Question 2 of 8: ARC-100 Reactor with Gas Cycle

Question 2: ARC-100 Reactor with Gas Cycle (10 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, recuperated Brayton cycle on carbon dioxide, heated by the same sodium loop as Question 1:

Find. The ideal and actual compressor-discharge and turbine-exhaust temperatures, all four recuperator boundary temperatures, the cycle efficiency and net output, and the reason a low pressure ratio suits this duty.

Approach. Take the isentropic temperature ratio from the pressure ratio and the printed specific heats, apply the machine efficiencies to get the real end states, let the two 10 °C terminal differences fix the recuperator boundaries, and then form the efficiency from net work over the heat the reactor actually has to supply.

1. Fix the isentropic temperature ratio. With the paper's own specific heats, $k = c_p/c_v = 0.844/0.655 = 1.2886$ and $R = c_p - c_v = 0.189$ kJ/kg·K, so $(k-1)/k = 0.2239$. The pressure ratio is $r_p = 21/7 = 3$, hence $$\phi = r_p^{(k-1)/k} = 3^{0.2239} = 1.2789$$ Note how gentle that is: the same pressure ratio on air ($k = 1.4$) would give 1.369. The heavy triatomic molecule is the reason the cycle works at all, and it is the first half of the answer to part (g).
2. Part (b) — compressor outlet, ideal then actual. Isentropically $T_{2s} = T_1\phi = 305.15 \times 1.2789 = 390.26$ K, i.e. $\boxed{T_{2s} = 117.1\ {}^{\circ}\text{C}}$. The compressor efficiency is defined on the ideal work, $\eta_C = (T_{2s}-T_1)/(T_2-T_1)$, so the real discharge is hotter: $$T_2 = T_1 + \frac{T_{2s} - T_1}{\eta_C} = 305.15 + \frac{85.11}{0.90} = 399.72\ \text{K} = \boxed{126.6\ {}^{\circ}\text{C}}$$
3. Part (c) — turbine exhaust, ideal then actual. Expanding back through the same ratio, $T_{4s} = T_3/\phi = 763.15/1.2789 = 596.71$ K, i.e. $\boxed{T_{4s} = 323.6\ {}^{\circ}\text{C}}$. Here the efficiency is defined the other way round, $\eta_T = (T_3-T_4)/(T_3-T_{4s})$, so the real exhaust is hotter than the ideal one as well: $$T_4 = T_3 - \eta_T (T_3 - T_{4s}) = 763.15 - 0.90 \times 166.44 = 613.36\ \text{K} = \boxed{340.2\ {}^{\circ}\text{C}}$$ Both irreversibilities push temperature up, and both therefore hurt: one raises the work the compressor demands, the other lowers the work the turbine delivers.
4. Part (d) — the four recuperator boundaries. The question defines the terminal differences explicitly. The cold stream leaves 10 °C below the hot stream's inlet, and the hot stream leaves 10 °C above the cold stream's inlet:$$T_{2'} = T_4 - 10 = 340.2 - 10 = \boxed{330.2\ {}^{\circ}\text{C}} \qquad T_{4'} = T_2 + 10 = 126.6 + 10 = \boxed{136.6\ {}^{\circ}\text{C}}$$ so the cold side runs 126.6 → 330.2 °C and the hot side 340.2 → 136.6 °C. Two checks fall out at once. The same gas, at the same flow and the same specific heat, passes both sides, so the two temperature changes must be equal — and they are, 203.6 K each. And the question prints a ‘return temperature’ of 331 °C and a ‘reject temperature’ of 136 °C, which are exactly these two results to the rounding of the paper. That agreement validates the whole chain from part (b) onward.
5. Part (e) — specific work and cycle efficiency. With constant specific heat every duty is $c_p\Delta T$: $w_T = 0.844 \times (763.15 - 613.36) = 126.43$ kJ/kg, $w_C = 0.844 \times (399.72 - 305.15) = 79.82$ kJ/kg, so the net specific work is $w_{net} = 46.61$ kJ/kg and the back-work ratio is $w_C/w_T = 0.631$. The reactor only has to supply the gap between the recuperator outlet and the turbine inlet, $q_{in} = 0.844 \times (763.15 - 603.36) = 134.87$ kJ/kg, therefore $$\eta_{th} = \frac{w_{net}}{q_{in}} = \frac{46.61}{134.87} = \boxed{34.6\ \%}$$ The energy balance closes independently: $q_{rej} = 0.844 \times (409.72 - 305.15) = 88.26$ kJ/kg and $q_{in} - q_{rej} = 46.61$ kJ/kg, the net work exactly.
6. Part (f) — net output, and the cross-check the reactor rating gives. At the stated flow, $$\dot{W}_{net} = \dot{m}\,w_{net} = 1937 \times 46.61 = 90\,280\ \text{kW} = \boxed{90.3\ \text{MW}}$$ with a gross turbine output of 244.9 MW against 154.6 MW absorbed by the compressor, and a recuperator duty of 332.9 MW — larger than the reactor itself, which is why the recuperator is the biggest component on the page-12 layout. The decisive confirmation is the heat input: $\dot{m}\,q_{in} = 1937 \times 134.87 = 261.2$ MW, which reproduces the 260 MW core rating of Question 1 to within 0.5 %. The mass flow the paper prints was chosen to match this reactor, so if that product had not landed on 260 MW an efficiency would have been applied in the wrong direction.
7. Part (g) — quantify why a low pressure ratio is right here. For an ideal regenerative cycle the efficiency is $$\eta_{regen} = 1 - \frac{T_1}{T_3}\,r_p^{(k-1)/k}$$ which falls as $r_p$ rises — the exact opposite of the simple cycle. At $r_p = 3$ it gives 48.9 %; at $r_p = 6$ only 40.3 %. The reason is visible on the T-s sketch: regeneration is only possible while the turbine exhaust is hotter than the compressor discharge, and raising $r_p$ closes that window from both ends at once.

Beyond the algebra, four practical arguments make the low ratio the right choice for a small modular reactor:

• It is what makes the recuperator worth having. At $r_p = 3$ the exhaust leaves at 340 °C while the compressor discharges at only 127 °C, so 203 K of the 361 K total heating is recovered internally and the reactor supplies only the top 160 K. Strip the recuperator out of this same machine and the efficiency collapses from 34.6 % to 15.2 %.
• The compression is cheap because the cycle sits near the critical point. Carbon dioxide is critical at 7.38 MPa and 31.0 °C, and the compressor inlet is 7 MPa / 32 °C — deliberately just past it, where the fluid is dense and the work to raise its pressure is small. A high pressure ratio would drive the compressor well away from that region and the back-work ratio, already 0.63, would climb further.
• The machinery becomes small and simple. A ratio of three needs a handful of stages rather than a fifteen-stage axial compressor, and because the whole cycle runs between 7 and 21 MPa the volumetric flow is tiny — a 90 MW turbomachine of this type is roughly the size of a desk. For a modular reactor that has to be factory-built and shipped, that is the point.
• It suits the heat source. The sodium delivers heat over a narrow band near 500 °C. A low-ratio recuperated cycle asks for its heat over a narrow band too (330 → 490 °C), so the sodium–carbon dioxide exchanger works against a small, nearly uniform temperature difference and stays compact.