Fluid Mechanics Fundamentals And Applications
Fluid Mechanics Fundamentals And Applications
3rd Edition
ISBN: 9780073380322
Author: Yunus Cengel, John Cimbala
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 5, Problem 90P

Water enters a hydraulic turbine through a 30-cm-diameter pipe at a rate of 0.6 m3/s and exits
through a 25-cm-diameter pipe. The pressure drop in the turbine is measured by a mercury manometer to be 1.2 in. For a combined turbine-generator efficiency of 83 percent determine the net electric power output. Disregard the effect of the kinetic energy correction factors.

Expert Solution & Answer
Check Mark
To determine

The net electric power output.

Answer to Problem 90P

The net electric power output is 60.334kW.

Explanation of Solution

Given information:

The initial diameter of the pipe is 30cm, the volume flow rate is 0.6m3/s, the final diameter of the pipe is 25cm, the pressure drop in a mercury manometer is 1.2m, and the combined efficiency is 83%.

Write the expression for the initial velocity of the water.

  V1=V˙( π D 1 2 4)  .......(I)

Here, the initial diameter of the pipe is D1 and the volume flow rate is V˙.

Write the expression for the final velocity of the water.

  V2=V˙( π D 2 2 4)  .......(II)

Here, the final diameter of the pipe is D2.

Write the expression for the pressure drop.

  Δp=ρHg×g×hHg  .......(III)

Here, the density of the mercury and water is ρHg, the drop in the pressure head is hHg, and the gravitational constant is g.

Write the expression for the turbine head using the Bernoulli's equation.

  P1ρg+α1V122g+z1+hpump=P2ρg+α2V222g+z2+htur+hLP1ρg+α1V122g+0+0=P2ρg+α2V222g+0+htur+0htur=P1ρgP2ρg+α( V 1 2 2g V 2 2 2g)htur=ΔPρg+α( V 1 2 2g V 2 2 2g)  .......(IV)

Here, the head loss is hL, the kinetic energy correction factor at initial level is α1, the kinetic energy correction factor at initial discharge is α2, the pressure at initial level of water is P1, the atmospheric pressure at initial discharge is P2, the elevation from ground to water level is z1, the elevation of water stream from nozzle to maximum height is z2, the density of the water is ρ, and the gravitational constant is g.

Write the expression for the net electric power output.

  W˙tur=ηturgenρV˙ghtur  .......(V)

Here, the turbine and generator efficiency is ηturgen.

Calculation:

Draw the diagram for a tank fitted with orifice.

  Fluid Mechanics Fundamentals And Applications, Chapter 5, Problem 90P

  Figure-(2)

Substitute 30cm for D1 and 0.6m3/s for V˙ in Equation (I).

  V1=( 0.6 m 3 /s )( π ( 30cm ) 2 4 )=( 0.6 m 3 /s )( π( 900 cm 2 ) 4 ( 1 m 2 10 4 cm 2 ))=8.48m/s

Substitute 25cm for D2 and 0.6m3/s for V˙ in Equation (II).

  V2=( 0.6 m 3 /s )( π ( 25cm ) 2 4 )=( 0.6 m 3 /s )( π( 625 cm 2 ) 4 ( 1 m 2 10 4 cm 2 ))=12.22m/s

Substitute 13600kg/m3 for ρHg, 9.81m/s2 for g and 1.2m for hHg in Equation (III).

  ΔP=(13600kg/ m 3)(9.81m/ s 2)(1.2m)=160×103kg/ms2( 1Pa 1kg/ m s 2 )=160×103Pa

Here, the gravitational constant is 9.81m/sec2.

Here, the density of the water is 1000kg/m3.

Substitute 160×103Pa for ΔP, 1000kg/m3 for ρ, 8.48m/s for V1, 12.22m/s for V2, 0 for α and 9.81m/sec2 for g in Equation (IV).

  htur=[ ( 160× 10 3 Pa ) ( 1000 kg/ m 3 )( 9.81 m/ sec 2 )+1.0( ( 8.48m/s ) 2 2( 9.81 m/ sec 2 ) ( 12.22m/s ) 2 2( 9.81 m/ sec 2 ) )]=16.30kg/ms2( 1Pa 1kg/ m s 2 )+1.0(3.66m7.61m)=16.30+1.0(3.95m)=12.35m

Substitute 12.35m for htur, 0.83 for ηturgen, 1000kg/m3 for ρ, 9.81m/sec2 for g and 0.6m3/s for V˙ in Equation (V).

  W˙tur=0.83(1000kg/ m 3)(0.6 m 3/s)(9.81 m/ sec 2)(12.35m)=60.334×103kgm2/sec3( kW 10 3 kg m 2 / sec 3 )=60.334kW

Conclusion:

Thus, the net electric power output is 60.334kW.

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