I have a brayton cycle and want to calculate the efficiency and the back work ratio. In my textbook it is stated that the heat added at process 3-4 happens at constant pressure (isobaric) and therefore we can write $q_{added}=h_4-h_3$. It is also stated that this result is calculated via the first law of thermodynamics using the steady-flow equation: $$\dot{m}(h_3+\frac{C_3^2}{2} + Z_3g)+\dot{Q}+\dot{W}=\dot{m}(h_4+\frac{C_4^2}{2} + Z_4g)$$ Then they note that we can assume that $C_3=C_4$ and $Z_3=Z_4$. Now the question is that I have no clue how they find the result from this equation. I would think that we could write: $$\dot{Q}+\dot{W}=\dot{m}(h_4-h_3)$$ $$q+w=h_4-h_3$$ and that since this is an isobaric process we can write $w=-\int_{V_3}^{V_4}p dv = -p\int_{V_3}^{V_4}dv=-p[V_4-V_3]$. But then we immediately find that $q=h_4+pV_4-[h_3+pV_3]$, which is not the desired result. I have been baffled with this so any help is greatly appreciated on how I can find the correct result.
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2 Answers
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Are you aware that the W in the open system version of the first law of thermodynamics does not include all the work, but only the shaft work? It omits the work done to push the fluid into and out of the control volume. The latter is included separately in the h's. In the heat exchange you are analyzing, the shaft work is zero. So the heat added per unit mass is just equal to the specific enthalpy change between the inlet and outlet of the exchanger.
Wasn't this important information covered when they taught you about the open system version of the 1st law?
answered May 28, 2018 at 14:56
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$\begingroup$ Thank you for your answer! I relooked at my textbook and there is no comment on splitting up the work in two parts. They just mention adding the flow work to the equation. Now that I look at it again, it is logical that is not extra work, but already included in the overall work. $\endgroup$– RubenCommented May 28, 2018 at 16:41
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- Everything you say is correct up until your substitution $w = \int P \text{d}v$. This formula represents the boundary work for closed systems; each of the devices in the Brayton cycle (compressor, turbine, two heat exchangers) is an open system.
- The standard approach to the Brayton cycle is to assume that $q = 0$ in the compressor and the turbine (as they are well-insulated and/or the expansion/compression is fast relative to heat transfer) and $w = 0$ in the heat exchangers (as there are no spinning shafts, pistons, or other devices which could do work on the material flowing through the device).
Making the substitution $w = 0$ gives you the desired result.
