Brayton Cycle

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In brayton cycle, the initial temperature and pressure before compression is less than the final temperature and pressure after work has been done by the hot gases.
If the final temperature and pressure is made less than the initial temperature and pressure, would it reduce the efficiency of the Brayton cycle?
 
First of all, in your statment saying that the final temperature and pressure are greater than the initial is only partially correct. The final temperature is greater than that of the initial temperature but on a TS diagram, lines of constant pressure follow parabolic slopes each of which P1 and P4 pass though. They both, however, are points on the same parabola, thus P1=P4 or the intial and final pressures are the same. Think of a turbojet, the pressure before any compression is (well lets assume in engine is stationary on a test stand) in this case the initial pressure is simply the ambient pressure. Well the exhaust gases are expelled into....yes the same atmosphere so at the exit the pressure the gas is expanding to is still the same as that of the inlet. If the engine was in flight, we'd simply need to find the new pressure using some basic isentropic formulas relating the mach number, the specific heat ratio, and the pressure ratios to determine whats called the stagnation pressure. Anyways in the IDEAL brayton cycle the efficiency is equal to 1 - (Tburnexit/Tturbexit) (gam is 1.4 for air). So if we reduce the exit temperature what we essentially have done is used the energy to "overpower" the compressor and turbine. This is WASTED power because the useful power extracted here lies in the kinetic energy of the exhaust gases. because the exit temperature is less, so is our exit velocity of the exhaust as Vexit = Mexit*sqrt(gam*R*Texit). So we have less velocity, also Thrust is equal to (mass flow rate air + mass flow rate fuel)*Vexit. (mdot fuel << mdot air), but i'll still leave it in for completeness. So a lower exit temperature means less thrust as well. Thus, it is clear that a lower exhaust temperature/pressure will DECREASE the thermal effeciency of the brayton cycle!
 
501
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abercrombiems02 said:
First of all, in your statment saying that the final temperature and pressure are greater than the initial is only partially correct. The final temperature is greater than that of the initial temperature but on a TS diagram, lines of constant pressure follow parabolic slopes each of which P1 and P4 pass though. They both, however, are points on the same parabola, thus P1=P4 or the intial and final pressures are the same. Think of a turbojet, the pressure before any compression is (well lets assume in engine is stationary on a test stand) in this case the initial pressure is simply the ambient pressure. Well the exhaust gases are expelled into....yes the same atmosphere so at the exit the pressure the gas is expanding to is still the same as that of the inlet. If the engine was in flight, we'd simply need to find the new pressure using some basic isentropic formulas relating the mach number, the specific heat ratio, and the pressure ratios to determine whats called the stagnation pressure. Anyways in the IDEAL brayton cycle the efficiency is equal to 1 - (Tburnexit/Tturbexit) (gam is 1.4 for air). So if we reduce the exit temperature what we essentially have done is used the energy to "overpower" the compressor and turbine. This is WASTED power because the useful power extracted here lies in the kinetic energy of the exhaust gases. because the exit temperature is less, so is our exit velocity of the exhaust as Vexit = Mexit*sqrt(gam*R*Texit). So we have less velocity, also Thrust is equal to (mass flow rate air + mass flow rate fuel)*Vexit. (mdot fuel << mdot air), but i'll still leave it in for completeness. So a lower exit temperature means less thrust as well. Thus, it is clear that a lower exhaust temperature/pressure will DECREASE the thermal effeciency of the brayton cycle!
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?
 
yes. Basically what happens here is that an extra amount of kinetic energy will be used up here to run the compressor/turbine assembly at a higher speed. This in turn adds more air into the engine, this is not alwasy good though. If we assume the cycle is ideal we are also assuming that the air/fuel mixture is at an optimum, meaning all the air reacts with the fuel. By over powering the compressor and turbine, this introduces extra air into the combustor. This air does not react in the chemical reaction and is simply wasted. Basically we are getting the same work out of the system, but we are using more air to get this same amount of work, when it could be done more efficiently by not overpowering the compressor/turnbine assembly. Alternatively if you want to take into account the kinetic energy of the unreacted air, and think of it as thrust, then just use the 2nd law and we can reason that while the compressor may have been overpowered with lets say 20 kJ of energy, then there is no way we are going to get 20kJ of thrust out of that air, especially if it doesnt react. Thus our thermal efficiency will still decrease.
 
501
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I understand. Thanks for the help
 
501
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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?
 
Last edited:

FredGarvin

Science Advisor
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The thermal efficiency of the cold air-standard ideal Brayton cycle is indeed a function of the pressure rise across the compressor. The thing to remember with the cold air ideal cycle is that specific heats are assumed constant. It is also assuming that, with the same heat rejection capability, an increase in pressure ratio increases the heat added to the cycle and thus an increase in efficiency.

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]
 

Astronuc

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Last edited by a moderator:
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Thank you for the replies. Why can't P4/P3 be made greater than P2/P1 so that the inlet temperature equals the exit nozzle temperature to get better efficiency like in turboprop or turboshaft engines?
 
501
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Please does anyone know? Why can't the burner:turbine-nozzle pressure ratio be made greater than compressor:inlet pressure ratio. Does it defy any law of thermodynamics?
 

Astronuc

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sid_galt said:
Thank you for the replies. Why can't P4/P3 be made greater than P2/P1 so that the inlet temperature equals the exit nozzle temperature to get better efficiency like in turboprop or turboshaft engines?
Do you mean the compressor inlet temp = turbine exit nozzle temp?

See this diagram - http://www.grc.nasa.gov/WWW/K-12/airplane/epr.html
 
501
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Yes. Every Temperature Entropy diagram I have seen for the Brayton cycle shows the exit temperature to be greater than the inlet temperature. Is it impossible for the exit temperature to be equal to or less than the inlet temperature?

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

FredGarvin

Science Advisor
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No. Simply because you can't have a perfect turbine extract all of the work that was put into the air by the compressor and combustion. Also, P4<P3. P4/P3 can never be greater than P2/P1.

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.
 
501
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I get it. I should have thought of that. Thank you.
 

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