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| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3472/CH7 | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '3472/CH7')
29 files changed, 1192 insertions, 0 deletions
diff --git a/3472/CH7/EX7.1/Example7_1.sce b/3472/CH7/EX7.1/Example7_1.sce new file mode 100644 index 000000000..116a3826e --- /dev/null +++ b/3472/CH7/EX7.1/Example7_1.sce @@ -0,0 +1,29 @@ +// 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 : 7.1 :
+// Page number 73
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+connected_load = 450.0*10**3 // Connected load(kW)
+maximum_demand = 250.0*10**3 // Maximum demand(kW)
+units_generated = 615.0*10**6 // Units generated per annum(kWh)
+
+// Calculations
+// Case(i)
+demand_factor = maximum_demand/connected_load // Demand factor
+// Case(ii)
+hours_year = 365.0*24 // Total hours in a year
+average_demand = units_generated/hours_year // Average demand(kW)
+load_factor = average_demand/maximum_demand*100 // Load factor(%)
+
+// Results
+disp("PART I - EXAMPLE : 7.1 : SOLUTION :-")
+printf("\nCase(i) : Demand factor = %.3f ", demand_factor)
+printf("\nCase(ii): Load factor = %.1f percent", load_factor)
diff --git a/3472/CH7/EX7.10/Example7_10.sce b/3472/CH7/EX7.10/Example7_10.sce new file mode 100644 index 000000000..574c4fb72 --- /dev/null +++ b/3472/CH7/EX7.10/Example7_10.sce @@ -0,0 +1,44 @@ +// 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 : 7.10 :
+// Page number 76
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+cap_installed = 210.0*10**3 // Installed capacity of the station(kW)
+capital_cost_kW = 1000.0 // Capital cost of station(Rs/kW)
+fixed_cost_per = 0.13 // Fixed cost = 13% * cost of investment
+variable_cost_per = 1.3 // Variable cost = 1.3*fixed cost
+LF_1 = 1.0 // Load factor
+LF_2 = 0.5 // Load factor
+
+// Calculations
+MD = cap_installed // Maximum demand(kW)
+hours_year = 365.0*24 // Total hours in a year
+capital_cost = capital_cost_kW*cap_installed // Capital cost of station(Rs)
+// Case(i) At 100% load factor
+fixed_cost_1 = capital_cost*fixed_cost_per // Fixed cost(Rs)
+variable_cost_1 = variable_cost_per*fixed_cost_1 // Variable cost(Rs)
+operating_cost_1 = fixed_cost_1+variable_cost_1 // Operating cost per annum(Rs)
+units_gen_1 = LF_1*MD*hours_year // Total units generated(kWh)
+cost_gen_1 = operating_cost_1*100/units_gen_1 // Cost of generation per kWh(Paise)
+// Case(ii) At 50% load factor
+fixed_cost_2 = capital_cost*fixed_cost_per // Fixed cost(Rs)
+units_gen_2 = LF_2*MD*hours_year // Total units generated(kWh)
+variable_cost_2 = variable_cost_1*units_gen_2/units_gen_1 // Variable cost(Rs)
+operating_cost_2 = fixed_cost_2+variable_cost_2 // Operating cost per annum(Rs)
+cost_gen_2 = operating_cost_2*100/units_gen_2 // Cost of generation per kWh(Paise)
+
+// Results
+disp("PART I - EXAMPLE : 7.10 : SOLUTION :-")
+printf("\nCost of generation per kWh at 100 percent load factor = %.2f paise", cost_gen_1)
+printf("\nCost of generation per kWh at 50 percent load factor = %.1f paise", cost_gen_2)
+printf("\nComment: As the load factor is reduced, cost of generation is increased\n")
+printf("\nNOTE: ERROR: (1) In problem statement, Capital cost of station must be Rs. 1000/kW, not Rs. 1000/MW")
+printf("\n (2) Calculation mistake in Total units generated in Case(i) in textbook")
diff --git a/3472/CH7/EX7.11/Example7_11.sce b/3472/CH7/EX7.11/Example7_11.sce new file mode 100644 index 000000000..5840303c2 --- /dev/null +++ b/3472/CH7/EX7.11/Example7_11.sce @@ -0,0 +1,31 @@ +// 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 : 7.11 :
+// Page number 76
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+MD = 100.0*10**3 // Maximum demand(kW)
+capital_cost = 200.0*10**6 // Capital cost(Rs)
+LF = 0.4 // Annual load factor
+cost_fueloil = 15.0*10**6 // Annual cost of fuel and oil(Rs)
+cost_tax = 10.0*10**6 // Cost of taxes, wages and salaries(Rs)
+interest = 0.15 // Interest and depreciation
+
+// Calculations
+hours_year = 365.0*24 // Total hours in a year
+units_gen = MD*LF*hours_year // Units generated per annum(kWh)
+fixed_charge = interest*capital_cost // Annual fixed charges(Rs)
+running_charge = cost_fueloil+cost_tax // Annual running charges(Rs)
+annual_charge = fixed_charge+running_charge // Total annual charges(Rs)
+cost_unit = annual_charge*100/units_gen // Cost per unit(Paise)
+
+// Results
+disp("PART I - EXAMPLE : 7.11 : SOLUTION :-")
+printf("\nCost per unit generated = %.f paise", cost_unit)
diff --git a/3472/CH7/EX7.12/Example7_12.sce b/3472/CH7/EX7.12/Example7_12.sce new file mode 100644 index 000000000..ac7162455 --- /dev/null +++ b/3472/CH7/EX7.12/Example7_12.sce @@ -0,0 +1,36 @@ +// 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)
diff --git a/3472/CH7/EX7.13/Example7_13.sce b/3472/CH7/EX7.13/Example7_13.sce new file mode 100644 index 000000000..21e42734d --- /dev/null +++ b/3472/CH7/EX7.13/Example7_13.sce @@ -0,0 +1,47 @@ +// 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 : 7.13 :
+// Page number 77
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+gen_expense = 850000.0 // Annual generation expense(Rs)
+fuel_expense = 2800000.0 // Annual fuel expense(Rs)
+trans_expense = 345000.0 // Annual transmission expense(Rs)
+dist_expense = 2750000.0 // Annual distribution expense(Rs)
+repair_expense = 300000.0 // Annual repairs,etc expense(Rs)
+unit_gen = 600.0*10**6 // Number of units generated per year(kWh)
+MD = 75.0*10**3 // Maximum demand(kW)
+gen = 0.9 // Fixed charges for generation
+fuel = 0.15 // Fixed charges for fuel
+transm = 0.85 // Fixed charges for transmission
+dist = 0.95 // Fixed charges for distribution
+repair = 0.5 // Fixed charges for repairs,etc
+loss_dist = 0.2 // Losses in transmission and distribution
+
+// Calculations
+fixed_gen = gen_expense*gen // Fixed charge on generation(Rs)
+running_gen = gen_expense*(1-gen) // Running charge on generation(Rs)
+fixed_fuel = fuel_expense*fuel // Fixed charge on fuel(Rs)
+running_fuel = fuel_expense*(1-fuel) // Running charge on fuel(Rs)
+fixed_trans = trans_expense*transm // Fixed charge on transmission(Rs)
+running_trans = trans_expense*(1-transm) // Running charge on transmission(Rs)
+fixed_dist = dist_expense*dist // Fixed charge on distribution(Rs)
+running_dist = dist_expense*(1-dist) // Running charge on distribution(Rs)
+fixed_repair = repair_expense*repair // Fixed charge on repairs,etc(Rs)
+running_repair = repair_expense*(1-repair) // Running charge on repairs,etc(Rs)
+fixed_charge = fixed_gen+fixed_fuel+fixed_trans+fixed_dist+fixed_repair // Total fixed charges(Rs)
+running_charge = running_gen+running_fuel+running_trans+running_dist+running_repair // Total running charges(Rs)
+fixed_unit = fixed_charge/MD // Fixed charges per unit(Rs)
+units_dist = unit_gen*(1-loss_dist) // Total number of units distributed(kWh)
+running_unit = running_charge*100/units_dist // Running charges per unit(Paise)
+
+// Results
+disp("PART I - EXAMPLE : 7.13 : SOLUTION :-")
+printf("\nTwo part tariff is Rs %.3f per kW of maximum demand plus %.3f paise per kWh", fixed_unit,running_unit)
diff --git a/3472/CH7/EX7.14/Example7_14.sce b/3472/CH7/EX7.14/Example7_14.sce new file mode 100644 index 000000000..0c51cd69f --- /dev/null +++ b/3472/CH7/EX7.14/Example7_14.sce @@ -0,0 +1,37 @@ +// 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 : 7.14 :
+// Page number 77
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+cap_installed = 100.0*10**3 // Installed capacity of the station(kW)
+capital_cost_kW = 1000.0 // Capital cost(Rs/kW)
+depreciation = 0.15 // Annual depreciation charge
+royalty_kW = 2.0 // Royalty per kW per year(Rs)
+royalty_kWh = 0.03 // Royalty per kWh per year(Rs)
+MD = 70.0*10**3 // Maximum demand(kW)
+LF = 0.6 // Annual load factor
+cost_salary = 1000000.0 // Annual cost of salaries,maintenance charges etc(Rs)
+cost_salary_per = 0.2 // Annual cost of salaries,maintenance charges etc charged as fixed charges
+
+// Calculations
+hours_year = 365.0*24 // Total hours in a year
+unit_gen = MD*LF*hours_year // Units generated/annum(kWh)
+capital_cost = cap_installed*capital_cost_kW // Capital cost of plant(Rs)
+depreciation_charge = depreciation*capital_cost // Depreciation charges(Rs)
+salary_charge = cost_salary_per*cost_salary // Cost on salaries, maintenance etc(Rs)
+fixed_charge = depreciation_charge+salary_charge // Total annual fixed charges(Rs)
+cost_kW_fixed = (fixed_charge/MD)+royalty_kW // Cost per kW(Rs)
+salary_charge_running = (1-cost_salary_per)*cost_salary // Annual running charge on salaries, maintenance etc(Rs)
+cost_kWh_running = (salary_charge_running/unit_gen)+royalty_kWh // Cost per kWh(Rs)
+
+// Results
+disp("PART I - EXAMPLE : 7.14 : SOLUTION :-")
+printf("\nGeneration cost in two part form is given by, Rs. (%.2f*kW + %.3f*kWh) ", cost_kW_fixed,cost_kWh_running)
diff --git a/3472/CH7/EX7.15/Example7_15.sce b/3472/CH7/EX7.15/Example7_15.sce new file mode 100644 index 000000000..51ce0fcf9 --- /dev/null +++ b/3472/CH7/EX7.15/Example7_15.sce @@ -0,0 +1,45 @@ +// 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")
diff --git a/3472/CH7/EX7.16/Example7_16.sce b/3472/CH7/EX7.16/Example7_16.sce new file mode 100644 index 000000000..219ab6581 --- /dev/null +++ b/3472/CH7/EX7.16/Example7_16.sce @@ -0,0 +1,43 @@ +// 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 : 7.16 :
+// Page number 78
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+MD = 75.0*10**3 // Maximum demand(kW)
+LF = 0.4 // Yearly load factor
+cost_capital = 60.0 // Capital cost(Rs/annum/kW)
+cost_kWh = 1.0 // Cost per kWh transmitted(Paise)
+charge_trans = 2000000.0 // Annual capital charge for transmission(Rs)
+charge_dist = 1500000.0 // Annual capital charge for distribution(Rs)
+diversity_trans = 1.2 // Diversity factor for transmission
+diversity_dist = 1.25 // Diversity factor for distribution
+n_trans = 0.9 // Efficiency of transmission system
+n_dist = 0.85 // Efficiency of distribution system
+
+// Calculations
+// Case(a)
+capital_cost = cost_capital*MD // Annual capital cost(Rs)
+fixed_charge_sub = capital_cost+charge_trans // Total fixed charges for supply to substation per annum(Rs)
+sum_MD_sub = MD*diversity_trans // Sum of all maximum demand of substation(kW)
+cost_kW_sub = fixed_charge_sub/sum_MD_sub // Yearly cost per kW demand at substation(Rs)
+running_cost_unit_sub = 1/n_trans // Running cost per unit supplied at substation(Paise)
+// Case(b)
+sum_MD_con = sum_MD_sub*diversity_dist // Sum of all maximum demand of consumer(kW)
+fixed_charge_con = capital_cost+charge_trans+charge_dist // Total fixed charges for supply to cosnumers(Rs)
