1. 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 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?

2. CRF(20,20%)= 0.205
Net annual generation = 18*10^6 kWh
Load factor=
Fuel required
18*10^6*3600= mf*cv*efficiency
Mf= kg
Cost= *3 = Rs
O&M cost= 0.4*18*10^6= Rs
ALCC= CO*CRF +FUEL+O&M
ALCC=
=11.03 crore
Cost of Electricity generation = /18*10^6=6.12 Rs/kWh
NPV
Annual income(A)= 18*10^6* Rs8/kWh= Rs
NPV= A/crf –(cost)
= / crore= 45 crore
B/C= 2.8
IRR= when d value NPV=0
IRR=57%

3. STOICHIOMETRY OF COMBUSTION
DATA :
Coal :
Ultimate analysis : Proximate : Analysis
C = % FC = %
H = % VM= %
S = % Ash = %
O =7.2 % MC = %
N= %
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

4. 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 :

5. 2. Combustion Hydrogen :
3. Combustion of Sulphur :
(a) Theoritical O2 Required :
 Theoretical air required for complete combustion is : kg/kg of fuel
(b) Actual Air [40% excess] :-
O2  N2  Air
kg  kg  kg
Air/Fuel = :1

6. 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=
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
0.1213
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
kg/kg
100.00
0.4347wet
 mdry
kg/kg
0.4374dry *

7. Adiabatic Flame Temperature :-
GCV = LCV+mH2O. Levap
= LCV x 2305
LCV = kJ/kg
= x 1.03x (Tad-35)
Tad = 2051C
Loss due to moisture in fuel
=Mmc*Δh*100/GCV
={4.2(25-Ta) (Tfg-25)}/27000
= kJ
1.338%
Heat loss due to moisture from H2
Mh2o*Δh*100/GCV
=9*H2*3011.8*100/27000
=4.116% or kJ

8. ESTIMATION OF BOILER EFFICIENCY
DIRECT METHOD :
1.2 Efficiency of Boiler :-
Efficiency= 38.25%
Dry Gas Loss : [Stack Loss]

9. 1.2 Efficiency of Boiler :-
Dry Gas Loss : [Stack Loss]