Chapter 14, Problem 122P

To determine

The efficiency of the turbine A.

The efficiency of the turbine B.

The actual efficiency of the turbine B by using Moody expected efficiency of the turbine.

Answer to Problem 122P

The efficiency of the turbine A is $0.92473$.

The efficiency of the turbine B is $0.92481$.

The actual efficiency of the turbine B is $0.92795$.

Explanation of Solution

Given information:

The diameter of the turbine A is $1.50\text{m}$and the number of rotation for the turbine A is $150\text{rpm}$. The volume flow rate for the turbine A is $162{\text{m}}^3/\text{s}$and the net head for the turbine A is $-90\text{m}$. The brake horsepower for the turbine is $132\text{MW}$. The net head for the turbine B is $110\text{m}$and the number of rotation for the turbine B is $120\text{rpm}$.

Write the expression for the dimensionless turbine A parameter.

  $C_{H,\text{A}}=\frac{g{H}_{\text{A}}}{{\omega }_{\text{A}}^2{D}_{\text{A}}^2}$   ....... (I)

Here, the dimension less parameter for turbine A is $C_{H,\text{A}}$, the acceleration due to gravity is $g$, the net head of the turbine A is $H_{\text{A}}$, the diameter of the turbine A is $D_{\text{A}}$and the angular acceleration of the turbine A is ${\omega }_{\text{A}}$.

Write the equation for the dimensionless turbine B parameter.

  $C_{H,\text{B}}=\frac{g{H}_{\text{B}}}{{\omega }_{\text{B}}^2{D}_{\text{B}}^2}$  ....... (II)

Here, the dimensionless parameter for the turbine B is $C_{H,\text{B}}$, the net head of the turbine B is $H_{\text{B}}$, the diameter of the turbine B is $D_{\text{B}}$and the angular acceleration of the turbine B is ${\omega }_{\text{B}}$.

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

  $C_{P,\text{A}}=\frac{bh{p}_{\text{A}}}{{\rho }_{\text{A}}{\omega }_{\text{A}}^3{D}_{\text{A}}^5}$   ....... (III)

Here, the affinity dimensional parameter for turbine A is $C_{P,\text{A}}$, the density for the turbine A is ${\rho }_{\text{A}}$and the brake horsepower for the turbine A is $bh{p}_{\text{A}}$.

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

  $C_{P,\text{B}}=\frac{bh{p}_{\text{B}}}{{\rho }_{\text{B}}{\omega }_{\text{B}}^3{D}_{\text{B}}^5}$  ....... (IV)

Here, the affinity dimensional parameter for turbine B is $C_{P,2}$, the density for the turbine B is ${\rho }_2$and the brake horsepower for the turbine B is $bh{p}_2$.

Equate Equation (I) and Equation (II).

  $\begin{array}{l}\frac{g{H}_{\text{A}}}{{\omega }_{\text{A}}^2{D}_{\text{A}}^2}=\frac{g{H}_{\text{B}}}{{\omega }_{\text{B}}^2{D}_{\text{B}}^2}\\ \frac{H_{\text{A}}}{{\omega }_{\text{A}}^2{D}_{\text{A}}^2}=\frac{H_{\text{B}}}{{\omega }_{\text{B}}^2{D}_{\text{B}}^2}\\ \frac{{\omega }_{\text{A}}^2}{{\omega }_{\text{B}}^2}=\frac{H_{\text{A}}{D}_{\text{B}}^2}{H_{\text{B}}{D}_{\text{A}}^2}\\ \frac{{\omega }_{\text{A}}}{{\omega }_{\text{B}}}={\left(\frac{H_{\text{A}}}{H_{\text{B}}}\right)}^{1/2}\left(\frac{D_{\text{B}}}{D_{\text{A}}}\right)\end{array}$  ....... (V)

Equate Equation (III) and Equation (IV).

