# Chapter 2: Transformers¶

### Example 2.1, Page number: 63¶

In :
from __future__ import division
import math

#Variable declaration
Pc=16                               #Core loss at Bmax=1.5 T
VIrms=20                            #Voltamperess for the core
Vrms=194                            #Rms induced voltage(V)

#Calculation:
pf=Pc/VIrms
a=math.acos(pf)
I=VIrms/Vrms
Ic=I*pf
Im=I*math.fabs(math.sin(a))

#Results:
print "Power factor = ", round(pf,1),"lagging"
print "The core-loss current,Ic =", round(Ic,3), "A rms"
print "The magnetising current,Im =", round(Im,2),"A rms"

Power factor =  0.8 lagging
The core-loss current,Ic = 0.082 A rms
The magnetising current,Im = 0.06 A rms


### Example 2.2, Page number: 67¶

In :
from __future__ import division

#Variable declarations:
k=5                                 #turns ratio,N1/N2
Z2=1+4j                             #Impedance of secondary side(ohm)
Vp=120                              #primary voltage(V)

#Calculations:
Z2p=k**2*(Z2)
I=Vp/Z2p
Is=k*I

#Results:
print "Primary current:",complex(round(I.real,2),round(I.imag,2)), "A rms"
print "Current in the short:",round(Is.real,2)+1j*round(Is.imag,2),"A"

Primary current: (0.28-1.13j) A rms
Current in the short: (1.41-5.65j) A


### Example 2.4, Page number: 74¶

In :
from __future__ import division
import cmath

#Variable declaration:
R1=0.72                                 #Resistance at high voltage side(ohm)
R2=0.70                                 #Resistance at low voltage side(ohm)
X1=0.92                                 #Reactance at high voltage side(ohm)
X2=0.90                                 #Reactance at low voltage side(ohm)
Zq=632+4370j                            #Impedance of exciting circuit(ohm)

#Calculations:
Req=R1+R2
Xeq=X1+X2
Vcd=2400*Zq/(Zq+complex(R1,X1))
V=complex(round(Vcd.real,2),round(Vcd.imag,3))

#Results:
print "Req:",Req,"ohm"," and  Xeq:",Xeq,"ohm"
print "Voltage at low voltage terminal:",V,"V"

Req: 1.42 ohm  and  Xeq: 1.82 ohm
Voltage at low voltage terminal: (2399.45+0.316j) V


### Example 2.5, Page number: 76¶

In :
from __future__ import division
import math

#Variable declaration:
Zf=0.30+.160j                           #Impedance of feeder(ohm)
Zeq=1.42+3.42j             #Equiv.impedance of transformer refd. to primary(ohm)
k=2400/240                              #turns ratio
P=50000                                 #power rating of the transformer(VA)
Vs=2400                                 #sending end vltage of feeder(V)

#Calculations:
I=P/2400                                #Rated current(A)
theta=math.acos(0.80)
Zt=Zf+Zeq                      #combned impedance of feeder & transformer(ohm)
R=Zt.real
X=Zt.imag
bc=I*X*math.cos(theta)-I*R*math.sin(theta)
ab=I*R*math.cos(theta)+I*X*math.sin(theta)
Ob=(Vs**2-bc**2)**0.5
V2=Ob-ab

#Results:
print "The voltage at the secondary terminals:",round(V2/10,0),"V\n"

The voltage at the secondary terminals: 233.0 V



### Example 2.6, Page number: 80¶

In :
from __future__ import division
import cmath
import math

#Variable declaration:
Vsc=48                          #voltage(V)
Isc=20.8                        #current(A)
Psc=617                         #power(W)

Vs=240                         #Voltage(V)
I=5.41                          #current(A)
P=186                           #power(W)
V2ph=2400                        #voltage at full load at high voltage side(V)
pf=0.8                          #lagging power factor at full load

#Calculations:
theta=math.acos(pf)
Zeqh=Vsc/Isc                    #subscript h refers to high voltage side
Reqh=Psc/Isc**2
Xeqh=math.sqrt(Zeqh**2-Reqh**2)
Ih=50000/V2ph
Pout=50000*pf
Pwind=Ih**2*Reqh
Ptloss=P+Pwind
e=(1-Ptloss/(Ptloss+Pout))*100
Iph=(50000/2400)*complex(math.cos(theta),math.sin(-theta))
V1ph=V2ph+Iph*complex(Reqh,Xeqh)
r=(round(abs(V1ph),2)-2400)*100/V2ph

#Results:
print "The efficiency of the transformer:",round(e,0),"%"
print "Volatge Regulation:",round(r,2),"%"

The efficiency of the transformer: 98.0 %
Volatge Regulation: 1.94 %


### Example 2.7, Page number: 82¶

In :
from __future__ import division

#Varaible declaration:
Vx=2400                                 #Voltage at low voltage side(V)
Vbc=2400                                 #Voltage across branch bc(V)
Vab=240                                #Voltage induced in winding ab(V)
Pl=803                                  #transformer losses(W)
pf=0.8                                  #Power factor of the transformer

