# Chapter5, Electric heating and welding¶

## Ex1 : Page 272¶

In [1]:
from sympy import symbols, solve
from math import sqrt, pi
#To determine the size and length of the wire

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

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=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.

The radius of wire = 1.706 mm
& length = 16.77 m

The Temperature of the element when the charge is cold is 1105 degree celsius


## Ex2 : Page 274¶

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.

The Temperature for the following configurations are:
Two Groups connected in series first and then in delta : 715.5 degree Celsius
Two Groups connected in series first and then in star : 478.1 degree Celsius
Two Groups connected in parallel first and then in star : 789.3 degree Celsius


## Ex3 : Page 278¶

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.

The Average kW input to the furnance is 1995.12 kW
The Arc Voltage is 46.89 V
The kVA input is 2335.96 kVA


## Ex4 : Page 284¶

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)

The Effciency of the High Frequency Induction Furnace is 54 percent


## Ex5 : Page 284¶

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)

The Equivalent Resistance of the cylinder is 259 * 10**-6 ohm
The Required Current in the primary is 1772.1 A


## Ex6 : Page 285¶

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.

The Power Factor and The Absorbed power are 0.832 lagging and 481 kW respectively.


## Ex7 : Page 289¶

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.

The Current And Voltage are 5 A and 999V respectively
For the New Voltage the frequency is 83.15 MHz


## Ex8 : Page 290¶

In [8]:
#To estimate the voltage and Current during heating
l=30  #Length
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

The Voltage and Current are 415 V and 10.39 A respectively