{
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"CHAPTER 4.2: HEATING AND WELDING"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.2.1, Page number 724-725"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"P = 15.0*10**3 #Power supplied(W)\n",
"V = 220.0 #Voltage(V)\n",
"T_w = 1000.0 #Temperature of wire(\u00b0C)\n",
"T_c = 600.0 #Temperature of charges(\u00b0C)\n",
"k = 0.6 #Radiatting efficiency\n",
"e = 0.9 #Emissivity\n",
"\n",
"#Calculation\n",
"rho = 1.016/10**6 #Specific resistance(ohm-m)\n",
"d_square = 4*rho*P/(math.pi*V**2) #d^2 in terms of l\n",
"T_1 = T_w+273 #Absolute temperature(\u00b0C)\n",
"T_2 = T_c+273 #Absolute temperature(\u00b0C)\n",
"H = 5.72*10**4*k*e*((T_1/1000)**4-(T_2/1000)**4) #Heat produced(watts/sq.m)\n",
"dl = P/(math.pi*H)\n",
"l = (dl**2/d_square)**(1.0/3) #Length of wire(m)\n",
"d = dl/l #Diameter of wire(m)\n",
"T_2_cold = 20.0+273 #Absolute temperature at the 20\u00b0C normal temperature(\u00b0C)\n",
"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(\u00b0C)\n",
"T_1_c = T_1_cold-273 #Temperature when charge is cold(\u00b0C)\n",
"\n",
"#Result\n",
"print('Diameter of the wire, d = %.3f cm' %(d*100))\n",
"print('Length of the wire, l = %.2f m' %l)\n",
"print('Temperature of the wire when charge is cold, T_1 = %.f \u00b0C absolute = %.f \u00b0C' %(T_1_cold,T_1_c))\n",
"print('\\nNOTE: Slight changes in the obtained answer from that of textbook is due to more precision here')"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Diameter of the wire, d = 0.312 cm\n",
"Length of the wire, l = 24.24 m\n",
"Temperature of the wire when charge is cold, T_1 = 1197 \u00b0C absolute = 924 \u00b0C\n",
"\n",
"NOTE: Slight changes in the obtained answer from that of textbook is due to more precision here\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.2.2, Page number 725"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"P = 15.0*10**3 #Power supplied(W)\n",
"V = 220.0 #Voltage(V)\n",
"T_w = 1000.0 #Temperature of wire(\u00b0C)\n",
"T_c = 600.0 #Temperature of charges(\u00b0C)\n",
"k = 0.6 #Radiatting efficiency\n",
"e = 0.9 #Emissivity\n",
"thick = 0.25/1000 #Thickness of nickel-chrome strip(m)\n",
"\n",
"#Calculation\n",
"rho = 1.016/10**6 #Specific resistance(ohm-m)\n",
"R = V**2/P #Resistance(ohm)\n",
"l_w = R*thick/rho #Length of strip in terms of w\n",
"T_1 = T_w+273 #Absolute temperature(\u00b0C)\n",
"T_2 = T_c+273 #Absolute temperature(\u00b0C)\n",
"H = 5.72*10**4*k*e*((T_1/1000)**4-(T_2/1000)**4) #Heat produced(watts/sq.m)\n",
"wl = P/(2*H)\n",
"w = (wl/l_w)**0.5 #Width of nickel-chrome strip(m)\n",
"l = w*l_w #Length of nickel-chrome strip(m)\n",
"\n",
"#Result\n",
"print('Width of nickel-chrome strip, w = %.3f cm' %(w*100))\n",
"print('Length of nickel-chrome strip, l = %.1f m' %l)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Width of nickel-chrome strip, w = 1.223 cm\n",
"Length of nickel-chrome strip, l = 9.7 m\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.2.3, Page number 726-727"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"R = 50.0 #Resistance of each resistor in oven(ohm)\n",
"n = 6.0 #Number of resistance\n",
"V = 400.0 #Supply voltage(V)\n",
"tap = 50.0 #Auto-transformer tapping(%)\n",
"\n",
"#Calculation\n",
"#Case(a)(i)\n",
"P_a_i = n*V**2/R*10**-3 #Power consumption for 6 elements in parallel(kW)\n",
