Two wattmeter method is used for measurement of power in a balanced three-phase load supplied from a balanced three-phase system. If one of the wattmeters reads half of the other (both positive), then the power factor of the load is

Understanding the Two-Wattmeter Method in 3-Phase Power Measurement

The two-wattmeter method is a common technique used to measure the total power consumed by a balanced three-phase load connected to a balanced three-phase supply. This method involves using two wattmeters connected in a specific way to capture the power flow in the system.

Wattmeter Readings in a Balanced 3-Phase System

For a balanced three-phase system, the readings of the two wattmeters ($W_1$ and $W_2$) can be expressed in terms of the line voltage ($V_L$), line current ($I_L$), and the phase angle difference ($\varphi$) between the phase voltage and the phase current for the load. The formulas are:

- $W_1=V_L{I}_L\cos (30^{\circ }-\varphi )$
- $W_2=V_L{I}_L\cos (30^{\circ }+\varphi )$

The total power ($P_{total}$) consumed by the load is the sum of the readings of the two wattmeters:

$P_{total}=W_1+W_2=\sqrt{3}{V}_L{I}_L\cos (\varphi )$

Analyzing the Given Condition

The problem states that one of the wattmeters reads half of the other, and both readings are positive. Let's assume $W_2$ is the larger reading and $W_1$ is the smaller reading. Therefore, we have:

$W_1=\frac{1}{2}{W}_2$

Substituting the formulas for $W_1$ and $W_2$ into this condition, we get:

$V_L{I}_L\cos (30^{\circ }-\varphi )=\frac{1}{2}{V}_L{I}_L\cos (30^{\circ }+\varphi )$

We can cancel out $V_L{I}_L$ from both sides, as they are non-zero for a power system:

$\cos (30^{\circ }-\varphi )=\frac{1}{2}\cos (30^{\circ }+\varphi )$

Calculating the Power Factor

Now, we expand the cosine terms using trigonometric identities:

$\cos (A-B)=\cos A\cos B+\sin A\sin B$

$\cos (A+B)=\cos A\cos B-\sin A\sin B$

Applying these identities with $A=30^{\circ }$ and $B=\varphi$:

$\cos (30^{\circ })\cos (\varphi )+\sin (30^{\circ })\sin (\varphi )=\frac{1}{2}(\cos (30^{\circ })\cos (\varphi )-\sin (30^{\circ })\sin (\varphi ))$

We know that $\cos (30^{\circ })=\frac{\sqrt{3}}{2}$ and $\sin (30^{\circ })=\frac{1}{2}$. Substituting these values:

$\frac{\sqrt{3}}{2}\cos (\varphi )+\frac{1}{2}\sin (\varphi )=\frac{1}{2}(\frac{\sqrt{3}}{2}\cos (\varphi )-\frac{1}{2}\sin (\varphi ))$

Multiply both sides by 2:

$\sqrt{3}\cos (\varphi )+\sin (\varphi )=\frac{\sqrt{3}}{2}\cos (\varphi )-\frac{1}{2}\sin (\varphi )$

Rearrange the terms to group $\cos (\varphi )$ and $\sin (\varphi )$:

$\sqrt{3}\cos (\varphi )-\frac{\sqrt{3}}{2}\cos (\varphi )=-\frac{1}{2}\sin (\varphi )-\sin (\varphi )$

Combine like terms:

$\frac{\sqrt{3}}{2}\cos (\varphi )=-\frac{3}{2}\sin (\varphi )$

Now, we find the ratio of $\sin (\varphi )$ to $\cos (\varphi )$ (which is $\tan (\varphi )$):

$\frac{\sin (\varphi )}{\cos (\varphi )}=\frac{\sqrt{3}/2}{-3/2}$

$\tan (\varphi )=-\frac{\sqrt{3}}{3}$

The angle $\varphi$ for which $\tan (\varphi )=-\frac{1}{\sqrt{3}}$ is $\varphi =-30^{\circ }$.

The power factor (PF) is given by $\cos (\varphi )$:

$PF=\cos (\varphi )=\cos (-30^{\circ })=\cos (30^{\circ })=\frac{\sqrt{3}}{2}$

Calculating the value:

$PF\approx \frac{1.732}{2}=0.866$

Since both wattmeter readings are positive, the power factor angle $\varphi$ must be between $-90^{\circ }$ and $+30^{\circ }$ for the larger reading and between $-30^{\circ }$ and $+90^{\circ }$ for the smaller reading. Our calculated angle of $-30^{\circ }$ satisfies this condition, leading to positive readings for both wattmeters.