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

Question 3 of 6: Feedwater Heaters and Condensers

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

Notes on this paper

Paper format. National Examinations, December 2014 — 07-Mec-B3 Energy Conversion and Power Generation. Three hours, closed book. Section A is calculative (Questions 1–4) and Section B is descriptive (Questions 5–6); a candidate answers three from Section A and one from Section B, four questions of 15 marks each constituting a complete 60-mark paper. Reference data for particular questions are bound in as attachment pages 9–13, reference formulae and constants as pages 14–17, and the Granet & Bluestein steam tables as pages 18–35. All six questions are solved here, because the set is a study resource rather than a three-hour sitting.

Reference texts. I. Granet and M. Bluestein, Thermodynamics and Heat Power, 6th ed. (the steam tables bound into this paper) · M. M. El-Wakil, Powerplant Technology (station heat balances, condensers, feedwater heating, nuclear and renewable plant) · Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, 9th ed. (Brayton and Rankine cycle analysis) · Y. A. Çengel, Heat and Mass Transfer, 6th ed. (surface-condenser and recuperator performance).

Property data. Every enthalpy, entropy and saturation temperature quoted below is read from the tables bound into this examination paper — General Constants on page 15 ($c_p$ and $c_v$ for helium, air and water), the Question 2 enthalpy table on page 10, the Koeberg condenser data sheet on page 11, the Belledune heat balance diagram on page 13, and Granet & Bluestein Tables A.1–A.4 on pages 19–35. Where a table entry has to be interpolated the interpolation is shown.

Question 3: Feedwater Heaters and Condensers (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: a top feedwater heater passing 350 kg/s of feedwater at 20 MPa from 230 °C to 280 °C, heated by 5 MPa steam that enters at 400 °C and drains as subcooled water at 240 °C. Part II: the Koeberg condenser data sheet on attachment page 11.

Part II — design data used from the Koeberg data sheet (exam page 11)
Quantity	Value
Steam flow rate	2996 t/h = 832.2 kg/s
Cooling water flow rate	141 000 t/h = 39 166.7 kg/s
Cooling water inlet / outlet temperature	13 °C / 24 °C
Steam inlet pressure / temperature	0.043 bar abs (4.3 kPa) / 30 °C
Terminal temperature difference	6 °C
Cooling surface area	57 426 m²

Find. Part I: the heating-steam flow. Part II: the design temperature profile with ΔT and θ identified, and the new profiles, ΔT and θ for each of the four disturbances.

Figure 3.1 — Key points around the top feedwater heater. Circled numbers are the four terminal states used in the energy balance: feedwater in and out inside the tubes at 20 MPa, heating steam in and drain out on the steam side at 5 MPa. The drain leaves 10 °C above the entering feedwater, which is the drain-cooler approach.

Approach. Part I is a single steady-flow energy balance across the heater with all four enthalpies read from the bound steam tables. Part II works from two relations only — $\dot{Q} = \dot{M}_w c_p \Delta T$ on the water side and $\dot{Q} = UA\theta$ on the transfer side — with $UA$ fixed by the design point and then perturbed as each question requires.

