Condenser Performance: 22-Mec-B3 Energy Conversion and Power Generation, May 2018, Q4

Question 4: 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. The Koeberg condenser data sheet on page 10 fixes the design point: cooling water enters at 13 °C and leaves at 24 °C while the steam condenses at 30 °C under 0.043 bar absolute, with a stated terminal temperature difference of 6 °C.

Design data read from the page-10 sheet
Steam flow	$2996\ \text{t/h}$
Cooling water flow	$141\,000\ \text{t/h}$
Cooling water inlet / outlet	$13\ ^\circ\text{C} \rightarrow 24\ ^\circ\text{C}$
Steam inlet pressure and temperature	$0.043\ \text{bar},\ 30\ ^\circ\text{C}$
Terminal temperature difference	$6\ ^\circ\text{C}$
Cooling surface area	$57\,426\ \text{m}^2$

Find. The design values of the cooling water rise $\Delta T$ and the mean temperature difference $\theta$, and the new profiles, terminal temperatures, $\Delta T$ and $\theta$ for each of the four off-design conditions.

Question 4 — the four required temperature profiles. In every panel the dotted lines are the design condition (water 13 → 24 °C, steam 30 °C) and the solid lines are the new condition; the blue line is cooling water along the tubes and the red line the condensing steam, which is isothermal because it is a phase change at constant pressure.

Approach. The condenser obeys two statements of the same duty, $\dot{Q} = M_w c_p \Delta T$ on the water side and $\dot{Q} = U A \theta$ on the transfer side. Each part changes exactly one factor, so $\Delta T$ and $\theta$ can be scaled directly with no iteration, and the condensing temperature simply floats to whatever the mean water temperature plus $\theta$ demands.

Design point and the two governing statements. From the data sheet, $$\Delta T = 24 - 13 = 11\ \text{K},\qquad \theta = 30 - \tfrac{13+24}{2} = \boxed{11.5\ \text{K}} .$$ The terminal temperature difference $30-24 = 6$ K reproduces the value printed on the sheet exactly, which confirms the three temperatures have been read correctly. The duty implied is $(141\,000\times10^3/3600) \times 4.19 \times 11 = 1805\ \text{MW}$, which over 2996 t/h of steam is 2169 kJ/kg, an exhaust quality of 0.89 — entirely consistent with a large wet-steam turbine, so the sheet is internally sound. The two statements that govern every part are $$\dot{Q} = M_w c_p \Delta T = U A \theta .$$
Part (a) — cooling water inlet raised to 18 °C. Nothing on the transfer side changes: the load, the water flow and $UA$ are all as designed, so both $\Delta T$ and $\theta$ keep their design values and the whole profile simply lifts by 5 K. $$\Delta T = 11\ \text{K},\qquad \theta = 11.5\ \text{K},$$ $$T_{w,\text{out}} = 18 + 11 = 29\ ^\circ\text{C},\qquad T_{\text{steam}} = \tfrac{18+29}{2} + 11.5 = \boxed{35\ ^\circ\text{C}} .$$ The back pressure rises from 4.25 kPa to the saturation pressure at 35 °C, 5.63 kPa, which is why summer river temperatures cost a station real output.
Part (b) — turbine load reduced to one quarter. The duty falls to a quarter while $M_w$, $U$ and $A$ are unchanged, so both $\Delta T$ and $\theta$ fall in the same proportion: $$\Delta T = 0.25 \times 11 = 2.75\ \text{K},\qquad \theta = 0.25 \times 11.5 = 2.875\ \text{K},$$ $$T_{w,\text{out}} = 13 + 2.75 = 15.75 \approx 16\ ^\circ\text{C},\qquad T_{\text{steam}} = \tfrac{13+15.75}{2} + 2.875 = 17.25 \approx \boxed{17\ ^\circ\text{C}} .$$ In practice the vacuum would be limited before this point by air ingress and by the minimum permissible exhaust temperature, but the ideal answer is the one asked for.
Part (c) — cooling water flow halved and $U$ cut to 70 %. Two factors change, and they act on different sides. Halving the water flow at constant duty doubles the water-side rise, while cutting $U$ to 70 % at constant duty and area raises the mean difference by $1/0.7$: $$\Delta T = \frac{11}{0.5} = 22\ \text{K},\qquad \theta = \frac{11.5}{0.7} = 16.43\ \text{K},$$ $$T_{w,\text{out}} = 13 + 22 = 35\ ^\circ\text{C},\qquad T_{\text{steam}} = \tfrac{13+35}{2} + 16.43 = 40.4 \approx \boxed{40\ ^\circ\text{C}} .$$ The condensing temperature has risen 10 K above design and the back pressure has more than doubled — the classic consequence of losing half a circulating water pump.
Part (d) — overall coefficient reduced 20 % by fouling. Only the transfer side changes, so the water-side rise is untouched and only $\theta$ grows: $$\Delta T = 11\ \text{K},\qquad \theta = \frac{11.5}{0.8} = 14.375\ \text{K},$$ $$T_{w,\text{out}} = 24\ ^\circ\text{C},\qquad T_{\text{steam}} = \tfrac{13+24}{2} + 14.375 = 32.9 \approx \boxed{33\ ^\circ\text{C}} .$$ Comparing (c) and (d) makes the diagnostic point the question is really after: fouling moves the steam line without moving the water line, whereas a flow reduction moves both. Watching which of the two happens is how an operator tells a fouled tube bundle from a sick pump.

Final results
Quantity	Symbol	Value
Design	$\Delta T,\ \theta$	11 K, 11.5 K; water 13 → 24 °C, steam 30 °C
(a) inlet 18 °C	$\Delta T,\ \theta$	11 K, 11.5 K; water 18 → 29 °C, steam 35 °C
(b) quarter load	$\Delta T,\ \theta$	2.75 K, 2.875 K; water 13 → 16 °C, steam 17 °C
(c) half flow, $U$ = 70 %	$\Delta T,\ \theta$	22 K, 16.43 K; water 13 → 35 °C, steam 40 °C
(d) $U$ down 20 %	$\Delta T,\ \theta$	11 K, 14.375 K; water 13 → 24 °C, steam 33 °C

Topic. Off-design response of a surface condenser at fixed geometry.

Key relations. $\dot{Q} = M_w c_p \Delta T = U A \theta$, with $\theta$ the difference between the condensing temperature and the mean water temperature; $T_{\text{steam}} = \tfrac{1}{2}(T_{w,\text{in}} + T_{w,\text{out}}) + \theta$; $p_{\text{cond}} = p_{\text{sat}}(T_{\text{steam}})$.

Why this works. A condenser has fixed tubes, so $A$ never changes and the machine has only two degrees of freedom. Whatever heat the turbine sends it must be carried away by the water, which fixes $\Delta T$, and must also be driven across the tube wall, which fixes $\theta$. The condensing temperature is not an input at all: it floats to whatever value makes the transfer equation balance, and the back pressure follows from the saturation table. That is why each part of this question can be answered by scaling — change one factor, see which of the two equations it sits in, and only the quantity in that equation moves.

Common pitfall. Holding the condensing temperature fixed and solving for something else; and forgetting that a load change scales $\theta$ as well as $\Delta T$, because it appears in both equations. Using the log mean temperature difference here is not wrong physics but is explicitly excluded by the question and makes the arithmetic far heavier for no extra marks.

Source. El-Wakil, Powerplant Technology, Ch. 6 (condensers and circulating water systems); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, Ch. 11.

Question 4 of 8: Condenser Performance

Question 4: Condenser Performance (10 marks)