22-Mec-B3 Energy Conversion and Power Generation · December 2014

Question 1 of 6: Gas Turbine Modular Helium Reactor

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

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 cycle with two-stage intercooled compression, a recuperator, a nuclear heat source and a single turbine, delivering 250 MW of electricity with no mechanical or electrical losses and negligible circuit pressure loss.

Find. The temperature at every numbered state, the thermodynamic cycle efficiency, the helium circulation rate that yields 250 MW, and the cooling-water flow that carries away the rejected heat.

Approach. Fix each compression and expansion with the isentropic relation for an ideal gas corrected by the machine efficiency, close the recuperator with the stated terminal differences, then apply the steady-flow energy equation to the whole plant to get efficiency, flow rate and heat rejection.

1. Part (a) — the temperature–entropy sketch. Figure 1.1 above is the required diagram: the two compressions 1→2 and 3→4 with intercooling 2→3 between them, recuperative heating 4→5, reactor heating 5→6, expansion 6→7, recuperative cooling 7→8 and precooling 8→1, with the 20 K recuperator terminal differences marked at both ends and the three isobars shown dashed. It is plotted from the temperatures computed in part (b) rather than sketched freehand, so the relative lengths of the legs are to scale — which is what makes the intercooling and recuperation savings visible at a glance.
2. Establish the helium property group. The specific-heat ratio and the isentropic exponent follow directly from the constants on page 15: $$k = \frac{c_p}{c_v} = \frac{5.193}{3.117} = 1.666, \qquad \frac{k-1}{k} = 0.3998$$ The gas constant \(R = c_p - c_v = 2.076\ \text{kJ}/\text{kg}\cdot\text{K}\) confirms the identification — helium is monatomic, so \(k \to 5/3\) as it must.
3. Part (b) — low-pressure compression, 1 → 2 (2 MPa to 4 MPa). The isentropic end state comes from the reference equation on page 16, \(T_2/T_1 = (p_2/p_1)^{(k-1)/k}\): $$T_{2s} = 303.15\left(\frac{4}{2}\right)^{0.3998} = 399.9\ \text{K} = 126.8\ ^{\circ}\text{C}$$ The machine efficiency then places the real state above it, because the irreversibility appears as extra temperature rise: $$T_2 = T_1 + \frac{T_{2s}-T_1}{\eta_c} = 303.15 + \frac{96.75}{0.90} = \boxed{410.7\ \text{K} = 137.6\ ^{\circ}\text{C}}$$
4. Part (b) — high-pressure compression, 3 → 4 (4 MPa to 7 MPa). The gas re-enters the second stage at 30 °C, which is the whole point of the intercooler. Working through the same pair of relations with a pressure ratio of 1.75, $$T_{4s} = 303.15\left(\frac{7}{4}\right)^{0.3998} = 379.2\ \text{K}, \qquad T_4 = 303.15 + \frac{76.01}{0.90} = \boxed{387.6\ \text{K} = 114.5\ ^{\circ}\text{C}}$$ Notice that the second stage raises the temperature by only 84.5 K against the first stage's 107.6 K even though it handles a comparable pressure ratio — that is the intercooling saving, and it is visible on Figure 1.1 as the shorter 3–4 leg.
5. Part (b) — turbine expansion, 6 → 7 (7 MPa to 2 MPa). Expansion runs the same relation backwards, and here the efficiency reduces the temperature drop: $$T_{7s} = 1123.15\left(\frac{2}{7}\right)^{0.3998} = 680.7\ \text{K}, \qquad T_7 = T_6 - \eta_t\,(T_6 - T_{7s})$$ $$T_7 = 1123.15 - 0.85\,(442.4) = \boxed{747.1\ \text{K} = 473.9\ ^{\circ}\text{C}}$$
