Question 4 of 6: Steam Plant Heat Rejection

Question 4: Steam Plant Heat Rejection (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. Part I, from the page-10 Koeberg condenser data sheet; Part II, from the question page and the page-12 chart:

Find. Part I: the design temperature profile and the new profiles, with ΔT and θ, for four changes of duty. Part II: the circulating-water flow, the evaporative loss in m³/GJ and m³/s, that loss as a percentage of the circulating flow, and the water consumed per kWh generated.

Approach. Two equations govern the condenser: the water-side energy balance $\dot Q = \dot m_w c_p \Delta T$ and the transfer equation $\dot Q = UA\,\theta$ written on the average temperature difference as the question permits. Each part changes one term and holds the rest, so ΔT and θ simply scale — no iteration is needed. Part II is a mass and energy balance on the tower with one chart reading.

1. Part I — the design condition (dotted profiles). From the page-10 data the cooling water rises $\Delta T = 24 - 13 = \boxed{11\ \text{K}}$ and the mean water temperature is $(13+24)/2 = 18.5\ ^\circ\text{C}$, so $$\theta = t_{steam} - \bar t_w = 30 - 18.5 = \boxed{11.5\ \text{K}}$$ The printed terminal temperature difference, $30 - 24 = 6$ K, and the printed condenser pressure of 0.043 bar (saturation temperature 30.0 °C) both confirm the data set is self-consistent. As a final check on the sheet, the water side removes $\dot m_w c_p \Delta T = 39\,167 \times 4.19 \times 11 = 1805$ MW, which over the printed 2996 t/h of steam is 2169 kJ/kg — the latent heat of steam at 30 °C times a quality of 0.89, exactly what a large low-pressure turbine exhausts.
2. Part I (a) — cooling water inlet raised to 18 °C. The turbine load, the water flow and the surface are all unchanged, so neither $\Delta T$ nor $\theta$ can change: $$\Delta T = 11\ \text{K} \Rightarrow t_{w,out} = 18 + 11 = 29\ ^\circ\text{C}, \qquad t_{steam} = \frac{18+29}{2} + 11.5 = \boxed{35\ ^\circ\text{C}}$$ The whole profile simply lifts by the 5 K added at inlet, and the back pressure rises with it — from 0.043 bar to about 0.056 bar — costing turbine output. This is why summer river or sea temperature is a first-order constraint on station output.
3. Part I (b) — turbine load reduced to one quarter. Now the duty falls to $0.25\dot Q$ while $\dot m_w$, $U$ and $A$ all hold, so both differences scale directly with the duty: $$\Delta T = 0.25 \times 11 = 2.75\ \text{K} \Rightarrow t_{w,out} = 13 + 2.75 \approx \boxed{16\ ^\circ\text{C}}$$ $$\theta = 0.25 \times 11.5 = 2.875\ \text{K} \Rightarrow t_{steam} = \frac{13 + 15.75}{2} + 2.875 \approx \boxed{17\ ^\circ\text{C}}$$ The condenser is now hugely oversurfaced for the duty, so the steam sits only a few degrees above the incoming water and the back pressure collapses — in practice it would be limited by air in-leakage and by the turbine's own exhaust-loss floor rather than by heat transfer.
4. Part I (c) — water flow halved and U reduced to 70 %%. Full load returns, but with half the flow the same duty needs twice the temperature rise, and with 70 %% of the coefficient it needs the driving difference divided by 0.7: $$\Delta T = \frac{11}{0.5} = 22\ \text{K} \Rightarrow t_{w,out} = 13 + 22 = \boxed{35\ ^\circ\text{C}}, \qquad \theta = \frac{11.5}{0.7} = 16.4\ \text{K}$$ $$t_{steam} = \frac{13+35}{2} + 16.4 = 40.4 \approx \boxed{40\ ^\circ\text{C}}$$ This is the worst of the four cases: the steam temperature rises 10 K, roughly doubling the back pressure to about 0.074 bar. Note the two effects compound — reducing pump flow to save auxiliary power also degrades the film coefficient inside the tubes, so the coefficient penalty is not independent of the flow reduction.
5. Part I (d) — overall coefficient reduced 20 %% by fouling. Load and flow are unchanged, so the water still rises 11 K to 24 °C, but the same duty must now be pushed through a poorer surface: $$\theta = \frac{11.5}{0.8} = 14.375\ \text{K} \Rightarrow t_{steam} = 18.5 + 14.4 = 32.9 \approx \boxed{33\ ^\circ\text{C}}$$ so the back pressure rises about 3 K worth, near 0.050 bar. Because this penalty is permanent until the tubes are cleaned, condenser fouling is monitored continuously through exactly this quantity — the difference between measured steam saturation temperature and mean water temperature is the plant's cleanliness factor.
6. Part II (a) — circulating water flow. All the rejected heat is carried by the water over its 10 K rise: $$\dot m_w = \frac{\dot Q}{c_p \Delta t} = \frac{1\,500\,000}{4.19 \times 10} = 35800\ \text{kg/s} \quad\Rightarrow\quad \dot V_w = \boxed{35.80\ \text{m}^3\text{/s}}$$ Note the 5 K approach between the 25 °C returning water and the 30 °C condensing steam, consistent with the terminal difference of a large condenser.
7. Part II (b) — evaporative loss from the chart. The page-12 chart is entered on wet bulb against dry bulb, so the 40 % relative humidity at 30 °C must first be converted: the adiabatic-saturation balance gives a wet bulb of 20.1 °C, which lands on the chart's own 40 % dash-dot line and so confirms the entry point. Reading the contour through (30 °C dry bulb, 20.1 °C wet bulb) gives $$\boxed{0.375\ \text{m}^3\text{/GJ rejected}}$$ A useful bound: if every joule left as latent heat the loss would be $1/h_{fg} = 10^6/2430.5/1000 = 0.411\ \text{m}^3\text{/GJ}$, so 0.375 corresponds to 91 % of the duty leaving as evaporation and the balance as sensible heating of the air — the right split for warm, fairly dry air.
8. Part II (c) and (d) — loss in m³/s and as a percentage. The plant rejects 1500 MJ/s = 1.5 GJ/s, so $$\dot V_{evap} = 0.375 \times 1.5 = \boxed{0.5625\ \text{m}^3\text{/s}}$$ $$\frac{\dot V_{evap}}{\dot V_w} = \frac{0.5625}{35.80} = \boxed{1.57\ \%}$$ A little over 1.5 % of the circulating flow is evaporated on each pass, which is the classic figure for a wet tower on a 10 K range. Make-up must also cover drift and blowdown, so the real intake is appreciably larger — typically half as much again once the cycles of concentration are set.
9. Part II (e) — water consumed per unit generated. Converting to hourly quantities against the 600 MW sent out, $$\frac{0.5625\ \text{m}^3\text{/s} \times 1000 \times 3600}{600\,000\ \text{kW}} = \boxed{3.375\ \text{L/kWh}}$$ Roughly three and a half litres per kilowatt-hour, or about 2 million litres an hour for the station. This single number is what makes once-through and wet-tower cooling a siting question as much as a thermodynamic one, and it is the figure air-cooled condensers trade efficiency to avoid.