = sin x P28 - h lPlg· Problem 2.14 A single coluIIIn manometer is cOIlller/ed /0 (I pipe comaillins a liquid of sp. gr. 0.9 as slloll'n ill Fig. 2./7. Find lite pressure ill the pipe if the orea of Ihe reserl'Oir is 100 limes Ihe area of lite rube for IIIe mmwme/er reading ShOIl"1I in Fig. 2.17. The specific grol'ity of mercury is 13.6. Solution. Gi ve n:

Sp. gr . of liquid in pipe.

St = 0.9

I"

PI = 900 kglm )

Density Sp. gr. of heavy liqu id.

S 2 = 13 .6

Density.

p, '" I1.6 x 1000

f

1

Area of reservoi r '" ~ '" 100 Area of right limb Hcigtlt of liquid. Ri se of mercury in right limb.

a

hi::: 20 ern '" 0.2 rn f".g . 2 .17

h Z =40un=0.4m

Let Using equation (2.9). we get

PA = Pressure in pipe

"

PA = -A IhlpzR - PIS! + hop,S • - - "IPIS 1 : X 0.4[13.6 x 1000 )( 9.81 - 900 x 9.81 J + 0.4 x [3.6 x 1000 )( 9.8 1 - 0.2 x 900 x 9.81 100

== 0.4 [1 3341 6 _ 8829 ) + 53366.4 _ 1765.8 100 = 533.664 + 53366,4 - 1765.8 Nlmz = 52 [34 Nlmz = 5.21 Niem I. .. 2 . 7

A ilS •

DIFFERENTIAL MANOMETERS

Diff.:rcmial manomcters arc th e d.:viees used for mcasuring th e differencc of pressures betwee n two poims in a pipe or in two differelll pipes. A differential manometer l'Ons iSls of a U· lUbe. coma in· in g a heavy liqu id. whose two ends are con nected to lhe poinls. whose difference of pressure is 10 be measured . Most commonly types of differential manometers arc: !. U·tube differential lIlano lll eter and 2. In ve n ed U·tu bc diffe rent ia l manometer. 2 . 7 . 1 U - tube Differential Manometer. Fig. 2.18 shuws the differential manumeters of U·tubc type.

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IL Pressure ilnd its Measurement

T , 1

, t , -- J" ,

t

T

(a)Two rHpes at differenllevels

Fig. 2.18

511

'1-- ' (b) A aJ'ld B ale at the sa me Ieve1

U-wbe difftr..nlial manomtll"l'$.

In Fig. 2.18 (ll). the tWO points A and 8 are at different level and also contai ns liquid s of different sr. g r. These point~ are C{lonecled 10 the U -luhe differential manometer. Le t th e pressure al A "fld B arc I'A and PB" Le\

,,=

Difference of mercury level in the U -tune.

Y'" Distance o f th e cCl1m: o f B, from the mercury level in the right limh.

x'" Distall(;c of the l-emrc o f A. from thc mercury level in the right limb. PI

= Density of liquid at A.

P 2 = Density of liquid al B. p & = Density of he avy liquid or ml. mllf (iii) f"fS5Jf 1"1" \'(,,;111;011 /01/0,,"5 (II/;lIb(l/;(' low, T nkf' Ihe tlnlsit)' of ed for the mea,urement of pressure ?

8. What is the differencc between U·tube differential manometers and inverted U-lUbe differential manometers? Where arc they used ? 9. Distinguish between manometcrs and mechanical gauges. What are tlie different types of mc.:hanical pressure gaugcs ? 10. 1)erive an expres.chanics Cakul~tc th~

pressure due to a col umn of 0.4

In

of (u) wuter. (b) an oil of >1>. gr. 0.9, and (coJ mercul)' of!>p. gr.

"

13.6. Ta~e density 0( .....:l1cr. p _ 1000- 3 ,

m

jAns. (a) 0.3924 Nlcm', (b) 0.353 NiemI, (el 5.33 Nlemll

,

-" The pressure imensily at a point in a fluid i.< g;,-en 4,'0 pressure pipes A and E. Pipe A contains cmbon tetrachloride baving 3 specific gmvity 1.594 under a pressure of 11 .772 Nlcm 1 and pipe R contains oil of '1'. gr. 0.8 undcr a pressure of I ].772 Nkm". The pipe A lies 2.5 m abo"e pipe 8 . Find the difference of pressure measured by merc"ry as fluid filling U·tubc. IAlls. 3 1.36 cm of mercury1 A diffcrcmi al manometer is conn«tcd at the two point.' A and R a., shown in I' ig. 2.25. Al fJ air pre"ure is 7.848 Nkml (abs .j. fi nd the absol ute pressure at A. IAns. 6.91 Nlcm1 j

OIL Sp.

gr. ~O.8

, -'" ,';-1 "'=

,~

i ,,~

1 -M ERCURY

Sp. g'.En.6

Fi g . 2.25

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OIL Sp .

T "''' = I

""

T

. '

~i~=

1.

