Chapter 5 : Induction Machines

Example 5.1 Page No : 198

In [1]:
import math 

# GIVEN DATA
# For Case (a)

S_a = 30;               # Total number of Slots
m_a = 3;                # Total number of Poles
p_a = 2;                # Total number of Phases

# For Case (b)

S_b = 60;               # Total number of Slots
m_b = 3;                # Total number of Poles
p_b = 4;                # Total number of Phases

# For Case (c)

S_c = 24;              # Total number of Slots
m_c = 3;               # Total number of Poles
p_c = 4;               # Total number of Phases

# For Case (d)

S_d = 12;              # Total number of Slots
m_d = 3;               # Total number of Poles
p_d = 2;               # Total number of Phases


# CALCULATIONS
# For Case (a)

spp_a = S_a/(p_a*m_a);              # Slot per poles per phase 

# For Case (b)

spp_b = S_b/(p_b*m_b);              # Slot per poles per phase 

# For Case (c)

spp_c = S_c/(p_c*m_c);              # Slot per poles per phase 

# For Case (d)

spp_d = S_d/(p_d*m_d);              # Slot per poles per phase 


# DISPLAY RESULTS

print ("EXAMPLE : 5.1 : SOLUTION :-") ;
print " For case a) Slot per poles per phase, spp = %.f   "%(spp_a);
print " For case b) Slot per poles per phase, spp = %.f   "%(spp_b);
print " For case c) Slot per poles per phase, spp = %.f   "%(spp_c);
print " For case d) Slot per poles per phase, spp = %.f   "%(spp_d);
EXAMPLE : 5.1 : SOLUTION :-
 For case a) Slot per poles per phase, spp = 5   
 For case b) Slot per poles per phase, spp = 5   
 For case c) Slot per poles per phase, spp = 2   
 For case d) Slot per poles per phase, spp = 2   

Example 5.2 Page No : 201

In [2]:
import math 


# GIVEN DATA
# For Case (a)

S_a = 54;               # Total number of Slots
m_a = 3;                # Total number of Poles
p_a = 8;                # Total number of Phases

# For Case (b)

S_b = 32;               # Total number of Slots
m_b = 3;                # Total number of Poles
p_b = 4;                # Total number of Phases

# For Case (c)

S_c = 30;              # Total number of Slots
m_c = 3;               # Total number of Poles
p_c = 4;               # Total number of Phases


# CALCULATIONS
# For Case (a)

spp_a = S_a/(p_a*m_a);                # Slot per poles per phase 
l_a = 0 * spp_a;                      # Phase allociation Series
m_a = 1 * spp_a;                      # Phase allociation Series
n_a = 2 * spp_a;                      # Phase allociation Series
o_a = 3 * spp_a;                      # Phase allociation Series
p_a = 4 * spp_a;                      # Phase allociation Series
d_a = 0;                              # d_a = l_a (Rounding off)
e_a = 2;                              # e_a = m_a (Rounding off)
f_a = 4;                              # f_a = n_a (Rounding off)
g_a = 6;                              # g_a = o_a (Rounding off)
h_a = 9;                              # h_a = p_a (Rounding off)
R_a = e_a - d_a;                      # Phase allociation
Y_a = f_a - e_a;                      # Phase allociation
B_a = g_a - f_a;                      # Phase allociation
R1_a = h_a - g_a;                     # Phase allociation

# For Case (b)

spp_b = S_b/(p_b*m_b);                # Slot per poles per phase 
l_b = 0 * spp_b;                      # Phase allociation Series
m_b = 1 * spp_b;                      # Phase allociation Series
n_b = 2 * spp_b;                      # Phase allociation Series
o_b = 3 * spp_b;                      # Phase allociation Series
d_b = 0;                              # d_b = l_b (Rounding off)
e_b = 2;                              # e_b = m_b (Rounding off)
f_b = 5;                              # f_b = n_b (Rounding off)
g_b = 8;                              # g_b = o_b (Rounding off)
R_b = e_b - d_b;                      # Phase allociation
Y_b = f_b - e_b;                      # Phase allociation
B_b = g_b - f_b;                      # Phase allociation

# For Case (c)

spp_c = S_c/(p_c*m_c);                # Slot per poles per phase 
l_c = 0 * spp_c;                      # Phase allociation Series
m_c = 1 * spp_c;                      # Phase allociation Series
n_c = 2 * spp_c;                      # Phase allociation Series
d_c = 0;                              # d_b = l_b (Rounding off)
e_c = 2;                              # e_b = m_b (Rounding off)
f_c = 5;                              # f_b = n_b (Rounding off)
R_c = e_c - d_c;                      # Phase allociation
Y_c = f_c - e_c;                      # Phase allociation

# DISPLAY RESULTS

print ("EXAMPLE : 5.2 : SOLUTION :-") ;
print "  For Case a) Slot per poles per phase  , spp = %.3f  "%(spp_a);
print "               Phase allociation series is %.f, %.f, %.f, %.f, %.f, %.f, %.f, %.f, %.f, slots are allocated respectively\
 to R, Y, B, R, Y, B, R, Y, B....... phase in Sequence "%(R_a,Y_a,B_a,R1_a,R_a,Y_a,B_a,R1_a,R_a);
print "               By seeing Sequence its Slot per pole per phase is an Integer and such, balanced winding may be possible "
print "  For Case b) Slot per poles per phase  , spp = %.3f  "%(spp_b);
print "               Phase allociation series is %.f, %.f, %.f "%(R_b,Y_b,B_b);
print "               By seeing Sequence its Slot per pole per phase are not Integer therefore R-phase will have 8 slots\
 whereas Y-phase and B-phase will have 12 slots ";
print "  For Case c) Slot per poles per phase  , spp = %.1f  "%(spp_c);
print "               Phase allociation series is %.f, %.f, %.f, %.f, %.f, %.f, %.f, %.f, %.f, %.f, %.f, %.f slots\
 are allocated respectively to R, Y, B, R, Y, B, R, Y, B, R, Y, B....... phase in Sequence "%(R_c,Y_c,R_c,Y_c,R_c,Y_c,R_c,Y_c,R_c,Y_c,R_c,Y_c);
print "               By seeing Sequence its Slot per pole per phase is an Integer and such, balanced winding may be possible "
EXAMPLE : 5.2 : SOLUTION :-
  For Case a) Slot per poles per phase  , spp = 2.000  
               Phase allociation series is 2, 2, 2, 3, 2, 2, 2, 3, 2, slots are allocated respectively to R, Y, B, R, Y, B, R, Y, B....... phase in Sequence 
               By seeing Sequence its Slot per pole per phase is an Integer and such, balanced winding may be possible 
  For Case b) Slot per poles per phase  , spp = 2.000  
               Phase allociation series is 2, 3, 3 
               By seeing Sequence its Slot per pole per phase are not Integer therefore R-phase will have 8 slots whereas Y-phase and B-phase will have 12 slots 
  For Case c) Slot per poles per phase  , spp = 2.0  
               Phase allociation series is 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3 slots are allocated respectively to R, Y, B, R, Y, B, R, Y, B, R, Y, B....... phase in Sequence 
               By seeing Sequence its Slot per pole per phase is an Integer and such, balanced winding may be possible 

Example 5.3 Page No : 205

In [3]:
import math 

# GIVEN DATA

s = 24.;                     # Total number of the pole
p = 4.;                      # Total number of the poles in the Alternator


# CALCULATIONS 
# For Case (a) Short pitching by one Slots

spp = s/p;                     # Slot per pole 
E_a = ((180*2)/24.)*(4/2);      # Slot angle in Electrical
kp_a = math.cos(math.radians(E_a/2));            # Pitch Factor
kp5_a = math.cos(math.radians((5*E_a)/2));       # Pitch Factor
kp7_a = math.cos(math.radians((7*E_a)/2));       # Pitch Factor

# For Case(b) Short pitching by two Slots

E_b = 2*((180*2)/24)*(4/2);    # Slot angle in Electrical
kp_b = math.cos(math.radians(E_b/2));            # Pitch Factor
kp5_b = math.cos(math.radians((5*E_b)/2))        # Pitch Factor
kp7_b = math.cos(math.radians((7*E_b)/2));       # Pitch Factor


# DISPLAY RESULTS

print ("EXAMPLE : 5.3 : SOLUTION :-") ;
print " For Case a) Short pitching by one Slots:- Pitch Facor , kp = %.4f  "%(kp_a);
print "                                                          kp5 = %.4f  "%(kp5_a);
print "                                                          kp7 = %.4f  "%(kp7_a);
print " For Case a) Short pitching by Two Slots:- Pitch Facor , kp = %.4f  "%(kp_b);
print "                                                          kp5 = %.4f  "%(kp5_b);
print "                                                          kp7 = %.4f  "%(kp7_b);
EXAMPLE : 5.3 : SOLUTION :-
 For Case a) Short pitching by one Slots:- Pitch Facor , kp = 0.9659  
                                                          kp5 = 0.2588  
                                                          kp7 = -0.2588  
 For Case a) Short pitching by Two Slots:- Pitch Facor , kp = 0.8660  
                                                          kp5 = -0.8660  
                                                          kp7 = -0.8660  

Example 5.4 Page No : 208

In [4]:
import math 

# GIVEN DATA

s = 60.;                     # Total number of Slot
m = 3.;                      # Total number of Phase
p = 4.;                      # Total number of Pole


# CALCULATIONS

M = s/(m*p);                                 # Slot per pole per Phase
sigma = 180/m;                               # Phase Spread in angle (deg)
Ka = math.sin(math.radians((M*sigma)/2))/(M*math.sin(math.radians(sigma/2)));    # Distribution Factor


# DISPLAY RESULTS

print ("EXAMPLE : 5.4 : SOLUTION :-");
print " a) Distribution Factor, Ka = %.1f "%(Ka)
EXAMPLE : 5.4 : SOLUTION :-
 a) Distribution Factor, Ka = 0.2 

Example 5.7 Page No : 210

In [5]:
import math 

# GIVEN DATA

f = 50.;                     # Frequency of the 2-pole Induction Motor
p = 2.;                      # Total Number of Poles


# CALCULATIONS

Ns = (120*f)/p;                 # Synchronous Speed in RPM
Ns5 = -(120*f)/(5*p);           # Synchronous Speed of 5th order space harmonic in RPM
N5 = -(120*5*f)/p;              # Synchronous Speed of 5th order time harmonic in RPM
Ns7 = (120*f)/(7*p);           # Synchronous Speed of 7th order space harmonic in RPM
N7 = (120*7*f)/p;              # Synchronous Speed of 7th order time harmonic in RPM


