Power Factor

All loads of the power plant can be modeled by a two-terminal network of passive elements (resistors, inductors, capacitors, without any energy sources) with complex impedance

$$Z=R+jX=\vert Z\vert e^{j\angle Z}=\vert Z\vert\angle Z$$

where $\vert Z\vert=\sqrt{R^2+X^2},\;\;\;\phi=\angle Z=\tan^{-1}(X/R)$.

As most of the loads are inductive (e.g., electric motors), i.e., $X=j\omega L$, the phase angle $\phi$ of the impedance is always positive. Assume the input voltage to the load network is:

$$\dot{V}=V_{rms}\angle 0,\;\;\;\;v(t)=\sqrt{2}V_{rms} \cos(\omega t)$$

then the current through the power transmission line and load can be found as:

$$\dot{I}=\frac{\dot{V}}{\vert Z\vert\angle \phi}=\frac{V_{rms}... ...\angle(-\phi),\;\;\;\;i(t)=\sqrt{2}I_{rms} \cos(\omega t-\phi)$$

where $I_{rms}=V_{rms}/\vert Z\vert$ is the effective current.

The instantaneous power of the load is defined as

$$p_{in}(t) \stackrel{\triangle}{=} v(t)\;i(t)=2V_{rms}I_{rms}\;\cos(\omega t)\;\cos(\omega t-\phi)$$

$$\stackrel{*}{=} 2V_{rms}I_{rms}\;\cos(\omega t)\;[\cos(\omega t)\cos\phi +\sin(\omega t)\sin\phi ]$$

$$= V_{rms}I_{rms}\cos\phi [2\cos^2(\omega t)] +V_{rms}I_{rms}\sin\phi [2\cos(\omega t)\sin(\omega t)]$$

$$= V_{rms}I_{rms}\cos\phi\; p(t)+V_{rms}I_{rms}\sin\phi\; q(t)$$

Here $p(t)\stackrel{\triangle}{=}2\cos^2(\omega t),\;\;\;\;\; q(t)\stackrel{\triangle}{=}2\cos(\omega t)\sin(\omega t)$

(* $\cos(u-v)=\cos u\;\cos v+\sin u\;\sin v$)

In particular, consider the average of $p(t)$ and $q(t)$ over a period $T$:

$$\frac{1}{T}\int_T p(t)dt=\int_T 2\cos^2(\omega t) dt =\frac{1}{T}\int_T [1+cos(2\omega t)] dt=1$$

$$\frac{1}{T}\int_T q(t)dt=\int_T 2\cos(\omega t)\sin(\omega t) dt =\frac{1}{T}\int_T \sin(2\omega t) dt=0$$

It is seen that the instantaneous power is composed of two components: the first term $p(t)=2\cos^2(\omega t)$ representing energy dissipation, and the second term $q(t)=2\cos(\omega t)\sin(\omega t)$ representing energy not dissipated but stored in the system. Instantaneous $p_{in}(t)$ can be both positive (for energy consumed by the load) and negative (for energy released by the load).

The average power over a period $T=2\pi /\omega$ is:

$$P_{av} \stackrel{\triangle}{=} \frac{1}{T}\int_0^T p_{in}(t) dt =V_{rms}I_{rms}\cos\phi \frac{1}{T}\int_0^T p(t) dt +V_{rms}I_{rms}\sin\phi \frac{1}{T}\int_0^T q(t) dt$$

$$\stackrel{*}{=} V_{rms}I_{rms}\cos\phi \; 1+V_{rms}I_{rms}\sin\phi \;0$$

$$\stackrel{**}{=} V_{rms}I_{rms}[\cos\phi\;1 +\sin\phi\; 0] =V_{rms}I_{rms}\;\cos\phi$$

• Apparent Power: $S\stackrel{\triangle}{=}V_{rms}I_{rms}$;
• Real power: $P\stackrel{\triangle}{=}V_{rms}I_{rms}\;\cos\phi=P_{av}$

  This is the first term (non-zero) of the average power expression, which is consumed by the load (dissipated, converted to heat);

• Reactive Power: $Q\stackrel{\triangle}{=}V_{rms}I_{rms}\;\sin\phi$

  This is the second term (zero) of the average power expression, which is not consumed but converted back and forth between the energy source and the energy storing elements (inductors and capacitors).

