Chapter 6:Transformer Principles

Example 6.1:Page number-343

In [2]:
import math
#given
a=50*(10**-4)
e=400
f=50
n1=500
n2=1000
#phym=bm*a
#case a
#e=4.44*f*n*bm*a
bm=(e)/float(4.44*f*n1*a)
print "bm=",format(bm,'.4f'),"Wb/m2"
#case b
e2=4.44*f*n2*bm*a
print "e2=",format(e2,'.1f'),"V"

bm= 0.7207 Wb/m2
e2= 800.0 V

Example 6.2

In [4]:
import math
#given
e=3300
f=50
n1=600
n2=80
bm=1.2
h=425
l=1.6
density=7400
loss=1.5
#case a
phym=e/float(4.44*f*n1)
csa=phym/bm
print "cross sectional area=",format(csa,'.5f'),"m2"
#case b
sv=(e*n2)/n1
print "secondary voltage on no load=",format(sv,'.1f'),"V"
#case c
mc=(h*l)/n1
print "primary magnetising current=",format(mc,'.3f'),"A"
#case d
v=l*csa
m=v*density
closs=m*loss
print "core loss=",format(closs,'.1f'),"W"

cross sectional area= 0.02065 m2
secondary voltage on no load= 440.0 V
primary magnetising current= 1.133 A
core loss= 366.7 W

Example 6.3:Page number-356

In [9]:
import math
#given
#as per step up tranformer
v1=220
v2=6600
f=50
vturn=2.5
kva=10000
#case a
a=v1/float(v2)
print a
#as per step down case b
v1=6600
v2=220
a=v1/v2
print a
#case c
#high voltage soil
n=v1/float(vturn)
print "number of turns of high voltage soil=",format(n,'.1f')
#low voltage soil
n1=v2/float(vturn)
print "number of turns of high voltage soil=",format(n1,'.1f')
#case d
i=kva/float(v1)
print "primary current as a step down transformer is=",format(i,'.3f'),"A"
#case e
i=kva/float(v2)
print "secondary current as a step down transformer is=",format(i,'.2f'),"A"

0.0333333333333
30
number of turns of high voltage soil= 2640.0
number of turns of high voltage soil= 88.0
primary current as a step down transformer is= 1.515 A
secondary current as a step down transformer is= 45.45 A

Example 6.4:Page number-357

In [14]:
import math
#given
rl=32
#let ratio of sides be a
rs=2
a=(2/float(32))
p=a**0.5
print "turns ratio for impedance machting is",format(p,'.2f')

turns ratio for impedance machting is 0.25

Example 6.5:Page number-364

In [15]:
import math
#given
n1=2200
n2=220
kva=100
f=50
r1=0.75
r2=0.0007
x2=0.0009
#case a
#subcase 1
#lv side leakage impedance=r2+jx2-->complex number
#hv side leakage impedance=r1+jx1
#hv side impedance referred to lv side is r1'+jx1'=(r1+jx1)/a**2=(0.0075+j0.0115)
#shunt branch resistance referred to lv side gc-jbm=(0.0035-j0.025)
#The equivqlent circuit is shown in the diagram
#subcase 2
#lv side impedance referred to hv side is r2'+jx2'=a**2*(r2+jx2)=(0.70+j0.90)ohm
#the magnetising admittance refferred to hv side (gc-jbm)/a**2=(0.000035-j0.00025)
#the equivalent circuit is as in figure
#case b
#for an approximate equivalent circuit the magnetised admittance is neglected from the exact circuit
#subcase 1
#equivalent impedance referred to lv side (r2+r1')+j(x2+x1')=(0.0145+j0.0205)ohm
#equivalent circuit is shown in figure
#subcase 2
#equivalent impedance referred to hv side is (r1+r2')+j(x1+x2')=(1.45+j2.05)ohm
#equivalent circuit is shown in figure

Example 6.6:Page number-369

In [17]:
import math
#case a
#from oc test data shunt admittances are determined as follows
#given
v1=200
i0=1
pc=100
yc=i0/float(v1)
print "yc=",format(yc,'.4f'),"S"
gc=pc/float(v1**2)
print "gc=",format(gc,'.4f'),"S"
bm=(((0.005**2)-(0.0025**2))**0.5)
print "bm=",format(bm,'.4f'),"S"
#from sc test data
p=85
isc=10
vsc=15
req=p/float(isc**2)
print "req=",format(req,'.4f'),"ohm"
zeq=vsc/float(isc)
print "zeq=",format(zeq,'.4f'),"ohm"
xeq=(((zeq**2)-(req**2))**0.5)
print "xeq=",format(xeq,'.4f'),"ohm"
#case b
a=0.5
#equivalent impedance parameters referred to lv side
re=(a**2)*req
print "req1=",format(re,'.4f'),"ohm"
xe=(a**2)*xeq
print "xeq1=",format(xe,'.4f'),"ohm"
ze=(a**2)*zeq
print "zeq1=",format(ze,'.4f'),"ohm"
#equivalent circuit referred to lv side is as shown

yc= 0.0050 S
gc= 0.0025 S
bm= 0.0043 S
req= 0.8500 ohm
zeq= 1.5000 ohm
xeq= 1.2359 ohm
req1= 0.2125 ohm
xeq1= 0.3090 ohm
zeq1= 0.3750 ohm

