Chapter 5, Problem 115P

A pump-storage plant uses a turbine to generate hydropower during the day. letting water from the upper reservoir flow through the turbine Into the lower reservoir as sketched. The plant then pumps water back up to the upper reservoir during the night. In a typical pump-storage plant, the same turbomachine is used as both the pump and the turbine, and is called a punip-furbine. The plant is profitable because the demand for electric power is much higher during the day than it is at night, and utility companies sell power at much lower prices at night to encourage customers to use the available power generation capacity and to avoid building new expensive power plants that would 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 S0.030/kVh at night and is willing to pay $0.1 20/kWh for power produced during the day. A pump-storage plant has a gross head of 90.0 in, and can handle a volume flow rate of 4.6 m³/s in either direction. The irreversible head losses in the system are estimated as 5.0 m in either direction at this flow rate. The efficiency of the combined pump-motor is 88 percent and that of the combined turbine-generator is 92 percent. The plant runs in pump mode for 10 hours each night and runs in turbine mode for 10 hours each day. It operates 340 days in a year. How much net revenue (S) does this pump-storage plant generate in one year?

To determine

The plant net revenue generated in a year.

Answer to Problem 115P

The plant net revenue generated in a year is $940987.5639\$/\text{year}$.

Explanation of Solution

Given information:

The gross head of plant is, volume flow rate is $4.6{\text{m}}^{\text{3}}/\text{s}$,irreversible head loss in either direction is $5\text{m}$, the efficiency of combine pump motor is $88\text{percent}$, efficiency of combine turbine generator is $92\text{percent}$, rate of electric power during night is $\$0.030/\text{kWh}$, rate of electric power during day is $\$0.120/\text{kWh},$ plant runs during day and night is $10\text{hrs}$ each and $340\text{days}$ in year.

Write the expression for net head during electricity generation by turbine.

  $H_{\text{net-g}}=H_{\text{gross}}-H_{\text{loss}}$   ....... (I)

Here, gross head is $H_{\text{gross}}$ and irreversibility head loss is $H_{\text{loss}}$.

Write the expression for net head during electricity consumption by pump.

  $H_{\text{net-p}}=H_{\text{gross}}+H_{\text{loss}}$   ....... (II)

Write the expression for power generated by turbine.

  $P_{\text{gen}}=\frac{\rho Qg{H}_{\text{net-g}}}{1000}\times {\eta }_{\text{tur-gen}}$   ....... (III)

Here, density of water is $\rho$, volume flow rate is $Q$, acceleration due to gravity is $g$ and efficiency of turbine generator is ${\eta }_{\text{tur-gen}}$.

Write the expression for power consume by pump.

  $P_{\text{pump}}=\frac{\rho Qg{H}_{\text{net-p}}}{1000\times {\eta }_{\text{pum-mot}}}$   ....... (IV)

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

Write the expression for cost of electricity produced during day in a year.

  $\begin{array}{c}{C}_{\text{prod}}=\left(P_{\text{gen}}\right)\times \left(10\text{h}/\text{day}\right)\times \left(\$0.120/\text{kWh}\right)\times \left(340\text{days}/\text{year}\right)\\ =\left(P_{\text{gen}}\times 10\times 0.120\times 340\right)\$/\text{kW}\cdot \text{year}\\ =P_{\text{gen}}\times 408\$/\text{kW}\cdot \text{year}\end{array}$   ....... (V)

Write the expression for cost of electricity consume during night in a year.

  $\begin{array}{c}{C}_{\text{con}}=\left(P_{\text{pump}}\right)\times \left(10\text{h}/\text{day}\right)\times \left(\$0.030/\text{kWh}\right)\times \left(340\text{days}/\text{year}\right)\\ =\left(P_{\text{pump}}\times 10\times 0.030\times 340\right)\$/\text{kW}\cdot \text{year}\\ =P_{\text{pump}}\times 102\$/\text{kW}\cdot \text{year}\end{array}$   ....... (VI)

Write the expression for net revenue of the pump-turbine plant.

