6. A heat pump is essentially a refrigerator turned backwards-it uses work to mo

Question

6. A heat pump is essentially a refrigerator turned backwards-it uses work to move heat from the cold reservoir (outside) to the hot reservoir (inside a building) Explain qualitatively why a heat pump is more efficient than a simple heater when it comes to heating a building. a. b. A building has an internal volume of 1500 m3, filled with air (which you may consider to be a diatomic ideal gas). The building HVAC system turns over this air once every 200 minutes-that is, 1/200th of the air in the building is replaced with outside air each minute. Assuming the temperature inside is 25.0 °C and the temperature outside is 0.00 °C, how much power is needed to run a Carnot cycle heat pump to heat the building? We can use the same heat pump, run backwards, to cool the building. How much power is needed to keep the same building at 25.0 °C if the outside temperature is 35.0 °C? [Answers: (a) explain (b) 223 W (c) 34.4 W c.

Explanation / Answer

Q1. Heat pump transfers heat from low temperature to high temperature by doing some work. If Q is the desirable heat then W= Q/COP (where COP >1) amount of work required. In electric heater for Q amount of heat , W(>Q) amount of electrical energy is required. So, a heat pump is more efficient than a simple heater.

Q2. Heat required /minute = volume of air x density x specific heat of air x temperature difference

= (1500/200) x 1.225 x 0.718 x (25-0)

= 164.91 kj

Heat required /s = 164.91/60 = 2.75 kJ

COP of Heat pump = (25+273)/(25-0)= 11.92

so power requred = 2.75 /11.92= 0.223 kW= 223W

Q3. Heat required to be rejected/minute = volume of air x density x specific heat of air x temperature difference

= (1500/200) x 1.225 x 0.718 x (35-25)

= 65.96625 kj

Heat rejected /s = 65.96625/60 = 1.031 kJ

COP of backward Heat pump = (25+273)/(35-25)= 29.8

so power requred = 1.031 /29.8= 0.0344 kW= 34.4 W

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