Question

slader A heat pump cycle is used to maintain the interior of a building at 15°C. At steady state, the heat pump receives energy by heat transfer from well water at 9°C and discharges energy by heat transfer to the building at a rate of 120,000 kJ/h. Over a period of 14 days, an electric meter records that 2000 kW · h of electricity is provided to the heat pump. Determine the following (13 points) : (a) the amount of energy that the heat pump receives over the 14-day period from the well water by heat transfer, in kJ; (b) the heat pump’s coefficient of performance.; EGMN 204 Spring 2020 4 (c) the coefficient of performance of a reversible heat pump cycle operating between hot and cold reservoirs at 15°C and 9°C.

Answer 1

Answer:

a. 33120MJ

b. 5.6

c. 48

Explanation:

∆U= 120,000 KJ/h

Since

1 day =  24 hrs

14 days =24 x  14 hrs

14 days = 336 hrs

∆U = 120000 x 336 KJ

=40320000KJ

K = 1000

M = 1000000

∆U = 403200 MJ

Work done

W= 2000 KW.h    ( 1 h = 3600 s)

W= 7200 MJ

According to the first law of thermodynamics

∆U = Q+W

Q= 40320 - 7200 MJ

1)Qa=33120 MJ

Coefficient of performance

COP = ∆U/W

COP= 40320 / 7200

2)COP = 5.6

3)COP of ideal heat pump

Th/(Th - Tl)

Th = 15°C

Tl = 9°C

Convert Celsius to Kelvin

273 + 15 = 288

288/(15-9)

288/6

48

COP= 48

Answer 2

Answer:

$a) Q_a = 33120 MJ$

b) COP = 5.6

c) COP = 48

Explanation:

We are given=

T = 14 days,

Lets convert 14 days to hours,we have:

14days * 24hrs/day

T = 336hrs

$Q_r= 120,000 KJ/h$

$Q_r = 120000KJ/hr * 336hrs = 40320MJ$

W = 2000 KW/hr

Therefore W will be

2000KW hr * 3600sec/hr = 7200000000 => 7200M

a) Let's use the first law of thermodynamics expression

$Q_r= Q_a+W$

Since we are to find energy received Qa, let's make Qa subject of the formula

Therefore

$Q_a = Q_r - W$

$Q_a= 40320MJ - 7200MJ$

$Q_a =33120MJ$

b) let's use the fornula

$COP = \frac{Qr}{W}$

$COP= \frac{40320}{7200}$

COP = 5.6

c) we are to find the Coefficient of Performance of reversible heat pump cycle at 15° and 9°

We use the formula

$COP = \frac{T_H}{T_H -T_L}$

$COP = \frac{15}{15 - 9}$

COP = 48