Question 1 of 6: Thermal Power Plant — resource and emission audit of a 6 × 600 MW coal fired station

Question 1: Thermal Power Plant — resource and emission audit of a 6 × 600 MW coal fired station (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. One 600 MW unit of a six-unit coal fired station, with the cycle, boiler, fuel, steam, cooling-water and atmospheric data listed below; the station runs base load at a 75 % annual capacity factor.

Find. The full-load heat, fuel, ash, carbon dioxide, steam, cooling-water and make-up-water rates for one unit and for the station, and then the five annual station quantities at a 75 % capacity factor.

Approach. Work outwards from the electrical output: the cycle efficiency fixes the heat the steam cycle must absorb, the boiler efficiency fixes the fuel heat behind it, an energy balance on the cycle fixes the condenser duty, and the fuel and cooling-water streams then follow from mass balances and from a boiler energy balance that includes the reheat pass.

1. Part (a)(i) — heat into the steam cycle. The steam cycle efficiency is defined on the heat the working fluid receives, so $$\dot{Q}_{cycle}=\frac{P_{e}}{\eta_{c}}=\frac{6.00\times10^{5}}{0.48}=\boxed{1.250\times10^{6}\ \text{kJ}\cdot\text{s}^{-1}}$$ that is 1 250 MW of heat absorbed by the steam per unit.
2. Part (a)(ii) — heat released by the fuel. The boiler efficiency is the fraction of the fuel heat that reaches the steam, so the fuel heat is larger again by $1/\eta_{b}$: $$\dot{Q}_{fuel}=\frac{\dot{Q}_{cycle}}{\eta_{b}}=\frac{1.250\times10^{6}}{0.90}=1.3889\times10^{6}\ \text{kJ}\cdot\text{s}^{-1}$$ The difference, $1.3889\times10^{6}-1.250\times10^{6}=138\,889\ \text{kJ}\cdot\text{s}^{-1}$, leaves as dry flue gas, moisture and radiation loss up the stack and never reaches the water side. The overall thermal efficiency of the unit is therefore $\eta_{c}\eta_{b}=0.48\times0.90=0.432$, or 43.2 %.
3. Part (a)(iii) — heat rejected to the cooling water. Applying the first law to the steam cycle alone, everything absorbed that does not leave as electricity leaves through the condenser: $$\dot{Q}_{rej}=\dot{Q}_{cycle}-P_{e}=1.250\times10^{6}-6.00\times10^{5}=\boxed{6.50\times10^{5}\ \text{kJ}\cdot\text{s}^{-1}}$$ The stack loss is deliberately excluded here: it is rejected to the atmosphere by the flue gas, not to the circulating water.
4. Part (b) — heat rate. Heat rate is the fuel heat charged against each unit of electricity sent out, so it is simply the reciprocal of the overall efficiency expressed per kilowatt hour: $$HR=\frac{3600}{\eta_{c}\eta_{b}}=\frac{3600}{0.432}=\boxed{8333\ \text{kJ}\cdot(\text{kW}\cdot\text{h})^{-1}}$$ The Matla data sheet on page 9 quotes 8350.6 kJ/kW.h at maximum continuous rating, so the assumed efficiencies reproduce a real station to within 0.2 %.
5. Part (c) — coal consumption. A mass balance on the fuel gives $$\dot{m}_{coal}=\frac{\dot{Q}_{fuel}}{CV}=\frac{1.3889\times10^{6}}{20\,000}=\boxed{69.44\ \text{kg}\cdot\text{s}^{-1}\ \text{per unit}}$$ For all six units, $6\times69.44=416.67\ \text{kg}\cdot\text{s}^{-1}$, and converting to the units the question asks for, $416.67\times3600/1000=\boxed{1500\ \text{Mg}\cdot\text{h}^{-1}}$. The data sheet entry “coal consumed at full load 1500 tons per hour” is an exact match, which is a useful check that the calorific value and the efficiencies are mutually consistent.
6. Part (d) — ash production. Ash is an inert 25 % of the coal, so it leaves the furnace at $\dot{m}_{ash}=0.25\times69.44=\boxed{17.36\ \text{kg}\cdot\text{s}^{-1}}$ per unit, which is $62.5\ \text{Mg}\cdot\text{h}^{-1}$ per unit and $0.25\times1500=\boxed{375\ \text{Mg}\cdot\text{h}^{-1}}$ for the whole station. Typically about 80 % of this leaves as fly ash in the gas stream and is collected in the precipitators, the remainder as bottom ash.
