Chapter 2.8, Problem 122RP

The demand for electric power is usually much higher during the day than it is at night, and utility companies often sell power at night at much lower prices to encourage consumers to use the available power generation capacity and to avoid building new, expensive power plants that will be used only a short time during peak periods. Utilities are also willing to purchase power produced during the day from private parties at a high price.

Suppose a utility company is selling electric power for $0.05/kWh at night and is willing to pay $0.12/kWh for power produced during the day. To take advantage of this opportunity, an entrepreneur is considering building a large reservoir 40 m above the lake level, pumping water from the lake to the reservoir at night using cheap power, and letting the water flow from the reservoir back to the lake during the day, producing power as the pump–motor operates as a turbine–generator during reverse flow. Preliminary analysis shows that a water flow rate of 2 m³/s can be used in either direction. The combined pump–motor and turbine–generator efficiencies are expected to be 75 percent each. Disregarding the frictional losses in piping and assuming the system operates for 10 h each in the pump and turbine modes during a typical day, determine the potential revenue this pump–turbine system can generate per year.

FIGURE P2–122

To determine

The potential revenue generated per year for the system.

Answer to Problem 122RP

The potential revenue generated per year for the system is $\underset{\_}{\text{\$66,839/year}}$.

Explanation of Solution

Calculate the minimum power required to pump water from the lower reservoir to the higher reservoir.

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

   $\begin{array}{l}=\dot{m}\Delta pe\\ =\dot{m}g\Delta z\\ =\rho \dot{V}g\Delta z\end{array}$ (I)

Here, the maximum power of turbine is ${\dot{W}}_{\text{max,turbine}}$, the minimum power of pump is ${\dot{W}}_{\min ,\text{pump}}$, ideal power is ${\dot{W}}_{\text{ideal}}$, the rate of change in the mechanical power in the rate form is $\Delta {\dot{E}}_{\text{mech}}$, mass flow rate of water is $\dot{m}$, the mechanical energy change of a fluid is $\Delta e_{\text{mech}}$, acceleration due to gravity is g, change in the potential energy per unit mass is $\Delta pe$, change in the height of a reservoir from the base of the lake is $\Delta z$, density of water is $\rho$, and volume flow rate of water is $\dot{V}$.

Calculate the actual pump electric power.

${\dot{W}}_{\text{pump,elect}}=\frac{{\dot{W}}_{\text{ideal}}}{{\eta }_{\text{pump-motor}}}$ (II)

Here, the efficiency of combined pump-motor is ${\eta }_{\text{pump-motor}}$.

Calculate the turbine electric power.

${\dot{W}}_{\text{turbine}}={\eta }_{\text{turbine-gen}}{\dot{W}}_{\text{ideal}}$ (III)

Here, the efficiency of combined turbine-geneator is ${\eta }_{\text{turbine-gen}}$.

Calculate the power consumption cost of the pump.

$\text{Cost}={\dot{W}}_{\text{pump,elect}}\Delta t\times \text{Unit}\text{price}$ (IV)

Here, change in time of each in the pump and turbine modes during a typical day is $\Delta t$.

Calculate the revenue generated by the turbine.

$\text{Revenue}={\dot{W}}_{\text{turbine}}\Delta t\times \text{Unit}\text{price}$ (V)

Calculate the net income per year.

$\text{Net}\text{income}=\text{Revenue}-\text{Cost}$ (VI)

Conclusion:

Substitute $1000{\text{kg/m}}^{\text{3}}$ for $\rho$, $2{\text{m}}^{\text{3}}/\text{s}$ for $\dot{V}$, $9.81{\text{m/s}}^{\text{2}}$ for g, and 40 m for $\Delta z$ in Equation (I).

$\begin{array}{c}{\dot{W}}_{\text{max,turbine}}=\left(1000{\text{kg/m}}^{\text{3}}\right)\left(2{\text{m}}^{\text{3}}/\text{s}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)\left(40\text{m}\right)\\ =\left(1000{\text{kg/m}}^{\text{3}}\right)\left(2{\text{m}}^{\text{3}}/\text{s}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)\left(40\text{m}\right)\left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\left(\frac{1\text{kW}}{1000\text{N}\cdot \text{m/s}}\right)\\ =784.8\text{kW}\end{array}$

Substitute 784.8 kW for ${\dot{W}}_{\text{ideal}}$ and 0.75 for ${\eta }_{\text{pump-motor}}$ in Equation (II).

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

Substitute 784.8 kW for ${\dot{W}}_{\text{ideal}}$ and 0.75 for ${\eta }_{\text{pump-motor}}$ in Equation (III).

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

Substitute 1046 kW for ${\dot{W}}_{\text{pump,elect}}$, $365\times 10\text{h/year}$ for $\Delta t$, and $0.05/kWh for Unit price in Equation (IV).

$\begin{array}{c}\text{Cost}=\left(1046\text{kW}\right)\left(365\times 10\text{h/year}\right)\times \left(\$0.05\text{/kWh}\right)\\ =\$190,968\text{/year}\end{array}$

Substitute 588.6 kW for ${\dot{W}}_{\text{turbine}}$, $365\times 10\text{h/year}$ for $\Delta t$, and $0.12/kWh for Unit price in Equation (V).

$\begin{array}{c}\text{Revenue}=\left(588.6\text{kW}\right)\left(365\times 10\text{h/year}\right)\times \left(\$0.12\text{/kWh}\right)\\ =\$257,807\text{/year}\end{array}$

Substitute $\$190,968\text{/year}$ for cost and $\$257,807\text{/year}$ for revenue in Equation (VI).

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

Thus, the potential revenue generated per year for the system is $\underset{\_}{\text{\$66,839/year}}$.