Question

A house that is losing heat to the 100C outdoors at an average rate of 50,000 kJ/h is maintained at 220C at all
times during a winter night for 10 h. The house is to be heated by 50 glass containers each containing 20 L
(1L=10-3 m3
) of water that is heated to 80oC during the day by a solar heater. A thermostat-controlled 15-kW
back-up electric resistance heater turns on whenever necessary to keep the house at 220C. Analyse and
determine the following:
i. If the average solar intensity is 650 W/m2

and the solar heater has an efficiency of 80%, calculate the

area of solar heater to heat 20L of water from 220C to 800C in 8 hours.
ii. How long did the electric heating system run at night?
iii. Keeping 2nd law of thermodynamics in mind, what is the maximum efficiency of the back-up heater?
iv. If the heat provided by the back-up heater is provided using Carnot heat pump, determine the
reduction in electricity consumption.

Answer 1

Total heat loss to surrounding = 50000 * 10 = 500000KJ

Total heat stored in water glasses = = 50 * 20 * 1 * 4.2 * (80-22) = 243600 KJ

1. For solar heater, energy stored in water = Incident energy * area of heater * efficiency * time
243600 = 0.65 * A * 0.8 * 8 *3600
Solving, A = 16.2660 m2

Thus, required area of solar heater is 16.2660 m2

2. Energy (Heat) output by electric heater = Total heat loss to surrounding - Total heat stored in water = 500000 - 243600 = 256400KJ

Heat output of electric heater = Rating * running time
256400 = 15 * running time
Thus running time = 17093.33 seconds = 284.88 min =4.7481 hours

Thus, the heater run for 4.7481 hours during the period.

3. As this is electric heater, it is not affected by temperature of surrounding. The maximum work output will be when total energy input to heater is converted into heat. Thus, maximum efficiency of heater will be 100% or 1.

4. For heat pump in given problem,
Th = room temperature = 22°C
Tc = outside temperature = 10°

Efficiency of carnot heat pump

The net heating effect is 256400 KJ.
Thus energy input = = 10431.24 KJ

Electricity saving = Energy input for electric heat - energy input for heat pump
= 256400 - 10431.24 = 245968.76 KJ = (245968.76 / 3600 ) kWh = 68.3424 kWh = 68.34 units of electricity.

Thus, if backup heat is provided by carnot heat pump instead of electric heater, 68.34 units of electricity are saved.

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