Question 1 of 6: Gas Turbine Modular Helium Reactor

Question 1: Gas Turbine Modular Helium Reactor (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. A closed helium Brayton plant with two-stage intercooled compression, a gas-to-gas recuperator, a nuclear heat source and a single turbine, operating between the terminal conditions listed on the question page.

Find. The temperature at each of the eight numbered points, the thermodynamic cycle efficiency, and the helium and cooling-water flow rates that deliver the specified 250 MW.

[Figure not reproduced: Figure 1.1 — the page-8 gas circuit redrawn, with the eight numbered state points used throughout the answer. See the official exam paper.]

Approach. Treat helium as an ideal gas of constant specific heats, apply the isentropic pressure–temperature relation with the stated machine efficiency to each of the two compressions and the one expansion, close the recuperator with its equal terminal differences, and only then scale the specific quantities up to the required 250 MW.

1. Part (a) — fix the gas properties before anything else. The constants sheet on page 13 gives both specific heats, so the ratio and the exponent follow directly:$$k=\frac{c_p}{c_v}=\frac{5.193}{3.116}=1.6666,\qquad R=c_p-c_v=5.193-3.116=2.077\ \text{kJ/kg}\cdot\text{K}$$$$\boxed{\frac{k-1}{k}=\frac{0.6666}{1.6666}=0.4000}$$The value is worth a moment's check, because every temperature in part (b) hangs off it. Helium is monatomic, so the theoretical ratio is exactly 5/3 = 1.6667, and the gas constant from the universal value is $R_o/M = 8.314/4.003 = 2.077$ kJ/kg·K — both agree with the sheet.
2. Part (a) — read the cycle off the sketch. Figure 1.2 above is the required diagram. Four features earn the marks. The two compressions 1–2 and 3–4 lean to the right of the vertical because the compressor is 90 % efficient, and so does the expansion 6–7 at 85 %. The intercooling 2–3 and the precooling 8–1 run along the 4 MPa and 2 MPa isobars back to 30 °C. The recuperator legs 4–5 (cold side, at 7 MPa) and 7–8 (hot side, at 2 MPa) are drawn parallel, separated vertically by exactly the 20 °C terminal difference at each end. Because the same helium at the same $\dot{m}c_p$ flows on both sides of the recuperator, those two legs must span the same temperature interval — that observation is what makes part (b) solvable in one line rather than by iteration.
3. Part (b) — low-pressure compression, 1 to 2. The isentropic relation between the two states gives the ideal delivery temperature, and the compressor efficiency then stretches the real temperature rise:$$T_{2s}=T_1\left(\frac{p_2}{p_1}\right)^{(k-1)/k}=303.15\left(\frac{4}{2}\right)^{0.4000}=400.0\ \text{K}$$$$T_2=T_1+\frac{T_{2s}-T_1}{\eta_c}=303.15+\frac{400.0-303.15}{0.90}=410.8\ \text{K}$$so $t_{2s}=126.8$ and $\boxed{t_2=137.6\ ^\circ\text{C}}$. Note that the efficiency is applied to the temperature rise, not to the absolute temperature.
4. Part (b) — high-pressure compression, 3 to 4. The intercooler returns the gas to 30 °C, so the second stage starts from the same inlet temperature as the first but works on the smaller pressure ratio 7/4:$$T_{4s}=303.15\left(\frac{7}{4}\right)^{0.4000}=379.2\ \text{K},\qquad T_4=303.15+\frac{379.2-303.15}{0.90}=387.6\ \text{K}$$giving $t_{4s}=106.0$ and $\boxed{t_4=114.5\ ^\circ\text{C}}$. Intercooling is what keeps this figure low: a single stage from 2 to 7 MPa would leave the gas near 250 °C, and every degree of that would have to be paid for in compressor work.
