An ideal gas at a flow rate of 10m3/min enters a compressor at 25 °c and 1 bar. it leaves at 1 mpa. during this process, heat is dissipated to the surroundings at a rate of 2100 w. you may that the surroundings to be at a constant temperature of 25 °c. the heat capacity of the ideal gas is given by c_p/r=2.00+0.040t where t in in units of k. answer the following assuming that the process is reversible, calculate the temperature at the compressor outlet. calculate the minimum power required to compress the gas. if the compressor efficiency is 70%, calculate the actual power needed. calculate the actual final temperature.

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):

Since P2 = 1 MPa and P1 = 1 bar = 0.1 Mpa, then . Then, the expression can be simplified to:

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:

Where

Q: Heat dissipated / input in the system

W: Work

Ein : Energy (input)

Eout: Energy (output)

By assuming that the process is reversible:

Assuming that the process in in steady state:

Since it is a steady workflow system, the equation can be rearranged in the following way:

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