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 14, Problem 123P

Calculate and compare the turbine specific speed for both the small (A) and large (B) turbines of Prob. 14-121. What kind of turbine are these most likely to be?

Expert Solution & Answer
Check Mark
To determine

The specific speed of the turbine A.

The specific speed of the turbine B.

The type of turbine.

Answer to Problem 123P

The specific speed of the turbine A is 1.18699.

The specific speed of the turbine B is 1.18700.

The type turbine is Francis turbine.

Explanation of Solution

Given information:

The diameter of the turbine A is 1.50mand the number of rotation for the turbine A is 150rpm. The volume flow rate for the turbine A is 162m3/sand the net head for the turbine A is 90m. The brake horsepower for the turbine is 132MW. The net head for the turbine B is 110mand the number of rotation for the turbine B is

Write the expression for the dimensionless turbine A parameter.

  CH,A=gHAωA2DA2  ....... (I)

Here, the dimension less parameter for turbine A is CH,A, the acceleration due to gravity is g, the net head of the turbine A is HA, the diameter of the turbine A is DAand the angular acceleration of the turbine A is ωA.

Write the expression for the dimensionless turbine B parameter.

  CH,B=gHBωB2DB2  ....... (II)

Here, the dimensionless parameter for the turbine B is CH,B, the net head of the turbine B is HB, the diameter of the turbine B is DBand the angular acceleration of the turbine B is ωB.

Write the expression for the affinity law dimensionless parameter for turbine A.

  CP,A=bhpAρAωA3DA5  ....... (III)

Here, the affinity dimensional parameter for turbine A is CP,A, the density for the turbine A is ρAand the brake horsepower for the turbine A is bhpA.

Write the expression for the affinity law dimensionless parameter for pump B.

  CP,B=bhpBρBωB3DB5  ....... (IV)

Here, the affinity dimensional parameter for turbine B is CP,2, the density for the turbine B is ρ2and the brake horsepower for the turbine B is bhp2.

Equate Equation (I) and Equation (II).

  gHAωA2DA2=gHBωB2DB2HAωA2DA2=HBωB2DB2ωA2ωB2=HADB2HBDA2ωAωB=( HA HB )1/2( D B D A)  ....... (V)

Equate Equation (III) and Equation (IV).

  bhpBρBωB3DB5=bhpAρAωA3DA5DB5DA5=bhpBρAωA3bhpAρBωB3DB5DA5=(bh p Bbh p A)( ρ A ρ B)( ωA ωB )3  ....... (VI)

Substitute ( H A H B)1/2(DBDA)for ωAωBin Equation (V).

  DB5DA5=(bh p Bbh p A)( ρ A ρ B)( ( H A H B ) 1/2 ( D B D A ))3DB5DA5=(bh p Bbh p A)( ρ A ρ B)( HA HB )3/2( DB DA )3  ....... (VII)

Write the expression for the specific speed of turbine A.

  NA=ωA(bh p A)1/2( ρ A)1/2(g H A)5/4  ....... (VIII)

Here, the specific speed of turbine A is NA.

Write the expression for the specific speed of turbine B.

  NB=ωB(bh p B)1/2( ρ B)1/2(g H B)5/4  ....... (IX)

Here, the specific speed of turbine B is NB.

Calculation:

Substitute 150rpmfor n˙Ain Equation (VIII).

  ωA=2π(150rpm)60=942.4777rpm( 1rad/ s 2 1rpm)60=15.7079rad/s2

Substitute 120rpmfor n˙Bin Equation (IX).

  ωB=2π(120rpm)60=753.9822rpm( 1rad/ s 2 1rpm)60=12.5663rad/s2

Substitute 12.5663rad/s2for ωB, 15.7079rad/s2for ωA, 1.50mfor DA, 90mfor HAand 110mfor HBin Equation (V).

  15.7079rad/s212.5663rad/s2=( 90m 110m)1/2( D B1.50m)1.24999=(0.90453)( D B1.50m)DB=1.24999×1.50m0.90453DB=2.072898m

Substitute 12.5663rad/s2for ωB, 15.7079rad/s2for ωA, 1.50mfor DA, 2.072898mfor DBand 162m3/sfor V˙Ain Equation (X).

  V˙B=(162m3/s)(12.5663 rad/ s2 15.7079 rad/ s2 )( 2.072898m 1.50m)3=(162m3/s)(0.819097)(2.639)=342m3/s

The density of the fluid for both turbine is zero.

Substitute ρAfor ρB, 12.5663rad/s2for ωB, 15.7079rad/s2for ωA, 1.50mfor DA, 2.072898mfor DBand 132MWfor bphAin Equation (VI).

  ( 2.072898m)5( 1.5m)5=(bh p B132MW)( ρ A ρ A)( 15.7079rad/s2 12.5663rad/s2 )338.2727m57.59375m5=(bh p B132MW)(1.953134)bhpB=(5.0400)(132MW)1.953134bhpB=340.6240MW

Substitute 90mfor HA, 998kg/m3for ρA, 15.7079rad/s2for ωA, 9.81m/s2for gand 340.6240MWfor bphBin Equation (VIII).

  NA=(15.7079 rad/ s2 )( 132MW)1/2( 998kg/ m 3)1/2( 9.81m/ s 2×90m)5/4=(15.7079 rad/ s2 )( 132MW( 1000000W 1MW ))1/2( 998kg/ m 3)1/2( 9.81m/ s 2×90m)5/4=180470.0312W1/2152038.8817 kg 1/2 m 7/1 /s 9/2 ( 1W 1/2 1 kg 1/2 m 7/1/ s 9/2)=1.18699

Substitute 110mfor HB, 998kg/m3for ρB, 12.5663rad/s2for ωB, 9.81m/s2for gand 340.6240MWfor bphB in Equation (X).

  NB=(12.5663 rad/ s2 )( 340.6240MW)1/2( 998kg/ m 3)1/2( 9.81m/ s 2×110m)5/4=(12.5663 rad/ s2 )( 340.6240MW( 1000000W 1MW ))1/2( 998kg/ m 3)1/2( 9.81m/ s 2×110m)5/4=231923.6546W1/2195385.5251 kg 1/2 m 7/1 /s 9/2 ( 1W 1/2 1 kg 1/2 m 7/1/ s 9/2)=1.18700

When the specific speed of both turbines is the same then its called Francis turbine, the value of the specific speed of turbine A is approximately equal to the value of the specific speed of the turbine B is. Then the type of turbine is Francis turbine.

Conclusion:

The specific speed of the turbine A is 1.18699.

The specific speed of the turbine B is 1.18700.

The type turbine is Francis turbine.

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Chapter 14 Solutions

Fluid Mechanics: Fundamentals and Applications

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