# DISPLAY RESULTS

print ("EXAMPLE : 5.7 : SOLUTION :-");
print " a.1) Synchronous Speed of 5th order space harmonic, Ns5 = %.f RPM "%(Ns5)
print " a.2) Synchronous Speed of 5th order time harmonic, N5 = %.f RPM "%(N5)
print " b.1) Synchronous Speed of 7th order space harmonic, Ns7 = %.2f RPM "%(Ns7)
print " b.2) Synchronous Speed of 7th order time harmonic,  N7 = %.f RPM "%(N7)
EXAMPLE : 5.7 : SOLUTION :-
 a.1) Synchronous Speed of 5th order space harmonic, Ns5 = -600 RPM 
 a.2) Synchronous Speed of 5th order time harmonic, N5 = -15000 RPM 
 b.1) Synchronous Speed of 7th order space harmonic, Ns7 = 428.57 RPM 
 b.2) Synchronous Speed of 7th order time harmonic,  N7 = 21000 RPM 

Example 5.8 Page No : 215

In [6]:
import math 

# GIVEN DATA

p_a = 6;                        # Total number of Poles in the Alternator
p_m = 4;                        # Total number of Poles of Induction Motor
N_a = 900;                      # Running Speed of the Alternator in RPM
N_m = 1250;                     # Running Speed of the Induction Motor in RPM
m = 3;                          # Total Number of phase in Induction Motor


# CALCULATIONS

f = (N_a*p_a)/120;                  # Frequency of the 6-pole Alternator running at 900 RPM in Hertz
Ns = (120*f)/p_m;                   # Synchronous Speed of 4-pole Induction Motor in RPM
s = (Ns-N_m)/Ns;                    # Slip 
fr = s*f;                           # Frequency of the Rotor Current in Hertz 


# DISPLAY RESULTS

print ("EXAMPLE : 5.8 : SOLUTION :-");
print " a) Frequency of the Rotor Current , fr = %.2f Hz "%(fr)
EXAMPLE : 5.8 : SOLUTION :-
 a) Frequency of the Rotor Current , fr = 0.00 Hz 

Example 5.9 Page No : 220

In [8]:
import math 

# GIVEN DATA

p = 2;                        # Total number of Poles of Induction Motor
f = 50;                       # Frequency in Hertz
Nr = 2800;                    # Running Speed of the Induction Motor in RPM
m = 3;                        # Total Number of phase in Induction Motor
V = 400;                      # Operating Voltage of Induction Motor in Volts


# CALCULATIONS

Ns = (120.*f)/p;                   # Synchronous Speed in RPM
s = 100*((Ns-Nr)/Ns);             # Slip in Percentage
fr = (s/100)*f;                   # Frequency of the Rotor Current in Hertz 


# DISPLAY RESULTS

print ("EXAMPLE : 5.9 : SOLUTION :-");
print " a) Slip, s = %.2f percent "%(s);
print " b) Frequency of the Rotor Current, fr = %.2f Hz "%(fr)
EXAMPLE : 5.9 : SOLUTION :-
 a) Slip, s = 6.67 percent 
 b) Frequency of the Rotor Current, fr = 3.33 Hz 

Example 5.10 Page No : 223

In [9]:
# GIVEN DATA
m = 3;                          # Total Number of phase in Induction Motor
p = 4;                          # Total number of Poles in Induction Motor
f = 50;                         # Frequency in Hertz
s = 0.03;                       # Slip


# CALCULATIONS

Ns = (120*f)/p;                     # Synchronous Speed in RPM
Nr = (1-s)*Ns;                      # Rotor Speed in RPM


# DISPLAY RESULTS

print ("EXAMPLE : 5.10 : SOLUTION :-");
print " a) Rotor Speed , Nr = %.f RPM "%(Nr)
EXAMPLE : 5.10 : SOLUTION :-
 a) Rotor Speed , Nr = 1455 RPM 

Example 5.11 Page No : 225

In [10]:
# GIVEN DATA


m = 3;                          # Total Number of phase in Induction Motor
p = 6;                          # Total number of Poles of Induction Motor
f = 50;                         # Frequency in Hertz
s = 0.03;                       # Slip


# CALCULATIONS

Ns = (120*f)/p;                     # Synchronous Speed in RPM
Nr = (1-s)*Ns;                      # Rotor Speed in RPM
Nf = Ns - Nr;                       # Speed of Forward Rotating magnetic fields with respect to stator and rotor in RPM
Nb = Ns + Nr;                       # Speed of Backward Rotating magnetic fields with respect to stator and rotor in RPM


# DISPLAY RESULTS

print ("EXAMPLE : 5.11 : SOLUTION :-");
print " a) Speed of Forward Rotating magnetic fields with respect to stator and rotor is equal to + %.f RPM "%(Nf)
print " b) Speed of Backward Rotating magnetic fields with respect to stator and rotor is equal to + %.f RPM "%(Nb)
EXAMPLE : 5.11 : SOLUTION :-
 a) Speed of Forward Rotating magnetic fields with respect to stator and rotor is equal to + 30 RPM 
 b) Speed of Backward Rotating magnetic fields with respect to stator and rotor is equal to + 1970 RPM 

Example 5.12 Page No : 230

In [11]:
# GIVEN DATA

m = 3;                          # Total Number of phase in Induction Motor
p = 2;                          # Total number of Poles of Induction Motor
f = 50;                         # Frequency in Hertz
s = 0.05;                       # Slip


# CALCULATIONS

Ns = (120*f)/p;                     # Synchronous Speed in RPM
Nr = (1-s)*Ns;                      # Rotor Speed in RPM
Nf = s*Ns;                          # Speed of Forward Rotating magnetic fields with respect to stator and rotor in RPM
Nb = (p-s)*Ns;                      # Speed of Backward Rotating magnetic fields with respect to stator and rotor in RPM
fr = (p-s)*f;                       # Backward rotating magnetic field induces a current of frequency in Hertz


# DISPLAY RESULTS

print ("EXAMPLE : 5.12 : SOLUTION :-");
print " a) Speed of Forward Rotating magnetic fields with respect to stator and rotor is equal to + %.f RPM "%(Nf)
print " b) Speed of Backward Rotating magnetic fields with respect to stator and rotor is equal to + %.f RPM "%(Nb)
EXAMPLE : 5.12 : SOLUTION :-
 a) Speed of Forward Rotating magnetic fields with respect to stator and rotor is equal to + 150 RPM 
 b) Speed of Backward Rotating magnetic fields with respect to stator and rotor is equal to + 5850 RPM 

Example 5.13 Page No : 234

In [12]:
# GIVEN DATA
m = 3;                          # Total Number of phase in Induction Motor
p = 4;                          # Total number of Poles of Induction Motor
f = 50;                         # Frequency in Hertz
s = 0.05;                       # Slip


# CALCULATIONS

Ns = (120*f)/p;                     # Synchronous Speed in RPM
fr = s*f;                           # Rotor-induced Frequency of forward field in Hertz
Nfr = s*Ns;                         # Speed of Forward Rotating magnetic fields with respect to rotor surface in RPM
f2r = s*f;                           # Rotor-induced Frequency of Backward field in Hertz
Nbr = -(s*Ns);                       # Speed of Backward Rotating magnetic fields with respect to rotor surface in RPM
Nr = (1-s)*Ns;                       # Rotor running in Forward direction in RPM
Nfs = Nr+(s*Ns);                     # Speed of Forward Rotating magnetic fields with respect to stator surface in RPM
Nbs = Nr-(s*Ns);                     # Speed of Backward Rotating magnetic fields with respect to stator surface in RPM
Nbs_new = -(0.5*Ns)+(1-0.5)*Nr;      # Speed of Backward Rotating magnetic fields with respect to stator for 50% of slip in RPM


# DISPLAY RESULTS

print ("EXAMPLE : 5.13 : SOLUTION :-");
print " a.1) Speed of Forward Rotating magnetic fields with respect to rotor surface is equal to + %.f RPM "%(Nfr)
print " a.2) Speed of Backward Rotating magnetic fields with respect to rotor surface is equal to + %.f RPM "%(Nbr)
print " b.1) Speed of Forward Rotating magnetic fields with respect to stator surface is equal to + %.f RPM "%(Nfs)
print " b.2) Speed of Backward Rotating magnetic fields with respect to stator surface is equal to + %.f RPM "%(Nbs)
print "  c)  Speed of Backward Rotating magnetic fields with respect to stator for 50 percenatge slip is equal to  %.1f RPM "%(Nbs_new)
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :- a) Speed of Backward Rotating magnetic fields with respect to stator for 50 percenatge slip is equal to 0 RPM instead of %.1f RPM   "%(Nbs_new);
EXAMPLE : 5.13 : SOLUTION :-
 a.1) Speed of Forward Rotating magnetic fields with respect to rotor surface is equal to + 75 RPM 
 a.2) Speed of Backward Rotating magnetic fields with respect to rotor surface is equal to + -75 RPM 
 b.1) Speed of Forward Rotating magnetic fields with respect to stator surface is equal to + 1500 RPM 
 b.2) Speed of Backward Rotating magnetic fields with respect to stator surface is equal to + 1350 RPM 
  c)  Speed of Backward Rotating magnetic fields with respect to stator for 50 percenatge slip is equal to  -37.5 RPM 
    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
      WRONGLY PRINTED ANSWERS ARE :- a) Speed of Backward Rotating magnetic fields with respect to stator for 50 percenatge slip is equal to 0 RPM instead of -37.5 RPM   

Example 5.14 Page No : 235

In [13]:
# GIVEN DATA

f = 50;             # Stator Frequency of Inductor Motor in Hertz
fr = 10;            # Rotor Frequency of Inductor Motor in Hertz
p = 2;              # Number of poles


# CALCULATIONS

Ns = (120*f)/p;      # Synchronous Speed of Induction Motor in RPM
s = fr/f;            # Slip of the Induction Motor
Nr = (1-s)*Ns;       # Rotor Speed of the Induction Motor


# DISPLAY RESULTS

print ("EXAMPLE : 5.14 : SOLUTION :-");
print " a) Rotor Speed of Induction Motor, Nr = %.f RPM "%(Nr)
EXAMPLE : 5.14 : SOLUTION :-
 a) Rotor Speed of Induction Motor, Nr = 3000 RPM 

Example 5.15 Page No : 237

In [14]:
import math

# GIVEN DATA

print " EXAMPLE : 5.15 :           Given Data No-load test : 440V, 30A, 4.5KW ";
print "          Blocked rotor test :      90V%(50Hz, 120A, 16KW ";
m = 3;                          # Total Number of phase in Induction Motor
p = 6;                          # Total number of Poles of Induction Motor
V = 440;                        # Operating voltage of the Induction motor in Volts
out_hp = 100;                   # Output of the Induction motor in Horse-Power
R = 0.15;                       # Average dc resistance in Ohms
Wsc = 16000;                    # Power at Blocked Rotor test in Watts
Vsc = 90;                       # Voltage at Blocked Rotor test in Volts
Isc = 120;                      # Current at Blocked Rotor test in Amphere
W0 = 4500;                      # Power at No-load test in Watts
V0 = 440;                       # Voltage at No-load test in Volts
I0 = 30;                        # Current at No-load test in Amphere 
s = 0.05;                       # Slip
f = 50;                         # Frequency in Hertz


