Chapter4:SINGLE-PHASE TRANSFORMER

Ex4.1:pg-332

In [5]:
n2=40.0               #Assigning values to parameters
n1=600.0
kva=50.0
e1=2200.0
e2=(e1*n2)/n1
i1=kva*1000/e1
i2=kva*1000/e2
print"The primary full load current is",round(i1,2),"A"
print"The secondary full load current is",round(i2,2),"A"
print"The secondary voltage at node is",round(e2,2),"Volts"

The primary full load current is 22.73 A
The secondary full load current is 340.91 A
The secondary voltage at node is 146.67 Volts

Ex4.2:pg-333

In [7]:
e1=3200         #Assigning values to parameters
f=50
bm=1.2
e2=400
n2=111
kva=80
n1=e1*n2/e2
i2=kva*1000/e2
a=e2/(4.44*f*n2*bm)
print"number of turns on primary windings is n1=",round(n1,2)
print"The secondary full load current is i2=",round(i2,2),"A"
print"The cross-sectional area is a=",round(a,4),"meter square"

number of turns on primary windings is n1= 888.0
The secondary full load current is i2= 200.0 A
The cross-sectional area is a= 0.0135 meter square

Ex4.3:pg-333

In [10]:
e1=6000        #Assigning values to parameters
f=50
e2=250
fm=0.06
n1=e1/(4.44*f*fm)
n2=e2/(4.44*f*fm)
print"number of turns on primary windings is",round(n1,2),"turns"
print"number of turns on secondary windings is",round(n2,3),"turns"

number of turns on primary windings is 450.45 turns
number of turns on secondary windings is 18.769 turns

Ex4.4:pg-334

In [16]:
f=50.0
n2=50.0               #Assigning values to parameters
n1=500.0
kva=25.0
e1=3000.0
k=n2/n1
i1=kva*1000/e1
i2=i1/k
e2=k*e1
fm=e1/(4.44*f*n1)
print"The primary full load current is",round(i1,2),"A"
print"The secondary full load current is",round(i2,2),"A"
print"The secondary emf is",round(e2,2),"Volts"
print"The maximum flux is",round(fm,3),"Wb"

The primary full load current is 8.33 A
The secondary full load current is 83.33 A
The secondary emf is 300.0 Volts
The maximum flux is 0.027 Wb

Ex4.5:pg-335

In [23]:
e1=230.0           #Assigning values to parameters
v1=e1
i0=5.0
t=math.acos(0.25)
n1=200.0
f=50.0
fm=e1/(4.44*f*n1)
w1=v1*i0*cos(t)
iu=i0*sin(t)
print"The maximum flux is",round(fm*1000,3),"mWb"
print"The core loss is",round(w1,2),"Watts"
print"The maximum current is",round(iu,2),"A"

The maximum flux is 5.18 mWb
The core loss is 287.5 Watts
The maximum current is 4.84 A

Ex4.6:pg-335

In [25]:
k=0.25          #Assigning values to parameters
sr=50
pr=sr/(k*k)
print"The Secondary resistance is",round(pr,2),"ohms"

The Secondary resistance is 800.0 ohms

Ex4.9:pg-338

In [26]:
wf=2500        #Assigning values to parameters
w6=0.6*0.6*wf
w5=0.5*0.5*wf
print"The copper loss at 60% full-load condition is",round(w6,2),"Watts"
print"The copper loss at 50% full-load conditionis",round(w5,2),"Watts"

The copper loss at 60% full-load condition is 900.0 Watts
The copper loss at 50% full-load conditionis 625.0 Watts

Ex4.10:pg-338

In [27]:
w7=1200          #Assigning values to parameters
wf=w7/(0.75*0.75)
w5=0.5*0.5*wf
print"The copper loss at 50% full-load condition is",round(w5,2),"Watts"

The copper loss at 50% full-load condition is 533.33 Watts

Ex4.11:pg-339

In [33]:
V=230.0;           #Assigning values to parameters
VA=350.0;
loss=110.0;
I0=VA/V;
pf=loss/VA;
Iw=I0*pf;
Iu=sqrt(I0**2-Iw**2);
print"Iron loss component of no load current",round(Iw,3),"A"
print"Magnatizing component of no load current",round(Iu,2),"A"
print"no load power factor",round(pf,3)

