Chapter 3, Problem 86RQ

To determine

The potential revenue generated by the pump-turbine system.

Explanation of Solution

Given:

Volumetric flow rate of the water $\left(\dot{V}\right)$ is $\text{2}{\text{m}}^3/\text{s}$.

Elevation difference of the water $\left(\Delta z\right)$ is $\text{40}\text{m}$.

Efficiency of the turbine-generator $\left({\eta }_{\text{turbine-gen}}\right)$ is $\text{0}\text{.75}$.

Efficiency of the pump-motor $\left({\eta }_{\text{pump-motor}}\right)$ is $\text{0}\text{.75}$.

Operating hours $\left(\Delta t\right)$ is $10\text{hr}/\text{year}$.

Unit cost of electric power produced at night is $\$0.05/\text{kWh}$.

Unit cost of electric power produced at day is $\$0.12/\text{kWh}$.

Density of the water $\left(\rho \right)$ is $\text{1000}\text{kg}/{\text{m}}^{\text{3}}$.

Acceleration due to gravity $\left(g\right)$ is $\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}$.

Calculation:

Calculate the minimum power required to pump water from the lower to higher reservoir $\left({\dot{W}}_{\min ,\text{pump}}\right)$.

  $\begin{array}{c}{\dot{W}}_{\min ,\text{pump}}={\dot{W}}_{\text{max,turbine}}\\ =\Delta {\dot{E}}_{\text{mech}}\\ =\dot{m}\Delta e_{\text{mech}}\\ =\dot{m}\Delta pe\\ =\dot{m}g\Delta z\end{array}$

  $\begin{array}{c}=\rho \dot{V}g\Delta z\\ =\left(\text{1000}\text{kg}/{\text{m}}^{\text{3}}\right)\left(2{\text{m}}^{\text{3}}/\text{s}\right)\left(\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}\right)\left(40\text{m}\right)\\ =\left(\text{1000}\text{kg}/{\text{m}}^{\text{3}}\right)\left(2{\text{m}}^{\text{3}}/\text{s}\right)\left(\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}\right)\left(40\text{m}\right)\left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\left(\frac{1\text{kW}}{1000\text{N}\cdot \text{m/s}}\right)\\ =784.8\text{kW}\end{array}$

Calculate the actual pump electric power $\left({\dot{W}}_{\text{pump,elect}}\right)$.

  $\begin{array}{c}{\dot{W}}_{\text{pump,elect}}=\frac{{\dot{W}}_{\text{ideal}}}{{\eta }_{\text{pump-motor}}}\\ =\frac{784.8\text{kW}}{0.75}\\ =1046\text{kW}\end{array}$

Calculate the actual turbine electric power $\left({\dot{W}}_{\text{turbine}}\right)$.

  $\begin{array}{c}{\dot{W}}_{\text{turbine}}={\eta }_{\text{turbine-gen}}{\dot{W}}_{\text{ideal}}\\ =\left(0.75\right)\left(784.8\text{kW}\right)\\ =588.6\text{kW}\end{array}$

Calculate the power consumption cost of the pump.

  $\begin{array}{c}\text{Cost}={\dot{W}}_{\text{pump,elect}}\Delta t\times \text{Unit}\text{price}\\ =\left(1046\text{kW}\right)\left(365\times 10\text{hr}/\text{year}\right)\left(\$0.05/\text{kWh}\right)\\ =\$190,968/\text{year}\end{array}$

Calculate the revenue generated by the turbine.

  $\begin{array}{c}\text{Revenue}={\dot{W}}_{\text{turbine}}\Delta t\times \text{Unit}\text{price}\\ =\left(588.6\text{kW}\right)\left(365\times 10\text{hr}/\text{year}\right)\left(\$0.05/\text{kWh}\right)\\ =\$257,807/\text{year}\end{array}$

Calculate the net income per year.

  $\begin{array}{c}\text{Net}\text{income}=\text{Revenue}-\text{Cost}\\ =\$257,807/\text{year}-\$190,968/\text{year}\\ =\$66,839/\text{year}\end{array}$

Thus, the potential revenue generated by the pump-turbine system is $\underset{\_}{\$66,839/\text{year}}$.