# Chapter 3 : Transformers¶

## Example 3.1 Page No : 196¶

In [1]:
# Given Data
kVA = 500.;                #rating
V1 = 11000.;                #primary voltage in volts
V2 = 400.;                  #secondary voltage in volts
N2 = 100.;                  #number of turns in secondary winding
f = 50.;                    #frequency in hertz

# Calculations and Results
N1 = (V1*N2)/V2;            #number of turns in primary winding
print "number of turns in primary  winding, N1 = %dturns"%(N1)
I1 = (kVA*1000)/V1;
I2 = (kVA*1000)/V2
print "primary current, I1 = %fA"%(I1)
print "secondary current, I2 = %fA"%(I2)
E1 = V1;
phi = E1/(4.44*f*N1)
print "maximium flux in the core = %fWb"%(phi)

number of turns in primary  winding, N1 = 2750turns
primary current, I1 = 45.454545A
secondary current, I2 = 1250.000000A
maximium flux in the core = 0.018018Wb


## Example 3.2 Page No : 196¶

In [2]:
# Given Data
V1 = 6600.;                #primary voltage in volts
V2 = 230.;                  #secondary voltage in volts
f = 50.;                    #frequency in hertz
Bm = 1.1;                    #flux density in Wb/m**2

# Calculations and Results
A = (25*25*10**(-4));         #area of the core in m**2
phi = Bm*A
print "flux = %fWb"%(phi)
E1 = V1;
E2 = V2;
N1 = E1/(4.44*f*phi);
N2 = E2/(4.44*f*phi);
print "number of turns in primary winding, N1 = %dturns"%(N1)
print "number of turns in secondary winding, N2 = %dturns"%(N2)

flux = 0.068750Wb
number of turns in primary winding, N1 = 432turns
number of turns in secondary winding, N2 = 15turns


## Example 3.3 Page No : 197¶

In [3]:
# Given Data
V1 = 230.;                #primary voltage in volts
f = 50.;                    #frequency in hertz
N1 = 100.;                  #number of primary turns
N2 = 400.;                  #number of secondary turns

# Calculations and Results
A = 250.*10**(-4);            #cross section area of core in m**2
print ("since at no-load E2 = V2")
E2 = (V1*N2)/N1;
print "induced secondary winding, E2 = %dV"%(E2);
phi = E2/(4.44*f*N2);
Bm = phi/A;
print "Maximium flux density in the core = %fWb/m**2"%(Bm)

since at no-load E2 = V2
induced secondary winding, E2 = 920V
Maximium flux density in the core = 0.414414Wb/m**2


## Example 3.4 Page No : 197¶

In [4]:
# Given Data
kVA = 40.;                #rating of the transformer
V1 = 2000.;                #primary side voltage in volts
V2 = 250.;                 #secondary side voltage in volts
R1 = 1.15;                #primary resistance in ohms
R2 = 0.0155;              #secondary resistance in ohms

# Calculations and Results
R = R2+(((V2/V1)**2)*R1)
print "Total resistance of the transformer in terms of the secondary winding = %fohms"%(R)
I2 = (kVA*1000)/V2;
print "Full load secondary current = %dA"%(I2)
print "Total copper loss on full load = %fWatts"%((I2)**2*R)

Total resistance of the transformer in terms of the secondary winding = 0.033469ohms
Full load secondary current = 160A
Total copper loss on full load = 856.800000Watts


## Example 3.5 Page No : 206¶

In [6]:
import math
# Given Data
I2 = 300.;                        #Secondary current in amperes
N1 = 1200.;                        #number of primary turns
N2 = 300.;                         #number of secondary turns
I0 = 2.5;                         #load current in amperes

# Calculations
I1 = (I2*N2)/N1;
phi0 = math.degrees(math.acos(0.2));
phi2 = math.degrees(math.acos(0.8));
I = math.sqrt(I1c**2+I1s**2);

# Results
print "primary power factor = %fdegrees"%(math.cos(math.radians(phi)));

primary power factor = 1.000000degrees


## Example 3.6 Page No : 207¶

In [7]:
import math
# Given Data
phi0 = math.degrees(math.acos(0.2))
I2 = 40;                 #secondary current in amperes
phi2 = math.degrees(math.acos(0.8))
r = 3;                    #ratio of primary and secondary turns
I1 = I2/r;