Iron loss component of no load current 0.478 A
Magnatizing component of no load current 1.44 A
no load power factor 0.314

Ex4.13:pg-354

In [35]:
r1=0.2          #Assigning values to parameters
x1=0.75
r2=0.05
x2=0.2
pf=0.8
e2=125.0
e1=250.0
t=math.acos(0.8)
k=e2/e1
kva=5.0
i2=kva*1000/e2
r02=r2+k*k*r1
x02=x2+k*k*x1
pr1=(i2*r02*cos(t)-i2*x02*sin(t))*100/e2
v2=e2-(e2*pr1/100)
print"The percentage regulation at full load 0.8 pf leading is",round(pr1,2)
print"The secondary terminal voltage is",round(v2,2),"Volts"

The percentage regulation at full load 0.8 pf leading is -4.88
The secondary terminal voltage is 131.1 Volts

Ex4.14:pg-355

In [37]:
r1=2.0        #Assigning values to parameters
r2=0.02
wi=412.0
pf=0.8
x=1.0
kva=50.0
e1=2300.0
e2=230.0
i2=kva*1000/e2
i1=kva*1000/e1
wcf=(i1*i1*r1)+(i2*i2*r2)
n1=x*kva*pf*100/((x*kva*pf)+(wi*0.001)+(x*x*wcf*0.001))
x=0.5
n2=x*kva*pf*100/((x*kva*pf)+(wi*0.001)+(x*x*wcf*0.001))
print"Efficiency at full node 0.8pf is",round(n1,2),"%"
print"Efficiency at half full node 0.8pf is",round(n2,2),"%"

Efficiency at full node 0.8pf is 94.56 %
Efficiency at half full node 0.8pf is 95.76 %

Ex4.15:pg-356

In [40]:
x=1.0            #Assigning values to parameters
kva=25.0
pf=0.8
wi=0.35
wcf=0.4
n1=x*kva*pf*100/((x*kva*pf)+(wi*0.001)+(x*x*wcf*0.001))
kva1=kva*(sqrt(wi/wcf))
nm=kva1*pf*100/((kva1*pf)+2*wi)
print"Load in KVA  is",round(kva1,3)
print"Maximum Efficency is",round(nm,2),"%"

Load in KVA  is 23.385
Maximum Efficency is 96.39 %

Ex4.16:pg-357

In [43]:
x=1.0        #Assigning values to parameters
kva=40.0
pf=0.8
wi=450.0
wcf=850.0
n1=x*kva*pf*100/((x*kva*pf)+(wi*0.001)+(x*x*wcf*0.001))
x=sqrt(wi/wcf)
n2=x*kva*pf*100/((x*kva*pf)+(2*wi*0.001))
kva1=kva*sqrt(wi/wcf)
print"Efficiency at full node 0.8pf is",round(x,4)
print"Maximum Efficency is",round(n2,2)
print"Load in KVA at which maximum occurs is",round(kva1,2)

Efficiency at full node 0.8pf is 0.7276
Maximum Efficency is 96.28
Load in KVA at which maximum occurs is 29.1

Ex4.17:pg-358

In [46]:
e1=2000.0        #Assigning values to parameters
e2=200.0
r1=2.3
x1=4.2
r2=0.025
x2=0.04
kva=20.0
i1=kva*1000/e1
i2=kva*1000/e2
k=e2/e1
r01=r1+r2/(k*k)
x01=x1+x2/(k*k)
r02=r2+k*k*r1
x02=x2+k*k*x1
print"The equivalent primary resistance is",round(r01,2),"ohms"
print"The equivalent primary reactance is",round(x01,2),"ohms"
print"The equivalent Secondary resistance is",round(r02,3),"ohms"
print"The equivalent Secondary reactance is",round(x02,3),"ohms"

The equivalent primary resistance is 4.8 ohms
The equivalent primary reactance is 8.2 ohms
The equivalent Secondary resistance is 0.048 ohms
The equivalent Secondary reactance is 0.082 ohms

