Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 3, Problem 3.159P
To determine

(a)

The absolute velocity V1.

Expert Solution
Check Mark

Answer to Problem 3.159P

The absolute velocity V1 is 21.87ft/s.

Explanation of Solution

Given information:

The flow rate of the pump is 4000gal/m . The temperature of the water is 20°C . The shaft rotation rate is 1750r/min . The radius of the impeller at section (1) is 6in and the radius of the pump at section (2) is 14in . The width of the blade is 1.75in . The tangential velocity at section (1) is 10ft/s and the tangential velocity at section (2) is 110ft/s.

Write the expression for the absolute velocity at section (1).

V1=Vt,12+Vn,12 …… (I)

Here, the absolute velocity is V1, the tangential velocity at section (1) is Vt,1 and the normal velocity is Vn,1.

Write the expression for the normal velocity at section (1).

Vn,1=QA1 …… (II)

Here, the flow rate is Q and the area at section (1) is A1.

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

A1=2πr1b1

Here, the width of the blade is b1.

Substitute 2πr1b1 for A1 in Equation (II).

Vn,1=Q2πr1b1 …… (III)

Calculation:

Substitute 4000gal/m for Q, 6in for r1 and 1.75in for b1 in Equation (III).

Vn,1= 4000 gal/m 2π 6in 1.75in= 4000 gal/m 1 ft 3 7.48gal 1m 60s 65.97 in 2 1 ft 2 144 in 2 =19.45ft/s

Substitute 19.45ft/s for Vn,1 and 10ft/s for Vt,1 in Equation (I).

V1= 10 ft/s 2+ 19.45 ft/s 2=478.3025 ft 2/ s 2=21.87ft/s

Conclusion:

The absolute velocity V1 is 21.87ft/s.

To determine

(b)

The absolute velocity V2.

Expert Solution
Check Mark

Answer to Problem 3.159P

The absolute velocity V2 is 110.31ft/s.

Explanation of Solution

Given information:

The flow rate of the pump is 4000gal/m . The temperature of the water is 20°C . The shaft rotation rate is 1750r/min . The radius of the impeller at section (1) is 6in and the radius of the pump at section (2) is 14in . The width of the blade is 1.75in . The tangential velocity at section (1) is 10ft/s and the tangential velocity at section (2) is 110ft/s.

Write the expression for the absolute velocity at section (2).

V2=Vt,22+Vn,22 …… (IV)

Here, the absolute velocity is V2, the tangential velocity at section (2) is Vt,2 and the normal velocity is Vn,2.

Write the expression for the normal velocity at section (2).

Vn,2=QA2 …… (V)

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

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

A2=2πr2b2

Here, the width of the blade is b2.

Substitute 2πr2b2 for A2 in Equation (V).

Vn,2=Q2πr2b2 …… (VI)

Calculation:

Substitute 4000gal/m for Q, 14in for r2 and 1.75in for b2 in Equation (VI).

Vn,2= 4000 gal/m 2π 14in 1.75in= 4000 gal/m 1 ft 3 7.48gal 1m 60s 153.94 in 2 1 ft 2 144 in 2 =8.3ft/s

Substitute 8.3ft/s for Vn,2 and 110ft/s for Vt,2 in Equation (VI).

V2= 110 ft/s 2+ 8.3 ft/s 2=12168.89 ft 2/ s 2=110.31ft/s

Conclusion:

The absolute velocity V2 is 110.31ft/s.

To determine

(c)

The horsepower required.

Expert Solution
Check Mark

Answer to Problem 3.159P

The horsepower required is 894.25hp.

Explanation of Solution

Given information:

The flow rate of the pump is 4000gal/m . The temperature of the water is 20°C . The shaft rotation rate is 1750r/min . The radius of the impeller at section (1) is 6in and the radius of the pump at section (2) is 14in . The width of the blade is 1.75in . The tangential velocity at section (1) is 10ft/s and the tangential velocity at section (2) is 110ft/s.

Write the expression for the Bernoulli’s equation between the section (1) and section (2) of the given system.

p1ρg+V122g+z1=p2ρg+V222g+z2hf p 1 p 2 ρg+ V 1 2 2g V 2 2 2g+z1z2=hf …… (VII)

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

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

p1p2=0

Since the datum is at the same level so,

z1z2=0

Substitute 0 for p1p2 and 0 for z1z2 in Equation (VII).

0 ρg+ V 1 2 2g V 2 2 2g+0=hfhf= V 2 2 V 1 2 2g …… (VIII)

Write the expression for the power loss due to friction head.

Ploss=ρQghf …… (IX)

Here, the power loss is Ploss.

Write the expression for the ideal horsepower developed.

Pideal=ρQr2Vt,2r1Vt,1ω …… (X)

Here, the ideal power is Pideal the density is ρ and angular revolutions is ω.

Write the expression for the horsepower required.

Pactual=Pideal+Ploss …… (XI)

Here, the power required is Pactual.

Calculation:

Substitute 21.87ft/s for V1, 110.31ft/s for V2 and 32.2ft/s2 for g in Equation (VIII).

hf= 110.31 ft/s 2 21.87 ft/s 2 2 32.2 ft/ s 2 =11689.99 ft 2/ s 264.4ft/ s 2=181.52ft

Substitute 1.94slug/ft3 for ρ, 4000gal/m for Q, 32.2ft/s2 for g and 181.52ft for hf in Equation (IX).

Ploss=1.94slug/ ft 34000gal/m32.2ft/ s 2181.52ft=1.94slug/ ft 34000gal/m 1 ft 3 7.48gal 1m 60s5844.94 ft 2/ s 2=101062.31lbft/s 1hp 550 lbft/s =183.75hp

Substitute 1.94slug/ft3 for ρ, 4000gal/m for Q, 14in for r2, 110ft/s for Vt,2, 6in for r1, 10ft/s for Vt,1, 1750r/min for ω in Equation (X).

Pideal=1.94slug/ ft 34000gal/m 14in 110 ft/s 6in 10 ft/s 1750r/min= 1.94 slug/ ft 3 4000 gal/m 1 ft 3 7.48gal 1m 60s × 14in 1ft 12in × 110 ft/s 6in 1ft 12in × 10 ft/s × 1750r/ min 2πrad 1rev 1min 60s =390801.31lbft/s 1hp 550 lbft/s =710.5hp

Substitute 183.75hp for Ploss and 710.5hp for Pideal in Equation (XI).

P=710.5hp+183.75hp=894.25hp

Conclusion:

The horsepower required is 894.25hp.

To determine

(d)

The comparison of actual and ideal horsepower.

Expert Solution
Check Mark

Answer to Problem 3.159P

The actual horsepower is greater than ideal horsepower.

Explanation of Solution

Given information:

The flow rate of the pump is 4000gal/m . The temperature of the water is 20°C . The shaft rotation rate is 1750r/min . The radius of the impeller at section (1) is 6in and the radius of the pump at section (2) is 14in . The width of the blade is 1.75in . The tangential velocity at section (1) is 10ft/s and the tangential velocity at section (2) is 110ft/s.

Write the ideal horsepower required.

Pideal=710.5hp

Write the actual horsepower required.

Pactual=894.25hp

Since due to frictional losses between the section (1) and section (2) there is an additional power required to overcome the friction.

Pactual>Pideal

Conclusion:

The actual horsepower is greater than ideal horsepower.

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Chapter 3 Solutions

Fluid Mechanics

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