FLUID MECHANICS FUND. (LL)-W/ACCESS
FLUID MECHANICS FUND. (LL)-W/ACCESS
4th Edition
ISBN: 9781266016042
Author: CENGEL
Publisher: MCG CUSTOM
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Chapter 8, Problem 148P
To determine

The percentage increase in net power output.

Expert Solution & Answer
Check Mark

Answer to Problem 148P

The percentage increase in net power output is 14.30%.

Explanation of Solution

Given information:

The diameter of the cast iron pipe is 0.35m and the length of the cast iron pipe is 200m. The head loss during pump is 140m and the density of water at 20°C is 998kg/m3. The viscosity of water at 20°C is 0.001002kg/ms and the combined turbine efficiency is 80%. The diameter of the pipe is tripled in order to reduced the pipe loss.

Write the expression for energy equation for a control volume.

  P1ρg+α1V122g+z1+hpump=P2ρg+α2V222g+z2+hturbine+hL  ...... (I)

Here, the pressure at point 1 is P1, the pressure at point 2 is P2, the height at 1 is z1, the height at 2 is z2, the velocity at point 1 is V1, the velocity at point 2 is V2,the head loss due to pump is hpump, the head loss due to turbine is hturbine, the kinetic energy correction factor at point 1 is α1, the kinetic energy correction factor at point 2, acceleration due to gravity is g, the density of fluid is ρ and the head loss in the pipe is hL.

Write the expression for average velocity.

  V=4V˙π(D)2....... (II)

Here, the volume flow rate is V˙, the diameter of the pipe is D and the relative velocity is V.

Write the expression for Reynolds number for flow in pipe.

  Re=ρVDμ...... (III)

Here, the Reynolds number for flow in pipe is ReA and the density of fluid is ρ and

Write the expression for Colebrook equation for pipe A.

  1f=2log(ε/D3.7+2.51Ref)...... (IV)

Here, the relative velocity through the pipe is ε/D and the friction factor of the pipe f.

Write the expression for the head loss in pipe.

  hL=fLV22gD..... (V)

Here, the head loss in the pipe is hL, the friction factor in pipe is fA, the length of the pipe is LA and the velocity ion the pipe is LB.

Write the expression for percentage increase in net power output.

  %increase=hturbinehturbine,ehturbine,e×100  ...... (VI)

Calculation:

Substitute Patm for P1, 0 for V1, Patm for P2, 140m for hpump and 0 for V2 in Equation (I).

  PEρg+α1(0)2g+0+140m=P atmρg+α2(0)2g+0+hturbine+hL140m=hturbine+hLhturbine+hL=140m....... (VII)

Substitute 0.35m for D and 0.55m3/s for V˙ in Equation (II).

  V=4( 0.55 m 3 /s )π ( 0.35m )2=2.2 m 3/s1.099×0.35m2=5.7194m/s

Substitute 998kg/m3 for ρ, 1.002×103kg/ms for μ and 0.35m for D in Equation (III).

  Re=( 998 kg/ m 3 )( 5.7194m/s )( 0.35m)1.002× 10 3kg/ms=1997.786kg/ms1.002× 10 3kg/ms=1993798.822

Refer the Table-3.1, "Roughness height for certain common Pipe materials" to obtain the value of the roughness height 0.26×103m corresponding through cast iron pipe.

Substitute 0.35m for D, 1993798.822 for Re and 0.26×103m for ε in Equation (IV).

  1f=2log( 0.26× 10 3 m/ 0.35m 3.7+ 2.51 1993798.822 f )=2log( 7.422857× 10 4 3.7+ 1.258× 10 6 f )=2log(2.006× 10 4+ 1.258× 10 6 f )=0.0185

Substitute 5.71m/s for D, 0.35m for D, 200m for L, 9.81m/s2 for g and 0.0185 for f in Equation (V).

  hL=( 0.0185)( 200m) ( 5.71m/s )22( 9.81m/ s 2 )( 0.35m)=120.635 m 3/s6.865 m 2/s=17.5724m

Substitute 17.5724m for hL in Equation (VII).

  hturbine,e+17.5724m=140mhturbine,e=140m-17.5724mhturbine,e=122.427m

Substitute (0.35×3)m for D and 0.55m3/s for V˙ in Equation (II).

  V=4( 0.55 m 3 /s )π ( 3×0.35m )2=2.2 m 3/s0.38484m2=0.6354m/s

Substitute 998kg/m3 for ρ, 1.002×103kg/ms for μ and (0.35×3)m for D in Equation (III).

  Re=( 998 kg/ m 3 )( 0.6354m/s )( 0.35m×3)1.002× 10 3kg/ms=665.835kg/ms1.002× 10 3kg/ms=664506.646

Substitute (0.35×3)m for D, 664506.646 for Re and 0.26×103m for ε in Equation (IV).

  1f=2log( 0.26× 10 3 m/ ( 0.35×3 )m 3.7+ 2.51 664506.646× f )=2log( 2.4722857× 10 4 3.7+ 3.777× 10 6 f )=2log(6.6924× 10 5+ 3.777× 10 6 f )=0.0155

Substitute 0.6354m/s for V, (0.35×3)m for D, 200m for L, 9.81m/s2 for g and 0.0155 for f in Equation (V).

  hL=( 0.0155)( 200m) ( 0.6354m/s )22( 9.81m/ s 2 )( 0.35×3m)=1.2515 m 3/s20.601 m 2/s=0.06074m

Substitute 0.06074m for hL in Equation (VII).

  hturbine+0.06074m=140mhturbine=140m0.06074mhturbine=139.9392m

Substitute 122.427m for hturbine,e and 139.9392m for hturbine in Equation (VI).

  %increase=139.9392m122.427m122.427m×100=17.5122m122.427m×100=14.30%

Conclusion:

The percentage increase in net power output is 14.30%.

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Chapter 8 Solutions

FLUID MECHANICS FUND. (LL)-W/ACCESS

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