7. Part (e) — carbon dioxide emission. Complete combustion of the carbon follows $\text{C}+\text{O}_{2}\rightarrow\text{CO}_{2}$, so each kilogram of carbon yields $44/12$ kilograms of carbon dioxide: $$\dot{m}_{CO_{2}}=c\,\dot{m}_{coal}\frac{44}{12}=0.60\times416.67\times\frac{44}{12}=916.7\ \text{kg}\cdot\text{s}^{-1}$$ that is 152.78 kg/s from each unit, and for the station $916.7\times3.6=\boxed{3300\ \text{Mg}\cdot\text{h}^{-1}}$ of carbon dioxide. Using the exact isotopic masses (12.011 and 44.009) changes this by less than 0.1 %.
8. Part (f) — steam flow from each boiler. The steam receives $\dot{Q}_{cycle}$ in two passes, the main evaporator and superheater from feedwater to main-steam conditions and the reheater from the cold-reheat to the hot-reheat condition, and the question directs that both passes carry the same mass flow: $$\dot{Q}_{cycle}=\dot{m}_{s}\left[(h_{1}-h_{fw})+(h_{rh,out}-h_{rh,in})\right]$$ From the supplied steam tables, $h_{1}=h(17\ \text{MPa},500\,{}^{\circ}\text{C})=3283.6$, $h_{fw}=h(20\ \text{MPa},280\,{}^{\circ}\text{C})=1231.5$, $h_{rh,out}=h(3\ \text{MPa},500\,{}^{\circ}\text{C})=3457.2$ and $h_{rh,in}=h(3\ \text{MPa},300\,{}^{\circ}\text{C})=2994.3\ \text{kJ}\cdot\text{kg}^{-1}$.
9. Part (f) continued — substitution. The two enthalpy rises are $3283.6-1231.5=2052.1$ and $3457.2-2994.3=462.9\ \text{kJ}\cdot\text{kg}^{-1}$, a total of $2515.0\ \text{kJ}\cdot\text{kg}^{-1}$ absorbed per kilogram of steam circulated. Hence $$\dot{m}_{s}=\frac{1.250\times10^{6}}{2515.0}=\boxed{497.0\ \text{kg}\cdot\text{s}^{-1}}$$ The reheat pass is only 18.4 % of the total heat absorption but it is what keeps the low-pressure exhaust dry, and the Matla data sheet figure of 453.4 kg/s of steam at maximum continuous rating is the same order, the difference reflecting that real station’s slightly different steam conditions.
10. Part (g) — circulating water flow. The condenser duty is carried away by a 12 K rise in the cooling water, so $$\dot{m}_{cw}=\frac{\dot{Q}_{rej}}{c_{p}\Delta T_{cw}}=\frac{6.50\times10^{5}}{4.190\times12}=12\,928\ \text{kg}\cdot\text{s}^{-1}$$ At a density of 1000 kg/m³ this is $\boxed{12.93\ \text{m}^{3}\cdot\text{s}^{-1}}$ per unit and, split between two 50 % duty pumps, $\boxed{6.46\ \text{m}^{3}\cdot\text{s}^{-1}}$ each. The data sheet lists the circulating water pumps at 6.46 m³/s each, an exact agreement that confirms both the rejection rate and the assumed temperature rise.
11. Part (h) — cooling tower make-up water. The evaporative-loss chart on page 10 is entered with the dry-bulb and wet-bulb temperatures of the site air. At $32\,{}^{\circ}\text{C}$ and 20 % relative humidity the humidity ratio is $0.622\,p_{w}/(p_{atm}-p_{w})=5.885\ \text{g}\cdot\text{kg}^{-1}$ of dry air, and the adiabatic-saturation (thermodynamic wet-bulb) temperature that this state implies is $17.0\,{}^{\circ}\text{C}$. Reading Fig. 7.138 at 32 °C dry bulb and 17 °C wet bulb gives an evaporative loss of about $0.385\ \text{m}^{3}$ per GJ rejected.
12. Part (h) continued — make-up rates. Expressing the rejection rate in the chart’s units, $\dot{Q}_{rej}=6.50\times10^{5}\times3600/10^{6}=2340\ \text{GJ}\cdot\text{h}^{-1}$ per unit, so $$\dot{V}_{mu}=0.385\times2340=\boxed{900.9\ \text{m}^{3}\cdot\text{h}^{-1}\ \text{per unit}}$$ and for the six-unit station $6\times900.9=\boxed{5405\ \text{m}^{3}\cdot\text{h}^{-1}}$. The data sheet quotes 1250 m³/min of evaporation for 1600 MW of towers, equivalent to about 940 m³/h per 600 MW unit, so the chart reading is confirmed to within about 4 %.
13. Part (i) — annual station requirements. A 75 % capacity factor is equivalent to $0.75\times8760=6570$ hours of full-load operation per year. Multiplying each station rate by that figure, the annual electricity production is $$E=N P_{e}\,t=6\times6.00\times10^{5}\times6570/10^{6}=\boxed{23\,652\ \text{GWh}}$$ and the annual coal, ash, make-up water and carbon dioxide quantities follow in the same way from the station rates of 1500, 375, 5405 and 3300 units per hour respectively, as collected in the results table.