In [1]:

```
import math
# Variables
f = 0.01 #flux in Wb
l = 1. #mean circumference in m
N = 1000. #tunrs
Ur = 1000. #relative permeability
# Calculations
Uo = 4*math.pi*(10**-7) #permeability of free space in H/m
a = 0.001 # cross section area in m**2
I = (f*l)/(N*Uo*Ur*a) # current in Amp. math.since f = A*T/(l/Uo*Ur*a)
# Results
print 'Thus Current required is %3.3f Amp'%(I)
```

In [2]:

```
import math
# Variables
f = 1.2*(10**-3) #flux in Wb
l = 1.4 #mean circumference in m
N = 500. #tunrs
# Calculations
Uo = 4*math.pi*(10**-7) #permeability of free space in H/m
a = 0.0012 # cross section area in m**2
I = 2 #current in Amp
Ur = (f*l)/(N*I*Uo*a) #relative permeability
# Results
print 'Thus the relative permeability of iron is %3.2f '%(Ur)
```

In [3]:

```
import math
# Variables
l = 0.4 #mean circumference in m
N = 200. #tunrs
Uo = 4*math.pi*(10**-7) #permeability of free space in H/m
a = 5*(10**-4) # cross section area in m**2
I = 6.4 #current in Amp
f = 0.8*(10**-3) #flux in Wb
# Calculations
fd = f/a #flux density in Wb/m**2
fi = I*N/l #Field intensity in AT/m
Ur = (f*l)/(N*I*Uo*a) #relative permeability
# Results
print 'i) The Flux density is %3.2f Wb/m**2 '%(fd)
print 'ii) The Field intensity is %3.2f AT/m '%(fi)
print 'iii) The Relative permeability of steel is %3.2f '%(Ur)
#The answer to part(iii) has a calculation error in the textbook, hence it doesn't match the answer here.
```

In [4]:

```
# Variables
Hl = 250. #Hysteresis loss per m**3 in J/cycle
V = 1./150 #Volume of specimen in m**3
N = 50. #No of cycles/sec
# Calculations
E = Hl*V*N #Energy loss per sec in J
Eh = (E*3600)/1000 #Energy loss per hour in kWh
# Results
print 'Thus Energy loss per hour is %3.2f kWh'%(Eh)
```

In [5]:

```
# Variables
P = 4. #no of poles
N = 1600. # Speed in rpm
f = P*N/120 #Frequency of magnetic reversal
V = 5400. #volume
d = 7.5 #density
# Calculations
m = (V*d)/1000 #Mass of armature in kg
L = 1.76 #Loss in W/kg
Cl = L*m #Core loss in Watts
# Results
print 'Thus Frequency of magnetic reversal is %3.2f c/s'%(f)
print ' and Core loss is %3.2f Watts'%(Cl)
```

In [6]:

```
# Variables
v = 76300. #volume in c.c
P = 8. # no of poles
N = 375. #rpm
f = P*N/120 #freqency in c/s
Bmax = 12000. #max. flux density in lines/cm**2
n = 0.002 #(assumed)
d = 7.8 #densityin gm/c.c
l = 1.7 #loss in watts per kg
# Calculations
Hl = n*v*f*(Bmax**1.6)*(10**-7) #Hysteresis loss in Watts
Al = v*d*l/1000 #Additional loss under particular running conditions
Tl = Hl+Al #total core loss
# Results
print 'Thus the total core loss is %4.0f Watts'%(Tl)
```

In [7]:

```
# Variables
m = 12000. #mass in gm
d = 7.5 #density of iron in gm/c.c
Hl = 3000. #Hysteresis loss per cc in ergs/cycle
N = 50. #No of cycles per sec
# Calculations
v = m/d #volume of specimen
E = v*Hl*N #Energy loss per cc in ergs
Eh = E/(10**10) #Energy loss per hour in kWh
# Results
print 'Thus the Loss in energy is %3.3f kWh'%(Eh)
```

In [8]:

```
# Variables
m = 10. #mass in kg
T1 = 20. #total loss in watts
f1 = 50. #frequency in c/s
T2 = 35. #total loss in watts
f2 = 75. #frequency in c/s
# Calculations
#both have same peak flux density
#total loss = hysteresis loss+ Eddy current loss
#all quantities except frequency are consmath.tant
#so Total loss = Af+Bf**2
#let c1 and c2 be consmath.tants such that total loss = c1*f + c2*f**2
c2 = (T2-(T1*f2/f1))/(f2**2-f1*f2)
c1 = (T1-c2*f1**2)/f1
k = c1/c2 #hysteresis loss/eddy current loss
H50 = T1*k/101 #hysteresis loss at 50 c/s
E50 = T1-H50 #eddy current loss at 50 c/s
# Results
print 'Thus hysteresis loss at 50 c/s is %3.1f Watts '%(H50)
print 'And Eddy current loss at 50c/s is %3.1f Watts'%(E50)
```