# Brayton cycle: heat calculation for isobaric processes

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. 1. 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.
2. 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.