Answer:
Temperature Outlet = 347.83 K = 74.65 C
Explanation:
We need to find the following:
1) Temperature at the compressor outlet.
2) Minimum power required to compress the gas.
3) Given efficiency of 70%, find power needed.
4) Calculate final temperature.
To find (1), let's consider the following expression (Equation #1):
[tex]\Delta S =\int\limits^{T1} _{T2} {\frac{C_{p} }{T} } \, dT - R*ln(\frac{P_{2} }{P_{1}} )[/tex]
Since P2 = 1 MPa and P1 = 1 bar = 0.1 Mpa, then [tex]\frac{P_{2} }{P_{1}} = 10[/tex]. Then, the expression can be simplified to:
[tex]8.3144*[2*ln(\frac{T_{2}}{T_{1}} ) + 0.04*(T_{2} - T_{1})] - 8.3144*ln(10) = 0[/tex]
To solve this expression for T2, use excel solver tool or iterations:
T2 = 347.833 K = 347.83 K = 74.65 C
I have also used Desmos to find the value that makes the expression zero.
Now, let's find the work needed to compress reversibly the system.
To find (2) let's consider the energy balance:
[tex]\Delta E_{Control Volume} = Q + W +E_{in} - E_{out}[/tex]
Where
Q: Heat dissipated / input in the system
W: Work
Ein : Energy (input)
Eout: Energy (output)
By assuming that the process is reversible:
[tex]\DeltaS = 0[/tex]Assuming that the process in in steady state:
[tex]\Delta E_{Control Volume} = 0 = Q + W +E_{in} - E_{out}[/tex]
Since it is a steady workflow system, the equation can be rearranged in the following way:
[tex] W = H_{out} - H_{in} - Q [/tex]
