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// 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 : 7.12 :
// Page number 76-77
clear ; clc ; close ; // Clear the work space and console
// Given data
cap_installed = 500.0 // Installed capacity of the station(MW)
CF = 0.45 // Capacity factor
LF = 0.6 // Annual laod factor
cost_fueloil = 10.0*10**7 // Annual cost of fuel,oil etc(Rs)
capital_cost = 10**9 // Capital cost(Rs)
interest = 0.15 // Interest and depreciation
// Calculations
// Case(i)
MD = cap_installed*CF/LF // Maximum demand(MW)
cap_reserve = cap_installed-MD // Reserve capacity(MW)
// Case(ii)
hours_year = 365.0*24 // Total hours in a year
units_gen = MD*10**3*LF*hours_year // Units generated per annum(kWh)
fixed_charge = interest*capital_cost // Annual fixed charges(Rs)
running_charge = cost_fueloil // Annual running charges(Rs)
annual_charge = fixed_charge+running_charge // Total annual charges(Rs)
cost_unit = annual_charge*100/units_gen // Cost per kWh generated(Paise)
// Results
disp("PART I - EXAMPLE : 7.12 : SOLUTION :-")
printf("\nCase(i) : Minimum reserve capacity of station = %.f MW", cap_reserve)
printf("\nCase(ii): Cost per kWh generated = %.f paise", cost_unit)
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