PREMIER TRADING CORPORATION
MEASUREMENT OF POWER IN THREE PHASE CIRCUIT
AIM : (a)
(b)
To measure power drawn by a 3 phase Resistive load using two wattmeters.
To calculate the power factor of the load from the readings of two wattmeters.
INSTRUMENTS REQUIRED :
S No.
1.
2.
3.
4.
Name
Wattmeter
Voltmeter
Ammeter
3 Phase Resistive Load
Type
Dynamometer
MI
MI
Resistive
Range
5/10A,125/250/500V
0-500V
0-10A
Quantity
2
1
1
1
THEORY
Power consumed by a 3 phase balanced or unbalanced load (star connected) can be
measured by using two Wattmeters properly connected in the load circuit. The current coils of the
wattmeters are connected in series with the load in any two lines, whereas the two pressure coils
are connected between these lines and the third line as shown in figure.
Schematic Diagram for Power measurement by two wattmeter method
The phasor diagram of this circuit, assuming balanced lagging load has been shown in
figure. As such, rms values of currents, IR, IY and IB are taken equal in magnitude and lagging by
an angle f with respect to its own phase voltage. Similarly, rms values of phase voltage are also
equal in magnitude but displaced by 120 O. The phase sequence has been assumed as RYB. Based
of the phasor diagram, power consumed and the power factor of load can be calculated from the
readings of two wattmeters W1 and W2 as explained below :(i)
Power consumed by the load :Current through the current coil of Wattmeter, W1 = IR
Voltage across the pressure coils of wattmeter, W1 = VRB
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Phase difference between IR and VRB (ref. phase diagram) = 30 - f
Hence, reading of wattmeter, W1 = IR VRB COS (30 - f)
W2 = IY VUB COS (30 - f)
Similarly, reading of wattmeter,
---------(i)
---------(ii)
Phasor diagram
Moreover, IR = IY = IB = IL (line current)
---------(iii)
Also, VRY = VRB = VBR = VL (line current)
---------(iv)
Substituting eqns (iii) and (iv) into eqns. (i) and (ii) and then adding these,
3 VL IL COS f = Power drawn by 3 ph. Load
W1 + W2 =
----------(v)
Hence, the sum of two wattmeter reading is equal to the total power by a 3 phase balanced load.
(ii)
Power factor of the load :Subtracting eqn. (ii) from eqn (i)
= VL IL COS (30 - f) – VL IL COS (30 + f)
= VL IL SIN f
W1 – W2
--------(vi)
Dividing eqn. (vi) be eqn. (v)
W1 – W2
Tan f =
3
W1 + W2
1
Thus, power factor of the load, =
1 + 3 (W1 – W2/W1 + W2)2
Hence, the power factor of the load can also be calculated from the observed readings of
the two wattmeters.
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CONCLUSIONS
Following important conclusions can be drawn from the above derivation, regarding the balanced
resistive load.
1.
When the power factor of the load is low (less that 0.5), the reading of wattmeter
W2 will be negative.
2.
When the power factor of the load is 0.5 lagging, reading of Wattmeter W2 will be
zero.
3.
When the power factor of the load is greater than 0.5, both the Wattmeter will
record positive readings.
4.
When the power factor of the load is unity, the readings of both the wattmeter will
be the same.
CIRCUIT DIAGRAM
The circuit diagram to perform the experiment has been shown as attached sheet. The
various details of the same have already been explained above.
WORKING OF CONTROL PANEL
1.
Connect the circuit as per panel layout diagram.
2.
Ensure that the output voltage of 3 phase Variac is at zero or low.
3.
Connect the Resistive load as per attached panel layout diagram. On the panel the
two wattmeter W1 & W2 and are to be externally connected as per connection
diagram.
The three-phase power measurement is conducted by connecting two wattmeters,
as shown in fig. The power value is indicated by the algebraical addition of the
indication value on the two wattmeters. When the power factor of the circuit being
measured is greater than 50%, both meters will indicated “positive” values. The
total load power is then calculated by the addition of these two values.
However, if the power factor of the circuit is less than 50%, one of the two
wattmeter will give a negative indication (the pointer will deflect to the left). If this
occurs, reverse the voltage connections of the meter with the negative deflection.
This and the meter should then indicate on the other meter, to obtain the total load
power.
In duel current range wattmeter for lower current range short the terminal B1 & B2
(series connection) & for higher current range short the terminal B1 -E1 & B2 -E2
(Parallel connection).
4.
Ensure that the resistive load is in OFF position (operated through rotary switch).
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5.
Switch-on 3 phase AC mains and increase the applied voltage till its rated value as
indicated by the voltmeter (415 V).
6.
Now switch ON the load using rotary switch. Now turn the rotary switch of the
resistive load for various load settings as indicated by the ammeter till its rated
load current.
7.
Take-down the readings of all the meters i.e. voltmeter, Ammeter & Wattmeter for
various load setting of resistive load.
8.
Decrease the load gradually by turning the rotary switch.
9.
Switch-off the supply and turn the rotary switch of the load.
OBSERVATIONS :
S No.
V
May be tabulated as follows.
I
W1
W2
W1 +W2
W1 -W2
COS f
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CONNECTION DIAGRAM FOR MEASUREMENT OF POWER BY TWO
WATTMETER METHOD
WATTMETERS
The three-phase power measurement is conducted by connecting two wattmeter, as shown
in above Fig. The power value is indicated by the algebraically addition of the indication value on
the two wattmeter. When the power factor of the circuit being measured is greater than 50%, both
meters will indicated “positive” values. The total load power is then calculated by the addition of
these two values.
However, if the power factor of the circuit is less than 50%, one of the two wattmeter will
give a negative indication (the pointer will deflect to the left). If this occurs, reverse the voltage
connections of the meter with the negative deflection. This and the meter should then indicate on
the other meter, to obtain the total load power.
In duel current range wattmeter for lower current range short the terminal B1 & B2 & for
higher current range short the terminal B1 & E2.
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