# Calculations
I = math.sqrt(I1c**2+I1s**2);

# Results
print "I1 = %fA"%(I)

I1 = 14.156879A


## Example 3.7 Page No : 208¶

In [8]:
import math

# Given Data
V1 = 230.;                #voltage in volts
f = 50.;                  #frequency of supply in hertz
N1 = 250.;                #number of primary turns
I0 = 4.5;                #no-load current in amperes

# Calculations and Results
phi0 = math.degrees(math.acos(0.25));
print "magnetimath.sing current, Im = %fA"%(Im);

print "Core loss = %dW"%(Pc)
print ("neglecting I**2R loss in primary winding at no-load")
E1 = V1;
phi = E1/(4.44*f*N1);
print "Maximium value of flux in the core = %fWb"%(phi)

magnetimath.sing current, Im = 4.357106A
Core loss = 258W
neglecting I**2R loss in primary winding at no-load
Maximium value of flux in the core = 0.004144Wb


## Example 3.8 Page No : 209¶

In [9]:
import math

# Given Data
I2 = 30.;                        #Secondary current in amperes
I0 = 2.;                         #load current in amperes
V1 = 660.;                        #primary voltage in volts
V2 = 220.;                        #secondary voltage in volts

# Calculations
I1 = (I2*V2)/V1;
phi0 = math.degrees(math.acos(0.225));
phi2 = math.degrees(math.acos(0.9));
I = math.sqrt(I1c**2+I1s**2);
phi = math.degrees(math.atan(I1s/I1c))

# Results
print "I1 = %fA"%(I)
print "primary power factor = %fdegrees"%(math.cos(math.radians(phi)));

I1 = 11.361713A
primary power factor = 0.831741degrees


## Example 3.9 Page No : 210¶

In [11]:
import math

# Given Data
phi_m = 7.5*10**(-3);                    #maximium flux
f = 50.;                            #frequecy in hertz
N1 = 144.;                         #number of primary turns
N2 = 432.;                         #number of secondary turns
kVA = 0.24;                        #rating of transformer

# Calculations and Results
E1 = (4.44*phi_m*f*N1)
V1 = E1;
print "V1 = %dV"%(V1)
I0 = (kVA*1000)/V1;
phi0 = math.degrees(math.acos(0.26));
print "Im = %fA"%(Im);
V2 = (E1*N2)/N1
print "V2 = %fV"%(V2)
print ("At a load of 1.2kVA and power factor of 0.8 lagging")
kVA = 1.2;
phi2 = math.degrees(math.acos(0.8));
I2 = (kVA*1000)/V2;
I = (I2*N2)/N1;
I = math.sqrt(I1c**2+I1s**2);
print "I1 = %fA"%(I);
print "primary power factor = %flagging"%(math.cos(math.radians(phi)))

V1 = 239V
Im = 0.966575A
V2 = 719.280000V
At a load of 1.2kVA and power factor of 0.8 lagging
I1 = 5.825933A
primary power factor = 0.844673lagging


## Example 3.10 Page No : 211¶

In [12]:
import math
# Given Data
V1 = 6600.;                    #primary voltage in volts
V2 = 240.;                     #secondary voltage in volts
kW1 = 10.;                     #power
phi1 = math.degrees(math.acos(0.8));
I2 = 50.;                       #current in amperes
kW3 = 5.;                       #power
phi2 = math.degrees(math.acos(0.7))
kVA = 8;                       #rating
phi4 = math.degrees(math.acos(0.6))

# Calculations and Results
I3 = (kW3*1000)/(1*V2);
I4 = (kVA*1000)/V2;
I5 = math.sqrt((Ih**2)+(Iv**2))
print "I5 = %dA"%(I5)
Ip = (I5*V2)/V1;
print "The current drawn by the primary from 6600Vmains is equal to, Ip = %fA"%(Ip);
phi = math.degrees(math.atan(Iv/Ih));

I5 = 124A
The current drawn by the primary from 6600Vmains is equal to, Ip = 4.516939A
power factor = 0.945934lagging


## Example 3.11 Page No : 212¶

In [13]:
# Given Data
kVA = 100.;                #rating of the tronsfromer
N1 = 400.;                   #number of primary turns
N2 = 80.;                    #number of secondary turns
R1 = 0.3;                    #primary resistance in ohms
R2 = 0.01;                    #secondary resistance in ohms
X1 = 1.1;                        #primary leakage reactance in ohs
X2 = 0.035;                     #secondary leakage reactance in ohms

