Chapter 2: Induction Motors(Additional problems)

Problem 1, Page 225

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

Synchronous speed = 750 r.p.m.

Problem 2, Page 225

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

Maximum speed = 3000 r.p.m.

Problem 3, Page 225

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

The actual motor speed is 365.62 r.p.m.

Problem 4, Page 225

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

Rotor frequency = 1.70 Hz
Slip = 3.40 %

Problem 5, Page 225

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

Total mechanical power = 48.5 kW

Problem 6, Page 225

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

Slip = 0.050
Torque developed = 127.99 N-m
Torque available = 113.9 N

Problem 7, Page 225

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

Total input to the rotor = 8.8 kW
Rotor copper loss = 351 W
Mechanical power developed = 8444.60 W

Problem 8, Page 226

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

Rotor current when slip is 4% is 11.3 A
Slip during maximum torque is 0.2
Rotor current during maximum torque is 204.1 A

Problem 9, Page 226

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

External resistance needed to be added is 0.9 ohms/phase

Problem 10, Page 226

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"

Required resistance =  0.59 ohms

Problem 11, Page 226

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

The number of poles are 20
Slip = 3.3 %
Slip if the rotor resistance is doubled = 6.7 %

Problem 12, Page 226

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

Starting torque is 40 % of full load torque

Problem 13, Page 226

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

The starting torque is 3.054 N-m
The maximum torque develped is 9.4 N-m

Problem 14, Page 226

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

Total torque = 40 N-m

Problem 15, Page 226

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

The external resistance to be connected per phase is 2.13 ohms
Total rotor copper loss = 7291 W
Ouput power of motor = 8333 W

Problem 16, Page 227

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

Slip = 0.068

Problem 17, Page 227

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

Rotor current per phase = 28.3 A

Problem 18, Page 227

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

Maximum torque at starting is 53 %
Maximum torque when slip is 3% is 10 %

Problem 20, Page 227

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

The ratio of starting torque to full load torque for the given cases are as below:
Part a: 0.50
Part a: 0.17
Part a: 0.28