"#Case(a)(ii)\n",
"P_each_a_ii = V**2/(R+R)*10**-3 #Power consumption in each group of 2 resistances in series(kW)\n",
"P_a_ii = n/2*P_each_a_ii #Power consumption for 3 groups(kW)\n",
"#Case(b)(i)\n",
"V_b_i = V/3**0.5 #Supply voltage against each resistance(V)\n",
"P_each_b_i = 2*V_b_i**2/R*10**-3 #Power consumption in each branch(kW)\n",
"P_b_i = n/2*P_each_b_i #Power consumption for 2 elements in parallel in each phase(kW)\n",
"#Case(b)(ii)\n",
"V_b_ii = V/3**0.5 #Supply voltage to any branch(V)\n",
"P_each_b_ii = V_b_ii**2/(R+R)*10**-3 #Power consumption in each branch(kW)\n",
"P_b_ii = n/2*P_each_b_ii #Power consumption for 2 elements in series in each phase(kW)\n",
"#Case(c)(i)\n",
"P_each_c_i = V**2/(R+R)*10**-3 #Power consumption by each branch(kW)\n",
"P_c_i = n/2*P_each_c_i #Power consumption for 2 elements in series in each branch(kW)\n",
"#Case(c)(ii)\n",
"P_each_c_ii = 2*V**2/R*10**-3 #Power consumption by each branch(kW)\n",
"P_c_ii = n/2*P_each_c_ii #Power consumption for 2 elements in parallel in each branch(kW)\n",
"#Case(d)\n",
"V_d = V*tap/100 #Voltage under tapping(V)\n",
"ratio_V = V_d/V #Ratio of normal voltage to tapped voltage\n",
"loss = ratio_V**2 #Power loss in terms of normal power\n",
"\n",
"#Result\n",
"print('Case(a): AC Single phase 400 V supply')\n",
"print(' Case(i) : Power consumption for 6 elements in parallel = %.1f kW' %P_a_i)\n",
"print(' Case(ii): Power consumption for 3 groups in parallel with 2 element in series = %.1f kW' %P_a_ii)\n",
"print('Case(b): AC Three phase 400 V supply with star combination')\n",
"print(' Case(i) : Power consumption for 2 elements in parallel in each phase = %.1f kW' %P_b_i)\n",
"print(' Case(ii): Power consumption for 2 elements in series in each phase = %.1f kW' %P_b_ii)\n",
"print('Case(c): AC Three phase 400 V supply with delta combination')\n",
"print(' Case(i) : Power consumption for 2 elements in series in each branch = %.1f kW' %P_c_i)\n",
"print(' Case(ii): Power consumption for 2 elements in parallel in each branch = %.1f kW' %P_c_ii)\n",
"print('Case(d): Power loss will be %.2f of the values obtained as above with auto-transformer tapping' %loss)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Case(a): AC Single phase 400 V supply\n",
" Case(i) : Power consumption for 6 elements in parallel = 19.2 kW\n",
" Case(ii): Power consumption for 3 groups in parallel with 2 element in series = 4.8 kW\n",
"Case(b): AC Three phase 400 V supply with star combination\n",
" Case(i) : Power consumption for 2 elements in parallel in each phase = 6.4 kW\n",
" Case(ii): Power consumption for 2 elements in series in each phase = 1.6 kW\n",
"Case(c): AC Three phase 400 V supply with delta combination\n",
" Case(i) : Power consumption for 2 elements in series in each branch = 4.8 kW\n",
" Case(ii): Power consumption for 2 elements in parallel in each branch = 19.2 kW\n",
"Case(d): Power loss will be 0.25 of the values obtained as above with auto-transformer tapping\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.2.4, Page number 728"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"w_brass = 1000.0 #Weight of brass(kg)\n",
"time = 1.0 #Time(hour)\n",
"heat_sp = 0.094 #Specific heat\n",
"fusion = 40.0 #Latent heat of fusion(kcal/kg)\n",
"T_initial = 24.0 #Initial temperature(\u00b0C)\n",
