Question

A house that is losing heat at a rate of 80,000 kj/h when the outside temperature drops to 15°c is to be heated by electric resistance heaters. If the house is to be maintained at 22°c at all times, determine the reversible work input for this process and the irreversibility

Answer 1

Hey mate,

◆ Answer-
WR = 0.5266 kW
i = 21.69 kW

◆ Explanation-
# Given-
Rate of loss of heat W = 80000 J/h = 22.22 kW
Present temperature T = 15 °C = 288 K
Desired temperature T' = 22° C = 295 K

# Solution-
Reversible work input by electric resistance heater is given by-
WR = W (1 - T/T')
WR = 22.22 (1 - 288/295)
WR = 22.22 (1 - 0.9763)
WR = 22.22 × 0.0237
WR = 0.5266 kW

Irreversibility is calculated by-
i = W - WR
i = 22.22 - 0.5266
i = 21.69 kW

Hope that is useful...