Chapter 1: Electric Heating

Example 1.1, Page 14

In [27]:
#Variable declaration
r1=100.;#in ohms
r2=r1;# in ohms
V=250.;# ac supply in volts

#Calculations&Results
rp = 1/(1/r1+1/r2);# equivalent resistance in ohms
pp=((V**2)/rp);#power drawn in watts
print "part (a) "
print "power drawn when elements are in parallel,(W)= %.f"%pp
rs=r1+r2;# equivalent resistance in ohms
ps=((V**2)/rs);#power drawn in watts
print "part (b) "
print "power drawn when elements are in series ,(W)=%.1f"%ps
part (a) 
power drawn when elements are in parallel,(W)= 1250
part (b) 
power drawn when elements are in series ,(W)=312.5

Example 1.2, Page 15

In [29]:
import math

#Variable declaration
P=2.5;#power in kW
V=240.;# in volts
K=1;#radiating efficiency
e=0.9;#emissivity
p=42.5*10**-6;# resistivity in ohm-cm
T1=1500.;#in dgree celsius
T2=450.;#in degree celsius

#Calculations
x=((math.pi*V**2)/(4*(p*10**-2)*P*10**3));#
H=((5.72*K*e)*(((T1+273)/100)**4-((T2+273)/100)**4));#
z=((P*10**3)/(math.pi*H))**2;#
l=(z*x)**(1./3);#length in meter
d=((math.sqrt(z))/l)*10**3;#diameter in mm

#Results
print "length in meter =%.2f"%l
print "diameter in mm =%.3f"%d
length in meter =4.79
diameter in mm =0.336

Example 1.3, Page 15

In [31]:
import math

#Variable declaration
V=440.;# volts
P=20.;#in kW
T1=1200.;#in degree celsius
T2=700.;# in degree celsius
K=0.6;#radiating efficiency
e=0.9;#emissivity
t=0.025;#thickness in mm
p=1.05*10**-6;#resisitivity in ohm - meter

#Calculations
Pp=(round(P*10**3))/3;#power per phase in watts
Pv= (V/math.sqrt(3));#phase voltage
R=Pv**2/Pp;#resistance of strip in ohms
x=((R*t*10**-3)/(p));#
H=((5.72*K*e)*(((T1+273)/100)**4-((T2+273)/100)**4));#in W/m^2
y=((Pp)/(H*2));#in m^2
w=math.sqrt(y/x)*10**3;#width in mm
l=x*w*10**-3;#length of strip in meter

#Results
print "width in mm =%.3f"%w
print "length of strip in meter =%.2f"%l
width in mm =11.084
length of strip in meter =2.55

Example 1.4, Page 16

In [32]:
#Variable declaration
a=6.;#area in m^2
l=a/6;#one side of tank in meter
V=l*l*l;#volume in m^2
e=90./100;#capacity
wh=6*e*1000;#water to be heated daily in kg
s=4200;#specific heat of water in J/Kg/degree celsius
t1=65;#in degree celsius
t2=20;#in degree celsius

#Calculations
hr=wh*s*(t1-t2)*10**-6;#heat required to raise the temperture of water
hr1=hr/3.6;#heat required in kWh
d=6.3;#difference in watts
l=((d*a*(t1-t2)*24)/1000);#losses from the surface of the tank in kWh
es=hr1+l;#energy supplied in kWh
lk=es/24;#loading in kW
ef=(hr1/es)*100;#efficiency of the tank in percentage

#Results
print "loading in kW = %.1f"%lk
print "efficiency of the tank in percentage = %.1f"%ef
loading in kW = 13.5
efficiency of the tank in percentage = 87.4

Example 1.5, Page 22

In [33]:
import math

#Variable declaration
sh=444;# specific heat of steel in J/Kg/°C
lh=37.25;#latent heat in kJ/kg
mp=1370;#melting point of steel °C
t1=19.1;#initial temperture in °C
e=0.5;#overall efficiency
ip=5700;#input current in amperes
rs=0.008;#resistance of transformer referred to secondary in ohms
rr=0.014;# recatance in ohms
m=4.3;# steel in tonnes

#Calculations
ers=((m*10**3*((sh*(mp-t1))+lh*10**3)));# energy required in joules
ersh=ers/(3.6*10**6);#energy required in kWh
ata=1;#time taken to melt steel in hours
ao=ersh/ata;#average output in kW
ai=ao/e;#average input in kW
vdr=ip*rs;#voltage drop due to resistance of furnace leads
vdr1=ip*rr;#voltage drop due to reactance of furnace leads
va=((ai*10**3)/(3*ip))-(vdr);#voltage resistive in nature
rac=va/ip;#arc resistance in Ω
oppv=math.sqrt((va+vdr)**2+vdr1**2);#open circuit phase voltage in volts
kvas=3*ip*oppv*10**-3;#total kVA drawn 
pf=((va+vdr)/oppv);#power factor 

