In [1]:

```
from sympy import symbols, solve
from math import sqrt, pi
#To determine the size and length of the wire
r=symbols('r') #Variable Value of radius
A=pi*(r**2) #Area of cross section of the wire
V=220 #Supply Voltage
P=20*(10**3) #Power input
#Temperatures
T1=1127 #Wire
T2=427 #Charge
R=(V**2)/P #Resistance of the wire
e=0.9 #emissivity constant
K=0.6 #Radiation Effciency
p=1.09*(10**-6) #Resistivity
l=R*A/p #Length in term of 'r'
H=5.72*e*K*((((T1+273)/100)**4)-(((T2+273)/100)**4)) #Heat dissipated per sq.m of the surface
CSA=pi*2*r*l #Curved surface area
CSAn=P/H #Numerical Value of Curved suraface area
X=CSA-CSAn #Polynomial to find 'r'
r=solve(X, r)#Numerical Value of radius
r=r[0] # taking real root only
print 'The radius of wire = %0.3f mm'%(r*1000)
l=CSAn/(2*pi*r) #Numerical Value of length
print '& length = %0.2f m\n'%(l)
#For charge temperature to be cold
Ti=25 #Cold Temperature
T=symbols('T') # value in degree C # Variable value of the element temperature
Hi=5.72*e*K*((((T+273)/100)**4)-(((Ti+273)/100)**4)) #Heat dissipated per sq.m of the surface
CSA=pi*2*r*l #Curved surface area
Y=Hi-H #polynomial to find the temperature of the element
#Roots of T must be real
T=solve(Y, T) #Numerical Value
T=T[1]
print 'The Temperature of the element when the charge is cold is %0.f degree celsius'%(T)
# Answer in the textbook is not accurate.
```

In [2]:

```
from __future__ import division
from math import sqrt
#To determine the various temperature by changing the connection of the resistance elements
#Note that the value in kelvin of the first case in the textbook is wrong
#P is directly proportion to V**2 and H is directly propostional to KT**4
#Different Temperatures for different configurations
T1=1125 #Temperature in First Case
from sympy import symbols, solve
T2,T3,T4=symbols('T2 T3 T4')
#Multiplying Factors to the square of voltages
V1=1 #Line to Line Voltage
V2=V1/2 #when connected in series first and then delta
V3=V1/(2*sqrt(3)) #when connected in series and then in star
V4=V1/(sqrt(3)) #When connected in parallel and in star
#To find the power loss in each case
Pow = lambda y:(y**2)
P1=Pow(V1)
P2=Pow(V2)
P3=Pow(V3)
P4=Pow(V4)
#To find the heat dissipated from each case
heatdiss = lambda y:(y**4)
H1=heatdiss(T1+273)
H2=heatdiss(T2+273)
H3=heatdiss(T3+273)
H4=heatdiss(T4+273)
#Polynomials to find the temperature in degree celsius
temp = lambda y,z:(P1/y)-(H1/z)
X2=temp(P2,H2)
X3=temp(P3,H3)
X4=temp(P4,H4)
#Temperature Numerical Value
T2=solve(X2, T2)
T3=solve(X3, T3)
T4=solve(X4, T4)
#Only to consider Real Roots
T2=T2[1]
T3=T3[1]
T4=T4[1]
print 'The Temperature for the following configurations are:'
print 'Two Groups connected in series first and then in delta : %0.1f degree Celsius'%T2
print 'Two Groups connected in series first and then in star : %0.1f degree Celsius'%(T3)
print 'Two Groups connected in parallel first and then in star : %0.1f degree Celsius'%(T4)
# Answer in the textbook are not accurate.
```

In [3]:

```
from __future__ import division
#To Determine the average KW input to the furnace
M=10*(10**3) #Mass of Steel Melted
t=2*3600 #Time Taken to Melt the steel
eff=50/100 #Overall Efficiency
I=9000 #Current Input
R=0.003 #Resistance
X=0.005 #Reactance
SH=0.12 #Specific Heat
LHF=8.89*(10**3) #Latent Heat of Fusion
Tm=1371 #Melting Point
Ti=20 #Room Temperature
Hm=M*LHF #Heat Required for melting
Hr=M*SH*(Tm-Ti)*1000 #Heat Required to raise the temperature
Ht=Hm+Hr #Total Amount of heat required
E=Ht*4.2/(3600) #Energy in Whr
P=E*3600/t #Power
Pa=P/eff #Actual Power Input to the Furnace
Vt=Pa/(3*I) #V Cos theta
#The Above voltage is the sum of arc drop and drop in resistance load
Va=Vt-(I*R) #Arc Drop
Vx=I*X #Reactance Drop
Vs=sqrt((Vt**2)+(Vx**2)) #Supply Voltage
S=3*Vs*I/1000 #KVA input
print 'The Average kW input to the furnance is %0.2f kW'%(Pa/1000)
print 'The Arc Voltage is %0.2f V'%(Va)
print 'The kVA input is %0.2f kVA'%(S)
#Answer in the textbook are not accurate.
```