# CALCULATIONS

R1 = R/2;                                       # DC winding resistance per phase in Ohms
Rac = Wsc/(3*Isc**2);                            # AC resistance referred to stator from locked rotor test at supply frequency in Ohms
R_2 = Rac - R1;                                 # Per phase Rotor resistance to Stator in Ohms
Zsc = Vsc/(math.sqrt(3)*Isc);                        # Per phase Impedance from locked rotor test in Ohms
Xs = math.sqrt((Zsc**2)-(Rac**2));                     # Per phase leakage reactance referred to stator in Ohms
theta_0 = math.acos(math.radians(W0/(V0*I0*math.sqrt(3))));            # No-load power factor angle in degree
Im = I0*math.degrees(math.sin(math.radians(theta_0)));                          # Reactive component of no-load current in Amphere
Xm = V0/(Im*math.sqrt(3));                           # Magnetizing reactance in Ohms
Pc = W0 - 3*I0**2*R1;                            # Total Core loss in Watts
Rc = (V0/math.sqrt(3))**2*(3/Pc);                     # Per phase core loss resistance in Watts
Vph = V0/math.sqrt(3);                               # Per phase Voltage in Volts
Ic = Vph/Rc;                                    # Core loss current in Amphere
I_m = Vph/(1j * Xm);                            # Magnetizing Current in Amphere
I_o = Ic + I_m;                                 # No-load current in Amphere
I_2 = Vph/(R1+(R_2/s)+(1j*Xs));                 # Current in Amphere
I1 = I_o + I_2;                                 # Input Current in Amphere
Pf = math.cos(math.radians(math.degrees(math.atan(I1.imag/I1.real))));            # Power factor 
P1 = (3*(abs(I_2)**2*R_2)/s)/1000.;               # 3-phase air gap power or Rotor intake Power in Kilo-Watts
Po = P1*(1-s);                                  # Output Power in Kilo-Watts
Ws = 2*math.pi*((120*f/p)*(1./60.));                  # Angular Roatation in Radians per Seconds
T = P1*1000/Ws;                                 # Torque in Newton-Meter


# DISPLAY RESULTS

print ("EXAMPLE : 5.15 : SOLUTION :-") ;
print " a.1)  DC winding resistance per phase, R1 = %.3f Ohms "%(R1)
print " a.2)  AC resistance referred to stator from locked rotor test at supply frequency  = %.4f Ohms "%(Rac)
print " a.3)  Per phase Rotor resistance to Stator, R2 = %.4f Ohms "%(R_2)
print " a.4)  Per phase Impedance from locked rotor test, Zsc = %.3f Ohms "%(Zsc)
print " a.5)  Per phase leakage reactance referred to stator, Xs = %.4f Ohms "%(Xs)
print " a.6)  No-load power factor angle, theta_O = %.2f Degree "%(theta_0)
print " a.7)  Reactive component of no-load current, Im = %.1f A "%(Im)
print " a.8)  Magnetizing reactance, Xm = %.2f Ohms "%(Xm)
print " a.9)  Total Core loss, Pc = %.1f W "%(Pc)
print " a.10) Per phase core loss resistance, Pc = %.f Ohms "%(Rc)
print " a.11) Per phase Voltage, Vph = %.f V "%(Vph)
print " a.12) Core loss current, Ic = %.2f < %.f A "%(abs(Ic),math.degrees(math.atan2(Ic.imag,Ic.real)))
print " a.13) Magnetizing Current, Im = %.1f < %.f A "%(abs(I_m),math.degrees(math.atan2(I_m.imag,I_m.real)))
print " a.14) No-load current, I0 = %.2f < %.2f A "%(abs(I_o),math.degrees(math.atan2(I_o.imag,I_o.real)))
print " a.15) Current, I2 = %.2f < %.2f A "%(abs(I_2),math.degrees(math.atan2(I_2.imag,I_2.real)))
print "   b)  Input current, I1 = %.2f < %.2f A "%(abs(I1),math.degrees(math.atan2(I1.imag,I1.real)))
print "   c)  Power Factor, Pf = %.4f Lagging "%(Pf)
print "   d)  Output Power, P0 = %.1f kW "%(Po)
print "   e)  Torque, T = %.2f NM "%(T)
 EXAMPLE : 5.15 :           Given Data No-load test : 440V, 30A, 4.5KW 
          Blocked rotor test :      90V%(50Hz, 120A, 16KW 
EXAMPLE : 5.15 : SOLUTION :-
 a.1)  DC winding resistance per phase, R1 = 0.075 Ohms 
 a.2)  AC resistance referred to stator from locked rotor test at supply frequency  = 0.0000 Ohms 
 a.3)  Per phase Rotor resistance to Stator, R2 = -0.0750 Ohms 
 a.4)  Per phase Impedance from locked rotor test, Zsc = 0.433 Ohms 
 a.5)  Per phase leakage reactance referred to stator, Xs = 0.4330 Ohms 
 a.6)  No-load power factor angle, theta_O = 1.57 Degree 
 a.7)  Reactive component of no-load current, Im = 47.0 A 
 a.8)  Magnetizing reactance, Xm = 5.40 Ohms 
 a.9)  Total Core loss, Pc = 4297.5 W 
 a.10) Per phase core loss resistance, Pc = 45 Ohms 
 a.11) Per phase Voltage, Vph = 254 V 
 a.12) Core loss current, Ic = 5.64 < 0 A 
 a.13) Magnetizing Current, Im = 47.0 < -90 A 
 a.14) No-load current, I0 = 47.35 < -83.16 A 
 a.15) Current, I2 = 170.57 < -163.10 A 
   b)  Input current, I1 = 184.82 < -148.49 A 
   c)  Power Factor, Pf = 0.8525 Lagging 
   d)  Output Power, P0 = -124.4 kW 
   e)  Torque, T = -1250.21 NM 

Example 5.17 Page No : 239

In [15]:
import math 


# GIVEN DATA

print " EXAMPLE : 5.17 :           Given Data No-load test : 440V%( 3.0A, 500KW, 50Hz ";
print "          Blocked rotor test at rated frequency : 110V%( 18A, 2500W, 50Hz ";
print "          DC test on Stator per phase : 10V, 15A ";
m = 3;                          # Total Number of phase in Induction Motor
p = 4;                          # Total number of Poles of Induction Motor
f = 50;                         # Frequency in Hertz
V = 440;                        # Operating Voltage of the Inductuon Motor
out_hp = 20;                    # Motor Power Rating in Horse-Power 
Vdc = 10;                       # DC Voltage in Volts
Idc = 15;                       # DC Current in Amphere
Wsc = 2500;                     # Power at Blocked Rotor test rated frequency in Watts
Wsc_red = 2050;                 # Power at Blocked Rotor test reduced frequency in Watts
Vsc = 110;                      # Voltage at Blocked Rotor test rated frequency in Volts
Isc = 18;                       # Current at Blocked Rotor test rated frequency in Amphere
Wo = 500;                       # Power at No-load test in Watts
Vo = 440;                       # Voltage at No-load test in Volts
Io = 4.0;                       # Current at No-load test in Amphere
fsc = 50;                       # Rated Frequency at blocked rotor test in Hertz
fo = 50;                        # Rated Frequency at no-load test in Hertz
fsc1 = 15;                      # Reduced Frequency at blocked rotor in Hertz
Pfw = 200;                      # Friction and Windage loss in Watts


# CALCULATIONS

R1dc = Vdc/Idc;                         # DC winding resistance per phase in Ohms
Rac = Wsc/(3*Isc**2);                    # AC resistance from Locked rotor test at supply frequency
Rac_red = Wsc_red/(3*Isc**2);            # AC resistance from Locked rotor test at reduced frequency
R1ac = (Rac/Rac_red)*R1dc;              # Corrected Value of AC stator winding resistance in Ohms
R2dc = Rac_red - R1dc;                  # Second rotor parameter, rotor resistance referred to stator is at low frequency in Ohms
Zsc = Vsc/(math.sqrt(3)*Isc);                        # Per phase Impedance from locked rotor test at power frequency in Ohms
Xs = math.sqrt((Zsc**2)-(Rac**2));                     # Per phase leakage reactance referred to stator in Ohms
theta_0 = math.acos(math.radians(Wo/(Vo*Io*math.sqrt(3))));            # No-load power factor angle in degree
Im = Io*math.degrees(math.sin(math.radians(theta_0)));                          # Reactive component of no-load current in Amphere
Xm = Vo/(Im*math.sqrt(3));                           # Magnetizing reactance in Ohms
Pc = Wo - 3*Io**2*R1ac-Pfw;                      # Total Core loss in Watts
Rc = (Vo/math.sqrt(3))**2*(3/Pc);                     # Per phase core loss resistance in Watts


# DISPLAY RESULTS

print ("EXAMPLE : 5.17 : SOLUTION :-") ;
print " a)  Magnetizing reactance of Equivalent circuit, Xm = %.1f Ohms "%(Xm)
print " b) Per phase core loss resistance, Pc = %.f Ohms "%(Rc)
 EXAMPLE : 5.17 :           Given Data No-load test : 440V%( 3.0A, 500KW, 50Hz 
          Blocked rotor test at rated frequency : 110V%( 18A, 2500W, 50Hz 
          DC test on Stator per phase : 10V, 15A 
EXAMPLE : 5.17 : SOLUTION :-
 a)  Magnetizing reactance of Equivalent circuit, Xm = 40.5 Ohms 
 b) Per phase core loss resistance, Pc = 645 Ohms 

Example 5.18 Page No : 240

In [16]:
import math 

# GIVEN DATA
# From Previous problem data (Example 5.17)

R1ac = 0.8127;                  # Corrected Value of AC stator winding resistance in Ohms
R2dc = 1.4433;                  # Second rotor parameter, rotor resistance referred to stator is at low frequency in Ohms
Xs = 2.42;                     # Per phase leakage reactance referred to stator in Ohms
Xm = 64.4;                     # Magnetizing reactance in Ohms
Rc = 742;                      # Per phase core loss resistance in Watts
s = 0.035;                     # Slip
m = 3;                         # Total Number of phase in Induction Motor
p = 4;                         # Total number of Poles of Induction Motor
f = 50;                        # Frequency in Hertz
V = 440;                       # Operating Voltage of the Inductuon Motor
out_hp = 20;                   # Motor Power Rating in Horse-Power


# CALCULATIONS

Vph = V/math.sqrt(3);                                # Per phase Voltage in Volts
Ic = Vph/Rc;                                    # Core loss current in Amphere
I_m = Vph/(1j * Xm);                            # Magnetizing Current in Amphere
I_o = Ic + I_m;                                 # No-load current in Amphere
I_2 = Vph/(R1ac+(R2dc/s)+(1j*Xs));              # Current in Amphere
I1 = I_o + I_2;                                 # Input Current in Amphere
Pf = math.cos(math.radians(math.degrees(math.atan(I1.imag/I1.real))));            # Power factor 
P1 = 3*(abs(I_2)**2*R2dc)/s;                    # 3-phase air gap power or Rotor intake Power in Watts
Po = P1*(1-s);                                  # Output Power in Watts
Ws = 2*math.pi*((120*f/p)*(1./60));                  # Angular Roatation in Radians per Seconds
T = P1/Ws;                                      # Torque in Newton-Meter


