Chapter 14, Problem 116P

Prove that the model turbine (Prob. 14—114) and the prototype turbine (Prob. 14—115) operate at homologous points by comparing turbine efficiency and turbine specific speed for both cases.

To determine

That the model turbine and the prototype turbine operate at homogenous points by comparing turbine efficiency and turbine specific speed for both cases.

Answer to Problem 116P

The efficiency of the prototype is same as that of the model turbine.

The specific speed of the prototype is same as that of the model turbine.

Therefore, the model turbine and the prototype turbine operate at homogenous points.

There are many reasons for this increase in the efficiency. The relative roughness of the surfaces of the prototype runner blades may be significantly smaller than that on the model turbine. The lager turbines have smaller tip clearances relative to blade diameter, therefore tip losses are less significant. Also, leakage is less significant in large turbines.

Explanation of Solution

Given information:

The temperature of the water in the turbine is $20°\text{C}$, the diameter of the model is $8\text{cm}$, the volume flow rate of the turbine is $25.5{\text{m}}^{\text{3}}/\text{h}$, the speed of the model is $1500\text{rpm}$, the net head of the model is $15\text{m}$, the net head of the prototype is $50\text{m,}$ and the power delivered by the model is $720\text{W}$.

The scale is assumed to be $1/5$, the density of water is $998\text{kg}/{\text{m}}^{\text{3}},$ and the acceleration due to gravity is $9.81\text{m}/{\text{s}}^{\text{2}}$.

Write the expression for the angular velocity of the model.

   ${\omega }_A=\frac{2\pi {\dot{n}}_A}{60}$    ...... (I)

Here, the speed of the model is ${\dot{n}}_A$.

Write the expression for the efficiency of the model turbine.

   ${\eta }_{turbine,A}=\frac{bh{p}_A}{\rho g{H}_A{\dot{V}}_A}$    ...... (II)

Here, the brake horse power of the model turbine is $bh{p}_A$, the density of water is $\rho$, the acceleration due to gravity is $g$, the net head of the model is $H_A,$ and the volume flow rate of the model is ${\dot{V}}_A$.

Write the expression for the turbine specific speed.

   $N_{st,A}=\frac{{\omega }_A{\left(bh{p}_A\right)}^{1/2}}{{\left(\rho \right)}^{{}^{1/2}}{\left(g{H}_A\right)}^{5/4}}$    ...... (III)

Here, the angular velocity of the model is ${\omega }_A$.

Write the expression for the angular velocity of the prototype using the turbine scaling law.

   ${\omega }_B={\omega }_A\sqrt{\frac{H_B}{H_A}}\left(\frac{D_A}{D_B}\right)$    ...... (IV)

Here, the diameter of the model is $D_A$, the net head of the model is $H_A$, the net head of the prototype is $H_B$, the angular speed of the model is ${\omega }_A,$ and the angular speed of the prototype is ${\omega }_B,$ and the diameter of the prototype is $D_B$.

Write the expression for the volume flow rate of the prototype using the turbine scaling law.

   ${\dot{V}}_B={\dot{V}}_A\left(\frac{{\omega }_B}{{\omega }_A}\right){\left(\frac{D_B}{D_A}\right)}^3$    ...... (V)

Here, the volume flow rate of the model is ${\dot{V}}_A$.

Write the expression for the brake horse power of the prototype using the turbine scaling law.

   $bh{p}_B=bh{p}_A{\left(\frac{{\omega }_B}{{\omega }_A}\right)}^3{\left(\frac{D_B}{D_A}\right)}^5$    ...... (VI)

Here, the brake horse power of the model is $bh{p}_A$.

Write the expression for the efficiency of the prototype turbine.

   ${\eta }_{turbine,B}=\frac{bh{p}_B}{\rho g{H}_B{\dot{V}}_B}$    ...... (VII)

Here, the brake horse power of the prototype turbine is $bh{p}_B$, the net head of the prototype is $H_B,$ and the volume flow rate of the prototype is ${\dot{V}}_B$.

Write the expression for the turbine specific speed.