#Calculations:
Vh=Vab+Vbc
Ih=50000/Vab
KVA=Vh*Ih/1000                          #Kva rating
P=pf*550000
e=(1-Pl/(P+Pl))*100

#Results:
print "Voltage ratings, Vh:",Vh,"V  ", "&  Vx:",Vx,"V"
print "KVA rating as an autotransformer:",KVA,"KVA"

Voltage ratings, Vh: 2640 V   &  Vx: 2400 V
KVA rating as an autotransformer: 550.0 KVA


### Example 2.8, Page number: 87¶

In :
from __future__ import division
import math

#Variable declaration:
Vl1=4160                      #line-to-line voltage at feeder's sending end(V)
Zf=0.30+.160j                           #Impedance of feeder(ohm)
Zeq=1.42+3.42j             #Equiv.impedance of transformer refd. to primary(ohm)
k=2400/240                              #turns ratio
P=50000                                 #power rating of the transformer(VA)
Vs=2400                                 #sending end vltage of feeder(V)

#Calculation:
#this problem can be treated on a single phase basis,
#and whole problem is similar to Ex 2.5.

I=P/2400                                #Rated current(A)
theta=math.acos(0.80)
Zt=Zf+Zeq                      #combned impedance of feeder & transformer(ohm)
R=Zt.real
X=Zt.imag
bc=I*X*math.cos(theta)-I*R*math.sin(theta)
ab=I*R*math.cos(theta)+I*X*math.sin(theta)
Ob=(Vs**2-bc**2)**0.5
V2=Ob-ab

#Results:
print "The line to line voltage:",round(Vload,0),"V line-to-line"

The line to line voltage: 233.0 V line-to-line


### Example 2.9, Page number: 89¶

In :
from __future__ import division
import math
import cmath

#Variable decclaration:
#All resistances, reactances, & impedances are on per phase basis
Req=1.42       #Series resist. of del-del transformer referred to 2400v side(ohm)
Xeq=1.82       #Series react. of del-del transformer referred to 2400v side(ohm)
Zs=0.17+0.92j                   #Equiv impedance of sending end transformer(ohm)
Xf=0.8j                          #Reactance of the feeder(ohm)
Vf=2400                         #Voltage of the feeder(V)
k=10                            #turns ratio(Vp/Vs)

#Calculations:
Zt=(complex(Req,Xeq)/3)+Zs+Xf
Ztot=complex(round(Zt.real,2),round(Zt.imag,2))
If=math.floor(Vf/(math.sqrt(3))/round(abs(Ztot),2))
I1=If/math.sqrt(3)
I2=I1*k
Ic=I2*math.sqrt(3)

#Results:
print "Short circuit current in the 2400 feeder, per phase wires:",round(Ic,1),"A"

Short circuit current in the 2400 feeder, per phase wires: 5720.0 A


### Example 2.10, Page number: 92¶

In :
from __future__ import division
import cmath
from math import *

#Variable declaration:
X1=143                             #Reactance of primary(ohm)
X21=164                           #Reactance of secondary ref. to primary(ohm)
Xm=163*10**3                           #Reactance of magnetising ckt(ohm)
R1=128                             #Resistance of primary(ohm)
R21=141                           #Resistane of secondary ref. to primary(ohm)
k=20                              #turns ratio(2400/120)
V1=2400                             #primary voltage(V)

#Calculations:
V2=(V1/k)*complex(0,Xm)/complex(R1,X1+Xm)
mag=abs(V2)
ph=degrees(cmath.phase(V2))

#Results:
print "Magnitude of V2:",round(mag,2),"V"
print "Phase of V2:",round(ph,3),"degrees"

Magnitude of V2: 119.89 V
Phase of V2: 0.045 degrees


### Example 2.11, Page number: 94¶

In :
from __future__ import division
import cmath

#variable declaration:
X1=44.8*10**-6                  #Reactance of the primary(ohm)
R1=10.3*10**-6                  #Resistance of the primary(ohm)
X21=54.3*10**-6          #Reactance of the secondary refd. to primary(ohm)
R21=9.6*10**-6                  #Resistance of secondary ref. to primary(ohm)
Xm=17.7*10**-3                  #Reactance of the magnetising ckt(ohm)
k=5/800                         #turms ratio(I2/I1)
I1=800                          #primary current(A)

#Calculations:
Zp=k**2*Zl
I2=I1*k*Xm*1j/(Zp+R21+(X21+Xm)*1j)
phase=cmath.phase(I2)

#Results:
print "Magnitude of current:",round(abs(I2),2),"A"
print "Phase of the current:",round(math.degrees(phase),3),"degrees"

Magnitude of current: 4.98 A
Phase of the current: 0.346 degrees


### Example 2.12, Page number: 97¶

In :
from __future__ import division
import math

#Variable declaration:
XL=0.040                                #Reactance at l.v side(ohm)
XH=3.75                                 #Reactance at h.v side(ohm)
Xm=114                                  #Magnetising reactance(ohm)
RL=0.76*10**-3                          #Resistance at l.v.side(ohm)
RH=0.085                                #Resistance at l.v.side(ohm)
VA_base=100*10**6                       #base VA
V_base=7.97*10**3                       #base voltage(V)