5. Part (b) — expansion through the turbine, 6 to 7. The turbine drops the full 7 MPa to 2 MPa from the reactor outlet temperature of 850 °C = 1123.15 K:$$T_{7s}=1123.15\left(\frac{2}{7}\right)^{0.4000}=680.5\ \text{K},\qquad T_7=T_6-\eta_t\,(T_6-T_{7s})=1123.15-0.85\,(1123.15-680.5)=746.9\ \text{K}$$so $t_{7s}=407.4$ and $\boxed{t_7=473.8\ ^\circ\text{C}}$. Here the efficiency multiplies the drop, because a real turbine delivers less than the isentropic machine; the exhaust is correspondingly hotter, which is precisely what makes recuperation worth doing.
6. Part (b) — close the recuperator with its terminal differences. At the hot end the turbine exhaust enters at $t_7$ and the cold stream leaves 20 °C below it; at the cold end the compressed gas enters at $t_4$ and the hot stream leaves 20 °C above it:$$t_5=t_7-20=473.8-20=453.8\ ^\circ\text{C},\qquad t_8=t_4+20=114.5+20=134.5\ ^\circ\text{C}$$$$\boxed{t_5=453.8\ ^\circ\text{C}\ \text{(reactor inlet)},\qquad t_8=134.5\ ^\circ\text{C}\ \text{(precooler inlet)}}$$The consistency check is immediate and should always be made: the hot-side drop $t_7-t_8 = 339.3$ K equals the cold-side rise $t_5-t_4 = 339.3$ K. That equality is not a coincidence — both sides carry the same mass flow of the same gas, so equal terminal differences force equal temperature spans. If the two numbers do not match, a machine efficiency has been applied in the wrong direction.
7. Part (c) — specific work and heat, then the cycle efficiency. Working per kilogram of helium keeps the arithmetic clean:$$w_t=c_p\,(T_6-T_7)=5.193\,(1123.15-746.90)=1953.9\ \text{kJ/kg}$$$$w_{c1}=5.193\,(410.75-303.15)=558.8\ \text{kJ/kg},\qquad w_{c2}=5.193\,(387.65-303.15)=438.8\ \text{kJ/kg}$$so the net work is $w_{net}=1953.9-558.8-438.8=956.3$ kJ/kg. The reactor supplies only the heat between the recuperator outlet and the turbine inlet:$$q_{in}=c_p\,(T_6-T_5)=5.193\,(1123.15-726.90)=2057.7\ \text{kJ/kg}$$$$\boxed{\eta_{cycle}=\frac{w_{net}}{q_{in}}=\frac{956.3}{2057.7}=0.4647\ \text{or}\ 46.5\,\%}$$The back-work ratio is 998/1954 = 0.51, which is high but normal for helium; the Carnot bound between 30 and 850 °C is 73.0 %, so the plant realises about two-thirds of the theoretical ceiling.
8. Part (d) — scale to the required output. The question states that there are no mechanical or electrical losses, so the electrical output equals the net shaft work:$$\dot{m}_{He}=\frac{P_e}{w_{net}}=\frac{250\,000}{956.3}\qquad\Rightarrow\qquad \boxed{\dot{m}_{He}=261.4\ \text{kg/s}}$$A quick sanity check on the size of the machine: at 7 MPa and 850 °C the helium density is about 3.7 kg/m³, so this flow is roughly 70 m³/s at turbine inlet — a large but entirely conventional single-casing machine.
9. Part (e) — heat rejection, and the cooling water it needs. Heat leaves the circuit in two places, the precooler (8 to 1) and the intercooler (2 to 3):$$\dot{Q}_{pre}=\dot{m}_{He}c_p\,(T_8-T_1)=261.4\times 5.193\times(407.65-303.15)=141\,866\ \text{kW}$$$$\dot{Q}_{ic}=\dot{m}_{He}c_p\,(T_2-T_3)=261.4\times 5.193\times(410.75-303.15)=146\,094\ \text{kW}$$$$\boxed{\dot{Q}_{rej}=141\,866+146\,094=287\,960\ \text{kW}}$$The whole-plant first law confirms it exactly: $\dot{m}_{He}q_{in}-P_e = 537\,960-250\,000=287\,960$ kW. Because the terminal difference is 15 °C at both ends of each rejection exchanger, the water temperature change must equal the helium temperature change, and the flow collapses to a single expression:$$\dot{m}_w=\frac{\dot{m}_{He}c_{p,He}\Delta T_{He}}{c_{p,w}\Delta T_{He}}=\dot{m}_{He}\frac{c_{p,He}}{c_{p,w}}=261.4\times\frac{5.193}{4.19}=324.0\ \text{kg/s}$$in each exchanger, whatever their duties. The water enters both at $30-15=15$ °C, leaving the precooler at $134.5-15=119.5$ °C and the intercooler at $137.6-15=122.6$ °C. Hence$$\boxed{\dot{m}_{w,total}=2\times 324.0=648.0\ \text{kg/s}=0.648\ \text{m}^3\text{/s}}$$