  $\begin{array}{l}\frac{bh{p}_{\text{B}}}{{\rho }_{\text{B}}{\omega }_{\text{B}}^3{D}_{\text{B}}^5}=\frac{bh{p}_{\text{A}}}{{\rho }_{\text{A}}{\omega }_{\text{A}}^3{D}_{\text{A}}^5}\\ \frac{D_{\text{B}}^5}{D_{\text{A}}^5}=\frac{bh{p}_{\text{B}}{\rho }_{\text{A}}{\omega }_{\text{A}}^3}{bh{p}_{\text{A}}{\rho }_{\text{B}}{\omega }_{\text{B}}^3}\\ \frac{D_{\text{B}}^5}{D_{\text{A}}^5}=\left(\frac{bh{p}_{\text{B}}}{bh{p}_{\text{A}}}\right)\left(\frac{{\rho }_{\text{A}}}{{\rho }_{\text{B}}}\right){\left(\frac{{\omega }_{\text{A}}}{{\omega }_{\text{B}}}\right)}^3\end{array}$   ....... (VI)

Substitute ${\left(\frac{H_{\text{A}}}{H_{\text{B}}}\right)}^{1/2}\left(\frac{D_{\text{B}}}{D_{\text{A}}}\right)$for $\frac{{\omega }_{\text{A}}}{{\omega }_{\text{B}}}$in Equation (V).

  $\begin{array}{l}\frac{D_{\text{B}}^5}{D_{\text{A}}^5}=\left(\frac{bh{p}_{\text{B}}}{bh{p}_{\text{A}}}\right)\left(\frac{{\rho }_{\text{A}}}{{\rho }_{\text{B}}}\right){\left({\left(\frac{H_{\text{A}}}{H_{\text{B}}}\right)}^{1/2}\left(\frac{D_{\text{B}}}{D_{\text{A}}}\right)\right)}^3\\ \frac{D_{\text{B}}^5}{D_{\text{A}}^5}=\left(\frac{bh{p}_{\text{B}}}{bh{p}_{\text{A}}}\right)\left(\frac{{\rho }_{\text{A}}}{{\rho }_{\text{B}}}\right){\left(\frac{H_{\text{A}}}{H_{\text{B}}}\right)}^{3/2}{\left(\frac{D_{\text{B}}}{D_{\text{A}}}\right)}^3\end{array}$   ....... (VII)

Write the equation for the angular velocity of the turbine A.

  ${\omega }_{\text{A}}=\frac{2\pi {\dot{n}}_{\text{A}}}{60}$  ....... (VIII)

Here, the number of rotation for turbine A is ${\dot{n}}_{\text{A}}$.

Write the equation for the angular velocity of the turbine B.

  ${\omega }_{\text{B}}=\frac{2\pi {\dot{n}}_{\text{B}}}{60}$  ....... (IX)

Here, the number of rotation for turbine B is ${\dot{n}}_{\text{B}}$.

Write the expression for the volume flow rate for the turbine B.

  $\begin{array}{l}\frac{{\dot{V}}_{\text{B}}}{{\dot{V}}_{\text{A}}}=\left(\frac{{\omega }_{\text{A}}}{{\omega }_{\text{B}}}\right){\left(\frac{D_{\text{B}}}{D_{\text{A}}}\right)}^3\\ {\dot{V}}_{\text{B}}={\dot{V}}_{\text{A}}\left(\frac{{\omega }_{\text{A}}}{{\omega }_{\text{B}}}\right){\left(\frac{D_{\text{B}}}{D_{\text{A}}}\right)}^3\end{array}$  ....... (X)

Write the expression for the efficiency of turbine A.

  ${\eta }_{\text{A}}=\frac{bh{p}_{\text{A}}}{\rho g{H}_{\text{A}}{\dot{V}}_{\text{A}}}$  ....... (XI)

Here, the efficiency of the turbine is ${\eta }_{\text{A}}$.

Write the expression for the efficiency of turbine B.

  ${\eta }_{\text{B}}=\frac{bh{p}_{\text{B}}}{\rho g{H}_{\text{B}}{\dot{V}}_{\text{B}}}$  ....... (XII)

Here, the efficiency of the turbine B is ${\eta }_{\text{B}}$.

Write the expression for the Moody efficiency correction equation.

