2. Figure 1 schematically shows a house that is heated by a geothermal heat pump

Question

2. Figure 1 schematically shows a house that is heated by a geothermal heat pump (HP) during the winter season in the Winnipeg area. The HP operates between two heat sources, whose temperatures can be treated as constants over the entire winter season: the underground soil temperature is TL = 12°C and the indoor environment temperature is TH = 22°C. The HP receives 12 kW heat from the underground soil (i.e., QL- 12 kW) and its nominal coefficient of performance (COP) is B-5. The HP operates continuously during the entire winter season. and Manitoba Hydro charges electric energy consumption at 7 cents per kWh (i.e., $0.07/kWh) 2 (a) Determine the rate of heat discharged by the HP to the indoor environment, Qu, in kW. (b) What is the expected monthly charge (call this C for charge) shown on the Hydro bill for the 31 days in January, in dollars? (c) What is the maximum possible COP (COPmax or ^max) value for a HP under the operating conditions described above? (d) One day, the owner of the house received a letter from a company advising him that he could reduce his January Hydro bill to $23 (for operating the HP) in order to maintain the required indoor environment temperature, if he replaces his existing HP with a new- generation high-technology HP designed by the company. What is your advice to the owner of the house? Rigorously justify your answer with calculations T,-22"C 0 Ground HP Electric power Figure 1: Figure for problem 2

Explanation / Answer

(1). Heat discharged by the HP (QH)

COP = QH / QH - QL

5 = QH / QH - 12

QH = 15 kW

(b) expected monthly charge

COP = QH / Wi/p

5 = 15 / Wi/p

Wi/p = 3 kW

work for one hour,

Wi/p = 3 * 1 = 3 kWh

work for January month,

Wi/p = 3 * 31 *24 = 2232 kWh

monthly bill at the rate of $0.07/kWh

= 2232 * 0.07 = $156.24

(c) the maximum possible COP of HP

(COP)max = TH / TH - TL

(COP)max = 295 / 295 - 285

(COP)max = 29.5

(d) reduction of $23 in hydro bill

owner of the house wants to reduce hydro bill by $23, for this purpose we need to identify how many hours we need to run HP to maintain the required indoor environment temperature. so lets calculate

bill = $156.24

reduced bill = 156.24 - 23 = $133.24

divided it to charge rate $0.07/kWh then we will have total power consumed in a month

133.24 / 0.07 = 1903.4 kWh

now we have to find how many hours we required to run HP to reduce the bill. let the hours are "x"

Wi/p * days * hours = 1903.4

3 * 31 * x = 1903.4

x = 20.46 hours

we need to run HP, to maintain the required indoor environment temperature 20.46 hours in a day to get required bill.

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