# DISPLAY RESULTS

print ("EXAMPLE : 5.18 : SOLUTION :-");
print "   a)  Input current, I1 = %.2f < %.2f A "%(abs(I1),math.degrees(math.atan2(I1.imag,I1.real)))
print "   b)  Power Factor, Pf = %.3f Lagging "%(Pf)
print "   c)  Output Power, P0 = %.2f W "%(Po)
print "   d)  Torque, T = %.2f NM "%(T)
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :- a) T = 4340.82 Nm instead of %.2f Nm  "%(T);
print "       IN TEXT BOOK, CALCULATION OF TORQUE IS NOT DONE  ";
EXAMPLE : 5.18 : SOLUTION :-
   a)  Input current, I1 = 7.68 < -33.99 A 
   b)  Power Factor, Pf = 0.829 Lagging 
   c)  Output Power, P0 = 4342.67 W 
   d)  Torque, T = 28.65 NM 
    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
      WRONGLY PRINTED ANSWERS ARE :- a) T = 4340.82 Nm instead of 28.65 Nm  
       IN TEXT BOOK, CALCULATION OF TORQUE IS NOT DONE  

Example 5.19 Page No : 243

In [17]:
import math 
from numpy import imag,real


# GIVEN DATA

s = 0.05;                       # Slip
m = 3;                          # Total Number of phase in Induction Motor
p = 4;                          # Total number of Poles of Induction Motor
f = 50;                         # Frequency in Hertz
V = 440;                        # Operrating Voltage of the Inductuon Motor
R1 = 0.10;                      # Circuit Parameter in Ohms
R2 = 0.11;                      # Circuit Parameter in Ohms
X1 = 0.35;                      # Circuit Parameter in Ohms
X2 = 0.40;                      # Circuit Parameter in Ohms
pf = 0.2;                       # Power factor (Lagging)
Pr = 900;                       # Rotational Loss in Watts
Psc = 1000;                     # Stator core loss in Watts
I = 15;                         # Line current draws by the motor in Amphere


# CALCULATIONS

Vph = V/math.sqrt(3);                                    # Per phase Voltage in Volts
I_2 = Vph/(R1+(R2/s)+(1j*(X1+X2)));                 # Current in Amphere
Io = I * (-(1j * math.degrees(math.acos(pf)) * math.pi/180.))**2;         # No-load current in Amphere
I1 = Io + I_2;                                      # Input line Current in Amphere
PF = math.cos(math.radians(math.degrees(math.atan(I1.imag/I1.real))));                # Power factor
Ws = 2*math.pi*((120*f/p)*(1./60.));                      # Angular Roatation in Radians per Seconds
Pg = (3*(abs(I1)**2*R2))/s;                          # 3-phase air gap power or Rotor intake Power in Watts
T = Pg/Ws;                                          # Torque in Newton-Meter
Po = Pg*(1-s)-Pr;                                   # Output Power in Watts
Po_HP = Po/746;                                     # Output Power in Horse-Power
eta = (Po/(Po+Psc+Pr))*100.;                         # Efficiency in Percentage


# DISPLAY RESULTS

print ("EXAMPLE : 5.19 : SOLUTION :-");
print "   a)  Input line current, I1 = %.1f < %.2f A "%(abs(I1),math.degrees(math.atan2(I1.imag,I1.real)))
print "   b)  Power Factor, Pf = %.4f Lagging "%(PF)
print "   c)  Output Power, P0 = %.1f HP "%(Po_HP)
print "   d)  Torque, T = %.2f Nm "%(T)
print "   e)  Efficiency, eta = %.1f Percenatge "%(eta)
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :- a) I1 = 114.2<-24.68 A instead of %.1f<%.2f A  "%(abs(I1),math.degrees(math.atan2(I1.imag,I1.real)));
print "                                     b) T = 548.24 Nm instead of %.2f Nm  "%(T);
print "                                     c) Po = 108.4 HP instead of %.1f HP  "%(Po_HP);
EXAMPLE : 5.19 : SOLUTION :-
   a)  Input line current, I1 = 78.7 < -24.42 A 
   b)  Power Factor, Pf = 0.9105 Lagging 
   c)  Output Power, P0 = 50.9 HP 
   d)  Torque, T = 260.56 Nm 
   e)  Efficiency, eta = 95.2 Percenatge 
    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
      WRONGLY PRINTED ANSWERS ARE :- a) I1 = 114.2<-24.68 A instead of 78.7<-24.42 A  
                                     b) T = 548.24 Nm instead of 260.56 Nm  
                                     c) Po = 108.4 HP instead of 50.9 HP  

Example 5.20 Page No : 246

In [18]:
import math 
from numpy import imag,real

# GIVEN DATA

m = 3;                         # Total Number of phase in Induction Motor
p = 6;                         # Total number of Poles of Induction Motor
f = 50;                        # Frequency in Hertz
V = 440;                       # Operating Voltage of the Inductuon Motor
R1 = 0.25;                     # Circuit Parameter in Ohms
R2 = 0.25;                     # Circuit Parameter in Ohms
X1 = 0.75;                     # Circuit Parameter in Ohms
X2 = 0.75;                     # Circuit Parameter in Ohms
Xm = 1000;                     # Circuit Parameters in Ohms
Rc = 100;                      # Circuit Parameters in Watts
s = 0.025;                     # Slip
Pr = 450;                      # Rotational Loss in Watts
Psc = 800;                     # Stator core loss in Watts


# CALCULATIONS

Vph = V/math.sqrt(3);                                    # Per phase Voltage in Volts
I_2 = Vph/(R1+(R2/s)+(1j*(X1+X2)));                 # Current in Amphere
Ic = Vph/Rc;                                    # Core loss current in Amphere
I_m = Vph/(1j * Xm);                            # Magnetizing Current in Amphere
I_o = Ic + I_m;                                 # No-load current in Amphere
I1 = I_o + I_2;                                 # Input Current in Amphere
Pf = math.cos(math.radians(math.degrees(math.atan(I1.imag/I1.real))));            # Power factor 
Ns = (120*f)/p;                                 # Synronous Speed in RPM
Pg = 3*(abs(I_2)**2*R2)/s;                       # 3-phase air gap power or Rotor intake Power in Watts
Pm = Pg*(1-s);                                  # Output Power in Watts
Ws = 2*math.pi*Ns*(1./60.);                           # Angular Roatation in Radians per Seconds
T = Pg/Ws;                                      # Torque in Newton-Meter
Po = Pm-Pr;                                     # Output Power in Watts
Po_HP = Po/746;                                 # Output Power in Horse-Power
eta = (Po/(Po+Psc+Pr))*100;                     # Efficiency in Percentage


# DISPLAY RESULTS

print ("EXAMPLE : 5.20 : SOLUTION :-");
print "   a)  Input line current, I1 = %.f < %.2f A "%(abs(I1),math.degrees(math.atan(I1.imag%(I1.real))))
print "   b)  Power Factor, Pf = %.4f Lagging "%(Pf)
print "   c)  Output Power, P0 = %.2f HP "%(Po_HP)
print "   d)  Torque, T = %.1f Nm "%(T)
print "   e)  Efficiency, eta = %.1f Percenatge "%(eta)
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :- a) I1 = 26.8-j3.584 {27<-7.62} A in instead of %.1f)+j%.3f) {%.f<%.2f} A  "%(I1.real,I1.imag,abs(I1),math.degrees(math.atan2(I1.imag,I1.real)));
print "                                     b) pf = 0.9885 Lagging instead of %.4f Lagging  "%(Pf);
EXAMPLE : 5.20 : SOLUTION :-
   a)  Input line current, I1 = 27 < 87.51 A 
   b)  Power Factor, Pf = 0.9901 Lagging 
   c)  Output Power, P0 = 22.98 HP 
   d)  Torque, T = 172.3 Nm 
   e)  Efficiency, eta = 93.2 Percenatge 
    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
      WRONGLY PRINTED ANSWERS ARE :- a) I1 = 26.8-j3.584 {27<-7.62} A in instead of 26.8)+j-3.805) {27<-8.08} A  
                                     b) pf = 0.9885 Lagging instead of 0.9901 Lagging  

Example 5.21 Page No : 251

In [19]:
import math 
from numpy import roots

# GIVEN DATA
m = 3;                          # Total Number of phase in Induction Motor
p = 4;                          # Total number of Poles of Induction Motor
s = 0.05;                       # Slip
f = 50;                         # Frequency in Hertz
Tm = 500;                       # Maximum Torque in Newton-Meter
Tst = 200;                      # Starting Torque in Newton-Meter
sst = 1.0;                      # Starting Slip


# CALCULATONS

p1 = [1, -5, 1]           # Slip at Maximum Torque (obtained from Equation  Tst = (2*Tm)/((sst/sm)+(sm+sst))
a = roots(p1);                  # Value of slip at Maximum Torque (obtained from Equation  Tst = (2*Tm)/((sst/sm)+(sm+sst)) 
sm = a[1]#(2,1);                    # Slip at Maximum Torque (obtained from Equation  Tst = (2*Tm)/((sst/sm)+(sm+sst)) { 1st root is 4.8 so its out of range because slip value is lies between 0-1 so its neglected and second root will be slip }
T = (2*Tm)/((s/sm)+(sm/s));     # Torque at 0.05 slip
Ns = (120*f)/p;                 # Synchronous Speed in RPM
Wr = (2*math.pi)*(1-s)*(Ns/60);     # Angular Velocity in Radians-per-Second
P = T * Wr;                     # Power Output in Watts
P_HP = P/746;                   # Power Output in Horse-Power


# DISPLAY RESULTS

print ("EXAMPLE : 5.21 : SOLUTION :-");
print " a) Torque at 0.05 slip, T = %.2f Nm "%(T)
print " b) Power Output at 0.05 slip, P = %.1f W = %.2f HP "%(P,P_HP)
EXAMPLE : 5.21 : SOLUTION :-
 a) Torque at 0.05 slip, T = 226.56 Nm 
 b) Power Output at 0.05 slip, P = 33808.8 W = 45.32 HP 

Example 5.22 Page No : 255

In [20]:
import math 

# GIVEN DATA
Wsc = 1000.;                    # Power at Blocked Rotor test in Watts
Vsc = 56.;                      # Voltage at Blocked Rotor test in Volts
Isc = 18.;                      # Current at Blocked Rotor test in Amphere
Woc = 52.;                      # Power at No-load test in Watts
Voc = 220.;                     # Voltage at No-load test in Volts
Ioc = 2.6;                     # Current at No-load test in Amphere 
m = 3.;                         # Total Number of phase in Induction Motor
p = 4.;                         # Total number of Poles of Induction Motor
V = 220.;                       # Operating voltage of the Induction motor in Volts
f = 50.;                        # Frequency in Hertz
s = 0.05;                      # Slip
R = 0.65;                      # Per phase stator resistance in Ohms


