Answer to Question #292386 in Mechanics | Relativity for bookaddict

Question

Pumped-storage hydroelectricity allows energy to be stored to prepare for periods of high demand. Let a high-level reservoir be 500 m above a low-level reservoir. What is the useful power output if the water from the high-level reservoir flows at a rate of 100 kg per minute and the turbine is 85% efficient?

Answer 1

Explanations & Calculation

• Energy received by the generator during a period of a minute is

$\qquad\qquad \begin{aligned} \small E&=\small mgh \\ \end{aligned}$

• Then the received power, which is the energy per unit time is

$\qquad\qquad \begin{aligned} \small P_r&=\small \frac{mgh}{t}=\frac{m}{t}.gh\\ \end{aligned}$

• Finally, the power output is

$\qquad\qquad \begin{aligned} \small P_{out}&=\small P_r\times\text{efficiency}\\ \small P_{out}&=\small \big(\frac{m}{t}\Big).gh\times\frac{85}{100} \end{aligned}$

• You can give it a try substituting the given values in the formula above. m/t is the rate of water flow: 100kg per minute.
• Per-minute needs to be converted into per second.