Chapter 25, Problem 47A

(a)

To determine

To Calculate:Rate at which falling water must supply energy to the turbine.

Answer to Problem 47A

Rate at which falling water must supply energy to the turbine is $441MW$

Explanation of Solution

Given:

Power output and efficiency are:

  $\begin{array}{l}{P}_{out}=375\text{MW}\\ eff=85\%\end{array}$

Formula used:

  $eff=\frac{P_{out}}{P_{in}}\times 100$

  $eff:$ Efficiency

  $P_{out}:$ Output power

  $P_{in}:$ Input power

Calculation:

Efficiency:

  $eff=\frac{P_{out}}{P_{in}}\times 100$

Rearranging the above formula to get $P_{in}$

  $P_{in}=P_{out}\times \frac{100}{eff}$

Substitute the values:

  $\begin{array}{c}{P}_{in}=\text{375MW}\times \frac{100}{85}\\ =\text{441MW}\end{array}$

Conclusion:

Hence, therate at which falling water must supply energy to the turbine $441\text{MW}$ .

(b)

To determine

To Calculate:Change in gravitational potential energy needed in each second.

Answer to Problem 47A

  $4.4\times 1{\text{0}}^{\text{8}}\text{J/s}$

Explanation of Solution

Given:

Power output and efficiency are:

  $\begin{array}{l}{P}_{out}=375\text{MW}\\ eff=85\%\end{array}$

Formula used:

  $eff=\frac{P_{out}}{P_{in}}\times 100$

  $eff:$ Efficiency

  $P_{out}:$ Output power

  $P_{in}:$ Input power

Calculation:

  $eff=\frac{P_{out}}{P_{in}}\times 100$

Rearranging the above formula to get $P_{in}$ ,

  $P_{in}=P_{out}\times \frac{100}{eff}$

Substitute the values in above formula,

  $\begin{array}{c}{P}_{in}=37\text{5MW×}\frac{100}{85}\\ =44\text{1MW}\\ =4.4\times 1{\text{0}}^{\text{8}}\text{J/s}\end{array}$

Conclusion:

Hence,change in gravitational potential energy needed in each second is

  $4.4\times {10}^{\text{8}}\text{J/s}$ .

(c)

To determine

To Calculate:Mass of the water that must pass through the turbine each second.

Answer to Problem 47A

  $2.0\times {10}^6\text{kg}$

Explanation of Solution

Given:

  $\begin{array}{l}GPE=4.4\times {10}^8\text{J}\\ g=9.8\text{N/kg}\\ h=22\text{m}\end{array}$

Formula used:

  $PE=mgh$

Where,

  $PE:$ Potential energy

  $m:$ Mass

  $g:$ Acceleration of gravity

  $h:$ Height

Calculation:

Rearrange the above formula to get mass

  $m=\frac{PE}{gh}$

Substituting the values in above formula

  $\begin{array}{c}m=\frac{\text{4}{\text{.4×10}}^{\text{8}}\text{J}}{\left(\text{9}\text{.8 N/kg}\right)\left(\text{22 m}\right)}\\ \text{=2}{\text{.0×10}}^{\text{6}}\text{kg}\end{array}$

Conclusion:

Hence,mass of the water that must pass through the turbine each second to supply the power is $2.0\times {10}^6\text{kg}\text{.}$