Non-Ideal Brayton Cycle with Duct Pressure Drop: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2016, Q3

Question 3: Non-Ideal Brayton Cycle with Duct Pressure Drop (equal value)

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Point	Description	$P$ (kPa)	$T$
1	compressor inlet	100	20 °C (293 K)
2	compressor exit	475	220 °C (494 K)
3	turbine inlet (after $\Delta P$)	461.3	870 °C (1143 K)
4	turbine exit	100	526 °C (799 K)

Figure 3 — T–s diagram of the non-ideal open Brayton cycle: 1→2 irreversible compression to 475 kPa, a small 13.7 kPa combustor pressure loss (2→3), 3→4 irreversible expansion to atmospheric 100 kPa. Dashed points 2s, 4s are the isentropic references.

Given. An open Brayton cycle: inlet 20 °C/100 kPa, compressor delivery 475 kPa (82 % efficient), turbine-inlet limit 870 °C (85 % efficient), and a 13.7 kPa combustor pressure drop. Find. Every state pressure and temperature, the net work, the thermal efficiency, and the back-work fraction.

Approach. Cold-air-standard analysis ($c_p=1.005$ kJ/kg·K, $k=1.4$): find each isentropic reference temperature from the pressure ratio, then apply the stated component efficiencies; the duct loss lowers the turbine-inlet pressure.

Compressor exit (point 2). Isentropic $T_{2s}=T_1(P_2/P_1)^{(k-1)/k}=293.15(4.75)^{0.2857}=457.5$ K; with $\eta_c=0.82$, $$T_2=T_1+\frac{T_{2s}-T_1}{\eta_c}=293.15+\frac{164.4}{0.82}=493.6\ \text{K}\ (220\,{}^\circ\text{C}),\ P_2=475\ \text{kPa}.$$
Turbine inlet (point 3). The 13.7 kPa combustor loss gives $P_3=475-13.7=461.3$ kPa at the metallurgical limit $T_3=1143.15$ K (870 °C).
Turbine exit (point 4). Isentropic $T_{4s}=T_3(P_4/P_3)^{(k-1)/k}=1143.15(100/461.3)^{0.2857}=738.6$ K; with $\eta_t=0.85$, $$T_4=T_3-\eta_t(T_3-T_{4s})=1143.15-0.85(404.6)=799.3\ \text{K}\ (526\,{}^\circ\text{C}),\ P_4=100\ \text{kPa}.$$
Specific works. $w_c=c_p(T_2-T_1)=1.005(200.5)=201.5$ kJ/kg; $w_t=c_p(T_3-T_4)=1.005(343.9)=345.6$ kJ/kg, so $$w_\text{net}=w_t-w_c=345.6-201.5=\boxed{144\ \text{kJ/kg}}.$$
Thermal efficiency and back-work. $q_\text{in}=c_p(T_3-T_2)=1.005(649.5)=652.8$ kJ/kg, giving $$\eta_\text{th}=\frac{144}{652.8}=\boxed{22.1\%},\qquad \frac{w_c}{w_t}=\frac{201.5}{345.6}=\boxed{58.3\%}.$$

Check

The back-work ratio of 58 % is high — characteristic of a low-efficiency gas turbine whose modest pressure ratio (4.75) and component irreversibilities leave little margin. Cold-air-standard constant specific heats are assumed; using temperature-dependent air-table $h$ and $p_r$ would raise $\eta$ by roughly 2–3 points but leaves the conclusions unchanged.

Quantity	Result
$T_2$ (compressor exit)	493.6 K (220 °C) @ 475 kPa
$T_3$ (turbine inlet)	1143 K (870 °C) @ 461.3 kPa
$T_4$ (turbine exit)	799.3 K (526 °C) @ 100 kPa
Compressor / turbine work	201.5 / 345.6 kJ/kg
Net work	≈ 144 kJ/kg
Thermal efficiency	≈ 22.1 %
Turbine work driving compressor	≈ 58.3 %

Concept & Theory

Topic. Non-ideal (irreversible) air-standard Brayton cycle with a combustor pressure drop.

Key relations. Isentropic $T_s/T=(P_s/P)^{(k-1)/k}$; compressor $\eta_c=(T_{2s}-T_1)/(T_2-T_1)$; turbine $\eta_t=(T_3-T_4)/(T_3-T_{4s})$; $w=c_p\Delta T$; back-work ratio $r_{bw}=w_c/w_t$; $\eta_\text{th}=w_\text{net}/q_\text{in}$.

Why this works. Each isentropic reference temperature comes purely from its pressure ratio; the efficiencies then map the ideal temperature change onto the real one (more work into the compressor, less out of the turbine). The duct loss matters only through the reduced turbine expansion ratio $P_3/P_4$. Because so much of the turbine output merely runs the compressor, the net work — and hence efficiency — is very sensitive to the component efficiencies.

Common pitfall. Expanding the turbine from 475 kPa instead of the reduced 461.3 kPa, or applying the compressor efficiency as a multiplier ($\eta_c\Delta T_s$) rather than a divisor.

Source. Çengel & Boles, Thermodynamics §9 (gas power cycles — deviation of actual gas-turbine cycles from idealized ones).

Question 3 of 8: Non-Ideal Brayton Cycle with Duct Pressure Drop

Question 3: Non-Ideal Brayton Cycle with Duct Pressure Drop (equal value)