~ ~Ei

~

WATER

Fig . 2.26

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IL Pressure ilnd its Measurement

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15. An in"crled differential manomcl~r containing an oil of sp. gr. 0 .9 is connected to find the diffuence of pressures at twO poinB of a pipe comaining waler. If the manometer rcading is 40 em, find the difference of prc«urcs. IAn_s, 392.4 N/m'l 16. In above ''is' 2.26 shows an in\'ened differential rna""mc!e, connected 10 lwo pipes A and B containing water. The fluid in manometer is nil of sp. gr. O.H. For the manometer readings shown in the figure. find the di fference of pressure head octween A and B. [,\US. 0 .2 6 m of waterl 17. If the atmospheric pressure at sea · lc"cI is 10.143 Nlcm l , dClcnninc the pressure at a height of 2000 m assl!ming that (he pressun: variation fol lows: (i) Hydrostatic law, and (U) lsothennallaw. The density of

air is given as 1.208 k8/mJ.

[,\ns. (i) 7.77 Niem I. (ii) 8.03 Nlem Jj

III, Calculate the pressure at a height of 8CXXl m abo,·c sea-level if the atmospheric pressure is 101 .3 kN/ml

19.

20.

2 1.

U.

2.1.

24 .

ZS .

Z6_

~ I

and tempemlUre is 15°C at the sea-level assuming (i) air is incompressible. (ii) pressure variation follows ~diabatic law. and (iii ) pressure vari"tion follows i>othermailaw. Take the dcn.ity of air at the sea -Ic,'el a. equal to 1.285 kg/m). Neglect "ariation of K with altitude. jAilS. (i) 607.5 Nlml. (ii) 31.5 kNlm l (iii) 37.45 kNlml ] Calculate the pressure and density of air at a hcight of 3000 m above sea-leve l where pressure and temperature of the air are 10.143 Niem I and IS oC res""clively. The lemperalUre lapse-rate is given as O.()(I6S" KIm. Take dCl1Sily of air at sea-level equal to 1285 kg/111 ). jAl1 S_ 6.896 Nfcml . 0.937 kg/1n J I An aeroplane is flying at an altitude of 4000 m. Calculate the pressure around the aeroplane. given the lapse-rate in the atmosphere as 0.0065°KJm. Neglect varimioll of 1/ with altitude. Take pressure and temperature at ground Ie"el as 10.143 Nlcm l and 15°C resp«tivcly. The density of air at ground level is gi"en as 1.285 kglnl '. [,\ ns. 6.038 NiemI] The atmospheric pressure at the sea"level is 101.3 kN/ml and the temperature is 15°C. Calcul"te the pressure 8CXXlm abo.'c sea-Iewi. aSSum ing Ii) air is incompressible. (ii) isothcnnal .·"riation of pressure and density. and (iii ) adiabatic variation of pressure and density. A,sume den,it y of air at sea-level a, 1.2!\5 kg/1n J . Ncgl""t varialion of '8' with ~Ititude. ]Ans. (i) 501.3 Nlm ' , (ii ) 37.45 kN/m '.liii) 31.5 kN111l1] An oil of sp. gr, is 0.8 under a pressure of 137.2 kN/ml (i) What is the pressure head cxpressed in melre of water ? (ii) What is the pressure head cxpressed in metre of oil? [AtlS. (i) 14 m. (ii) 17.5 m] The atmospheric pressure at the sea-level is 101.3 kN/m l and tC11l""ralUrC is I soc. Calculate the pressure 8000 m above sea-level. assumi ng (i) isolhennal variatio!1 of pressure and dcnsity. and (ii) adiabatic variation of pressure and density. Assume density of air at sea-level as 1.285 kg/m J . Neglect variation of 'g' with altitude . Derive the fonnula that you may use. IAII". U) 37.45 kNI11l'. (ii) 31.5 kN111l'1 What are the gauge pres.,ure and absolute pre.",ure at a point 4 m below the frec surface of a liquid of specific gravity 1.53_ if atmospheric pre.ure is equivalent 10 750 mm of mereury. [,\Ils. 60037 Nfm ' and 160099 N/m ' ] Find the gauge pressure and absolute pressure in Nltn ' at a point 4 In below the free surface of a liquid of sp. gr. 1.2. if the almospheric prcssure is equivalent to 7~ mm of tnCrcury [Ans. 47088 NI11l '; 147150 NI11l ' j II tank contains a liquid of specific gravity 0.8 Find the absolute pressure and gauge prcssure at a point. which is 2 m below the free surface of lhe liquid. The atmospheric pressure head is equivalent 10 7(1) mm oi" mercury. [Ans. 11 7092 Nf11l' ; 15696 Nfml j

I~

~ I

IL

I I

Ii

... 3. 1

INTRODUCTION

This chapter de~ls with the fluids (i.e .. liqu ids ~nd gases) d' -J

,

bh' -36

-

1--' ----0<

"

Triangle

,/1\i . ~

x~

•

-

3

-bh

bh'

" COllld ..

