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If the final temperature and pressure is made less than the initial temperature and pressure, would it reduce the efficiency of the Brayton cycle?

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- Thread starter sid_galt
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If the final temperature and pressure is made less than the initial temperature and pressure, would it reduce the efficiency of the Brayton cycle?

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abercrombiems02 said:

One question - by overpower, do you mean more kinetic energy has been deposited into the turbine-compressor system instead of going into the exit velocity of the gases?

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I understand. Thanks for the help

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One more question. The efficiency of a brayton cycle depends on the ratio of temperature after heat addition to temperature after work extraction. However the book I have here says that this ratio equal to the ratio of temperature after pressure increase through compressor and initial temperature.

I don't understand how. Can someone help?

I don't understand how. Can someone help?

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- #7

FredGarvin

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I had to go back to my books for this one, but here is a progression for getting to this point, starting with the efficiency definition you are familiar with:

[tex]\eta = \frac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_2)}[/tex]

[tex]\eta = \frac{c_p(T_3 - T_4)-c_p(T_2 - T_1)}{c_p(T_3 - T_2)}[/tex]

[tex]\eta = 1 - \frac{(T_4 - T_1)}{(T_3 - T_2)}[/tex]

[tex]\eta = 1 - \frac{T_1}{T_2} (\frac{T_4 / T_1 - 1}{T_3 / T_2 - 1})[/tex]

Since we know that for isentropic, ideal gases:

[tex]T_2 = T_1 (\frac{p_2}{p_1})^{(k-1)/k}[/tex] and

[tex]T_4 = T_3 (\frac{p_4}{p_3})^{(k-1)/k} = T_3 (\frac{p_1}{p_2})^{(k-1)/k}[/tex]

We can now say [tex]\frac{T_4}{T_1} = \frac{T_3}{T_2}[/tex]

Therefore the efficiency can reduce to [tex]\eta = 1 - \frac{T_1}{T_2}[/tex]

And finally....

[tex]\eta = 1 - \frac{1}{(p_2/p_1)^{(k-1)/k}}[/tex]

- #8

Astronuc

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Supplementing Fred's excellent post -

http://www.grc.nasa.gov/WWW/K-12/airplane/brayton.html

http://en.wikibooks.org/wiki/Applications_(Engineering_Thermodynamics)#Gas_Turbine_Cycle_.28or_Joule-Brayton_Cycle.29 [Broken]

And nice little article on gas turbines - http://en.wikipedia.org/wiki/Gas_turbine

http://www.grc.nasa.gov/WWW/K-12/airplane/brayton.html

http://en.wikibooks.org/wiki/Applications_(Engineering_Thermodynamics)#Gas_Turbine_Cycle_.28or_Joule-Brayton_Cycle.29 [Broken]

And nice little article on gas turbines - http://en.wikipedia.org/wiki/Gas_turbine

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- #11

Astronuc

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Do you mean the compressor inlet temp = turbine exit nozzle temp?sid_galt said:

See this diagram - http://www.grc.nasa.gov/WWW/K-12/airplane/epr.html

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Also shouldn't the Engine Pressure Ratio as given in the link be ideally equal to 1?

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FredGarvin

Science Advisor

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Also, look at this version of thermal efficiency:

[tex]\eta = \frac{c_p(T_3 - T_4)-c_p(T_2 - T_1)}{c_p(T_3 - T_2)}[/tex]

In your question, you would have to have, at best, T4 = T1. What does that do to your efficiency? It makes it 1. If T4 < T1 you would have an efficiency > 1.

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I get it. I should have thought of that. Thank you.

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