diff options
Diffstat (limited to '3472/CH40')
| -rw-r--r-- | 3472/CH40/EX40.1/Example40_1.sce | 39 | ||||
| -rw-r--r-- | 3472/CH40/EX40.2/Example40_2.sce | 36 | ||||
| -rw-r--r-- | 3472/CH40/EX40.3/Example40_3.sce | 55 | ||||
| -rw-r--r-- | 3472/CH40/EX40.4/Example40_4.sce | 31 | ||||
| -rw-r--r-- | 3472/CH40/EX40.5/Example40_5.sce | 30 | ||||
| -rw-r--r-- | 3472/CH40/EX40.6/Example40_6.sce | 36 | ||||
| -rw-r--r-- | 3472/CH40/EX40.7/Example40_7.sce | 44 | ||||
| -rw-r--r-- | 3472/CH40/EX40.8/Example40_8.sce | 44 |
8 files changed, 315 insertions, 0 deletions
diff --git a/3472/CH40/EX40.1/Example40_1.sce b/3472/CH40/EX40.1/Example40_1.sce new file mode 100644 index 000000000..38b4614c8 --- /dev/null +++ b/3472/CH40/EX40.1/Example40_1.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 IV : UTILIZATION AND TRACTION
+// CHAPTER 2: HEATING AND WELDING
+
+// EXAMPLE : 2.1 :
+// Page number 724-725
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+P = 15.0*10**3 // Power supplied(W)
+V = 220.0 // Voltage(V)
+T_w = 1000.0 // Temperature of wire(°C)
+T_c = 600.0 // Temperature of charges(°C)
+k = 0.6 // Radiatting efficiency
+e = 0.9 // Emissivity
+
+// Calculations
+rho = 1.016/10**6 // Specific resistance(ohm-m)
+d_square = 4*rho*P/(%pi*V**2) // d^2 in terms of l
+T_1 = T_w+273 // Absolute temperature(°C)
+T_2 = T_c+273 // Absolute temperature(°C)
+H = 5.72*10**4*k*e*((T_1/1000)**4-(T_2/1000)**4) // Heat produced(watts/sq.m)
+dl = P/(%pi*H)
+l = (dl**2/d_square)**(1.0/3) // Length of wire(m)
+d = dl/l // Diameter of wire(m)
+T_2_cold = 20.0+273 // Absolute temperature at the 20°C normal temperature(°C)
+T_1_cold = (H/(5.72*10**4*k*e)+(T_2_cold/1000)**4)**(1.0/4)*1000 // Absolute temperature when charge is cold(°C)
+T_1_c = T_1_cold-273 // Temperature when charge is cold(°C)
+
+// Results
+disp("PART IV - EXAMPLE : 2.1 : SOLUTION :-")
+printf("\nDiameter of the wire, d = %.3f cm", d*100)
+printf("\nLength of the wire, l = %.2f m", l)
+printf("\nTemperature of the wire when charge is cold, T_1 = %.f°C absolute = %.f°C \n", T_1_cold,T_1_c)
+printf("\nNOTE: Slight changes in the obtained answer from that of textbook is due to more precision here")
diff --git a/3472/CH40/EX40.2/Example40_2.sce b/3472/CH40/EX40.2/Example40_2.sce new file mode 100644 index 000000000..1f8a7f3d3 --- /dev/null +++ b/3472/CH40/EX40.2/Example40_2.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 IV : UTILIZATION AND TRACTION
+// CHAPTER 2: HEATING AND WELDING
+
+// EXAMPLE : 2.2 :
+// Page number 725
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+P = 15.0*10**3 // Power supplied(W)
+V = 220.0 // Voltage(V)
+T_w = 1000.0 // Temperature of wire(°C)
+T_c = 600.0 // Temperature of charges(°C)
+k = 0.6 // Radiatting efficiency
+e = 0.9 // Emissivity
+thick = 0.25/1000 // Thickness of nickel-chrome strip(m)
+
+// Calculations
+rho = 1.016/10**6 // Specific resistance(ohm-m)
+R = V**2/P // Resistance(ohm)
+l_w = R*thick/rho // Length of strip in terms of w
+T_1 = T_w+273 // Absolute temperature(°C)
+T_2 = T_c+273 // Absolute temperature(°C)
+H = 5.72*10**4*k*e*((T_1/1000)**4-(T_2/1000)**4) // Heat produced(watts/sq.m)
+wl = P/(2*H)
+w = (wl/l_w)**0.5 // Width of nickel-chrome strip(m)
+l = w*l_w // Length of nickel-chrome strip(m)
+
+// Results
+disp("PART IV - EXAMPLE : 2.2 : SOLUTION :-")
