Answer:
Detailed solution is given in the attached diagram
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
Answer:
a)COP=5.01
b)
KW
c)COP=6.01
d)
Explanation:
Given
= -12°C,
=40°C
For refrigeration
We know that Carnot cycle is an ideal cycle that have all reversible process.
So COP of refrigeration is given as follows
,T in Kelvin.
a)COP=5.01
Given that refrigeration effect= 15 KW
We know that 
RE is the refrigeration effect
So
5.01=
b) KW
For heat pump
So COP of heat pump is given as follows
,T in Kelvin.
c)COP=6.01
In heat pump
Heat rejection at high temperature=heat absorb at low temperature+work in put
Given that 
KW
We know that 
d)
This statement is b which is true: hope this helped
Answer: 15%
Explanation:
Frequency of obtaining = 63/420 = 0.15 = 15%
