Measurement of Power in a Three-phase Circuit

 

In the previous lesson, the phase and line currents for balanced delta-connected load fed from a three-phase supply, along with the expression for total power, are presented. In this lesson, the measurement of total power in a three-phase circuit, both balanced and unbalanced, is discussed. The connection diagram for two-wattmeter method, along with the relevant phasor diagram for balanced load, is described.

Keywords: power measurement, two-wattmeter method, balanced and unbalanced loads, star- and delta-connections.

After going through this lesson, the students will be able to answer the following questions:

 

1. How to connect the two-wattmeter to measure the total power in a three-phase circuit - both balanced and unbalanced?

 

2. Also how to find the power factor for the case of the above balanced load, from the reading of the two-wattmeter, for the two types of connections - star and delta?

 

Two-wattmeter Method of Power Measurement in a Three-phase Circuit

Fig. 20.1 Connection diagram for tw

The connection diagram for the measurement of power in a three-phase circuit using two wattmeters, is given in Fig. 20.1. This is irrespective of the circuit connection - star or delta. The circuit may be taken as unbalanced one, balanced type being only a special case. Please note the connection of the two wattmeters. The current coils of the wattmeters, 1 & 2, are in series with the two phases, R&B, with the pressure or voltage •••••RcLcRaLaLbRbIRN1IYN1IBN1W2Y B R W1VRY•

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coils being connected across YR− and YB− respectively. Y is the third phase, in which no current coil is connected. 2W

If star-connected circuit is taken as an example, the total instantaneous power consumed in the circuit is,

NBNBNYNYNRNRviviviW′′′′′′⋅+⋅+⋅=

Each of the terms in the above expression is the instantaneous power consumed for the phases. From the connection diagram, the current in, and the voltage across the respective (current, and pressure or voltage) coils in the wattmeter, are and . So, the instantaneous power measured by the wattmeter, is, 1WNRi′NYNRRYvvv′′−=1W

()NYNRNRRYNRvviviW′′′′−⋅=⋅=1

Similarly, the instantaneous power measured by the wattmeter, is,

()NYNBNBBYNBvviviW′′′′−⋅=⋅=2

The sum of the two readings as given above is,

()()(NBNRNYNBNBNRNRNYNBNBNYNRNRiivvivivvivviWW′′′′′′′′′′′′′+⋅−⋅+⋅=−⋅+−⋅=+21

Since, or, 0=++′′′NBNYNRiii()NBNRNYiii′′′+=

Substituting the above expression for in the earlier one, NYi′

NYNYNBNBNRNRviviviWW′′′′′′⋅+⋅+⋅=+21

If this expression is compared with the earlier expression for the total instantaneous power consumed in the circuit, they are found to be the same. So, it can be concluded that the sum of the two wattmeter readings is the total power consumed in the three-phase circuit, assumed here as a star-connected one. This may also be easily proved for delta-connected circuit. As no other condition is imposed, the circuit can be taken as an unbalanced one, the balanced type being only a special case, as stated earlier.

 

Phasor diagram for a three-phase balanced star-connected circuit

The phasor diagram using the two-wattmeter method, for a three-phase balanced star-connected circuit is shown in Fig. 20.2. Please refer to the phasor diagrams shown in the figures 18.4 &18.6b. As given in lesson No. 18, the phase currents lags the respective phase voltages by pφφ=, the angle of the load impedance per phase. The angle, φ is taken as positive for inductive load. Also the neutral point on the load () is same as the neutral point on the source (), if it is assumed to be connected in star. The voltage at that point is zero (0). N′N

The reading of the first wattmeter is,

()()()ϕϕ+°⋅⋅⋅=+°⋅⋅=⋅⋅=30cos330cos,cos1ppRNRYRNRYRNRYIVIVIVIVW

The reading of the second wattmeter is,

()()()ϕϕ−°⋅⋅⋅=−°⋅⋅=⋅⋅=30cos330cos,cos2ppBNBYBNBYBNBYIVIVIVIVW

The line voltage, leads the respective phase voltage, by , and the phase voltage, leads the phase current, by RYVRNV°30RNVRNIφ. So, the phase difference between & is RYVRNI(φ+°30. Similarly, the phase difference between & in the second case, can be found and also checked from the phasor diagram. BYVBNI

 

V_BY VRYVYNIYNVBNVRNIRNΦ Φ Φ Φ-30° IBN 30°+Φ30°30°Fig. 20.2 Phasor diagram for two-wattmeter method of power measurement in a three-phase system with balanced star-connected load VYB30°

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The sum of the two wattmeter readings is,

()()[]φϕϕcos30cos2330cos30cos321⋅°⋅⋅⋅⋅=−°++°⋅⋅⋅=+ppppIVIVWW

φφcos3cos3⋅⋅⋅=⋅⋅⋅=LLppIVIV

So, () is equal to the total power consumed by the balanced load. 21WW+

This method is also valid for balanced delta-connected load, and can be easily obtained. The phasor diagram for this case is shown in the example No. 20.2.