Ex4.18:pg-359

In [49]:
x=1.0        #Assigning values to parameters
kva=20.0
pf=0.8
wi=450.0
wcf=900.0
n1=x*kva*pf*100/((x*kva*pf)+(wi*0.001)+(x*x*wcf*0.001))
x=sqrt(wi/wcf)
n2=x*kva*pf*100/((x*kva*pf)+(2*wi*0.001))
print"Efficiency at full node 0.8pf is",round(n1,2),"%"
print"Maximum Efficency is",round(n2,2),"%"
print"Load  at which maximum occurs is",round(x,3)

Efficiency at full node 0.8pf is 92.22 %
Maximum Efficency is 92.63 %
Load  at which maximum occurs is 0.707

Ex4.20:pg-361

In [55]:
nm=98.0        #Assigning values to parameters
x=0.5
kva=200.0
pf=1.0
wi=1000*((x*kva*pf*100/nm)/2-(x*kva*pf)/2)
wcu=wi
wcf=wcu/(0.5*0.5)
n1=(x*kva*pf*100)/((x*kva*pf)+(wi*0.001)+(x*x*wcf*0.001))
x=0.75
n2=(x*kva*pf*100)/((x*kva*pf)+(wi*0.001)+(x*x*wcf*0.001))
print"The core loss is",round(wi/1000,4),"kWatts"
print"Efficiency at full node 0.8pf is",round(n1,2)
print"Efficiency at 75% full node 0.8pf is",round(n2,2)

The core loss is 1.0204 kWatts
Efficiency at full node 0.8pf is 98.0
Efficiency at 75% full node 0.8pf is 97.84

Ex4.21:pg-362

In [61]:
r1=0.3 #Assigning values to parameters
r2=0.01
x1=1.1
x2=0.035
kva=100
v1=2200
e1=v1
n1=400.0
n2=80.0
k=n2/n1
r01=r1+r2/(k*k)
x01=x1+x2/(k*k)
z01=sqrt(r01*r01+x01*x01)
e2=k*e1
i2=kva*1000/e2
r02=k*k*r01
x02=k*k*x01
pr1=(i2*r02*cos(t)-i2*x02*sin(t))*100/e2
v2=e2-(e2*pr1/100)
print"The equivalent primary resistance is z01=",round(z01,2),"ohms"
print"The percentage voltage regulation at full load 0.8 pf leading is x02=",round(x02,3),"ohms"
print"The secondary terminal voltage is v2=",round(v2,2),"volts"

The equivalent primary resistance is z01= 2.05 ohms
The percentage voltage regulation at full load 0.8 pf leading is x02= 0.079 ohms
The secondary terminal voltage is v2= 446.77 volts

Ex4.22:pg-363

In [18]:
E2=20.0;         #Assigning values to parameters
E1=1000.0;
kva=5.0;
I2=kva*1000/E2;
K=E2/E1;
R01=4.4
R02=K*K*R01;
X01=8.98
X02=K*K*X01;
pf=0.8
percentreg=(I2*R02*pf+I2*X02*sqrt(1-pf*pf))*100/E2;
print"Percentage maximum regulation is=",round(percentreg,2)
wi=90
I1=kva*1000/E1
Wcf=I1*I1*R01
kvam=kva*sqrt(wi/Wcf)
print"kva at maximum Efficency is kvam=",round(kvam,2)

Percentage maximum regulation is= 4.45
kva at maximum Efficency is kvam= 4.52

Ex4.23:pg-365

In [50]:
v1=200.0                  #Assigning values to parameters
i0=0.7
w=70.0
k=400/200
t=math.acos(w/(v1*i0))
iw=i0*cos(t)
iu=i0*sin(t)
r0=v1/iw
x0=v1/iu
vsc=15.0
i2=10.0
w=85.0
r02=w/(i2*i2)
z02=vsc/i2
x02=sqrt(z02*z02-r02*r02)
r01=r02/(k*k)
x01=x02/(k*k)
e2=400.0
i2=5*1000/(0.8*e2)
v2=e2-i2*r02*cos(t)-i2*x02*sin(t)
print"The secondary Voltage is v2=",round(v2,2),"volts"
#the answer of v2 in the book is wrong,because in the book ,the values of cos(t) & sin(t) are wrong.

