# Chapter : Transformers¶

## Example 1 : pg 66¶

In [2]:
# Example 4.1;NUMBER OF TURNS
#calculate the number of turns
# given :
e1=2200.;#voltage in volts
f=50.;#frequency in Hz
e2=220.;#voltage in volts
fd=1.6;#magnetic field in Tesla
a=3600.;#area in mm**2
#calculations
n1=(e1/(4.44*f*fd*a*10**-6));#number  of turns
n2=(e2/(4.44*f*fd*a*10**-6));#number of turns
#results
print "number of primary winding turns are",round(n1)
print "number of secondary winding turns are",round(n2)
number of primary winding turns are 1720.0
number of secondary winding turns are 172.0

## Example 2 : pg 68¶

In [3]:
# Example 4.2;
#calculate the components of no load currents,magnetising and working components of exciting current
from math import sqrt
# given
print "part (a)"
pv=11000.;#primary voltage
#calculations and results
Iw=nlw/pv;#current in amperes
Io=0.6;#current in amperes
Imu=sqrt(Io**2-Iw**2);#current in amperes
print "iron loss current is, (A)=",round(Iw,3)
print "magnetising component is, (A)=",round(Imu,3)
pf=0.2;#power factpr
Io=0.5;#current in amperes
Iw=Io*(pf);#current in amperes
Imu=Io*sqrt(1-pf**2);#magnetising component in amperes
print " part (b)"
print "iron loss current is, (A)=",Iw
print "magnetising component  is, (A)=",round(Imu,3)
part (a)
iron loss current is, (A)= 0.182
magnetising component is, (A)= 0.572
part (b)
iron loss current is, (A)= 0.1
magnetising component  is, (A)= 0.49

## Example 3 : pg 68¶

In [4]:
# Example 4.3;current
#calculate the current
from math import sqrt, acos, cos
# given
pf1=0.866;#power factor
pf2=0.1736;#power factor
#calculations
ir=120.;#current in amperes
n2=110;#number of turns
n1=440.;#number of turns
i2d=(n2/n1)*ir;#current in amperes
io=5.;#current in amperes
aioi2=ph2-ph1;#change in angle in degree
i1=sqrt(io**2+i2d**2+(2*io*i2d*cos(aioi2)));#current in amperes
#results
print "current is, (A)=",round(i1,2)
current is, (A)= 33.43

## Example 4 : pg 69¶

In [5]:
# Example 4.4;core losses
#calculate the core losses
# given
f=50.;#frquency in Hz
hl=650.;#hystresis loss
edl=400.;#eddy current loss
#calculations
A=hl/f;#parameter
B=edl/f**2;#parameter
Ph=A*2*f;#loss in watts
Pe=B*(2*f)**2;#loss in watts
pt=Ph+Pe;#total loss in watts
#results
print "total core losses is,(W) = ",pt
total core losses is,(W) =  2900.0

## Example 5 : pg 71¶

In [6]:
# Example 4.5;
#calculate the efficiency and load for maximum efficiency
from math import sqrt
# given
cl=125.;#copper losses
il=457.;#iron losses
pf=0.8;#power factor
kba=30.;#loss
#calculations and results
print "part (a)"
print "full load efficiency at 0.8 pf is,(%)=",round(fle,3)
lme=kba*sqrt(il/fcl);#variable
pfl=(lme/kba)*100;#percentage of full load on which efficiency will be maximum
print "part (b)"
print "percentage of full load on which efficiency will be maximum is,(%)=",round(pfl,3)
part (a)
full load efficiency at 0.8 pf is,(%)= 96.165
part (b)
percentage of full load on which efficiency will be maximum is,(%)= 95.603

## Example 6 : pg 73¶

In [7]:
# Example 4.6;
#calculate the all day efficiency
#given
ef=0.98;#efficiency in %
kva=15;#kVA
pf=1;#power factor
#calculations
op=kva*pf;#output power in kW
ip=op/ef;#input power in kW
loss=ip-op;#loss in kW
cl=(loss*10**3)/2;#copper loss in W
il=cl;#iron loss in W
t1=12;#time in hours
p1=2;#power in kW
pf1=0.5;#power factor
y1=(p1)/pf1;#kVA
il1=il*t1;#loss in Wh
cl1=cl*((y1)/kva)**2*t1;#copper loss in Wh
top1=p1*t1;#kWht1=12;#time in hours
t2=6;#time in hours
p2=12;#power in kW
pf2=0.8;#power factor
y2=(p2)/pf2;#kVA
il2=il*t2;#iron loss in Wh
cl2=cl*((y2/kva)**2)*t2;#copper loss in Wh
top2=p2*t2;#kWh
t3=6;#time in hours
il3=il*t3;#iron loss Wh
tol=top1+top2;#iron loss kWh
til=(il1+il2+il3)*10**-3;#total iron loss in kWh
tcl=(cl1+cl2)*10**-3;#total copper loss in kWh
#results
all day efficiency is,(%)= 95.31

## Example 7 : pg 75¶

In [8]:
# Example 4.7;iron losses
#calculate the iron losses
from math import sqrt
#given
kva=200;#kVA
pf=0.8;#power factor
rflo=kva*pf;#kW
ef=0.96;#efficiency
#calculations
ip=rflo/ef;#kW
tl=ip-rflo;#kW
e2=800;#volts
e1=6600;#volts
n21=((e2/sqrt(3))/e1);#turn ratiom
r1=4;#ohms
r2=0.05;#ohms
roe=(r1)*n21**2+r2;#ohms
fli=((kva*10**3)/(sqrt(3)*e2));#amperes
fcl=3*fli**2*roe;#kW
il=tl-(fcl)*10**-3;#kW
#results
print "iron losses is,(kW)=",round(il,2)
iron losses is,(kW)= 2.32

