Click to Chat
1800-2000-838
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
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. ______________________________________________________________________________ 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. _________________________________________________________________________ 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 =√2gh 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 =√2gh, the speed of the water is v =√2gh =√2g(d+h_{2}) Therefore, the speed of the water emerges from the point C is √2g(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.
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?
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}
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
= [(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.
______________________________________________________________________________
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.
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.
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.
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} - 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
= ½ ρv^{2}A
Therefore, the resultant upward force on the CC is ½ ρv^{2}A.
_________________________________________________________________________
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?
From Torricelli’s law, the speed of the water emerges from a depth h through a hole in a tank is
v =√2gh
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.
(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 =√2gh, the speed of the water is
=√2g(d+h_{2})
Therefore, the speed of the water emerges from the point C is √2g(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_{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
= (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.
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.