Chapter 5 : Single Phase Transformers

Example 5.1 Page No : 5.5

In [1]:
# Variables
kVA = 250.;				#kVA
V1 = 11000.;				#V(Primary voltage)
V2 = 400.;				#V(secondary voltage)
f = 50.;				#Hz
N2 = 80.;				#no. of turns in secondary

# Calculations and Results
Ifl1 = kVA*1000/V1;				#A(Full load primay current)
Ifl2 = kVA*1000/V2;				#A(Full load secondary current)
print ("Part(a)");
print "Full load primary current(A) %.2f"%Ifl1
print "Full load secondary current(A) : %.2f"%Ifl2

print ("Part(b)");
N1 = N2*V1/V2;				#no. of turns in secondary
print "No. of turns in primary : %.2f"%N1
print ("Part(c)");
fi_m = V2/(4.44*N2*f);				#Wb
print "Maximum value of flux(mWb) : %2.f"%(fi_m*1000)
Part(a)
Full load primary current(A) 22.73
Full load secondary current(A) : 625.00
Part(b)
No. of turns in primary : 2200.00
Part(c)
Maximum value of flux(mWb) : 23

Example 5.2 Page No : 5.6

In [2]:
import math 

# Variables
N1 = 480;				#no. of turns in primary
N2 = 90;				#no. of turns in secondary
lfp = 1.8;				#m(length of flux path)
ag = 0.1;				#mm(airgap)
Flux = 1.1;				#T(flux density)
MF = 400;				#A/m(Magnetic flux)
c_loss = 1.7;				#W/kg
f = 50;				#Hz
d = 7800;				#kg/m**3(density of core)
V = 2200;				#V(potential difference)

# Calculations and Results
#Part (a)
fi_m = V/(4.44*N1*f);				#Wb
A = fi_m/Flux;				#m**2(Cross sectional area)
print "(a) Cross sectional area(m**2) : %.2f"%A
#Part (b)
Vnl2 = V*N2/N1;				#V(2ndary voltage on no load)
print "(b) 2ndary voltage on no load(V) : %.f"%Vnl2

#Part (c)
Fm1 = MF*lfp;				#A(Magnetootive force for the core)
Fm2 = Flux/(4*math.pi*10**-7)*ag*10**-3;				#A(Magnetootive force for airgap)
Fm = Fm1+Fm2;				#A(Total magnetomotive force)
Imax = Fm/N1;				#A(maximum value of magnetizing current)
Iom = Imax/math.sqrt(2);				#A(rms current)
v = lfp*A;				#m**3(Volume of core)
m = v*d;				#kg(Mass of core)
coreLoss = c_loss*m;				#W(Core Loss)
Io1 = coreLoss/V;				#A(Core loss component of curent)
Io = math.sqrt(Iom**2+Io1**2);				#A(no load current)
print "(c) Primary current on no load(A) : %.2f"%Io

pf = Io1/Io;				#lagging pf on no load
print "(c) Power factor(lagging) on no load : %.2f"%pf
(a) Cross sectional area(m**2) : 0.02
(b) 2ndary voltage on no load(V) : 412
(c) Primary current on no load(A) : 1.21
(c) Power factor(lagging) on no load : 0.17

Example 5.3 Page No : 5.8

In [9]:
# Variables
N1 = 1000;				#no. of turns in primary
N2 = 200;				#no. of turns in secondary
I0 = 3;				#A
pf0 = 0.2;				#lagging power factor
I2 = 280;				#A(2ndary current)
pf2 = 0.8;				#lagging power factor

# Calculations and Results
I2dash = I2*N2/N1;				#A
cosfi0 = pf0;cosfi2 = pf2;sinfi0 = math.sqrt(1-cosfi0**2);sinfi2 = math.sqrt(1-cosfi2**2);
I1_cosfi1 = I2dash*cosfi2+I0*cosfi0;				#A
I1_sinfi1 = I2dash*sinfi2+I0*sinfi0;				#A
I1 = math.sqrt(I1_cosfi1**2+I1_sinfi1**2);				#A
print "Primary current(A) : %.1f"%I1

fi1 = math.degrees(math.atan(I1_sinfi1/I1_cosfi1));				#degree
pf1 = math.cos(math.radians(fi1));				#lagging
print "Primary power factor(lagging) : %.2f"%pf1
Primary current(A) : 58.3
Primary power factor(lagging) : 0.78