Chapter 5:Chemical and Heating Effects of Electric Current

Example 5.1: Page 74:

In [1]:
from __future__ import division
import math

#given  data  :
t=200;#  time  in  sec
M=111.83;#  silver  in  mg
I=0.5;#  current  in  A

#calculations:
Z=(M/(I*t*1000))*1000#  electro-chemical-equivalent

#Results
print  "E.C.E,Z(mg/C)  =  ",Z

E.C.E,Z(mg/C)  =   1.1183

Example 5.2: page 74:

In [2]:
from __future__ import division
import math

#given  data  :
Z=0.329*10**-3#  IN  g/C
I=1    #  in  amperes
t=90*60#  in  seconds

#calculation:
M=Z*I*t#  in  grams  
A=200#area  in  centimete  square
S=8.9#density  in  g/cc
T=(M)/(2*A*S)#thickness  in  cm

#Results
print  "thickness  of  copper  in  cm  is", round(T,6)

thickness  of  copper  in  cm  is 0.000499

Example 5.3: Page 76:

In [3]:
from __future__ import division
import math

#given data:
w=15  #  in  kg
t1=15#  in  degree  celsius
t2=100#in  degree  celsius
t=25  #  time  in  minutes
I=10  #  in  ampere
n=85  #efficiency  of  conversion  in  percentage

#calculations:
ho=w*(t2-t1)#output  heat  required  in  kcal
R=((ho*4187*100)/(I**2*t*60*n))#  resistance  in  ohms

#Results
print  "resistance  in  ohms",R

resistance  in  ohms 41.87

Example 5.4: page 76:

In [4]:
from __future__ import division
import math

#given data:
w=20  #  in  kg
t1=10#  in  degree  celsius
t2=90#in  degree  celsius
t=2*3600+19*60+34#  time  in  seconds
I=4    #  in  ampere
n=80  #efficiency  of  conversion  in  percentage

#calculations:
ho=w*(t2-t1)#output  heat  required  in  kcal
V=((ho*4187*100)/(I*t*n))#  POTENTIAL  DROP  IN  VOLTS

#Results
print  "potential  drop  across  heater  element  in  volts  is", V

potential  drop  across  heater  element  in  volts  is 250.0