Chapter 8 : Q factor Power and Power Factor

Example8_1,pg 234

In [11]:
# Q factor of coil

import math
#Variable declaration
fr= 400.0*10**3                   #resonance frequency
C = 400.0*10**-12                 #tuned capacitance
R = 10.0                          #resistance of coil
n = 40.0                          #Cp=nC

#Calculations
Cp=n*(100.0/400.0)*10**-12      
L=(1.0/(4*(math.pi**2)*(fr**2)*(C+Cp)))
Q=2*math.pi*fr*(L/R)

#Result
print("Inductance:\nL = %f mH"%(L*1000))
print("Observed Q-factor:")
print("Q = %.2f "%Q)
# Aanswer do not match with the book
Inductance:
L = 0.386133 mH
Observed Q-factor:
Q = 97.05 

Example8_2,pg 240

In [13]:
# truncation error

import math
#Variable declaration
fs=50*10**3                      #sampling rate
delt=2.0                         #summation interval
f=50.0                           #signal frequency

#Calculations
n=(fs/delt)                      #value of samples for 2s
maxer1=100.0/(2*n)               #max error for synchronous case
maxer2=(100.0/(2*fs*delt*math.sin((2*math.pi*f)/fs)))

#Result
print("max error for synchronous case:")
print("maxer1 = %.3f%% \n"%maxer1)
print("max error for asynchronous case:")
print("maxer2 = %.2f%% "%maxer2)
max error for synchronous case:
maxer1 = 0.002% 

max error for asynchronous case:
maxer2 = 0.08% 

Example8_3,pg 258

In [26]:
# find ratio errror and phase angle

import math
#Variable declaration
#assume no iron loss and magnetizing current=1% of 10A, i.e 0.01A
Xs=1.884                   #reactance of secondary
Rs=0.5                     #resistance of secondary
Xm=2.0                     #reactance of meter
Rm=0.4                     #reactance of meter
Im=0.01                    #magnetizing current
n2=10
n1=1

#Calculations
B=math.atan((Xs+Xm)/(Rs+Rm))
#nominal ratio (n2/n1)=10/1
R=n2+((Im*math.sin(B))/n1) #actual impedance
R1=0.0097                  #practical impedance
perer=(R1/R)*100           #percentage error
theta=((Im*math.cos(B))/n2)

#Result
print("percentage error = %.3f%% \n"%perer)
print("phase angle:")
print("theta = %.5f rad"%(math.floor(theta*10**5)/10**5))
percentage error = 0.097% 

phase angle:
theta = 0.00022 rad

Example8_4,pg 499

In [29]:
# inductor Q factor and resistance

import math
#Variable declaration
Vc=100.0                   #voltage across capacitor
Vi=12.0                    #input voltage
f=100.0                    #frequency of operation
Vl=100.0                   #Vc=Vl at resonance
Ir=5.0                     #current at resonance

#Calculations
Q=(Vc/Vi)                  #Q-factor
Xl=(Vl/Ir)                 #inductive reactance
L=(Xl/(2*math.pi*f))       #inductance
Rl=(Xl/Q)                  #resistance

#Result
print("Inductance of coil:")
print("L = %.1f mH\n"%(L*1000))
print("Q-factor:")
print("Q = %.2f\n"%Q)
print("Resistance of coil:")
print("Rl = %.1f ohm"%Rl)
Inductance of coil:
L = 31.8 mH

Q-factor:
Q = 8.33

Resistance of coil:
Rl = 2.4 ohm

Example8_5,pg 499

In [32]:
# actual Q factor and resistance

import math
# Variable declaration
#when switch is open
C1=0.011*10**-6                           #capacitance-1
Q1=10.0                                   #Q-factor-1
#when switch is closed
C2=0.022*10**-6                           #capacitance-2
Q2=100.0                                  #Q-factor-2

#Calculations
Qac=((Q1*Q2)/(Q1-Q2))*((C1-C2)/C1)        #actual Q-factor
Rp=((Q1*Q2)/(Q2-Q1))*(1/(2*math.pi*C2))       #parallel resistance