"melt_point = 920.0 #Melting point of brass(\u00b0C)\n",
"n = 0.65 #Efficiency\n",
"\n",
"#Calculation\n",
"heat_req = w_brass*heat_sp*(melt_point-T_initial) #Heat required to raise the temperature(kcal)\n",
"heat_mel = w_brass*fusion #Heat required for melting(kcal)\n",
"heat_total = heat_req+heat_mel #Total heat required(kcal)\n",
"energy = heat_total*1000*4.18/(10**3*3600*n) #Energy input(kWh)\n",
"power = energy/time #Power(kW)\n",
"\n",
"#Result\n",
"print('Amount of energy required to melt brass = %.f kWh' %energy)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Amount of energy required to melt brass = 222 kWh\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.2.5, Page number 728-729"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"V_2 = 12.0 #Secondary voltage(V)\n",
"P = 30.0*10**3 #Power(W)\n",
"PF = 0.5 #Power factor\n",
"\n",
"#Calculation\n",
"I_2 = P/(V_2*PF) #Secondary current(A)\n",
"Z_2 = V_2/I_2 #Secondary impedance(ohm)\n",
"R_2 = Z_2*PF #Secondary resistance(ohm)\n",
"sin_phi = (1-PF**2)**0.5\n",
"X_2 = Z_2*sin_phi #Secondary reactance(ohm)\n",
"h = R_2/X_2\n",
"H_m = h #Height up to which the crucible should be filled to obtain maximum heating effect in terms of H_c\n",
"\n",
"#Result\n",
"print('Height up to which the crucible should be filled to obtain maximum heating effect, H_m = %.3f*H_c ' %H_m)\n",
"print('\\nNOTE: ERROR: Calculation mistake in textbook solution and P is 30 kW not 300 kW')"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Height up to which the crucible should be filled to obtain maximum heating effect, H_m = 0.577*H_c \n",
"\n",
"NOTE: ERROR: Calculation mistake in textbook solution and P is 30 kW not 300 kW\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.2.6, Page number 732"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"l = 10.0 #Length of material(cm)\n",
"b = 10.0 #Breadth of material(cm)\n",
"t = 3.0 #Thickness of material(cm)\n",
"f = 20.0*10**6 #Frequency(Hz)\n",
"P = 400.0 #Power absorbed(W)\n",
"e_r = 5.0 #Relative permittivity\n",
"PF = 0.05 #Power factor\n",
"\n",
"#Calculation\n",
"e_0 = 8.854*10**-12 #Absolute permittivity\n",
"A = l*b*10**-4 #Area(Sq.m)\n",
"C = e_0*e_r*A/(t/100) #Capacitace of parallel plate condenser(F)\n",
"X_c = 1.0/(2*math.pi*f*C) #Reactance of condenser(ohm)\n",
"phi = math.acos(PF)*180/math.pi #\u03a6(\u00b0)\n",
"R = X_c*math.tan(phi*math.pi/180) #Resistance of condenser(ohm)\n",
"V = (P*R)**0.5 #Voltage necessary for heating(V)\n",
"I_c = V/X_c #Current flowing in the material(A)\n",
"\n",
"#Result\n",
"print('Voltage necessary for heating, V = %.f V' %V)\n",
"print('Current flowing in the material, I_c = %.2f A' %I_c)\n",
"print('\\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here & approximation in textbook')"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Voltage necessary for heating, V = 2076 V\n",
"Current flowing in the material, I_c = 3.85 A\n",
"\n",
"NOTE: Changes in the obtained answer from that of textbook is due to more precision here & approximation in textbook\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.2.7, Page number 732-733"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"l = 4.0 #Length of material(cm)\n",
"b = 2.0 #Breadth of material(cm)\n",
"t = 1.0 #Thickness of material(cm)\n",
"l_e = 20.0 #Length of area(cm)\n",