+cost_kW_con = fixed_charge_con/sum_MD_con // Yearly cost per kW demand on consumer premises(Rs)
+running_cost_unit_con = running_cost_unit_sub/n_dist // Running cost per unit supplied to consumer(Paise)
+
+// Results
+disp("PART I - EXAMPLE : 7.16 : SOLUTION :-")
+printf("\nCase(a): Yearly cost per kW demand at the substations = Rs. %.2f ", cost_kW_sub)
+printf("\n Cost per kWh supplied at the substations = %.2f paise\n", running_cost_unit_sub)
+printf("\nCase(b): Yearly cost per kW demand at the consumer premises = Rs. %.2f ", cost_kW_con)
+printf("\n Cost per kWh supplied at the consumer premises = %.3f paise", running_cost_unit_con)
diff --git a/3472/CH7/EX7.17/Example7_17.sce b/3472/CH7/EX7.17/Example7_17.sce new file mode 100644 index 000000000..b6595d582 --- /dev/null +++ b/3472/CH7/EX7.17/Example7_17.sce @@ -0,0 +1,39 @@ +// 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 : 7.17 :
+// Page number 79
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+kVA_tariff_hv = 60.0 // HV supply per kVA per annum(Rs)
+kWh_tariff_hv = 3.0/100 // HV supply per kWh annum(Rs)
+kVA_tariff_lv = 65.0 // LV supply per kVA per annum(Rs)
+kWh_tariff_lv = 3.3/100 // LV supply per kWh annum(Rs)
+cost_equip_kVA = 50.0 // Cost of transformers and switchgear per kVA(Rs)
+loss_full_load = 0.02 // Full load transformation loss
+fixed_charge_per = 0.2 // Fixed charges per annum
+no_week = 50.0 // Number of working weeks in a year
+
+// Calculations
+rating_equip = 1000/(1-loss_full_load) // Rating of transformer and switchgear(kVA)
+cost_equip = cost_equip_kVA*rating_equip // Cost of transformers and switchgear(Rs)
+fixed_charge = fixed_charge_per*cost_equip // Fixed charges per annum on HV plant(Rs)
+X = poly(0,"X") // Number of working hours per week
+units_consumed = (no_week*X)*1000.0 // Yearly units consumed by load
+total_units = units_consumed/(1-loss_full_load) // Total units to be paid on HV supply
+// Case(a)
+annual_cost_hv = (kVA_tariff_hv*rating_equip)+(kWh_tariff_hv*cost_equip*X)+fixed_charge // Annual cost(Rs)
+// Case(b)
+annual_cost_lv = (kVA_tariff_lv*1000.0)+(kWh_tariff_lv*units_consumed) // Annual cost(Rs)
+p = annual_cost_hv-annual_cost_lv // Finding unknown value i.e working hours in terms of X
+x = roots(p) // Finding unknown value i.e working hours
+
+// Results
+disp("PART I - EXAMPLE : 7.17 : SOLUTION :-")
+printf("\nAbove %.1f working hours per week the H.V supply is cheaper ", x)
diff --git a/3472/CH7/EX7.18/Example7_18.sce b/3472/CH7/EX7.18/Example7_18.sce new file mode 100644 index 000000000..d3a7851d1 --- /dev/null +++ b/3472/CH7/EX7.18/Example7_18.sce @@ -0,0 +1,61 @@ +// 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 : 7.18 :
+// Page number 79-80
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+load_1 = 10.0*10**3 // Load per annum(kVA)
+time_1 = 1800.0 // Time(hours)
+load_2 = 6.0*10**3 // Load per annum(kVA)
+time_2 = 600.0 // Time(hours)
+load_3 = 0.25*10**3 // Load per annum(kVA)
+time_3 = 400.0 // Time(hours)
+rating_trans = 10.0*10**3 // Transformer rating(kVA)
+pf = 0.8 // Lagging power factor
+n_fl_A = 98.3/100.0 // Full load efficiency of transformer A
+n_fl_B = 98.8/100.0 // Full load efficiency of transformer B
+loss_A = 70.0 // Core loss at rated voltage of transformer A(kW)
+loss_B = 40.0 // Core loss at rated voltage of transformer B(kW)
+cost_A = 250000.0 // Cost of transformer A(Rs)
+cost_B = 280000.0 // Cost of transformer B(Rs)
+interest_per = 0.1 // Interest and depreciation charges
+cost_energy_unit = 3.0 // Energy costs per unit(Paise)
+
+// Calculations
+// Transformer A
+output_A = rating_trans*pf // kW output at full load(kW)
+input_A = output_A/n_fl_A // Input at full load(kW)
+cu_loss_fl_A = input_A-output_A-loss_A // Copper loss at full load(kW)
+cu_loss_2_A = (load_2/load_1)**2*cu_loss_fl_A // Copper loss at 6 MVA output(kW)
+cu_loss_3_A = (load_3/load_1)**2*cu_loss_fl_A // Copper loss at 0.25 MVA output(kW)
+ene_iron_loss_A = loss_A*(time_1+time_2+time_3) // Energy consumed due to iron losses(kWh)
+ene_cu_loss_A = time_1*cu_loss_fl_A+time_2*cu_loss_2_A+time_3*cu_loss_3_A // Energy consumed due to copper losses(kWh)
+total_loss_A = ene_iron_loss_A+ene_cu_loss_A // Total loss per annum(kWh)
+cost_energy_A = cost_energy_unit/100*total_loss_A // Energy cost per annum due to losses(Rs)
+// Transformer B
+output_B = rating_trans*pf // kW output at full load(kW)
+input_B = output_B/n_fl_B // Input at full load(kW)
+cu_loss_fl_B = input_B-output_B-loss_B // Copper loss at full load(kW)
+cu_loss_2_B = (load_2/load_1)**2*cu_loss_fl_B // Copper loss at 6 MVA output(kW)
+cu_loss_3_B = (load_3/load_1)**2*cu_loss_fl_B // Copper loss at 0.25 MVA output(kW)
+ene_iron_loss_B = loss_B*(time_1+time_2+time_3) // Energy consumed due to iron losses(kWh)
+ene_cu_loss_B = time_1*cu_loss_fl_B+time_2*cu_loss_2_B+time_3*cu_loss_3_B // Energy consumed due to copper losses(kWh)
+total_loss_B = ene_iron_loss_B+ene_cu_loss_B // Total loss per annum(kWh)
+cost_energy_B = cost_energy_unit/100*total_loss_B // Energy cost per annum due to losses(Rs)
+diff_capital = cost_B-cost_A // Difference in capital costs(Rs)
+annual_charge = interest_per*diff_capital // Annual charge due to this amount(Rs)
+diff_cost_energy = cost_energy_A-cost_energy_B // Difference in energy cost per annum(Rs)
+cheap = diff_cost_energy-annual_charge // Cheaper in cost(Rs)
+
+// Results