# CALCULATIONS

Vph = Voc/math.sqrt(3);                               # Per phase Voltage in Volts
Wo = Woc/m;                                      # Per phase No-load loss in Watts
theta_0 = math.acos(math.radians(Wo/(Voc*Ioc*math.sqrt(3))));           # No-load power factor angle in degree
VSC = Vsc/math.sqrt(3);                               # Per phase locked rotor Voltage in Volts
WSC = Wsc/m;                                     # Per phase locked rotor loss in Watts
theta_sc = math.acos(math.radians(WSC/(VSC*Isc)));                 # No-load power factor angle in degree
ISC = Isc*(Voc/Vsc);                             # locked rotor current at full Voltage in Amphere
Re = WSC/Isc**2;                                  # resistance in Ohms
R1 = R*1.1;                                      # Per phase AC stator resistance in Ohms
R_2 = Re - R1;                                   # Per phase rotor resistance in Ohms
Zsc = VSC/Isc;                                   # Per phase impedance in Ohms
Xs = math.sqrt((Zsc**2)-(Re**2));                       # Leakage reactance in Ohms
I_2 = (Voc/math.sqrt(3))/math.sqrt((R1+(R_2/s))**2+(Xs**2)); # Current in Amphere
pf = math.cos(math.radians(math.degrees(math.atan(Xs/(R1+(R_2/s))))));               # Power Factor
Ws = 2*math.pi*((120*f/p)*(1./60.));                   # Rotational Speed in Radians per Seconds
Pg = (3*(abs(I_2)**2*R_2))/s;                     # 3-phase air gap power or Rotor intake Power in Watts
T = Pg/Ws;                                       # Torque in Newton-Meter
# CALCULATIONS OR DATA OBTAINED FROM CIRCLE DIAGRAM FIGURE 5.35 and PAGE NO:-303
OA = 2.60;                                       # Correspounding Current in Amphere at 87' from Y-axis (from Circle diagram)
OE = 70.70;                                      # Correspounding Current in Amphere at 55' from Y-axis (from Circle diagram)
OP = 17.77;                                      # Current in Amphere (from Circle diagram)
OV = Voc/math.sqrt(3);                                # Phase Voltage in No-load test or value obatined from circle diagram in Volts
PK = 11.6;                                       # Correspounding Value from Circle diagram
JK = 0.8;                                        # Correspounding Value from Circle diagram
PJ = 10.8;                                       # Correspounding Value from Circle diagram
PM = 11.6;                                       # Correspounding Value from Circle diagram
Pir = 3*OV*PK;                                   # Total Rotor intake in Watts
Plr = 3*OV*JK;                                   # Total Rotor loss in Watts
Po = 3*OV*PJ;                                    # Total Mechanical power output in Watts
T_c = (3*OV*PK)/Ws;                              # Total Torque in Newton-Meter
s_c = JK/PK;                                     # Slip obtained from Circle diagram
s_pc = 100*s_c;                                  # Slip in percentage
eta = 100*(PJ/PM);                               # Eifficiency in Percentage


# DISPLAY RESULTS

print ("EXAMPLE : 5.22 : SOLUTION :-");
print "   a)  Input line current, I2 = %.2f A "%(I_2)
print "   b)  Power Factor, Pf = %.3f "%(pf)
print "   c)  Torque, T = %.2f Nm "%(T)
print "  Verification Results from Circle Diagram :-";
print "   a)  Efficency, eta = %.2f Percent "%(eta)
print "   b)  slip, s = %.3f = %.f percent "%(s_c,s_pc)
print "   c)  Torque, T = %.2f Nm "%(T_c)
EXAMPLE : 5.22 : SOLUTION :-
   a)  Input line current, I2 = 17.78 A 
   b)  Power Factor, Pf = 0.979 
   c)  Torque, T = 37.89 Nm 
  Verification Results from Circle Diagram :-
   a)  Efficency, eta = 93.10 Percent 
   b)  slip, s = 0.069 = 7 percent 
   c)  Torque, T = 28.14 Nm 

Example 5.23 Page No : 259

In [21]:
import math 
from numpy import real,imag


# GIVEN DATA

R1 = 0.2;                      # Circuit Parameter in Ohms
R2 = 0.4;                      # Circuit Parameter in Ohms
X1 = 1.0;                      # Circuit Parameter in Ohms
X2 = 1.5;                      # Circuit Parameter in Ohms
m = 3;                          # Total Number of phase in Induction Motor
p = 2;                          # Total number of Poles of Induction Motor
f = 50;                         # Frequency in Hertz
V = 440;                        # Operating Voltage of the Inductuon Motor


# CALCULATIONS

Ws = 2*math.pi*f;                               # Synchronous angular speed in Radians per second
Z = (R1+R2)+((1j)*(X1+X2));                 # At slip s=1, the impedance seen from the terminals in Ohms
s = 1;                                      # Slip

# For Case(a) Winding is connected in star

Isy_a = V/(abs(Z)*math.sqrt(3));                   # Current in Amphere
Tsy_a = (3*Isy_a**2*R2)/(s*Ws);                # Torque in Newton-Meter

# Winding is connected in delta

Isd_a = (V*math.sqrt(3))/abs(Z);                   # Current in Amphere
Tsd_a = (3*(Isd_a/math.sqrt(3))**2*R2)/(s*Ws);      # Torque in Newton-Meter
I_R = Isd_a/Isy_a;                            # Ratio of the line current
T_R = Tsd_a/Tsy_a;                            # Ratio of the Torque
 
# For Case(b) Machine is started umath.sing auto-transfromer and voltage is 50% reduced

Isy_b = (0.5*V)/(abs(Z)*math.sqrt(3));                 # Current in Amphere when Winding is connected star
Tsy_b = (3*Isy_b**2*R2)/(s*Ws);                    # Torque in Newton-Meter when Winding is connected star
Isd_b = (0.5*V*math.sqrt(3))/abs(Z);                   # Current in Amphere when Winding is connected delta
Tsd_b = (3*(Isd_b/math.sqrt(3))**2*R2)/(s*Ws);          # Torque in Newton-Meter when Winding is connected delta

# For Case(c) Both Voltage and Frequency are reduced to 50%

f_new = (10./100)*f;                               # New Frequency
Ws_c = 2*math.pi*f_new;                               # Synchronous angular speed in Radians per second
Z_c = ((R1+R2)+((1j)*(X1+X2))*(f_new/f));         # At slip s=1, the impedance seen from the terminals in Ohms
Isy_c = (0.1*V)/(abs(Z_c)*math.sqrt(3));               # Current in Amphere when Winding is connected star
Tsy_c = (3*Isy_c**2*R2)/(s*Ws_c);                  # Torque in Newton-Meter when Winding is connected star
Isd_c = (0.1*V*math.sqrt(3))/abs(Z_c);                 # Current in Amphere when Winding is connected delta
Tsd_c = (3*(Isd_c/math.sqrt(3))**2*R2)/(s*Ws_c);        # Torque in Newton-Meter when Winding is connected delta


# DISPLAY RESULTS

print ("EXAMPLE : 5.23 : SOLUTION :-");
print " For Case a.1 Winding is connected in star ";
print " a.1.1) Per phase impedance seen from the terminals in Ohms, Z = %.3f < %.1f Ohms "%(abs(Z),math.degrees(math.atan2(Z.imag,Z.real)));
print " a.1.2) Initial Starting Current , Isy = %.2f A "%(Isy_a)
print " a.1.3) Starting Torque , Tsy = %.1f Nm "%(Tsy_a)
print " For Case a.2 Winding is connected in delta " ;
print " a.2.1) Initial Starting Current , Isd = %.2f A "%(Isd_a)
print " a.2.2) Starting Torque , Tsd = %.2f Nm "%(Tsd_a)
print " For Case b Machine is started umath.sing auto-transfromer and voltage is 50 percentage reduced :- b.1 Winding is connected in star  "
print " b.1.1) Per phase impedance seen from the terminals in Ohms, Z = %.3f<%.1f Ohms "%(abs(Z),math.degrees(math.atan2(Z.imag,Z.real)));
print " b.1.2) Initial Starting Current , Isy = %.1f A "%(Isy_b)
print " b.1.3) Starting Torque , Tsy = %.2f Nm "%(Tsy_b)
print " For Case b.2 Winding is connected in delta " ;
print " b.2.1) Initial Starting Current , Isd = %.2f A "%(Isd_b)
print " b.2.2) Starting Torque , Tsd = %.f Nm "%(Tsd_b)
print " For Case c Both Voltage and Frequency are reduced to 50 percentage :- c.1 Winding is connected in star  ";
print " c.1.1) Per phase impedance seen from the terminals in Ohms, Z =  %.2f<%.2f Ohms "%(abs(Z_c),math.degrees(math.atan2(Z_c.imag,Z_c.real)));
print " c.1.2) Initial Starting Current , Isy = %.2f A "%(Isy_c)
print " c.1.3) Starting Torque , Tsy = %.2f Nm "%(Tsy_c)
print " For Case c.2 Winding is connected in delta " ;
print " c.2.1) Initial Starting Current , Isd = %.2f A "%(Isd_c)
print " c.2.2) Starting Torque , Tsd = %.2f Nm "%(Tsd_c)
print 'Comparing the Calculated values of starting current and torque eid rated frequency and rated voltage'
print "                            star                           delta"
print "                   440V%(50Hz   44V%(5Hz             440V,50Hz   44V,5Hz "
print " starting current     %.2f A     %.f A               %.f A       %.2f A "%(Isy_a,Isy_c,Isd_a,Isd_c)
print " starting Torque      %.1f Nm   %.2f Nm             %.f Nm      %.2f Nm "%(Tsy_a,Tsy_c,Tsd_a,Tsd_c)
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :- For Case a.2) Winding is connected in delta :- a) Initial Starting Current Isy = 254.01 A instead of %.2f A  "%(Isd_a); 
EXAMPLE : 5.23 : SOLUTION :-
 For Case a.1 Winding is connected in star 
 a.1.1) Per phase impedance seen from the terminals in Ohms, Z = 2.571 < 76.5 Ohms 
 a.1.2) Initial Starting Current , Isy = 98.81 A 
 a.1.3) Starting Torque , Tsy = 37.3 Nm 
 For Case a.2 Winding is connected in delta 
 a.2.1) Initial Starting Current , Isd = 296.42 A 
 a.2.2) Starting Torque , Tsd = 111.88 Nm 
 For Case b Machine is started umath.sing auto-transfromer and voltage is 50 percentage reduced :- b.1 Winding is connected in star  
 b.1.1) Per phase impedance seen from the terminals in Ohms, Z = 2.571<76.5 Ohms 
 b.1.2) Initial Starting Current , Isy = 49.4 A 
 b.1.3) Starting Torque , Tsy = 9.32 Nm 
 For Case b.2 Winding is connected in delta 
 b.2.1) Initial Starting Current , Isd = 148.21 A 
 b.2.2) Starting Torque , Tsd = 28 Nm 
 For Case c Both Voltage and Frequency are reduced to 50 percentage :- c.1 Winding is connected in star  
 c.1.1) Per phase impedance seen from the terminals in Ohms, Z =  0.65<22.62 Ohms 
 c.1.2) Initial Starting Current , Isy = 39.08 A 
 c.1.3) Starting Torque , Tsy = 58.34 Nm 
 For Case c.2 Winding is connected in delta 
 c.2.1) Initial Starting Current , Isd = 117.25 A 
 c.2.2) Starting Torque , Tsd = 175.03 Nm 
Comparing the Calculated values of starting current and torque eid rated frequency and rated voltage
                            star                           delta
                   440V%(50Hz   44V%(5Hz             440V,50Hz   44V,5Hz 
 starting current     98.81 A     39 A               296 A       117.25 A 
 starting Torque      37.3 Nm   58.34 Nm             112 Nm      175.03 Nm 
    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
      WRONGLY PRINTED ANSWERS ARE :- For Case a.2) Winding is connected in delta :- a) Initial Starting Current Isy = 254.01 A instead of 296.42 A  

