Question

You are installing a heat pump whose COP is half the COP of a reversible heat pump. You plan to use the pump on chilly winter nights to increase the air temperature in your bedroom. Your bedroom's dimensions are 5.35 m ? 3.6 m ? 2.7 m. The air temperature should increase from 63° F to 68° F. The outside temperature is 35° F, and the temperature at the air handler in the room is 112° F. If the pump's electric-power consumption is 787 W, how long will you have to wait for the room's air to warm if the specific heat of air is 1.005 kJ/(kg · C°)? Assume you have good window draperies and good wall insulation so that you can neglect the release of heat through windows, walls, ceilings, and floors. Also assume that the heat capacity of the floor, ceiling, walls, and furniture are negligible. (Assume 1.293 kg/m3 for the density of air.)

Answer 1

We assume that we have good window draperies and good wall insulation so that you can neglect the release of heat through windows, walls, ceilings, and floors.

And the heat capacity of the floor, ceiling, walls, and furniture are negligible.

Density of air = 1.293 kg/m³

Volume of the bedroom, v_c = ( 5.35 × 3.6 × 2.7 ) m = 52.003 m3

The air temperature increase from 63° F to 68° F.

∆T = 68°F - 63°F = 5°F

Outside temperature, T_out = 35°F = 275 K

temperature at the air handler in the room is

T_handler = 112°F = 317 K

specific heat of air , C = 1.005 kJ/kgK= 1005 J/kgK

the pump's electric-power consumption P = 787 WW

Now,

The coefficient of performance ( COP) of the heat pump is given by