I I

Ii

~ I

IL 172

Fluid Mechanics

Plull/'

c.G. from lhe

sur/au

1M",

I

Area

Momem of i"alia

/IIu"",,,,1 of

I'lbouIIITI axis lUlSS;IIK

i,jfrtiu aboul

rhrough C G. lInil

base (lu)

puralleilO b" se

(/01

3 Circle

)/ ,

Gj 1 ~

j"

.w'

d

--

.1'=-

2

4

.w' -64

-

4. Trapezium

fzL~IG~ I

"

x_(2l1+b)~ IHb

3

(u +b )

---d

2

1 1 (u +4ab +b )Xh1

36(o+b)

-

If

Pro blem 3 .1 A ,ec/(lIIg/liar plane .mrface is 2 III wide ami 3 In deep. II lie.! in I'errical plane in W -

0'

32L 6 1 t 18 - 0

7935.9 16 (7848 +T)

I I

[

7848 (7848 + T)

(7848 +

T)] >

15871.8

0

Ii

~ I

IL Buoyancy and Floa tation 7935.9

16 (7848 + T )

151 1

7848 + 7848 + T , 0 (7848 +T) 15871.8

,-79"3;;',.;9,,-~'6;-':..;7~84,,,8 + (7848 + T) 16 (7&48 + T)

>

0

15871.8

~-"I~'7C;6~:n",+ (7848 + T) > 0 J6 (7848+T)

15871.8

(7848 + T)

:>

15871.8

,,-~''~7~63~2"" 16(7848 + T)

(7&48 + n 2:> 117632 x 1587 1.8

0'

16.0 :> 11 6689473.5 :> ( 10802.))!

7848 + T:> 10802.3 T:>

10802.3 - 7848

:>

2954. 3 N. A ns .

Th.e force in lhe chain IllUS! be at least 2954.3 N so that the cy lindric al buoy can be !;cpt in vert ic al position. An s. Pro blem 4.17 A solid coile floats in warer ",itl, irs apex downwards. Determill e Ille lew/ apex ollgle of cone jar .liable equilibrium. Tlw specific 8WI';I)' of lile material of the coile is gil'el l O.S. Solullo n. Given: Spo gr. of cone == 0.8 P == 0.8 x 1000 == SOO kg/m l DensiTy uf I:one.

/

D == Dia. of the cone d == Dia. of cone ,H water leve l

Co<

29

+

,

CAN 0' C ONE AT

= Apex angl e o f cOil e

w" ER LINE

H == Hei ght o f C()nc

D

I'

•I

II = De pth of cone in water

For tllc I:onc. thc

di st~ncc

G'" Centre of gravi ty of the cone B = Centrc uf buoya ncy of Th c conc of ce nTre of gravity from the ~[lcx A is

AC = t height of conc '" tH also

AB", t dcpth of cone in wa1cr '"

Volume of water di splaced '" Volume of cone

= 1-

:. Weight of cone

=800 xgx tx TtR'" xH

Now from JiAEF.

Similarly.

I I

t It? x II X TtR 2 X II

EF

Ian 9= -

t"

"

, Fig. 4.2 1

il

R

=-

EA H R=Hta n 9

r",htan6

Ii

~ I

IL 1152 :.

Fluid Mechani cs Wei~hl

of cOile

t

.. Weight of water displaced", ]()OO x g x

x II'! X II

. .000"""x""' '" lOOOxgx tXI[(htan 9) , X/I= -1 For equilibrium Wcigllt of l"OIlC

3.0

'" Weight of water displaced 3.0

3.0

800xlt=

or

'X'O'ITfX~h" _'""'''_',>.

1000

x

II)

1000 X11.1or!!'" = (1000)'"

H3 =

800

/,

800

For stable cljuilibrium. Meta -centric height GM should be positive. But GM is given by GM= -

/

•

-8G

where 1= M.O.! . of cone al wateT-lil'" = ~ d~ 64 . In, V = Volume of con" In water =- - d" x h

34

/

IT d ' l "3)("4 ' IT d ,- X "

V=64

I x3 d 2 3d' 3 ' 3r' =x - ==x(2, r = - 16

=

II

161,

1611

4 II

3 (/l lan6)1 4

I':

1/

r=I,[an91

=t illa ,,2e and

BG '" AG - AB = fH - til'" f (H - II) GM =

til tanl e -

t(ll - II)

For stable equilibrium GM should be positive or t ht3n!e -1 (1I - 11»O

or

htan 2e>(H _ h)

or

ur

htan!e - (If - h»O htan 2 e + h>H

hltan!a+ 11>11 So<

!!.. = ('OOO)iIJ = 1.077 h

800

, I scc~ e > 1.077 or cos' 0 > - - '" 0.9285

cos

e > 0.9635

1.077

0> 15° 30' or 20>31 ° Apex angle (20) s hould be aileasl 3 1°. Am.

