Three Phase Systems
9.11
295
MEASUREMENT OF POWER
Star-Connected Balanced Load with Neutral Point
Power can be measured in this case by connecting a single wattmeter with its
current coil in one line and the pressure coil between the line and the neutral
point as shown in Fig. 9.16. The reading of the wattmeter thus connected, gives
the power per phase.
Total power = 3 ´ power per phase
= 3 ´ wattmeter reading
R
L
W
R 3-phase
balanced
load
N
3-phase
ac supply
R
L
R
L
Y
B
Fig. 9.16
Measurement of power by single wattmeter
Measurement of Power by Two Wattmeter Method
Power in a 3-phase three wire system, with balanced or unbalanced load can be
measured by using two wattmeters. The load may be star or delta connected.
The current coils of the two wattmeters are connected in any of the two lines
and the pressure coils are connected between these lines and the third line, as
shown in Fig. 9.17. Let eRN,, eYN and eBN be the voltages across the three phases of the load and iR, iY and iB the currents flowing in the three lines.
Total instantaneous power in the load = eRN iR + eYN iY + eBN iB
Instantaneous current through the current coil of wattmeter W1 = iR
Instantaneous voltage across the pressure coil of wattmeter W1
eRB = eRN – eBN
Instantaneous power measured by wattmeter W1 = iR ´ (eRN – eBN)
= iR ´ eRB
Instantaneous current through current coil of wattmeter W2 = iY
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296
R
iR
W1
eRN
Z1
3-phase
load
3-phase
ac supply
N
Z2
eYN
W2
Y
B
Fig. 9.17
Z3
eBN
iY
iB
Measurement of power in 3-phase, three wire system
Instantaneous voltage across the pressure coil of wattmeter W2
eYB = eYN – eBN
Instantaneous power measured by W2 = iY ´ (eYN – eBN)
= iY ´ eYB
Total instantaneous power = eRN iR + eYN iY + eBN iB
Moreover, for 3-phase, 3-wire system, iR + iY + iB = 0
(9.18)
\
iB = – (iR + iY)
Substituting the value of iB in Eq. (9.18)
Total instantaneous power = eRN iR + eYN iY + eBN (– iR – iY)
= iR (eRN – eBN) + iY (eYN – eBN)
= Power measured by W1 + Power measured
by W2
Hence at any instant, the total power is equal to the sum of the two wattmeter
readings. However, the inertia of the moving system of the wattmeters causes it
to indicate the mean value of power taken over a cycle. Thus the sum of the
readings of the two wattmeters gives the average value of the total power fed to
the three phase load. No particular conditions have been assumed, while deriving the above relationship, hence, the result holds good for balanced as well as
unbalanced load. Though a star-connected load has been assumed in the above
derivation, but the same result will be obtained by taking a delta-connected
load.
Three Phase Systems
297
Measurement of Power by Three Wattmeter Method
In case the supply is 3-phase, 4-wire system, then for unbalanced loads the two
wattmeter method cannot be used as there is a current flowing in the neutral. For
such a system, three wattmeters are to be used to measure the total
3-phase power. Each wattmeter is connected with its current coil in series with
the line and the pressure coil connected between the phase and neutral as shown
in Fig. 9.18.
Total Power = PR + PY + PB
= WR + WY + WB
Thus, the addition of the readings of three wattmeters will give the total
power consumed by the load.
R
IR
WR
3-phase
ac supply
VRN
3-f
unbalanced
load
N
N
WY
IY
Y
VBN
VYN
WB
B
Fig. 9.18
IB
Measurement of power by three wattmeter method
9.12 MEASUREMENT OF POWER AND POWER FACTOR
IN THREE-PHASE SYSTEM WITH BALANCED LOAD
USING TWO WATTMETER METHOD
Figure 9.17 shows 3-phase star-connected balanced load, supplied from a
3-phase supply system with two wattmeters properly connected in the circuit to
measure the input power to the load. Let IR, IY and IB be the rms values of the
currents in the lines and VRN, VYN and VBN the rms values of voltages across the
phases. The phasor diagram of such a circuit with an assumption of lagging current has been shown in Fig. 9.19. The load being balanced, currents IR, IY and IB
are taken equal in magnitude and lagging by an angle f with respect to its own
phase voltage. Similarly, phase voltage VRN, VYN and VBN are equal in magnitude
but displaced by 120°, the phase sequence being RYB.
