The structure of a house is such that it loses heat at a rate of 5400 kJ/h per degree Cdifference between the indoors and outdoors. A heat pump that requires a power input of 6 kW isused to maintain this house at 21 C. Determine the lowest outdoor temperature for which the heatpump can meet the heating requirements of this house

Tl = - 13.3°C

the lowest outdoor temperature is - 13.3°C

Explanation:

Given that;

Temperature of Th = 21°C = 21 + 273 = 294 K

the rate at which heat lost is Qh = 5400 kJ/h°C

the power input to heat pump Wnet = 6 kw

The COP of a reversible heat pump depends on the temperature limits in the cycle only, and is determined by;

COPhp = Th/(Th - Tl)

COPhp = Qh/Wnet

Qh/Wnet = Th/(Th -Tl)

the amount of heat loss is expressed as

Qh = 5400/3600(294 - Tl)

the temperature of sink

( 5400/3600(294 - Tl)) / 6 = 294 / ( 294 - Tl)

now solving the equation

Tl = 259.7 - 273

Tl = - 13.3°C

so the lowest outdoor temperature is - 13.3°C

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