Example 6.8:Page number-376

In [1]:
import math
#case a
#transformer output=0.01x1000cos(angle)W
#loss=10xScos(angle)
#transformer efficiency n=(10xScos(angle)/(10xScos(angle)+pc+0.0001x2Pcu))
print "n=((10xScos(angle))/(10xScos(angle)+pc+0.0001x2Pcu))"
%matplotlib inline
import matplotlib.pyplot as plt
x1=20.5
x2=30
x3=40
x4=50
x5=60.5
x6=70
x7=80
x8=90
x9=100
x10=110
y1=94.3
y2=95
y3=96
y4=96.5
y5=96.8
y6=96.9
y7=97
y8=97
y9=97
y10=97
plt.plot([x1,x2,x3,x4,x5,x6,x7,x8,x9,x10],[y1,y2,y3,y4,y5,y6,y7,y8,y9,y10],marker='o',color='b',label='0.65')
p1=120.5
p2=30
p3=40
p4=50
p5=70
p6=80
p7=90
p8=100
p9=110
q1=95.3
q2=86
q3=96.7
q4=97.2
q5=97.5
q6=97.5
q7=97.5
q8=97.5
q9=97.5
plt.plot([p1,p2,p3,p4,p5,p6,p7,p8,p9],[q1,q2,q3,q4,q5,q6,q7,q8,q9],marker='o',color='g',label='0.8')
x2=[20.5,30,40,50,70,80,90,100,110]
y2=[96.2,96.6,97.4,97.6,98,98,98,98,98]
plt.plot(x2,y2,label='pf=1')
plt.xlabel('% of full load')
plt.ylabel('% efficiency')
plt.legend()
plt.show()

n=((10xScos(angle))/(10xScos(angle)+pc+0.0001x2Pcu))

Example 6.10:Page number-378

In [14]:
import math
#at unity power factor
op=15000
n=0.98
i=op/float(n)
print i
loss=i-op
print loss
pc=float(loss)/2000 #actually division by 2 but value given only to make pc 0.153 instead of 153
t=pc*24 #iron loss in a day
toteng=20+96+108 #sum of energy outputs
engloss=0.109+1.224+1.632 #sum of energy losses
n=toteng/float(engloss+toteng+t)
print n

15306.122449
306.12244898
0.971216989926

Example 6.11:Page number-381

In [16]:
import math
kva=10000
pf=0.8
iloss=75
closs=150
a=0.5
#case a
po=kva*pf
loss=75+150
n=po/float(po+loss)
print n
#case b
i2=(10*1000)/(200)
i1=i2+((10*1000)/400)
kvar=(600*50)/1000
print kvar
po=30*0.8
n=1-(0.225/24)
print n

0.9726443769
30
0.990625

Example 6.12:Page number-382

In [22]:
import math
#case 1
#2300 winding used as secondary
#given and derived
st=150
v1=13800
v2=2300
a=(v1-v2)/v2
b=a+1
sat=(6*150)/5
print "sat=",format(sat,'.1f'),"Kva"
#case 2
v1=13.8
v2=11.5
a=(v1-v2)/v2
sat=((1+a)/a)*150
print "sat=",format(sat,'.1f'),"kva"

sat= 180.0 Kva
sat= 900.0 kva

Example 6.13:Page number-391

In [23]:
import math
#given and 1.732 is the value of root 3
v=6600
i=10
n=15
#case a
v2l=v/n
print "v2l=",format(v2l,'.1f'),"V"
i1p=10/1.732
i2p=i1p*n
print "i2p=",format(i2p,'.1f'),"A"
i2l=n*i1p*1.732
print "i2l=",format(i2l,'.1f'),"A"
#case b
v2p=v/(n*1.732)
print "v2p=",format(v2p,'.1f'),"V"
v2l=v2p*1.732
print "v2l=",format(v2l,'.1f'),"V"
i2l=i2p=n*i
print "i2p=i2l=",format(i2p,'.1f'),"A"
#case c
v2p=v/n
print "v2p=",format(v2p,'.1f'),"V"
v2l=(v*1.732)/n
print "v2l=",format(v2l,'.1f'),"V"
i1p=i/1.732
i2p=i2l=(n*i1p)
print "i2p=",format(i2p,'.1f'),"A"
#case d
v1p=v/1.732
v2p=v2l=v/(n*1.732)
print "v2p=",format(v2p,'.1f'),"V"
i1p=10
i2p=i1p*n
print "i2p=",format(i2p,'.1f'),"A"
i2l=i2p*1.732
print "i2l=",format(i2l,'.1f'),"A"

v2l= 440.0 V
i2p= 86.6 A
i2l= 150.0 A
v2p= 254.0 V
v2l= 440.0 V
i2p=i2l= 150.0 A
v2p= 440.0 V
v2l= 762.1 V
i2p= 86.6 A
v2p= 254.0 V
i2p= 150.0 A
i2l= 259.8 A

Example 6.14

In [24]:
import math
#given
hp=75
v=415
n=0.9
pf=0.85
op=75*746 #since its horse power
ip=op/n
ilv=ip/(1.732*v*pf) #line current on low voltage start side
a=(6600*1.732)/415 #given in question
ihv=ilv/a
print "ihv=",format(ihv,'.5f'),"A"

ihv= 3.69402 A
In [ ]: