Air-Standard Two-Turbine Cycle with a Gas-to-Gas Heat Exchanger: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, Undated paper, Q3

Question 3: Air-Standard Two-Turbine Cycle with a Gas-to-Gas Heat Exchanger

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. The MAIN air stream (1→2→3→4) runs through turbine 1, the COLD side of the heat exchanger, and turbine 2; a SEPARATE hot air stream (5→6) supplies the heat exchanger's HOT side and is not otherwise connected to the main stream. Cold-air-standard properties are used throughout ($c_p=1.005$ kJ/kg·K, $k=1.4$) since only ordinary air is named and no gas table accompanies this paper.

Given data
State	Stream	$T$	$P$
1	main, turbine 1 in	1400 K	20 bar
2	main, turbine 1 out / HX cold-in	1100 K	5 bar
3	main, HX cold-out / turbine 2 in	? (find)	4.5 bar
4	main, turbine 2 out	980 K	1 bar
5	hot, HX hot-in	1480 K	1.35 bar, 1200 kg/min
6	hot, HX hot-out	1200 K	1 bar

[Figure not reproduced: Two turbines and a gas-to-gas heat exchanger, as printed in the source exam. See the official exam paper or the cited reference text.]

As printed: the main stream runs turbine 1 → heat exchanger (cold side, 2→3) → turbine 2; a separate hot stream runs the exchanger's other coil, 5→6.

Find. The heat-exchanger exit temperature $T_3$, the power $\dot W_{t2}$, the isentropic efficiency of each turbine, and the heat exchanger's effectiveness.

Approach. Get the (unstated) main-stream mass flow from turbine 1's own energy balance, then use it with the hot stream's known duty ($\dot m_5,T_5,T_6$) to close the heat exchanger's energy balance for $T_3$; turbine 2's power follows from its own energy balance once $T_3$ is known. Isentropic efficiency for each turbine compares its actual $\Delta T$ to the ideal gas isentropic $\Delta T_s=T_{in}(P_{out}/P_{in})^{(k-1)/k}$ at its own pressure ratio; effectiveness compares the actual heat exchanger duty to the maximum possible for the two streams' capacity rates.

Main-stream mass flow from turbine 1. $$\dot W_{t1}=\dot m_{main}c_p(T_1-T_2)\ \Rightarrow\ \dot m_{main}=\frac{10{,}000}{1.005\times(1400-1100)}$$ $$\boxed{\dot m_{main}=33.17\text{ kg/s}}$$
Heat exchanger duty from the hot stream. $\dot m_5=1200/60=20.0$ kg/s: $$\dot Q_{HX}=\dot m_5c_p(T_5-T_6)=20.0\times1.005\times(1480-1200)$$ $$\boxed{\dot Q_{HX}=5628\text{ kW}}$$
Heat exchanger exit temperature $T_3$. Same duty heats the cold (main) stream from $T_2$: $$T_3=T_2+\frac{\dot Q_{HX}}{\dot m_{main}c_p}=1100+\frac{5628}{33.17\times1.005}$$ $$\boxed{T_3=1268.8\text{ K}\ (995.7\,{}^{\circ}\text{C})}$$
Power from turbine 2. $$\dot W_{t2}=\dot m_{main}c_p(T_3-T_4)=33.17\times1.005\times(1268.8-980)$$ $$\boxed{\dot W_{t2}=9628\text{ kW}}$$
Isentropic efficiency, turbine 1. $$T_{2s}=T_1\left(\frac{P_2}{P_1}\right)^{(k-1)/k}=1400\times\left(\frac{5}{20}\right)^{0.2857}=942.1\text{ K}$$ $$\eta_{t1}=\frac{T_1-T_2}{T_1-T_{2s}}=\frac{1400-1100}{1400-942.1}$$ $$\boxed{\eta_{t1}=0.655\ (65.5\%)}$$
Isentropic efficiency, turbine 2. $$T_{4s}=T_3\left(\frac{P_4}{P_3}\right)^{(k-1)/k}=1268.8\times\left(\frac{1}{4.5}\right)^{0.2857}=825.6\text{ K}$$ $$\eta_{t2}=\frac{T_3-T_4}{T_3-T_{4s}}=\frac{1268.8-980}{1268.8-825.6}$$ $$\boxed{\eta_{t2}=0.652\ (65.2\%)}$$ The two turbines' isentropic efficiencies agree to within 0.3 percentage points despite being derived from entirely independent pressure ratios — a strong check that the split-node reading above (state 3 built from the hot-stream energy balance) is the intended one.
Heat exchanger effectiveness. $C_{main}=33.17\times1.005=33.33$ kW/K, $C_{hot}=20.0\times1.005=20.10$ kW/K, so $C_{min}=C_{hot}$: $$\dot Q_{max}=C_{min}(T_{5}-T_{2})=20.10\times(1480-1100)=7638\text{ kW}$$ $$\varepsilon=\frac{\dot Q_{HX}}{\dot Q_{max}}=\frac{5628}{7638}$$ $$\boxed{\varepsilon=0.737\ (73.7\%)}$$

Schematic T–s trace (entropy shown relative, ideal-gas $\Delta s=c_p\ln(T_2/T_1)-R\ln(P_2/P_1)$): turbine 1 drops steeply 1→2, the cold side of the heat exchanger rises 2→3 while the hot side falls 5→6, and turbine 2 drops 3→4.

Question 3 — results
Quantity	Value
Main-stream mass flow	33.17 kg/s
Heat-exchanger exit temperature $T_3$	1268.8 K (995.7°C)
Turbine 2 power $\dot W_{t2}$	9628 kW
Isentropic efficiency, turbine 1 / turbine 2	65.5% / 65.2%
Heat-exchanger effectiveness $\varepsilon$	73.7%

Topic. Air-standard Brayton-type power system with a regenerative gas-to-gas heat exchanger fed by a separate hot stream.

Key relations. Ideal-gas steady-flow energy balance $\dot W=\dot mc_p\Delta T$; isentropic ideal-gas relation $T_{2s}/T_1=(P_2/P_1)^{(k-1)/k}$; heat-exchanger effectiveness $\varepsilon=\dot Q_{actual}/[C_{min}(T_{h,in}-T_{c,in})]$.

Why this works. No pressures are given for the hot 5→6 stream that would let its own isentropic behaviour be checked, and none is needed: the problem is fully closed by TWO independent energy balances (turbine 1 for $\dot m_{main}$, then the heat exchanger for $T_3$) before anything downstream is touched, which is why solving strictly in the printed order (mass flow → $T_3$ → $\dot W_{t2}$ → efficiencies → effectiveness) never stalls on a missing datum.

Common pitfall. Assuming the two streams mix physically inside the heat exchanger (they do not — it is two separate coils, confirmed by the figure) and trying to force a single combined state; and forgetting that turbine 2's pressure ratio uses $P_3=4.5$ bar, not $P_2=5$ bar, since the heat exchanger imposes its own (unstated) small pressure drop between 2 and 3.

Source. Çengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 9 (Brayton cycle, regeneration) & Ch. 5 (isentropic efficiency).

Question 3 of 8: Air-Standard Two-Turbine Cycle with a Gas-to-Gas Heat Exchanger

Question 3: Air-Standard Two-Turbine Cycle with a Gas-to-Gas Heat Exchanger