Chapter 14, Problem 86EP

To determine

The swirl angles ${\alpha }_2$ and ${\alpha }_1$.

Whether the turbine have forward or swirl flow.

The power output in $\text{hp}$.

The net head in $\text{ft}$.

Answer to Problem 86EP

The swirl angles ${\alpha }_2$ is $4.845°$ and ${\alpha }_1$ is $35.48°$.

The turbine will have a forward swirl.

The power output is $2.01\times {10}^5\text{hp}$.

The net head is $170.47\text{ft}$.

Explanation of Solution

Given information:

The gross head is $90\text{m}$, the volume flow rate of water is $4.7\times {10}^6\text{gpm}$, the speed of the runner is $120\text{rpm}$, the runner blade angle ${\beta }_2$ is $82°$, the runner blade angle ${\beta }_1$ is $46°$.

Write the expression for the angular velocity of turbine.

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

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

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

   $V_{2,n}=\frac{\dot{\vee }}{2\pi r_2{b}_2}$ .   ...... (II)

Here, the volume flow rate of water is $\dot{\vee }$, the radius at inlet is $r_2,$ and the breadth at the inlet is $b_2$.

Write the expression for the tangential velocity component at inlet.

   $V_{2,t}=\omega r_2-\frac{V_{2,n}}{\tan {\beta }_2}$    ...... (III)

Here, the radius at the inlet is $r_2$ and the runner blade angle at inlet is ${\beta }_2$.

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

   ${\alpha }_2={\tan }^{-1}\left(\frac{V_{2,t}}{V_{2,n}}\right)$    ...... (IV)

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

   $V_{1,n}=\frac{\dot{\vee }}{2\pi r_1{b}_1}$ .   ...... (V)

Write the expression for the tangential velocity component at outlet.

   $V_{1,t}=\omega r_1-\frac{V_{1,n}}{\tan {\beta }_1}$    ...... (VI)

Here, the radius at the outlet is $r_2$ and the runner blade angle at outlet is ${\beta }_1$.

Here, the volume flow rate of water is $\dot{\vee }$, the radius at outlet is $r_1,$ and the breadth at the outlet is $b_1$.

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

   ${\alpha }_1={\tan }^{-1}\left(\frac{V_{1,t}}{V_{1,n}}\right)$    ...... (VII)

Write the expression for the shaft output power.

   ${\dot{W}}_{shaft}=\rho \omega \dot{\vee }\left(r_2{V}_{2,t}-r_1{V}_{1,t}\right)$    ...... (VIII)

Here, the density of water is $\rho$.

Write the expression for the net head.

   $H=\frac{{\dot{W}}_{shaft}}{\rho g\dot{\vee }}$    ...... (IX)

Calculation:

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

   $\begin{array}{c}\omega =\frac{2\pi \left(120\text{rpm}\right)}{60}\\ =12.56\text{rpm}\left(\frac{\text{1}\text{rad}/\text{s}}{1\text{rpm}}\right)\\ =12.56\text{rad}/\text{s}\end{array}$

Substitute $6.60\text{ft}$ for $r_2$, $2.60\text{ft}$ for $b_2$ and $4.7\times {10}^6\text{gpm}$ for $\dot{\vee }$ in Equation (II).

   $\begin{array}{c}{V}_{2,n}=\frac{4.7\times {10}^6\text{gpm}}{2\pi \left(6.60\text{ft}\right)\left(2.60\text{ft}\right)}\\ =\frac{4.7\times {10}^6\text{gpm}\left(\frac{1{\text{ft}}^{\text{3}}/\text{s}}{448.8\text{gpm}}\right)}{2\pi \left(6.60\text{ft}\right)\left(2.60\text{ft}\right)}\\ =\frac{10472.37{\text{ft}}^{\text{3}}/\text{s}}{2\pi \left(6.60\text{ft}\right)\left(2.60\text{ft}\right)}\\ =97.12\text{ft}/\text{s}\end{array}$

Substitute $6.60\text{ft}$ for $r_2$, $12.56\text{rad}/\text{s}$ for $\omega$, $82°$ for ${\beta }_2,$ and $97.12\text{ft}/\text{s}$ for $V_{2,n}$ in Equation (III).

   $\begin{array}{c}{V}_{2,t}=12.56\text{rad}/\text{s}\left(6.60\text{ft}\right)-\frac{97.12\text{ft}/\text{s}}{\tan 82°}\\ =82.89\text{ft}/\text{s}-13.64\text{ft}/\text{s}\\ =69.25\text{ft}/\text{s}\end{array}$

Substitute $69.25\text{ft}/\text{s}$ for $V_{2,t}$ and $97.12\text{ft}/\text{s}$ for $V_{2,n}$ in Equation (IV).

   $\begin{array}{c}{\alpha }_2={\tan }^{-1}\left(\frac{69.25\text{ft}/\text{s}}{97.12\text{ft}/\text{s}}\right)\\ ={\tan }^{-1}\left(0.71\right)\\ =35.48°\end{array}$

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

Substitute $4.40\text{ft}$ for $r_1$, $7.20\text{ft}$ for $b_1$ and $4.7\times {10}^6\text{gpm}$ for $\dot{\vee }$ in Equation (V).

   $\begin{array}{c}{V}_{1,n}=\frac{4.7\times {10}^6\text{gpm}}{2\pi \left(4.40\text{ft}\right)\left(7.20\text{ft}\right)}\\ =\frac{4.7\times {10}^6\text{gpm}\left(\frac{1{\text{ft}}^{\text{3}}/\text{s}}{448.8\text{gpm}}\right)}{2\pi \left(4.40\text{ft}\right)\left(7.20\text{ft}\right)}\\ =\frac{10472.37{\text{ft}}^{\text{3}}/\text{s}}{2\pi \left(4.40\text{ft}\right)\left(7.20\text{ft}\right)}\\ =52.609\text{ft}/\text{s}\end{array}$

Substitute $4.40\text{ft}$ for $r_1$, $12.56\text{rad}/\text{s}$ for $\omega$, $46°$ for ${\beta }_1,$ and $52.609\text{ft}/\text{s}$ for $V_{1,n}$ in Equation (VI).

