In [1]:

```
#Variable declaration
P = 8 #number of poles
f = 50 #Hz
#Calculations
Ns = (120*f)/P
#Result
print "Synchronous speed = %d r.p.m."%Ns
```

In [2]:

```
#Variable declaration
P = 2 #assumption - number of poles
f = 50 #Hz
#Calculations
Ns = (120*f)/P
#Result
print "Maximum speed = %d r.p.m."%Ns
```

In [3]:

```
#Variable declaration
Pa = 6 #number of poles in alternator
N = 1000 #rpm
Pi = 16 #number of poles in induction motor
S = 2.5 #slip
#Calculations
f = (N*Pa)/120 #Hz
Ns = (120*f)/Pi #rpm
Nr = Ns - ((S*Ns)/100)
#Results
print "The actual motor speed is %.2f r.p.m."%Nr
```

In [4]:

```
#Variable declaration
P = 14. #no. of poles
f = 50. #Hz
N = 414. #r.p.m.
#calculations
Ns = (120*f)/P #rpm
S = ((Ns-N)/Ns)*100
fr = (S*f)/100
#Results
print "Rotor frequency = %.2f Hz"%fr
print "Slip = %.2f %%"%S
#Incorrect answers in the textbook
```

In [5]:

```
#Variable declaration
Pi = 50 #motor input - KW
S = 3./100 #slip - %
#Calculations
Lc = Pi*S #rotor copper loss
Pm = Pi-Lc
#Result
print "Total mechanical power = %.1f kW"%Pm
```

In [6]:

```
import math
#Variable declaration
P = 4 #no. of poles
m = 3
f = 50 #Hz
n = 85 #efficiency %
Po = 17. #useful output power kW
Ls = 900 #stator losses W
Lwf = 1100 #windage&friction losses W
Lrc = 1. #rotor copper loss kW
#Calculations
Pi = Po/n*100 #input power
Tl = Pi-Po #total losses kW
Lc = Tl - Ls - Lwf #copper loss W
Pir = Pi-Ls #input to rotor kW
S = Lrc/Pi
Ns = 120*f/P #rpm
N = Ns-(S*Ns) #rpm
wT = 19.1*1000 #W
Td = (wT*60)/(math.pi*2*N)
Ta = (Po*1000*60)/(2*math.pi*N)
#Results
print "Slip = %.3f"%S
print "Torque developed = %.2f N-m"%Td
print "Torque available = %.1f N"%Ta
#Answers vary due to rounding-off errors
```

In [7]:

```
import math
#Variable declaration
P = 6 #no. of poles
f = 60 #Hz
T = 70 #N-m
N = 1152. #rpm
#Calculations
#Part a
Pi = (2*math.pi*1200*T)/60/1000 #input to rotor W
Ns = (120.*f)/P #rpm
S = (Ns-N)/Ns
#Part b
Lrc = S*Pi*1000 #W
Td = (2*math.pi*N*T)/60
#Part c
Pm = Td #W
#Results
print "Total input to the rotor = %.1f kW"%Pi
print "Rotor copper loss = %d W"%Lrc
print "Mechanical power developed = %.2f W"%Pm
#answers vary due to rounding-off errors
```

In [8]:

```
import math
#Variable declaration
Vs = 100 #voltage between two slip rings V
R2 = 0.2 #ohms/phase
Xo = 1 #ohm/phase
S = 0.04 #slip
#Calculations
Eo = Vs/math.sqrt(3)
I2 = (S*Eo)/math.sqrt(R2**2+(S*Xo)**2)
#During maximum torque,
Sm = R2/Xo
Ir = (Sm*Eo)/math.sqrt(S**2+(S*Xo)**2)
#Results
print "Rotor current when slip is 4%% is %.1f A"%I2
print "Slip during maximum torque is %.1f"%Sm
print "Rotor current during maximum torque is %.1f A"%Ir #incorrect answer in the textbook
```

In [9]:

```
#Variable declaration
S = 1 #slip at the time of starting
X2 = 1 #ohms/pase
X1 = 0.1 #internal rotor resistance ohms/phase
#Calculations
#Total resistance required for maximum torque is 1 Ohm
Re = X2-X1
#Result
print "External resistance needed to be added is %.1f ohms/phase"%Re
```

In [1]:

```
import sympy
from sympy.solvers import solve
from sympy import Symbol
#Variable declaration
T = 162.8 #maximum torque Nw-m
N = 1365 #speed rpm
f = 50 #Hz
P = 4 #no. of poles
R2 = 0.2 #ohms/phase
#Calculations
Ns = 120*f/P
S = Ns-N/Ns #slip at maximum torque
X2 = R2/S #ohms
Th = T/2 #half of maximum torque
R2 = Symbol('R2')
x = solve((Th*R2**2-18.1*40*R2+400),R2) #solving the equation
#Results
print "Required resistance = ",round(x[0],2),"ohms"
```

In [13]:

```
#Variable declaration
N = 290. #speed of motor at full load rpm
f = 50 #Hz
#Calculations
P = 20 #since motors running at full load
Ns= 120*f/P
S1 = (Ns-N)/Ns*100
#since S is directly proportinal to R2,
#when R2 is doubled, S is also doubled
S = 2*S1
#Results
print "The number of poles are %d"%P
print "Slip = %.1f %%"%S1
print "Slip if the rotor resistance is doubled = %.1f %%"%S
```

