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Chapter 9: Q.61 (page 230)

A hydroelectric power plant uses spinning turbines to transform the kinetic energy of moving water into electric energy with 80% efficiency. That is, 80% of the kinetic energy becomes electric energy. A small hydroelectric plant at the base of a dam generates 50MW of electric power when the falling water has a speed of 18m/s. What is the water flow rate鈥攌ilograms of water per second鈥攖hrough the turbines?

Short Answer

Expert verified

The water flow rate-kilograms of water per second through the turbines is3.85脳105kg/s.

Step by step solution

01

Content Introduction

From work energy theorem, work done (W)by all the forces is equal to change in kinetic energy (鈭�K.E).

W=鈭�KE...................(1)

Assuming water starts from rest, hence, change in kinetic energy is,

鈭�KE=12mvf2-12mvi2鈭�KE=12mv2

02

Content Explanation

As 80%of kinetic energy is converted into electrical energy. Available kinetic energy is

鈭�KE'=80%鈭�KE鈭�KE'=(0.80)12mv2

Therefore, rate of water flow is

role="math" localid="1647801359773" P=WtP=鈭�KE'tP=(0.80)12mv2tmt=2P(0.80)v2mt=2(50MW)(18m/s2)(0.80)mt=2(50MW)(106W)1MW(18)2(0.80)mt=1000脳105(324)(0.80)=3.85脳105kg/s

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