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?

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 $1.50\text{m}$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

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 expression 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 expression for the specific speed of turbine A.

  $N_{\text{A}}=\frac{{\omega }_{\text{A}}{\left(bh{p}_{\text{A}}\right)}^{1/2}}{{\left({\rho }_{\text{A}}\right)}^{1/2}{\left(g{H}_{\text{A}}\right)}^{5/4}}$  ....... (VIII)

Here, the specific speed of turbine A is $N_{\text{A}}$.

Write the expression for the specific speed of turbine B.

  $N_{\text{B}}=\frac{{\omega }_{\text{B}}{\left(bh{p}_{\text{B}}\right)}^{1/2}}{{\left({\rho }_{\text{B}}\right)}^{1/2}{\left(g{H}_{\text{B}}\right)}^{5/4}}$  ....... (IX)

Here, the specific speed of turbine B is $N_{\text{B}}$.

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 ${\rho }_{\text{A}}$, $15.7079\text{rad}/{\text{s}}^2$for ${\omega }_{\text{A}}$, $9.81\text{m}/{\text{s}}^2$for $g$and $340.6240\text{MW}$for $bp{h}_{\text{B}}$in Equation (VIII).

  $\begin{array}{c}{N}_{\text{A}}=\frac{\left(15.7079\text{rad}/{\text{s}}^2\right){\left(132\text{MW}\right)}^{1/2}}{{\left(998\text{kg}/{\text{m}}^3\right)}^{1/2}{\left(9.81\text{m}/{\text{s}}^2\times 90\text{m}\right)}^{5/4}}\\ =\frac{\left(15.7079\text{rad}/{\text{s}}^2\right){\left(132\text{MW}\left(\frac{1000000\text{W}}{1\text{MW}}\right)\right)}^{1/2}}{{\left(998\text{kg}/{\text{m}}^3\right)}^{1/2}{\left(9.81\text{m}/{\text{s}}^2\times 90\text{m}\right)}^{5/4}}\\ =\frac{180470.0312{\text{W}}^{1/2}}{152038.8817{\text{kg}}^{1/2}\cdot {\text{m}}^{7/1}/{\text{s}}^{9/2}\left(\frac{1{\text{W}}^{1/2}}{1{\text{kg}}^{1/2}\cdot {\text{m}}^{7/1}/{\text{s}}^{9/2}}\right)}\\ =1.18699\end{array}$

Substitute $110\text{m}$for $H_{\text{B}}$, $998\text{kg}/{\text{m}}^3$for ${\rho }_{\text{B}}$, $12.5663\text{rad}/{\text{s}}^2$for ${\omega }_{\text{B}}$, $9.81\text{m}/{\text{s}}^2$for $g$and $340.6240\text{MW}$for $bp{h}_{\text{B}}$ in Equation (X).

  $\begin{array}{c}{N}_{\text{B}}=\frac{\left(12.5663\text{rad}/{\text{s}}^2\right){\left(340.6240\text{MW}\right)}^{1/2}}{{\left(998\text{kg}/{\text{m}}^3\right)}^{1/2}{\left(9.81\text{m}/{\text{s}}^2\times 110\text{m}\right)}^{5/4}}\\ =\frac{\left(12.5663\text{rad}/{\text{s}}^2\right){\left(340.6240\text{MW}\left(\frac{1000000\text{W}}{1\text{MW}}\right)\right)}^{1/2}}{{\left(998\text{kg}/{\text{m}}^3\right)}^{1/2}{\left(9.81\text{m}/{\text{s}}^2\times 110\text{m}\right)}^{5/4}}\\ =\frac{231923.6546{\text{W}}^{1/2}}{195385.5251{\text{kg}}^{1/2}\cdot {\text{m}}^{7/1}/{\text{s}}^{9/2}\left(\frac{1{\text{W}}^{1/2}}{1{\text{kg}}^{1/2}\cdot {\text{m}}^{7/1}/{\text{s}}^{9/2}}\right)}\\ =1.18700\end{array}$

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.