Question

(b) at a school science exhibition, a student designed a model of a hydroelectric power station shown in fig. 2.1 (not to scale). (i) describe the main energy changes which take place when the tap is opened. (ii) the 20 w lamp operates normally when the tap is opened and water flows out at 3 kgs⁻¹. calculate the 1. potential energy lost by the water per unit time, 2. efficiency of the model.

Explanation:

Step1: Identify energy - change for (i)

When the tap is opened, water in the reservoir has gravitational potential energy due to its height. As it falls, this potential energy is converted into kinetic energy. The moving water then turns the turbine of the generator, converting kinetic energy into electrical energy which lights up the lamp.

Step2: Calculate potential - energy loss per unit time for (ii)1

The formula for gravitational potential energy is $E_p = mgh$. The rate of mass flow of water is $\frac{dm}{dt}=3\ kg/s$, $g = 9.8\ m/s^2$ and $h = 1.5\ m$. The potential - energy loss per unit time $\frac{dE_p}{dt}=\frac{dm}{dt}gh$.
Substitute the values: $\frac{dE_p}{dt}=3\times9.8\times1.5 = 44.1\ J/s = 44.1\ W$.

Step3: Calculate efficiency for (ii)2

The power output of the system is the power of the lamp, $P_{out}=20\ W$, and the power input is the potential - energy loss per unit time $P_{in}=\frac{dE_p}{dt}=44.1\ W$.
The efficiency $\eta=\frac{P_{out}}{P_{in}}\times100\%$.
Substitute the values: $\eta=\frac{20}{44.1}\times100\%\approx45.4\%$.

Answer:

(i) Gravitational potential energy of water in the reservoir is converted into kinetic energy as water falls. Then kinetic energy of water is converted into electrical energy by the generator which is then converted into light and heat energy by the lamp.
(ii)1. The potential energy lost by the water per unit time is $44.1\ W$.

1. The efficiency of the model is approximately $45.4\%$.