blob: 51ce0fcf987604c1f6793e14726efbf1e05f76a3 (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
|
// A Texbook on POWER SYSTEM ENGINEERING
// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
// DHANPAT RAI & Co.
// SECOND EDITION
// PART I : GENERATION
// CHAPTER 7: TARIFFS AND ECONOMIC ASPECTS IN POWER GENERATION
// EXAMPLE : 7.15 :
// Page number 78
clear ; clc ; close ; // Clear the work space and console
// Given data
cap_installed = 100.0*10**3 // Installed capacity of station(kW)
cost_gen = 30.0 // Generating cost per annum(Rs/kW)
cost_fixed = 4000000.0 // Fixed cost per annum(Rs)
cost_fuel = 60.0 // Cost of fuel(Rs/tonne)
calorific = 5700.0 // Calorific value of fuel(kcal/kg)
rate_heat_1 = 2900.0 // Plant heat rate at 100% capacity factor(kcal/kWh)
CF_1 = 1.0 // Capacity factor
rate_heat_2 = 4050.0 // Plant heat rate at 50% capacity factor(kcal/kWh)
CF_2 = 0.5 // Capacity factor
// Calculations
cost_fixed_kW = cost_fixed/cap_installed // Fixed cost per kW(Rs)
cost_fixed_total = cost_gen+cost_fixed_kW // Fixed cost per kW capacity(Rs)
average_demand_1 = CF_1*cap_installed // Average demand at 100% capacity factor(kW)
average_demand_2 = CF_2*cap_installed // Average demand at 50% capacity factor(kW)
hours_year = 365.0*24 // Total hours in a year
unit_gen_1 = CF_1*hours_year // Energy generated per annum with average demand of 1 kW(kWh)
unit_gen_2 = CF_2*hours_year // Energy generated per annum with average demand of 0.5 kW(kWh)
cost_kWh_fixed_1 = cost_fixed_total*100/unit_gen_1 // Cost per kWh due to fixed charge with 100% CF(Paise)
cost_kWh_fixed_2 = cost_fixed_total*100/unit_gen_2 // Cost per kWh due to fixed charge with 50% CF(Paise)
kg_kWh_1 = rate_heat_1/calorific // Weight(kg)
kg_kWh_2 = rate_heat_2/calorific // Weight(kg)
cost_coal_1 = kg_kWh_1*cost_fuel*100/1000.0 // Cost due to coal at 100% CF(Paise/kWh)
cost_coal_2 = kg_kWh_2*cost_fuel*100/1000.0 // Cost due to coal at 50% CF(Paise/kWh)
cost_total_1 = cost_kWh_fixed_1+cost_coal_1 // Total cost per unit with 100% CF(Paise)
cost_total_2 = cost_kWh_fixed_2+cost_coal_2 // Total cost per unit with 50% CF(Paise)
// Results
disp("PART I - EXAMPLE : 7.15 : SOLUTION :-")
printf("\nOverall generating cost per unit at 100 percent capacity factor = %.3f paise", cost_total_1)
printf("\nOverall generating cost per unit at 50 percent capacity factor = %.3f paise\n", cost_total_2)
printf("\nNOTE: Slight changes in obtained answer from that of textbook answer is due to more precision here")
|
