# Q on Steady Flow Open systems and Stagnation properties Watch

1. I want to check my understanding of a question I'm doing that I do not have the answers to.

In the question, you are first asked to find some missing quantities after air flows into a turbine assuming the process is isentropic and initial velocity is 0. The next part of the question then asks you how would the stagnation temperature, static temperature and discharge velocity be different if the process is adiabatic but irreversible.

The way I think about it: the power output has to be lower than before. So mCp(T02-TO1) has to be smaller in magnitude. I think initial stagnation temperature stays the same as the initial conditions have not changed so discharge stagnation temperature has to increase.

Since stagnation temperature = static temperature + Velocity squared/2Cp
Either static temperature or velocity has to increase as well. Since it is irreversible, the discharge velocity has to be less than before, so the discharge static temperature has to have increased.

If the above is correct, I just don't quite understand why the temperature rises when the output power is lower. Does that mean there is more energy losses as heat to the surroundings rather than converted to kinetic energy?
2. Bump for above, I really need help please.

Another thing I cannot understand, is why the equation on the change of specific enthalpy (h2-h1)=Cp(T2-T1) applies for all processes when it is derived using an equation that only works for constant pressure? (W=p(V2-V1)?

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Updated: January 14, 2015
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