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"
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"
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"
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"
from __future__ import division
import cmath
import math
#Variable declaration:
#short ckt test readings:
Vsc=48 #voltage(V)
Isc=20.8 #current(A)
Psc=617 #power(W)
#Open ckt test readings:
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),"%"
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"
print "full-load efficiency:", round(e,2),"%"
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
Vload=V2/k
#Results:
print "The line to line voltage:",round(Vload,0),"V line-to-line"
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"
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"
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)
Zl=2.5+0j #Impedance ofthe load(ohm)
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"
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"
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"
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"
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)
Vl=438 #load voltage(V)
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)
Vpl=Vl/Vb #per unit load voltage(V)
Ppl=Pl/Pb #per unit load power
Ipl=Ppl/Vpl #per unit load current(A)
Vpp=Vpl+Ipl*Zpn #high side voltage of the transformer(V)
#Results:
print "The high side voltage:",round(abs(Vpp*Vp),0),"V"