Chapter 15, Problem 49P

(a)

To determine

The energy $\left(\text{kWh}\right)$ is needed to do the given amount of work each night.

Answer to Problem 49P

The energy needed is $1.32\times {10}^6\text{kWh}$ .

Explanation of Solution

Given:

The given height of the lake is 135m above the turbines at a rate of

  $1.00\times {10}^5\text{kg}/\text{s}$ for $10.0\text{h}$ at night.

Formula used:

The potential energy formula is given by,

  ${\mathrm{U}}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{w}}=\mathrm{m}\mathrm{g}\mathrm{h}$

  $\mathrm{U}$is a potential energy,

  $\mathrm{m}$ is mass, $\mathrm{g}$ is gravity and $\mathrm{h}$ is a height.

Calculation:

Substitute the values in the formula

  $\begin{array}{c}{\mathrm{U}}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{w}}=\mathrm{m}\mathrm{g}\mathrm{h}\\ =\left(\left(1.00\times {10}^5\raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\right)\left(10.0\text{h}\right)\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(135\text{m}\right)\\ =1.323\times 10^9\text{Wh}\\ \approx 1.32\times {10}^6\text{kWh}\end{array}$

Conclusion:

Therefore, the energy needed is $1.32\times {10}^6\text{kWh}$ .

(b)

To determine

The average power output.

Answer to Problem 49P

The average of power output is $7.1\times {10}^4\text{kW}$ .

Explanation of Solution

Given:

The energy is released during $14\text{h}$ day, at $75\mathrm{\%}$ efficiency.

Formula used:

Efficiency is given as,

  $\mathrm{\eta }=\frac{\text{Output}}{\text{Input}}$

Calculation:

The average power output is,

  $\begin{array}{c}\mathrm{P}=\mathrm{\eta }\times \text{input}\\ \text{=}\frac{75}{100}\times \frac{1.323\times {10}^6\text{kWh}}{14\text{h}}\\ =7.08\times {10}^4\text{kW}\\ \approx 7.1\times {10}^4\text{kW}\end{array}$

Conclusion:

Therefore, the average of power output is $7.1\times {10}^4\text{kW}$ .