Isentropic efficiency of a adiabatic compressor

Question

(a) What is the isentropic efficiency of this compressor?

Saturated water vapor at 0.3000 MPa enters an adiabatic compressor and exits as steam at 3.000MPa at 450.0 C

(b) What is the mass flow rate if the input power is 920.0 kW?

I have solved part b but part a is where I’m having difficulty.

Answer 1

Given:

a) & b)

State 1:

- Saturated Water Vapor ( x = 1.0 )

- P = 0.3000 MPa

State 2:

- Steam (Superheated Vapor)

- P = 3.000 MPa

- T = 450.0°C

b) dW_c/dt = 920.0 kW

Find:

a) η_c = ? %

b) dm/dt = ? kg/s

Solution:

Using the givens and your resource for looking up specific enthalpy ( h ) and specific entropy ( s ) of water in your textbook's tables and/or online table resources, you'll be able to narrow values down to their respective description.

Because the quality is 1.0, you do not need to worry yourself with the fluid or fluid-gas values in your tables for saturated water, you need only to look at the gaseous values ( h_g and s_g )

Based off of the values in the givens:

h₁ = h_g = 2724.9 kJ/kg

s₁ = s_g = 6.9916 kJ/ ( kg * K )

h_2a = 3351.60 kJ/kg

I am going to denote 2a being the Actual state that State 2 reaches and 2s as the iSentropic state that State 2 would have reached if the efficiency were an optimistic 100%.

The equation for the isentropic efficiency for devices like compressors is as follows:

η_c = ( h_2s - h₁ ) / ( h_2a - h₁ )

Everything else in the above has been found except for h_2s

To find this out, you must consider this: if the compressor was to be 100% efficient, there would be zero change in entropy and would be a reversible compressor (call me if you can ever create that, we can make mad money but alas, the universal laws don't allow it)

So, the state properties that state 2s is at is when the specific entropies between state 1 and state 2s are equal (synonymously, the difference between their entropies is zero, same end result, different description)

Doing some linear interpolation, you can find that the specific enthalpy of state 2s is:

h_2s ≈ 3279.30 kJ/kg

Because all units will cancel out, I won't write " kJ/kg " multiple times in the efficiency equation

η_c = ( h_2s - h₁ ) / ( h_2a - h₁ ) = ( 3279.3 - 2724.9 ) / ( 3351.6 - 2724.9 )

∴ η_c ≈ 0.885 ≈ 88.5%

I hope this helps! Message me in the comments if you have any questions, comments, or concerns!

Since you said you did part b, I won't post my solution; however, message me in the comments as well if you'd like me to go thru it to double-check your numbers!