# Brayton Cycle Question



## Viper5 (Apr 7, 2017)

I'm working a Brayton Cycle problem, but I'm getting two different answers when trying to solve and I can't figure out why. 

Given:

Pressure Ratio = 8:1

Combustor Temp prior to Turbine = 1253K     App 23.S determines p_r prior to turbine =284.3

Trying to find Temp after Turbine

Method 1: Since isentropic, use pressure ratio [p_r after turbine = (p_r prior to turbine) / 8]  and App. 23 S to yield p_r after turbine = 35.54 and T after turbine = 740K

Method 2: Since isentropic, use isentropic relations.  T after turbine = (T prior to turbine) * (pressure ratio)^(.4/1.4) =692K.

So which is right?


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## P-E (Apr 7, 2017)

I love this stuff but it is beer day.   I'll get back to you.


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## JHW 3d (Apr 7, 2017)

Does the problem state "treat as ideal gas" or "use air tables"? I think that assumption may produce differing results.

For reference, see MERM13 example 29.1. Calculations for TD produce *821K* in one method, and (the equivalent of) *863K* in the other.


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## Viper5 (Apr 7, 2017)

JHW 3d said:


> Does the problem state "treat as ideal gas" or "use air tables"? I think that assumption may produce differing results.
> 
> For reference, see MERM13 example 29.1. Calculations for TD produce *821K* in one method, and (the equivalent of) *863K* in the other.


It doesn't specify to treat as ideal gas or use air tables. What's interesting is that the solution I'm looking at and the example you referenced both have about the same value for T_B when using the alternative method to compare. It's only T_D that has a much larger percent difference.  

So which method is the "correct" one if given on a test?


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## JHW 3d (Apr 8, 2017)

If I had to guess, either the exam question would suggest a method or the results would be such that either method result would be close enough to avoid confusion.

But to answer your question ("which one is _more_ correct?"), I do not know.


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## MikeGlass1969 (Apr 8, 2017)

Method 2:  My reasoning.  After injecting fuel into a combustor you can no longer assume the air to be an ideal gas.  You will have the excess air and the products of combustion at the inlet of the turbine.


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## Viper5 (Apr 8, 2017)

MikeGlass1969 said:


> Method 2:  My reasoning.  After injecting fuel into a combustor you can no longer assume the air to be an ideal gas.  You will have the excess air and the products of combustion at the inlet of the turbine.


I don't think this explains away the discrepancy. Both methods assume only air.


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## MikeGlass1969 (Apr 8, 2017)

Viper5 said:


> I don't think this explains away the discrepancy. Both methods assume only air.


Also true..  Did a little more reading and the combustor could just be a heat exchanger.    I guess both are valid methods.  Answers are within 6% of each other.


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## P-E (Apr 8, 2017)

Both methods are okay.  Method 2 assumes ideal gas which assumes Van der Waal's forces are negligible.  As pressure increases air becomes less of an ideal gas.  Method 1 uses air tables.  This should be more accurate.

Don't forget to use the expansion efficiency if that is given.


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## Audi Driver P.E. (Apr 10, 2017)

MERM13 says (last para page 29-4) you can use the air tables_ IF_ the isentropic efficiencies are known.  So, unless you're either able to calculate it, or it's given, I don't think it's proper to use air tables for this.  But your answers are pretty close either way.


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## Viper5 (Apr 12, 2017)

Audi driver said:


> MERM13 says (last para page 29-4) you can use the air tables_ IF_ the isentropic efficiencies are known.  So, unless you're either able to calculate it, or it's given, I don't think it's proper to use air tables for this.  But your answers are pretty close either way.


Does this mean it would be acceptable to use the T actual (from Eqn 29.21 or Eqn 29.23) and substitute into isentropic relations (Eqn 29.2 through Eqn 29.19)?


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## Viper5 (Apr 12, 2017)

Or must you only use isentropic temperatures for the isentropic relations?


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## amats42 (Apr 14, 2017)

Would you mind posting a photo of the problem? I was trying to figure out when to use isentropic equations versus air tables and my answers were always within a small margin. I don't know if I've seen 6% differences on the questions I tried both methods on.

Also, I thought the air table would be (Pr2/Pr1)=(P2/P1) and thus Pr2=Pr1*pressure ratio, not divided as in your OP.


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## Viper5 (Apr 14, 2017)

amats42 said:


> Would you mind posting a photo of the problem? I was trying to figure out when to use isentropic equations versus air tables and my answers were always within a small margin. I don't know if I've seen 6% differences on the questions I tried both methods on.
> 
> Also, I thought the air table would be (Pr2/Pr1)=(P2/P1) and thus Pr2=Pr1*pressure ratio, not divided as in your OP.


I have just accepted the marginal difference. As long as you know that isentropic values are used to find actual values (when efficiency applies) then I think it will be fine.  

If a problem arises in which a particular method is required to yield the correct answer, then I will take note and ask NCEES. However, I highly doubt this will be necessary. 

I did notice the error in my equation from op, but don't know how to edit on this forum.


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