# Calculations and Results
Rr2 = (((N1/N2)**2)*R2)
print "R2 = %f ohms"%(Rr2);
Xx2 = (((N1/N2)**2)*X2);
print "X2 = %f ohms"%(Xx2);
Ze = math.sqrt((R1+Rr2)**2+(X1+Xx2)**2);
print "Equivqlent impedence = %f"%(Ze);

R2 = 0.250000 ohms
X2 = 0.875000 ohms
Equivqlent impedence = 2.050152


## Example 3.12 Page No : 216¶

In [14]:
# Given Data
f = 50.;                #frequency in hertz
r = 6.;                #turns ratio
R1 = 0.90;             #primary resistance in ohms
R2 = 0.03;              #secondary resistance in ohms
X1 = 5.;                 #primary reactance in ohms
X2 = 0.13;              #secondary reactance in ohms

# Calculations and Results
Re = (R1+(R2*r**2));
print "equivalent resistance reffered to primary, Re = %fohms"%(Re);
Xe = (X1+(X2*r**2));
print "equivalent reactance reffered to primary, Xe = %fohms"%(Xe);
Ze = math.sqrt(Re**2+Xe**2);
print "equivalent impedance reffered to primary, Ze = %fohms"%(Ze);
Ii2 = r*I2;
print "secondary current reffered to primary side = %fA"%(Ii2);
print "a)Voltage to be applied to the high voltage side = %dvolts"%(Ii2*Ze);
print "b)Power factor = %f"%(Re/Ze);

equivalent resistance reffered to primary, Re = 1.980000ohms
equivalent reactance reffered to primary, Xe = 9.680000ohms
equivalent impedance reffered to primary, Ze = 9.880425ohms
secondary current reffered to primary side = 1200.000000A
a)Voltage to be applied to the high voltage side = 11856volts
b)Power factor = 0.200396


## Example 3.13 Page No : 216¶

In [17]:
# Given Data
R1 = 0.21;                    #primary resistance in ohms
X1 = 1.;                        #primary reactance in ohms
R2 = 2.72*10**(-4);            #secondary resistance in ohms
X2 = 1.3*10**(-3);              #secondary reactanced in ohms
V1 = 6600.;                      #primary voltage in volts
V2 = 250.;                        #secondary voltage in volts

# Calculations and Results
r = V1/V2;                         #turns ratio
Re = R1+(r**2*R2);
print "Equivalent resistance referred to primary side = %fohms"%(Re);
Xe = X1+(r**2*X2);
print "Equivalent reactance referred to primary side = %fohms"%(Xe);
Ze = math.sqrt(Re**2+Xe**2);
print "equivalent impedance reffered to primary, Ze = %fohms"%(Ze);
V = 400.;                        #voltage in volts
I1 = V/Ze;
print "I1 = %f"%(I1);
print "Power input = %fW"%(I1**2*Re);

Equivalent resistance referred to primary side = 0.399573ohms
Equivalent reactance referred to primary side = 1.906048ohms
equivalent impedance reffered to primary, Ze = 1.947480ohms
I1 = 205.393656
Power input = 16856.612924W


## Example 3.14 Page No : 217¶

In [18]:
# Given Data
N1 = 90.;                #number of primary turns
N2 = 180.;                #number of secondary turns
R1 = 0.067;              #primary resistance in ohms
R2 = 0.233;              #secondary resistance in ohms

# Calculations and Results
print "Primary winding resistance referred to secondary side = %fohms"%((R1*N2/N1)**2)
print "secondary winding resistance referred to primary side = %fohms"%((R2*N1/N2)**2)
print "Total resistance of the transformer refferred to primary side = %fohms"%((((R1*N2/N1)**2)+R2*N2/N1)**2)

Primary winding resistance referred to secondary side = 0.017956ohms
secondary winding resistance referred to primary side = 0.013572ohms
Total resistance of the transformer refferred to primary side = 0.234213ohms