Current through the current coil of wattmeter W1 = IR
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298
–V B
N
VRB
VRN
IR
f
VYB
IB
VBN
Fig. 9.19
f
f
30°
IY
N
–V B
VYN
Phasor diagram for measurement of power
Voltage across the pressure coil of wattmeter W1 = VRN – VBN
= VRB
Referring to the phasor diagram, Fig. 9.19, phase difference between
IR and VRB = (30 – f)
(9.19)
Thus, the reading of wattmeter W1 = IR VRB cos (30 – f)
Current through the current coil of wattmeter W2 = IY
Voltage across the pressure coil of wattmeter W2 = VYN – VBN = VYB
Referring to the phasor diagram (Fig. 9.19), the phase difference between the
current IY and voltage VYB = (30 + f).
Thus, reading of wattmeter W2 = IY VYB cos (30 + f)
As the load is balanced, IR = IY = IB = IL
VRY = VYB = VBR = VL
Substituting these in Eqs 9.19 and 9.20,
W1 = VL IL cos (30 – f)
W2 = VL IL cos (30 + f)
Adding Eqs. (9.21 and 9.22)
W1 + W2 = VL IL cos (30 – f) + VL IL cos (30 + f)
= VL IL [cos 30° cos f + sin 30° sin f + cos 30° cos f
– sin 30° sin f]
(9.20)
(9.21)
(9.22)
= 3 VL IL cos f
= Total power input to a balanced load.
Hence, the sum of the readings indicated by the two wattmeters connected as
per Fig. 9.17 is equal to the total power drawn by a 3-phase balanced load.
Now,
W1 + W2 = 3 VL IL cos f
(9.23)
Three Phase Systems
Subtracting Eq. (9.22) from Eq. (9.21),
W1 – W2 = VL IL cos (30 – f) – [VL IL cos (30 + f)]
= VL IL [cos 30° cos f + sin 30° sin f – cos 30° cos f
+ sin 30° sin f]
W1 – W2 = VL IL sin f
Dividing Eq. (9.24) by Eq. (9.23),
W1 - W2
=
W1 + W2
tan f =
Power factor, cos f =
=
(9.24)
V L I L sin f
3 V L I L cos f
3
FG W - W IJ
HW +W K
1
2
1
2
(9.25)
1
1
=
sec f
cos f =
299
sec 2 f
1
1 + tan 2 f
1
L W - W OP
1 + 3M
NW + W Q
1
2
1
2
2
(9.26)
Hence, power factor can be calculated using Eq. (9.26) from the readings of
two wattmeters. Equation (9.26) may also be put in the following form,
1
cos f =
2
W
1- 2
W1
1+ 3
W2
1+
W1
F
GG
GH
cos f =
I
JJ
JK
1
L1 - r OP
1+ 3 M
N1 + r Q
2
(9.27)
W2
W1
A graph of power factor vs the ratio of wattmeter reading using Eq. (9.27),
has been drawn in Fig. 9.20 for r varying from +1 to –1. This graph can also be
used for finding out the value of power factor, corresponding to a particular ratio
of wattmeter readings.
The following important conclusions can be drawn from the graph of power
factor plotted in Fig. 9.20.
where r = ratio of wattmeter readings =
(i) Reading of wattmeter W2 is zero, when the load power factor is 0.5 lagging, i.e. f = 60°.
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300
1.0
0.8
Ratio of wattmeter reading
0.6
0.4
0.2
0
–0.2
–0.4
–0.6
–0.8
–1.0
Fig. 9.20
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Power factor
Relation between pf and ratio of wattmeter readings
(ii) Reading of wattmeter W2 is negative, for the load power factor less than
0.5 lagging, i.e. for f > 60°. In such a case it is necessary to reverse the
connections to either the current or pressure coil, in order to measure the
power registered by wattmeter W2. However, the reading thus obtained
must be taken as negative, while calculating the total power and the
power factor.