+disp("PART I - EXAMPLE : 7.18 : SOLUTION :-")
+printf("\nTransformer B is cheaper by Rs. %.f per year \n", cheap)
+printf("\nNOTE: ERROR: Full load efficiency for transformer B is 98.8 percent, not 98.3 percent as given in problem statement")
+printf("\n Changes in obtained answer from that of textbook answer is due to more precision")
diff --git a/3472/CH7/EX7.19/Example7_19.sce b/3472/CH7/EX7.19/Example7_19.sce new file mode 100644 index 000000000..531d82932 --- /dev/null +++ b/3472/CH7/EX7.19/Example7_19.sce @@ -0,0 +1,41 @@ +// 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 : 7.19 :
+// Page number 80-81
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+fixed_cost = 4.0*10**4 // Fixed cost of plant(Rs)
+salvage_value = 4.0*10**3 // Salvage value(Rs)
+n = 20.0 // Useful life(years)
+r = 0.06 // Sinking fund depreciation compounded annually
+
+// Calculations
+n_2 = n/2 // Halfway of useful life(years)
+// Case(a)
+total_dep_A = fixed_cost-salvage_value // Total depreciation in 20 years(Rs)
+dep_10_A = total_dep_A/2 // Depreciation in 10 years(Rs)
+value_10_A = fixed_cost-dep_10_A // Value at the end of 10 years(Rs)
+// Case(b)
+P_B = fixed_cost // Capital outlay(Rs)
+q_B = (salvage_value/fixed_cost)**(1/n) // q = (1-p)
+value_10_B = P_B*(q_B)**n_2 // Value at the end of 10 years(Rs)
+// Case(c)
+P_C = fixed_cost // Capital cost of plant(Rs)
+P__C = salvage_value // Scrap value(Rs)
+Q_C = P_C-P__C // Cost of replacement(Rs)
+q_C = Q_C/(((1+r)**n-1)/r) // Yearly charge(Rs)
+amount_dep = q_C*((1+r)**n_2-1)/r // Amount deposited at end of 10 years(Rs)
+value_10_C = P_C-amount_dep // Value at the end of 10 years(Rs)
+
+// Results
+disp("PART I - EXAMPLE : 7.19 : SOLUTION :-")
+printf("\nCase(a): Valuation halfway through its life based on Straight line depreciation method = Rs %.1e ", value_10_A)
+printf("\nCase(b): Valuation halfway through its life based on Reducing balance depreciation method = Rs %.2e ", value_10_B)
+printf("\nCase(c): Valuation halfway through its life based on Sinking fund depreciation method = Rs %.2e ", value_10_C)
diff --git a/3472/CH7/EX7.2/Example7_2.sce b/3472/CH7/EX7.2/Example7_2.sce new file mode 100644 index 000000000..e27794c52 --- /dev/null +++ b/3472/CH7/EX7.2/Example7_2.sce @@ -0,0 +1,23 @@ +// 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 : 7.2 :
+// Page number 73
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+maximum_demand = 480.0*10**3 // Maximum demand(kW)
+LF = 0.4 // Annual load factor
+
+// Calculation
+hours_year = 365.0*24 // Total hours in a year
+energy_gen = maximum_demand*LF*hours_year // Total energy generated annually(kWh)
+
+// Results
+disp("PART I - EXAMPLE : 7.2 : SOLUTION :-")
+printf("\nTotal energy generated annually = %.5e kWh", energy_gen)
diff --git a/3472/CH7/EX7.20/Example7_20.sce b/3472/CH7/EX7.20/Example7_20.sce new file mode 100644 index 000000000..517fa00dc --- /dev/null +++ b/3472/CH7/EX7.20/Example7_20.sce @@ -0,0 +1,38 @@ +// 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 : 7.20 :
+// Page number 81
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+h = 30.0 // Mean head(m)
+area_catch = 250.0 // Catchment area(Square km)
+average_rain = 1.25 // Average rainfall per annum(m)
+utilized_rain = 0.7 // Rainfall utilized
+LF = 0.8 // Expected load factor
+n_turbine = 0.9 // Mechanical efficiency of turbine
+n_gen = 0.95 // Efficiency of generator
+
+// Calculations
+water_avail = utilized_rain*area_catch*10**6*average_rain // Water available(m^3)
+sec_year = 365.0*24*60*60 // Total seconds in a year
+Q = water_avail/sec_year // Quantity available per second(m^3) i.e Discharge(m^3/sec)
+w = 1000.0 // Density of water(kg/m^3)
+n = n_turbine*n_gen // Overall efficiency
+P = 0.736/75*Q*w*h*n // Average output of generator units(kW)
+rating_gen = P/LF // Rating of generator(kW)
+rating_gen_each = rating_gen/2.0 // Rating of each generator(kW)
+rating_turbine = rating_gen/2*(1/(0.736*n_gen)) // Rating of each turbine(metric hp)
+
+// Results
+disp("PART I - EXAMPLE : 7.20 : SOLUTION :-")
+printf("\nChoice of units are:")
+printf("\n 2 generators each having maximum rating of %.f kW ", rating_gen_each)
+printf("\n 2 propeller turbines each having maximum rating of %.f metric hp \n", rating_turbine)
+printf("\nNOTE: Changes in obtained answer from that of textbook answer is due to more precision here')
diff --git a/3472/CH7/EX7.21/Ex7_21.png b/3472/CH7/EX7.21/Ex7_21.png Binary files differnew file mode 100644 index 000000000..4dee85be4 --- /dev/null +++ b/3472/CH7/EX7.21/Ex7_21.png diff --git a/3472/CH7/EX7.21/Example7_21.sce b/3472/CH7/EX7.21/Example7_21.sce new file mode 100644 index 000000000..844fd63fd --- /dev/null +++ b/3472/CH7/EX7.21/Example7_21.sce @@ -0,0 +1,94 @@ +// 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 : 7.21 :
+// Page number 81-82
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+t0 = 0.0 // Time 12 morning
+l0 = 4.0 // Load at 12 morning(kW*1000)
+t1 = 1.0 // Time 1 a.m
+l1 = 3.5 // Load at 1 a.m(kW*1000)
+t2 = 2.0 // Time 2 a.m
+l2 = 3.0 // Load at 2 a.m(kW*1000)
+t3 = 3.0 // Time 3 a.m
+l3 = 3.0 // Load at 3 a.m(kW*1000)
+t4 = 4.0 // Time 4 a.m
+l4 = 3.5 // Load at 4 a.m(kW*1000)
+t5 = 5.0 // Time 5 a.m
+l5 = 3.0 // Load at 5 a.m(kW*1000)
+t6 = 6.0 // Time 6 a.m
+l6 = 6.0 // Load at 6 a.m(kW*1000)
+t7 = 7.0 // Time 7 a.m
+l7 = 12.5 // Load at 7 a.m(kW*1000)