  ${\eta }_{\text{B1}}=1-\left(1-{\eta }_{\text{A}}\right){\left(\frac{D_{\text{A}}}{D_{\text{B}}}\right)}^{1/5}$  ....... (XIII)

Write the expression for the actual efficiency of turbine B.

  $\eta =\frac{2}{3}\left({\eta }_{\text{B1}}-{\eta }_{\text{A}}\right)+{\eta }_{\text{B}}$  ....... (XIV)

Here, the efficiency increase for turbine B is $\eta$.

Calculation:

Substitute $150\text{rpm}$for ${\dot{n}}_{\text{A}}$in Equation (VIII).

  $\begin{array}{c}{\omega }_{\text{A}}=\frac{2\pi \left(150\text{rpm}\right)}{60}\\ =\frac{942.4777\text{rpm}\left(\frac{1\text{rad}/{\text{s}}^2}{1\text{rpm}}\right)}{60}\\ =15.7079\text{rad}/{\text{s}}^2\end{array}$

Substitute $120\text{rpm}$for ${\dot{n}}_{\text{B}}$in Equation (IX).

  $\begin{array}{c}{\omega }_{\text{B}}=\frac{2\pi \left(120\text{rpm}\right)}{60}\\ =\frac{753.9822\text{rpm}\left(\frac{1\text{rad}/{\text{s}}^2}{1\text{rpm}}\right)}{60}\\ =12.5663\text{rad}/{\text{s}}^2\end{array}$

Substitute $12.5663\text{rad}/{\text{s}}^2$for ${\omega }_{\text{B}}$, $15.7079\text{rad}/{\text{s}}^2$for ${\omega }_{\text{A}}$, $1.50\text{m}$for $D_{\text{A}}$, $90\text{m}$for $H_{\text{A}}$and $110\text{m}$for $H_{\text{B}}$in Equation (V).

  $\begin{array}{l}\frac{15.7079\text{rad}/{\text{s}}^2}{12.5663\text{rad}/{\text{s}}^2}={\left(\frac{90\text{m}}{110\text{m}}\right)}^{1/2}\left(\frac{D_{\text{B}}}{1.50\text{m}}\right)\\ 1.24999=\left(0.90453\right)\left(\frac{D_{\text{B}}}{1.50\text{m}}\right)\\ D_{\text{B}}=\frac{1.24999\times 1.50\text{m}}{0.90453}\\ D_{\text{B}}=2.072898\text{m}\end{array}$

Substitute $12.5663\text{rad}/{\text{s}}^2$for ${\omega }_{\text{B}}$, $15.7079\text{rad}/{\text{s}}^2$for ${\omega }_{\text{A}}$, $1.50\text{m}$for $D_{\text{A}}$, $2.072898\text{m}$for $D_{\text{B}}$and $162{\text{m}}^3/\text{s}$for ${\dot{V}}_{\text{A}}$in Equation (X).

  $\begin{array}{c}{\dot{V}}_{\text{B}}=\left(162{\text{m}}^3/\text{s}\right)\left(\frac{12.5663\text{rad}/{\text{s}}^2}{15.7079\text{rad}/{\text{s}}^2}\right){\left(\frac{2.072898\text{m}}{1.50\text{m}}\right)}^3\\ =\left(162{\text{m}}^3/\text{s}\right)\left(0.819097\right)\left(2.639\right)\\ =342{\text{m}}^3/\text{s}\end{array}$

The density of the fluid for both turbine is zero.

Substitute ${\rho }_{\text{A}}$for ${\rho }_{\text{B}}$, $12.5663\text{rad}/{\text{s}}^2$for ${\omega }_{\text{B}}$, $15.7079\text{rad}/{\text{s}}^2$for ${\omega }_{\text{A}}$, $1.50\text{m}$for $D_{\text{A}}$, $2.072898\text{m}$for $D_{\text{B}}$and $132\text{MW}$for ${\text{bph}}_{\text{A}}$in Equation (VI).