## Example 8 : pg 78¶

In [9]:
# Example 4.8;
#calculate the resistance,reactances and impedances and copper losses
from math import sqrt
#given
r1=3.45;#ohms
r2=0.009;#ohms
x1=5.2;#ohms
x2=0.015;#ohms
kva=100.;#kVA
e1=8800.;#volts
e2=440.;#volts
#calculations
i1=(kva*10**3)/e1;#in amperes
i2=(kva*10**3)/e2;#in amperes
k=e2/e1;#transformation ratio
ro1=r1+(r2/k**2);#ohms
xo1=x1+(x2/k**2);#ohms
ro2=r2+(k**2*r1);#ohms
xo2=k**2*xo1;#ohms
zo1=sqrt(ro1**2+xo1**2);#ohms
zo2=sqrt(ro2**2+xo2**2);#ohms
tcl=i1**2*r1+i2**2*r2;#in watts
tcl1=i1**2*ro1;#in watts
tcl2=i2**2*ro2;#in watts
#results
print "part (a) "
print "equivalent resistance referred to the primary is,(Ohm)=",ro1
print "equivalent reactance referred to the primary is,(Ohm)=",xo1
print "equivalent resistance referred to the secondary is,(Ohm)=",ro2
print "equivalent reactance referred to the secondary is,(Ohm)=",xo2
print "equivalent impedance referred to the primary is,(Ohm)=",round(zo1,3)
print "equivalent impedance referred to the secondary is,(Ohm)=",round(zo2,3)
print "part (b) "
print "total copper losses considering individual resistance is,(W)=",round(tcl,3)
print "total copper losses consdering equivalent resistance (for primary) is,(W)=",round(tcl1,3)
print "total copper losses consdering equivalent resistance (for secondary) is,(W)=",round(tcl2,3)
#copper losses are calculated wrong in the textbook
part (a)
equivalent resistance referred to the primary is,(Ohm)= 7.05
equivalent reactance referred to the primary is,(Ohm)= 11.2
equivalent resistance referred to the secondary is,(Ohm)= 0.017625
equivalent reactance referred to the secondary is,(Ohm)= 0.028
equivalent impedance referred to the primary is,(Ohm)= 13.234
equivalent impedance referred to the secondary is,(Ohm)= 0.033
part (b)
total copper losses considering individual resistance is,(W)= 910.382
total copper losses consdering equivalent resistance (for primary) is,(W)= 910.382
total copper losses consdering equivalent resistance (for secondary) is,(W)= 910.382

## Example 9 : pg 82¶

In [11]:
# Example 4.9;
#calculate the parameter of primary side ,regulation and efficiency
from math import sqrt
#given
po=100.;#watts
v1=200.;#volts
io=1;#amperes
#calculations and results
ocpf=po/(v1*io);#open circuit power factor
sinpf=sqrt(1-ocpf**2);#
im=io*sinpf;#in amperes
iw=io*ocpf;#current in amperes
rm=v1/iw;#ohms
xm=v1/im;#in ohms
vs=15.;#volts
ia=10.;#amperes
zo2=vs/ia;#in ohms
wa=85.;#watts
ro2=wa/(ia)**2;#ohms
e2=400.;#volts
e1=200.;#volts
k=e2/e1;#transformation ratio
zo1=zo2/k**2;#ohms
ro1=ro2/k**2;#ohms
xo1=sqrt(zo1**2-ro1**2);#ohms
print " part (a)"
print "magnetising component of no load current (Im) is,(A)=",round(im,2)
print "working component of no load current (Iw) is,(A)=",iw
print "resistance for primary side  (Rm) is,(Ohm)=",round(rm,2)
print "reactance for primary ohms (Xm) is,(Ohm)=",round(xm,2)
print "impedence for primary side (X01) is,(Ohm)=",round(xo1,2)
print "part (b)"
kva=4000;#kVA
i2=kva/e2;#in amperes
xo2=sqrt(zo2**2-ro2**2);#ohms
pf=0.8;# power factor
vlag=i2*(ro2*pf+xo2*sqrt(1-pf**2));#in volts
prld=(vlag*po)/e2;#
vlag1=i2*(ro2*pf-xo2*sqrt(1-pf**2));#in volts
prld1=(vlag1*po)/e2;#
print "percentage regulation on lagging load is,(%)=",round(prld,2)
print "part (c)"
cl=85;#copper losses
pf=0.8;#power factor
flo=kva*pf;#efficiency
effl=flo/(flo+fll);#efficiency
hll=(1./2)**2*cl+nloss;#loss in watts
op=(1./2)*kva*pf;#ouput power in watts
print "efficiency at full load is,(%)=",round(effl*100,2)
print "efficiency at half load is,(%)=", round(efhl*100,2)
part (a)
magnetising component of no load current (Im) is,(A)= 0.87
working component of no load current (Iw) is,(A)= 0.5
resistance for primary side  (Rm) is,(Ohm)= 400.0
reactance for primary ohms (Xm) is,(Ohm)= 230.94
impedence for primary side (X01) is,(Ohm)= 0.31
part (b)
percentage regulation on lagging load is,(%)= 3.55