I I

Ii

~ I

IL 153 1

Buoyancy and Floatati on Problem 4 .18

di(lmeler

f)

A cOile of Ifleciftc gra\'it)' S. is floating ill wmer willi its apex dowlI.",ard,~. II has a

(md l'erlin,/Iu:ighl H. Show Ilwl jor stah/e equilibrium of the COli I' H <

!.. [V i. 5;;; ]"1, 2 2

S

Solution. Given: Dia. of oone =D Hciglll o f cone'" H Sp. gr. o f cone = S Let G", CemTC of gravity of cone

B '" Ccmrc of

PLAN OF CONE AT WATER LINE

bUOYHIlCy

~ D _I

29 = Arcx angle

T

= Apex of the cone II'" Depth of immersion J = Dia. of cmw at Waler surface

A

Then

"

1

,

AG:2H 4

fig. 4.22

3 A8= - II 4 Also weight of cone =Weigh t of water displaced. I' I , lOOOSxgX,lTWX H : IOOOxgx-rrr'xh

3

or

'2 sn-H=,h

SR!H II = - ,-

B"

R" ,

t~ne = - = -

H

"

N=Hmne.r=ltlan8

SX(Hlan6)l xH II '"

(1/ tan

er,

SX II l xtan 1 exfI SH J " : __ It" tan " e I,,,, (SH1)1fJ = Sill H

I,:

Distancc.

,,1

or h 3 =SH 3 ... (1)

8G '" AG -An

J 4

3 3 (H - b)= -3H(- .1/3 S /I) 4 4 4

=-H - - b= -

'" 2 HI I _SI Il I

Also

... (2)

4 I", M.O. Inenia of the plan of body at wate r surface

'" 3... It 64

.

V", Volume o f cu ne III water'"

I I

, In: , 3In: x 4 x iI + x h '" 34 iI+

IH.S

III

I

Ii

~ I

IL 1154

Fluid Mechani cs

-

J

V

=

~d' 64

.----o"'c----c 1 x

3

It

d

H.S'1l

16.H.slll

4

Now Mcl

4.8

EXPERIMENTAL METHOD Of DETERMINATION OF META -CENTRIC HEIGHT

The mcla-cc ntric heigh t of a noati ng vessel can be determined, provided we know th e (emre of grav it y of the Ooaling vc~scl. Lei"", is a known we ig hl placed over the ce ntre of the VCS1;e J as show n in Fig. 4.23 (iI) and the vesse l is floating.

w

(a) Floating body

(b) Tilted body

Fig. 4.23 M l!l d-o\:.t o. / -v

s

Fig. S.l S

'0

Pottwtiallinn for souru.

Ii

~ I

IL 1216

Fluid Mechanics

Pressure distribution in a plant source flow The pressure distri bution in n pl~nc source flow can be obtained with the tion. Let us assume that the plane of the now is horiwnt,d. In that case the for two poin ts of now. LeI p" prc~sure at a (lOint I which is at a radius r from Ihe source al II,: velocity al point I Po" pressure al point 2, whirl! is al a large dist ance away from Ihe be zero at point 2. [Refer \0 equation (5.39)] Apply ing Bernoulli's equation. we get

(p- Po) pg

, p .u,

help uf Bernoulli's equadmum head will be S,

'" .!L log " , and 2,

due to sin~ il is g iven as 4', = - q log,r,. The equation for re~ultanl potential function (9) will be the - l /t sum of these two potential function. 4'=4'1+4',

(-q)

=.!L log, r l + log, r 2 l/t l /t

I I

Ii

~ I

IL Kinematics of Flow and Ideal Flow

'" ..!L 21(

110&,' 1 - log,'21""

.!L log ,

(.'L)

223 1

...(5 .46)

21 t ,l.

To pron' that n'Sult lml strt'!lnl lines will be circular arc pass ing through source and s ink The resultant stre am fUllcti o n is give n by equ ati on (5.45) as

-q.a .~ -­

"

For a given stream lin e l¥ '" co nstant. In the above eq uati on tile te rm .!L is al so constant. T liis

"

mea ns th ai (9 2 - e,l Of ang le 0: will also be constant fo r vario us positio ns o f P in the pl ane. To s,uisfy ttJi s. th e locus of P mu st be a (;ircJc with AB as chord . hav ing its ce ntre on y- ax is. as shown in Fi g. 5 .40. Consider th e cquali on (5.45) ag ain as

- q 21\:

-q (EI1 - EI,) 2/f

. ~ - "~ -

v

= ...2...(6 1 - O

(e, -a!)=

"

2 lTIjI

q Tak ing tange nt 10 both s ides. we ge t

1= lan (2:1tf )

HIn (9 1 - 9 2

B",

tan 9 1 '"

-

and

'-

x +a

Suhstilut ing Ih e values o f Ian 9 1 and Ian

- )-' - - - '(x+ a)

(.I· - a)

= Ian

1+ -' - .--'(.1+ (1) (X- II )

9~

tan 9,

or

-

tan

0;

,".( i)

1+l an9 , .1an8.

ta n 9 , = - Y• x- a

... (S A M )

in equ al ion (i).

(2'.)

,

q

y (x-o))' (.1'+0) = Ian ( " . ) , l x'

a + .vl

, ,: I

"

0'

,"0 ' I-- , x·, - a"" +

I I

y- = -

2a)" COl

q (".)