  $C_{\text{net}}=\left(C_{\text{prod}}\$/\text{year}-C_{\text{con}}\$/\text{year}\right)$   ....... (VII)

Calculation:

Substitute $90\text{m}$ for $H_{\text{gross}}$ and $5\text{m}$ for $H_{\text{loss}}$ in Equation (I).

  $\begin{array}{c}{H}_{\text{net-g}}=90\text{m}-5\text{m}\\ =85\text{m}\end{array}$

Refer table "properties of saturated water" to obtain the density of water as $998\text{Kg}/{\text{m}}^{\text{3}}$.

Substitute $85\text{m}$ for $H_{\text{net-g}}$, $4.6{\text{m}}^{\text{3}}/\text{s}$ for $Q$, $998\text{Kg}/{\text{m}}^{\text{3}}$ foe $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.92$ for ${\eta }_{\text{tur-gen}}$ in equation (III).

  $\begin{array}{c}{P}_{\text{gen}}=\left(998\text{Kg}/{\text{m}}^{\text{3}}\right)\left(4.6{\text{m}}^{\text{3}}/\text{s}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(85\text{m}\right)\times \left(0.92\right)\\ =\left(4590.8\text{Kg}/\text{s}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(85\text{m}\right)\\ =3521795.4\text{kg}\cdot {\text{m}}^2/{\text{s}}^{\text{3}}\times \frac{1\text{kW}}{\text{1000}\text{kg}\cdot {\text{m}}^2/{\text{s}}^{\text{3}}}\\ =3521.7954\text{kW}\end{array}$

Substitute $3521.7954\text{kW}$ for $P_{\text{gen}}$ in Equation (V).

  $\begin{array}{c}{C}_{\text{prod}}=3521.7954\text{kW}\times 408\$/\text{kW}\cdot \text{year}\\ =1436892.5613\$/\text{year}\end{array}$

Substitute $90\text{m}$ for $H_{\text{gross}}$ and $5\text{m}$ for $H_{\text{loss}}$ in Equation (II).

  $\begin{array}{c}{H}_{\text{net-p}}=90\text{m}+5\text{m}\\ =95\text{m}\end{array}$

Substitute $95\text{m}$ for $H_{\text{net-g}}$, $4.6{\text{m}}^{\text{3}}/\text{s}$ for $Q$, $998\text{Kg}/{\text{m}}^{\text{3}}$ foe $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.88$ for ${\eta }_{\text{pum-mot}}$ in equation (IV).

  $\begin{array}{c}{P}_{\text{con}}=\frac{\left(998\text{Kg}/{\text{m}}^{\text{3}}\right)\left(4.6{\text{m}}^{\text{3}}/\text{s}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(95\text{m}\right)}{0.88}\\ =\frac{\left(4590.8\text{Kg}/\text{s}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(95\text{m}\right)}{0.88}\\ =4861813.7\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}\frac{\text{1}\text{kW}}{1000\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}}\\ =4861.8137\text{1}\text{kW}\end{array}$

Substitute, $4861.8137\text{kWh}$ for $P_{\text{con}}$ in equation (VI).

  $\begin{array}{c}{C}_{\text{con}}=4861.8137\text{kW}\times 102\$/\text{kW}\cdot \text{year}\\ =495904.9974\$/\text{year}\end{array}$

Substitute $1436892.5613\$/\text{year}$ for $C_{\text{prod}}$ and $495904.9974\$/\text{year}$ for $C_{\text{con}}$ in Equation (VII).

  $\begin{array}{c}{C}_{\text{net}}=\left(1436892.5613\$/\text{year}-495904.9974\$/\text{year}\right)\\ =940987.5639\$/\text{year}\end{array}$

Conclusion:

The plant net revenue generated in a year is $940987.5639\$/\text{year}$.