Air enters the compressor of a gas-turbine plant at ambient conditions of
100 kPa and 25°C with a low velocity and exits at 1 MPa and 347°C with a
velocity of 90 m/s. The compressor is cooled at a rate of 1500 kJ/min, and
the power input to the compressor is 250 kW. Determine the mass flow rate of
air through the compressor.
4 comments:
The equation I use is:
mdot = -(Qdot-Wshaftdot)/(deltaEnthalpySpecif)
Using this equation that leaves our change in Kinetic Energy, I get .76 kg/s (deviates .06 from the answer key)...if this is right, why is change in kinetic energy not considered here? Initially there is a low velocity and then when it exits, there is a velocity of 90m/s. So there is a change in velocity and therefore a change in Kinetic Energy...right? What am I forgetting?
Yoda 8:13 AM
Low velocity is a cute way to say v1=0. But it is NOT OK to neglect the Ekin at the outlet !
So, you are doing the problem correctly, except you cannot cancel the delta-Ekin term in the 1st Law.
I have a quick question, qhy do we use deltaH rather than deltaU?
greenpepper7 5:33 PM
That is a tricky question. Here are 2 popular forms of the 1st Law for open systems:
Qdot-Wdottotal = mdot*{dU^+dE^kin+dE^pot}
Qdot-Wdotshaft = mdot*{dH^+dE^kin+dE^pot}
Wdottotal = Wdotshaft+Wdotflow
Take the 1st form of the 1st Law and move the wdotflow over the left-hand side of the equation. When it gets there, it combines with dU^ and becomes dH^.
So, both forms mean the same thing and both always apply.
It is generally more USEFUL to use the latter form because you can look up H^ and flow work is built in without making you do any extra calculations. Basically, we use it because it is usually easier. We do not have to lookup both U^ and V^ and then calculate Wdotflow separately.
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