## Example 3.15 Page No : 217¶

In [19]:
# Given Data
kVA = 30.;                #rating of the transformer
V1 = 6000.;                #primary voltage in volts
V2 = 230.;                #secondary voltage in volts
R1 = 10.;                #primary resistance in ohms
R2 = 0.016;              #secondary resistance in ohms
Xe = 23.;                    #total reactance reffered to the primary

# Calculations and Results
phi = math.degrees(math.acos(0.8));                #lagging
Re = (R1+((V1/V2)**2*R2))
print "equivalent resistance, Re = %fohms"%(Re)
I2dash = (kVA*1000)/V1;
V2dash = 5847;
print "percentage regulation = %fpercent"%(Reg)

equivalent resistance, Re = 20.888469ohms
percentage regulation = 2.609097percent


## Example 3.16 Page No : 218¶

In [20]:
# Given Data
kVA = 10.;                #rating of the transformer
V1 = 2000.;                #primary voltage in volts
V2 = 400.;                #secondary voltage in volts
R1 = 5.5;                #primary voltage in ohms
R2 = 0.2;                #secondary voltage in ohms
X1 = 12.;                  #primary reactance in ohms
X2 = 0.45;                #secondary reactance in ohms

# Calculations and Results
#assuming  (V1/V2) = (N1/N2)
Re = R2+(R1*(V2/V1)**2);
print "equivalent resistance referred to the secondary = %fohms"%(Re);
Xe = X2+(X1*(V2/V1)**2);
print "equivalent reactance referred to the secondary = %fohms"%(Xe);
Ze = math.sqrt(Re**2+Xe**2);
print "equivalent impedance referred to the secondary = %fohms"%(Ze);
phi = math.degrees(math.acos(0.8));
Vl = 374.5;
print "Voltage across the full load and 0.8 p.f lagging = %fV"%(Vl);
reg = ((V2-Vl)*100)/Vl;
print "percentage voltage regulation = %f percent"%(reg);

equivalent resistance referred to the secondary = 0.420000ohms
equivalent reactance referred to the secondary = 0.930000ohms
equivalent impedance referred to the secondary = 1.020441ohms
Voltage across the full load and 0.8 p.f lagging = 374.500000V
percentage voltage regulation = 6.809079 percent


## Example 3.17 Page No : 219¶

In [21]:
# Given Data
kVA = 80.;                #rating of the transformer
V1 = 2000.;                #primary voltage in volts
V2 = 200.;                 #secondary voltage in volts
f = 50.;                    #frequency in hertz
Id = 8.;                    #impedence drop
Rd = 4.;                    #resistance drop

# Calculations and Results
phi = math.degrees(math.acos(0.8))
I2Ze = (V2*Id)/100;
I2Re = (V2*Rd)/100;
I2Xe = math.sqrt(I2Ze**2-I2Re**2)
print "percentage regulation = %fpercent"%(reg)
pf = I2Xe/math.sqrt(I2Re**2+I2Xe**2)
print "Power factor for zero regulation = %fleading)"%(pf)

percentage regulation = 7.356922percent
Power factor for zero regulation = 0.866025leading)


## Example 3.19 Page No : 225¶

In [22]:
# Given Data
kVA = 50.;                        #rating of the transformer
V1 = 3300.;                        #open circuit primary voltage
Culoss = 540.;                    #copper loss from short circuit test
coreloss = 460.;                  #core loss from open circuit test
V1sc = 124.;                      #short circuit primary voltage in volts
I1sc = 15.4;                      #short circuit primary current in amperes
Psc = 540.                         #short circuit primary power in watts

# Calculations and Results
phi = math.degrees(math.acos(0.8))
print "From the open-circuit test, core-loss = %dW"%(coreloss);
print "From short circuit test, copper loss = %dW"%(Culoss);
print "The efficiency at full-load and 0.8 lagging power factor = %f"%(effi);
Ze = V1sc/I1sc;
Re = Psc/I1sc**2;
Xe = math.sqrt(Ze**2-Re**2);
V2 = 3203;
phi2 = math.degrees(math.acos(0.8));
phie = math.degrees(math.acos(Culoss/(V1sc*I1sc)));
print "Voltage regulation = %dpercent"%(reg)

From the open-circuit test, core-loss = 460W
From short circuit test, copper loss = 540W
The efficiency at full-load and 0.8 lagging power factor = 97.560976
Voltage regulation = 3percent


