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 14, Problem 86EP
To determine

The swirl angles α2 and α1.

Whether the turbine have forward or swirl flow.

The power output in hp.

The net head in ft.

Expert Solution & Answer
Check Mark

Answer to Problem 86EP

The swirl angles α2 is 4.845° and α1 is 35.48°.

The turbine will have a forward swirl.

The power output is 2.01×105hp.

The net head is 170.47ft.

Explanation of Solution

Given information:

The gross head is 90m, the volume flow rate of water is 4.7×106gpm, the speed of the runner is 120rpm, the runner blade angle β2 is 82°, the runner blade angle β1 is 46°.

Write the expression for the angular velocity of turbine.

   ω=2πn˙60    ...... (I)

Here, the speed of the runner is n˙.

Write the expression for the normal component of velocity at the inlet.

   V2,n=˙2πr2b2 .   ...... (II)

Here, the volume flow rate of water is ˙, the radius at inlet is r2, and the breadth at the inlet is b2.

Write the expression for the tangential velocity component at inlet.

   V2,t=ωr2V2,ntanβ2    ...... (III)

Here, the radius at the inlet is r2 and the runner blade angle at inlet is β2.

Write the expression for the angle for tangential velocity component at inlet.

   α2=tan1(V 2,tV 2,n)    ...... (IV)

Write the expression for the normal component of velocity at the outlet.

   V1,n=˙2πr1b1 .   ...... (V)

Write the expression for the tangential velocity component at outlet.

   V1,t=ωr1V1,ntanβ1    ...... (VI)

Here, the radius at the outlet is r2 and the runner blade angle at outlet is β1.

Here, the volume flow rate of water is ˙, the radius at outlet is r1, and the breadth at the outlet is b1.

Write the expression for angle of the tangential velocity component at outlet.

   α1=tan1(V 1,tV 1,n)    ...... (VII)

Write the expression for the shaft output power.

   W˙shaft=ρω˙(r2V2,tr1V1,t)    ...... (VIII)

Here, the density of water is ρ.

Write the expression for the net head.

   H=W˙shaftρg˙    ...... (IX)

Calculation:

Substitute 120rpm for n˙ in Equation (I).

   ω=2π( 120rpm)60=12.56rpm( 1rad/s 1rpm)=12.56rad/s

Substitute 6.60ft for r2, 2.60ft for b2 and 4.7×106gpm for ˙ in Equation (II).

   V2,n=4.7× 106gpm2π( 6.60ft)( 2.60ft)=4.7× 106gpm( 1 ft 3 /s 448.8gpm )2π( 6.60ft)( 2.60ft)=10472.37 ft 3/s2π( 6.60ft)( 2.60ft)=97.12ft/s

Substitute 6.60ft for r2, 12.56rad/s for ω, 82° for β2, and 97.12ft/s for V2,n in Equation (III).

   V2,t=12.56rad/s(6.60ft)97.12ft/stan82°=82.89ft/s13.64ft/s=69.25ft/s

Substitute 69.25ft/s for V2,t and 97.12ft/s for V2,n in Equation (IV).

   α2=tan1( 69.25 ft/s 97.12 ft/s )=tan1(0.71)=35.48°

Therefore, the runner leading angle at the inlet is 35.48°.

Substitute 4.40ft for r1, 7.20ft for b1 and 4.7×106gpm for ˙ in Equation (V).

   V1,n=4.7× 106gpm2π( 4.40ft)( 7.20ft)=4.7× 106gpm( 1 ft 3 /s 448.8gpm )2π( 4.40ft)( 7.20ft)=10472.37 ft 3/s2π( 4.40ft)( 7.20ft)=52.609ft/s

Substitute 4.40ft for r1, 12.56rad/s for ω, 46° for β1, and 52.609ft/s for V1,n in Equation (VI).

   V1,t=12.56rad/s(4.40ft)52.609ft/stan46°=55.264ft/s50.803ft/s=4.46ft/s

Substitute 4.46ft/s for V1,t and 52.609ft/s for V1,n in Equation (VII).

   α1=tan1( 4.46 ft/s 52.609 ft/s )=tan1(0.0.8)=4.845°

Therefore, the runner leading angle at the outlet is 4.845°.

Refer to Table A-3E, “Properties of saturated water” to obtain the value of density (ρ) as 998kg/m3 at 20°C.

Substitute 62.30lbm/ft3 for ρ, 12.56rad/s for ω, 4.7×106gpm for ˙, 6.60ft for r2, 69.246ft/s for V2,t, 4.40ft for r1, and 4.46ft/s for V1,t in Equation (VIII).

   W˙shaft=( ( 62.30 lbm/ ft 3 )( 12.56 rad/s )( 4.7× 10 6 gpm ) ×( ( 6.60ft )( 69.246 ft/s )( 4.40ft )( 4.46 ft/s ) ))=( ( 62.30 lbm/ ft 3 )( 12.56 rad/s )( 4.7× 10 6 gpm( 1 ft 3 /s 448.8gpm ) ) ×( ( 6.60ft )( 69.246 ft/s )( 4.40ft )( 4.46 ft/s ) ))=( ( 62.30 lbm/ ft 3 )( 12.56 rad/s )( 10472.37 ft 3 /s ) ×( ( 6.60ft )( 69.246 ft/s )( 4.40ft )( 4.46 ft/s ) ))=3.584128×109lbmft2/s3

   W˙shaft=3.584128×109lbmft2/s3( 1 Btu/s 25051.6 lbm ft 2 / s 3 )=1.4306×105Btu/s( 1.41hp 1 Btu/s )=2.01×105hp

Therefore, the power output is 2.01×105hp.

Substitute 3.584128×109lbmft2/s3 for W˙shaft, 62.30lbm/ft3 for ρ, 32.2ft/s2 for g and 4.7×106gpm for ˙ in Equation (IX).

   H=3.584128× 109lbm ft 2/ s 3( 62.30 lbm/ ft 3 )( 32.2 ft/ s 2 )( 4.7× 10 6 gpm)=3.584128× 109lbm ft 2/ s 3( 62.30 lbm/ ft 3 )( 32.2 ft/ s 2 )( 4.7× 10 6 gpm( 1 ft 3 /s 448.8gpm ))=3.584128× 109lbm ft 2/ s 3( 62.30 lbm/ ft 3 )( 32.2 ft/ s 2 )( 10472.37 ft 3 /s )=170.47ft

Therefore, the net head is 170.47ft.

Conclusion:

Therefore, the swirl angles α2 is 4.845° and α1 is 35.48° .

The turbine will have a forward swirl.

The power output is 2.01×105hp and the net head is 170.47ft.

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Chapter 14 Solutions

Fluid Mechanics Fundamentals And Applications

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