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

Question 4 of 6: Belledune Heat Balance Diagram

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 4: Belledune Heat Balance Diagram (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. The manufacturer's heat balance diagram for Belledune Unit 2 (Toshiba, 600 MVA generator) at 100 % load, 430 000 kW generator output, with pressures in MPa absolute, flows in kg/h, temperatures in °C and enthalpies in kJ/kg.

Readings taken from the heat balance diagram (exam page 13)
Stream	Flow (kg/h)	Enthalpy (kJ/kg)	Other
Main steam to HP turbine	1 287 309	3397.5	16.65 MPa, 538.0 °C
Final feedwater to boiler	1 287 309	1231.4	280.0 °C
Steam through the reheater	1 039 160	3002.7 → 3542.5	2.87 MPa, 538.0 °C hot reheat
HP extraction to top heater	127 729	3175.9	6.72 MPa, 404.1 °C
HP turbine exhaust (cold reheat)	1 141 480	3002.7	3.19 MPa, 305.7 °C
HP gland and valve-stem leak-offs	12 220 + 3940 + 1600 + 340	3397.5	= 18 100 kg/h
Steam to feed-pump turbine (BFPT)	50 140	3068.9 → 2432.6	0.542 MPa in, 0.0048 MPa out
Feed pump	1 287 309	Δh = 24.45	P_D = 19.8 MPa, ρ = 912 kg/m³
Deaerator / pump suction pressure	—	—	0.554 MPa

Find. The cycle efficiency on boiler heat input, the HP turbine shaft power, the steam power into and the shaft power out of the feed-pump drive, the hydraulic power the pump delivers, and hence the pump efficiency.

Figure 4.1 — Control volumes used to read the diagram. Each part of the question is one boundary: the boiler and reheater together for (a), the HP casing alone for (b), the feed-pump turbine for (c), and the feed pump itself for (d), (e) and (f). The panel beneath lists every value taken off attachment page 13.

Reading the diagram. The HP turbine mass balance on the diagram closes exactly (1 287 309 − 18 100 = 1 141 480 + 127 729), the cold-reheat header closes exactly (1 141 480 = 1 039 160 + 101 910 + 410), the printed heater-casing saturation temperatures agree with the steam tables (6.31 MPa → 278.9 °C, 0.554 MPa → 155.7 °C), and the answer to part (d) below reproduces the 8741 kW printed at the feed-pump-turbine coupling to within 0.02 %.

Approach. Every part is one control volume and one steady-flow energy balance. Flows are in kg/h throughout the diagram, so each is divided by 3600 to give kW directly from kJ/kg.