## Example 3.20 Page No : 226¶

In [23]:
import math

# Given Data
kVA = 100.;
V1 = 6600.;                #primary voltage in volts
V2 = 330.;                  #secondary voltage in volts
f = 50.;                    #frequency in hertz
V1sc = 100.;                      #short circuit primary voltage in volts
I1sc = 10.;                      #short circuit primary current in amperes
Psc = 436.;                         #short circuit primary power in watts

# Calculations and Results
Ze = V1sc/I1sc;
Re = Psc/I1sc**2;
phi = math.degrees(math.acos(0.8));
Xe = math.sqrt(Ze**2-Re**2);
print "Total resistance = %fohms"%(Re);
print "Total impedence = %fohms"%(Ze)
Il = (kVA*1000)/V1;
print "full voltage current, V1 = %dV"%(V1dash)

Total resistance = 4.360000ohms
Total impedence = 10.000000ohms
full voltage current, V1 = 6735V


## Example 3.21 Page No : 227¶

In [26]:
import math

# Given Data
V2 = 500.;                #secondary voltage in volts
V1 = 250.;              #primary voltage in short circuit test in volts
I0 = 1.;                #current in short circuit test in amperes
P = 80.;                #core loss in watt
Psc = 100.;                #power in short circuit test in watts
Vsc = 20.;                 #short circuit voltage in volts
Isc = 12.;                 #short circuit current in amperes

# Calculations and Results
phi0 = math.degrees(math.acos(P/(V1*I0)));
print "From open circuit test , math.cos(phi0) = %f"%(math.cos(phi0));
print "Loss component of no-load current, Ic = %fA"%(Ic)
Im = math.sqrt(I0**2-Ic**2);
print "Magnetising current, Im = %fA"%(Im);
Rm = V1/Ic;
Xm = V1/Im;
Re = Psc/(Isc**2);
Ze = Vsc/Isc;
Xe = math.sqrt(Ze**2-Re**2);
print "Equvalent resistance referred to secondary = %fohms"%(Re);
print "Equvalent reactance referred to secondary = %fohms"%(Xe);
print "Equvalent impedance referred to secondary = %fohms"%(Ze);
K = V2/V1;                            #turns ratio
print "Equvalent resistance referred to primary = %fohms"%(Re/K**2);
print "Equvalent reactance referred to primary = %fohms"%(Xe/K**2);
print "Equvalent impedance referred to primary = %fohms"%(Ze/K**2);
V = 500;                        #output in volts
I = 10;                         #output current in amperes
phi = math.degrees(math.acos(0.80));
print "Effiency = %fpercent"%(effi);

From open circuit test , math.cos(phi0) = -0.606173
Loss component of no-load current, Ic = 0.320000A
Magnetising current, Im = 0.947418A
Equvalent resistance referred to secondary = 0.694444ohms
Equvalent reactance referred to secondary = 1.515099ohms
Equvalent impedance referred to secondary = 1.666667ohms
Equvalent resistance referred to primary = 0.173611ohms
Equvalent reactance referred to primary = 0.378775ohms
Equvalent impedance referred to primary = 0.416667ohms
Effiency = 96.398447percent


## Example 3.22 Page No : 231¶

In [30]:
import math

# Given Data
kVA = 200.;                #Rating of the transformer
Pin = 3.4;                #power input to two transformer in watt
Pin2 = 5.2;
coreloss = Pin;           #core loss of two transformers

# Calculations and Results
phi = math.degrees(math.acos(0.8));
print "Core loss of two transformer = %fkW"%(Pin)
print "Core loss of each transformer = %fkW"%(Pin/2)
print "Full load copper loss of the two transformer = %fkW"%(Pin2)
print "Therefore, full load copper loss of each transformer = %fkW"%(Pin2/2);
print "Full load efficiency at 0.8 p.f. lagging = %fpercent"%(effi);

Core loss of two transformer = 3.400000kW
Core loss of each transformer = 1.700000kW
Full load copper loss of the two transformer = 5.200000kW
Therefore, full load copper loss of each transformer = 2.600000kW
Full load efficiency at 0.8 p.f. lagging = 97.382836percent


## Example 3.24 Page No : 233¶

In [31]:
# Given Data
kVA = 50.;                          #rating of the transformer
V1 = 6360.;                          #primary voltage rating
V2 = 240.;                            #secondary voltage rating
pf = 0.8
coreloss = 2;                      #core loss in kilo watt from open circuit test
Culoss = 2;                        #copper loss at secondary current of 175A
I = 175.;                            #current in amperes

