Optimistic Analysis of Heat Pump Cycle
To calculate the amount of energy received by the heat pump from well water over a 14-day period, we can use the first law of thermodynamics. Given that the heat pump receives energy from well water at 9°C and discharges energy to the building at a rate of 120,000 kJ/h, we need to find the total energy received and the coefficient of performance.
Calculation:
Given data:
Energy received by the heat pump from well water (Qr): 120,000 kJ/h
Electricity provided to the heat pump over 14 days (W): 2000 kW · h
a) The amount of energy received by the heat pump from well water over the 14-day period:
We convert 14 days to hours: 14 days x 24 hrs/day = 336 hrs
Qr = 120,000 kJ/h x 336 hrs = 40,320,000 kJ = 40,320 MJ
W = 2,000 kW · h x 3,600 s = 7,200,000 J = 7,200 MJ
Using the first law of thermodynamics formula, Q received = Qr - W:
Q received = 40,320 MJ - 7,200 MJ = 33,120 MJ
Therefore, the amount of energy received by the heat pump from well water over the 14-day period is 33,120 MJ.
b) The coefficient of performance (COP) of the heat pump:
We use the formula COP = Qr / W to calculate the COP.
COP = 40,320 MJ / 7,200 MJ = 5.6
c) The coefficient of performance of a reversible heat pump cycle at 15°C and 9°C:
We use the formula COP = TH / (TH - TL) to calculate the COP of the reversible heat pump cycle.
TH = 15°C, TL = 9°C
Convert Celsius to Kelvin: TH = 288 K, TL = 282 K
COP = 288 K / (288 K - 282 K) = 48