Respuesta :
Answer:
a. 33120MJ
b. 5.6
c. 48
Explanation:
∆U= 120,000 KJ/h
Since
1 day = Â 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:
[tex] a) Q_a = 33120 MJ [/tex]
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
[tex] Q_r= 120,000 KJ/h [/tex]
[tex] Q_r = 120000KJ/hr * 336hrs = 40320MJ [/tex]
W = 2000 KW/hr
Therefore W will be
2000KW hr * 3600sec/hr = 7200000000 => 7200M
a) Let's use the first law of thermodynamics expression
[tex] Q_r= Q_a+W [/tex]
Since we are to find energy received Qa, let's make Qa subject of the formula
Therefore
[tex] Q_a = Q_r - W [/tex]
[tex] Q_a= 40320MJ - 7200MJ [/tex]
[tex] Q_a =33120MJ [/tex]
b) let's use the fornula
[tex] COP = \frac{Qr}{W} [/tex]
[tex] COP= \frac{40320}{7200} [/tex]
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
[tex] COP = \frac{T_H}{T_H -T_L} [/tex]
[tex] COP = \frac{15}{15 - 9} [/tex]
COP = 48