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 96P

Water to a residential area is transported at a rate of 1.5 m3/s via 70-cm-internal-diameter concrete pipes with a surface roughness of 3 mm and a total length of 1500 m. In order to reduce pumping power requirements, it is proposed to line the interior surfaces of the concrete pipe with 2-cm-thick petroleum-based lining that has a surface roughness thickness of 004 mm, this is a concern that the reduction of pipe diameter to 66 cm and the increase in average velocity may offset any gains. Taking p = 1000 kg/m3 and v = 1 × 10-6 m2/s for water, determine the percent increase or decrease in the pumping power requirements due to pipe frictional losses as a result of lining time concrete pipes.

Expert Solution & Answer
Check Mark
To determine

The percentage change in pumping power requirement due to frictional losses.

Answer to Problem 96P

The percentage decrease in pumping power requirement due to frictional losses is 54.157%.

Explanation of Solution

Given information:

The density of the water is 1000kg/m3, kinematic viscosity of the water is 1×106m2/s, diameter of the pipe without line interior surface is 70cm, diameter of the pipe with line interior surface is 66cm, roughness of the line interior concrete pipe surfaces is 0.04mm, roughness of the concrete pipe surfaces is 3mm, volume flow rate of water is 1.5m3/s, and length of the pipe is 1500m.

Write the expression for the Reynolds number.

  Re=vDυ........... (I)

Here, Reynolds number is Re, velocity of the water is v, diameter of the pipe is D, and kinematic viscosity of the water is υ.

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 plastic pipe surfaces is ε.

Write the expression for the head loss through the pipe.

  hf=fLv22gD........... (IV)

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

Write the expression for the pressure drop through the pipe.

  ΔP=ρghf........... (V)

Here, the pressure drop through the pipe is ΔP and density of the water is ρ.

Write the expression for the pumping power requirement to maintain the flow rate of water.

  W˙pump=Q˙ΔP........... (VI)

Here, pumping power requirement to maintain the flow rate of water is W˙pump.

Substitute ρghf for ΔP and π4D2v for Q˙ in Equation (VI).

  W˙pump=(π4D2v)(ρghf)=π4D2vρghf........... (VII)

Write the expression for the pumping power for without line interior surface pipe.

  ( W ˙ pump)withoutline=π4ρg(vhfD2)withoutline........... (VIII)

Here, subscript for without line interior pipe is without line.

Write the expression for the pumping power for with line interior surface pipe.

  ( W ˙ pump)withline=π4ρg(vhfD2)withline........... (IX)

Here, subscript for with line interior pipe is with line.

Write the expression for the percentage change in power due to frictional losses.

  m=( ( W ˙ pump ) withline ( W ˙ pump ) withoutline)×100%........... (X)

Here percentage change in power due to frictional losses is m.

Substitute π4ρg(vhfD2)withoutline for ( W ˙ pump)withoutline and π4ρg(vhfD2)withline for ( W ˙ pump)withline in Equation (X).

  m=( π 4 ρg ( v h f D 2 ) withline π 4 ρg ( v h f D 2 ) withoutline )×100%=( ( v h f D 2 ) withline ( v h f D 2 ) withoutline )×100%........... (XI)

Calculation:

Substitute 1.5m3/s for Q˙, 70cm for D and vwithoutline for v in Equation (II).

  (1.5 m 3/s)=π4(70cm)2(v)withoutline(1.5 m 3/s)=π4(( 70cm)[ 1m 100cm ])2(v)withoutline(v)withoutline=3.897m/s

Substitute 3.897m/s for v, 1×106m2/s for υ, 70cm for D and Rewithoutline for Re in Equation (I).

  Rewithoutline=( 3.897m/s )( 70cm)( 1× 10 6 m 2 /s )=( 3.897m/s )( 70cm)[ 1m 100cm]( 1× 10 6 m 2 /s )=2.728×106

Substitute 2.728×106 for Re, 3mm for ε, 70cm for D and fwithoutline for f in Equation (III).

  1 f withoutline =2log( ( 3mm ) ( 70cm ) 3.7+ 2.51 ( 2.728× 10 6 ) f withoutline )1 f withoutline =2log( ( 3mm )[ 1cm 10mm ] ( 70cm ) 3.7+ 2.51 ( 2.728× 10 6 ) f withoutline )fwithoutline=0.029

Substitute 0.029 for f, 1500m for L, 3.897m/s for v, 9.81m/s2 for g, 70cm for D and (hf)withoutline for hf in Equation (IV).

  ( h f)withoutline=( 0.029)( 1500m) ( 3.897m/s )22( 9.81m/ s 2 )( 70cm)=( 0.029)( 1500m) ( 3.897m/s )22( 9.81m/ s 2 )( 70cm)[ 1m 100cm]=48.1m

Substitute 1.5m3/s for Q˙, 66cm for D and vwithline for v in Equation (II).

  (1.5 m 3/s)=π4(66cm)2(v)withline(1.5 m 3/s)=π4(( 66cm)[ 1m 100cm ])2(v)withline(v)withline=4.384m/s

Substitute 4.384m/s for v, 1×106m2/s for υ, 66cm for D and Rewithline for Re in Equation (I).

  Rewithline=( 4.384m/s )( 66cm)( 1× 10 6 m 2 /s )=( 4.384m/s )( 66cm)[ 1m 100cm]( 1× 10 6 m 2 /s )=2.893×106

Substitute 2.893×106 for Re, 0.04mm for ε, 66cm for D and fwithline for f in Equation (III).

  1 f withline =2log( ( 0.04mm ) ( 66cm ) 3.7+ 2.51 ( 2.893× 10 6 ) f withline )1 f withline =2log( ( 0.04mm )[ 1cm 10mm ] ( 66cm ) 3.7+ 2.51 ( 2.893× 10 6 ) f withline )fwithline=0.0117

Substitute 0.0117 for f, 1500m for L, 4.384m/s for v, 9.81m/s2 for g, 66cm for D and (hf)withline for hf in Equation (IV).

  ( h f)withline=( 0.0117)( 1500m) ( 4.384m/s )22( 9.81m/ s 2 )( 66cm)=( 0.0117)( 1500m) ( 4.384m/s )22( 9.81m/ s 2 )( 66cm)[ 1m 100cm]=26.048m

Substitute 3.897m/s for vwithoutline, 70cm for Dwithoutline, 48.1m for (hf)withoutline, 4.384m/s for vwithline, 66cm for Dwithline and 26.048m for (hf)withline in Equation (XI).

  m=( ( 4.384m/s )( 26.048m ) ( 66cm ) 2 ( 3.897m/s )( 48.1m ) ( 70cm ) 2 )×100%=( ( 4.384m/s )( 26.048m ) ( 66cm ) 2 ( 3.897m/s )( 48.1m ) ( 70cm ) 2 )×100%=54.157%

Conclusion:

The percentage decrease in pumping power requirement due to frictional losses is 54.157%.

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Chapter 8 Solutions

Fluid Mechanics: Fundamentals and Applications

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