Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 8, Problem 76EP
To determine

The elevation of the higher reservoir for the flow rate of 10ft3/min.

Expert Solution & Answer
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Answer to Problem 76EP

The elevation of the higher reservoir for the flow rate of 10ft3/min is 12.655ft.

Explanation of Solution

Given information:

The temperature of the water is 70°F, density of the water is 62.3lbm/ft3, dynamic viscosity of the water is 6.556×104lbm/fts, loss coefficient at entrance is 0.03, loss coefficient for elbow is 0.3, loss coefficient for valve is 0.2, loss coefficient at exit is 1.0, length of the pipe is 60ft, diameter of the pipe is 2in, roughness of the cast iron pipe surfaces is 8.5×104ft, and volume flow rate of the water is 10ft3/min.

Write the expression for the Reynolds number.

  Re=ρvDμ........... (I)

Here, Reynolds number is Re, density of the water is ρ, velocity of the water is v, diameter of the pipe is D, and dynamic viscosity of the water is μ.

Write the expression for the volume flow rate of the water.

  Q˙=π4D2v........... (II)

Here, the volume flow rate of the water is Q˙.

Write the expression for the friction factor for turbulent flow.

  1f=2log(εD3.7+2.51Ref)........... (III)

Here, friction factor for turbulent flow is f and roughness of the copper pipe surfaces is ε.

Write the expression for the major head loss through the pipe.

  hf,major=fLv22gD........... (IV)

Here, length of the pipe is L, gravitational acceleration is g, and the major head loss through the pipe is hf,major.

Write the expression for the minor head loss through the pipe.

  hf,minor=KL( v 2 2g)........... (V)

Here, minor head loss through the pipe is hf,minor and sum of all loss coefficients is KL.

Write the expression for the sum of all loss coefficients.

  KL=KL,entrance+4KL,elbow+KL,valve+KL,exit........... (VI)

Here, loss coefficient at entrance is KL,entrance, loss coefficient for elbow is KL,elbow, loss coefficient for valve is KL,valve, and loss coefficient at exit is KL,exit.

Write the expression for the total head loss in the pipe.

  hf=hf,major+hf,minor........... (VII)

Here, total head loss in the pipe is hf.

Substitute KL( v 2 2g) for hf,minor and fLv22gD for hf,major in Equation (VI).

  hf=fLv22gD+KL( v 2 2g )=( fLD+ K L )v22g........... (VIII)

Write the expression for the Bernoulli Equation.

  P1ρg+V122g+Z1=P2ρg+V222g+Z2+hf........... (IX)

Here, pressure at point 1 is P1, velocity at point 1 in v1, density of the fluid is ρ, gravitational acceleration is g, total head loss in pipe is hf, elevation of point 1 is Z1, pressure at point 2 is P2, velocity at point 2 in v2, and elevation of point 2 is Z2.

Calculation:

Substitute 10ft3/min for Q˙ and 2in for D in Equation (II).

  (10 ft 3/min)=π4(2in)2v(10 ft 3/min)[1min60s]=π4(( 2)[ 1ft 12in ])2vv=7.64ft/s

Substitute 62.3lbm/ft3 for ρ, 7.64ft/s for v, 6.556×104lbm/fts for μ and 2in for D in Equation (I).

  Re=( 62.3 lbm/ ft 3 )( 7.64 ft/s )( 2in)( 6.556× 10 4 lbm/ fts )=( 62.3 lbm/ ft 3 )( 7.64 ft/s )( 2in)[ 1ft 12in]( 6.556× 10 4 lbm/ fts )=60700

Since value of the Reynolds number is greater than 4000, therefore the flow is turbulent.

Substitute 60700 for Re, 8.5×104ft for ε and 2in for D in Equation (III).

  1f=2log( ( 8.5× 10 4 ft ) ( 2in ) 3.7+ 2.51 ( 60700 ) f )12f=log( ( 8.5× 10 4 ftt ) ( 2in )[ 1ft 12in ] 3.7+ 2.51 ( 60700 ) f )12f=log( ( 8.5× 10 4 ft ) ( 0.166ft ) 3.7+ 2.51 ( 60700 ) f )f=0.0320

Substitute 0.03 for KL,entrance, 0.3 for KL,elbow, 0.2 for KL,valve, and 1.0 for KL,exit in Equation (VI).

  KL=0.03+4×0.3+0.2+1.0=0.03+1.2+0.2+1.0=2.43

Substitute Patm for P1, Patm for P2, 0 for v1, 0 for v2 and 0 for Z2 in Equation (IX).

  P atmρg+ ( 0 )22g+Z1=P atmρg+ ( 0 )22g+0+hfZ1=hf........... (X)

Substitute 0.032 for f, 2.43 for KL, 60ft for L, 7.64ft/s for v, 32.17ft/s2 for g, and 2in for D in Equation (VIII).

  hf=( ( 0.032 )( 60ft ) ( 2in )+2.43) ( 7.64 ft/s )22( 32.17 ft/ s 2 )=( ( 0.032 )( 60ft ) ( 2in )[ 1ft 12in ]+2.43) ( 7.64 ft/s )22( 32.17 ft/ s 2 )=(11.52+2.43) ( 7.64 ft/s )22( 32.17 ft/ s 2 )=12.655ft

Substitute 12.655ft for hf in Equation (X).

  Z1=12.655ft

Conclusion:

The elevation of the higher reservoir for the flow rate of 10ft3/min is 12.655ft.

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Chapter 8 Solutions

Fluid Mechanics: Fundamentals and Applications

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