**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* = ½ *v*^{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* = *gh*

Here, acceleration due to gravity of the Earth is *g* and height of the window from the basement is *h*.

The volume rate of the flow of water from the hose through the window is

*R* = *vA*

Here, cross-sectional area of the hose is *A* and speed of the water flow is *v*.

The mass rate of the flow of water is

*R*_{m} = *ρR*

Here, 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* = *vA* gives

*R* = *v* (*πr*^{2})

Using the above values in the equation *R*_{m} = *ρR*

gives

*R*_{m} = *ρv* (*πr*^{2})

Substitute the values of *K*, *U* and *R*_{m} in the equation *P* = (*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} for *g*, 2.90 m for *h*, 1000 kg/m^{3} for *ρ* and 9.70 mm for *r* in the above equation gives

P = [(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 *DD* attached to its end as shown in the below figure. When air of density *ρ* is blown through the tube, the disk attracts the card *CC*. Let the area of the card be *A* and let *v* be the average air speed between the card and the disk. Calculate the resultant upward force on *CC*. Neglect the card’s weight; assume that *v*_{0}<<*v*, where *v*_{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*+ ½ *ρv*^{2} + *ρgh* = constant

Here, pressure of the flow of air is *p*, density of the air is *ρ*, speed of the air in the hollow tube is *v*, acceleration due to gravity of the earth at the point of observation is *g* and position of the flow of air from the ground is *h*.

The resultant upward force on the card *CC* is equal to the difference in the pressure times the area of the card *CC*.

**Solution:-**

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

*p*_{1}+ ½ *ρv*_{0}^{2} + *ρgy*_{1} = constant

Here, pressure of the flow of air just before it leaves the hollow tube is *p*_{1}, speed of the air in the hollow tube is *v*_{0}, and position of the flow of air from the ground before the air leaves the hollow tube is *y*_{1}.

On the upper surface of the card *CC*, the total energy of the flow of air is

*p*_{2}+ ½ *ρv*^{2} + *ρgy*_{2} = constant

Here, pressure of the flow of air above the car *CC* is *p*_{2}, speed of the air above the card *CC* is *v*, and position of the flow of air above the card *CC* from the ground is *y*_{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}+ ½ *ρv*_{0}^{2} + *ρgy*_{1} = *p*_{2}+ ½ *ρv*^{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 neglecting* v*_{0 }in the equation *p*_{1}+ ½ *ρv*_{0}^{2} + *ρgy*_{1} = *p*_{2}+ ½ *ρv*^{2} + *ρgy*_{2}

Gives

*p*_{1}+ ½ *ρv*_{0}^{2} + *ρgy*_{1} = *p*_{2}+ ½ *ρv*^{2} + *ρgy*_{2}

*p*_{1} - *p*_{2} = ½ *ρv*^{2}

Now, the resultant upward force on the card *CC* is

*F* = (*p*_{1}-*p*_{2}) *A*

Substitute ½ *ρv*^{2} for* p*_{1}-*p*_{2} in the above equation gives

*F* = (*p*_{1}-*p*_{2}) *A*

= ½ *ρv*^{2}*A*

Therefore, the resultant upward force on the *CC* is ½ *ρv*^{2}*A*.

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**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 density *ρ* and negligible viscosity. *(a)* With what speed does the liquid emerge from the tube at *C*? *(b)* What is the pressure in the liquid at the topmost point *B*? *(c)* What is the greatest possible height *h* that a siphon may lift water?

**Concept:-**

From Torricelli’s law, the speed of the water emerges from a depth *h* through a hole in a tank is

*v* =√2*gh*

Here, acceleration due to gravity is *g*.

The pressure at the depth *h* is

*p* = *p*_{0}+*ρgh*

Here, pressure at the surface of the water is *p*_{0} and density of the water is *ρ*.

**Solution:-**

(a) Water will flow in the pipe and emerges from the point* C*. It is considered that water flows from depth equal to the *d*+*h*_{2}.

Here, the equivalent depth of the water is

*h* = *d*+* h*_{2}

From equation* v* =√2*gh*, the speed of the water is

*v* =√2*gh*

=√2*g*(*d*+*h*_{2})

Therefore, the speed of the water emerges from the point *C* is √2*g*(*d*+*h*_{2}).

(b) The pressure at the point *C* is

*p*_{C} = *ρg* (*h*_{1}+*d*+*h*_{2})

Now, the pressure at the point *B* is

*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 *B* is *p*_{0} - *ρg* (*h*_{1}+*d*+*h*_{2}).

(c) Now, the maximum possible height is obtained as

*p*_{0} = *ρgh*

*h* = *p*_{0}/*ρg*

Substitute 1.01×10^{5} kg/m.s^{2} for *p*_{0}, 1000 kg/m^{3} for *ρ* and 9.8 m/s^{2} for *g* in 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.