Determination of power factor for the balanced load

The difference of the two wattmeter readings is,

()()[]φϕϕsin30sin2330cos30cos312⋅°⋅⋅⋅⋅=+°−−°⋅⋅⋅=−ppppIVIVWW

φsin3⋅⋅⋅=ppIV

If the two sides is multiplied by 3, we get

()φφsin3sin3312⋅⋅⋅=⋅⋅⋅=−⋅LLppIVIVWW

From the two expressions, we get,

φtan311212⋅=+−WWWW or, ⎥⎦⎤⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛+−⋅=−121213tanWWWWφ

The power factor, φcos of the balanced load can be obtained as given here, using two wattmeter readings.

()⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛+−⋅+=⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛+−⋅+=+=221212211311311tan11cosyyWWWWφφ

where, 21WWy=

The two relations, φcosand φsin can also be found as,

ppLLIVWWIVWW⋅⋅+=⋅⋅+=33cos2121φ and ppLLIVWWIVWW⋅⋅−=⋅−=3sin1212φ

Comments on Two Wattmeter Readings

When the balanced load is only resistive (°=0φ), i.e. power factor (0.1cos=φ), the readings of the two wattmeters ()(866.030cos21veWW+=°∝=), are equal and positive.

Before taking the case of purely reactive (inductive/capacitive) load, let us take first lagging power factor as (5.0cos=φ), i.e. °+=60φ. Under this condition,

0.01=W, as 0.090cos)6030(cos1=°=°+°∝W, and

)(866.030cos)6030(cos2veW+=°=°−°∝ is positive.

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It may be noted that the magnitudes of the phase or line voltage and also phase current are assumed to be constant, which means that the magnitude of the load impedance (inductive) is constant, but the angle, φ varies as stated.

As the lagging power factor decreases from 1.0 to 0.5, with φ increasing from to , the reading of the first wattmeter, decreases from a certain positive value to zero (0). But the reading of the second wattmeter, increases from a certain positive value to positive maximum, as the lagging power factor is decreased from 1.0 to , with °0°+601W2W)30cos(866.0°=φ increasing from to °0°+30. As the lagging power factor decreases from 0.866 to 0.5, with φ increasing from °+30 to °+60, the reading of the second wattmeter, decreases from positive maximum to a certain positive value. It may be noted that, in all these cases, , with both the readings being positive. 2W12WW>

If the lagging power factor is 0.0 (°+=90φ), the circuit being purely inductive, the readings of the two wattmeters (5.0120cos)9030(cos21−=°=°+°∝−=WW) are equal and opposite, i.e., is negative and is positive. The total power consumed is zero, being the sum of the two wattmeter readings, as the circuit is purely inductive. This means that, as the lagging power decreases from 0.5 to 0.0, with 1W2Wφ increasing from °+60 to , the reading of the first wattmeter, decreases from zero (0) to a certain negative value, while the reading of the second wattmeter, decreases from a certain positive value to lower positive one. It may be noted that °+901W2W12WW>, which means that the total power consumed, i.e., (21WW+) is positive, with only being negative. The variation of two wattmeter readings as stated earlier, with change in power factor (or phase angle) is now summarized in Table 20.1. The power factor [pf] (1Wφcos=) is taken as lagging, the phase current lagging the phase voltage by the angle, φ (taken as positive (+ve)), as shown for balanced star-connected load in Fig. 20.2. The circuit is shown in Fig. 20.1. All these are also valid for balanced delta-connected load.

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Sl.  No. Power factor [pf]  (Phase angle) Wattmeter readings (W) Remarks
		1W						2W
1.	pf = unity [1.0] (°=0φ )		+ve		+ve	21WW=
2.	0.5 < pf < 1.0  (°>>°060φ)		+ve		+ve	21WW>
3.	pf = 0.5  (°=60φ )		+ve		zero (0.0)	Total power = 1W
4.	0.0 < pf < 0.5  (°>>°0690φ)		+ve		-ve	21WW> (Total power = +ve)
5.	pf = zero [0.0]  (°=90φ )		+ve		-ve	21WW= (Total power = zero (0.0))