# 249586 Chap3B 1819

```Power & Energy
Measurements
Chapter 3
Part B
3-Phase Power System
Line current
Line voltage
Vs1=√2Vs(rms)sin(wt)
Z1
Phase
voltage
Phase
current
Vs2=√2Vs(rms)sin(wt-120º)
Vs3=√2Vs(rms)sin(wt-240º)
3-phase
Generator
Reference: http://en.wikipedia.org/wiki/Three-phase_electric_power
Z2
Z3
3-phase
2
Phase difference between the
phase voltage
3
3-Phase Voltage and Current Relationship
For star (Y) connected Load
VL = 3 VP
IL = IP
For Mesh () connected Load
VL = VP
IL = 3 IP
Example 3B.1:
A balanced star-connected load of (8+j6)Ω per phase is connected to a balanced
3-phase 400V supply. Find the phase voltage and the line current?
Solution 3B.1
The impedance per each phase is: Z p  82  62  10
Vp  VL / 3  400 / 3  231V
I p  Vp / Z p  231/10  23.1A
I L  I p  23.1A
4
Power Measurement in 3-Phase circuits
Following methods are used for measuring power in 3-load:
1. Single Wattmeter - Balanced case (equal impedances).
2. Three Wattmeters - Unbalanced case
3. Two Wattmeters - Balanced or Unbalanced case
One Wattmeter Method:
In this method, the total power is three times the power
consumed by one of the load elements.
PT = 3 × P1
Wattmeter
M
VA
IA
L
C
V
VAN
ZA
N
5
Three Wattmeters Method:
In this case, the sum of the readings of the three wattmeters
will give the total power consumed by the three-phase load.
PT = P1 + P2 + P3
IA
M
L
C
V
VA
ZA
VAN
N
VB
VC
IB
IC
M
M
ZC
L
C
V
VBN
C
V
VCN
ZB
L
6
Two Wattmeters Method:
In this case, the sum of the readings of the two wattmeters
will give the total power consumed by the three-phase load.
PT = PW1 + PW2
IA M
L
VA
W1
ZA
VAB
IB
VC
VB
ZB
ZC
VCB
IC
M
L
W2
7
3-Phase Diagram
The voltage VAB = VAN – VBN and VCB = VCN – VBN are indicated in
the phase diagram.
VAB
VAN
IA
30o

IC
VCB
30
o
VBC

VCN
VBN
The phase angle between
VAB and VAN is 300. If
current lags behind their
respective phase voltage
by , then the phase
angle between iA and vAB
is (300+).
IB
VCA
8
Two Wattmeters Method
The reading of first wattmeter W1 and second wattmeter W2 :
PW 1  VAB I A cos  AB  A  Real{v AB iA* }
PW 2  VCB I C cos CB C  Real{vCB iC* }
Derivation !!!
Thus, the power measured (PW1+PW2) is the sum of real power
consumed in the three phases.
PW 1  PW 2  Real{vAN i*A}  Real{vBN iB* }  Real{vCN iC* }
 PZ A  PZB  PZC
(Proven!!!)
9
Reading of wattmeter W1:
P1 = VAB IA cos AB-A = VL IL cos (  30º)
VAB
Reading of wattmeter W2:
P2 = VCB IC cos CB-C = VL IL cos ( – 30º)
VAN
IA
30o

IC
VCB
30
o
VBC

VCN
VBN
For balanced system, the magnitudes:
VAB = VCB = VCA = VL and
IA = IB = IC = IL
IB
VCA
10
The sum of the two reading of two wattmeters':
P1 + P2 = VL IL cos (  30 )  VL IL cos (  30 )
= 3 VL IL . cos
cos(A+B)
= cosAcosB - sinAsinB
The difference between the readings of two wattmeters':
P2 - P1 = VL IL cos (  30 )  VL IL cos (  30 )
= VL IL . sin
power factor
OR
11
Example 3B.2
On a 3-phase balanced-delta connected load supplied at
240V a.c., the two Wattmeters readings are: –1710 and
3210 watts. Find the power factor (P.F) and the current.
Solution 3B.2:
cosφ = 0.1733
P1 + P2 = 3 VL IL . cos OR P2 – P1 = VL IL . sin
IL = 20.82A
12
Exercise 3.2
(a) A balanced star-connected load of (3+j4)Ω per phase
is connected to a balanced 3-phase 200V supply. Find
the phase voltage and the line current.