#Result
print "average input in kW = %.2f"%ai
print "arc voltage in volts = %.1f"%va
print "arc resistance in Ω = %.4f"%rac
print "pf of the current drawn from the supply (lagging) = %.4f"%pf
print "total kVA drawn in kVA = %.f"%kvas
average input in kW = 1521.84
arc voltage in volts = 43.4
arc resistance in Ω = 0.0076
pf of the current drawn from the supply (lagging) = 0.7445
total kVA drawn in kVA = 2044

Example 1.6, Page 23

In [34]:
import math

#Variable declaration
sh=0.12;# specific heat of steel in kcal/Kg/°C
lh=8.89;#latent heat in kcal/kg
mp=1370;#melting point of steel °C
t1=19.1;#initial temperture in °C
e=0.5;#overall efficiency
ip=5700;#input current in amperes
rs=0.008;#resistance of transformer referred to secondary in ohms
rr=0.014;# recatance in ohms
m=4.3;# steel in tonnes

#Calculations
ers=((m*10**3*((sh*(mp-t1))+lh)));# energy required in joules
ersh=ers/(860);#energy required in kWh
ata=1;#time taken to melt steel in hours
ao=ersh/ata;#average output in kW
ai=ao/e;#average input in kW
vdr=ip*rs;#voltage drop due to resistance of furnace leads
vdr1=ip*rr;#voltage drop due to reactance of furnace leads
va=((ai*10**3)/(3*ip))-(vdr);#voltage resistive in nature
rac=va/ip;#arc resistance in Ω
oppv=math.sqrt((va+vdr)**2+vdr1**2);#open circuit phase voltage in volts
kvas=3*ip*oppv*10**-3;#total kVA drawn 
pf=((va+vdr)/oppv);#power factor 

#Result
print "average input in kW =%.f"%ai
print "arc voltage in volts =%.1f"%va
print "arc resistance in Ω =%.5f"%rac
print "pf of the current drawn from the supply (lagging) =%.4f"%pf
print "total kVA drawn in kVA =%.1f"%kvas
average input in kW =1710
arc voltage in volts =54.4
arc resistance in Ω =0.00954
pf of the current drawn from the supply (lagging) =0.7816
total kVA drawn in kVA =2187.7

Example 1.7, Page 29

In [35]:
import math

#Variable declaration
sh=0.1;# specific heat of steel in kcal/Kg/°C
lh=26.67;#latent heat in kcal/kg
mp=555;#melting point of steel °C
t1=35;#initial temperture in °C
e=0.8;#overall efficiency
ip=5700;#input current in amperes
rs=0.008;#resistance of transformer referred to secondary in ohms
rr=0.014;# recatance in ohms
m=2;# steel in tonnes

#Calculations
ers=((m*10**3*((sh*(mp-t1))+lh)));# energy required in joules
ersh=ers/(860);#energy required in kWh
ata=1;#time taken to melt steel in hours
ao=ersh/ata;#average output in kW
ai=ao/e;#average input in kW
vdr=ip*rs;#voltage drop due to resistance of furnace leads
vdr1=ip*rr;#voltage drop due to reactance of furnace leads
va=((ai*10**3)/(3*ip))-(vdr);#voltage resistive in nature
rac=va/ip;#arc resistance in Ω
oppv=math.sqrt((va+vdr)**2+vdr1**2);#open circuit phase voltage in volts
kvas=3*ip*oppv*10**-3;#total kVA drawn 
pf=((va+vdr)/oppv);#power factor 
rf=ai/ata;# in kW

#Result
print "rating of furnance  in kW =%.1f"%rf
rating of furnance  in kW =228.7

Example 1.8, Page 29

In [36]:
#Variable declaration
sh=880;# specific heat of steel in J/Kg/°C
lh=32000;#latent heat in J/kg
mp=660;#melting point of steel °C
t1=15;#initial temperture in °C
ip=5700;#input current in amperes
rs=0.008;#resistance of transformer referred to secondary in ohms
rr=0.014;# recatance in ohms
m=1.8;# IN KG

#Calculations
ers=((m*((sh*(mp-t1))+lh)));# energy required in joules
ersh=ers/(3.6*10**6);#energy required in kWh
TM=10.;#TIME TO MELT IN MINS
ip=5;#input of the furnance in kW
ei=(ip)*(TM/60);#energy input in kWh
n=(ersh/ei)*100;#efficiency of furnance in percentage

#Result
print "efficiency of furnance in percentage =%.f"%n
efficiency of furnance in percentage =36

