The Two Wattmeter Method

Electrical Circuit Analysis > Polyphase Systems

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The power delivered to a three-phase, three-wire, $\Delta$- or Y-connected, balanced or unbalanced load can be found using only two watt-meters if the proper connection is employed and if the wattmeter readings are interpreted properly. The basic connections of this two-wattmeter method are shown in Fig. 1. One end of each potential coil is connected to the same line. The current coils are then placed in the remaining lines.

Fig. 1: Two-wattmeter method for a $\Delta$- or a Y-connected load.

Fig. 2: Alternative hookup for the two-wattmeter method.

The connection shown in Fig. 2 will also satisfy the requirements. A third hookup is also possible, but this is left to the reader as an exercise.

The total power delivered to the load is the algebraic sum of the two wattmeter readings. For a balanced load, we will now consider two methods of determining whether the total power is the sum or the difference of the two wattmeter readings.

The first method to be described requires that we know or be able to find the power factor (leading or lagging) of any one phase of the load. When this information has been obtained, it can be applied directly to the curve of Fig. 3.

Fig. 3: Determining whether the readings obtained using the two-wattmeter method should be added or subtracted.

The curve in Fig. 3 is a plot of the power factor of the load (phase) versus the ratio $P_l /P_h$, where $P_l$ and $P_h$ are the magnitudes of the lower- and higher-reading watt-meters, respectively. Note that for a power factor (leading or lagging) greater than 0.5, the ratio has a positive value. This indicates that both wattmeters are reading positive, and the total power is the sum of the two wattmeter readings; that is, $P_T = P_l + P_h$. For a power factor less than 0.5 (leading or lagging), the ratio has a negative value. This indicates that the smaller-reading wattmeter is reading negative, and the total power is the difference of the two wattmeter readings; that is, $P_T = P_h - P_l$

For a balanced system, since

$$P_T = P_h \pm P_l = \sqrt{3}E_LI_L \cos \theta_{I_\phi}^{V_\phi}$$

the power factor of the load (phase) can be found from the wattmeter readings and the magnitude of the line voltage and current:

$$ \bbox[10px,border:1px solid grey]{F_p = \cos \theta_{I_\phi}^{V_\phi} = { P_h \pm P_l \over \sqrt{3} E_L I_L}}$$

Example 1: 8 For the unbalanced Delta-connected load of Fig. 4 with two properly connected watt-meters:
a. Determine the magnitude and angle of the phase currents.
b. Calculate the magnitude and angle of the line currents.
c. Determine the power reading of each wattmeter.
d. Calculate the total power absorbed by the load.
e. Compare the result of part (d) with the total power calculated using the phase currents and the resistive elements.

Fig. 4: Example 1.

Solution:
a.

$$\begin{aligned}\mathbf{I}_{a b} &=\frac{\mathbf{V}_{a b}}{\mathbf{Z}_{a b}}=\frac{\mathbf{E}_{A B}}{\mathbf{Z}_{a b}}=\frac{208 \mathrm{~V} \angle 0^{\circ}}{10 \Omega \angle 0^{\circ}}=\mathbf{2 0 . 8} \mathbf{A} \angle \mathbf{0}^{\circ} \\\mathbf{I}_{b c} &=\frac{\mathbf{V}_{b c}}{\mathbf{Z}_{b c}}=\frac{\mathbf{E}_{B C}}{\mathbf{Z}_{b c}}=\frac{208 \mathrm{~V} \angle-120^{\circ}}{15 \Omega+j 20 \Omega}=\frac{208 \mathrm{~V} \angle-120^{\circ}}{25 \Omega \angle 53.13^{\circ}} \\&=\mathbf{8 . 3 2} \mathbf{A} \angle-173.13^{\circ} \\\mathbf{I}_{c a} &=\frac{\mathbf{V}_{c a}}{\mathbf{Z}_{c a}}=\frac{\mathbf{E}_{C A}}{\mathbf{Z}_{c a}}=\frac{208 \mathrm{~V} \angle+120^{\circ}}{12 \Omega+j 12 \Omega}=\frac{208 \mathrm{~V} \angle+120^{\circ}}{16.97 \Omega \angle-45^{\circ}} \\&=\mathbf{1 2 . 2 6} \mathbf{A} \angle \mathbf{1 6 5 ^ { \circ }}\end{aligned}$$

$$\begin{aligned}\mathbf{I}_{A a} &=\mathbf{I}_{a b}-\mathbf{I}_{c a} \\&=20.8 \mathrm{~A} \angle 0^{\circ}-12.26 \mathrm{~A} \angle 165^{\circ} \\&=20.8 \mathrm{~A}-(-11.84 \mathrm{~A}+j 3.17 \mathrm{~A}) \\&=20.8 \mathrm{~A}+11.84 \mathrm{~A}-j 3.17 \mathrm{~A}=32.64 \mathrm{~A}-j 3.17 \mathrm{~A} \\&=32.79 \mathrm{~A} \angle-5.55^{\circ}\\ \mathbf{I}_{B b} &=\mathbf{I}_{b c}-\mathbf{I}_{a b} \\&=8.32 \mathrm{~A} \angle-173.13^{\circ}-20.8 \mathrm{~A} \angle 0^{\circ} \\&=(-8.26 \mathrm{~A}-j 1 \mathrm{~A})-20.8 \mathrm{~A} \\&=-8.26 \mathrm{~A}-20.8 \mathrm{~A}-j 1 \mathrm{~A}=-29.06 \mathrm{~A}-j 1 \mathrm{~A} \\&=\mathbf{2 9 . 0 8} \mathbf{A} \angle-178.03^{\circ} \\\mathbf{I}_{C c} &=\mathbf{I}_{c a}-\mathbf{I}_{b c} \\&=12.26 \mathrm{~A} \angle 165^{\circ}-8.32 \mathrm{~A} \angle-173.13^{\circ} \\&=(-11.84 \mathrm{~A}+j 3.17 \mathrm{~A})-(-8.26 \mathrm{~A}-j 1 \mathrm{~A}) \\&=-11.84 \mathrm{~A}+8.26 \mathrm{~A}+j(3.17 \mathrm{~A}+1 \mathrm{~A})=-3.58 \mathrm{~A}+j 4.17 \mathrm{~A} \\&=\mathbf{5 . 5} \mathbf{A} \angle \mathbf{1 3 0 . 6 5}\end{aligned}$$

$$ \begin{aligned} P_{T} &=P_{1}+P_{2}=6788.35 \mathrm{~W}+379.1 \mathrm{~W} \\ &=7167.45 \mathrm{~W} \end{aligned} $$

$$ \begin{aligned} P_{T}&=\left(I_{a b}\right)^{2} R_{1}+\left(I_{b c}\right)^{2} R_{2}+\left(I_{c a}\right)^{2} R_{3}\\ &=(20.8 \mathrm{~A})^{2} 10 \Omega+(8.32 \mathrm{~A})^{2} 15 \Omega+(12.26 \mathrm{~A})^{2} 12 \Omega \\&=4326.4 \mathrm{~W}+1038.34 \mathrm{~W}+1803.69 \mathrm{~W} \\&=\mathbf{7 1 6 8 . 4 3} \mathbf{W}\end{aligned}$$

(The slight difference is due to the level of accuracy carried through the calculations.)

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