Question 3 of 8: Condenser Performance

Question 3: Condenser Performance (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. From the page-13 Koeberg data sheet, the design point of the condenser:

Find. The design profile, and then for each of four changed conditions the new cooling-water rise ΔT, the new mean temperature difference θ, the new water outlet temperature and the new condensing temperature, each plotted against the design profile.

Approach. Two equations govern everything: the water-side sensible balance $\dot{Q} = \dot{m}_w c_p \Delta T$ and the surface rating $\dot{Q} = UA\theta$. Because $A$ never changes, each part alters exactly one factor, so ΔT and θ each scale by a single ratio and no iteration is needed. The condensing temperature is not an input — it floats to whatever value the surface requires, and the back pressure follows it.

1. Establish the design point and confirm the data sheet is self-consistent. The cooling water rises $\Delta T = 24 - 13 = 11$ K, its mean temperature is 18.5 °C, so the mean temperature difference is $\theta = 30 - 18.5 = 11.5$ K and the terminal difference at the outlet end is $30 - 24 = 6$ K, exactly as the sheet prints. The duty follows from the water side, $\dot{Q} = 39\,167 \times 4.19 \times 11 = 1.805 \times 10^6$ kW, i.e. 1805 MW. Divided by the steam flow that is 2169 kJ/kg given up per kilogram of steam, which against $h_{fg} = 2429$ kJ/kg at 30 °C implies a turbine exhaust quality of 0.89 — entirely normal for the last stage of a saturated-steam nuclear turbine, so the sheet is internally consistent. The printed 0.043 bar is $p_{sat}$ at 30 °C, which closes the loop.
2. Write the two scaling rules once. From $\dot{Q} = \dot{m}_w c_p \Delta T = UA\theta$ with $A$ fixed,$$\Delta T = \frac{\dot{Q}}{\dot{m}_w c_p} \propto \frac{\dot{Q}}{\dot{m}_w} \qquad \theta = \frac{\dot{Q}}{UA} \propto \frac{\dot{Q}}{U}$$ and the condensing temperature is then read off as $T_s = \tfrac{1}{2}(T_{w1} + T_{w2}) + \theta$. Every part below is one substitution into these three relations.
3. Part (a) — cooling water inlet raised to 18 °C. Neither the duty, the water flow nor $U$ has changed, so $\Delta T$ and $\theta$ both hold at their design values. The whole profile simply translates upward by 5 K: the water leaves at $18 + 11 = 29$ °C, its mean is 23.5 °C, and $$T_s = 23.5 + 11.5 = \boxed{35\ {}^{\circ}\text{C}}$$ The back pressure rises from 4.3 to 5.6 kPa. This is the summer condition, and it costs output: 5 K of extra condensing temperature is worth roughly 1.5 % of turbine work.
4. Part (b) — turbine load reduced to a quarter. The duty falls to a quarter with the load while the water flow and $U$ are held, so both $\Delta T$ and $\theta$ scale by 0.25: $\Delta T = 2.75$ K and $\theta = 2.875$ K. The water leaves at $13 + 2.75 = 15.75 \approx 16$ °C, its mean is 14.4 °C, and $$T_s = 14.375 + 2.875 = 17.25 \approx \boxed{17\ {}^{\circ}\text{C}}$$ at a back pressure of about 1.9 kPa. In practice a real plant would not be allowed to go this deep — below about 3 kPa the LP exhaust annulus chokes and air in-leakage dominates — so part load is normally run with fewer circulating-water pumps, which is case (c).
5. Part (c) — cooling water flow halved and $U$ down to 70 %. The duty is unchanged (full load), so halving the flow doubles the rise, $\Delta T = 11/0.5 = 22$ K, and the degraded coefficient widens the mean difference to $\theta = 11.5/0.7 = 16.43$ K. The water now leaves at $13 + 22 = 35$ °C, its mean is 24 °C, and $$T_s = 24 + 16.43 = 40.4 \approx \boxed{40\ {}^{\circ}\text{C}}$$ The back pressure has risen to about 7.4 kPa, and the terminal difference has collapsed to 5 K. Note that the two effects compound: the flow reduction alone would give 35 °C and the $U$ reduction alone about 34 °C, but together they give 40 °C. This is why taking circulating-water pumps out of service at full load is normally interlocked.
6. Part (d) — $U$ reduced 20 % by fouling. Full duty and full water flow, so $\Delta T$ stays at 11 K and the outlet stays at 24 °C; only the surface has got worse, so $\theta = 11.5/0.8 = 14.375$ K and $$T_s = 18.5 + 14.375 = 32.9 \approx \boxed{33\ {}^{\circ}\text{C}}$$ at about 5.0 kPa. The cooling-water profile is identical to the design one and only the steam line lifts — which is exactly the diagnostic signature of fouling, and the reason condenser performance is monitored by $\theta$ (or by the cleanliness factor $U/U_{design}$) rather than by back pressure alone.