The secondary Voltage is v2= 376.64 volts

Ex4.24:pg-366

In [21]:
wi=1000.0                #Assigning values to parameters
kva=50.0
e1=2200.0
ifl=kva*1000/e1
x=1.0
pf=0.8
wcf=(ifl/20)*(ifl/20)*500
n1=x*kva*pf*100/((x*kva*pf)+(wi*0.001)+(x*x*wcf*0.001))
x=sqrt(wi/wcf)
n2=x*kva*pf*100/((x*kva*pf)+(2*wi*0.001))
print"Efficiency at full node 0.8pf is n1=",round(n1,3)
print"Maximum Efficency is n2=",round(n2,2)
print"Load  at which maximum occurs is x=",round(x,2)

Efficiency at full node 0.8pf is n1= 96.048
Maximum Efficency is n2= 96.14
Load  at which maximum occurs is x= 1.24

Ex4.25:pg-367

In [22]:
kva=5.0                #Assigning values to parameters
e2=400.0
r02=0.85
x02=1.236
i2f=kva*1000/e2
t=math.acos(0.8)
pr1=(i2f*r02*cos(t)+i2f*x02*sin(t))*100/e2
pr2=(i2f*r02*cos(t)-i2f*x02*sin(t))*100/e2
print"The percentage regulation at full load 0.8 pf lagging is",round(pr1,2)
print"The percentage regulation at full load 0.8 pf leading is",round(pr2,2)

The percentage regulation at full load 0.8 pf lagging is 4.44
The percentage regulation at full load 0.8 pf leading is -0.19

Ex4.26:pg-369

In [23]:
cl=(10.0/12)*(10.0/12)*100          #Assigning values to parameters
op=500*10*0.8
il=80.0
eff=op*100/(op+il+cl)
print"The efficiency is eff=",round(eff,2)

The efficiency is eff= 96.4

Ex4.27:pg-370

In [24]:
kw=15          #Assigning values to parameters
t=math.acos(0.8)
kva=kw/cos(t)
x=kva/25
wcf=500
cl1=0.75*0.75*wcf
kw=20
t=math.acos(0.9)
kva=kw/cos(t)
x=kva/25
cl2=x*x*500
kw=10
t=math.acos(0.9)
kva=kw/cos(t)
x=kva/25
cl3=x*x*500
tec=cl1*6+cl2*10+cl3*4
tei=400*24
eo=330000
n=eo*100/(eo+tei+tec)
print"The efficiency is n=",round(n,2),"%"

The efficiency is n= 95.48 %

Ex4.28:pg-371

In [35]:
kw=400.0          #Assigning values to parameters
pf=0.8
kva=kw/pf
cl1=4.5
kw=300.0
pf=0.75
kva=kw/pf
cl2=(kva/500)*(kva/500)*4.5
kw=100.0
pf=0.8
kva=kw/pf
cl3=(kva/500)*(kva/500)*4.5
cl4=0
tec=cl1*6+cl2*10+cl3*4+cl4*4
tei=84.0
eo=5800.0
n=eo*100/(eo+tei+tec)
print"The efficiency is n=",round(n,2),"%"

The efficiency is n= 97.63 %

Ex4.29:pg-372

In [36]:
nm=0.98          #Assigning values to parameters
kva=15.0
x=1.0
pf=1.0
wi=((x*kva*pf/nm)/2-(x*kva*pf)/2)
wcu=wi
kw=2.0
pf=0.5
kva=kw/pf
cl1=(kva/15)*(kva/15)*wi
kw=12.0
pf=0.8
kva=kw/pf
cl2=0.153
kw=18.0
pf=0.9
kva=kw/pf
cl3=(kva/15)*(kva/15)*wi
tec=cl1*12+cl2*6+cl3*6
tei=3.672
eo=204.0
n=eo*100/(eo+tei+tec)
print"The efficiency is n=",round(n,2)

The efficiency is n= 96.98

Ex4.30:pg-374

In [15]:
cl1=1.5          #Assigning values to parameters
cl2=0.5*0.5*cl1
tec=cl1*3+cl2*4
tei=36
eo=500
n=eo*100/(eo+tei+tec)
print"The efficiency is n=",round(n,2)

The efficiency is n= 92.25