7-127 An adiabatic steady-flow device compresses argon at 200 kPa and 27 C to 2 MPa. If the argon leaves this compressor at 550°C, what is the isentropic efficiency of the compressor?

Answer

To find the isentropic efficiency of the compressor, we need to follow several steps, including calculating the actual and isentropic enthalpy changes involved in the compression of argon gas.

1. Find Properties of Argon:

The pressures are given as:

2. Locate States on the Thermodynamic Table:

3. Calculate the Actual Work Done: [ \text{Actual Work, } W_{actual} = h_2 - h_1 ] where ( h_2 ) (actual enthalpy at exit) can be looked up using the outlet temperature and pressure (which is 550°C).

4. Calculate the Isentropic Work Done: [ \text{Isentropic Work, } W_{isentropic} = h_{2s} - h_1 ] Here, ( h_{2s} ) is the enthalpy of the argon at the outlet pressure (P2) and at the isentropic condition.

5. Calculate Isentropic Efficiency: The isentropic efficiency ( \eta ) of the compressor is defined as: [ \eta = \frac{W_{actual}}{W_{isentropic}} = \frac{h_2 - h_1}{h_{2s} - h_1} ]

6. Plug in Values: You will need to look up the enthalpy values for argon at the specified states to compute ( h_1 ), ( h_2 ), and ( h_{2s} ). Once you have these values, substitute them into the formula for efficiency.

This analysis requires property data for argon, which can be found in engineering thermodynamics textbooks or software that computes properties.