Answer:
T₂ =93.77 °C
Explanation:
Initial temperature ,T₁ =27°C= 273 +27 = 300 K
We know that
Absolute pressure = Gauge pressure + Atmospheric pressure
Initial pressure ,P₁ = 300+1=301 kPa
Final pressure ,P₂= 367+1 = 368 kPa
Lets take temperature=T₂
We know that ,If the volume of the gas is constant ,then we can say that
Now by putting the values in the above equation we get
The temperature in °C
T₂ = 366.77 - 273 °C
T₂ =93.77 °C
Answer:
I am gonna say it is AMERICAN
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:
Increased, 5%
Explanation:
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