# Calculations and Results
I2 = (kVA*1000)/V2;
print "Full load secondary current, I2 = %fA"%(I2);
effi = (kVA*pf*100)/((kVA*pf)+coreloss+(Culoss*(I2/I)**2))
print "Efficiency = %fpercent"%(effi)

Full load secondary current, I2 = 208.333333A
Efficiency = 89.217075percent


## Example 3.25 Page No : 234¶

In [32]:
# Given Data
kVA = 500.;                #rating of the transformer
R1 = 0.4;                #resistance in primary winding inohms
R2 = 0.001;              #resistance in secondary winding in ohms
V1 = 6600.;                #primary voltahe in volts
V2 = 400.;                 #secondary voltage in volts
ironloss = 3.;              #iron loss in kilowatt
pf = 0.8;                  #power factor lagging

# Calculations and Results
I1 = (kVA*1000)/V1;
print "Primary winding current = %fA"%(I1);
I2 = (I1*V1)/V2;
print "Secondary winding current = %fA"%(I2);
Culoss = ((I1**2*R1)+(I2**2*R2));
print "Copper losses in the two winding = %fWatts"%(Culoss);
effi = (kVA*pf*100)/((kVA*pf)+ironloss+(Culoss/1000));
print "Efficiency at 0.8 p.f = %fpercent"%(effi);

Primary winding current = 75.757576A
Secondary winding current = 1250.000000A
Copper losses in the two winding = 3858.184114Watts
Efficiency at 0.8 p.f = 98.314355percent


## Example 3.26 Page No : 234¶

In [33]:
# Given Data
kVA = 400.;                    #rating of the transformer
ironloss = 2.;                #iron loss in kilowatt
pf = 0.8;                     #power factor
kW = 240.;                    #load in kilowatt

# Calculations and Results
kVA1 = kW/pf;
print ("Efficiency is maximium when,core-loss = copper-loss")
coreloss = ironloss;
print ("Maximium efficiency occurs at 240kw,0.8 power factor,i.e., at 300kVA load")
Cl300 = coreloss;
Cl400 = (Cl300*(kVA/kVA1)**2);
pf1 = 0.71;            #power factor for full load
effi = (kVA*pf1*100)/((kVA*pf1)+coreloss+Cl400);
print "Efficiency at full-load and 071 power factor = %dpercent"%(effi);
pf2 = 1                    #maximium efficiency occurs at unity power factor
MAXeffi = (kVA1*pf2*100)/((kVA1*pf2)+coreloss+Cl300)
print "Maximium efficiency at 300kVA and unity power factor = %fpercent"%(MAXeffi);

Efficiency is maximium when,core-loss = copper-loss
Maximium efficiency occurs at 240kw,0.8 power factor,i.e., at 300kVA load
Efficiency at full-load and 071 power factor = 98percent
Maximium efficiency at 300kVA and unity power factor = 98.684211percent


## Example 3.27 Page No : 235¶

In [34]:
# Given Data
kVA = 40.;                    #rating of the transformer
coreloss = 450.;               #core-loss in watts
Culoss = 800.;                 #copper loss in watt
pf = 0.8;                     #power factor of the load

# Calculations and Results
FLeffi = (kVA*pf*100)/((kVA*pf)+((coreloss+Culoss)/1000));
print ("For maximium efficiency, Core loss = copper loss")
Culoss2 = coreloss;            #for maximium efficiency
n = math.sqrt(Culoss2/Culoss);
kVA2 = n*kVA;                  #load for maximium efficiency
MAXeffi = (kVA2*pf*100)/((kVA2*pf)+((coreloss+Culoss2)/1000));
print "Value of maximium efficiency = %fpercent"%(MAXeffi);

Full-load efficiency = 96.240602percent
For maximium efficiency, Core loss = copper loss
Value of maximium efficiency = 96.385542percent


## Example 3.28 Page No : 236¶

In [35]:
# Given Data
kVA = 50.;                    #rating of the transformers
I1 = 250.;                    #primary current in amperes
Re = 0.006;                  #total resistance referred to the primary side
ironloss = 200.;                #iron loss in watt

