# Chapter 7 : Electrothermal energy conversion¶

## Example 1 : pg 135¶

In [1]:
#Example  7.1# width and length
#calculate the length and width
from math import sqrt
#given:
vph=400.;#phase voltage in volts
n=3.;#number of phase
kw=36.;#power in kW
#calculations
r=((vph**2)/(n*((kw*10**3)/n)));#resistance in ohms
p=1.016*10**-6;#resitivity
t=0.3;#thickness in mm
x=(((r*t*10**-3)/(p)));#variable
t1=1000;#initial temperature in degree celsius
t1k=273+t1;#initial temperature in kelvin
t2=650;#final temperature in degree celsius
t2k=273+t2;#final temperature in kelvin
h=((3*10**4)*((t1k/1000)**4-(t2k/1000)**4));#W/m**2
y=((kw*10**3)/(3*2*h));#variable
l=sqrt(x*y);#length in meter
w=y/l;#width in meter
#results
print "length is,(m)=",round(l,3)
print "width is,(mm)=",round(w*10**3,3)

length is,(m)= 16.201
width is,(mm)= 12.345


## Example 2 : pg 138¶

In [2]:
#Example  7.2#
#calculate the power required
#given:
l=0.2;#length in meter
w=0.1;#width in meter
th=25.;#thickness in mm
#calculations
vw=l*w*th*10**-3;#volume in m**3
ww=600.;#weight of wood in kg/m**3
ww1=vw*ww;#weight of wood kg
shw=1500.;#specific heat of wood in J/kg/degree celsius
t=200.;#temperature in degree celsius
rg=t*shw*ww1;#energy in joules
h=(rg/(3.6*10**3));#Wh
t=15.;#time in minutes
pr=h*(60./t);#power required in Watt
#results
print "power required is,(W)=",pr

power required is,(W)= 100.0


## Example 3 : pg 140¶

In [3]:
#Example  7.3# voltage and current
#calculate the current and voltage
from math import sqrt, acos, pi
l=0.2;#length meter
w=0.1;#width in meter
th=25;#thickness in mm
#calculations
vw=l*w*th*10**-3;#volume of wood in m**3
ww=600;#weight of wood in kg/m**3
ww1=vw*ww;#weight of wood kg
shw=1500.;#specific heat of wood  in J/kg/degree celsius
t=200.;#temperature in degree celsius
rg=t*shw*ww1;#energy in joules
h=(rg/(3.6*10**3));#Wh
t=15.;#time in minutes
pr=h*(60./t);#power required in Watt
eo=8.854*10**-12;#permittivity constant
er=5.;#permittivity of wood

voltage is ,(V)= 424.0