In [4]:

```
#To determine the effciency of a high frequency induction furnance
t=10*60 #Time Taken to rise temperature in seconds
M=1.815 #Mass of aluminium melted
Pi=5*(10**3) #Power Input
Ti=15 #Initial Temperature
Tm=660 #Melting Point of Al
SHAl=0.212 #Specific heat of Al
LHFAl=76.8*(10**3) #Laten Heat of fusin in Cal/Kg
Hm=M*LHFAl #Heat required to melt Al
Htr=SHAl*M*1000*(Tm-Ti) #Heat required to raise the temperature
HTot=Hm+Htr #Total Heat Required
HToth=HTot*3600/t #Heat required per hour
Po=HToth*4.2/3600 #Power Output
eff=Po*100/Pi #Efficiency
print 'The Effciency of the High Frequency Induction Furnace is %0.f percent '%(eff)
```

In [5]:

```
from math import pi
#To Determine the equivalent resistance of the charge and current
f=960 #Frequency
N1=20 #Primary Turns
N2=1 #Secondary is Single Turn
Pi=325*(10**3) #Power Input
Di=45 #Internal Diameter
l=50 #Depth of the charge
#Assumptions
p=200*(10**-6) #Resistivity
M=1 #For Molten Steel
t=(1/(2*pi))*sqrt(p*(10**9)/(M*f)) #Depth of penentration of the current
A=t*l #Effective Area
Dm=Di+t #Mean Diameter
Dmcf=pi*Dm #Mean Length of current flow
Rc=p*Dmcf/A #Resistance of the Cylinder
Is=sqrt(Pi/Rc) #Current flowing through secondary
Ip=Is*N2/N1 #Primary Current
print 'The Equivalent Resistance of the cylinder is %0.f * 10**-6 ohm '%(Rc/(10**-6))
print 'The Required Current in the primary is %g A'%(Ip)
```

In [6]:

```
from math import exp, degrees, cos, atan, acos, sin
#To Detemine power absorbed and the power factor
Vs=15 #Secondary Voltage
P=500*(10**3) # Power Taken
pfs=0.6 #Power Factor
Is=P/(Vs*pfs) #Secondary Current
#Taking Current as Reference voltage will be
t=degrees(acos(pfs)) #Power Factor Angle
Vsp=Vs*(complex(cos(t*pi/180),sin(t*pi/180))) #Phasor Secondary Voltage
R=Vsp/Is #Impedance
#if the resistance is doubled, The Total impedance doubles, Considering Vs as reference
R2=R.real+R
I2=Vs/R2 #New Current
pfn=cos(atan((I2.imag)/(I2.real))) #power factor of new current
Pab=Vs*abs(I2)*pfn/1000 #Power Absorbed
print 'The Power Factor and The Absorbed power are %0.3f lagging and %0.f kW respectively.'%(pfn,Pab)
#Answer in the textbook are not accurate.
```

In [7]:

```
from math import tan
#To determine the Voltage Required and Current Drawn
t=2*(10**-2) #Thickness
A=150*(10**-4) #Area of the slab
Er=4 #Relative Permittivity
pf=0.04 #Power Factor
f=30*(10**6) #Frequency of supply
w=2*pi*f #Angular Frequency
P=200 #Power Required
Eo=8.854*(10**-12) #Permittivity of free space
C=Er*Eo*A/t #Capacitance
Xc=1/(C*w) #Capacitative Reactance
phi=degrees(acos(pf)) #power factor angle
R=tan(phi*pi/180)*Xc #Resistance
V=sqrt(P*R) #Voltage
I1=V/R #Current
Ic=V/Xc #Curent through the Capacitor
It=sqrt((I1**2)+(Ic**2)) #Total Current
Vn=600 #Limited Voltage
Rn=(Vn**2)/P #New Resistance
wn=tan(phi*pi/180)/(C*Rn) #New Angular Frequency
fn=wn/(2*pi) #New Frequnency
print 'The Current And Voltage are %0.f A and %0.fV respectively'%(It,V)
print 'For the New Voltage the frequency is %0.2f MHz'%((fn/(10**6)))
#No rounding off. Accurate Answers here.
```

In [8]:

```
#To estimate the voltage and Current during heating
l=30 #Length
b=15 #Breadth
t=2 #Thickness
t1=20 #Initial Temperature
t2=180 #Final Temperature
T=10*60 #Time Period in Seconds
f=40*(10**6) #Frequency of supply
w=2*pi*f #Angular Frequency
SH=0.35 #Specific Heat Of Wood
Er=5 #Relative Permitivity
Eo=8.854*(10**-12) # Permitivity of free space
pf=0.05 #Power Factor
Eff=90/100 #Efficiency
p=0.55 #Density
A=l*b #Area of the wooden board
W=(A*t)*p/1000 #Weight of wood in kilograms
H=W*SH*(t2-t1) #Heat required to raise temperature
E=H*4.2/3600 #Energy in kWhr
P=E*3600/T #In kilowatts
AP=P*1000/Eff #Actual Power
C=Eo*Er*A*(10**-2)/t #Capacitance
phi=degrees(acos(pf)) #Power Factor Angle
Del=(90-phi)*pi/180 #In Radian
V=sqrt(AP/(w*C*Del)) #Voltage
I=V*w*C #Current
print 'The Voltage and Current are %0.f V and %0.2f A respectively'%(V,I)
#Answer in the textbook are not accurate.
```