Part I — feedwater enthalpies at 20 MPa (Table A.4). The compressed-liquid table gives 949.3 kJ/kg at 220 °C and 1040.0 kJ/kg at 240 °C, so the 230 °C inlet is the mean of the two, and 280 °C is a direct entry: $$h_{fw,in} = \tfrac{1}{2}(949.3 + 1040.0) = 994.65\ \text{kJ/kg}, \qquad h_{fw,out} = 1230.6\ \text{kJ/kg}$$
Heater duty. The feedwater side is a straightforward sensible-heat pickup: $$\dot{Q} = \dot{M}_{fw}(h_{fw,out} - h_{fw,in}) = 350 \times 235.95 = \boxed{82\,583\ \text{kW} \approx 82.6\ \text{MW}}$$ That is the whole of the heat transfer, and the steam side must supply exactly it.
Steam-side enthalpies at 5 MPa. The inlet is superheated, so Table A.3 applies, and the drain is subcooled liquid, so Table A.4 applies. Note that the saturation temperature at 5 MPa is 264.0 °C, so a drain at 240 °C is indeed subcooled and the heater must include a drain cooler: $$h_{s,in} = 3195.7\ \text{kJ/kg}\ (5\ \text{MPa},\ 400\ ^{\circ}\text{C}), \qquad h_{s,out} = 1037.5\ \text{kJ/kg}\ (5\ \text{MPa},\ 240\ ^{\circ}\text{C})$$
Part I — required steam flow. Equating the two sides of the energy balance, $$\dot{M}_s = \frac{\dot{Q}}{h_{s,in} - h_{s,out}} = \frac{82\,583}{3195.7 - 1037.5} = \frac{82\,583}{2158.2} = \boxed{38.3\ \text{kg/s}}$$ So the heater bleeds about 11 % of the feedwater flow — a plausible figure for a top heater, and a useful check that no enthalpy has been read from the wrong table.
Part II — fix the design point. The data sheet is internally consistent, which is worth confirming before relying on it: the steam is at 0.043 bar, whose saturation temperature is the quoted 30 °C, and the terminal temperature difference of $30 - 24 = 6\ ^{\circ}\text{C}$ is exactly the printed value. The design cooling-water rise and mean temperature difference are therefore $$\Delta T = 24 - 13 = 11\ ^{\circ}\text{C}, \qquad \theta = 30 - \tfrac{1}{2}(13+24) = 11.5\ ^{\circ}\text{C}$$
Design duty and $UA$. With $\dot{M}_w c_p = 39\,166.7 \times 4.19 = 164\,108\ \text{kW}/^{\circ}\text{C}$, $$\dot{Q} = 164\,108 \times 11 = 1805\ \text{MW}, \qquad UA = \frac{\dot{Q}}{\theta} = \frac{1\,805\,190}{11.5} = 156\,973\ \text{kW}/^{\circ}\text{C}$$ As an independent check, dividing the duty by the steam flow gives 2169 kJ/kg, implying an exhaust quality of 0.89 — exactly what a large low-pressure turbine delivers, so the two sides of the data sheet agree.
Figure 3.2 — The design temperature profile along the tubes: cooling water rising 13 → 24 °C, condensing steam flat at 30 °C, ΔT = 11 °C, θ = 11.5 °C and a terminal difference of 6 °C at the water outlet.
The two governing relations, rearranged for each case. Everything that follows comes from $$\Delta T = \frac{\dot{Q}}{\dot{M}_w c_p}, \qquad \theta = \frac{\dot{Q}}{UA}, \qquad T_{steam} = T_{cw,in} + \tfrac{1}{2}\Delta T + \theta$$ The last relation is simply the definition of θ as the gap between the condensing steam and the average water temperature, which is what the question directs us to use in place of the log-mean difference.
(a) Cooling water inlet raised to 18 °C. Neither the load nor the water flow nor $U$ changes, so $\dot{Q}$, ΔT and θ are all unchanged and the whole profile simply translates upward by 5 °C: $$\Delta T = 11\ ^{\circ}\text{C}\ (18 \to 29\ ^{\circ}\text{C}), \qquad \theta = 11.5\ ^{\circ}\text{C}, \qquad T_{steam} = 18 + 5.5 + 11.5 = \boxed{35\ ^{\circ}\text{C}}$$ The back pressure rises from 4.3 kPa to the saturation pressure at 35 °C, 5.6 kPa, and the turbine loses output accordingly. This is why summer river temperature, not condenser design, often sets a station's peak-season capability.
(b) Turbine load reduced to one quarter. Now the duty falls to a quarter while the water flow and $UA$ hold: $$\Delta T = \tfrac{11}{4} = 2.75 \approx 3\ ^{\circ}\text{C}\ (13 \to 16\ ^{\circ}\text{C}), \qquad \theta = \tfrac{11.5}{4} = 2.9 \approx 3\ ^{\circ}\text{C}$$ $$T_{steam} = 13 + 1.4 + 2.9 = 17.25 \approx \boxed{17\ ^{\circ}\text{C}}$$ Both differences scale linearly with load, so the condenser pressure collapses to about 2 kPa. In practice the vacuum would be limited by air in-leakage and by the exhaust-hood spray long before this point, but on the stated basis the profile is as computed.
(c) Cooling water flow halved and $U$ reduced to 70 %. Full turbine load, half the water and a poorer coefficient: $$\Delta T = \frac{1\,805\,190}{0.5 \times 164\,108} = 22\ ^{\circ}\text{C}\ (13 \to 35\ ^{\circ}\text{C}), \qquad \theta = \frac{1\,805\,190}{0.70 \times 156\,973} = 16.4 \approx 16\ ^{\circ}\text{C}$$ $$T_{steam} = 13 + 11 + 16.4 = 40.4 \approx \boxed{40\ ^{\circ}\text{C}}$$ This is much the worst of the four cases, because the halved flow and the degraded coefficient push in the same direction; the back pressure roughly doubles to 7.4 kPa. Note that $U$ falls with flow because the tube-side coefficient scales roughly as velocity to the power 0.8, so halving the flow really does drag $U$ down with it — the question's 70 % is consistent with that exponent.
(d) Overall coefficient reduced 20 % by fouling. Load and water flow are unchanged, so only θ moves: $$\Delta T = 11\ ^{\circ}\text{C}\ (13 \to 24\ ^{\circ}\text{C}), \qquad \theta = \frac{1\,805\,190}{0.80 \times 156\,973} = 14.4 \approx 14\ ^{\circ}\text{C}$$ $$T_{steam} = 13 + 5.5 + 14.4 = 32.9 \approx \boxed{33\ ^{\circ}\text{C}}$$ The cooling-water profile is identical to the design case and only the steam line lifts — which is exactly the diagnostic signature of fouling, and the reason operators watch the terminal difference rather than the outlet temperature to decide when to clean tubes.

