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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 : 7.24 :
// Page number 84
clear ; clc ; close ; // Clear the work space and console
// Given data
cap_installed = 30.0*10**3 // Rating of each generators(kW)
no = 4.0 // Number of installed generators
MD = 100.0*10**3 // Maximum demand(kW)
LF = 0.8 // Load factor
cost_capital_kW = 800.0 // Capital cost per kW installed capacity(Rs)
depreciation_per = 0.125 // Depreciation,etc = 12.5% of capital cost
cost_operation = 1.2*10**6 // Annual operation cost(Rs)
cost_maintenance = 600000.0 // Annual maintenance cost(Rs)
fixed_maintenance = 1.0/3 // Fixed cost
variable_maintenance = 2.0/3 // Variable cost
cost_miscellaneous = 100000.0 // Miscellaneous cost(Rs)
cost_fuel_kg = 32.0/1000 // Cost of fuel oil(Rs/kg)
calorific = 6400.0 // Calorific value of fuel(kcal/kg)
n_gen = 0.96 // Generator efficiency
n_thermal = 0.28 // Thermal efficiency of turbine
n_boiler = 0.75 // Boiler efficiency
n_overall = 0.2 // Overall thermal efficiency
// Calculations
// Case(a)
rating_turbine = cap_installed/(n_gen*0.736) // Rating of each steam turbine(metric hp)
// Case(b)
avg_demand_B = LF*MD // Average demand(kW)
hours_year = 365.0*24 // Total hours in a year
energy_B = avg_demand_B*hours_year // Annual energy produced(kWh)
// Case(c)
steam_consumption_C = (0.8+3.5*LF)/LF // Average steam consumption(kg/kWh)
// Case(d)
LF_D = 1.0 // Assumption that Load factor for boiler
steam_consumption_D = (0.8+3.5*LF_D)/LF_D // Steam consumption(kg/kWh)
energy_D = cap_installed*1.0 // Energy output per hour per set(kWh)
evaporation_cap = steam_consumption_D*energy_D // Evaporation capacity of boiler(kg/hr)
// Case(e)
total_invest = no*cap_installed*cost_capital_kW // Total investment(Rs)
capital_cost = depreciation_per*total_invest // Capital cost(Rs)
maintenance_cost = fixed_maintenance*cost_maintenance // Maintenance cost(Rs)
fixed_cost_total = capital_cost+maintenance_cost // Total fixed cost per annum(Rs)
var_maintenance_cost = variable_maintenance*cost_maintenance // Variable part of maintenance cost(Rs)
input_E = energy_B/n_overall // Input into system per annum(kWh)
weight_fuel = input_E*860/calorific // Weight of fuel(kg)
cost_fuel = weight_fuel*cost_fuel_kg // Cost of fuel(Rs)
variable_cost_total = cost_operation+var_maintenance_cost+cost_miscellaneous+cost_fuel // Total variable cost per annum(Rs)
cost_total_E = fixed_cost_total+variable_cost_total // Total cost per annum(Rs)
cost_kWh_gen = cost_total_E/energy_B*100 // Cost per kWh generated(Paise)
// Results
disp("PART I - EXAMPLE : 7.24 : SOLUTION :-")
printf("\nCase(a): Rating of each steam turbine = %.f metric hp", rating_turbine)
printf("\nCase(b): Energy produced per annum = %.3e kWh", energy_B)
printf("\nCase(c): Average steam consumption per kWh = %.1f kg/kWh", steam_consumption_C)
printf("\nCase(d): Evaporation capacity of boiler = %.f kg/hr", evaporation_cap)
printf("\nCase(e): Total fixed cost = Rs %.2e ", fixed_cost_total)
printf("\n Total variable cost = Rs %.2e ", variable_cost_total)
printf("\n Cost per kWh generated = %.2f paise\n", cost_kWh_gen)
printf("\nNOTE: Changes in obtained answer from that of textbook answer is due to more precision here')
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