Chapter 5, Problem 84P

Water enters a hydraulic turbine through a 30-cm-diameter pipe at a rate of 0.6 m³/s and exits
through a 25-cm-diameter pipe. The pressure drop in the turbine is measured by a mercury manometer to be 1.2 in. For a combined turbine-generator efficiency of 83 percent determine the net electric power output. Disregard the effect of the kinetic energy correction factors.

To determine

The net electric power output.

Answer to Problem 84P

The net electric power output is $60.334\text{kW}$.

Explanation of Solution

Given information:

The initial diameter of the pipe is $30\text{cm}$, the volume flow rate is $0.6{\text{m}}^{\text{3}}/\text{s}$, the final diameter of the pipe is $25\text{cm}$, the pressure drop in a mercury manometer is $1.2\text{m,}$ and the combined efficiency is $83\%$.

Write the expression for the initial velocity of the water.

  $V_1=\frac{\dot{V}}{\left(\frac{\pi D_1^2}{4}\right)}$  .......(I)

Here, the initial diameter of the pipe is $D_1$ and the volume flow rate is $\dot{V}$.

Write the expression for the final velocity of the water.

  $V_2=\frac{\dot{V}}{\left(\frac{\pi D_2^2}{4}\right)}$  .......(II)

Here, the final diameter of the pipe is $D_2$.

Write the expression for the pressure drop.

  $\Delta p={\rho }_{Hg}\times g\times h_{Hg}$  .......(III)

Here, the density of the mercury and water is ${\rho }_{Hg}$, the drop in the pressure head is $h_{Hg},$ and the gravitational constant is $g$.

Write the expression for the turbine head using the Bernoulli's equation.

  $\begin{array}{l}\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1^2}{2g}+z_1+h_{pump}=\frac{P_2}{\rho g}+{\alpha }_2\frac{V_2^2}{2g}+z_2+h_{tur}+h_L\\ \frac{P_1}{\rho g}+{\alpha }_1\frac{V_1^2}{2g}+0+0=\frac{P_2}{\rho g}+{\alpha }_2\frac{V_2^2}{2g}+0+h_{tur}+0\\ h_{tur}=\frac{P_1}{\rho g}-\frac{P_2}{\rho g}+\alpha \left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)\\ h_{tur}=\frac{\Delta P}{\rho g}+\alpha \left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)\end{array}$  .......(IV)

Here, the head loss is $h_L$, the kinetic energy correction factor at initial level is ${\alpha }_1$, the kinetic energy correction factor at initial discharge is ${\alpha }_2$, the pressure at initial level of water is $P_1$, the atmospheric pressure at initial discharge is $P_2$, the elevation from ground to water level is $z_1$, the elevation of water stream from nozzle to maximum height is $z_2$, the density of the water is $\rho ,$ and the gravitational constant is $g$.

Write the expression for the net electric power output.

  ${\dot{W}}_{tur}={\eta }_{tur-gen}\rho \dot{V}g{h}_{tur}$  .......(V)

Here, the turbine and generator efficiency is ${\eta }_{tur-gen}$.

Calculation:

Draw the diagram for a tank fitted with orifice.

  

  Figure-(2)

Substitute $30\text{cm}$ for $D_1$ and $0.6{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$ in Equation (I).

  $\begin{array}{c}{V}_1=\frac{\left(0.6{\text{m}}^{\text{3}}/\text{s}\right)}{\left(\frac{\pi {\left(30\text{cm}\right)}^2}{4}\right)}\\ =\frac{\left(0.6{\text{m}}^{\text{3}}/\text{s}\right)}{\left(\frac{\pi \left(900{\text{cm}}^2\right)}{4}\left(\frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)\right)}\\ =8.48\text{m}/\text{s}\end{array}$

Substitute $25\text{cm}$ for $D_2$ and $0.6{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$ in Equation (II).

  $\begin{array}{c}{V}_2=\frac{\left(0.6{\text{m}}^{\text{3}}/\text{s}\right)}{\left(\frac{\pi {\left(25\text{cm}\right)}^2}{4}\right)}\\ =\frac{\left(0.6{\text{m}}^{\text{3}}/\text{s}\right)}{\left(\frac{\pi \left(625{\text{cm}}^2\right)}{4}\left(\frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)\right)}\\ =12.22\text{m}/\text{s}\end{array}$

Substitute $13600\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{Hg}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $1.2\text{m}$ for $h_{Hg}$ in Equation (III).

  $\begin{array}{c}\Delta P=\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(1.2\text{m}\right)\\ =160\times {10}^3\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\left(\frac{1\text{Pa}}{\text{1}\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}\right)\\ =160\times {10}^3\text{Pa}\end{array}$

Here, the gravitational constant is $9.81{\text{m}/\sec }^2$.

Here, the density of the water is $1000\text{kg}/{\text{m}}^{\text{3}}$.

Substitute $160\times {10}^3\text{Pa}$ for $\Delta P$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $8.48\text{m}/\text{s}$ for $V_1$, $12.22\text{m}/\text{s}$ for $V_2$, $0$ for $\alpha$ and $9.81{\text{m}/\sec }^2$ for $g$ in Equation (IV).

  $\begin{array}{c}{h}_{tur}=\left[\begin{array}{l}\frac{\left(160\times {10}^3\text{Pa}\right)}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81{\text{m}/\sec }^2\right)}+\\ 1.0\left(\frac{{\left(8.48\text{m}/\text{s}\right)}^2}{2\left(9.81{\text{m}/\sec }^2\right)}-\frac{{\left(12.22\text{m}/\text{s}\right)}^2}{2\left(9.81{\text{m}/\sec }^2\right)}\right)\end{array}\right]\\ =16.30\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\left(\frac{1\text{Pa}}{\text{1}\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}\right)+1.0\left(3.66\text{m}-7.61\text{m}\right)\\ =16.30+1.0\left(-3.95\text{m}\right)\\ =12.35\text{m}\end{array}$

Substitute $12.35\text{m}$ for $h_{tur}$, $0.83$ for ${\eta }_{tur-gen}$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81{\text{m}/\sec }^2$ for $g$ and $0.6{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$ in Equation (V).

  $\begin{array}{c}{\dot{W}}_{tur}=0.83\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.6{\text{m}}^{\text{3}}/\text{s}\right)\left(9.81{\text{m}/\sec }^2\right)\left(12.35\text{m}\right)\\ =60.334\times {10}^3\text{kg}\cdot {\text{m}}^2/{\sec }^{\text{3}}\left(\frac{\text{kW}}{{\text{10}}^3\text{kg}\cdot {\text{m}}^2/{\sec }^{\text{3}}}\right)\\ =60.334\text{kW}\end{array}$

Conclusion:

Thus, the net electric power output is $60.334\text{kW}$.