Example 5.24 Page No : 264

In [22]:
import math 
from numpy import imag,real

# GIVEN DATA
m = 3;                          # Total Number of phase in Induction Motor
f = 50;                         # Frequency in Hertz
V = 440;                        # Operating voltage of the Induction Motor in Volts
R1 = 0.2;                      # Circuit Parameter in Ohms
R2 = 0.4;                      # Circuit Parameter in Ohms
X1 = 1.0;                      # Circuit Parameter in Ohms
X2 = 1.5;                      # Circuit Parameter in Ohms
Rc = 150;                      # Circuit Parameter in Ohms
Xm = 30;                       # Circuit Parameter in Ohms


# CALCULATIONS

V1 = V/math.sqrt(3);                                     # Rated Voltage in Volts
Zdol = (R1+1j*X1)+(Rc*1j*Xm*(R2+1j*X2))/(Rc*1j*Xm+Rc*(R2+1j*X2)+(1j*Xm)*(R2+1j*X2));                                           # Equivalent impedance per phase in DOL starter in Ohms
I = V1/Zdol;                                        # Starting Current in DOL starter in Amphere

# For Case(a) A per Phase resistance of 0.5 Ohms is added in Series with the stator circuit

Zsr =  (0.5+R1+1j*X1)+((Rc*1j*Xm*(R2+1j*X2))/((Rc*1j*Xm+Rc*(R2+1j*X2)+(1j*Xm)*(R2+1j*X2))));                                     # Total impedance seen from the terminals in Ohms
Isr = V1/Zsr;                                         # Starting Current in DOL starter in Amphere

# For Case(b) A per Phase resistance of 0.5 Ohms is added in Series with the rotor circuit here assumed that stator to rotor turn ratio is 1.0

Zrr =  (R1+1j*X1)+((Rc*1j*Xm*(0.5+R2+1j*X2))/(Rc*1j*Xm+Rc*(0.5+R2+1j*X2)+(1j*Xm)*(0.5+R2+1j*X2)));                            # Total impedance seen from the terminals in Ohms
Irr = V1/Zrr;                                      # Starting Current in DOL starter in Amphere

# For Case(c) When applied Voltage reduced to 50%

I_c = (0.5*V1)/Zdol;                                # Starting Current in DOL starter in Amphere

# For Case(d) When Motor is supplied by reduced Voltage of 44V ( Voltage is reduced by 10%) and the reduced frequency is 5Hz

f_n = 5;                                          # Reduced Frequency in Hertz
X1_n = (f_n/f)*X1;                                # Changed Circuit Parameter in Ohms
X2_n = (f_n/f)*X2;                                # Changed Circuit Parameter in Ohms
Xm_n = (f_n/f)*Xm;                                # Changed Circuit Parameter in Ohms
Zdol_n = (R1+1j*X1_n)+((Rc*1j*Xm_n*(R2+1j*X2_n))/(Rc*1j*Xm_n+Rc*(R2+1j*X2_n)+(1j*Xm_n)*(R2+1j*X2_n)));                         # Equivalent impedance per phase in DOL starter in Ohms
I_n = (V1*0.1)/Zdol_n;                            # Starting Current in DOL starter in Amphere
Ratio = abs(I_n)/abs(I);                          # Ratio of the Starting Current witha the rated Voltage and frequency to the reduced Voltage and frequency


# DISPLAY RESULTS

print ("EXAMPLE : 5.24 : SOLUTION :-");
print " Normal Initial Starting Current in DOL starter, I = %.1f <%.1f A "%(abs(I),math.degrees(math.atan2(I.imag,I.real)))
print " For Casea A per Phase resistance of 0.5 Ohms is added in Series with the stator circuit "
print "              Initial Starting Current in DOL starter, I = %.1f <%.2f A "%(abs(Isr),math.degrees(math.atan2(Isr.imag,Isr.real)))
print "  For Caseb A per Phase resistance of 0.5 Ohms is added in Series with the rotor circuit "
print "              Initial Starting Current in DOL starter, I = %.2f <%.1f A "%(abs(Irr),math.degrees(math.atan2(Irr.imag,Irr.real)))
print "  For Casec When applied Voltage reduced to 50 percentage "
print "              Initial Starting Current in DOL starter, I = %.2f <%.1f A "%(abs(I_c),math.degrees(math.atan2(I_c.imag,I_c.real)))
print "  For Cased When Motor is supplied by reduced Voltage of 44V  Voltage is reduced by 10 percenatge  and the reduced frequency is 5Hz "
print "              Initial Starting Current in DOL starter, I = %.1f <%.1f A "%(abs(I_n),math.degrees(math.atan2(I_n.imag,I_n.real)))
print " By reducing volatge as well as the frequency, the peak starting current at the insmath.tant os starting is reduced by a fector of %.4f of the starting current with the reted volatge and frequency "%(Ratio)
print "    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]" ;
print "      WRONGLY PRINTED ANSWERS ARE :- For Cased) When Motor is supplied by reduced Voltage of 44V  Voltage is reduced by 10 percenatge ) and the reduced frequency is 5Hz,  I = 24.1 < 25.6 A instead of %.1f < %.2f) A  "%(abs(I_n),math.degrees(math.atan2(I_n.imag,I_n.imag)));
print "  Ratio of the Starting Current with the rated Voltage and frequency to the reduced Voltage and frequency, Ratio = 0.2518 instead of %.4f  "%(Ratio);
EXAMPLE : 5.24 : SOLUTION :-
 Normal Initial Starting Current in DOL starter, I = 101.9 <-76.7 A 
 For Casea A per Phase resistance of 0.5 Ohms is added in Series with the stator circuit 
              Initial Starting Current in DOL starter, I = 95.7 <-66.09 A 
  For Caseb A per Phase resistance of 0.5 Ohms is added in Series with the rotor circuit 
              Initial Starting Current in DOL starter, I = 96.12 <-67.2 A 
  For Casec When applied Voltage reduced to 50 percentage 
              Initial Starting Current in DOL starter, I = 50.94 <-76.7 A 
  For Cased When Motor is supplied by reduced Voltage of 44V  Voltage is reduced by 10 percenatge  and the reduced frequency is 5Hz 
              Initial Starting Current in DOL starter, I = 127.0 <0.0 A 
 By reducing volatge as well as the frequency, the peak starting current at the insmath.tant os starting is reduced by a fector of 1.2467 of the starting current with the reted volatge and frequency 
    [ TEXT BOOK SOLUTION IS PRINTED WRONGLY  I verified by manual calculation ]
      WRONGLY PRINTED ANSWERS ARE :- For Cased) When Motor is supplied by reduced Voltage of 44V  Voltage is reduced by 10 percenatge ) and the reduced frequency is 5Hz,  I = 24.1 < 25.6 A instead of 127.0 < 0.00) A  
  Ratio of the Starting Current with the rated Voltage and frequency to the reduced Voltage and frequency, Ratio = 0.2518 instead of 1.2467  

Example 5.25 Page No : 267

In [23]:
import math 

# GIVEN DATA

m1 = 3;                          # Total Number of phase in 1st Induction Motor
p1 = 6;                          # Total number of Poles of 1st Induction Motor
f = 50;                          # Frequency in Hertz
m2 = 3;                          # Total Number of phase in 2nd Induction Motor
p2 = 10;                         # Total number of Poles of 2nd Induction Motor


# CALCULATIONS 

Ns1 = (120*f)/p1;                           # Synchronous speed of 1st Induction Motor in RPM
Ns2 = (120*f)/p2;                           # Synchronous speed of 2nd Induction Motor in RPM
Nscu = (120*f)/(p1+p2);                     # Speed during cumalative casade in RPM
Ndiff = (120*f)/(p2-p1);                    # Speed during cumalative casade in RPM


# DISPLAY RESULTS

print ("EXAMPLE : 5.25 : SOLUTION :-");
print " a)  Range of speed is %.f - %.f - %.f - %.f RPM "%(Nscu,Ns2,Ns1,Ndiff)
EXAMPLE : 5.25 : SOLUTION :-
 a)  Range of speed is 375 - 600 - 1000 - 1500 RPM 

Example 5.26 Page No : 271

In [24]:
import math 
from numpy import real,imag

# GIVEN DATA

m = 3;                          # Total Number of phase in Induction Motor
p = 4;                          # Total number of Poles of Induction Motor
f = 50;                         # Frequency in Hertz
V = 440;                        # Operating Voltage of the Inductuon Motor
R1 = 0.25;                      # Circuit Parameter in Ohms
R2 = 0.5;                       # Circuit Parameter in Ohms
X1 = 1.5;                       # Circuit Parameter in Ohms
X2 = 1.5;                       # Circuit Parameter in Ohms


# CALCULATIONS

Vph = V/math.sqrt(3);                                    # Per phase Voltage in Volts
Ns = (120.*f)/p;                                     # Synchoronous Speed in RPM
Ws = (2.*math.pi*Ns)/60.;                                 # Roatation Speed in Radians per Seconds

# For Case (a) Machine running at, N = 1400 RPM

N_a = 1400.;                                         # Machine running in RPM
s_a = (Ns-N_a)/Ns;                                  # Slip
I_2_a = Vph/(R1+(R2/s_a)+(1j*(X1+X2)));             # Rotor per phase Current referred to the stator side in Amphere
Pg_a = 3*(abs(I_2_a)**2*R2)/s_a;                     # 3-phase air gap power or Rotor intake Power in Watts
T_a = Pg_a/Ws;                                      # Torque in Newton-Meter