6. Part (b) — close the recuperator on its terminal differences. The turbine exhaust heats the compressor delivery in counter-flow, and the question fixes a 20 °C difference at each end. At the hot end the cold stream leaves 20 K below the entering hot stream; at the cold end the hot stream leaves 20 K above the entering cold stream: $$T_5 = T_7 - 20 = \boxed{727.1\ \text{K} = 453.9\ ^{\circ}\text{C}}, \qquad T_8 = T_4 + 20 = \boxed{407.6\ \text{K} = 134.5\ ^{\circ}\text{C}}$$ This pair is only self-consistent because both sides of the recuperator carry the same mass flow of the same gas, so equal terminal differences force equal temperature changes. That is worth checking rather than assuming: the hot side falls \(747.1 - 407.6 = 339.5\) K and the cold side rises \(727.1 - 387.6 = 339.5\) K, so the recuperator energy balance closes exactly.
7. Assemble the specific work terms. With constant specific heats every energy term is \(c_p\,\Delta T\): $$w_t = c_p(T_6 - T_7) = 5.193(376.06) = 1952.9\ \text{kJ/kg}$$ $$w_{c1} = 5.193(107.55) = 558.5\ \text{kJ/kg}, \qquad w_{c2} = 5.193(84.45) = 438.5\ \text{kJ/kg}$$ The compressors absorb just over half the turbine output, which is characteristic of a gas cycle and is the reason the back-work ratio matters so much in this plant. The net specific work is $$w_{net} = 1952.9 - 558.5 - 438.5 = 955.9\ \text{kJ/kg}$$
8. Cycle efficiency (part c). The reactor only has to lift the gas from the recuperator outlet, state 5, to the turbine inlet, state 6: $$q_{in} = c_p(T_6 - T_5) = 5.193(396.06) = 2056.9\ \text{kJ/kg}$$ $$\eta_{th} = \frac{w_{net}}{q_{in}} = \frac{955.9}{2056.9} = \boxed{0.465 = 46.5\ \%}$$ That figure is a useful sanity check in its own right: the published design efficiency of the GT-MHR is in the high forties, so a hand calculation landing at 46.5 % with realistic machine efficiencies is behaving properly.
9. Helium flow rate (part d). There are no mechanical or electrical losses, so the net cycle work is delivered whole to the terminals: $$\dot{M}_{He} = \frac{P_e}{w_{net}} = \frac{250\,000}{955.9} = \boxed{261.5\ \text{kg/s}}$$
10. Rate of heat rejection (part e, first half). Heat leaves the circuit in two exchangers, the precooler between states 8 and 1 and the intercooler between states 2 and 3: $$\dot{Q}_{pre} = \dot{M}_{He}\,c_p (T_8 - T_1) = 261.5 \times 5.193 \times 104.45 = 141.9\ \text{MW}$$ $$\dot{Q}_{ic} = \dot{M}_{He}\,c_p (T_2 - T_3) = 261.5 \times 5.193 \times 107.55 = 146.1\ \text{MW}$$ $$\dot{Q}_{rej} = 141.9 + 146.1 = 288.0\ \text{MW}$$ The whole-plant first law is the check that this is right: \(\dot{M}_{He} q_{in} - P_e = 538.0 - 250.0 = 288.0\) MW, which agrees exactly.
11. Cooling-water flow (part e, second half). A 15 °C terminal difference at each end of a rejection exchanger fixes the water inlet at \(30 - 15 = 15\ ^{\circ}\text{C}\) and the water outlet 15 K below the entering helium. Equal terminal differences at both ends again force the water temperature change to equal the helium temperature change, so each exchanger needs $$\dot{M}_w = \frac{\dot{M}_{He} c_{p,He}}{c_{p,w}} = \frac{261.5 \times 5.193}{4.190} = 324.1\ \text{kg/s}$$ Explicitly, for the precooler the water runs 15 → 119.5 °C and \(\dot{M}_w = 141\,890/(4.190 \times 104.45) = 324.1\) kg/s; for the intercooler it runs 15 → 122.6 °C for the same 324.1 kg/s. The total requirement is $$\dot{M}_{w,total} = 2 \times 324.1 = \boxed{648.2\ \text{kg/s} \approx 0.648\ \text{m}^3/\text{s}}$$