Power In 3Φ Systems

POWER IN 3Φ SYSTEMS

Measurement of three phase power by one wattmeter method

Limitation of this method is that it cannot be applied on unbalanced load. So under this condition we have I₁ – I₂,- I₃= I and V₁=V₂=V₃=V. Diagram is shown below:

Two switches are given which are marked as 1-3 and 1-2, by closing the switch 1-3 we get reading of wattmeter as P1 = V1I, cos (30-4)= cos(30-6)=√3×VI. Similarly the reading of wattmeter when switch 1-2 is closed is P2 = V12, cos (30+6) = √3×VI cos(30+Φ).

Total power is P1 + P2 = 3 VI cos 0.

Three-Wattmeter Method of Three Phase Power Measurement

Three Wattmeter Method is employed to measure power in a 3 phase, 4 wire system. However, this method can also be employed in a 3 phase, 3 wire delta connected load, where power consumed by each load is required to be determined separately.

The connections for star connected loads for measuring power by three wattmeter method is shown below.

The peasure coil of all the three wattmeters namely W1, W2 and W, are connected to a common terminal known as the neutral point. The product of the phase current and line voltage represents as phase power and is recorded by individual wattmeter.

The total power in a three wattmeter method of power measurement is given by the algebraic sum of the readings of three wattmeters, i.e.,

Total Power P = W₁ + W₂+W₃

where

W₁ = V₁I₁                      

W₂ = V₂I₂

Except for 3 phase, 4 wire unbalanced load, 3 phase power can be measured by using only two wattmeter method.

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 section, 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 delta?

 

Two-Wattmeter Method of Power Measurement in a Three- Phase Circuit

The connection diagram for the measurement of power in a three-phase circuit using two wattmeters, is given in Figure 1.62. 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 and 2, are in series with the two phases, R & B, with the pressure or voltage coils being connected across R - Y and B - Y respectively. Y is the third phase, in which no current coil is connected.

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

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, W, are i So, the instantaneous power measured by the wattmeter, W1 is:

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 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 Figure 1.63, the phase currents lags the respective phase voltages by Φ=Φp, the angle of the load impedance per phase. The angle, Φ is taken as as the neutral point on the source (N), if it is assumed to be connected in star. The voltage at that point is zero (0).

The line voltage line voltage, VRY leads the respective phase voltage, VRN by 30°, and the phase voltage, VRN leads the phase current, IRN by . So, the phase difference between V and

Comments on Two Wattmeter Readings

When the balanced load is only 1 nly resistive (Φ = 0^o), i.e., power factor (cos Φ = 1.0), the readings of the two wattmeters (W₁ =W₂ x cos 30° =0.866 (+ve)), are equal and positive.

Before taking the case of purely reactive (inductive/capacitive) load, let us take first lagging power factor as (cos p=0.5), i.e., $=+60°. Under this condition,

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, o varies as stated.

As the lagging power factor decreases from 1.0 to 0.5, with o increasing from 0° to +60°, the reading of the first wattmeter, W, decreases from a certain positive value to zero (0). But the reading of the second wattmeter, W, increases from a certain positive value to positive maximum, as the lagging power factor is decreased from 1.0 to 0.866 (= cos 30°), with increasing from 0° to +30°. As the lagging power factor decreases from 0.866 to 0.5, with increasing from +30° to +60°, the reading of the second wattmeter, W2 decreases from positive maximum to a certain positive value. It may be noted that, in all these cases, W2> W1, with both the readings being positive.

If the lagging power factor is 0.0 ( = +90°), the circuit being purely inductive, the readings of the two wattmeters (W1 =- W2 x cos (30° +90°) = cos 120° = -0.5) are equal and opposite, i.e., W, is negative and W2 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 o increasing from +60° to +90°, the reading of the first wattmeter, W, decreases from zero (0) to a certain negative value, while the reading of the second wattmeter, W, decreases from a certain positive value to lower positive one. It may be noted that W2>|W|, which means that the total power consumed, i.e., (W1 + W2) is positive, with only W, being negative. The variation of two wattmeter readings as stated earlier, with change in power factor (or phase angle) is now summarized in Table 1.1. The power factor [pf] (= cos () is taken as lagging, the phase current lagging the phase voltage by the angle 6) (taken as positive), as shown for balanced star-connected load in Figure 1.64. The circuit is shown in Figure 1.62. All these are also also valid for balanced delta-connected load.