#Result
print("actual Q-factor:")
print("Qac = %.2f \n"%Qac)
print("parallel resistance:")
print("Rp = %.f M-ohm"%(Rp/10**6))
actual Q-factor:
Qac = 11.11 

parallel resistance:
Rp = 80 M-ohm

Example8_6,pg 499

In [12]:
# find Q factor

import math
#Variable declaration
Cr=0.01*10**-6               #capacitance at resonance
Cu=0.014*10**-6              #capacitance at upper half
Cl=0.008*10**-6              #capacitance at lower half

#Calculations
Qac=((2*Cr)/(Cu-Cl))         #actual Q-factor

#Result
print("actual Q-factor:")
print("Qac = %.2f "%Qac)
actual Q-factor:
Qac = 3.33 

Example8_7,pg 499

In [33]:
# find lag

import math
#Variable declaration
V=10.0                    #v=10sin6280t
I=1.0                     #current peak
P=3.1                     #active power

#Calculations
phi=math.acos((P*2)/V)    #phase in radian
w=6280.0                  #v=10sin6280t
lag=(phi/w)               #lag

#Result
print("lag = %.2f ms"%(lag*10**3))
lag = 0.14 ms

Example8_8,pg 500

In [36]:
# find truncation error

import math
#Variable declaration
V=4.0                   #peak voltage
I=0.4                   #peak current
f=1*10**3               #operating frequency
fs=40*10**3             #sampling rate
delt=2.2                #time interval

#Calculations
phi=((2*math.pi*f)/fs)  #phase 
Et=(V*I*phi)/(4*math.pi*f*delt*math.sin(phi))

#Result
print("truncation error:")
print("Et = %.1f * 10^-6 "%(Et*10**6))
truncation error:
Et = 58.1 * 10^-6 

Example8_9,pg 500

In [37]:
# find frequency of PF meter

import math
#Variable declaration
ar=1.0                     #gain of rectifier
nc=40.0                    #turns ratio (1:40)
Vm=4.0                     #peak load voltage
PF=0.85                    #power factor

#Calculations
f=(1/math.pi)*ar*Vm*nc*PF #frequency

#Result
print("frequency of digital power meter:")
print("f = %.1f Hz"%f)
frequency of digital power meter:
f = 43.3 Hz

Example8_10,pg 500

In [52]:
# calculate ratio error and phase angle

import math
#Variable declaration
Rp=94.0                     #primary resistance
Xp=64.3                     #primary reactance
Rs=0.85*10**2               #secondary resistance
Im=31*10**-3                #magnetizing current
PF=0.4                      #power factor
n=10.0                      #PT ratio
Is=1.0                      #load current
Vs=110.0                    #n=(Vp/Vs)

#Calculations
B=math.acos(PF)
beta = math.floor(math.sin(B)*10)/10
R=Rp+Rs                     #total resistance
nerr=n+((((Is/n)*((R*PF)+(Xp*beta)))+Im*Xp)/Vs)
theta=((PF*(Xp/n))-(beta*(R/n))-(Im*Rp))/(Vs*n)

#Result
print("ratio error:")
print("nerr = %.3f\n"%nerr)
print("phase angle:")
print("theta = %.3f"%theta)
#Answer for theta do not match with the book
ratio error:
nerr = 10.136

phase angle:
theta = -0.015

Example8_11,pg 500

In [56]:
# calculate ratio error and phase angle

import math
#Variable declaration
n=20.0                          #(Vs/Is)
Is=5.0                          #n=(Vs/Is)
Vs=100.0                        #n=(Vs/Is)
N=0.25                          #resistance to reactance ratio
Bur=15.0                        #burden of CT=15VA (rating)
IL=0.13                         #iron loss
Im=1.3                          #magnetizing current

#Calculations
V=(Bur/Is)                      #voltage rating
B=math.atan(N)                  #cos(B)-> power factor
#B=B*(180/math.pi)               #conversion into degree
I=(Bur/Vs)                      #current rating
I1=(IL/I)
Rac=0.23                        #actual value
R=n+((I1*math.cos(B)+Im*math.sin(B))/Is)
theta=((Im*math.cos(B)-I1*math.sin(B))/Vs)
nerr=-(Rac/R)*100               #ratio error

# Result
print("ratio error:")
print("nerr = %.3f%%\n "%nerr)
print("phase angle \n")
print("theta = %.4f° "%theta)
ratio error:
nerr = -1.137%
 
phase angle 

theta = 0.0105°