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:
The source temperature is 1248 R.
Explanation:
Second law efficiency of the engine is the ratio of actual efficiency to the maximum possible efficiency that is reversible efficiency.
Given:
Temperature of the heat sink is 520 R.
Second law efficiency is 60%.
Actual thermal efficiency is 35%.
Calculation:
Step1
Reversible efficiency is calculated as follows:
![\eta_{II}=\frac{\eta_{a}}{\eta_{rev}}](https://tex.z-dn.net/?f=%5Ceta_%7BII%7D%3D%5Cfrac%7B%5Ceta_%7Ba%7D%7D%7B%5Ceta_%7Brev%7D%7D)
![0.6=\frac{0.35}{\eta_{rev}}](https://tex.z-dn.net/?f=0.6%3D%5Cfrac%7B0.35%7D%7B%5Ceta_%7Brev%7D%7D)
![\eta_{rev}=0.5834](https://tex.z-dn.net/?f=%5Ceta_%7Brev%7D%3D0.5834)
Step2
Source temperature is calculated as follows:
![\eta_{rev}=1-\frac{T_{L}}{T}](https://tex.z-dn.net/?f=%5Ceta_%7Brev%7D%3D1-%5Cfrac%7BT_%7BL%7D%7D%7BT%7D)
![\eta_{rev}=1-\frac{520}{T}](https://tex.z-dn.net/?f=%5Ceta_%7Brev%7D%3D1-%5Cfrac%7B520%7D%7BT%7D)
![0.5834=1-\frac{520}{T}](https://tex.z-dn.net/?f=0.5834%3D1-%5Cfrac%7B520%7D%7BT%7D)
T = 1248 R.
The heat engine is shown below:
Thus, the source temperature is 1248 R.
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