It may be noted that, if the power factor is leading (= negative (-ve)), the circuit being capacitive, the readings of the two wattmeters change with the readings interchanging, i.e., W, taking the value of W2, and vice versa. All the points as stated earlier, remain valid, with the comments as given earlier. The first one (#1) in Table 1.1 is a special case, neither lagging, nor leading, with pf = 1.0. But in second one (#2), both readings remain +ve, with W, <W2. Same is the case in fourth one (#4), where W1 is -ve and W2 is +ve, with W1| <W2, total power being positive (+ve). For third case (#3), W1 = 0.0 and W2 is +ve, with W2. For last (fifth) case (#5), W1 is - -ve and W2 is +ve, with W1| = |W2l, total power being zero (0.0).

 

Power Measurement using One Wattmeter only for a balanced load

The circuit diagram for measuring power for a balanced three-phase load is shown in Figure 1.65.

The only assumption made is that, either the neutral point on the load or source side is available. The wattmeter measures the power consumed for one phase only, and the reading is W=V_p I_p cosΦ.

The total power is three times the above reading, as the circuit is balanced. So, the load must be star-connected and of course balanced one, with the load neutral point being available. The load may also be delta-connected balanced one, if the neutral pinpoint on the source side is available. Otherwise for measuring total power for delta-connected balanced load using one wattmeter only, the connection diagram is given in Figure 1.66. The wattmeter as stated earlier, measures power for one phase only, with the total power consumed may be obtained by multiplying it by three.

Example:

Calculate the readings of the two wattmeters (W, and W2) connected to measure the total power for a balanced star-connected load shown in Figure 1.67, fed from a three- phase, 400 V balanced supply with phase sequence as R-Y-B. The load impedance per phase is (20+j15) Ω Also find the line and phase currents, power factor, total power, total reactive VA and total VA.

The power factor of the load is cos 0=cos 36.87° 0.8 lagging, with Φ=+36.87°, as the load is inductive.

Total VA =3.Vp Ip =3×231.0×9.24=6.403k VA

This can be taken as √3.VI =√3×400×9.24=6.403 k VA

Total power =3. Vp Vp Ip coso =3x231.0×9.24×0.8=5.123k W

Total reactive VA=3. Vp Ip sino=3×231.0×9.24 sin 36.87° = 3.842k VAR

The readings of the two wattmeters are:

Example:

Calculate the readings of the wattmeter (W) connected as shown in Figure 1.68 (a). The load is the same, as in Figure 1.68 (b) i.e., balanced star-connected one, with impedance of (20+j5) Ω per phase, fed from a three-phase, 400 V, balanced supply, with the phase sequence as R-Y-B.

Solution

The steps are not repeated here, but taken from previous example.

The phasor diagram is shown in Figure 1.68(b).

The phase voltage, V_RN is taken as reference.

Example

Calculate the readings of the two wattmeters (W₁ & W₂) connected to measure the total power for a balanced delta-connected load shown in Figure 1.18(a), fed from a three- phase, 200 V balanced supply with phase sequence as R-Y-B. The load impedance per phase is (14 - j14) Ω. Also find the line and phase currents, power factor, total power, total reactive VA and total VA.

The power factor of the load is cos 0=cos 45° =0.707 leading, with Φ=-45°, as the load is capacitive.

As the phase currents are balanced, the magnitude of the line current is√3 times the magnitude of the phase current, and the value is:

It has been shown that the line current, IR lags the corresponding phase current, IRY by 30°.

The sum of two readings is (3.38 +0.906) = 4.286 kW, which is same as the total power computed earlier.

Alternatively, the phase current, Ipy can be taken as reference, with the corresponding phase voltage, Vry leading the current by the angle of the load impedance, Φ45°. So, the phase current and voltage are:

The measurement of power using two wattmeters for load (unbalanced or balanced), fed from a balanced three-phase supply is discussed. Also presented are the readings of the two wattmeters for balanced load, along with the determination of the load power factor from the two readings, and some comments on the way, the two readings vary with change in power factor of the load, with the magnitude of the load impedance remaining constant.

The difference between the star and delta connection are given below in the tabulated form.