# 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