   $N_{st,B}=\frac{{\omega }_B{\left(bh{p}_B\right)}^{1/2}}{{\left(\rho \right)}^{{}^{1/2}}{\left(g{H}_B\right)}^{5/4}}$    ...... (VIII)

Here, the angular velocity of the prototype is ${\omega }_B$.

Calculation:

Substitute $1500\text{rpm}$ for ${\dot{n}}_A$ in the Equation (I).

   $\begin{array}{c}{\omega }_A=\frac{2\pi \left(1500\text{rpm}\right)}{60}\\ =\frac{9424.77}{60}\\ =157.08\text{rad}/\text{s}\end{array}$

Substitute $720\text{W}$ for $bh{p}_A$, $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $15\text{m}$ for $H_A,$ and $25.5{\text{m}}^{\text{3}}/\text{h}$ for ${\dot{V}}_A$ in the Equation (II).

   $\begin{array}{c}{\eta }_{turbine,A}=\frac{720\text{W}}{\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(15\text{m}\right)\left(25.5{\text{m}}^{\text{3}}/\text{h}\right)}\\ =\frac{720\text{W}}{\left(146855.7\text{kg}/\text{m}\cdot {\text{s}}^2\right)\times \left(25.5{\text{m}}^{\text{3}}/\text{h}\left(\frac{1\text{h}}{3600\text{s}}\right)\right)}\\ =\frac{720\text{W}}{\left(146855.7\text{kg}/\text{m}\cdot {\text{s}}^2\right)\times \left(7.08334\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\right)}\\ =\frac{720\text{W}\left(\frac{1\text{kg}\cdot {\text{m}}^2/{\text{s}}^3}{1\text{W}}\right)}{\left(1040.22854\text{kg}\cdot {\text{m}}^2/{\text{s}}^3\right)}\end{array}$

   $\begin{array}{c}{\eta }_{turbine,A}=0.692\times 100\%\\ =69.2\%\end{array}$

Substitute $720\text{W}$ for $bh{p}_A$, $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $15\text{m}$ for $H_A,$ and $157.08\text{rad}/\text{s}$ for ${\omega }_A$ in the Equation (III).

   $\begin{array}{c}{N}_{st,A}=\frac{\left(157.08\right){\left(720\text{W}\right)}^{1/2}}{{\left(998\text{kg}/{\text{m}}^{\text{3}}\right)}^{{}^{1/2}}{\left(9.81\text{m}/{\text{s}}^{\text{2}}\times 15\text{m}\right)}^{5/4}}\\ =\frac{4214.8987{\left(\text{W}\right)}^{1/2}}{16189.875{\left(\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}\right)}^{1/2}}\\ =\frac{4214.8987{\left(\text{W}\left(\frac{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}}{1\text{W}}\right)\right)}^{1/2}}{16189.875{\left(\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}\right)}^{1/2}}\\ =0.2603\end{array}$

Substitute $50\text{m}$ for $H_B$, $15\text{m}$ for $H_A$, $1500\text{rpm}$ for ${\omega }_A,$ and $1/5$ for $\left(\frac{D_A}{D_B}\right)$ in the Equation (IV).

   $\begin{array}{c}{\omega }_B=1500\text{rpm}\sqrt{\frac{50\text{m}}{15\text{m}}}\left(\frac{1}{5}\right)\\ =1500\text{rpm}\times 1.82574\left(\frac{1}{5}\right)\\ =547.72\text{rpm}\end{array}$

Substitute $25.5{\text{m}}^{\text{3}}/\text{h}$ for ${\dot{V}}_A$, $1500\text{rpm}$ for ${\omega }_A$, $547.72\text{rpm}$ for ${\omega }_B$ and $5/1$ for $\left(\frac{D_B}{D_A}\right)$ in the Equation (V).