"b_e = 2.0 #Breadth of area(cm)\n",
"dis = 1.6 #Distance of separation of electrode(cm)\n",
"f = 20.0*10**6 #Frequency(Hz)\n",
"P = 80.0 #Power absorbed(W)\n",
"e_r1 = 5.0 #Relative permittivity\n",
"e_r2 = 1.0 #Relative permittivity of air\n",
"PF = 0.05 #Power factor\n",
"\n",
"#Calculation\n",
"e_0 = 8.854*10**-12 #Absolute permittivity\n",
"A_1 = (l_e-l)*b_e*10**-4 #Area of one electrode(sq.m)\n",
"A_2 = l*b*10**-4 #Area of material under electrode(sq.m)\n",
"d = dis*10**-2 #Distance of separation of electrode(m)\n",
"d_1 = t*10**-2 #(m)\n",
"d_2 = (d-d_1) #(m)\n",
"C = e_0*((A_1*e_r2/d)+(A_2/((d_1/e_r1)+(d_2/e_r2)))) #Capacitance(F)\n",
"X_c = 1.0/(2*math.pi*f*C) #Reactance(ohm)\n",
"phi = math.acos(PF)*180/math.pi #\u03a6(\u00b0)\n",
"R = X_c*math.tan(phi*math.pi/180) #Resistance(ohm)\n",
"V = (P*R)**0.5 #Voltage applied across electrodes(V)\n",
"I_c = V/X_c #Current through the material(A)\n",
"\n",
"#Result\n",
"print('Voltage applied across electrodes, V = %.f V' %V)\n",
"print('Current through the material, I_c = %.1f A' %I_c)\n",
"print('\\nNOTE: ERROR: Calculation mistake in the textbook solution')"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Voltage applied across electrodes, V = 2188 V\n",
"Current through the material, I_c = 0.7 A\n",
"\n",
"NOTE: ERROR: Calculation mistake in the textbook solution\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.2.8, Page number 736-737"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"weight = 3000.0 #Weight of steel(kg)\n",
"I = 5000.0 #Current(A)\n",
"V_arc = 60.0 #Arc voltage(V)\n",
"R_t = 0.003 #Resistance of transformer(ohm)\n",
"X_t = 0.005 #Reactance of transformer(ohm)\n",
"heat_sp = 0.12 #Specific heat of steel\n",
"heat_latent = 8.89 #Latent heat of steel(kilo-cal/kg)\n",
"t_2 = 1370.0 #Melting point of steel(\u00b0C)\n",
"t_1 = 18.0 #Initial temperature of steel(\u00b0C)\n",
"n = 0.6 #Overall efficiency\n",
"\n",
"#Calculation\n",
"R_arc_phase = V_arc/I #Arc resistance per phase(ohm)\n",
"IR_t = I*R_t #Voltage drop across resistance(V)\n",
"IX_t = I*X_t #Voltage drop across reactance(V)\n",
"V = ((V_arc+IR_t)**2+IX_t**2)**0.5 #Voltage(V)\n",
"PF = (V_arc+IR_t)/V #Power factor\n",
"heat_kg = (t_2-t_1)*heat_sp+heat_latent #Amount of heat required per kg of steel(kcal)\n",
"heat_total = weight*heat_kg #Heat for 3 tonnes(kcal)\n",
"heat_actual_kcal = heat_total/n #Actual heat required(kcal)\n",
"heat_actual = heat_actual_kcal*1.162*10**-3 #Actual heat required(kWh)\n",
"P_input = 3*V*I*PF*10**-3 #Power input(kW)\n",
"time = heat_actual/P_input*60 #Time required(min)\n",
"n_elect = 3*V_arc*I/(P_input*1000)*100 #Electrical efficiency(%)\n",
"\n",
"#Result\n",
"print('Time taken to melt 3 metric tonnes of steel = %.f minutes' %time)\n",
"print('Power factor of the furnace = %.2f ' %PF)\n",
"print('Electrical efficiency of the furnace = %.f percent' %n_elect)\n",
"print('\\nNOTE: ERROR: Calculation and substitution mistake in the textbook solution')"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time taken to melt 3 metric tonnes of steel = 53 minutes\n",
"Power factor of the furnace = 0.95 \n",
"Electrical efficiency of the furnace = 80 percent\n",
"\n",
"NOTE: ERROR: Calculation and substitution mistake in the textbook solution\n"
]
}
],
"prompt_number": 1
}
],
"metadata": {}
}
]
}