22-Mec-B3 Energy Conversion and Power Generation · December 2014
Question 2 of 6: Steam Cycle and Coal Fired Plant
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 2: Steam Cycle and Coal Fired Plant (15 marks)
Given. Part I supplies the seven state enthalpies of a regenerative Rankine cycle directly, so no steam-table work is needed; Part II supplies a chain of three efficiencies and the coal and cooling-water properties for a 600 MW station.
Part I — state points from the attachment table (exam page 10)
Point
Pressure (MPa)
Temperature (°C)
Enthalpy (kJ/kg)
Condition
1 — condenser outlet
0.004
29
121
saturated water
2 — condensate pump delivery
0.6
29
122
subcooled water
3 — heater outlet
0.6
159
671
saturated water
4 — feed pump delivery
6
160
677
subcooled water
5 — boiler outlet / throttle
6
400
3177
superheated steam
6 — turbine extraction
0.6
159
2662
wet mixture
7 — turbine exhaust
0.004
29
1970
wet mixture
Find. Part I: the fraction of throttle steam bled to the heater and the resulting cycle efficiency. Part II: the cooling-water mass and volume flows at the permitted 11 °C rise, and the coal burn and ash production per day.
Figure 2.1 — The regenerative cycle on temperature–entropy axes, plotted from the entropy of each tabulated state. Points 5, 6 and 7 lie on one vertical because the expansion is isentropic; 1–2 and 3–4 are the two pumps, which barely move the state at this scale; 2–3 is the mixing that takes place inside the direct-contact heater.
Approach. Part I is an energy balance on the direct-contact heater to find the bleed fraction, followed by a work-and-heat accounting per kilogram of throttle steam. Part II is a chain of efficiencies from the coal to the terminals, with the condenser duty taken as the difference between the heat delivered to the steam cycle and the shaft work it produces.
Part I (b) — balance the direct-contact heater. Take one kilogram of throttle steam and let \(y\) be the fraction bled at 0.6 MPa. The heater receives \(y\) kg of extraction steam at \(h_6\) and \((1-y)\) kg of condensate at \(h_2\), and complete mixing delivers saturated water at \(h_3\):
$$y\,h_6 + (1-y)h_2 = h_3$$
$$y = \frac{h_3 - h_2}{h_6 - h_2} = \frac{671 - 122}{2662 - 122} = \frac{549}{2540} = \boxed{0.216}$$
So 21.6 % of the throttle steam never reaches the condenser. The physical reading is that raising 1 kg of water from 29 °C to 159 °C costs 549 kJ, and each kilogram of bled steam brings 2540 kJ of that with it.
Part I (c) — turbine work per kilogram of throttle steam. The full kilogram expands from 5 to 6, but only the surviving fraction carries on to the condenser:
$$w_t = (h_5 - h_6) + (1-y)(h_6 - h_7) = 515 + 0.784(692) = 1057.4\ \text{kJ/kg}$$
The 515 kJ/kg is common to every kilogram; the second term is the work the bled steam gives up by not expanding to the condenser, and it is precisely what regeneration trades away.
Pump work. The condensate extraction pump handles only the unextracted fraction, the boiler feed pump the whole flow:
$$w_p = (1-y)(h_2 - h_1) + (h_4 - h_3) = 0.784(1) + 6 = 6.8\ \text{kJ/kg}$$
$$w_{net} = 1057.4 - 6.8 = 1050.6\ \text{kJ/kg}$$
Pump work is well under one per cent of turbine work here, which is typical of a steam cycle and is why it is so often (wrongly) neglected.
Part I (c) — heat input and efficiency. The boiler receives feedwater at state 4, not at condenser conditions, and that is the whole benefit of the heater:
$$q_{in} = h_5 - h_4 = 3177 - 677 = 2500\ \text{kJ/kg}$$
$$\eta_{cycle} = \frac{w_{net}}{q_{in}} = \frac{1050.6}{2500} = \boxed{0.420 = 42.0\ \%}$$
For comparison, the same throttle and condenser conditions without the heater give \(\eta = (3177-1970-6)/(3177-127) = 39.4\ \%\), so the single heater is worth about 2.6 percentage points.
Part II — establish the efficiency chain. The three stated efficiencies act in series from the coal to the terminals:
$$\eta_{overall} = \eta_{cycle}\,\eta_{boiler}\,\eta_{gen} = 0.41 \times 0.94 \times 0.96 = 0.370$$
$$\dot{Q}_{fuel} = \frac{P_e}{\eta_{overall}} = \frac{600}{0.370} = 1621.7\ \text{MW}$$
Working forward again to check: the boiler passes \(0.94 \times 1621.7 = 1524.4\) MW to the steam, the cycle converts \(0.41 \times 1524.4 = 625.0\) MW to shaft work, and the generator delivers \(0.96 \times 625.0 = 600.0\) MW. The chain closes.
Part II (a) — condenser duty and cooling water. The condenser rejects the heat that entered the steam cycle and did not leave as shaft work; the 25 MW of generator loss goes to the generator's own coolers, not to the circulating water:
$$\dot{Q}_{rej} = \dot{Q}_{cycle} - W_{shaft} = 1524.4 - 625.0 = 899.4\ \text{MW}$$
Taking the full permitted rise so that the flow is the minimum that satisfies the licence condition,
$$\dot{M}_w = \frac{\dot{Q}_{rej}}{c_{p,w}\,\Delta T} = \frac{899\,390}{4.19 \times 11} = \boxed{19\,514\ \text{kg/s}}$$
$$\dot{V}_w = \frac{\dot{M}_w}{\rho} = \frac{19\,514}{1025} = \boxed{19.04\ \text{m}^3/\text{s}}$$
The water leaves at 13 + 11 = 24 °C. This is the standard trap in the question: crediting the condenser with the boiler stack loss as well would give \(1621.7 - 600 = 1021.7\) MW and overstate the cooling water by 14 %.
Part II (b) — coal and ash. The coal flow follows from the calorific value as received:
$$\dot{M}_{coal} = \frac{\dot{Q}_{fuel}}{CV} = \frac{1\,621\,690}{35\,000} = \boxed{46.33\ \text{kg/s}}$$
Over a day at full load,
$$M_{coal,day} = 46.33 \times 86\,400 = 4.003 \times 10^6\ \text{kg} = \boxed{4003\ \text{t/day}}$$
$$M_{ash,day} = 0.06 \times 4003 = \boxed{240.2\ \text{t/day}}$$
Both figures are worth a moment's reflection: 4000 t/day is roughly forty rail cars a day of coal in and about two and a half of ash out, which is what sets the size of the coal yard, the ash lagoon and the rail connection long before any thermodynamics is argued about.