---t=.:..:'-=1,__ F ig. 5.41 (d )

Ii

~ I

IL 1224

Fluid Mechanics

, "

.\--a + y" + 2(1},l:ol

.l!+ /

(2"' q 1')

=0

COl (2:W) -(/ = 0 ,[2+ / + 2ay COl(2:'V ) + al COIl (2:"') _al +

2a)'

CO,2

(2:1JI ) _al= 0

[ Adding and subt ract ing I' ! cot! (2:1¥)]

... (5.4 7)

The above is the eq uat ion of a cin:lc· with centre on y·axis M a distance of± a COl (2:"' ) from Ih e origi n. The radius of the cin:lc will be a cusc' +,.' ) _ 2),2

(.ol +l) ,

I I

.I

1

+ (X-'

+)' ') -

Ii

~ I

IL 1232

Fluid Mechanics

= irr ;, [i" ~ y' )1=irr [ix;:~;)' 1 Su bstitutin g the va lu es o f )l = 5 m'ls, x" 0 .5 and y" 1.0, we ge t tile veloc ity co mpone nts as

u= _L[ (XlXl +-if i'l = _ ~[ O.5l -ll'1= _ 2- 0.7~ + IT 2 J't

and

2 1t (0.5'

= - 0.382

2It 1.25"

\'" ;n [(x::"/ )!]'" n[(O~5~ :'~l)l[ ] '" ~[~] ,,- 0.509 5 2

Res ultant ve loc ity,

V=

Ju

1

- ~(- 0.382 i

+ v"

+ ( - O.509 )" = O.63fi m/s. ADS.

( ii) Value of strewn lime/ion m poilll P ,

I

l it

1.25

-- x == _ 0 .636 m1/s. A DS.

Solution in polar co-ordinates

e

The above qu estion ca n also be done in r. (i.e.. polar ) co-ord inalCs. The s tream func tion in r. 9 co-ord inates is give n by equatio n (550) as )l si n 9 1jI =

-- x --

2rr

--.(i)

,

and vel oc it y contlx mc nt s in radial and tangC llli a l di rec ti o ns arc given as

" : ~ x ,)IjI= ~ ~ [_ LSi lle] ,

ae rae

r

'" ..!. x r

,,'

21t

r

(_L] x ..!.~(sin 9) J9 lit

r

[ .: L2rr

is a constant te rm and also r is conSlanl w. r. t. e]

=- -x...., " I cos e 21t r·

... (Ii)

"" '[" ';0 ']

ue : - - : - - - - - ar ar 21t r = _ ( _ _11

21t

sin e]~ [~]= ~ sine (-I ) . -

ar '

21t

, !

[.: ~

sin e

21t

,.

= - - ' -,-

I I

~~i;eiSaCOnSlanl w.r. I. '] ... (iii)

Ii

~ I

IL Kinematics of Flow and Ideal Flow

233 1

1.",5 ~.f " + Y _ ~0.5 " +1 _","" y ' 894 and cos , x O.5 sin , =-:~=O. =-:~= 0 .44 7

Now

r=

r

Substituti ng the values of

r,

r

,,1.25

"1.25

a and cos e in above cqumions (i), (ii) and (iii), we gel J.I sin e 5 0.894 z ' l I = - - - - = - - x r.= =- O.636 m Is. Ans. sin

21t

r

J.I 21t

I

21t

If,'" - - x ----,- x cos ,"

,, 1.25

e '" - - 5

I x - - x 0.447 = - 0.2845 IlI/s 21t (1.25)

J.I sin 9 5 0.894 lie'" - - x -,- =- - x - - = - 0.569 rnls

lUlU

21t

r

211

1.25

Resu ltant velocity.

= J(-Q.2845)! + (-0.569)2 = 0.636 S. 11.3

m'.~.

AilS.

A Plane Source in a Uniform flow (flow Past a Half-Body), Fig. 5.46 (d) shows

a unifonn flow of velocity U and Fig. 5.46 (b) shows a source fluw of strength q. When this unifonn flow is flowing over the sourc'01 Length. L = 2 x x, = 2 x 1.076 = 2.152 nl. Width. B '" 2 x y""" ... (i) Let us now lind Ihe value of Y"",. Usin g equ3lion (5.64). we gel

y

Y"",,"'eharge through equal to the su m of the discliarges through tlie different rectangular notclies.

I I

st~ppcd

notch is

Ii

~ I

IL 363 1

Notches and Weirs Co nsider a ste ppe d no tc h as show n in Fig. 8 .6.

Lei H I" Height o f water abo ve the ercS( of nOlc h !. Ll

"

Length of notch I.

H1. L2 and H3_ LJ arc corre sponding v:tlucs for notdlcS 2 and 3 respect ive ly.

f-- L' --1

Cd " Co-effi cie nt of di sc harge fo r all no tc hes Total di Sl: hargc Q '" Q 1 + Q2 + QJ

r-- l , I•

2

r;;: 31'1 .Yl Q= "] xCd x LI x ,,2g IHI -Hl 1

_I

L,

•

I

Thr Itrpped notch .

Fig. 8.6

... (8 .5)

Problem 8 .8 = 0.62.