+printf("\nWidth of nickel-chrome strip, w = %.3f cm", w*100)
+printf("\nLength of nickel-chrome strip, l = %.1f m", l)
diff --git a/3472/CH40/EX40.3/Example40_3.sce b/3472/CH40/EX40.3/Example40_3.sce new file mode 100644 index 000000000..557f98f57 --- /dev/null +++ b/3472/CH40/EX40.3/Example40_3.sce @@ -0,0 +1,55 @@ +// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART IV : UTILIZATION AND TRACTION
+// CHAPTER 2: HEATING AND WELDING
+
+// EXAMPLE : 2.3 :
+// Page number 726-727
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+R = 50.0 // Resistance of each resistor in oven(ohm)
+n = 6.0 // Number of resistance
+V = 400.0 // Supply voltage(V)
+tap = 50.0 // Auto-transformer tapping(%)
+
+// Calculations
+// Case(a)(i)
+P_a_i = n*V**2/R*10**-3 // Power consumption for 6 elements in parallel(kW)
+// Case(a)(ii)
+P_each_a_ii = V**2/(R+R)*10**-3 // Power consumption in each group of 2 resistances in series(kW)
+P_a_ii = n/2*P_each_a_ii // Power consumption for 3 groups(kW)
+// Case(b)(i)
+V_b_i = V/3**0.5 // Supply voltage against each resistance(V)
+P_each_b_i = 2*V_b_i**2/R*10**-3 // Power consumption in each branch(kW)
+P_b_i = n/2*P_each_b_i // Power consumption for 2 elements in parallel in each phase(kW)
+// Case(b)(ii)
+V_b_ii = V/3**0.5 // Supply voltage to any branch(V)
+P_each_b_ii = V_b_ii**2/(R+R)*10**-3 // Power consumption in each branch(kW)
+P_b_ii = n/2*P_each_b_ii // Power consumption for 2 elements in series in each phase(kW)
+// Case(c)(i)
+P_each_c_i = V**2/(R+R)*10**-3 // Power consumption by each branch(kW)
+P_c_i = n/2*P_each_c_i // Power consumption for 2 elements in series in each branch(kW)
+// Case(c)(ii)
+P_each_c_ii = 2*V**2/R*10**-3 // Power consumption by each branch(kW)
+P_c_ii = n/2*P_each_c_ii // Power consumption for 2 elements in parallel in each branch(kW)
+// Case(d)
+V_d = V*tap/100 // Voltage under tapping(V)
+ratio_V = V_d/V // Ratio of normal voltage to tapped voltage
+loss = ratio_V**2 // Power loss in terms of normal power
+
+// Results
+disp("PART IV - EXAMPLE : 2.3 : SOLUTION :-")
+printf("\nCase(a): AC Single phase 400 V supply")
+printf("\n Case(i) : Power consumption for 6 elements in parallel = %.1f kW", P_a_i)
+printf("\n Case(ii): Power consumption for 3 groups in parallel with 2 element in series = %.1f kW", P_a_ii)
+printf("\nCase(b): AC Three phase 400 V supply with star combination")
+printf("\n Case(i) : Power consumption for 2 elements in parallel in each phase = %.1f kW", P_b_i)
+printf("\n Case(ii): Power consumption for 2 elements in series in each phase = %.1f kW", P_b_ii)
+printf("\nCase(c): AC Three phase 400 V supply with delta combination")
+printf("\n Case(i) : Power consumption for 2 elements in series in each branch = %.1f kW", P_c_i)
+printf("\n Case(ii): Power consumption for 2 elements in parallel in each branch = %.1f kW", P_c_ii)
+printf("\nCase(d): Power loss will be %.2f of the values obtained as above with auto-transformer tapping", loss)
diff --git a/3472/CH40/EX40.4/Example40_4.sce b/3472/CH40/EX40.4/Example40_4.sce new file mode 100644 index 000000000..e315d65dc --- /dev/null +++ b/3472/CH40/EX40.4/Example40_4.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 IV : UTILIZATION AND TRACTION
+// CHAPTER 2: HEATING AND WELDING
+
+// EXAMPLE : 2.4 :
+// Page number 728
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+w_brass = 1000.0 // Weight of brass(kg)
