In [1]:

```
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)
k = 0.6 #Radiatting efficiency
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')
```

In [1]:

```
#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)
k = 0.6 #Radiatting efficiency
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)
```

In [1]:

```
#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)
```

In [1]:

```
#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)
```

In [1]:

```
#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')
```

In [1]:

```
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')
```

In [1]:

```
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')
```

In [1]:

```
#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')
```