#Calculations:
#for l.v side
VA_base=100*10**6                       #base VA
V_base=7.97*10**3                       #base voltage(V)
Rbase1=Xbase1=V_base**2/VA_base

#for h.v side:
VA_base=100*10**6                       #base VA
V_base=79.7*10**3                       #base voltage(V)
Rbase2=Xbase2=V_base**2/VA_base

XL_pu=XL/Xbase1
XH_pu=XH/Xbase2
Xm_pu=Xm/Xbase1
RL_pu=RL/Rbase1
RH_pu=RH/Rbase2
K_pu=1                                  #per unit utrns ratio

#Results:
print "The per unit parameters are:"
print "XL_pu =",round(XL_pu,3),"p.u"
print "XH_pu =",round(XH_pu,4),"p.u"
print "Xm_pu =",math.ceil(Xm_pu),"p.u"
print "RL_pu =",round(RL_pu,4),"p.u"
print "XL_pu =",round(RH_pu,4),"p.u"
print "Turns ratio =",K_pu,"p.u"

The per unit parameters are:
XL_pu = 0.063 p.u
XH_pu = 0.059 p.u
Xm_pu = 180.0 p.u
RL_pu = 0.0012 p.u
XL_pu = 0.0013 p.u
Turns ratio = 1 p.u


### Example 2.13, Page number: 98¶

In :
from __future__ import division
import cmath

#Variable declaration:
Ic=5.41                             #Exciting current ref. to low volt. side(A)
k=10                                #turns ratio(N1/N2=2400/240)
Vbh=2400                            #base voltage at primary side(V)
Vbl=240                             #base voltage at secondary side(V)
Ibh=20.8                            #base current at primary side(A)
Ibl=208                             #base current at secondary side(A)
Z=1.42+1.82j                    #Equiv.impedance ref.to high voltage side(ohm)

#Calculations:
Zbh=Vbh/Ibh
Zbl=Vbl/Ibl
Icl=Ic/Ibl
Ich=Ic/(Ibh*k)
Zl=Z/(k**2*Zbl)
Zh=Z/Zbh

#Results:
print "Per unit exciting current on low volt. sides:",round(Icl,3,),"A"
print "Per unit exciting current on high volt. sides:",round(Ich,3),"A"
print "per unit equiv.impedance at low volt. sides:",round(Zl.real,4)+round(Zl.imag,4)*1j,"ohm"
print "per unit equiv.impedance at high voltage sides:",round(Zh.real,4)+round(Zh.imag,4)*1j,"ohm"

Per unit exciting current on low volt. sides: 0.026 A
Per unit exciting current on high volt. sides: 0.026 A
per unit equiv.impedance at low volt. sides: (0.0123+0.0158j) ohm
per unit equiv.impedance at high voltage sides: (0.0123+0.0158j) ohm


### Example 2.14, Page number: 100¶

In :
from __future__ import division
from math import *

#Variable declaration:
Vb=24000               #Base voltage of secondary of sending end transformer(V)
Z=0.17+0.92j      #Impedance of sending end transformer ref. to 2400V side(ohm)
P=150                       #Power rating of the transformer(KVA)
V=2400                      #Primary voltage of sending end transformer(v)
Ztot=0.64+2.33j                 #Total series impedance(ohm)

#Calculations:
Zb=V**2/(P*10**3)
Ztotb=Ztot/Zb
Vsb=1                           #Vs in terms of per unit values
Isc=Vsb/abs(Ztotb)                   #Short current in per unit values(A)
Ib1=P*10**3/(sqrt(3)*2400)       #base current of the feeder at 2400V side(A)
If=Ib1*Isc
Ib2=P*10**3/(sqrt(3)*240)
Iscs=Isc*Ib2                     #short ckt current at 2400V afeeder side (A)

#Results:
print "Short circuit current in 2400 feeder:",round(Iscs/10**3,2),"KA"

Short circuit current in 2400 feeder: 5.73 KA


### Example 2.15, Page number: 102¶

In :
from __future__ import division

#Variable declaration:
P=250*10**3                            #power rating of transformer(KVA)
Vp=2400                             #primary volatge(V)
Vs=460                              #secondary voltage(V)
Pb=100*10**3                           #new base power of transformer(KVA)
Vb=460                              #new base voltage(V)
Z=0.026+0.12j                       #series impedance on its own base(ohm)
Pl=95*10**3                         #power drawn by the load(kW)

#Calculations:
Zbo=Vs**2/P                        #base impedance for the transformer(ohm)
Zbn=Vb**2/Pb            #base impedance for the transformer at 100KVA base(ohm)
Zpn=Z*Zbo/Zbn                      #base impedance at 100KVA base(ohm)

The high side voltage: 2313.0 V