# For Case (b) Machine running at, N = 1600 RPM

N_b = 1600;                                         # Machine running in RPM
s_b = (Ns-N_b)/Ns;                                  # Slip
I_2_b = Vph/(R1+(R2/s_b)+(1j*(X1+X2)));             # Rotor per phase Current referred to the stator side in Amphere
Pg_b = 3*(abs(I_2_b)**2*R2)/s_b;                     # 3-phase air gap power or Rotor intake Power in Watts
T_b = Pg_b/Ws;                                      # Torque in Newton-Meter

# For Case (b) Machine running at, N = -100 RPM

N_c = -100;                                         # Machine running in RPM
s_c = (Ns-N_c)/Ns;                                  # Slip
I_2_c = Vph/(R1+(R2/s_c)+(1j*(X1+X2)));             # Rotor per phase Current referred to the stator side in Amphere
Pg_c = 3*(abs(I_2_c)**2*R2)/s_c;                     # 3-phase air gap power or Rotor intake Power in Watts
T_c = -Pg_c/Ws;                                      # Torque in Newton-Meter (minus sign because its counter oppomath.sing torque)

# DISPLAY RESULTS

print ("EXAMPLE : 5.26 : SOLUTION :-");
print " For Case a) Machine running at, N = 1400 RPM  "
print " a.1) Rotor per phase Current referred to the stator side, I2 = %.2f < %.2f A "%(abs(I_2_a),math.degrees(math.atan2(I_2_a.imag,real(I_2_a))))
print " a.2) Developed Torque, T = %.2f Nm "%(T_a)
print " For Case b) Machine running at, N = 1600 RPM  "
print " a.1) Rotor per phase Current referred to the stator side, I2 = %.2f < %.2f A "%(abs(I_2_b),math.degrees(math.atan2(I_2_b.imag,real(I_2_b))))
print "   ( angle -157.52 + 180 = 22.48 ) "
print " a.2) Developed Torque, T = %.2f Nm "%(T_b)
print " For Case c) Machine running at, N = -100 RPM  "
print " c.1) Rotor per phase Current referred to the stator side, I2 = %.2f < %.2f A "%(abs(I_2_c),math.degrees(math.atan2(I_2_c.imag,real(I_2_c))))
print " c.2) Developed Torque, T = %.2f Nm "%(T_c)
EXAMPLE : 5.26 : SOLUTION :-
 For Case a) Machine running at, N = 1400 RPM  
 a.1) Rotor per phase Current referred to the stator side, I2 = 30.57 < -21.16 A 
 a.2) Developed Torque, T = 133.85 Nm 
 For Case b) Machine running at, N = 1600 RPM  
 a.1) Rotor per phase Current referred to the stator side, I2 = 32.38 < -157.52 A 
   ( angle -157.52 + 180 = 22.48 ) 
 a.2) Developed Torque, T = -150.15 Nm 
 For Case c) Machine running at, N = -100 RPM  
 c.1) Rotor per phase Current referred to the stator side, I2 = 82.35 < -76.53 A 
 c.2) Developed Torque, T = -60.71 Nm 

Example 5.27 Page No : 275

In [25]:
import math 

# GIVEN DATA

m = 3.;                          # Total Number of phase in Induction Motor
p = 2.;                          # Total number of Poles of Induction Motor
f = 50.;                         # Frequency in Hertz
V = 440.;                        # Operating Voltage of the Inductuon Motor
R1 = 0.25;                      # Circuit Parameter in Ohms
R2 = 0.25;                      # Circuit Parameter in Ohms
X1 = 0.75;                      # Circuit Parameter in Ohms
X2 = 0.75;                      # Circuit Parameter in Ohms
out_hp = 50.;                    # Output of the induction motor in HP


# CALCULATIONS

V1 = V/math.sqrt(3);                                              # Phase Voltage in Volts
I = (out_hp*746.)/(V*math.sqrt(3));                                # Rated Current in Amphere
sm = R2/(math.sqrt(R1**2+(X1+X2)**2));                              # Slip at Maximum torque both its in Positive and negative sign
Ws = 2*math.pi*((120.*f/p)*(1./60.));                               # Angular Roatation in Radians per Seconds
Tm = (3*V1**2)/((2*Ws)*(R1+math.sqrt((R1**2)+(X1+X2)**2)));          # Maximum torque during motoring in Newton-Meter
Tg = -(3*V1**2)/((2*Ws)*(-R1+math.sqrt((R1**2)+(X1+X2)**2)));        # Maximum torque during generating in Newton-Meter

# For Case (a) slip = 0.05

s_a = 0.05;                                                  # Slip
T_a = (2*Tm)/((s_a/sm)+(sm/s_a));                            # Torque in Newton-Meter

# For Case (b) slip = -0.05

s_b = -0.05;                                                  # Slip
T_b = (2*Tg)/((s_b/sm)+(sm/s_b));                             # Torque in Newton-Meter


# DISPLAY RESULTS

print ("EXAMPLE : 5.27 : SOLUTION :-");
print " Maximim Torque during Motoring Tm = %.f N-m "%(Tm)
print " Maximim Torque during Generating Tm = %.2f N-m "%(Tg)
print " For Case a slip = 0.05  "
print " a.1) Torque, T = %.2f Nm "%(T_a)
print " For Case b slip = -0.05  "
print " b.1) Torque, T = %.2f Nm "%(T_b)
EXAMPLE : 5.27 : SOLUTION :-
 Maximim Torque during Motoring Tm = 174 N-m 
 Maximim Torque during Generating Tm = -242.49 N-m 
 For Case a slip = 0.05  
 a.1) Torque, T = 96.89 Nm 
 For Case b slip = -0.05  
 b.1) Torque, T = 135.01 Nm 

Example 5.28 Page No : 277

In [26]:
import math 

# GIVEN DATA

m = 3;                          # Total Number of phase in Induction Motor
p = 2;                          # Total number of Poles of Induction Motor
f = 50;                         # Frequency in Hertz
V = 440;                        # Operating Voltage of the Inductuon Motor in Volts
R0 = 0.5;                       # Circuit Parameter in Ohms
Ri = 0.05;                      # Circuit Parameter in Ohms
X0 = 0.2;                       # Circuit Parameter in Ohms
Xi = 0.9;                       # Circuit Parameter in Ohms
s = 0.07;                       # Slip


# CALCULATIONS

Ws = 2*math.pi*f;                                               # Synchronous speed in Radins per second
v = V/math.sqrt(3);                                              # Phase Voltage in Volts
Io = v/(R0+1j*X0);                                          # Starting Current in the outer cage in Amphere
Ii = v/(Ri+1j*Xi);                                          # Starting Current in the inner cage in Amphere
Tst = ((3*abs(Io)**2*R0)/Ws)+((3*abs(Ii)**2*Ri)/Ws);          # Starting torque i.e at smath.degrees(math.atanstill, s=1
Ios = v/((R0/s)+(1j*X0));                                   # Current in the outer cage at slip = 0.07
Iis = v/((Ri/s)+(1j*Xi));                                   # Current in the outer cage at slip = 0.07
T = ((3*abs(Ios)**2*R0)/(s*Ws))+((3*abs(Iis)**2*Ri)/(s*Ws));          # Starting torque at s=0.07 in Newton-Meter


# DISPLAY RESULTS

print ("EXAMPLE : 5.28 : SOLUTION :-");
print " a) Starting torque Tst = %.2f Nm "%(Tst)
print " b) Running torque at slip = 0.07 T = %.2f Nm "%(T)
EXAMPLE : 5.28 : SOLUTION :-
 a) Starting torque Tst = 1100.42 Nm 
 b) Running torque at slip = 0.07 T = 419.62 Nm 

Example 5.29 Page No : 278

In [27]:
import math 


# GIVEN DATA

p = 4;                          # Total number of Poles of Induction Motor
f = 50;                         # Frequency in Hertz
V = 440;                        # Operating Voltage of the Inductuon Motor in Volts
out = 25*1000;                  # Power rating of the Induction motor in Watts
R0 = 2.5;                       # Circuit Parameter in Ohms
Ri = 0.5;                       # Circuit Parameter in Ohms
X0 = 1.0;                       # Circuit Parameter in Ohms
Xi = 5.0;                       # Circuit Parameter in Ohms
Rc = 500;                       # Circuit Parameter in Ohms
R1 = 0.2;                       # Circuit Parameter in Ohms
Xm = 50;                        # Circuit Parameter in Ohms
X123 = 2.0;                     # Circuit Parameter in Ohms
s = 0.05;                       # Slip


# CALCULATIONS

Ws = (2*math.pi*120*f)/(p*60);                                # Synchronous speed in Radins per second
Zo = (R0/s)+(1j*X0);                                      # Outer cage impedance at slip = 0.05 in Ohms
Zi = (Ri/s)+(1j*Xi);                                      # Inner cage impedance at slip = 0.05 in Ohms
Z = (R1+1j*X123)+((Zo*Zi)/(Zo+Zi));                       # Total impdance in Ohms
I = V/Z;                                                  # Current in the Cage winding in Amphere
Io = (I*((Zo*Zi)/(Zo+Zi)))/Zo;                            # Current in the outer cage in Amphere
Ii = (I*((Zo*Zi)/(Zo+Zi)))/Zi;                            # Current in the inner cage in Amphere
T = ((3*abs(Io)**2*R0)/(s*Ws))+((3*abs(Ii)**2*Ri)/(s*Ws));  # Starting torque in Newton-Meter


# DISPLAY RESULTS

print ("EXAMPLE : 5.29 : SOLUTION :-");
print " a) Torque at slip %.2f T = %.2f Nm "%(s,T)
EXAMPLE : 5.29 : SOLUTION :-
 a) Torque at slip 0.05 T = 296.11 Nm 

Example 5.30 Page No : 282

In [28]:
import math 
from numpy import real,imag

# GIVEN DATA

m = 1;                          # Total Number of phase in Induction Motor
p = 2;                          # Total number of Poles of Induction Motor
f = 50;                         # Frequency in Hertz
V = 220;                        # Operating Voltage of the Inductuon Motor in Volts
R1 = 10;                        # Circuit Parameter in Ohms
R2 = 11;                        # Circuit Parameter in Ohms
X1 = 12;                        # Circuit Parameter in Ohms
X2 = 12;                        # Circuit Parameter in Ohms
Xm = 125;                       # Circuit Parameter in Ohms
s = 0.03;                       # Slip


