Answer:
a) Ql=33120000 kJ
b) COP = 5.6
c) COPreversible= 29.3
Explanation:
a) of the attached figure we have:
HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:
W=Qh-Ql
Ql=Qh-W
where W=2000 kWh
Qh=120000 kJ/h

b) The coefficient of performance is:

c) The coefficient of performance of a reversible heat pump is:

Th=20+273=293 K
Tl=10+273=283K
Replacing:

C is your answers!!!!!$3&2)//
Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