There are three methods which
are used for the measurement of three phase power in three phase circuits. The
three methods are:

- Three Wattmeters Method

- Two Wattmeters Method

- Single Wattmeter Method

In this post, we will
discuss on the Three Phase Power Measurement using Two Wattmeters.

**Two Wattmeters Method:**

In this method, a three
phase balanced voltage is to a balanced three phase load where the current in
each phase is assumed lagging by an angle of Ø behind the corresponding phase
voltage.

The schematic diagram for the measurement of three phase power using two wattmeters is shown below.

The schematic diagram for the measurement of three phase power using two wattmeters is shown below.

From the figure, it is
obvious that current through the Current Coil (CC) of Wattmeter W

_{1}= I_{R}, current though Current Coil of wattmeter W_{2}= I_{B}whereas the potential difference seen by the Pressure Coil (PC) of wattmeter W_{1}= V_{RB}(Line Voltage) and potential difference seen by Pressure Coil of wattmeter W_{2}= V_{BY}. The phasor diagram of the above circuit is drawn by taking VR as reference phasor as shown below.
From the above phasor
diagram,

Angle between the current I

_{R}and voltage V_{RB}= (30° - Ø)
Angle between current I

_{Y}and voltage V_{YB}= (30° + Ø)
Therefore, Active power
measured by wattmeter W

_{1}= V_{RB}I_{R}Cos (30° - Ø)
Similarly, Active power
measured by wattmeter W

_{2}= V_{YB}I_{Y}Cos(30° + Ø)
As the load is balanced,
therefore magnitude of line voltage will be same irrespective of phase taken
i.e. V

_{RY}, V_{YB}and V_{RB}all will have same magnitude. Also for Star / Y connection line current and phase current are equal, say I_{R }= I_{Y}= I_{B}= I
Let V

_{RY}= V_{YB}= V_{RB}= V_{L}_{}

Therefore,

W

_{1}= V_{RB}I_{R}Cos (30° - Ø)
= V

_{L}ICos(30° - Ø)
In the same manner,

W

_{2}= V_{L}ICos(30° + Ø)
Hence, total power measured
by wattmeters for the balanced three phase load is given as,

W = W

_{1}+ W_{2}_{}

= V

_{L}I×Cos(30° - Ø) + V_{L}I×Cos(30° + Ø)
= V

_{L}I [Cos(30° - Ø) + Cos(30° + Ø)]
= 2V

_{L}I×Cos30°CosØ ……………….[ CosC + CosD = 2Cos(C+D)/2×Cos(C-D)/2 ]
Now, suppose you are asked
to find the power factor of the load when individual power measured by the
wattmeters are given, then we should proceed as

Similarly,

W

_{1}– W_{2}= V_{L}I×Cos(30° - Ø) + V_{L}I×Cos(30° + Ø)
= V

_{L}I [Cos(30° - Ø) + Cos(30° + Ø)]
= 2V

_{L}I×Sin30°SinØ ………[ CosC - CosD = 2Sin(C+D)/2×Sin(D-C)/2 ]
=
V

_{L}ISinØ
Hence,

W

_{1}– W_{2}= V_{L}ISinØ ………………………………(2)
Dividing equation (2) by equation
(1),

Hence,

From the above equation, we can find the value
of Ø and hence the power factor Cos Ø
of the load.

Hope you understand the method of
measurement of three phase power using two wattmeter method. Now we will
consider three cases and will observe the how the individual wattmeter measures
the power in each case.

**Case1: When the power factor of load is unity.**

As the power factor of load is unity,
hence Ø = 0

Therefore,

Power measured by first wattmeter W

_{1}= V_{L}I Cos(30° - 0)
= V

_{L}I Cos30°
= 0.866 V

_{L}I
Power measured by second wattmeter W

_{2}= V_{L}I Cos(30° + 0)
= V

_{L}I Cos30°
= 0.866 V

_{L}I**Thus we see that, when the power factor of load is unity then both the wattmeter reads the same value.**

**Case2: When power factor of load is 0.5 lagging.**

As power factor is 0.5 hence CosØ = 0.5
i.e. Ø = 60°

Therefore,

Power measured by first wattmeter W

_{1}= V_{L}I Cos(30° - 60°)
= V

_{L}I Cos30° ……[Cos (-Ɵ) = CosƟ ]
= 0.5 V

_{L}I
Power measured by second wattmeter W

_{2}= V_{L}I Cos(30° + 60°)
= V

_{L}I Cos90°
= 0

**Thus we see that, when power factor of load is 0.5 lagging then power is only measured by first wattmeter and reading of second wattmeter is ZERO.**

**Case3: When power factor of load is zero.**

As power factor of load is
zero, hence CosØ = 0 i.e. Ø = 90°

Therefore,

Power measured by first wattmeter W

_{1}= V_{L}I Cos(30° - 90°)
= V

_{L}I Cos60°
= 0.5 V

_{L}I
Power measured by second wattmeter W

_{2}= V_{L}I Cos(30° + 90°)
= -V

_{L}I Cos60°
= -0.5 V

_{L}I
Thus we see that, when power factor of
load is zero then one wattmeter reads +ive while second wattmeter reads –ive.
As second wattmeter reads –ive hence wattmeter won’t read anything practically,
therefore for second wattmeter we need to interchange the leads of either
Pressure Coil or Current Coil so that second wattmeter may read value. As we
have interchanged the connection of leads of either PC or CC, hence second
wattmeter will read +ive but while calculating the total power measured we must
take the reading of second wattmeter as –ive.

**It shall be noted that when**

**60° <Ø < 90°, reading of one wattmeter will be positive while the reading of second wattmeter will be negative.**

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## 2 comments:

this is really helpful.

Good

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