Constant and variable specific heat assumptions

AI Thread Summary
The discussion revolves around calculating power output and temperature changes in a gas-powered turbine and compressor system, specifically addressing the use of constant versus variable specific heats. The calculations for the turbine's power output yield approximately 2.21 Watts, while the compressor's outlet temperature is determined to be around 176.72 degrees Celsius. The challenge lies in calculating the mass flow rate of ambient air leaving the intercooler, which requires using variable specific heats from tables. The heat load of the intercooler is calculated as 1.67 Watts, based on enthalpy changes. The distinction between constant and variable specific heats is clarified, emphasizing that constant values are selected from tables, while variable values are directly read from enthalpy data.
Luchekv
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1. Homework Statement
A gas (treated as air) powered turbine provides power to a compressor which then sends the compressed air through an intercooler (heat exchanger).

Turbine:
Gas enters @ 0.03kg/s, 370 degrees Celsius
Gas leaves @ 300 degrees Celsius and a "lower pressure"

Compressor:
Draws fresh air @ 0.015kg/s, 30 degrees Celsius, 100kPa
Air leaves at a "higher pressure"

Heat exchanger:
The intercooler cools the compressed air down to 70 degrees celsius
Ambient air flowing through/across the intercooler is @ 30 degrees Celsius, 100kPa
Ambient air leaving the intercooler is @ 60 degrees Celsius

Q1.) Calc Power output of turbine. Constant avg spec heats Cp=1.005 @ 300K
Q2.) Calc temp of air leaving compressor. Const avg spec heats Cp = 1.005 @ 300K
Q3.) Calc mass flow rate of the ambient air leaving the intercooler (not the compressed air going to the engine) - Use variable spec heats for extracted from tables


2. Homework Equations
Q*-W*=m*[h2-h1+(V22-V12)/2+g(z2-z1]
Δh=Cp(T2-T1)



3. The Attempt at a Solution
Q1.) -W*=m*[h2-h1]
First I found Cp at Tavg=335 Celsius or 608K..Cp=1.05324
So then
-W=m*(Cp(T1-T2))
-W=0.03(1.05324(370-300) =2.2118Watts

Q2.)
-W*=m*[h2-h1]
-W=m*(Cp(T2-T1))
2.2118=0.015(1.005(T2-30C)
T2= 176.72 Celsius

Q3.) I'm absolutely stumped with this one..Ive gone to the "Ideal-gas properties of air" table in the book and picked the enthalpy values for T1 and T2...but don't know where to go from there.

I know I'm lacking with units, just focusing on the method and will go back and straighten that out after.
 
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You know the mass flow passing the compressor as well as the inlet temperture before the intercooler and the outlet temperture after the intercooler. What happens with the energy extracted from this mass flow?
 
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I forgot to add in for Q3. "Use variable spec heats for air extracted from tables" I have updated this. I'm not quite sure what you mean, here is a diagram of the intercooler..we are trying to solve for air at the number 3.

ic.png
 
What is the heat load of the inter cooler, based on the flow rate and temperatures of the working gas?
 
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By heat load do you mean Q?
 
Luchekv said:
By heat load do you mean Q?
Yes
 
Well, I know Q =m*c*(delta T). 'c' being specific heat, but I don't see how I could apply that to this situation..
 
Luchekv said:
Well, I know Q =m*c*(delta T). 'c' being specific heat, but I don't see how I could apply that to this situation..
It's ##\dot{m}Δh##. You know ##\dot{m}##, and you can look up the h values in your table.
 
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m* from intercooler: 0.015
Temp before entering intercooler = 176.72 C ~ 450K
Air is then cooled to 70 C ~ 340K

Assuming variable spec heats I go to the "Ideal-gas properties of air" table and read of h for the said temps
h1 =451.8
h2 = 340.42

So Q=0.015(h1-h2)
Q= 1.6707W

Now would I substitute that load and do the same thing with the enthalpies but for the ambient air stream?
 
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Luchekv said:
m* from intercooler: 0.015
Temp before entering intercooler = 176.72 C ~ 450K
Air is then cooled to 70 C ~ 340K

Assuming variable spec heats I go to the "Ideal-gas properties of air" table and read of h for the said temps
h1 =451.8
h2 = 340.42

So Q=0.015(h1-h2)
Q= 1.6707W

Now would I substitute that load and do the same thing with the enthalpies but for the ambient air stream?
Yes
 
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You're an amazing teacher. Thank you!

Slightly off topic question...when it says assume constant spec heats...we go to the tables and we pick a value to use...but for variable we just go ahead and read the values for enthalpy directly..why is that?
 

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