In [14]:

```
#Variable declarartion
S = 2.5/100 #slip
#Isc=4*If
#Calculations
'''
Ts Isc ^2
-- = --- * slip
Tf If
'''
Ts_by_Tf = 4**2*S*100
#Result
print "Starting torque is %d %% of full load torque"%Ts_by_Tf
```

In [15]:

```
import math
#Variable declaration
f = 50 #Hz
P = 6 #no. of poles
S = 1 #slip
Eo = 62.8 #phase voltage V
m = 3.
Xo = 6 #rotor reactance ohms
R2 = 1 #resistance ohms
#Calculations
Ns = 120*f/P #rpm
ws = 2*math.pi*Ns/60 #rad/sec
K = m/ws
Ts = (K*S*Eo**2*R2)/((R2**2)+(S*Xo)**2) #at starting
#For maximum torque
R2 = 1 #ohm
X2 = 6 #ohms
S = R2/X2
Tmax = (K*Eo**2)/(2*X2)
#Result
print "The starting torque is %.3f N-m"%Ts
print "The maximum torque develped is %.1f N-m"%Tmax
#Incorrect soltion in the textbook
```

In [16]:

```
import math
#Variable declaration
Vl = 200. #line voltage of stator V
t = 0.67 #turn ratio
Ns = 1500. #rpm
R2 = 0.1 #resistance ohms
S = 0.04 #slip
X = 0.9 #reactance ohms
m = 3. #number of phases
#Calculations
Vp = Vl/math.sqrt(3) #phase voltage of stator V
E = Vp*t
ws = (2*math.pi*Ns)/60
K = m/ws
T = (K*S*E**2*R2)/(R2**2+(S*X)**2)
#Result
print "Total torque = %d N-m"%T
#Incorrect answer in the textbook
```

In [17]:

```
#Variable declaration
P = 8 #number of pos
f = 50. #Hz
S = 0.04 #slip
Po = 15*1000 #oput power W
R2 = 0.2 #rotor resistance Ohms
X = 1.5 #rotor reactance Ohms
N = 400 #ohms
#Calculations
Ns = (120*f)/P #rpm
S_dash = (Ns-N)/Ns
R2_dash = (R2*S_dash)/S #ohms
R = R2_dash-R2 #resistance to be added
Pi = Po/(1-S) #W
Lc = S_dash*Pi #rotor copper loss W
P = Pi - Lc #ouput power W
#Results
print "The external resistance to be connected per phase is %.2f ohms"%R
print "Total rotor copper loss = %d W"%Lc
print "Ouput power of motor = %d W"%P
#Incorrect answers in the textbook
```

In [18]:

```
#Variable declaration
n = 0.8 #efficiency
O = 20*735.5 #load/output
#Calculations
i = O/n #input W
Tl = i - O #total losses W
#since total losses is a sum of rotor copper loss+stator copper loss+iron loss+mechanical loss
K = (Tl*3)/10 #rotor copper loss W
Ri = O+(K/3)+K #input to the rotor W
S = K/Ri #slip
#Result
print "Slip = %.3f"%S
```

In [19]:

```
import math
#Variable declaration
Pi = 20 #input power KW
N = 960. #speed rpm
f = 50 #Hz
P = 6 #number of poles
R2 = 1./3 #resistance ohms
#Calculations
Ns = 1000.
S = (Ns-N)/Ns #slip
Lrc = S*Pi*1000 #rotor copper loss W
Lrc_ph = Lrc/3 #rotor copper loss per phase
I2 = math.sqrt(Lrc_ph/R2)
#Result
print "Rotor current per phase = %.1f A"%I2
```

In [20]:

```
#Variable declaration
P = 4 #number of poles
f = 50. #Hz
N = 645. #speed rpm
R2 = 0.04 #rotor resistance ohms
#Calculations
Ns = (120*f)/P #rpm
S = (Ns-N)/Ns #slip
X2 = R2/S #rotor reactance ohms
a = R2/X2
St_by_mt = (2*a)/(a*2+1) #starting torque/maximum torque
S = 0.03
Flt_by_mt = (2*a*S)/(a**2+S**2) #full load torque/maximum torque
#Results
print "Maximum torque at starting is %d %%"%(St_by_mt*100)
print "Maximum torque when slip is 3%% is %d %%"%(Flt_by_mt*100)
#Value of Ns ic wrongly calculated in the textbook. Hence the answers vary
```

In [21]:

```
#Variable declaration
R2 = 0.5 #rotor resistance ohms
X2 = 5 #rotor reactance ohms
Tm_by_tfl = 2.5 #max torque/full load torque
#Calculations
#Tfl = Tm/2.5 = KE^2/2.5*2*X2
#Part a
Tst_by_Tfl_a = (25*R2)/(R2**2+X2**2)
#Part b
Tst_by_Tfl_b = (25*R2)/(3*(R2**2+X2**2))
#Part c
#Rotor voltage at starting = 0.75E2
Tst_by_Tfl_c = (0.75**2*25*R2)/(R2**2+X2**2)
#Results
print "The ratio of starting torque to full load torque for the given cases are as below:"
print "Part a: %.2f"%Tst_by_Tfl_a
print "Part a: %.2f"%Tst_by_Tfl_b
print "Part a: %.2f"%Tst_by_Tfl_c
```