+t8 = 8.0 // Time 8 a.m
+l8 = 14.5 // Load at 8 a.m(kW*1000)
+t9 = 9.0 // Time 9 a.m
+l9 = 13.5 // Load at 9 a.m(kW*1000)
+t10 = 10.0 // Time 10 a.m
+l10 = 13.0 // Load at 10 a.m(kW*1000)
+t11 = 11.0 // Time 11 a.m
+l11 = 13.5 // Load at 11 a.m(kW*1000)
+t113 = 11.50 // Time 11.30 a.m
+l113 = 12.0 // Load at 11.30 am(kW*1000)
+t12 = 12.0 // Time 12 noon
+l12 = 11.0 // Load at 12 noon(kW*1000)
+t123 = 12.50 // Time 12.30 noon
+l123 = 5.0 // Load at 12.30 noon(kW*1000)
+t13 = 13.0 // Time 1 p.m
+l13 = 12.5 // Load at 1 p.m(kW*1000)
+t133 = 13.50 // Time 1.30 p.m
+l133 = 13.5 // Load at 1.30 p.m(kW*1000)
+t14 = 14.0 // Time 2 p.m
+l14 = 14.0 // Load at 2 p.m(kW*1000)
+t15 = 15.0 // Time 3 p.m
+l15 = 14.0 // Load at 3 p.m(kW*1000)
+t16 = 16.0 // Time 4 p.m
+l16 = 15.0 // Load at 4 p.m(kW*1000)
+t163 = 16.50 // Time 4.30 p.m
+l163 = 18.0 // Load at 4.30 p.m(kW*1000)
+t17 = 17.0 // Time 5 p.m
+l17 = 20.0 // Load at 5 p.m(kW*1000)
+t173 = 17.50 // Time 5.30 p.m
+l173 = 17.0 // Load at 5.30 p.m(kW*1000)
+t18 = 18.0 // Time 6 p.m
+l18 = 12.5 // Load at 6 p.m(kW*1000)
+t19 = 19.0 // Time 7 p.m
+l19 = 10.0 // Load at 7 p.m(kW*1000)
+t20 = 20.0 // Time 8 p.m
+l20 = 7.5 // Load at 8 p.m(kW*1000)
+t21 = 21.0 // Time 9 p.m
+l21 = 5.0 // Load at 9 p.m(kW*1000)
+t22 = 22.0 // Time 10 p.m
+l22 = 5.0 // Load at 10 p.m(kW*1000)
+t23 = 23.0 // Time 11 p.m
+l23 = 4.0 // Load at 11 p.m(kW*1000)
+t24 = 24.0 // Time 12 morning
+l24 = 4.0 // Load at 12 morning(kW*1000)
+
+// Calculations
+t = [t0,t1,t2,t3,t4,t5,t6,t7,t8,t9,t10,t11,t12,t13,t14,t15,t16,t17,t18,t19,t20,t21,t22,t23,t24]
+l = [l0,l1,l2,l3,l4,l5,l6,l7,l8,l9,l10,l11,l12,l13,l14,l15,l16,l17,l18,l19,l20,l21,l22,l23,l24]
+a = gca() ;
+a.thickness = 2 // sets thickness of plot
+plot(t,l,'ro-') // Plot of Chronological load curve
+T = [0,0.5,1,1.5,2.5,4.5,6,7,9,9.5,10,11,12,13,15.5,18.5,20.5,23.5,24] // Solved time
+L = [20,18,17,15,14.5,14,13.5,13,12.5,12,11,10,7.5,6,5,4,3.5,3,3] // Solved load
+plot(T,L,'--mo') // Plot of load duration curve
+a.x_label.text = 'Time & No. of hours' // labels x-axis
+a.y_label.text = 'Load in 10^3 kW' // labels y-axis
+xtitle("Fig E7.2 . Plot of Chronological load curve and load duration curve")
+xset('thickness',2) // sets thickness of axes
+xstring(17.5,17,'Chronological load curve')
+xstring(1.1,17,'Load duration curve')
+
+// Results
+disp("PART I - EXAMPLE : 7.21 : SOLUTION :-")
+printf("\nThe chronological load curve and the load duration curve is shown in the Figure E7.2\n")
+printf("\nNOTE: The time is plotted in 24 hours format')
diff --git a/3472/CH7/EX7.22/Example7_22.sce b/3472/CH7/EX7.22/Example7_22.sce new file mode 100644 index 000000000..96aa6f505 --- /dev/null +++ b/3472/CH7/EX7.22/Example7_22.sce @@ -0,0 +1,36 @@ +// 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 : 7.22 :
+// Page number 82
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+MD = 20.0*10**3 // Maximum demand(kW)
+LF = 0.6 // Load factor
+CF = 0.48 // Plant capacity factor
+UF = 0.8 // Plant use factor
+
+// Calculations
+// Case(a)
+avg_demand = LF*MD // Average demand(kW)
+ene_daily = avg_demand*24.0 // Daily energy produced(kWh)
+// Case(b)
+cap_installed = avg_demand/CF // Installed capacity(kW)
+cap_reserve = cap_installed-MD // Reserve capacity(kW)
+// Case(c)
+max_ene_C = cap_installed*24.0 // Maximum energy that could be produced daily(kWh)
+// Case(d)
+max_ene_D = ene_daily/UF // Maximum energy that could be produced daily as per schedule(kWh)
+
+// Results
+disp("PART I - EXAMPLE : 7.22 : SOLUTION :-")
+printf("\nCase(a): Daily energy produced = %.f kWh", ene_daily)
+printf("\nCase(b): Reserve capacity of plant = %.f kW", cap_reserve)
+printf("\nCase(c): Maximum energy that could be produced daily when plant runs at all time = %.f kWh", max_ene_C)
+printf("\nCase(d): Maximum energy that could be produced daily when plant runs fully loaded = %.f kWh", max_ene_D)
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')
diff --git a/3472/CH7/EX7.24/Example7_24.sce b/3472/CH7/EX7.24/Example7_24.sce new file mode 100644 index 000000000..840dedbac --- /dev/null +++ b/3472/CH7/EX7.24/Example7_24.sce @@ -0,0 +1,68 @@ +// 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')
diff --git a/3472/CH7/EX7.25/Ex7_25.png b/3472/CH7/EX7.25/Ex7_25.png Binary files differnew file mode 100644 index 000000000..b4434614f --- /dev/null +++ b/3472/CH7/EX7.25/Ex7_25.png diff --git a/3472/CH7/EX7.25/Example7_25.sce b/3472/CH7/EX7.25/Example7_25.sce new file mode 100644 index 000000000..e280847ac --- /dev/null +++ b/3472/CH7/EX7.25/Example7_25.sce @@ -0,0 +1,65 @@ +// 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 : 7.25 :
+// Page number 85
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+w1 = 1.0 // Week 1
+Q1 = 200.0 // Discharge during week 1(m^2/sec)
+w2 = 2.0 // Week 2
+Q2 = 300.0 // Discharge during week 2(m^2/sec)
+w3 = 3.0 // Week 3
+Q3 = 1100.0 // Discharge during week 3(m^2/sec)
+w4 = 4.0 // Week 4
+Q4 = 700.0 // Discharge during week 4(m^2/sec)
+w5 = 5.0 // Week 5
+Q5 = 900.0 // Discharge during week 5(m^2/sec)
+w6 = 6.0 // Week 6
+Q6 = 800.0 // Discharge during week 6(m^2/sec)
+w7 = 7.0 // Week 7
+Q7 = 600.0 // Discharge during week 7(m^2/sec)
+w8 = 8.0 // Week 8
+Q8 = 1000.0 // Discharge during week 8(m^2/sec)
+w9 = 9.0 // Week 9
+Q9 = 500.0 // Discharge during week 9(m^2/sec)