  $\begin{array}{l}\frac{{\left(2.072898\text{m}\right)}^5}{{\left(1.5\text{m}\right)}^5}=\left(\frac{bh{p}_{\text{B}}}{132\text{MW}}\right)\left(\frac{{\rho }_{\text{A}}}{{\rho }_{\text{A}}}\right){\left(\frac{15.7079\text{rad}/{\text{s}}^2}{12.5663\text{rad}/{\text{s}}^2}\right)}^3\\ \frac{38.2727{\text{m}}^5}{7.59375{\text{m}}^5}=\left(\frac{bh{p}_{\text{B}}}{132\text{MW}}\right)\left(1.953134\right)\\ bh{p}_{\text{B}}=\frac{\left(5.0400\right)\left(132\text{MW}\right)}{1.953134}\\ bh{p}_{\text{B}}=340.6240\text{MW}\end{array}$

Substitute $90\text{m}$for $H_{\text{A}}$, $998\text{kg}/{\text{m}}^3$for ${\rho }_{\text{A}}$, $162{\text{m}}^3/\text{s}$for ${\dot{V}}_{\text{A}}$, $9.81\text{m}/{\text{s}}^2$for $g$and $132\text{MW}$for $bp{h}_{\text{A}}$in Equation (XI).

  $\begin{array}{c}{\eta }_{\text{A}}=\frac{132\text{MW}}{\left(998\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(90\text{m}\right)\left(162{\text{m}}^3/\text{s}\right)}\\ =\frac{132\text{MW}\left(\frac{1000000\text{W}}{1\text{MW}}\right)}{142743740.4\text{kg}\cdot {\text{m}}^2/{\text{s}}^3\left(\frac{1\text{W}}{1\text{kg}\cdot {\text{m}}^2/{\text{s}}^3}\right)}\\ =0.92473\end{array}$

Substitute $110\text{m}$for $H_{\text{B}}$, $998\text{kg}/{\text{m}}^3$for ${\rho }_{\text{B}}$, $342{\text{m}}^3/\text{s}$for ${\dot{V}}_{\text{A}}$, $9.81\text{m}/{\text{s}}^2$for $g$and $340.6240\text{MW}$for $bp{h}_{\text{B}}$in Equation (XII).

  $\begin{array}{c}{\eta }_{\text{B}}=\frac{340.6240\text{MW}}{\left(998\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(110\text{m}\right)\left(342{\text{m}}^3/\text{s}\right)}\\ =\frac{340.6240\text{MW}\left(\frac{1000000\text{W}}{1\text{MW}}\right)}{368314095.6\text{kg}\cdot {\text{m}}^2/{\text{s}}^3\left(\frac{1\text{W}}{1\text{kg}\cdot {\text{m}}^2/{\text{s}}^3}\right)}\\ =0.92481\end{array}$

Substitute $0.92473$for ${\eta }_{\text{A}}$, $1.50\text{m}$for $D_{\text{A}}$and $2.072898\text{m}$for $D_{\text{B}}$in Equation (XIII).

  $\begin{array}{c}{\eta }_{\text{B1}}=1-\left(1-0.92473\right){\left(\frac{1.50\text{m}}{2.072898\text{m}}\right)}^{1/5}\\ =1-\left(0.07527\right)\left(0.937351\right)\\ =1-0.0705544\\ =0.929445\end{array}$

Substitute $0.929445$for ${\eta }_{\text{B1}}$, $0.92473$for ${\eta }_{\text{A}}$and $0.92481$for ${\eta }_{\text{B}}$in Equation (XIV).

  $\begin{array}{c}\eta =\frac{2}{3}\left(0.929445-0.92473\right)+0.92481\\ =0.003143+0.92481\\ =0.92795\end{array}$

The efficiency of the turbine B is increased too. The relative roughness on the surface of the prototypes runner blades may be significantly smaller than the model turbine. The larger turbine has smaller tips clearance relative to the blade diameter. The smaller turbines have larger tips clearance relative to the blade diameter. So the leakage is less in the larger turbine as compared to the leakage in the smaller turbine.

Conclusion:

The efficiency of the turbine A is $0.92473$.

The efficiency of the turbine B is $0.92481$.

The actual efficiency of the turbine B is $0.92795$.