Fig. 8.7 sho ws

II

stepped lIolc/'. Find lilt' discharg/' l"rOJ/gll Ihe notch

il e d/o r 1111

sec/ion

Solution. Given:

T L I = 40 em. ~" 80 em.

LJ = 120 Clll

"I

= 50 + 30 + 15 =95 em.

Hl '" 80 em. HJ

fl,

50cm

= SO em.

Cd = 0.62

-~ Iscm

T

_ ,.

Total di scharge. Q '" Q, + Ql + QJ where

~40cm""'1

80cm _

120 cm

. ,

Fig. 8.7

2 r->:: QI ="3 X C" X L1x .;2g

J.i2

IHI

- H!

.lI2

1

= ~ x 0.62 x 40 )( J2 x 98 1 x [95 31l - 80Jl'1 1 3

= 732.261925.94 - 7 1'i.54 1 == 154067 cm3/s '" 154.067 li tis f'>:: 1I1 1I1 2 Q2 ""3xCJ xL2 X,,2 g xlH2 - HJ 1

" ~XO.62X80 X J2X98 1 3

x [80 312 _ 50 312 1

== 1464.52(7 15.54 - 353.55( cm3/s= 5301 41 cl1I 3ls" 530.144 li tis

"

~ 3

x 0.62 x 120 x

J2 x 98 1 x 50ll.l " 77677 1 cm 3,s" 776 .77 1 litis

Q = Q 1 + Q 2 + Q 3 = 154.067 + 530 . 144 + 776.77 1 = 1460.98 ]iUs. ADS.

I I

Ii

~ I

IL 1364

Fluid Mechanics

.. B.8

EffECT ON DISCHARGE OVER A NOTCH OR WEIR DUE TO ERROR IN THE MEASUREMENT OF HEAD

For all accurate value of the discharge over a weir or notc h. an accurate measurement o f he ad ove r the weir or notch is ve ry esse ntial as the disch arg e ove r a triangular !lotch is proportional 10 HYl and in case of rectang ular no tch i1 is proportional \0 HY2 . A s mall error in thc measurement o f head. will affect lhe di.: 15 C d' tan T,,2 g

'" KH

xH

YJ.

SIZ

... (iii)

8 e Cd' tan - ,J2i IS 2

where K '" -

Differe nli~ting equ~tion

(iii). we gel

dQ= K~H3I2 XdH

.. .( i,')

2

K~

1/ 311

dll

Di viding (il') by (iii ). we get dQ '" _.2co",,~JIl Q K fi Equation (8.7) show s thaI :111 error of 1% in a triangul~r weir or notch.

I I

5 dH 2 H

III c ~ suring

H will produce 2.5% e rror in

...(8.7) disch~rg c

over

Ii

~ I

IL Notches and Weirs

365 1

Problem 8 .9 A recrangular 1I(1Icii 40 em hlllg is IIsed for lIIea~'llfing a dj.~c1wrge of 30 Iilre.1 per second. All error of 1.5 mm was made. wllile medsurillg the head ol'a Ihe notcll. Calcula te IIII' pefcelilage error in II,e discharge. Take Cd '" 0.60. Solu tion. Given: Length of notch. L=40cm Dischilrgc. Q = 30 litis = 30000 c11l 3fs Error in head. dH= l.5mm = O.IScm Cd = 0.60 Let the hdght of water over rectangu lar notch = Ii Ttw discharge through a r;:c(angular notch is given by (8.1 )

2

Q=")XCd XLx.[fiXH 3I!

or

,

30000

"'

=..:. x 0.60 x 40 x .J2 x 98 1 x 11Y2 3

H3Il =

3 x 300Xl 2 x.6O x 40x

.J2 x 98 1

= 42.33

II : (42 .33)Z13 = 12.16cm

Usi ng eq uatio n (8.6). we gel dQ

Q

= ~dH =~X~ =O.OI85= 2 H

2

1.85%. A ns.

12.16

Problem 8 .10 A righl ·angled V.notch is IHed Jor me(lSllring (/ di~Tll(lrge of JO li/res/s. All e rror of /.5 "'''' "'a5 ",ade "'bi/e ",,,aJ'uring Ill" ile"d Ol"er IIIe n"'cll. C,,/culale Ille percell/age eTTor in lile disdwrge. Take Cd '" 0.61. Soluti on. Give n : Angle of V-no tc h. e '" 900 Dis.:harg'" Error in head.

Q '" 30 litis'" 30000 cm 1/s dH:: l.5mm",0. 15cm

Cd:: 0.62 Let the he~d over the V- notch:: H The d ischarge Q through a triangu lar notch is given by eq uation (8.2)

8

e

fi":"

Q'" - Cd ' tan - x ,,2g x H 15 2

30000 ::

Y.!

.! x 0.62 x tan ( _90_°) x , "lex"9","1 x HYl. 15

2

:: -8 X .62 x 1 x 44.29 X H Y2 15 30000 x 15 :: 2048.44 8 x .62 x 44.29 H '" (2048 .44):!I·~:: 2 1. 1 1 ern

I I

Ii

~ I

IL 1366

Fluid Mechanics

Usi ng equatio n (8 .7). we get dQ '"

Q

~ dH 2 11

'" 2.5 x 0.15 '" 0 .0 1776 '" 1.77 % . 21.11

AD S.