+time = 1.0 // Time(hour)
+heat_sp = 0.094 // Specific heat
+fusion = 40.0 // Latent heat of fusion(kcal/kg)
+T_initial = 24.0 // Initial temperature(°C)
+melt_point = 920.0 // Melting point of brass(°C)
+n = 0.65 // Efficiency
+
+// Calculations
+heat_req = w_brass*heat_sp*(melt_point-T_initial) // Heat required to raise the temperature(kcal)
+heat_mel = w_brass*fusion // Heat required for melting(kcal)
+heat_total = heat_req+heat_mel // Total heat required(kcal)
+energy = heat_total*1000*4.18/(10**3*3600*n) // Energy input(kWh)
+power = energy/time // Power(kW)
+
+// Results
+disp("PART IV - EXAMPLE : 2.4 : SOLUTION :-")
+printf("\nAmount of energy required to melt brass = %.f kWh", energy)
diff --git a/3472/CH40/EX40.5/Example40_5.sce b/3472/CH40/EX40.5/Example40_5.sce new file mode 100644 index 000000000..480c63bfd --- /dev/null +++ b/3472/CH40/EX40.5/Example40_5.sce @@ -0,0 +1,30 @@ +// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART IV : UTILIZATION AND TRACTION
+// CHAPTER 2: HEATING AND WELDING
+
+// EXAMPLE : 2.5 :
+// Page number 728-729
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+V_2 = 12.0 // Secondary voltage(V)
+P = 30.0*10**3 // Power(W)
+PF = 0.5 // Power factor
+
+// Calculations
+I_2 = P/(V_2*PF) // Secondary current(A)
+Z_2 = V_2/I_2 // Secondary impedance(ohm)
+R_2 = Z_2*PF // Secondary resistance(ohm)
+sin_phi = (1-PF**2)**0.5
+X_2 = Z_2*sin_phi // Secondary reactance(ohm)
+h = R_2/X_2
+H_m = h // Height up to which the crucible should be filled to obtain maximum heating effect in terms of H_c
+
+// Results
+disp("PART IV - EXAMPLE : 2.5 : SOLUTION :-")
+printf("\nHeight up to which the crucible should be filled to obtain maximum heating effect, H_m = %.3f*H_c \n", H_m)
+printf("\nNOTE: ERROR: Calculation mistake in textbook solution and P is 30 kW not 300 kW")
diff --git a/3472/CH40/EX40.6/Example40_6.sce b/3472/CH40/EX40.6/Example40_6.sce new file mode 100644 index 000000000..01c586c05 --- /dev/null +++ b/3472/CH40/EX40.6/Example40_6.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 IV : UTILIZATION AND TRACTION
+// CHAPTER 2: HEATING AND WELDING
+
+// EXAMPLE : 2.6 :
+// Page number 732
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+l = 10.0 // Length of material(cm)
+b = 10.0 // Breadth of material(cm)
+t = 3.0 // Thickness of material(cm)
+f = 20.0*10**6 // Frequency(Hz)
+P = 400.0 // Power absorbed(W)
+e_r = 5.0 // Relative permittivity
+PF = 0.05 // Power factor
+
+// Calculations
+e_0 = 8.854*10**-12 // Absolute permittivity
+A = l*b*10**-4 // Area(Sq.m)
+C = e_0*e_r*A/(t/100) // Capacitace of parallel plate condenser(F)
+X_c = 1.0/(2*%pi*f*C) // Reactance of condenser(ohm)
+phi = acosd(PF) // Φ(°)
+R = X_c*tand(phi) // Resistance of condenser(ohm)
+V = (P*R)**0.5 // Voltage necessary for heating(V)
+I_c = V/X_c // Current flowing in the material(A)
+
+// Results
+disp("PART IV - EXAMPLE : 2.6 : SOLUTION :-")
+printf("\nVoltage necessary for heating, V = %.f V", V)
+printf("\nCurrent flowing in the material, I_c = %.2f A\n", I_c)
+printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here & approximation in textbook")
diff --git a/3472/CH40/EX40.7/Example40_7.sce b/3472/CH40/EX40.7/Example40_7.sce new file mode 100644 index 000000000..33bac80b8 --- /dev/null +++ b/3472/CH40/EX40.7/Example40_7.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 IV : UTILIZATION AND TRACTION
