# Brayton Cycle Efficiency

Hi, I'm trying to calculate the efficiency for the Brayton Cycle. The end result is supposed to be:

$$\eta = \frac{C_p(T_3-T_4) - C_p(T_2-T_1)}{C_p(T_3-T_2)}$$

Where the denominator is from $Q_h$.

However, when I calculate $Q_h$, I also get a component from work:

$$dQ_h = C_p dT + pdV$$

$$Q_h = C_p(T_3-T_2) + p_{max}(V_3-V_2)$$

Why are you able to ignore the work component when calculating the added heat?

What you are considering heat given in your question is actually a wrong form of the internal energy change equation, $dU=C_p dT-pdV$(in this case, not +pdV). The work component and the added heat are separate entities which combine to give the change in internal energy . The added heat in this case is only $C_p dT$. Hence, the work component doesn't come into play.