   $\begin{array}{c}{V}_{1,t}=12.56\text{rad}/\text{s}\left(4.40\text{ft}\right)-\frac{52.609\text{ft}/\text{s}}{\tan 46°}\\ =55.264\text{ft}/\text{s}-50.803\text{ft}/\text{s}\\ =4.46\text{ft}/\text{s}\end{array}$

Substitute $4.46\text{ft}/\text{s}$ for $V_{1,t}$ and $52.609\text{ft}/\text{s}$ for $V_{1,n}$ in Equation (VII).

   $\begin{array}{c}{\alpha }_1={\tan }^{-1}\left(\frac{4.46\text{ft}/\text{s}}{52.609\text{ft}/\text{s}}\right)\\ ={\tan }^{-1}\left(\mathrm{0.0.8}\right)\\ =4.845°\end{array}$

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 $\left(\rho \right)$ as $998\text{kg}/{\text{m}}^{\text{3}}$ at $20°\text{C}$.

Substitute $62.30\text{lbm}/{\text{ft}}^{\text{3}}$ for $\rho$, $12.56\text{rad}/\text{s}$ for $\omega$, $4.7\times {10}^6\text{gpm}$ for $\dot{\vee }$, $6.60\text{ft}$ for $r_2$, $69.246\text{ft}/\text{s}$ for $V_{2,t}$, $4.40\text{ft}$ for $r_1,$ and $4.46\text{ft}/\text{s}$ for $V_{1,t}$ in Equation (VIII).

   $\begin{array}{c}{\dot{W}}_{shaft}=\left(\begin{array}{l}\left(62.30\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(12.56\text{rad}/\text{s}\right)\left(4.7\times {10}^6\text{gpm}\right)\\ \times \left(\left(6.60\text{ft}\right)\left(69.246\text{ft}/\text{s}\right)-\left(4.40\text{ft}\right)\left(4.46\text{ft}/\text{s}\right)\right)\end{array}\right)\\ =\left(\begin{array}{l}\left(62.30\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(12.56\text{rad}/\text{s}\right)\left(4.7\times {10}^6\text{gpm}\left(\frac{1{\text{ft}}^{\text{3}}/\text{s}}{448.8\text{gpm}}\right)\right)\\ \times \left(\left(6.60\text{ft}\right)\left(69.246\text{ft}/\text{s}\right)-\left(4.40\text{ft}\right)\left(4.46\text{ft}/\text{s}\right)\right)\end{array}\right)\\ =\left(\begin{array}{l}\left(62.30\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(12.56\text{rad}/\text{s}\right)\left(10472.37{\text{ft}}^{\text{3}}/\text{s}\right)\\ \times \left(\left(6.60\text{ft}\right)\left(69.246\text{ft}/\text{s}\right)-\left(4.40\text{ft}\right)\left(4.46\text{ft}/\text{s}\right)\right)\end{array}\right)\\ =3.584128\times {10}^9\text{lbm}\cdot {\text{ft}}^2/{\text{s}}^{\text{3}}\end{array}$

   $\begin{array}{c}{\dot{W}}_{shaft}=3.584128\times {10}^9\text{lbm}\cdot {\text{ft}}^2/{\text{s}}^{\text{3}}\left(\frac{1\text{Btu}/\text{s}}{25051.6\text{lbm}\cdot {\text{ft}}^2/{\text{s}}^{\text{3}}}\right)\\ =1.4306\times {10}^5\text{Btu}/\text{s}\left(\frac{1.41\text{hp}}{1\text{Btu}/\text{s}}\right)\\ =2.01\times {10}^5\text{hp}\end{array}$

Therefore, the power output is $2.01\times {10}^5\text{hp}$.

Substitute $3.584128\times {10}^9\text{lbm}\cdot {\text{ft}}^2/{\text{s}}^{\text{3}}$ for ${\dot{W}}_{shaft}$, $62.30\text{lbm}/{\text{ft}}^{\text{3}}$ for $\rho$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$ and $4.7\times {10}^6\text{gpm}$ for $\dot{\vee }$ in Equation (IX).

   $\begin{array}{c}H=\frac{3.584128\times {10}^9\text{lbm}\cdot {\text{ft}}^2/{\text{s}}^{\text{3}}}{\left(62.30\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(4.7\times {10}^6\text{gpm}\right)}\\ =\frac{3.584128\times {10}^9\text{lbm}\cdot {\text{ft}}^2/{\text{s}}^{\text{3}}}{\left(62.30\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(4.7\times {10}^6\text{gpm}\left(\frac{1{\text{ft}}^{\text{3}}/\text{s}}{448.8\text{gpm}}\right)\right)}\\ =\frac{3.584128\times {10}^9\text{lbm}\cdot {\text{ft}}^2/{\text{s}}^{\text{3}}}{\left(62.30\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(10472.37{\text{ft}}^{\text{3}}/\text{s}\right)}\\ =170.47\text{ft}\end{array}$

Therefore, the net head is $170.47\text{ft}$.

Conclusion:

Therefore, the swirl angles ${\alpha }_2$ is $4.845°$ and ${\alpha }_1$ is $35.48°$ .

The turbine will have a forward swirl.

The power output is $2.01\times {10}^5\text{hp}$ and the net head is $170.47\text{ft}$.