Air flows from a large reservoir in which the pressure and temperature are 1 MPa and 30°C, respectively, through a convergent–di
2 answers:
Answer:
The back pressure for a normal shock to appear at the exit to the duct is 375.959 kPa
Explanation:
Here we have
A₁/A* = 3 = Which gives
Mₐ₁ = 2.637
P₀₁ = = 1 MPa
∴ P₁ = 47298.69 Pa
P₂/P₁ shock = = 7.949
∴ P₂ = 47298.69 Pa× 7.949 = 375959.457 Pa
Therefore, for a normal shock to appear at the exit to the duct we have the back pressure = 375959.457 Pa = 375.959 kPa
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Answer:
The work furnished by the compressor is
The minimum work required for the state to change is
Explanation:
The explanation to these solution is on the first, second , third and fourth uploaded image respectively
Answer:
W= 8120 KJ
Explanation:
Given that
Process is isothermal ,it means that temperature of the gas will remain constant.
T₁=T₂ = 400 K
The change in the entropy given ΔS = 20.3 KJ/K
Lets take heat transfer is Q ,then entropy change can be written as
Now by putting the values
Q= 20.3 x 400 KJ
Q= 8120 KJ
The heat transfer ,Q= 8120 KJ
From first law of thermodynamics
Q = ΔU + W
ΔU =Change in the internal energy ,W=Work
Q=Heat transfer
For ideal gas ΔU = m Cv ΔT]
At constant temperature process ,ΔT= 0
That is why ΔU = 0
Q = ΔU + W
Q = 0+ W
Q=W= 8120 KJ
Work ,W= 8120 KJ