A biomass based 5 MW power plant operates on fuel wood A biomass based 5 MW power plant operates on fuel wood. The capital cost is Rs 25 crores. The cost of the fuel is Rs 3 per kg. The CV of biomass is 15000 kJ/kg. the plant efficiency is 25%. The other O&M cost is Rs 0.40/kWh. The net annual generation is 18 GWh. Assuming a 20 year life and discount rate 20% compute the annualized life cycle cost of the power plant. Compute cost of electricity generated in Rs/kWh. If a special price of Rs 8/kWh is offered to “green power”, compute the Net Present Value and the benefit to cost ratio for the company. Is this a viable investment? What is the internal rate of return for this project? What is plant load factor?
CRF(20,20%)= 0.205 Net annual generation = 18*10^6 kWh Load factor= 0.4109 Fuel required 18*10^6*3600= mf*cv*efficiency Mf=17280000kg Cost= 17280000*3 = 51840000 Rs O&M cost= 0.4*18*10^6= 7200000 Rs ALCC= CO*CRF +FUEL+O&M ALCC= 51250000+51840000+7200000 =11.03 crore Cost of Electricity generation = 110290000/18*10^6=6.12 Rs/kWh NPV Annual income(A)= 18*10^6* Rs8/kWh= 144000000 Rs NPV= A/crf –(cost) =144000000/0.202 -25crore= 45 crore B/C= 2.8 IRR= when d value NPV=0 IRR=57%
STOICHIOMETRY OF COMBUSTION DATA : Coal : Ultimate analysis : Proximate : Analysis C = 64.0 % FC = 50.19 % H = 4.1 % VM=30.11 % S = 1.7 % Ash = 7.7 % O =7.2 % MC = 12.0 % N= 1.3 % MC = 12.0 % Excess Air= 40 % Ash = 9.7 % Gross CV=27 MJ/kg. Design capacity: 4 tonnes/h, Feed water flow rate: 1450 kg/h, Steam pressure: 8.2 bar (abs) and steam temperature is 179° C. Enthlpy of steam= 2786 kJ/kg. Water preheat temperature: 65 C, Enthalpy of feed water = 293 kJ/kg, Coal consumption: 350kg/h Cpflue_gas=1.030 kJ/kg-K, Flue gas temp= 350 C, Ambient temp=35 C, Lfg= 2305 kJ/kg, Cp of steam= 1.88 kJ/kg-K
Calculate: (i) theoretical air required (ii) Actual air required (iii) Composition of flue gas (iv) heat loss due to moisture in fuel (v) heat carried away by flue gas (vi) heat loss due to moisture from burning hydrogen (vii) adiabatic flame temperature (viii) boiler efficiency by direct method Assume 40% excess air SOLUTION :- 1. Combustion of Carbon :
2. Combustion Hydrogen : 3. Combustion of Sulphur : (a) Theoritical O2 Required : Theoretical air required for complete combustion is : 8.604 kg/kg of fuel (b) Actual Air [40% excess] :- O2 N2 Air 2.7706 kg 9.2792 kg 12.0456 kg Air/Fuel = 12.0456 :1
Composition on mass basis % (c) Flue Gas Composition (With 40% Excess Air) : Constituent Mass kg/kg of fuel Composition on mass basis % Mol. Mass Moles Composition on volume/mole Basis % Wet Basis Dry basis Wet. Dry basis CO2 2.346 18.11 18.82 44 2.346/44= 0.0533 12.18 12.99 {MC {H2O 0.12 0.369 - H2O]fg 0.489 3.677 0.00 18 0.2717 6.204 SO2 0.034 0.2625 0.272 64 0.000531 0.1213 0.129 O2]ex 0.7916* 6.113 6.35 32 0.0247 5.64 6.02 {N2}fuel {N2}air 0.013 6.6625 N2]fg 9.288 71.72 74.54 28 0.3317 75.80 80.86 mwet 12.9486 kg/kg 100.00 0.4347wet mdry 12.4596 kg/kg 0.4374dry *2.8462-2.033
Adiabatic Flame Temperature :- GCV = LCV+mH2O. Levap 2700 0 = LCV+0.489 x 2305 LCV = 25872.86 kJ/kg 25872.86 = 12.4596 x 1.03x (Tad-35) Tad = 2051C Loss due to moisture in fuel =Mmc*Δh*100/GCV ={4.2(25-Ta)+2442.8+1.88(Tfg-25)}/27000 =361.416 kJ 1.338% Heat loss due to moisture from H2 Mh2o*Δh*100/GCV =9*H2*3011.8*100/27000 =4.116% or 1111.35 kJ
ESTIMATION OF BOILER EFFICIENCY DIRECT METHOD : 1.2 Efficiency of Boiler :- Efficiency= 38.25% Dry Gas Loss : [Stack Loss]
1.2 Efficiency of Boiler :- Dry Gas Loss : [Stack Loss]