+// CHAPTER 2: HEATING AND WELDING
+
+// EXAMPLE : 2.7 :
+// Page number 732-733
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+l = 4.0 // Length of material(cm)
+b = 2.0 // Breadth of material(cm)
+t = 1.0 // Thickness of material(cm)
+l_e = 20.0 // Length of area(cm)
+b_e = 2.0 // Breadth of area(cm)
+dis = 1.6 // Distance of separation of electrode(cm)
+f = 20.0*10**6 // Frequency(Hz)
+P = 80.0 // Power absorbed(W)
+e_r1 = 5.0 // Relative permittivity
+e_r2 = 1.0 // Relative permittivity of air
+PF = 0.05 // Power factor
+
+// Calculations
+e_0 = 8.854*10**-12 // Absolute permittivity
+A_1 = (l_e-l)*b_e*10**-4 // Area of one electrode(sq.m)
+A_2 = l*b*10**-4 // Area of material under electrode(sq.m)
+d = dis*10**-2 // Distance of separation of electrode(m)
+d_1 = t*10**-2 // (m)
+d_2 = (d-d_1) // (m)
+C = e_0*((A_1*e_r2/d)+(A_2/((d_1/e_r1)+(d_2/e_r2)))) // Capacitance(F)
+X_c = 1.0/(2*%pi*f*C) // Reactance(ohm)
+phi = acosd(PF) // Φ(°)
+R = X_c*tand(phi) // Resistance(ohm)
+V = (P*R)**0.5 // Voltage applied across electrodes(V)
+I_c = V/X_c // Current through the material(A)
+
+// Results
+disp("PART IV - EXAMPLE : 2.7 : SOLUTION :-")
+printf("\nVoltage applied across electrodes, V = %.f V", V)
+printf("\nCurrent through the material, I_c = %.1f A\n", I_c)
+printf("\nNOTE: ERROR: Calculation mistake in the textbook solution")
diff --git a/3472/CH40/EX40.8/Example40_8.sce b/3472/CH40/EX40.8/Example40_8.sce new file mode 100644 index 000000000..6556eaee1 --- /dev/null +++ b/3472/CH40/EX40.8/Example40_8.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 IV : UTILIZATION AND TRACTION
+// CHAPTER 2: HEATING AND WELDING
+
+// EXAMPLE : 2.8 :
+// Page number 736-737
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+weight = 3000.0 // Weight of steel(kg)
+I = 5000.0 // Current(A)
+V_arc = 60.0 // Arc voltage(V)
+R_t = 0.003 // Resistance of transformer(ohm)
+X_t = 0.005 // Reactance of transformer(ohm)
+heat_sp = 0.12 // Specific heat of steel
+heat_latent = 8.89 // Latent heat of steel(kilo-cal/kg)
+t_2 = 1370.0 // Melting point of steel(°C)
+t_1 = 18.0 // Initial temperature of steel(°C)
+n = 0.6 // Overall efficiency
+
+// Calculations
+R_arc_phase = V_arc/I // Arc resistance per phase(ohm)
+IR_t = I*R_t // Voltage drop across resistance(V)
+IX_t = I*X_t // Voltage drop across reactance(V)
+V = ((V_arc+IR_t)**2+IX_t**2)**0.5 // Voltage(V)
+PF = (V_arc+IR_t)/V // Power factor
+heat_kg = (t_2-t_1)*heat_sp+heat_latent // Amount of heat required per kg of steel(kcal)
+heat_total = weight*heat_kg // Heat for 3 tonnes(kcal)
+heat_actual_kcal = heat_total/n // Actual heat required(kcal)
+heat_actual = heat_actual_kcal*1.162*10**-3 // Actual heat required(kWh)
+P_input = 3*V*I*PF*10**-3 // Power input(kW)
+time = heat_actual/P_input*60 // Time required(min)
+n_elect = 3*V_arc*I/(P_input*1000)*100 // Electrical efficiency(%)
+
+// Results
+disp("PART IV - EXAMPLE : 2.8 : SOLUTION :-")
+printf("\nTime taken to melt 3 metric tonnes of steel = %.f minutes", time)
+printf("\nPower factor of the furnace = %.2f ", PF)
+printf("\nElectrical efficiency of the furnace = %.f percent\n", n_elect)
+printf("\nNOTE: ERROR: Calculation and substitution mistake in the textbook solution")
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