A heat engine that pumps water out of an underground mine accepts 500 kJ of heat
ID: 1829105 • Letter: A
Question
A heat engine that pumps water out of an underground mine accepts 500 kJ of heat and produces 200 kJ of work. How much heat does it reject? A heat engine with a thermal efficiency of 40% rejects 1000 kJ/kg of heat. How much heat does it receive? A coal burning steam power plant produces a net power of 300 MW with an overall efficiency of 32%. The heating value of the coal is 28,000 kJ/kg. Determine the amount (mass) of coal consumed during a 24 hour period. The coefficient of performance of a residential heat pump is 1.6. Calculate the heating effect in kJ/s, this heat pump will produce when it consumes 2 kW of electric power. A food freezer is to produce a 5 kW cooling effect and its COP is 1.3. How many kW of power will this refrigerator require for operation? An automotive air conditioner produces a 1-kW cooling effect while consuming 0.75 kW of power. What is the rate at which heat is rejected? Refrigerant R-134a enters the condenser of a residential heat pump at 800 kPa and 35 C at a rate of 0.018 kg/s and leaves at 800 kPa as a saturated liquid. If the compressor consumes 1.2 kW of power determine the COP of the heat pump and the rate of heat absorption from outside air.Explanation / Answer
1) Qin = input heat =500 kj
W = work = 200 k j
Q2 = reject heat = Qin -W = 300 kj
2) n = efficiency = 40 % = 0.4
Q2 = reject heat = 400 kj/kg
Qin = input heat
as we know :: n = (Qin-Q2)/Qin
Qin = Q2/(1-n) = 400/0.6 =666.66 kj/kg
3) Q0 = 300 MW
n = efficincy = 32 % = 0.32
hence Qin = input heat = Qo/n = 937.5 MW
C = heat value of coal = 28 Mj/kg
Heat produce by m kg of coal :: m*C
C*m = 937.5 *24*3600
m = 28928*10^2 kg
mass of coal =28928*10^2 kg