Example 1.9, Page 30

In [38]:
import math

#Variable declaration
vs=10;#secondary voltage in volts
p=500;#power drawn in kW
pf=0.5;
ir=(p*10**3)/pf;#secondary current in amperes

#Calculations
zs=vs/ir;#impedence of secondary circuit in ohms
rs=zs*pf;#resistance of secondary circuit in ohms
res=zs*(math.sqrt(1-pf**2));#rectancetance of secondary circuit in ohms
rs1=2*rs;# resistacne when hearth is full in Ω
res1=res;# reactance when hearth is full in Ω
zs1=(math.sqrt(rs1**2+res1**2));#impedance of secondary circuit in Ω
pf1=rs1/zs1;#power factor
is1=vs/zs1;#secondary current in amperes
pd=is1**2*rs1*10**-4;#power drawn in kW

#Results
print "power factor is =%.3f"%pf1
print "power drawn in kW =%.3f"%pd
power factor is =0.756
power drawn in kW =571.429

Example 1.10, Page 30

In [39]:
import math

#Variable declaration
vs=10;#secondary voltage in volts
p=400;#power drawn in kW
pf=0.6;#

#Calculations
is1=(p*10**3)/pf;#secondary current in amperes
zs=vs/is1;#impedence of secondary circuit in ohms
rs=zs*pf;#resistance of secondary circuit in ohms
res=zs*(math.sqrt(1-pf**2));#rectancetance of secondary circuit in ohms
x=(rs)/res;#height 

#Result
print "maximum heat will be obtained with the height of charge as 3/4 of height of hearth =%.2f"%x
maximum heat will be obtained with the height of charge as 3/4 of height of hearth =0.75

Example 1.11, Page 32

In [40]:
import math

#Variable declaration
p=5*10**-7;#specific resistance in Ω-m
rp=1;#relative permeability
dp=0.0015;#depth of penetration in mter

#Calculations
f=((p*10**7)/((rp*(dp)**2)*4*(math.pi)**2))*10**-3;#frequency in kHz

#Result
print "frequency in kHz =%.3f"%f
frequency in kHz =56.290

Example 1.12, Page 36

In [41]:
#Variable declaration
l=0.5;#length in meter
b=0.25;#breadh in meter
h=0.02;#in meter
t1=25;# temperture °C
t2=125;# temperture °C
t=10;#time in minutes
f=30;#frequency in 30 MHz
w=600;#weight of the wood in kg/m^3
sh=1500;#specific heat in J/Kg/°C
e=50;#efficiency

#Calculations
vp=l*b*h;#volume in m^3
wp=vp*w;#weight of plywood in kg
hr=sh*wp*(t2-t1);#heat required in joules
hrt=(hr/(3600));#heat required to raise the temperture of plywood in Wh
pu=hrt/(1./6);#power utilized in watts
pi=(pu/e)*100;#power input required in percentage

#Result
print "power input required ,(W)= %.f"%pi
power input required ,(W)= 750

Example 1.13, Page 37

In [42]:
import math

#Variable declaration
vl=600;#in volts
p=200;#power absorbed in watts
pf=0.05;#power factor
f=30*10**6;#frequency in Hz
ep=8.854*10**-12;#constant
er=5;#
a=150;# in cm^2
t=0.02;# in meter

#Calculations
c=((ep*er*a*10**-4)/t);#capacitance in farads
vr=(math.sqrt(p/(2*math.pi*f*c*pf)));#voltage is required in volts
i=p/(vr*pf);#current in amperes
f2=((f*(vr/vl)**2))*10**-6;#frequency in Mhz

#Results
print "voltage in volts =%.f"%vr
print "current in amperes =%d"%i
print "frequency in MHz=%.1f"%f2
voltage in volts =799
current in amperes =5
frequency in MHz=53.3

Example 1.14, Page 37

In [43]:
import math

#Variable declaration
pf=0.04;#power factor
p=1000;#in watts
f=10*10**6;# in MHz
a1=.004;#area in m^2
a2=0.001;#area in m^2
t=0.02;#thickness in meter
t1=.01;#thicknes sin meter
t2=t-t1;#thickness in meter
ep=8.854*10**-12;#constant in F/m
er=5;#relative permittivity of plywood
er1=1;#relative permittivity in air

#Calculations&Results
c=(ep*(((a1*er1)/t)+(a2/((t1/er)+(t2/er1)))));#capacitance in farads
vr=(math.sqrt(p/(2*math.pi*f*c*pf)));#voltage is required in volts
print "part (a)"
print "voltage across the electrodes in volts =%.f"%vr
i=p/(vr*pf);#current in amperes
print "part (b)"
print "cureent in amperes is =%.3f"%i
part (a)
voltage across the electrodes in volts =12594
part (b)
cureent in amperes is =1.985