Simple Brayton Cycle, Constant Specific Heats: 04-BS-10, May 2018, Q7

Question 7: Simple Brayton Cycle, Constant Specific Heats (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $T_1=310$ K (compressor inlet), $T_3=900$ K (turbine inlet), $r_p=P_2/P_1=8$, $\dot m=460$ kg/s, $\eta_c=0.80$, $\eta_t=0.86$. Constant specific heats at room temperature: $c_p=1.005$ kJ/kg·K, $k=1.4$. Simple (non-regenerative) cycle.

Find. $\dot W_{net}$ [MW].

Approach

With constant specific heats the isentropic temperature ratios reduce to closed-form power-law expressions in $r_p$; compute the isentropic compressor/turbine work directly, divide/multiply by the respective isentropic efficiencies to get actual work, and take the difference weighted by the mass flow rate.

Isentropic compressor exit and actual work. $$T_{2s}=T_1\,r_p^{(k-1)/k}=310\times8^{0.2857}=\boxed{561.6\ \text{K}}.$$ $$w_{c,s}=c_p(T_{2s}-T_1)=1.005\times(561.6-310)=252.8\ \text{kJ/kg}.$$ $$w_{c,a}=\frac{w_{c,s}}{\eta_c}=\frac{252.8}{0.80}=\boxed{316.0\ \text{kJ/kg}}.$$
Isentropic turbine exit and actual work. $$T_{4s}=\frac{T_3}{r_p^{(k-1)/k}}=\frac{900}{8^{0.2857}}=\boxed{496.8\ \text{K}}.$$ $$w_{t,s}=c_p(T_3-T_{4s})=1.005\times(900-496.8)=405.2\ \text{kJ/kg}.$$ $$w_{t,a}=\eta_t\,w_{t,s}=0.86\times405.2=\boxed{348.5\ \text{kJ/kg}}.$$
Net specific work and net power output. $$w_{net}=w_{t,a}-w_{c,a}=348.5-316.0=32.4\ \text{kJ/kg}.$$ $$\dot W_{net}=\dot m\,w_{net}=460\times32.4=14{,}924\ \text{kW}=\boxed{14.9\ \text{MW}}.$$

Quantity	Result
$\dot W_{net}$	14.9 MW

Concept & Theory

Topic. Simple (non-regenerative) Brayton cycle with non-ideal compressor and turbine, evaluated with the cold-air-standard (constant specific heat) assumption.

Key relations. Constant-$k$ isentropic temperature ratio: $T_{2s}/T_1=r_p^{(k-1)/k}$, $T_{4s}/T_3=r_p^{-(k-1)/k}$. Actual work from isentropic efficiency: $w_{c,a}=w_{c,s}/\eta_c$, $w_{t,a}=\eta_t w_{t,s}$. Net specific work $w_{net}=w_{t,a}-w_{c,a}$.

Why this works. "Using constant specific heats at room temperature" is an explicit instruction to use the cold-air-standard shortcut ($c_p$, $k$ fixed at their 300 K values) rather than the variable-property air tables used elsewhere on this paper (e.g. Q4 and Q6) — this simplifies every isentropic step to a closed-form power-law temperature ratio instead of a root-solve against the standard-state entropy function.

Common pitfall. Forgetting that with non-ideal (non-isentropic) components, the compressor work must be divided by $\eta_c$ (it costs more actual work to reach the same pressure ratio) while the turbine work must be multiplied by $\eta_t$ (it delivers less actual work than the isentropic ideal) — swapping these operations is a very common sign error.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 9 (Brayton Cycle, cold-air-standard).

Question 7 of 9: Simple Brayton Cycle, Constant Specific Heats

Question 7: Simple Brayton Cycle, Constant Specific Heats (15 marks)

Approach