Figure 3.3 — The four disturbed profiles (solid) against the design profile (dotted). Reading them together shows the diagnostic value of the plot: a raised inlet translates the whole picture, a load reduction shrinks both differences, restricted flow steepens the water line and lifts the steam line together, and fouling lifts the steam line alone.

Question 3 — final results
Case	Cooling water	ΔT	θ	Steam temperature
Part I — heater duty 82 583 kW	Heating steam flow = 38.3 kg/s (feedwater 350 kg/s, 994.65 → 1230.6 kJ/kg; steam 3195.7 → 1037.5 kJ/kg)
Part II design	13 → 24 °C	11 °C	11.5 °C	30 °C (4.3 kPa)
(a) inlet raised to 18 °C	18 → 29 °C	11 °C	11.5 °C	35 °C (5.6 kPa)
(b) load to one quarter	13 → 16 °C	3 °C	3 °C	17 °C (1.9 kPa)
(c) half flow, U to 70 %	13 → 35 °C	22 °C	16 °C	40 °C (7.4 kPa)
(d) U reduced 20 % (fouling)	13 → 24 °C	11 °C	14 °C	33 °C (5.0 kPa)

Topic. Feedwater-heater energy balances and off-design surface-condenser performance.

Key relations. Heater balance $\dot{M}_{fw}\Delta h_{fw} = \dot{M}_s \Delta h_s$; condenser duty $\dot{Q} = \dot{M}_w c_p \Delta T = UA\theta$; mean-difference form $T_{steam} = T_{cw,in} + \tfrac{1}{2}\Delta T + \theta$; terminal temperature difference $TTD = T_{steam} - T_{cw,out}$; $p_{sat} = f(T_{steam})$ from Table A.1.

Why this works. A condenser is unusual among heat exchangers in that one stream changes phase, so its temperature is constant along the tubes and is not an independent variable at all: it is whatever value makes $UA\theta$ equal the heat the turbine is dumping. That single observation turns every part of this question into arithmetic. The water side sets ΔT from the duty and the flow, the transfer side sets θ from the duty and $UA$, and the steam temperature is simply what sits θ above the mean water temperature. Using the average difference rather than the log-mean is a legitimate simplification here because the ratio of the terminal differences never exceeds about two, where the two means agree within a few per cent.

Common pitfall. The most common error in Part II is assuming the steam temperature is fixed by the turbine, and then trying to force the water outlet to move instead. The turbine fixes the heat flow, not the back pressure; the condenser fixes the back pressure. In Part I the usual slip is treating the drain as saturated liquid at 5 MPa (264 °C, 1154.2 kJ/kg) rather than reading the subcooled entry at 240 °C, which understates the steam-side enthalpy drop and overstates the steam flow by about 5 %.

Source. El-Wakil, Powerplant Technology, Ch. 6 (condensers and cooling systems) and Ch. 2 (feedwater heating); Çengel, Heat and Mass Transfer, 6th ed., Ch. 11 (heat exchangers, the effectiveness–NTU and mean-temperature-difference methods); Granet & Bluestein Tables A.1–A.4.

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