Problem 8.11 Tire /!"lItl ojwMer Ol'e, II Irilmgu/ar II oleil of lingle 60' is 50 em ami co -effieien, of discharge is 0.62. Ti,e flow measured by il is 10 be wililill all accuracy of 1.5% up or dowl!. Find Ihe limiting )'a/ues of the Iwail. So lution. Gi ve n : Angle o f V-no tch , H~ad

o f waler,

9 = 60° H "'SO cm Cd'" 0 .62

dQ '"

±

1.5'll '" ± 0 .0 15

Q The di sc harge Q over a triang ular no tc h is

8 Q'" -15

8

=-

15

c d .,f2i Carl x 0.62 x

, -::;- H Y2

_

Jz x 981

60· x tan x (50)Y.! 2

'" 14.64 x 0.5773 x 17677.67 '" 149405.86 C111 3fs Now appl yi ng eq uati on (8.7). we get dH dQ = 5 dH ur :t.O I 5=2.5 Q 2 1/ if

.015 2.5 T h", lim iting va lues of th e he ad

" '

tlH /I

-

.015 :± - -

2.5

.015 2.5

dH ::!: - - x H =::J::. - - x50=:!:0.3

" 1/ ± till " 50 ± 0.3 '" 50.3 e m. 49.7 elll = 50.3 e m und 49.7 e m. AilS.

.. 8 .9. (a)

TIME REQUIRED TO EMPTY A RESERVOIR OR A TANK WITH A RECTANGULAR WEIR OR NOTCH

Consid er a rese rvo ir o r tank o f uniform cross·section al are a A. A rec ta ng ul ar weir or notch is prov ided in one of its si des. L = Length of crest of the wei r o r notc h Cd = Co·efficien t of d i>'~ 1 = 5 1.1 5 [0 .3605 - .0005361 = 18.412 m l/s. AilS.

I I

Ii

~ I

IL Notches and Weirs Problem 8.26

All Ogee weir 5 metres 101ig lIa,\" a Ilead of 40 em of wilier.

If Cd '"

381 1

0.6, filld Ille

disellarge ora Ille weir.

Solution. Given:

L=51O

Len gth of we ir. H~ad

o f wala.

H= 4Qc rn '" DAD 111

Cd '" 0.6 Disdlargc over Ogee we ir is g iv en by equation (8.2 1) as

,

Q= ~ 3

X Cd X L

x

.ffi X HY!.

'" ~ x 0.60 x 5.0 x

J2 x 9.81

x (O.4)3i'l '" 2..2409 nh~. A n s.

Problem 8.27 Tlte /JeiglllS o/ . . . aler 011 I/le IIps/reom (mil downs/rem" side of (l sllb·merged we ir of J", /Om

.g

=

l-

2 xIO· x9.81 900x9.8 1

+ 20 = 22.22 + 20 = 42.22 m

(:~ + z Jis higher at A and Ilene" flow lakes place from A 10 B. An s.

(ii) Ral., o f flow. T he Joss of press ure IIcad for viscous flow lh rougll circular pipe is g iven by

I I

Ii

~ I

IL Viscous Flow For a vertical pi pe

395 1

/1/ '" Loss o f pc izo mclric head '" ( : ; +ZA

24.45 "

u ==

"'

J-(:;

+28) '" 66.67 -42.22 '" 24.45 III

32 xO.l5xu x20.0 i

900 x 9.8 [ x (.02)" 24.45 x 900 x 9.81 x.()()()..I

32xO.l Sx20.0

== 0.889 ,. 0.9 mls.

The Rey no lds number stlOuld be calcul ated. If Reyno ld s number is less llian 2000. the flow w i ll be

laminar and th e above ex pression for loss of pressure head for laminar flow can be used. Now Reynolds number where

~

p" 900 k g/Ill) and V",

pVD

--

"

w

=900x O.9x.02 = 108 0.15 As Rey nolds number is less than 2000. th e flow is laminar. " average velocity x area Rate of flow Rey nolds number

It "II X "4

,

1t

D-=O.9X "4

X

'3

(.02)- m Is: 2.827 x 10-

43

10

Is

'" 0.2827 lil res/s. Ans. Problem 9 .5 A fluid of "iscosily 0.7 Nsf",' (lnd specific grlll';/Y 1.3 is flo .... ing through" cireulM pipe of diameter 100 "'III, The maximum silear Siress 01 Iile pipe wall is gi,'en OJ' 196.2 N/", 2, find (i) Iile pre~'sufcity distribution for lurbulent now in pipes. Why this

vciocity distribution is called universal ? 7. What is a velocity derect ? Dcrive an expression for velocity defect in pipc •. g, How wou ld you distinguish between hydrodynamically smooth and rough boundarie s?