+w10 = 10.0 // Week 10
+Q10 = 400.0 // Discharge during week 10(m^2/sec)
+w11 = 11.0 // Week 11
+Q11 = 500.0 // Discharge during week 11(m^2/sec)
+w12 = 12.0 // Week 12
+Q12 = 700.0 // Discharge during week 12(m^2/sec)
+w13 = 13.0 // Week 13
+Q13 = 100.0 // Discharge during week 13(m^2/sec)
+no_week = 13.0 // Total weeks of discharge
+
+// Calculations
+Q_average = (Q1+Q2+Q3+Q4+Q5+Q6+Q7+Q8+Q9+Q10+Q11+Q12+Q13)/no_week // Average weekly discharge(m^3/sec)
+// Hydrograph
+W = [0,w1,w1,w2,w2,w3,w3,w4,w4,w5,w5,w6,w6,w7,w7,w8,w8,w9,w9,w10,w10,w11,w11,w12,w12,w13,w13,w13]
+Q = [200,Q1,Q2,Q2,Q3,Q3,Q4,Q4,Q5,Q5,Q6,Q6,Q7,Q7,Q8,Q8,Q9,Q9,Q10,Q10,Q11,Q11,Q12,Q12,Q13,Q13,Q13,0]
+a = gca()
+a.thickness = 2 // sets thickness of plot
+plot(W,Q) // Plotting hydrograph
+q = Q_average
+w = [0,w1,w2,w3,w4,w5,w6,w7,w8,w9,w10,w11,w12,w13,14]
+q_dash = [q,q,q,q,q,q,q,q,q,q,q,q,q,q,q] // Plotting average weekly discharge
+plot(w,q_dash,'r--')
+a.x_label.text = 'Time(week)' // labels x-axis
+a.y_label.text = 'Q(m^3/sec)' // labels y-axis
+xtitle("Fig E7.4 . Plot of Hydrograph")
+xset('thickness',2) // sets thickness of axes
+xstring(13,560,'Q_av')
+xstring(12.02,110,'Q_min')
+xstring(2.02,1110,'Q_max')
+
+// Results
+disp("PART I - EXAMPLE : 7.25 : SOLUTION :-")
+printf("\nThe hydrograph is shown in the Figure E7.4")
+printf("\nAverage discharge available for the whole period = %.f m^3/sec", Q_average)
diff --git a/3472/CH7/EX7.26/Ex7_26.png b/3472/CH7/EX7.26/Ex7_26.png Binary files differnew file mode 100644 index 000000000..d02304619 --- /dev/null +++ b/3472/CH7/EX7.26/Ex7_26.png diff --git a/3472/CH7/EX7.26/Example7_26.sce b/3472/CH7/EX7.26/Example7_26.sce new file mode 100644 index 000000000..70cfd3823 --- /dev/null +++ b/3472/CH7/EX7.26/Example7_26.sce @@ -0,0 +1,78 @@ +// 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 : 7.26 :
+// Page number 85-86
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+Q1 = 1100.0 // Discharge in descending order(m^3/sec)
+Q2 = 1000.0 // Discharge(m^3/sec)
+Q3 = 900.0 // Discharge(m^3/sec)
+Q4 = 800.0 // Discharge(m^3/sec)
+Q5 = 700.0 // Discharge(m^3/sec)
+Q6 = 600.0 // Discharge(m^3/sec)
+Q7 = 500.0 // Discharge(m^3/sec)
+Q8 = 400.0 // Discharge(m^3/sec)
+Q9 = 300.0 // Discharge(m^3/sec)
+Q10 = 200.0 // Discharge(m^3/sec)
+Q11 = 100.0 // Discharge(m^3/sec)
+no_week = 13.0 // Total weeks of discharge
+h = 200.0 // Head of installation(m)
+n_overall = 0.88 // Overall efficiency of turbine and generator
+w = 1000.0 // Density of water(kg/m^3)
+
+// Calculations
+n1 = 1.0 // Number of weeks for 1100 discharge(m^3/sec)
+n2 = 2.0 // Number of weeks for 1000 and above discharge(m^3/sec)
+n3 = 3.0 // Number of weeks for 900 and above discharge(m^3/sec)
+n4 = 4.0 // Number of weeks for 800 and above discharge(m^3/sec)
+n5 = 6.0 // Number of weeks for 700 and above discharge(m^3/sec)
+n6 = 7.0 // Number of weeks for 600 and above discharge(m^3/sec)
+n7 = 9.0 // Number of weeks for 500 and above discharge(m^3/sec)
+n8 = 10.0 // Number of weeks for 400 and above discharge(m^3/sec)
+n9 = 11.0 // Number of weeks for 300 and above discharge(m^3/sec)
+n10 = 12.0 // Number of weeks for 200 and above discharge(m^3/sec)
+n11 = 13.0 // Number of weeks for 100 and above discharge(m^3/sec)
+P1 = n1/no_week*100 // Percentage of total period for n1
+P2 = n2/no_week*100 // Percentage of total period for n2
+P3 = n3/no_week*100 // Percentage of total period for n3
+P4 = n4/no_week*100 // Percentage of total period for n4
+P5 = n5/no_week*100 // Percentage of total period for n5
+P6 = n6/no_week*100 // Percentage of total period for n6
+P7 = n7/no_week*100 // Percentage of total period for n7
+P8 = n8/no_week*100 // Percentage of total period for n8
+P9 = n9/no_week*100 // Percentage of total period for n9
+P10 = n10/no_week*100 // Percentage of total period for n10
+P11 = n11/no_week*100 // Percentage of total period for n11
+P = [0,P1,P2,P3,P4,P5,P6,P7,P8,P9,P10,P11]
+Q = [Q1,Q1,Q2,Q3,Q4,Q5,Q6,Q7,Q8,Q9,Q10,Q11] // Plotting flow duration curve
+a = gca() ;
+a.thickness = 2 // sets thickness of plot
+plot(P,Q,'ro-')
+a.x_label.text = 'Percentage of time' // labels x-axis
+a.y_label.text = 'Q(m^3/sec)' // labels y-axis
+xtitle("Fig E7.5 . Plot of Flow-duration curve")
+xset('thickness',2) // sets thickness of axes
+xgrid(4)
+Q_1 = 1.0 // Discharge(m^3/sec)
+P_1 = 0.736/75*w*Q_1*h*n_overall // Power developed for Q_1(kW)
+Q_av = 600.0 // Average discharge(m^3/sec). Obtained from Example 1.7.25
+P_av = P_1*Q_av/1000.0 // Average power developed(MW)
+Q_max = Q1 // Maximum discharge(m^3/sec)
+P_max = P_1*Q_max/1000.0 // Maximum power developed(MW)
+Q_10 = 1070.0 // Discharge for 10% of time(m^3/sec). Value is obtained from graph
+P_10 = P_1*Q_10/1000.0 // Installed capacity(MW)
+
+// Results
+disp("PART I - EXAMPLE : 7.26 : SOLUTION :-")
+printf("\nFlow-duration curve is shown in the Figure E7.5")
+printf("\nMaximum power developed = %.f MW", P_max)
+printf("\nAverage power developed = %.f MW", P_av)
+printf("\nCapacity of proposed station = %.f MW \n", P_10)
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here & approximation in textbook solution")
diff --git a/3472/CH7/EX7.3/Example7_3.sce b/3472/CH7/EX7.3/Example7_3.sce new file mode 100644 index 000000000..9f863200e --- /dev/null +++ b/3472/CH7/EX7.3/Example7_3.sce @@ -0,0 +1,40 @@ +// 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 : 7.3 :
