Ideal gas turbine with regenerator at 1400 K

Data: ideal (isentropic) compressor and turbine, $p_1 = 100$ kPa, $T_1 = 300$ K, $r_p = 10$, $T_3 = 1400$ K, $c_p = 1.005$ kJ/kg K, $\gamma = 1.4$; $r_p^{0.2857} = 10^{0.2857} = 1.9307$.

Simple ideal Brayton cycle

$$\begin{aligned} T_2 &= T_1 r_p^{0.2857} = 300 \times 1.9307 = 579.21\ \text{K} \\ T_4 &= \frac{T_3}{r_p^{0.2857}} = \frac{1400}{1.9307} = 725.13\ \text{K} \\ W_c &= 1.005(579.21 - 300) = 280.61\ \text{kJ/kg} \\ W_t &= 1.005(1400 - 725.13) = 678.25\ \text{kJ/kg} \\ W_{net} &= 678.25 - 280.61 = 397.64\ \text{kJ/kg} \\ Q_s &= 1.005(1400 - 579.21) = 824.89\ \text{kJ/kg} \\ \eta &= \frac{397.64}{824.89} = 0.4821 = 48.21\% \end{aligned}$$

Check: $\eta = 1 - \dfrac{1}{r_p^{(\gamma-1)/\gamma}} = 1 - \dfrac{1}{1.9307} = 48.21\%$.

With regenerator ($\varepsilon = 0.8$)

$T_4 = 725.13$ K $> T_2 = 579.21$ K, so regeneration is useful.

$$\begin{aligned} T_x &= T_2 + \varepsilon(T_4 - T_2) = 579.21 + 0.8(145.92) = 695.94\ \text{K} \\ Q_s' &= 1.005(1400 - 695.94) = 707.58\ \text{kJ/kg} \\ \eta' &= \frac{W_{net}}{Q_s'} = \frac{397.64}{707.58} = 0.5620 = 56.20\% \end{aligned}$$

Net work is unchanged; only the fuel heat is reduced.

Increment in efficiency

$$\Delta\eta = 56.20 - 48.21 = 7.99\ \text{percentage points}$$

(a relative increase of $7.99/48.21 = 16.6\%$).

Answer: Efficiency without regenerator = 48.21 %; with 80 % regenerator = 56.20 %; increase ≈ 7.99 % (points).