diff options
Diffstat (limited to '3472/CH7/EX7.23/Example7_23.sce')
-rw-r--r-- | 3472/CH7/EX7.23/Example7_23.sce | 74 |
1 files changed, 74 insertions, 0 deletions
diff --git a/3472/CH7/EX7.23/Example7_23.sce b/3472/CH7/EX7.23/Example7_23.sce new file mode 100644 index 000000000..93ff6f367 --- /dev/null +++ b/3472/CH7/EX7.23/Example7_23.sce @@ -0,0 +1,74 @@ +// 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 : 7.23 :
+// Page number 83-84
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+cap_3sets = 600.0 // Capacity of 3 generators(kW)
+no_3 = 3.0 // Number of sets of 600 kW
+cap_4thset = 400.0 // Capacity of 4th generator set(kW)
+no_4 = 1.0 // Number of sets of 400 kW
+MD = 1600.0 // Maximum demand(kW)
+LF = 0.45 // Load factor
+cost_capital_kW = 1000.0 // Capital cost per kW installed capacity(Rs)
+cost_annual_per = 0.15 // Annual cost = 15% of capital cost
+cost_operation = 60000.0 // Annual operation cost(Rs)
+cost_maintenance = 30000.0 // Annual maintenance cost(Rs)
+fixed_maintenance = 1.0/3 // Fixed cost
+variable_maintenance = 2.0/3 // Variable cost
+cost_fuel_kg = 40.0/100 // Cost of fuel oil(Rs/kg)
+cost_oil_kg = 1.25 // Cost of lubricating oil(Rs/kg)
+calorific = 10000.0 // Calorific value of fuel(kcal/kg)
+oil_consum = 1.0/400 // Consumption of lubricating oil. 1kg for every 400kWh generated
+fuel_consum = 1.0/2 // Consumption of fuel. 1kg for every 2kWh generated
+n_gen = 0.92 // Generator efficiency
+heat_lost = 1.0/3 // Heat lost in the fuel to cooling water
+theta = 11.0 // Difference of temperature between inlet and outlet(°C)
+
+// Calculations
+// Case(a)
+rating_3set_A = cap_3sets/n_gen // Rating of first 3 sets(kW)
+rating_4th_A = cap_4thset/n_gen // Rating of 4th set(kW)
+// 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)
+total_invest = (no_3*cap_3sets+cap_4thset*no_4)*cost_capital_kW // Total investment(Rs)
+annual_cost = cost_annual_per*total_invest // Annual cost(Rs)
+maintenance_cost = fixed_maintenance*cost_maintenance // Maintenance cost(Rs)
+fixed_cost_total = annual_cost+maintenance_cost // Total fixed cost per annum(Rs)
+fuel_consumption = energy_B*fuel_consum // Fuel consumption(Kg)
+cost_fuel = fuel_consumption*cost_fuel_kg // Cost of fuel(Rs)
+oil_consumption = energy_B*oil_consum // Lubrication oil consumption(Kg)
+cost_oil = oil_consumption*cost_oil_kg // Cost of Lubrication oil(Rs)
+var_maintenance_cost = variable_maintenance*cost_maintenance // Variable part of maintenance cost(Rs)
+variable_cost_total = cost_fuel+cost_oil+var_maintenance_cost+cost_operation // Total variable cost per annum(Rs)
+cost_total_D = fixed_cost_total+variable_cost_total // Total cost per annum(Rs)
+cost_kWh_gen = cost_total_D/energy_B*100 // Cost per kWh generated(Paise)
+// Case(c)
+n_overall = energy_B*860/(fuel_consumption*calorific)*100 // Overall efficiency(%)
+// Case(d)
+weight_water_hr = heat_lost*fuel_consumption/(hours_year*theta)*calorific // Weight of cooling water required(kg/hr)
+weight_water_min = weight_water_hr/60.0 // Weight of cooling water required(kg/min)
+capacity_pump = weight_water_min*MD/avg_demand_B // Capacity of cooling water pump(kg/min)
+
+// Results
+disp("PART I - EXAMPLE : 7.23 : SOLUTION :-")
+printf("\nCase(a): Rating of first 3 sets of diesel engine = %.f kW", rating_3set_A)
+printf("\n Rating of 4th set of diesel engine = %.f kW", rating_4th_A)
+printf("\nCase(b): Annual energy produced = %.1e kWh", energy_B)
+printf("\nCase(c): Total fixed cost = Rs %.f ", fixed_cost_total)
+printf("\n Total variable cost = Rs %.f ", variable_cost_total)
+printf("\n Cost per kWh generated = %.f paise", cost_kWh_gen)
+printf("\nCase(d): Overall efficiency of the diesel plant = %.1f percent", n_overall)
+printf("\nCase(e): Quantity of cooling water required per round = %.2e kg/hr = %.f kg/min", weight_water_hr,weight_water_min)
+printf("\n Capacity of cooling-water pumps under maximum load = %.f kg/min \n", capacity_pump)
+printf("\nNOTE: Changes in obtained answer from that of textbook answer is due to more precision here')
|