   $\begin{array}{c}{\dot{V}}_B=\left(25.5{\text{m}}^{\text{3}}/\text{h}\right)\left(\frac{547.722\text{rpm}}{1500\text{rpm}}\right){\left(\frac{5}{1}\right)}^3\\ =\left(25.5{\text{m}}^{\text{3}}/\text{h}\right)\left(0.365148\right)\left(125\right)\\ =1163.90925{\text{m}}^{\text{3}}/\text{h}\end{array}$

Substitute $720\text{W}$ for $bh{p}_A$

   $1500\text{rpm}$ for ${\omega }_A$, $547.72\text{rpm}$ for ${\omega }_B,$ and $5/1$ for $\left(\frac{D_B}{D_A}\right)$ in the Equation (VI).

   $\begin{array}{c}bh{p}_B=720\text{W}{\left(\frac{547.72\text{rpm}}{1500\text{rpm}}\right)}^3{\left(\frac{5}{1}\right)}^5\\ =720\text{W}\left(0.0486857\right)\left(3125\right)\\ =109542.825\text{W}\left(\frac{1\text{kW}}{1000\text{W}}\right)\\ =109.5428\text{kW}\end{array}$

Substitute $109542.825\text{W}$ for $bh{p}_B$, $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $50\text{m}$ for $H_B,$ and $1163.90925{\text{m}}^{\text{3}}/\text{h}$ for ${\dot{V}}_B$ in the Equation (VII).

   $\begin{array}{c}{\eta }_{turbine,B}=\frac{109542.825\text{W}}{\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(50\text{m}\right)\left(1163.90925{\text{m}}^{\text{3}}/\text{h}\right)}\\ =\frac{109542.825\text{W}}{\left(489519\text{kg}/\text{m}\cdot {\text{s}}^2\right)\times \left(25.5{\text{m}}^{\text{3}}/\text{h}\left(\frac{1\text{h}}{3600\text{s}}\right)\right)}\\ =\frac{109542.825\text{W}}{\left(489519\text{kg}/\text{m}\cdot {\text{s}}^2\right)\times \left(7.08334\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}\right)}\\ =\frac{109542.825\text{W}\left(\frac{1\text{kg}\cdot {\text{m}}^2/{\text{s}}^3}{1\text{W}}\right)}{\left(3467.429\text{kg}\cdot {\text{m}}^2/{\text{s}}^3\right)}\end{array}$

   $\begin{array}{c}{\eta }_{turbine,B}=0.692\times 100\%\\ =69.2\%\end{array}$

Substitute $109542.825\text{W}$ for $bh{p}_B$, $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $50\text{m}$ for $H_B,$ and $547.72\text{rpm}$ for ${\omega }_B$ in the Equation (VIII).

   $\begin{array}{c}{N}_{st,B}=\frac{\left(547.72\text{rpm}\right){\left(109542.825\text{W}\right)}^{1/2}}{{\left(998\text{kg}/{\text{m}}^{\text{3}}\right)}^{{}^{1/2}}{\left(9.81\text{m}/{\text{s}}^{\text{2}}\times 50\text{m}\right)}^{5/4}}\\ =\frac{\left(547.72\text{rpm}\left(\frac{2\pi }{60}\right)\right){\left(109542.825\text{W}\right)}^{1/2}}{{\left(998\text{kg}/{\text{m}}^{\text{3}}\right)}^{{}^{1/2}}{\left(9.81\text{m}/{\text{s}}^{\text{2}}\times 50\text{m}\right)}^{5/4}}\\ =\frac{18983.626{\left(\text{W}\right)}^{1/2}}{489519{\left(\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}\right)}^{1/2}}\\ =\frac{18983.626{\left(\text{W}\left(\frac{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}}{1\text{W}}\right)\right)}^{1/2}}{489519{\left(\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}\right)}^{1/2}}\end{array}$

   $N_{st,B}=0.2603$

Conclusion:

The efficiency of the prototype is same as that of the model turbine.

The specific speed of the prototype is same as that of the model turbine.

Therefore, the model turbine and the prototype turbine operate at homogenous points.

There are many reasons for this increase in the efficiency. The relative roughness of the surfaces of the prototype runner blades may be significantly smaller than that on the model turbine. The lager turbines have smaller tip clearances relative to blade diameter, therefore, tip losses are less significant. Also, leakage is less significant in large turbines.