Solved Examples on Flow of Liquids and Viscosity:-

Question 1:-Water is pumped steadily out of a flooded basement at a speed of 5.30 m/s through a uniform hose of radius 9.70 mm. The hose passes out through a window 2.90 m above the water line. How much power is supplied by the pump?

Concept:-The kinetic energy of the water per unit mass when it leaves from the uniform hose through the window is

K= ½v^{2}Here, mass of the flow of water is and speed of the water flow is

v.The corresponding potential energy per unit mass of the flow of water through the window is

U=ghHere, acceleration due to gravity of the Earth is

gand height of the window from the basement ish.The volume rate of the flow of water from the hose through the window is

R=vAHere, cross-sectional area of the hose is

Aand speed of the water flow isv.The mass rate of the flow of water is

R_{m}=ρRHere, density of the water is

ρ.The power supplied by the pump is given by

P= (K+U)R_{m}

Solution:-The cross-sectional area of the hose is

A=πr^{2}Here, radius of the hose is

r.Insert the values of in the equation

R=vAgives

R=v(πr^{2})Using the above values in the equation

R_{m}=ρRgives

R_{m}=ρv(πr^{2})Substitute the values of

K,UandR_{m}in the equationP= (K+U)R_{m}gives

P= [(1/2)v^{2}+gh] [ρv(πr^{2})]To obtain the power supplied by the pump, substitute 5.30 m/s for

v, 9.8 m/s^{2}forg, 2.90 m forh, 1000 kg/m^{3}forρand 9.70 mm forrin the above equation givesP = [(1/2)

v^{2}+gh] [ρv(πr^{2})]= [(1/2) (5.30 m/s)

^{2}+ (9.8 m/s)^{2}(2.90 m)] [(1000 kg/m^{3}) (5.30 m/s) (3.14) (9.70 mm)^{2}(10^{-3}m/1 mm)^{2}]= (14.045 m

^{2}/s^{2}+ 28.42 m^{2}/s^{2}) (1.56584578 kg/s)= (66.494 kg.m

^{2}/s) [1 W/(1kg.m^{2}/s)]= 66.494 W

Rounding off to three significant figures, the power supplied by the pump is 66.5 W.

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Question 2:-A hollow tube has a disk

DDattached to its end as shown in the below figure. When air of densityρis blown through the tube, the disk attracts the cardCC. Let the area of the card beAand letvbe the average air speed between the card and the disk. Calculate the resultant upward force onCC. Neglect the card’s weight; assume thatv_{0}<<v, wherev_{0}is the air speed in the hollow tube.

Concept:-According to Bernoulli’s equation, the total energy of the steady flow of air in the hollow tube remains constant.

It is given as

p+ ½ρv^{2}+ρgh= constantHere, pressure of the flow of air is

p, density of the air isρ, speed of the air in the hollow tube isv, acceleration due to gravity of the earth at the point of observation isgand position of the flow of air from the ground ish.The resultant upward force on the card

CCis equal to the difference in the pressure times the area of the cardCC.

Solution:-Apply Bernoulli’s equation just before it leaves the hollow tube, the total energy of the flow of air is

p_{1}+ ½ρv_{0}^{2}+ρgy_{1}= constantHere, pressure of the flow of air just before it leaves the hollow tube is

p_{1}, speed of the air in the hollow tube isv_{0}, and position of the flow of air from the ground before the air leaves the hollow tube isy_{1}.On the upper surface of the card

CC, the total energy of the flow of air is

p_{2}+ ½ρv^{2}+ρgy_{2}= constantHere, pressure of the flow of air above the car

CCisp_{2}, speed of the air above the cardCCisv, and position of the flow of air above the cardCCfrom the ground isy_{2}.The total energy for the flow of air remains the same when it flows from the hollow tube to the card CC.

Thus,

p_{1}+ ½ρv_{0}^{2}+ρgy_{1}=p_{2}+ ½ρv^{2}+ρgy_{2}The position of the flow of air is same for the same flow of air.

Thus,

y_{1}=y_{2}Also, it is given that

v_{0}<<v.So, the speed in the hollow tube

v_{0}can be neglected.Insert

y_{1}=y_{2}and neglectingv_{0 }in the equationp_{1}+ ½ρv_{0}^{2}+ρgy_{1}=p_{2}+ ½ρv^{2}+ρgy_{2}Gives

p_{1}+ ½ρv_{0}^{2}+ρgy_{1}=p_{2}+ ½ρv^{2}+ρgy_{2}

p_{1}-p_{2}= ½ρv^{2}Now, the resultant upward force on the card

CCis

F= (p_{1}-p_{2})ASubstitute ½

ρv^{2}forp_{1}-p_{2}in the above equation gives

F= (p_{1}-p_{2})A= ½

ρv^{2}ATherefore, the resultant upward force on the

CCis ½ρv^{2}A._________________________________________________________________________

Question 3:-A siphon is a device for removing liquid from a container that is not to be tipped. It operates as shown in below figure. The tube must initially be filled, but once this has been done the liquid will flow until its level drops below the tube opening at

A. The liquid has densityρand negligible viscosity.(a)With what speed does the liquid emerge from the tube atC?(b)What is the pressure in the liquid at the topmost pointB?(c)What is the greatest possible heighththat a siphon may lift water?

Concept:-From Torricelli’s law, the speed of the water emerges from a depth

hthrough a hole in a tank is

v=√2ghHere, acceleration due to gravity is

g.The pressure at the depth

his

p=p_{0}+ρghHere, pressure at the surface of the water is

p_{0}and density of the water isρ.

Solution:-(a) Water will flow in the pipe and emerges from the point

C. It is considered that water flows from depth equal to thed+h_{2}.Here, the equivalent depth of the water is

h=d+h_{2}From equation

v=√2gh, the speed of the water is

v=√2gh=√2

g(d+h_{2})Therefore, the speed of the water emerges from the point

Cis √2g(d+h_{2}).(b) The pressure at the point

Cis

p_{C}=ρg(h_{1}+d+h_{2})Now, the pressure at the point

Bis

p_{B}=p_{0}–p_{C}Insert

p_{C}=ρg(h_{1}+d+h_{2}) in the above equation gives

p_{B}=p_{0}–p_{C}=

p_{0}-ρg(h_{1}+d+h_{2})Therefore, the pressure at the point

Bisp_{0}-ρg(h_{1}+d+h_{2}).(c) Now, the maximum possible height is obtained as

p_{0}=ρgh

h=p_{0}/ρgSubstitute 1.01×10

^{5}kg/m.s^{2}forp_{0}, 1000 kg/m^{3}forρand 9.8 m/s^{2}forgin the above equation gives

h=p_{0}/ρg= (1.01×10

^{5}kg/m.s^{2})/( 1000 kg/m^{3}) (9.8 m/s^{2})= 10.306 m

Rounding off to three significant figures, the greatest possible height that the siphon may lift water is 10.3 m.