ARC-100 Reactor with Steam Cycle: 22-Mec-B3 Energy Conversion and Power Generation, December 2019, Q1

Question 1: ARC-100 Reactor with Steam 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. The reactor, the two sodium loops and the water cycle, all adiabatic to the surroundings:

Quantity	Symbol	Value
Reactor core thermal power	$\dot{Q}$	260 MW
Primary sodium, reactor inlet / outlet	$T$	355 / 510 °C
Secondary sodium, exchanger inlet / outlet	$T$	355 / 500 °C
Sodium specific heat	$c_{p,\mathrm{Na}}$	1.230 kJ/kg°C
Turbine inlet	$p_3,\ T_3$	5 MPa, 490 °C
Turbine exhaust pressure	$p_4$	0.005 MPa
Feedwater pump outlet temperature	$T_2$	33 °C
Turbine internal efficiency	$\eta_T$	85 %

Find. The secondary sodium and steam mass flow rates, the enthalpy at every numbered state of the cycle, the turbine power, the thermodynamic efficiency, and a reasoned list of the improvements available to this plant.

Part (a). State 1 saturated liquid in the condenser hotwell at 0.005 MPa; 1→2 feed pumping to 5 MPa (33 °C at outlet); 2→3 sodium-heated steam generation to 490 °C; 3→4 expansion in the turbine, with the vertical dashed line the isentropic end state 4s and the sloping solid line the actual expansion at 85 % internal efficiency; 4→1 condensation. The cycle carries no feedwater heating and no reheat, which is exactly what part (f) is about.

Approach. Close a steady-flow energy balance on each sodium loop to size its flow, close the same balance on the steam generator to size the steam flow, then work the four cycle states from the steam tables with the turbine efficiency applied to the isentropic enthalpy drop.

Part (b) — size the secondary sodium loop from the core duty. All 260 MW crosses each heat exchanger without loss, so for a single-phase liquid metal $\dot{Q} = \dot{m}_{\mathrm{Na}}\,c_{p,\mathrm{Na}}\,(T_{\mathrm{hot}} - T_{\mathrm{cold}})$. The secondary loop swings 500 − 355 = 145 °C, so $$\dot{m}_{\mathrm{Na,2}} = \frac{260\,000}{1.230 \times 145} = \boxed{1458\ \text{kg/s}}$$ As a check on the reading of the two loop temperatures, the primary loop swings 510 − 355 = 155 °C and therefore carries $260\,000/(1.230 \times 155) = 1364$ kg/s — a smaller flow across a larger temperature difference, as it must be for the same duty.
Fix the two ends of the water cycle before sizing the steam flow. At the condenser pressure of 0.005 MPa the saturation temperature is 32.87 °C, with $h_f = 137.75$ kJ/kg, $h_{fg} = 2422.98$ kJ/kg, $s_f = 0.4762$ and $s_{fg} = 7.9176$ kJ/kg·K. State 1 is that saturated liquid, so $h_1 = 137.75$ kJ/kg. The feed pump raises it to 5 MPa; because liquid water is nearly incompressible the pump work is $w_p = v_f\,\Delta p = 0.0010053 \times (5000 - 5) = 5.02$ kJ/kg and $h_2 = 137.75 + 5.02 = 142.77$ kJ/kg. That enthalpy at 5 MPa corresponds to 33.0 °C, which reproduces the feedwater pump outlet temperature the question prints — a free confirmation that the condenser pressure and the pump duty are consistent.
Part (b) — size the steam flow from the steam-generator balance. At 5 MPa and 490 °C the superheat table gives $h_3 = 3411.3$ kJ/kg and $s_3 = 6.9477$ kJ/kg·K. The steam generator must take the feedwater from state 2 to state 3 using the whole core output, so $$\dot{m}_s = \frac{\dot{Q}}{h_3 - h_2} = \frac{260\,000}{3411.3 - 142.77} = \boxed{79.55\ \text{kg/s}}$$
Part (c) — the isentropic end state of the expansion. An ideal turbine would hold $s_{4s} = s_3 = 6.9477$ kJ/kg·K down to 0.005 MPa. That entropy lies inside the dome, so the quality follows from $x_{4s} = (s_3 - s_f)/s_{fg} = (6.9477 - 0.4762)/7.9176 = 0.8174$ and $h_{4s} = 137.75 + 0.8174 \times 2422.98 = 2118.2$ kJ/kg.
Part (c) — apply the internal efficiency to get the real exhaust. The internal efficiency is defined on the enthalpy drop, $\eta_T = (h_3 - h_4)/(h_3 - h_{4s})$, so the actual state is $$h_4 = h_3 - \eta_T (h_3 - h_{4s}) = 3411.3 - 0.85 \times 1293.1 = \boxed{2312.2\ \text{kJ/kg}}$$ The friction that costs 15 % of the work reappears as entropy, so the exhaust is drier than the ideal one: $x_4 = (2312.2 - 137.75)/2422.98 = 0.8974$, i.e. 10.3 % moisture at the last stage. That is right at the customary erosion limit of about 10–12 % and is one of the points part (f) turns on.
Part (d) — turbine power. The specific work is the actual enthalpy drop, $w_T = h_3 - h_4 = 3411.3 - 2312.2 = 1099.1$ kJ/kg, so $$\dot{W}_T = \dot{m}_s\,w_T = 79.55 \times 1099.1 = 87\,434\ \text{kW} = \boxed{87.4\ \text{MW}}$$ This is shaft power at the turbine coupling; the generator and its excitation would take a further 1–2 % before the number became station output.
Part (e) — thermodynamic efficiency of the cycle. The feed pump is the only parasitic load inside the cycle boundary, $\dot{W}_p = \dot{m}_s w_p = 79.55 \times 5.02 = 399$ kW, so the net output is 87 434 − 399 = 87 035 kW and $$\eta_{th} = \frac{\dot{W}_T - \dot{W}_p}{\dot{Q}} = \frac{87\,035}{260\,000} = \boxed{33.5\ \%}$$ Ignoring the pump would give 33.6 %, so the pump is worth only a tenth of a point here — unremarkable in a cycle whose feedwater is barely above ambient, but it is the term that grows once feedwater heating is added.
Part (f) — measure the result against its ceiling first. Between the same two end temperatures a Carnot engine would return $1 - (32.87 + 273.15)/(490 + 273.15) = 59.9\ \%$, so this cycle achieves 56 % of its thermodynamic ceiling. The gap is not turbine friction — that costs about five points — it is the enormous irreversibility of heating 33 °C feedwater with 500 °C sodium. Every improvement worth listing attacks either that mismatch or the wet end of the expansion.