# Calculations and Results
Culoss = (I1**2*Re);           #copper loss in watt
pf = 0.8;                      #power factor lagging
print "Full-load copper loss = %fW"%(Culoss);
TL1 = ((Culoss+ironloss)/1000);
print "Total loss on full load = %fkW"%(TL1);
TL2 = ((((Culoss*(1/2)**2))+ironloss)/1000)
print "Total loss on half load = %fkW"%(TL2);
effi1 = (kVA*pf*100)/((kVA*pf)+TL1);
print "Efficiency at full load, 0.8 power factor lagging = %f percent"%(effi1)
effi2 = ((kVA/2)*pf*100)/(((kVA/2)*pf)+TL2);
print "Efficiency at half load, 0.8 power factor lagging = %f percent"%(effi2)

Full-load copper loss = 375.000000W
Total loss on full load = 0.575000kW
Total loss on half load = 0.200000kW
Efficiency at full load, 0.8 power factor lagging = 98.582871 percent
Efficiency at half load, 0.8 power factor lagging = 99.009901 percent


## Example 3.29 Page No : 237¶

In [38]:
# Given Data
kVA = 10.;                    #rating of the transformers
V1 = 400.;                    #primary voltage in volts
V2 = 200.;                    #secondary voltage in volts
f = 50.;                      #frequency in hertz

# Calculations and Results
MAXeffi = 0.96;              #maximium efficiency
output1 = (kVA*0.75);        #output at 75% of full load
input1 = (output1/MAXeffi);
print "Input at 75percent of full load = %fkW"%(input1);
TL = input1-output1;
print "Total losses = %fkW"%(TL);
Pi = TL/2;
Pc = TL/2;
print ("Maximiunm efficiency occurs at 3/4th of full load")
Pc = Pi/(3./4)**2;
print "Thus, total losses on full load = %fW"%((Pc+Pi)*1000);
pf = 0.8;               #power factor lagging
effi = (kVA*pf*100)/((kVA*pf)+(Pc+Pi));
print "Efficiency on full load. 0.8 power factor lagging = %fpercent"%(effi)

Input at 75percent of full load = 7.812500kW
Total losses = 0.312500kW
Maximiunm efficiency occurs at 3/4th of full load
Thus, total losses on full load = 434.027778W
Efficiency on full load. 0.8 power factor lagging = 94.853849percent


## Example 3.30 Page No : 237¶

In [39]:
# Given Data
kVA = 500.;                    #rating of the transformers
V1 = 3300.;                    #primary voltage in volts
V2 = 500.;                    #secondary voltage in volts
f = 50.;                      #frequency in hertz
MAXeffi = 0.97;
x = 0.75;                    #fraction of full load for maximium efficiency
pf1 = 1.;

# Calculations and Results
output1 = (kVA*x*pf1*1000);
print "Output at maximium efficiency = %dwatts"%(output1);
losses = ((1/MAXeffi)-1)*output1;
print "Thus, at maximium efficiency,  lossses = %fW"%(losses)
Culoss = losses/2;
print "Copper losses at 75percent of full load = %dW"%(Culoss);
CulossFL = Culoss/x**2;
print "Copper losses at full load = %dW"%(CulossFL);
Re = CulossFL/(kVA*1000);
Ze = 0.1;                        #equivalent impedence per unit
Xe = math.sqrt(Ze**2-Re**2);
phi = math.degrees(math.acos(0.8));
print "percentage regulation = %f percent"%(reg);

Output at maximium efficiency = 375000watts
Thus, at maximium efficiency,  lossses = 11597.938144W
Copper losses at 75percent of full load = 5798W
Copper losses at full load = 10309W
percentage regulation = 7.520562 percent


## Example 3.32 Page No : 240¶

In [40]:
# Given Data
V1 = 230.;                    #primary voltage of auto-transformer
V2 = 75.;                     #secondary voltage of auto-transformer

# Calculations
r = (V1/V2);                  #ratio of primary to secondary turns
I2 = 200.;                    #load current in amperes
I1 = I2/r;

# Results
print "Primary current, I1 = %fA"%(I1);
print "Load current, I1 = %fA"%(I2);
print "cirrent flowing through the common portion of winding = %fA"%(I2-I1);
print "Economy in saving in copper in percentage = %fpercent"%(100/r);

Primary current, I1 = 65.217391A