+// Page number 73
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+cap_baseload = 400.0*10**3 // Installed capacity of base load plant(kW)
+cap_standby = 50.0*10**3 // Installed capacity of standby unit(kW)
+output_baseload = 101.0*10**6 // Annual baseload station output(kWh)
+output_standby = 87.35*10**6 // Annual standby station output(kWh)
+peakload_standby = 120.0*10**3 // Peak load on standby station(kW)
+hours_use = 3000.0 // Hours of standby station use/year(hrs)
+
+// Calculations
+// Case(i)
+LF_1 = output_standby*100/(peakload_standby*hours_use) // Annual load factor(%)
+hours_year = 365.0*24 // Total hours in a year
+CF_1 = output_standby*100/(cap_standby*hours_year) // Annual capacity factor(%)
+// Case(ii)
+peakload_baseload = peakload_standby // Peak load on baseload station(kW)
+LF_2 = output_baseload*100/(peakload_baseload*hours_use) // Annual load factor on baseload station(%)
+hours_year = 365.0*24 // Total hours in a year
+CF_2 = output_baseload*100/(cap_baseload*hours_year) // Annual capacity factor on baseload station(%)
+
+// Results
+disp("PART I - EXAMPLE : 7.3 : SOLUTION :-")
+printf("\nCase(i) : Standby Station")
+printf("\n Annual load factor = %.2f percent", LF_1)
+printf("\n Annual capacity factor = %.2f percent\n", CF_1)
+printf("\nCase(ii): Base load Station")
+printf("\n Annual load factor = %.2f percent", LF_2)
+printf("\n Annual capacity factor = %.2f percent\n", CF_2)
+printf("\nNOTE: Incomplete solution in the textbook") ;
diff --git a/3472/CH7/EX7.4/Example7_4.sce b/3472/CH7/EX7.4/Example7_4.sce new file mode 100644 index 000000000..f61f3e733 --- /dev/null +++ b/3472/CH7/EX7.4/Example7_4.sce @@ -0,0 +1,26 @@ +// 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 : 7.4 :
+// Page number 74
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+MD = 500.0 // Maximum demand(MW)
+LF = 0.5 // Annual load factor
+CF = 0.4 // Annual capacity factor
+
+// Calculations
+hours_year = 365.0*24 // Total hours in a year
+energy_gen = MD*LF*hours_year // Energy generated/annum(MWh)
+plant_cap = energy_gen/(CF*hours_year) // Plant capacity(MW)
+reserve_cap = plant_cap-MD // Reserve capacity of plant(MW)
+
+// Results
+disp("PART I - EXAMPLE : 7.4 : SOLUTION :-")
+printf("\nReserve capacity of plant = %.f MW", reserve_cap)
diff --git a/3472/CH7/EX7.5/Example7_5.sce b/3472/CH7/EX7.5/Example7_5.sce new file mode 100644 index 000000000..5531672cb --- /dev/null +++ b/3472/CH7/EX7.5/Example7_5.sce @@ -0,0 +1,36 @@ +// 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 : 7.5 :
+// Page number 74
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+load_1 = 150.0 // Load supplied by station(MW)
+load_2 = 120.0 // Load supplied by station(MW)
+load_3 = 85.0 // Load supplied by station(MW)
+load_4 = 60.0 // Load supplied by station(MW)
+load_5 = 5.0 // Load supplied by station(MW)
+MD = 220.0 // Maximum demand(MW)
+LF = 0.48 // Annual load factor
+
+// Calculations
+// Case(a)
+hours_year = 365.0*24 // Total hours in a year
+units = LF*MD*hours_year // Number of units supplied annually
+// Case(b)
+sum_demand = load_1+load_2+load_3+load_4+load_5 // Sum of maximum demand of individual consumers(MW)
+diversity_factor = sum_demand/MD // Diversity factor
+// Case(c)
+DF = MD/sum_demand // Demand factor
+
+// Results
+disp("PART I - EXAMPLE : 7.5 : SOLUTION :-")
+printf("\nCase(a): Number of units supplied annually = %.2e units", units)
+printf("\nCase(b): Diversity factor = %.3f ", diversity_factor)
+printf("\nCase(c): Demand factor = %.3f = %.1f percent", DF,DF*100)
diff --git a/3472/CH7/EX7.6/Example7_6.sce b/3472/CH7/EX7.6/Example7_6.sce new file mode 100644 index 000000000..5e64fb484 --- /dev/null +++ b/3472/CH7/EX7.6/Example7_6.sce @@ -0,0 +1,32 @@ +// 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 : 7.6 :
+// Page number 74
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+power_del_1 = 1000.0 // Power delivered by station(MW)
+time_1 = 2.0 // Time for which power is delivered(hours)
+power_del_2 = 500.0 // Power delivered by station(MW)
+time_2 = 6.0 // Time for which power is delivered(hours)
+days_maint = 60.0 // Maintenance days
+max_gen_cap = 1000.0 // Maximum generating capacity(MW)
+
+// Calculations
+energy_sup_day = (power_del_1*time_1)+(power_del_2*time_2) // Energy supplied for each working day(MWh)
+days_total = 365.0 // Total days in a year
+days_op = days_total-days_maint // Operating days of station in a year
+energy_sup_year = energy_sup_day*days_op // Energy supplied per year(MWh)
+hours_day = 24.0 // Total hours in a day
+working_hours = days_op*hours_day // Hour of working in a year
+LF = energy_sup_year*100/(max_gen_cap*working_hours) // Annual load factor(%)
+
+// Results
+disp("PART I - EXAMPLE : 7.6 : SOLUTION :-")
+printf("\nAnnual load factor = %.1f percent", LF)
diff --git a/3472/CH7/EX7.7/Example7_7.sce b/3472/CH7/EX7.7/Example7_7.sce new file mode 100644 index 000000000..773007c4d --- /dev/null +++ b/3472/CH7/EX7.7/Example7_7.sce @@ -0,0 +1,33 @@ +// 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 : 7.7 :
+// Page number 74
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+load_industry = 750.0 // Industrial consumer load supplied by station(MW)
+load_commercial = 350.0 // Commercial establishment load supplied by station(MW)