(a) Boiler heat input. The boiler must raise the main steam from final-feedwater enthalpy and, separately, must resuperheat the cold reheat flow: $$\dot{Q}_{boiler} = \frac{\dot{M}_{ms}}{3600}(h_{ms} - h_{fw}) + \frac{\dot{M}_{rh}}{3600}(h_{hrh} - h_{crh})$$ $$= \frac{1\,287\,309}{3600}(3397.5 - 1231.4) + \frac{1\,039\,160}{3600}(3542.5 - 3002.7)$$ $$= 774\,568 + 155\,802 = \boxed{930\,370\ \text{kW}}$$ The reheater is worth noticing: it contributes 17 % of the total heat input, which is why reheat cycles need it accounted for explicitly rather than folded into the main-steam term.
(a) Steam cycle efficiency. Measured from boiler heat input to the generator terminals, $$\eta_{cycle} = \frac{P_e}{\dot{Q}_{boiler}} = \frac{430\,000}{930\,370} = \boxed{0.462 = 46.2\ \%}$$ Equivalently a heat rate of $3600/0.4622 = 7789$ kJ/kWh. For a 1990s subcritical reheat unit this is a good number, and it is high precisely because it excludes the boiler's own losses — the coal-to-terminals figure for the same station would be nearer 38 %.
(b) HP turbine — sort the flows first. Of the 1 287 309 kg/h admitted, 18 100 kg/h leaves through the glands and valve stems without expanding, so it does no work and leaves at throttle enthalpy. The remaining 1 269 209 kg/h splits into the 127 729 kg/h extraction to the top heater and the 1 141 480 kg/h exhaust to the reheater. That the two independent readings balance exactly is the check that these are the right numbers off a sideways-printed diagram.
(b) HP turbine shaft power. Energy in minus energy out across the casing: $$P_{HP} = \frac{(\dot{M}_{ms} - \dot{M}_{leak})h_{ms} - \dot{M}_{ext}h_{ext} - \dot{M}_{exh}h_{crh}}{3600}$$ $$= \frac{1\,269\,209(3397.5) - 127\,729(3175.9) - 1\,141\,480(3002.7)}{3600} = \boxed{133\,045\ \text{kW} \approx 133.0\ \text{MW}}$$ Grouping the same balance by expansion section gives an identical result and is a useful cross-check: the full flow drops 221.6 kJ/kg to the extraction point and the surviving flow drops a further 173.2 kJ/kg to exhaust, that is $1\,269\,209(221.6) + 1\,141\,480(173.2)$ over 3600, again 133.0 MW. The HP casing therefore produces just under a third of the unit's output.
(c) Steam power into the feed-pump turbine. The BFPT is a small back-pressure-free machine exhausting to the main condenser: $$P_{BFPT} = \frac{\dot{M}_{bfpt}}{3600}(h_{in} - h_{out}) = \frac{50\,140}{3600}(3068.9 - 2432.6) = 13.928 \times 636.3 = \boxed{8862\ \text{kW}}$$
(d) Shaft power into the feed pump. The diagram prints the pump's enthalpy rise directly, and that rise — being the actual enthalpy rise, not the ideal one — already contains the pump's own losses: $$P_{shaft} = \frac{\dot{M}_{fw}}{3600}\,\Delta h = \frac{1\,287\,309}{3600}(24.45) = 357.586 \times 24.45 = \boxed{8743\ \text{kW}}$$ The diagram itself prints 8741 kW at the coupling, so the reading and the arithmetic agree to 0.02 %. The 8862 → 8743 kW step also tells us the drive train (turbine coupling, gear and bearings) is 98.7 % efficient, which is the right order for a direct-coupled feed-pump turbine.
(e) Hydraulic power delivered by the pump. The useful output is the pressure rise across the pump times the volume flow. The pump takes suction from the deaerator at 0.554 MPa and discharges at P_D = 19.8 MPa: $$P_{hyd} = \frac{\dot{M}_{fw}}{3600}\cdot\frac{\Delta p}{\rho} = 357.586 \times \frac{(19.8 - 0.554)\times 10^{3}}{912} = 357.586 \times 21.104 = \boxed{7547\ \text{kW}}$$ The ideal specific work is 21.10 kJ/kg against an actual enthalpy rise of 24.45 kJ/kg, and the 3.35 kJ/kg difference is the irreversible heating of the feedwater inside the pump — a real effect worth roughly 0.8 °C of temperature rise.
(f) Feedwater pump efficiency. Useful hydraulic output over shaft input: $$\eta_{pump} = \frac{P_{hyd}}{P_{shaft}} = \frac{7547}{8743} = \boxed{0.863 = 86.3\ \%}$$ That is a typical figure for a large multi-stage barrel feed pump at its design point, and the fact that it lands in the expected 82–88 % band is the final confirmation that the diagram has been read correctly.

Question 4 — final results
Part	Quantity	Result
(a)	Boiler heat input (main steam + reheat)	930 370 kW (774 568 + 155 802)
(a)	Steam cycle efficiency	46.2 % (heat rate 7789 kJ/kWh)
(b)	HP turbine shaft power	133 045 kW ≈ 133.0 MW
(c)	Steam power into the feed-pump turbine	8862 kW
(d)	Shaft power into the feed pump	8743 kW (diagram prints 8741 kW)
(e)	Hydraulic power out of the feed pump	7547 kW (Δp = 19.246 MPa)
(f)	Feedwater pump efficiency	86.3 %

Topic. Reading a manufacturer's turbine heat balance diagram — the working document of station performance engineering.

Key relations. $\dot{Q} = \dot{M}\Delta h$ with $\dot{M}$ in kg/h converted by dividing by 3600; boiler duty as main-steam plus reheat terms; turbine shaft power as $\sum \dot{M}_{in}h_{in} - \sum \dot{M}_{out}h_{out}$; pump hydraulic power $\dot{M}\Delta p/\rho$; pump efficiency $\eta = P_{hyd}/P_{shaft}$; heat rate $= 3600/\eta$.

Why this works. A heat balance diagram is an already-solved mass and energy balance: every stream carries its pressure, temperature, enthalpy and flow, and the drawing's internal consistency is what makes it trustworthy. The engineer's task is to draw the right boundary. Choose the boiler and reheater and you get heat input; choose the HP casing and you get its shaft power; choose the pump and you get its efficiency. The one subtlety is the gland and valve-stem leak-offs. They are admitted at throttle conditions and taken away at throttle enthalpy without expanding, so they must be removed from the working flow — but they are not lost from the cycle, since they return downstream, which is why they still appear in the boiler's 1 287 309 kg/h.

Common pitfall. Omitting the reheat term from part (a) inflates the efficiency from 46.2 % to 55.5 %, an error so large it should be caught by inspection. In part (b) the usual slip is letting the leak-off flow expand through the casing, which overstates the HP power by about 1.5 MW. In part (e) the pressure rise is required, not the discharge pressure; using 19.8 MPa outright instead of 19.8 − 0.554 raises the apparent pump efficiency to 88.8 % and quietly credits the pump with work it never had to do.

Source. El-Wakil, Powerplant Technology, Ch. 2 and Ch. 5 (turbine heat balances, feedwater heating trains, feed-pump drives); Granet & Bluestein, Thermodynamics and Heat Power, 6th ed., Ch. 9 and Tables A.1–A.4.

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