Air-Standard Brayton Cycle — Two-Stage Turbine with Reheat, Variable Specific Heats: 04-BS-10, December 2014, Q2

Question 2: Air-Standard Brayton Cycle — Two-Stage Turbine with Reheat, Variable Specific Heats (20 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. Compressor inlet (state 1): $T=300$ K, $P=100$ kPa; compressor exit (state 2): $P=400$ kPa, $\eta_c=0.90$. Turbine 1 inlet (state 3): $P=400$ kPa, $T=1200$ K; expands to reheat pressure 200 kPa. Turbine 2 inlet (state 5, after reheat): $T=1200$ K at 200 kPa; expands to 100 kPa (state 6). Each turbine stage $\eta_t=0.88$. Variable specific heats (air evaluated via high-accuracy ideal-gas-limit properties in place of a printed $s^\circ(T)$ table). $T_0=300$ K, taken as the heat-rejection reference for the reversible-work comparison. Note: the source question lists the second determine-item as "(a)" a second time (typo); it is treated here as part (b) thermal efficiency, following on from part (a) net work, with the remaining two items as (c) net reversible work and (d) second-law efficiency.

Find. (a) $w_{net}$ [kJ/kg]; (b) $\eta_{th}$; (c) $w_{rev}$ [kJ/kg]; (d) $\eta_{II}$.

Fig. Q2 — T–s state points for the two-stage reheated Brayton cycle (1→2 compressor, 2→3 combustor, 3→4 turbine 1, 4→5 reheater, 5→6 turbine 2). No saturation dome for an ideal-gas working fluid.

Approach

Use variable-specific-heat air properties (an ideal-gas-limit equation of state standing in for the classical printed $s^\circ(T)$/air table) with root-finding on temperature to satisfy each isentropic pressure ratio, then apply the given component efficiencies to get the actual states. Heat is added in two places (the primary combustor 2→3 and the reheater 4→5); net work and both efficiency figures follow directly, with the second-law comparison referencing the same $T_0=300$ K used to reject heat in an equivalent reversible engine.

Compressor (1→2). Inlet $h_1=426.30$ kJ/kg (300 K). Isentropic exit at 400 kPa: $T_{2s}=171.4\ ^\circ$C, $h_{2s}=572.61$ kJ/kg, giving isentropic work $w_{c,s}=h_{2s}-h_1=146.31$ kJ/kg. Actual: $$w_{c,a}=\frac{w_{c,s}}{\eta_c}=\frac{146.31}{0.90}=\boxed{162.56\ \text{kJ/kg}},\qquad T_2=187.4\ ^\circ\text{C},\ \ h_2=h_1+w_{c,a}=588.86\ \text{kJ/kg}.$$
Turbine 1 (3→4), 400→200 kPa. Isentropic exit $T_{4s}=737.6\ ^\circ$C, giving $w_{t1,s}=219.4$ kJ/kg. Actual: $$w_{t1,a}=\eta_t\,w_{t1,s}=0.88\times219.4=\boxed{193.11\ \text{kJ/kg}},\qquad T_4=760.6\ ^\circ\text{C},\ \ h_4=1211.10\ \text{kJ/kg}.$$
Turbine 2 (5→6), 200→100 kPa. Because the second stage starts at the same inlet temperature (1200 K) and has the same pressure ratio (2) as the first stage, it does identical work: $T_{6s}=737.6\ ^\circ$C, $w_{t2,s}=219.4$ kJ/kg, and $$w_{t2,a}=\eta_t\,w_{t2,s}=\boxed{193.11\ \text{kJ/kg}},\qquad T_6=760.6\ ^\circ\text{C}.$$
Net specific work (part a). $$w_{net}=(w_{t1,a}+w_{t2,a})-w_{c,a}=386.22-162.56=\boxed{223.65\ \text{kJ/kg}}.$$
Total heat input and thermal efficiency (part b). Heat is added in the combustor (2→3) and reheater (4→5), with $h_3=h_5=1404.21$ kJ/kg (same $T=1200$ K): $$q_{in}=(h_3-h_2)+(h_5-h_4)=815.35+193.11=\boxed{1008.46\ \text{kJ/kg}}.$$ $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{223.65}{1008.46}=\boxed{0.2218\ (22.2\%)}.$$
Net reversible work (part c). The maximum work extractable from the same heat input $q_{in}$, referenced to the dead state $T_0=300$ K, is its exergy content. The question does not name a source reservoir, so (stating the assumption as the paper's Note 1 requires) the heat is taken to be supplied from a reservoir at the cycle's maximum temperature, $T_H=1200$ K: $$w_{rev}=q_{in}\left(1-\frac{T_0}{T_H}\right)=1008.46\times\left(1-\frac{300}{1200}\right)=\boxed{756.34\ \text{kJ/kg}}.$$
Second-law efficiency (part d). $$\eta_{II}=\frac{w_{net}}{w_{rev}}=\frac{223.65}{756.34}=\boxed{0.2957\ (29.6\%)}.$$

Quantity	Result
(a) $w_{net}$	223.65 kJ/kg
(b) $\eta_{th}$	0.2218 (22.2%)
(c) $w_{rev}$	756.34 kJ/kg
(d) $\eta_{II}$	0.2957 (29.6%)

Concept & Theory

Topic. Air-standard Brayton cycle with a single compressor stage, two-stage turbine expansion with reheat between stages, and variable specific heats.

Key relations. Variable-cp isentropic process: $s^\circ(T_2)-s^\circ(T_1)=R\ln(P_2/P_1)$, solved for $T_2$ by root-finding rather than a closed-form $T_2/T_1=(P_2/P_1)^{(k-1)/k}$ (which assumes constant $k$). Exergy of a heat input from a source at $T_H$: $x_{in}=q_{in}(1-T_0/T_H)$.

Why this works. Because both turbine stages share the same inlet temperature (1200 K) and the same pressure ratio (2), they perform identical work — solving one stage and reusing the result avoids a redundant second root-find. Reheat raises the average temperature of the second heat-addition step and boosts turbine work, but (unlike a regenerative cycle) it does not recover any of the compressor-exit-to-combustor temperature gap, which is why this cycle's thermal efficiency is markedly lower than a regenerative version of the same hardware would achieve.

Common pitfall. Assuming this arrangement has a regenerator (heat exchanger recovering turbine-exhaust heat to preheat the compressor discharge) by analogy with similar-looking gas-turbine problems — the question describes only a two-stage turbine with reheat, no regenerator, so all of $h_3-h_2$ must come from the combustor at full fuel cost, not from a preheated air stream.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 9 (Gas Power Cycles) & Ch. 8 (Exergy).

Question 2 of 9: Air-Standard Brayton Cycle — Two-Stage Turbine with Reheat, Variable Specific Heats

Question 2: Air-Standard Brayton Cycle — Two-Stage Turbine with Reheat, Variable Specific Heats (20 marks)

Approach