(iii) The reading of wattmeter W2 will be positive, when the load power factor is greater than 0.5 lagging, i.e. f < 60°.
(iv) Both the wattmeters will indicate the same readings, when the power factor of the load is unity, i.e. f = 0.
Examples on Power Measurement
Example 9.8 Two wattmeters connected to measure the power input to a
3-phase circuit indicate 15 kW and 1.5 kW respectively, the latter reading being
obtained after reversing the current coil connections. Calculate the power and
power factor of the load.
Solution:
Reading of first wattmeter, W1 = 15 kW
Reading of second wattmeter has been obtained after reversing the current coil
connections. As such this reading is really negative, because of reversing the current coil connections.
Thus reading of second wattmeter, W2 = – 1.5 kW
Total power fed to the load = W1 + W2
= 15 + (– 1.5) = 13.5 kW
Three Phase Systems
Power factor of the load can be calculated from the above reading by first finding out tan f, that is tangent of the power factor angle.
tan f =
3
W1 - W2
W1 + W2
=
3
15 - ( - 15
.)
15 + ( - 15
.)
=
3´
16.5
= 2.117
.
135
Hence power factor angle, f = 64.7°
Power factor of the load, cos f = 0.427
Example 9.9 Two wattmeters have been used to measure the power input to
a 150 kW, 440 V, 3-phase slip ring induction motor running at full load. The
wattmeter readings are 115 kW and 50 kW. Calculate (i) the input to the motor,
(ii) power factor of the motor, (iii) line current drawn by the motor and (iv) efficiency of the motor.
Solution:
(i) Power input to the motor = W1 + W2 = 115 + 50 = 165 kW
(ii) tan f =
3
W1 - W2
=
W1 + W2
3
115 - 50
= 0.682
115 + 50
Power factor angle, f = 34.3°
Power factor, cos f = 0.826
(iii) Power input to the motor =
=
Thus line current, IL =
3 ´ VL ´ IL ´ cos f
3 ´ 440 ´ IL ´ 0.826
165 ´ 10 3
= 262.1 A
3 ´ 440 ´ 0.826
(iv) Output of the motor = 150 kW
Input to the motor = 165 kW
output
150
=
165
input
= 0.909 = 90.9%
Thus, the efficiency of the induction motor =
Example 9.10 A balanced star-connected load is supplied from a symmetrical, 3-phase, 440 V, 50 Hz supply system. The current in each phase is 20 A and
lags behind its phase voltage by an angle 40°. Calculate (i) phase voltage, (ii) load
parameters, (iii) total power and (iv) readings of two wattmeters, connected in the
load circuit to measure the total power.
Solution: Figure 9.17 shows a balanced star-connected load connected across a
3-phase supply.
(i) Line voltage, VL = 440 V
V
440
Phase voltage in star-connected circuit, Vph = L =
= 254 V
3
3
301
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302
(ii) Current in each phase = 20 A
Impedance of the load per phase, Zph =
Zph =
Vph
254
=
= 12.7 W
Iph
20
R2 + X 2 = 12.7
(i)
Moreover, the current in each phase lags behind its voltage by 40°, hence
X
tan 40° =
R
or Inductive reactance of the load, X = 0.839 R
Substituting the value of X in Eq. (i)
R2 + (0.839)2 R2 = (12.7)2
12.7
Resistance of the load, R =
1 + (0.839 )2
= 9.73 W
Inductive reactance of the load, X = 0.839 ´ 9.73 = 8.16 W
(iii) Power consumed by each phase = VPh Iph cos f
= 254 ´ 20 ´ cos 40°
= 3891.5 W
Total power = 3 ´ 3891.5 = 11674.5 W
= 11.6745 kW
(iv) Total power = W1 + W2 (reading of two wattmeters)
Thus
W1 + W2 = 11.6745
W - W2
Also
tan 40° = 3 1
W1 + W2
or
W 1 – W2 =
(ii)
11.6745 ´ 0.839
3
W1 – W2 = 5.656
Solving Eqs (ii and iii), W1 = 8.665 kW
W2 = 3.009 kW
(iii)
Example 9.11 Power input to a 3-phase 440 V, 37.3 kW induction motor,
whose efficiency and power factor are respectively 88 percent and 0.82, is to be
measured by two wattmeter method. Find the reading of both the wattmeters and
the full load line current drawn by the motor.