+load_power = 10.0 // Domestic power load supplied by station(MW)
+load_light = 50.0 // Domestic light load supplied by station(MW)
+MD = 1000.0 // Maximum demand(MW)
+kWh_gen = 50.0*10**5 // Number of kWh generated per year
+
+// Calculations
+// Case(i)
+sum_demand = load_industry+load_commercial+load_power+load_light // Sum of max demand of individual consumers(MW)
+diversity_factor = sum_demand/MD // Diversity factor
+// Case(ii)
+hours_year = 365.0*24 // Total hours in a year
+average_demand = kWh_gen/hours_year // Average demand(MW)
+LF = average_demand/MD*100 // Load factor(%)
+
+// Results
+disp("PART I - EXAMPLE : 7.7 : SOLUTION :-")
+printf("\nCase(i) : Diversity factor = %.2f ", diversity_factor)
+printf("\nCase(ii): Annual load factor = %.f percent", LF)
diff --git a/3472/CH7/EX7.8/Example7_8.sce b/3472/CH7/EX7.8/Example7_8.sce new file mode 100644 index 000000000..1b603987c --- /dev/null +++ b/3472/CH7/EX7.8/Example7_8.sce @@ -0,0 +1,42 @@ +// 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 : 7.8 :
+// Page number 74-75
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+load_domestic = 15000.0 // Domestic load supplied by station(kW)
+diversity_domestic = 1.25 // Diversity factor of domestic load
+DF_domestic = 0.7 // Demand factor of domestic load
+load_commercial = 25000.0 // Commercial load supplied by station(kW)
+diversity_commercial = 1.2 // Diversity factor of commercial load
+DF_commercial = 0.9 // Demand factor of commercial load
+load_industry = 50000.0 // Industrial load supplied by station(kW)
+diversity_industry = 1.3 // Diversity factor of industrial load
+DF_industry = 0.98 // Demand factor of industrial load
+diversity_factor = 1.5 // Overall system diversity factor
+
+// Calculations
+// Case(a)
+sum_demand = load_domestic+load_commercial+load_industry // Sum of max demand of individual consumers(MW)
+MD = sum_demand/diversity_factor // Maximum demand
+// Case(b)
+MD_domestic = load_domestic*diversity_domestic // Maximum domestic load demand(kW)
+connected_domestic = MD_domestic/DF_domestic // Connected domestic load(kW)
+MD_commercial = load_commercial*diversity_commercial // Maximum commercial load demand(kW)
+connected_commercial = MD_commercial/DF_commercial // Connected commercial load(kW)
+MD_industry = load_industry*diversity_industry // Maximum industrial load demand(kW)
+connected_industry = MD_industry/DF_industry // Connected industrial load(kW)
+
+// Results
+disp("PART I - EXAMPLE : 7.8 : SOLUTION :-")
+printf("\nCase(a): Maximum demand = %.f kW", MD)
+printf("\nCase(b): Connected domestic load = %.1f kW", connected_domestic)
+printf("\n Connected commercial load = %.1f kW", connected_commercial)
+printf("\n Connected industrial load = %.1f kW", connected_industry)
diff --git a/3472/CH7/EX7.9/Example7_9.sce b/3472/CH7/EX7.9/Example7_9.sce new file mode 100644 index 000000000..532b32629 --- /dev/null +++ b/3472/CH7/EX7.9/Example7_9.sce @@ -0,0 +1,54 @@ +// 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 : 7.9 :
+// Page number 75-76
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+MD = 10000.0 // Maximum demand(kW)
+load_1 = 2000.0 // Load from 11 PM-6 AM(kW)
+t_1 = 7.0 // Time from 11 PM-6 AM(hour)
+load_2 = 3500.0 // Load from 6 AM-8 AM(kW)
+t_2 = 2.0 // Time from 6 AM-8 AM(hour)
+load_3 = 8000.0 // Load from 8 AM-12 Noon(kW)
+t_3 = 4.0 // Time from 8 AM-12 Noon(hour)
+load_4 = 3000.0 // Load from 12 Noon-1 PM(kW)
+t_4 = 1.0 // Time from 12 Noon-1 PM(hour)
+load_5 = 7500.0 // Load from 1 PM-5 PM(kW)
+t_5 = 4.0 // Time from 1 PM-5 PM(hour)
+load_6 = 8500.0 // Load from 5 PM-7 PM(kW)
+t_6 = 2.0 // Time from 5 PM-7 PM(hour)
+load_7 = 10000.0 // Load from 7 PM-9 PM(kW)
+t_7 = 2.0 // Time from 7 PM-9 PM(hour)
+load_8 = 4500.0 // Load from 9 PM-11 PM(kW)
+t_8 = 2.0 // Time from 9 PM-11 PM(hour)
+
+// Calculations
+energy_gen = (load_1*t_1)+(load_2*t_2)+(load_3*t_3)+(load_4*t_4)+(load_5*t_5)+(load_6*t_6)+(load_7*t_7)+(load_8*t_8) // Energy generated during 24 hours(kWh)
+LF = energy_gen/(MD*24.0) // Load factor
+no_units = 3.0 // Number of generating set
+cap_1 = 5000.0 // Capacity of first generating unit(kW)
+cap_2 = 3000.0 // Capacity of second generating unit(kW)
+cap_3 = 2000.0 // Capacity of third generating unit(kW)
+cap_reserve = cap_1 // Reserve capacity(kW) i.e largest size of generating unit
+cap_installed = cap_1+cap_2+cap_3+cap_reserve // Installed capacity(kW)
+cap_factor = energy_gen/(cap_installed*24.0) // Plant capacity factor
+cap_plant = cap_3*t_1+(cap_3+cap_2)*t_2+(cap_2+cap_1)*t_3+cap_2*t_4+(cap_2+cap_1)*t_5+(cap_3+cap_2+cap_1)*t_6+(cap_3+cap_2+cap_1)*t_7+cap_1*t_8 // Capacity of plant running actually(kWh)
+use_factor = energy_gen/cap_plant // Plant use factor
+
+// Results
+disp("PART I - EXAMPLE : 7.9 : SOLUTION :-")
+printf("\nNumber of generator units = %.f", no_units)
+printf("\nSize of generator units required are %.f kW, %.f kW and %.f kW", cap_1,cap_2,cap_3)
+printf("\nReserve plant capacity = %.f kW", cap_reserve)
+printf("\nLoad factor = %.2f = %.f percent", LF,LF*100)
+printf("\nPlant capacity factor = %.4f = %.2f percent", cap_factor,cap_factor*100)
+printf("\nPlant use factor = %.3f = %.1f percent", use_factor,use_factor*100)
+printf("\n\nNOTE: Capacity of plant is directly taken & operating schedule is not displayed here")
+
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