Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 3, Problem 3.142P
To determine

(a)

The differential equation for the free surface height during draining with respect to time.

Expert Solution
Check Mark

Answer to Problem 3.142P

The differential equation for the free surface height during draining with respect to time is dh(t)h(t)=D24D14dt.

Explanation of Solution

Given information:

The diameter of the cylindrical tank is D . The initial height of the liquid is h0 . The losses are negligible. The stopper is removed at time t=0.

The Figure-(1) shows a cylinder of diameter D and height h0.

Fluid Mechanics, Chapter 3, Problem 3.142P

Figure-(1)

Write the expression for the Bernoulli’s equation between the section (1) and section (2) of the given system for no losses in the system.

p1ρg+V122g+z1=p2ρg+V222g+z2dh(t)dt( p 1 p 2ρg)+( V 1 22g V 2 22g)+(z1z2)=dh(t)dt …… (I)

Here, the pressure at section (1) is p1, the velocity at section (1) is V1, the height of the section (1) from the datum is z1, the pressure at section (2) is p2, the velocity at section (2) is V2, the height of the section (2) from the datum is z2, the density of water is ρ, and the acceleration due to gravity is g.

Since the pressure at the section (1) and section (2) is atmospheric, so the difference between the pressures is zero.

p1p2=0

Substitute 0 for p1p2 in Equation (I).

(0ρg)+( V 1 22g V 2 22g)+(z1z2)=dh(t)dtdh(t)dt=( V 1 22g V 2 22g)+(z1z2) …… (II)

Write the expression for the flow rate at section (1).

Q=A1V1 …… (III)

Here, the area at section (1) is A1 and the velocity at section (1) is V1.

Write the expression for the area at section (1).

A1=π4D12

Here, the diameter at section (1) is D1.

Substitute π4D12 for A1 in Equation (III).

Q=π4D12V1 …… (IV)

Write the expression for the flow rate at section (2).

Q=A2V2 …… (V)

Here, the flow rate is Q, the jet area at section (2) is A2 and the velocity at section (2) is V2.

Write the expression for the area at section (2).

A2=π4D22

Here, the diameter at section (2) is D2.

Substitute π4D22 for A2 in Equation (V).

Q=π4D22V2 …… (VI)

Since the flow rate at both the sections is equal, it satisfies the continuity equation.

Substitute π4D12V1 for Q in Equation (V).

π4D12V1=π4D22V2D12V1=D22V2V1=D22D12V2 …… (VII)

Write the expression for velocity at section (2).

V2=2gh(t)

Here, the water level is h.

Substitute 2gh(t) for V2 in Equation (VII).

V1=2gh(t)(D22D12)

Substitute 2gh(t)(D22D12) for V1, 2gh(t) for V2, h(t) for z1 and 0 for z2 in Equation (II).

dh(t)dt=( {2gh(t)( D 2 2 D 1 2 )} 22g (2gh( t) ) 22g)+(h(t)0)dh(t)dt=2gh(t)2g( D 2 4 D 1 41)+h(t)dh(t)dt=h(t)(( D24 D14 1)+1)dh(t)h(t)=D24D14dt

The negative sign is due to decrease in the head level in the tank.

Conclusion:

The differential equation for the free surface height during draining with respect to time is dh(t)h(t)=D24D14dt.

To determine

(b)

The expression for the time t0 to drain the entire tank.

Expert Solution
Check Mark

Answer to Problem 3.142P

The expression for the time t0 to drain the entire tank is

Explanation of Solution

Given information:

The diameter of the cylindrical tank is D . The initial height of the liquid is h0 . The losses are negligible. The stopper is removed at time t=0.

Write the differential equation for the free surface height.

dh(t)h(t)=D24D14dt

Integrate the above expression on both the sides.

dh( t ) h( t )=D24D14dtln{h(t)}=D24D14t+C …… (VIII)

Apply boundary conditions to Equation (VIII).

At t=0 ; h(t)=h0.

Substitute h0 for h(t) in Equation (VIII).

ln{h0}=D24D14×0+CC=lnh0

Substitute lnh0 for C in Equation (VIII).

ln{h0}=D24D14t+lnh0ln{h(t)}lnh0=D24D14th(t)h0=e( D2D1 )4th(t)=h0e( D2D1 )4t

Since the tank is empty after time t0.

Conclusion:

The force per unit width of the water is 68248.19N/m.

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Chapter 3 Solutions

Fluid Mechanics

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