Solution:
Power output of 3-phase induction motor = 37.3 kW
Efficiency of the motor = 88 %
output
37.3
Power input to the motor =
=
= 42.386 kW
0.88
efficiency
Power input when measured by two wattmeter method = W1 + W2
where, W1 and W2 are the readings of the two wattmeters.
Hence
W1 + W2 = 42.386
Power factor of the motor, cos f = 0.82
Power factor angle, f = cos–1 0.82 = 34.91°
(i)
Three Phase Systems
303
In power measurement by two wattmeter method,
W - W2
tan f = 3 1
W1 + W2
W1 - W2
42.386
42.386 ´ 0.698
W1 – W2 =
= 17.081 kW
3
tan 34.91° =
Thus,
3
(ii)
Solving Eqs (i and ii)
2 W1 = 42.386 + 17.081
59.467
W1 =
= 29.734 kW
2
W2 = 42.386 – 29.734 = 12.652 kW
Hence the readings of two wattmeters are 29.734 and 12.652 kW.
power inpu t
Full load line current drawn by the motor =
3 ´ VL ´ cos f
42.386 ´ 10 3
3 ´ 440 ´ 0.82
= 67.8 A
=
Example 9.12 Three inductive coils, each having a resistance of 20 W and
reactance of 15 W are connected (i) in star (ii) in delta, across a 3-phase, 400 V,
50 Hz supply. Calculate in each case, the readings on two wattmeters connected
in the circuit to measure the power input. Also determine the phase and line currents in each case.
Solution:
(i) Three coils with resistance 20 W and reactance 15 W each, are connected
in star configuration.
V
400
= 231 V
Phase voltage, Vph = L =
3
3
Impedance of each coil, i.e. of each phase, Zph =
(20 )2 + (15)2
= 25 W
Phase current =
Vph
231
=
= 9.24 A
Zph
25
Line current = phase current (for star–connected circuits)
= 9.24 A
Power factor of the circuit, cos f =
Power input to the circuit, P =
Rph
20
=
= 0.8
Zph
25
3 ´ VL ´ IL cos f
= 3 ´ 400 ´ 9.24 ´ 0.8 = 5121 W
When two wattmeters are connected in the circuit to measure the power
input to the circuit, then,
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304
Power input = W1 + W2 (W1 and W2 are the readings of two wattmeters)
Hence
W1 + W2 = 5121
(i)
Power factor of the circuit as calculated from the parameters of the circuit,
cos f = 0.8
f = 36.87°
or
However, when power is measured by two wattmeters
W1 - W2
tan f = 3
W1 + W2
tan 36.87° =
or
W1 – W2 =
3
W1 - W2
5121
0.75 ´ 5121
= 2217.5
3
(ii)
Solving Eqs (i and ii)
5121 + 2217.5
W1 =
= 3669.25 W
2
W2 = 1451.75 W
Hence the readings of two wattmeters in case of star connection are
3669.25 and 1451.75 W.
(ii) Now the same coils are connected in mesh across a 3-phase, 400 V, 50 Hz
supply.
Thus, phase voltage = 400 V (equals line voltage)
Impedance per phase, Zph = 25 W
400
Current per phase =
= 16 A
25
Line current =
3 ´ phase current (in delta-connected circuits)
= 3 ´ 16 = 27.72 A
Power factor of the delta connected circuit = 0.8 (remains same)
Power input to the circuit =
=
Thus,
3 VL IL cos f
3 ´ 400 ´ 27.72 ´ 0.8 = 15362 W
W¢1 + W¢2 = 15362
W ¢- W2¢
tan f = 3 1
W1¢+ W2¢
0.75 =
W1¢ – W¢2 =
3
(iii)
W1¢- W2¢
15362
0.75 ´ 15362
3
W1¢ – W¢2 = 6652.5
Solving Eqs (iii and iv),
W¢1 = 11007.75 W
W¢2 = 4315.25 W
Readings on two wattmeters are, 11007.75 and 4315.25 W.
(iv)