# A 5.5-kW water heater operates at 240 V. i. Should the heater circuit have a 20A or a 30A circuit breaker? [2 marks]

Question

A 5.5-kW water heater operates at 240 V.

i. Should the heater circuit have a 20A or a 30A circuit breaker? [2 marks]

ii. Assuming 85% efficiency, how long will the heater take to heat the water in

a 0.2085-m3 tank from 20°C to 80°C? The density of water is 1,000 kg/m3

and the specific heat capacity is 4,200 J/kg/°C. ​

in progress 0
1 year 2021-09-05T05:25:46+00:00 1 Answers 5 views 0

The time taken is  $$t = 11.23 \ sec$$

Explanation:

From the question we are told

The power the water is $$P = 5.5KW = 5.5 *10^ {3} W$$

The the voltage  of the heater is  is  $$V = 240 V$$

The volume of the heater  $$Z = 0.2085\ m^3$$

The specific heat of water is  $$c_w = 4200 J /kg/^oC$$

The initial temperature is $$T_1 = 20^oC$$

The final temperature is $$T_2 = 80^oC$$

The density of water is  $$\rho_w = 1000 \ kg/m^3$$

The current of the heater is mathematically represented as

$$I = \frac{P}{V }$$

substituting values we have

$$I = \frac{5.5 *10^{3}}{240}$$

$$I = 22.91 \ A$$

So since the current produced is greater than 20 A hence the heater current rating would be 30A

The quantity of heat required to heat the water is mathematically represented as

$$Q = m c_w \Delta T$$

Where m is the mass which is mathematically evaluated as

$$m = \rho_w * Z$$

$$m =1000 * 0.2085$$

$$m =208.5 \ kg$$

Therefore

$$Q = 208.5 * 4.200 * (80 – 20)$$

$$Q = 52542 J$$

Since the heater has an  efficiency then the heat generate by the heater is

$$Q_h = \frac{52542}{0.85}$$

$$Q_h =61814.1 J$$

Now Power is mathematically  represented as

$$P = \frac{Q}{t}$$

Where t is the time taken for heater to heat the water

=>      $$t = \frac{Q}{P}$$

Substituting values

$$t = \frac{61814.1}{5.5*10^{3}}$$

$$t = 11.23 \ sec$$