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Chapter 15: Q54P (page 412)

(II) Energy may be stored by pumping water to a high reservoir when demand is low and then releasing it to drive turbines during peak demand. Suppose water is pumped to a lake 115 m above the turbines at a rate of\({\bf{1}}{\bf{.00 \times 1}}{{\bf{0}}{\bf{5}}}\;{\bf{kg/s}}\)for 10.0 h at night. (a) How much energy (kWh) is needed to do this each night? (b) If all this energy is released during a 14-h day, at 75% efficiency, what is the average power output?

Short Answer

Expert verified

(a) The amount of energy needed to pump the water each night is\(1.13 \times {106}\;{\rm{kWh}}\).

(b) The average power output is \(60\;{\rm{MW}}\).

Step by step solution

01

Understanding the energy required for water

In this problem, the energy required to pump the water to the lake will be equal to the gravitational potential energy of the water because no dissipative force is acting on the water.

02

Given data

The height above the turbine is \(h = 115\;{\rm{m}}\)

The rate of flow of water is \(m = 1.00 \times {105}\;{\rm{kg/s}}\)

The time is \(t = 10\;{\rm{h}}\)

The energy is released for the time \({t_{\rm{e}}} = 14\;{\rm{h}}\)

The efficiency is \(\eta = 75\% \)

03

(a) Evaluation of the energy needed to pump water

The relation to calculate the potential energy is given by:

\(PE = mgh\)

Here, g is the acceleration due to gravity.

Substitute the values in the above expression.

\(\begin{array}{l}PE = \left( {1.00 \times {{10}5}\;{\rm{kg/s}}} \right)\left( {10\;{\rm{h}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}2}} \right)\left( {115\;{\rm{m}}} \right)\\PE = 1.13 \times {109}\;{\rm{Wh}} \times \frac{{1\;{\rm{kWh}}}}{{{{10}3}\;{\rm{Wh}}}}\\PE = 1.13 \times {106}\;{\rm{kWh}}\end{array}\)

Thus, the amount of energy needed to pump the water each night is \(1.13 \times {106}\;{\rm{kWh}}\).

04

(b) Evaluation of the average power output

The relation to calculate the power output is given by:

\(P = \frac{{PE \times \eta }}{{{t_{\rm{e}}}}}\)

Substitute the values in the above expression.

\(\begin{array}{l}P = \frac{{\left( {1.13 \times {{10}6}\;{\rm{kWh}} \times {\rm{0}}{\rm{.75}}} \right)}}{{14\;{\rm{h}}}}\\P = 60000\;{\rm{kW}}\\P = 60\;{\rm{MW}}\end{array}\)

Thus, the average power output is\(60\;{\rm{MW}}\).

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Most popular questions from this chapter

On a very hot day, could you cool your kitchen by leaving the refrigerator door open?

(a) Yes, but it would be very expensive.

(b) Yes, but only if the humidity is below 50%.

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(d) No, the heat exhausted by the refrigerator into the room is more than the heat the refrigerator takes out of the room.

Question: (II) How much less per year would it cost a family to operate a heat pump that has a coefficient of performance of 2.9 than an electric heater that costs \(2000 to heat their home for a year? If the conversion to the heat pump costs \)15,000, how long would it take the family to break even on heating costs? How much would the family save in 20 years?

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