Question

7.4 A coal-fired 930-MW power plant is operating under the following conditions: Coal carbon content C = 80%, coal ash content A = 6%, and heating value LHV = 30 MJ/kg Cooling water temperature difference Atew = 10.5 K Plant electrical efficiency ne is 0.43 and capacity factor CF is 0.82. Boiler efficiency n = 0.92 ESP efficiency nesp = 99% Calculate (a) the plant fuel rate, (b) the hourly and annual CO, emission rates, (c) the specific CO2 emissions per MWh of electric power output, (d) the condenser heat rejection rate and cooling water rate, (e) the annual yield of ash, and (f) the annual emissions of particulates.

Answer 1

Answer Griven - Pelectric = 930 Hal coal carbon content) - 80% Ash content a= 6%. how Heating Value = 30 MI Differ. cooling wathe temp. otw = 10.5k na electric eff = 0.43 CF (Capacity facter) = 0.82 Mespa 99%. 6= 0.92 (Boiler efficiency) Plant fuel reale Energy offel Qf = pene - 23.00 - 2162.8 MT 6.43 plant Hourly fuel rale 3600 Of 259.536tln 3600 X 21628 30 xlooo LHV Hourly & annual CO2 emission, respectively mco2= 3.67x cx mf - 3.67 x 0.8% 259.536 = 762 tine mor, y = 8670 X CFx meo, = 8670 X 0.82 x 762 ; 5417 36 2.8 tl ye

per of co MWh electric power Specific emission 8coz: mco2 = 762 X 1600 930 MW 5 819,35 kg 9.35 h how w Pel Q6= Boiler useful heat Heat Condenser Heat Rejection rate Condenser | acd= ob- Pel = P4.no - pee = 2162.8x 0.92- 930 Pcd = 1059.716 MI J Cooling water rate McW CPW = 1.16 wh/kgk hgk (specific I heat 1059.776 = 87 8106 kg 1.168 10.5 2 Qcd Cow stw € annual yield of ash mash=8760 CF mq A 2 8760 x0.82% 259.536x0.06 a 111,858 metric tons /ys Annual particulate matter enussions Mpm = mash (1-"Esp) 111,858 (1-0.99) ; 1118.58 metric tons lys