# CHAPTER 4.2: HEATING AND WELDING¶

## Example 4.2.1, Page number 724-725¶

In :
import math

#Variable declaration
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)
e = 0.9            #Emissivity

#Calculation
rho = 1.016/10**6                                                  #Specific resistance(ohm-m)
d_square = 4*rho*P/(math.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/(math.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)

#Result
print('Diameter of the wire, d = %.3f cm' %(d*100))
print('Length of the wire, l = %.2f m' %l)
print('Temperature of the wire when charge is cold, T_1 = %.f °C absolute = %.f °C' %(T_1_cold,T_1_c))
print('\nNOTE: Slight changes in the obtained answer from that of textbook is due to more precision here')

Diameter of the wire, d = 0.312 cm
Length of the wire, l = 24.24 m
Temperature of the wire when charge is cold, T_1 = 1197 °C absolute = 924 °C

NOTE: Slight changes in the obtained answer from that of textbook is due to more precision here


## Example 4.2.2, Page number 725¶

In :
#Variable declaration
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)
e = 0.9                #Emissivity
thick = 0.25/1000      #Thickness of nickel-chrome strip(m)

#Calculation
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)

#Result
print('Width of nickel-chrome strip, w = %.3f cm' %(w*100))
print('Length of nickel-chrome strip, l = %.1f m' %l)

Width of nickel-chrome strip, w = 1.223 cm
Length of nickel-chrome strip, l = 9.7 m


## Example 4.2.3, Page number 726-727¶

In :
#Variable declaration
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(%)

#Calculation
#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

#Result
print('Case(a): AC Single phase 400 V supply')
print('         Case(i) :  Power consumption for 6 elements in parallel = %.1f kW' %P_a_i)
print('         Case(ii):  Power consumption for 3 groups in parallel with 2 element in series = %.1f kW' %P_a_ii)
print('Case(b): AC Three phase 400 V supply with star combination')
print('         Case(i) :  Power consumption for 2 elements in parallel in each phase = %.1f kW' %P_b_i)
print('         Case(ii):  Power consumption for 2 elements in series in each phase = %.1f kW' %P_b_ii)
print('Case(c): AC Three phase 400 V supply with delta combination')
print('         Case(i) :  Power consumption for 2 elements in series in each branch = %.1f kW' %P_c_i)
print('         Case(ii):  Power consumption for 2 elements in parallel in each branch = %.1f kW' %P_c_ii)
print('Case(d): Power loss will be %.2f of the values obtained as above with auto-transformer tapping' %loss)

Case(a): AC Single phase 400 V supply
Case(i) :  Power consumption for 6 elements in parallel = 19.2 kW
Case(ii):  Power consumption for 3 groups in parallel with 2 element in series = 4.8 kW
Case(b): AC Three phase 400 V supply with star combination
Case(i) :  Power consumption for 2 elements in parallel in each phase = 6.4 kW
Case(ii):  Power consumption for 2 elements in series in each phase = 1.6 kW
Case(c): AC Three phase 400 V supply with delta combination
Case(i) :  Power consumption for 2 elements in series in each branch = 4.8 kW
Case(ii):  Power consumption for 2 elements in parallel in each branch = 19.2 kW
Case(d): Power loss will be 0.25 of the values obtained as above with auto-transformer tapping


## Example 4.2.4, Page number 728¶

In :
#Variable declaration
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

#Calculation
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)

#Result
print('Amount of energy required to melt brass = %.f kWh' %energy)

Amount of energy required to melt brass = 222 kWh


## Example 4.2.5, Page number 728-729¶

In :
#Variable declaration
V_2 = 12.0         #Secondary voltage(V)
P = 30.0*10**3     #Power(W)
PF = 0.5           #Power factor

#Calculation
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

#Result
print('Height up to which the crucible should be filled to obtain maximum heating effect, H_m = %.3f*H_c ' %H_m)
print('\nNOTE: ERROR: Calculation mistake in textbook solution and P is 30 kW not 300 kW')

Height up to which the crucible should be filled to obtain maximum heating effect, H_m = 0.577*H_c

NOTE: ERROR: Calculation mistake in textbook solution and P is 30 kW not 300 kW


## Example 4.2.6, Page number 732¶

In :
import math

#Variable declaration
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

#Calculation
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*math.pi*f*C)            #Reactance of condenser(ohm)
phi = math.acos(PF)*180/math.pi      #Φ(°)
R = X_c*math.tan(phi*math.pi/180)    #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)

#Result
print('Voltage necessary for heating, V = %.f V' %V)
print('Current flowing in the material, I_c = %.2f A' %I_c)
print('\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here & approximation in textbook')

Voltage necessary for heating, V = 2076 V
Current flowing in the material, I_c = 3.85 A

NOTE: Changes in the obtained answer from that of textbook is due to more precision here & approximation in textbook


## Example 4.2.7, Page number 732-733¶

In :
import math

#Variable declaration
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

#Calculation
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*math.pi*f*C)                              #Reactance(ohm)
phi = math.acos(PF)*180/math.pi                        #Φ(°)
R = X_c*math.tan(phi*math.pi/180)                      #Resistance(ohm)
V = (P*R)**0.5                                         #Voltage applied across electrodes(V)
I_c = V/X_c                                            #Current through the material(A)

#Result
print('Voltage applied across electrodes, V = %.f V' %V)
print('Current through the material, I_c = %.1f A' %I_c)
print('\nNOTE: ERROR: Calculation mistake in the textbook solution')

Voltage applied across electrodes, V = 2188 V
Current through the material, I_c = 0.7 A

NOTE: ERROR: Calculation mistake in the textbook solution


## Example 4.2.8, Page number 736-737¶

In :
#Variable declaration
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

#Calculation
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(%)

#Result
print('Time taken to melt 3 metric tonnes of steel = %.f minutes' %time)
print('Power factor of the furnace = %.2f ' %PF)
print('Electrical efficiency of the furnace = %.f percent' %n_elect)
print('\nNOTE: ERROR: Calculation and substitution mistake in the textbook solution')

Time taken to melt 3 metric tonnes of steel = 53 minutes
Power factor of the furnace = 0.95
Electrical efficiency of the furnace = 80 percent

NOTE: ERROR: Calculation and substitution mistake in the textbook solution