Part (f) — how the efficiency of this plant could be improved. The list below is ordered by what each measure would actually buy on this machine.

Regenerative feedwater heating (the single largest gain). Feedwater entering the steam generator at 33 °C is heated by sodium at 500 °C; that temperature mismatch destroys more availability than every other loss in the plant combined. Bleeding steam from two or three turbine stages into closed heaters and a deaerator to raise the feedwater to roughly 180–220 °C typically lifts a cycle of this size by four to six percentage points, because the heat that is added is added at a higher mean temperature and the bled steam has already done work.
Reheat. Expanding to about 1 MPa, returning the steam to the sodium loop to be reheated to 490 °C and then completing the expansion raises the mean temperature of heat addition and, just as importantly, dries the exhaust. It is worth two to four points and it removes the 10.3 % exhaust moisture that currently limits blade life. The sodium is available at 500 °C, so a reheater costs the cycle nothing in peak temperature.
Higher steam pressure and temperature. The sodium leaves the intermediate exchanger at 500 °C, so steam at 490 °C already uses a 10 K approach and there is nothing left in temperature. Pressure is the free variable: raising the throttle from 5 to 12–16 MPa raises the mean temperature of heat addition, though it needs reheat alongside it or the exhaust becomes unacceptably wet.
Better turbine internal efficiency. Eighty-five per cent is modest for a machine of this size; a well-staged multi-cylinder turbine reaches 88–90 %, worth roughly one and a half points directly.
A lower condenser pressure, if the site allows it. The condenser already sits at 0.005 MPa (32.9 °C), which needs cooling water in the low twenties. On a cold-water site 0.004 MPa is attainable and worth about half a point, but on a warm-water or air-cooled site the pressure would move the other way.
Replacing the cycle rather than tuning it. With sodium available at 500 °C the supercritical carbon-dioxide cycle of Question 2 reaches a comparable efficiency in a fraction of the volume and with no water-steam reaction hazard next to the sodium — which is precisely why the paper asks both questions about the same reactor.

Quantity	Result
(b) Secondary sodium mass flow rate	1458 kg/s
(b) Steam mass flow rate	79.55 kg/s
(c) $h_1$ — saturated liquid, 0.005 MPa	137.75 kJ/kg
(c) $h_2$ — feed-pump outlet, 5 MPa / 33 °C	142.77 kJ/kg
(c) $h_3$ — turbine inlet, 5 MPa / 490 °C	3411.3 kJ/kg
(c) $h_{4s}$ — isentropic exhaust ($x_{4s} = 0.8174$)	2118.2 kJ/kg
(c) $h_4$ — actual exhaust ($x_4 = 0.8974$)	2312.2 kJ/kg
(d) Turbine power output	87.4 MW
(e) Thermodynamic efficiency of the cycle	33.5 %
(f) Carnot bound between the same end temperatures	59.9 %

Concept & Theory

Topic. Simple Rankine cycle on a liquid-metal-cooled small modular reactor.

Key relations. $\dot{Q} = \dot{m}c_p\Delta T$ for each sodium loop; $w_p = v_f\,\Delta p$; $\eta_T = (h_3-h_4)/(h_3-h_{4s})$; $x = (s - s_f)/s_{fg}$; $\eta_{th} = (w_T - w_p)/q_{in}$; Carnot bound $1 - T_{cond}/T_{max}$.

Why this works. A liquid metal has no phase change and a nearly constant specific heat, so each sodium loop is a pure sensible-heat transport problem and its flow follows from one energy balance. The water side is fixed by two independent properties at each state: pressure and quality at the condenser, pressure and temperature at the throttle, and pressure and entropy at the ideal exhaust. The internal efficiency then converts the ideal enthalpy drop into the real one, and because the friction is dissipated into the steam rather than lost from it, the real exhaust always sits to the right of the ideal one and is drier. The whole cycle efficiency is bounded by Carnot between the condenser temperature and the peak steam temperature, and how far below that bound it falls is a direct measure of how badly the heat addition is spread over temperature — which is why the answer to part (f) is dominated by feedwater heating.

Common pitfall. Two mistakes recur. The first is sizing the sodium loop on the primary temperature difference (155 K) when the question asks for the secondary loop (145 K), which understates the flow by 6 %. The second is applying the turbine efficiency in the wrong direction — dividing rather than multiplying by $\eta_T$ — which produces an exhaust enthalpy below the isentropic one and a quality that violates the second law.

Source. Granet & Bluestein, Thermodynamics and Heat Power, ch. 9 (vapour power cycles) and the bound steam tables; Çengel & Boles, Thermodynamics: An Engineering Approach, §10.2–10.6; El-Wakil, Powerplant Technology, ch. 2 and ch. 10 (liquid-metal-cooled reactors).

Question 1 of 8: ARC-100 Reactor with Steam Cycle

Question 1: ARC-100 Reactor with Steam Cycle (10 marks)