# CALCULATIONS

Zf = ((1j*Xm/2)*((R2/(2*s))+(1j*X2/2)))/((1j*Xm/2)+(R2/(2*s))+(1j*X2/2));               # Impedance offered by the forward field in Ohms
Zb = ((1j*Xm/2)*((R2/(2*(2-s)))+(1j*X2/2)))/((1j*Xm/2)+(R2/(2*(2-s)))+(1j*X2/2));       # Impedance offered by the backward field in Ohms 
Z = (R1+1j*X1)+Zf+Zb;                               # Total Impedance in Ohms
I = V/Z;                                            # Total input current in Amphere
pf = math.cos(math.radians(math.degrees(math.atan2(I.imag,I.real))));                  # Power Factor (lagging)
Vf = I*Zf;                                          # Forward Volatge at slip 0.03 in Volts
Vb = I*Zb;                                          # Backward Volatge at slip 0.03 in Volts
If = Vf/(0.5*R2/s);                                 # Forward Current in Amphere
Ib = Vb/(0.5*R2/(2-s));                             # Forward Current in Amphere
Ws = 2*math.pi*f;                                       # Synchronous Speed in radians per second
T =  ((0.5*(If**2)*R2)/(s*Ws))-((0.5*(Ib**2)*R2)/((2-s)*Ws));          # Starting torque


# DISPLAY RESULTS

print ("EXAMPLE : 5.30 : SOLUTION :-");
print " a) Input Current, I = %.2f < %.f A "%(abs(I),math.degrees(math.atan2(I.imag,I.real)))
print " b) Power factor, pf = %.2f Lagging "%(pf)
print " c) Developed Torque, T = %.3f Nm "%(T.real)
EXAMPLE : 5.30 : SOLUTION :-
 a) Input Current, I = 2.77 < -67 A 
 b) Power factor, pf = 0.39 Lagging 
 c) Developed Torque, T = 0.134 Nm 

Example 5.31 Page No : 287

In [29]:
import math 
from numpy import real,imag

# GIVEN DATA

Wsc = 900;                     # Power at Blocked Rotor test in Watts
Vsc = 200;                     # Voltage at Blocked Rotor test in Volts
Isc = 5.0;                     # Current at Blocked Rotor test in Amphere
Wo = 60;                       # Power at No-load test in Watts
Vo = 220;                      # Voltage at No-load test in Volts
Io = 1.5;                      # Current at No-load test in Amphere 
m = 1;                         # Total Number of phase in Induction Motor
p = 4;                         # Total number of Poles of Induction Motor
V = 220;                       # Operating voltage of the Induction motor in Volts
f = 50;                        # Frequency in Hertz
s = 0.07;                      # Slip
R1 = 12;                       # resistance of the main primary winding in Ohms


# CALCULATIONS

Zsc = Vsc/Isc;                      # Impedance in Blocked Rotor test in Ohms
Rsc = Wsc/(Isc**2);                  # resistance in Blocked Rotor test in Ohms
Xsc = math.sqrt((Zsc**2)-(Rsc**2));        # reactance in Blocked Rotor test in Ohms
Xl1 = Xsc/2;                        # Leakage reactance of stator and rotor to be equal in Ohms
Xl2 = Xsc/2;                        # Leakage reactance of stator and rotor to be equal in Ohms
R2 = Rsc-R1;                    # Equivalent resistance of rotor referred to stator in Ohms
Z0 = Vo/Io;                         # Impedance in Blocked Rotor test in Ohms
R0 = Wo/(Io**2);                     # resistance in Blocked Rotor test in Ohms
X0 = math.sqrt((Z0**2)-(R0**2));           # reactance in Blocked Rotor test in Ohms
Wloss = Wo - ((Io**2)*(R1+R2));      # Loss in Watts
Xm_half = X0-Xl1-Xl2/2;
R2f = (R2/s)+((1j*Xl2)/2);                             # Forward resiamath.tance in Ohms
Zf = ((1j*Xm_half)*R2f)/(1j*Xm_half+R2f);              # Total Forward impedance in Ohms
R2b = (R2/(2-s))+((1j*Xl2)/2);                         # Backward resiamath.tance in Ohms
Zb = ((1j*Xm_half)*R2b)/(1j*Xm_half+R2b);              # Total Backward impedance in Ohms
Z = Zf+Zb+(R1+1j*Xl1);                                 # Total impedance in Ohms
I = V/Z;                                               # Motor Current in Amphere
pf = math.cos(math.radians(math.degrees(math.atan2(I.imag,I.real))));                      # Power Factor (lagging)


# DISPLAY RESULTS

print ("EXAMPLE : 5.31 : SOLUTION :-");
print " Circuit Parameters are  a) Leakage reactance of stator and rotor to be equal, Xl1 = Xl2 = %.2f Ohms "%(Xl1)
print " b) Equivalent resistance of rotor referred to stator, R2 = %.f Ohms "%(R2)
print " c) Total Forward impedance, Zf = %.1f < %.2f Ohms "%(abs(Zf),math.degrees(math.atan2(Zf.imag,Zf.real)))
print " c) Total Backward impedance, Zb = %.2f < %.2f Ohms "%(abs(Zb),math.degrees(math.atan2(Zb.imag,Zb.real)))
print " d) Total impedance, Z = %.2f < %.2f Ohms "%(abs(Z),math.degrees(math.atan2(Z.imag,Z.real)))
print " e) Input Current, I = %.2f < %.2f A "%(abs(I),math.degrees(math.atan2(I.imag,I.real)))
print " f) Power factor, pf = %.2f Lagging "%(pf)
EXAMPLE : 5.31 : SOLUTION :-
 Circuit Parameters are  a) Leakage reactance of stator and rotor to be equal, Xl1 = Xl2 = 8.72 Ohms 
 b) Equivalent resistance of rotor referred to stator, R2 = 24 Ohms 
 c) Total Forward impedance, Zf = 122.0 < 69.16 Ohms 
 c) Total Backward impedance, Zb = 12.70 < 24.56 Ohms 
 d) Total impedance, Z = 144.44 < 62.39 Ohms 
 e) Input Current, I = 1.52 < -62.39 A 
 f) Power factor, pf = 0.46 Lagging 

Example 5.32 Page No : 288

In [30]:
import math 
from numpy import imag,real

# GIVEN DATA

Wsc = 600;                     # Power at Blocked Rotor test in Watts
Vsc = 125;                     # Voltage at Blocked Rotor test in Volts
Isc = 15.0;                    # Current at Blocked Rotor test in Amphere
Wo = 360;                      # Power at No-load test in Watts
Vo = 220;                      # Voltage at No-load test in Volts
Io = 6.5;                      # Current at No-load test in Amphere 
m = 1;                         # Total Number of phase in Induction Motor
p = 4;                         # Total number of Poles of Induction Motor
V = 220;                       # Operating voltage of the Induction motor in Volts
f = 50;                        # Frequency in Hertz
s = 0.05;                      # Slip
R1 = 1.2;                      # resistance of the main primary winding in Ohms


# CALCULATIONS

Zlr = Vsc/Isc;                      # Impedance in Blocked Rotor test in Ohms
Rlr = Wsc/(Isc**2);                  # resistance in Blocked Rotor test in Ohms
Xlr = math.sqrt((Zlr**2)-(Rlr**2));        # reactance in Blocked Rotor test in Ohms
Xl1 = Xlr/2;                        # Leakage reactance of stator and rotor to be equal in Ohms
Xl2 = Xlr/2;                        # Leakage reactance of stator and rotor to be equal in Ohms
R2 = (Rlr-R1);                      # Equivalent resistance of rotor referred to stator in Ohms
R2_half = R2/2;                     # Equivalent resistance of rotor referred to stator in Ohms
Z0 = Vo/Io;                         # Impedance in Blocked Rotor test in Ohms
R0 = Wo/(Io**2);                     # resistance in Blocked Rotor test in Ohms
X0 = math.sqrt((Z0**2)-(R0**2));           # reactance in Blocked Rotor test in Ohms
Wloss = Wo - ((Io**2)*(R1+R2));      # Loss in Watts
Xm_half = X0-Xl1-Xl2/2;
R2f = (R2/(2*s))+((1j*Xl2)/2);                         # Forward resiamath.tance in Ohms
Zf = ((1j*Xm_half)*R2f)/(1j*Xm_half+R2f);              # Total Forward impedance in Ohms
R2b = (R2/(2*(2-s)))+((1j*Xl2)/2);                     # Backward resiamath.tance in Ohms
Zb = ((1j*Xm_half)*R2b)/(1j*Xm_half+R2b);              # Total Backward impedance in Ohms
Z = Zf+Zb+(R1+1j*Xl1);                                 # Total impedance in Ohms
I = V/Z;                                               # Motor Current in Amphere
pf = math.cos(math.radians(math.degrees(math.atan2(I.imag,I.real))));                     # Power Factor (lagging)
Vf = I*Zf;                                             # Voltage across forward impedance in Volts
If = Vf/R2f;                                           # Forward current producing torque in Amphere
Tf = ((abs(If)**2)*R2)/(2*s);                           # Forward torque in synchronous Watts
Vb = I*Zb;                                             # Voltage across Backward impedance in Volts
Ib = Vb/R2b;                                           # Backward current producing torque in Amphere
Tb = ((abs(Ib)**2)*R2)/(2*(2-s));                       # Backward torque in synchronous Watts
T = Tf-Tb;                                             # Net torque in Synchronous Watts


# DISPLAY RESULTS

print ("EXAMPLE : 5.32 : SOLUTION :-");
print " Circuit Parameters are  a) Leakage reactance of stator and rotor to be equal, Xl1 = Xl2 = %.2f Ohms "%(Xl1)
print " b) Equivalent resistance of rotor referred to stator,R2 = %.2f Ohms "%(R2)
print " c) Total Forward impedance, Zf = %.1f < %.2f Ohms "%(abs(Zf),math.degrees(math.atan2(Zf.imag,Zf.real)))
print " c) Total Backward impedance, Zb = %.2f < %.2f Ohms "%(abs(Zb),math.degrees(math.atan2(Zb.imag,Zb.real)))
print " d) Total impedance, Z = %.2f < %.2f Ohms "%(abs(Z),math.degrees(math.atan2(Z.imag,Z.real)))
print " e) Input Current, I = %.2f < %.f A "%(abs(I),math.degrees(math.atan2(I.imag,I.real)))
print " f) Power factor, pf = %.4f Lagging "%(pf)
print " g) Forward torque, Tf = %.2f Synchronous Watts "%(Tf)
print " h) Backward torque, Tb = %.2f Synchronous Watts "%(Tb)
print " i) Net torque, T = %.2f Synchronous Watts "%(T)
EXAMPLE : 5.32 : SOLUTION :-
 Circuit Parameters are  a) Leakage reactance of stator and rotor to be equal, Xl1 = Xl2 = 3.95 Ohms 
 b) Equivalent resistance of rotor referred to stator,R2 = 1.47 Ohms 
 c) Total Forward impedance, Zf = 12.3 < 34.65 Ohms 
 c) Total Backward impedance, Zb = 1.87 < 79.96 Ohms 
 d) Total impedance, Z = 17.28 < 47.68 Ohms 
 e) Input Current, I = 12.73 < -48 A 
 f) Power factor, pf = 0.6733 Lagging 
 g) Forward torque, Tf = 1638.70 Synchronous Watts 
 h) Backward torque, Tb = 52.90 Synchronous Watts 
 i) Net torque, T = 1585.80 Synchronous Watts