9. Obta in an expression for the "c!ocity distrib ution for turbulent flow in smooth pipes. III . Show that "eloc.ily distribution for turbu lent flow through rough pipc is gi "en by

-"

~

5.75

log l~

(,-Ik) + 8.5

". where u.", shear velocity.)," distance from pipe wall. k., roughness factor. II . Obtain an expression for wlodty distribution in temts of ""crage velocity for (

X

,

VI- '" pA,V1- - pA1VIV1

'" pA 2 [V/ - VI V,I ... (iii) Now ncl force acting on the control volume in the direction of now must be eq ual co Ihe rate of change of mom~ntum or change of momentum iX'r secoJld. Hence equaling (Ii) and (iii) (PI - p z)A j '" pA1[Vl' - VI Vl l

Dividing by g on bolh sides. we have PI - Pl.

pg

I I

1', '" ,V",'~~ --:V""VL' = 'V./_~CV""VLl. or -P I - -'--'g

pgpg

g

Ii

~ I

IL Flow Through Pipes Substituting the va lue of

(li -Ji2) pg

pg

473 1

in equation (0. we get

v,--

2 ' 2 V, V, +V,'"- V,' V,, - V,V, V," V; , ' + - , -'- = ' . . g

2g2g

28

V , V' 'V V = ' + ' - - 1;

2,

II ,

=

(ltj _ V, )'

2,

.

. .. ( I 1.5)

I 1.4 .2 Lou of Head due to Sudden Contraction . Consider a liquid flowing in II pipe which has a sudden COll trn clion in area as shown in Fig. 11.2. Co nsider two sect io ns I- I and 2-2 before and af!cr (;ontraction. As Ihe liquid flows from large pipe 10 small er pipe. Ihe area of flow goes on decreasing and beco mes minimum ,11 a sect ion C-C as shown in Fig. 11.2. This sect ion C-C is called Vcna-comracta. After section e·c, 11 sudden cnl~rgcmc n l of the area takes place. The loss of head duc 10 slldd~ n contraction is actually due to sudden ClllaT)! Cmcnl from Vcna -cnntracla to smaller pipe. Lei

Ac '" Area o f now at section C-C V,. '" Velocily of flow .11 se

,

. ---

-

Fi g. ]1.25

4/ xLxV 1 4x .OO5 x 4000 x V " The head loss du e to fri cti on in pipe A B = -'--;cc-;;- - or 20 = 0.6)(2x9.8 1

d x2g

v= :. Disc harg:' ...".,..

T'10m _____ ~~~~N.: ___ '.:a~~ ____ L -vr" a

C

30m

Fig. 11.28

Apply ing BcmooJ!i's equation to poillts D and C

[

'C·

ZD+POJ=4: +h pg

( Zv +

J,

OO ",,',,,,V,C' =30 + 4jxL.,xV/ :: 30+ -,4C ""5CX"8COO ilJ x2g O.l5x2x9.81

~; ) = 30.0 + 5.436 V/

... (iii)

Adding (il :md (ii), we hav e 60 - 40 = 2.711\ V12 + 5.09

V/ + 5.09 V/

20:: 2.718

or

VJ ... ( i v)

Adding (i) and (iii). we have 60 :: 2.71 R V I1 + 30.0 + 5.436 Vi

or

60 - 30:: 30 = 2.7 18 1'1 2 + 5.436 Also from cOll1inuily cqu~lion. we have Q,:: Ql + Q,l

"41t ill 2 X VI :: "41t

"' "'

2

2

iI! V! + 1

"4J!

D.3 V 1 '" O.2 V1 + 0.15

And from (v) ,

.. .{v)

222

d j V1 ur ill VI :: til V1 + tl3 V1 2

X

VJ or .09\', :: .04 V2 + .0225 V]

2.7 18 v/ 5.09

... ( vii)

30 - 2.7 18Vll 5.4 36

... (I'iii)

20

Now from (il'),

1

V/

Substituting the value of V" and VJ in (I'i), we gel

0.09 VI'" .04

20 - 2.7 18V/ + .0225 5.09

30

2.7 18 ~2

5.436

Squari ng bolh sides. we gel (0.09 V ) 2 '" (.04)2 X I

(20 -5.09 2.718 VI?] + (0.0225)2 X 30 - 2.718 V/ + 2 x.04 5.436 2~7~18~"CLll "v x .0225 X \1,~20,--~ 5.09

I I

Ii

~ I

IL Flow Through Pipes

529 1

.0081 VI! '" JlO628 - .000854 VI! + .00279 - .000253 VI! + .0018 .0081 VI! + .000854 VI! + .000253 V i l '" .00628 + .00279 + .0018", .01087 .009207 V12 ", .01087

or

or or

VI ==

' g08~7~ '" 1.0&6 mls J~.O~ .009207

Substituting this value o f VI in (1·;1) and (I'iii)

V = 2

,lo20,--~c2".7~'~8cXC~"-ll

=

5.09

30

2.718x 1.086

Jo2e8). Hence Ihis pipe will gel Iwo corm:li'ms. Afler Ihe Iwo corm:li{ms. lhe r"sul!;.",( flow in pipe 80 is ncgali"e in I(JOp OC8. Ik nce Ihe di ....i:I;on of flnw will be anlidockwisc in pipe nD for loop DCn.

37.4

,. ,

The di