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Brayton cycle efficiency and optimum pressure ratio

Short answer

For ideal Brayton cycle, thermal efficiency is η = 1 - 1/r_p^((γ-1)/γ). The pressure ratio for maximum specific net work at given T_max and T_min is r_p,opt = (T_max/T_min)^(γ/[2(γ-1)]).

Draw the p-v and T-s diagrams of the ideal Brayton cycle and derive an expression for its thermal efficiency in terms of the pressure ratio. Also derive the pressure ratio for maximum specific net work output for given maximum and minimum temperatures.

Ideal Brayton cycle

Processes: 1-2 isentropic compression; 2-3 constant-pressure heat addition; 3-4 isentropic expansion; 4-1 constant-pressure heat rejection.

 T-s diagram                      p-v diagram
 T                                p
 |             3                  |  2 ___________ 3
 |            /|                  |   \             \
 |           / |                  |    \             \
 |   p2     /  |                  |     \             \
 |         2   |                  |      1 ___________ 4
 |         |   4                  +------------------------ v
 |         |  /   p1
 |         | /
 |         1
 +-------------------- s

On the T-s diagram 1-2 and 3-4 are vertical (isentropic); on the p-v diagram 2-3 and 4-1 are horizontal (constant pressure).

For a perfect gas with constant cpc_p, and pressure ratio rp=p2/p1=p3/p4r_p = p_2/p_1 = p_3/p_4:

qin=cp(T3−T2),qout=cp(T4−T1)q_{in} = c_p (T_3 - T_2), \qquad q_{out} = c_p (T_4 - T_1) η=1−qoutqin=1−T4−T1T3−T2\eta = 1 - \frac{q_{out}}{q_{in}} = 1 - \frac{T_4 - T_1}{T_3 - T_2}

From the isentropic relations, T2T1=T3T4=rp(γ−1)/γ\dfrac{T_2}{T_1} = \dfrac{T_3}{T_4} = r_p^{(\gamma-1)/\gamma}, hence T4T1=T3T2\dfrac{T_4}{T_1} = \dfrac{T_3}{T_2} and T4−T1T3−T2=T1T2\dfrac{T_4 - T_1}{T_3 - T_2} = \dfrac{T_1}{T_2}:

η=1−1rp(γ−1)/γ\boxed{\eta = 1 - \frac{1}{r_p^{(\gamma-1)/\gamma}}}

The efficiency depends only on the pressure ratio and rises with it.

Pressure ratio for maximum net work

Let x=rp(γ−1)/γx = r_p^{(\gamma-1)/\gamma}, T2=T1xT_2 = T_1 x, T4=T3/xT_4 = T_3/x:

wnet=cp(T3−T4)−cp(T2−T1)=cp[T3(1−1x)−T1(x−1)]w_{net} = c_p (T_3 - T_4) - c_p (T_2 - T_1) = c_p \left[T_3\left(1 - \frac1x\right) - T_1(x - 1)\right] dwnetdx=cp[T3x2−T1]=0  ⇒  x2=T3T1\frac{dw_{net}}{dx} = c_p\left[\frac{T_3}{x^2} - T_1\right] = 0 \;\Rightarrow\; x^2 = \frac{T_3}{T_1} rp,opt=(T3T1)γ/[2(γ−1)]\boxed{r_{p,opt} = \left(\frac{T_3}{T_1}\right)^{\gamma/[2(\gamma-1)]}}

At this ratio T2=T4=T1T3T_2 = T_4 = \sqrt{T_1 T_3} and the maximum work is

wnet,max=cp(T3−T1)2w_{net,max} = c_p\left(\sqrt{T_3} - \sqrt{T_1}\right)^2

The efficiency keeps rising above this pressure ratio, but the net work falls and becomes zero at x=T3/T1x = T_3/T_1. Practical gas turbines are therefore designed at a pressure ratio between the maximum-work value and this limit, usually closer to the maximum-work value.

✓ Answer checked by Claude Opus 5.5 (high effort)

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