(b) In a 2 wattmeter method of measurement of 3-phase
power, if the total power is 240kW and the power
factor is 0.8, calculate the reading of each wattmeter.
13
14
Variation in Wattmeter Readings
The readings of the two wattmeters are:
P1 = VL IL cos (  30 )
and
P2 = VL IL cos (  30 )
Both meters indicate equal and positive
readings: P1 = P2 = VL IL cos 30
P1 = VL IL cos (30+60) = 0

0º
0 <  < 60
60º 60 <  < 90
P. F =
cos
1
1 > P.F> 0.5
0.5
0.5 >P.F> 0
P1
+Ve
+Ve
0
-Ve
P2
+Ve
P1 = P2
+Ve
P1  P2
+Ve
+Ve
Both meters have positive readings
but not equal: P1  P2
What
happen if
 = 90º?
Readings for both meters:
P2 is +ve but P1 is -ve
15
Wattmeter Errors
 Power Loss in Pressure Coil and Fixed coil
 Moving coil Inductance: the moving coil current will tend
to lag the supply voltage.
 Moving coil Capacitance: the moving coil current will
tend to lead the supply voltage. (will not be discussed)
 Eddy Current: alternating magnetic field of the fixed coil
induces eddy current in the solid metal parts near the fixed
coil. These eddy currents set up their own magnetic field and
thus effect the magnitude and phase of the magnetic field
causing the deflection. (will not be discussed)
 Stray magnetic fields: electrodynamic wattmeter has to be
shielded for protection.
16
Error due to Moving coil Inductance
Let RM , LM and RS are the moving coil resistance, inductance and
the series connected resistance respectively.
Total resistance of pressure coil circuit is RP = RM + RS.
Hence, the impedance of the moving coil:
Wattmeter
ZM
IF
And the current:
k1iL iM
vL
 ikM1i=L
| Z M | 
ZM
XM

S
O
U
R
C
E
L IL
M
C
V
RM + j wLM
L
O
A
D
VL
RS
IM
RP
17
iM is lagging behind the vL by an angle of
  tan 1
wLM
Rp
Thus, the angle between the load current iL and iM will be ( - β).
The deflecting torque is given as:
Td  k1iL iM
vL
 k1iL
| Z M | 
VL = VM


IM
( )
vL
Td  k1iL
cos   kd iL vL' cos 
RP 
IL
Moving coil phase diagram
where vL'  2VL sin(wt   )
and iL  2I L sin(wt   )
cos  
RP
| ZM |
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The average value of deflecting torque:
1 T '
Td ( average )  kd cos   iL vL dt  kd I LVL cos  cos(   )
T 0
This gives the power metered by the wattmeter due to inductance
of the moving coil.
If the inductance LM of moving coil circuit were zero, then at all
frequencies, the phase angle between voltage vL and current iM,
namely  would be zero and average wattmeter deflection torque
would be equal to:
Td ( average )  kd I LVL cos 
which gives the true power without error.
19
With the former as metered reading and the latter as the true power that
should have been metered:
kd I LVL cos 
True Power
cos 


Metered Power kd I LVL cos  cos(   ) cos  cos(   )
Applying correction, the true power that should have been metered is:
True Power 
The term
cos 
 Metered Power
cos  cos(   )
cos 
cos  cos(   )
is called as the correction factor and the error is


cos 
Error  Metered Power  True Power  Metered Power  1 

 cos  cos(   ) 
% Error 
 cos  cos(   ) 
Metered Power  True Power
100%  
 1 100%
True Power
cos 


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Example 3B.3
The inductive reactance of the moving coil circuit of a
dynamometer wattmeter is 0.4% of its resistance at
normal frequency and the capacitance is negligible.
Calculate the percentage error and correction factor
due to reactance for loads at 0.7071 and 0.5 P.F.
lagging
21
Solution 3B.3
22
Example 3B.4
A dynamometer type wattmeter is rated at 10A and
100V with a full-scale reading of 1000W. The
inductance of the moving coil circuit is 5mH and its
total series resistance is 3000Ω. If the voltage drop
across the fixed coil of the wattmeter is negligible,
what is the error in the wattmeter at the full rated voltamperes at zero power factor?
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Solution 3B.4
Watmeter reading = VLILcosβcos(- β)
= (100)(10)cos0.03ºcos(90º- 0.03º)
= 0.5236W
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