The definitions of real power $P=V_{rms}I_{rms} \cos\phi$ and reactive power $Q=V_{rms}I_{rms} \sin\phi$ above suggest that the power $S$ can be treated a complex variable:

$${\bf S}\stackrel{\triangle}{=}\dot{V}\dot{I}^* =V_{rms}\angle... ...}I_{rms}\angle \phi=V_{rms}I_{rms}(\cos\phi+j\;\sin\phi) =P+jQ$$

i.e.,

$$\left\{ \begin{array}{l} \vert{\bf S}\vert=\sqrt{P^2+Q^2}=V_{... ...}=S \\ \angle {\bf S}=\tan^{-1}(Q/P)=\phi \end{array} \right.$$

Substituting $\dot{V}=\dot{I}Z$ into the equation, we have

$${\bf S}\stackrel{\triangle}{=}\dot{V}\dot{I}^*=IZI^*=I^2_{rms}(R+jX) =I^2_{rms}R+j I^2_{rms}X$$

Comparing the two expressions for ${\bf S}$ above, we get:

$$\left\{ \begin{array}{l} P=I^2_{rms}R=V_{rms}I_{rms}\cos\phi=... ..._{rms}X=V_{rms}I_{rms}\sin\phi=S\;\sin\phi \end{array} \right.$$

Now we see that the real power $P=I^2_{rms}R$ is the power dissipated by the resistive component $R$ of the load, while the reactive power $Q=I^2_{rms}X$ is stored in and released from the reactive component (inductiv $L$ or capacitive $C$) of the load.

Improvement of Power Factor

The Power factor is defined as

$$\lambda\stackrel{\triangle}{=}\cos \phi < 1$$

which not only measures the phase difference between the voltage and current in the system, but also determines the ratio between the real power ( $P=V_{rms}I_{rms} \cos\phi$) transmitted from power generation to power consumption; and the reative power ( $Q=V_{rms}I_{rms} \sin\phi$) not contributing to power transmission. To increase the efficiency of the power transmission system, i.e., to deliver as much real power $P$ to the consumer as needed while minimizing the current (limited by the power transmission system), it is desirable to maximize the power factor $\lambda=\cos\phi$ by reducing $\phi$. As most loads are inductive (due to the coils in electric motors), the power factor can be reduced by using the shunt capacitor to cancel the inductance in the system.

Example: Assume the impedance of the inductive load is $Z=R+j\omega L$ with phase $\phi=tan^{-1} (\omega L/R)$. Find $C$ of the shunt capacitor so that $\phi=0$ and $\lambda=\cos \phi=1$.

The most straight forward way is to add a shunt capacitor in series with the inductive load, so that the inductance can be canceled by the capacitance:

$$j\omega L+\frac{1}{j\omega C}=0$$

Solving this we get $C=1/\omega^2L$. However, note that now the overall load becomes a series RCL circuit at resonance with quality factor:

$$Q=\frac{1}{R}\sqrt{\frac{L}{C}}$$

and the magnitude of the voltage across the inductor and the capacitor is $Q$ times that across the resistor:

$$\dot{V}_L=jQ\dot{V},\;\;\;\;\;\dot{V}_C=-jQ\dot{V}$$

When $X_L=\omega L \gg R$, $Q$ could be very large and the high voltages across $L$ and $C$ may be harmful.

As an alternative, the shunt capacitor $C$ can be added in parallel to the load, now the overall load becomes

$$Z'=(R+j\omega L)\; \vert\vert \;(1/j\omega C) =\frac{(R+j\om... ...L)+1/j\omega C} =\frac{R+j\omega L}{j\omega CR-\omega^2 LC+1}$$

For the new phase angle $\angle Z'$ to be zero, we need to have

$$\tan^{-1}\frac{\omega L}{R}=\tan^{-1}\frac{\omega RC}{1-\omeg... ...e.}\;\;\; \frac{\omega L}{R}=\frac{\omega RC}{1-\omega^2 LC}$$

which can be solved for $C$ to get

$$C=\frac{L}{R^2+\omega^2 L^2} =\frac{1}{R^2/L+\omega^2 L}<\frac{1}{\omega^2 L}$$

Note that another benefit of this method is that the required $C$ is smaller than the series approach.

Sometimes an unrealistically large capacitance is needed if the power factor has to be increased to 1 ($\phi=0$), while it is actually acceptable for the improved power factor to be less than 1, e.g., 0.9. In this case, the phase angle of the load is

$$\angle Z'=\tan^{-1}\frac{\omega L}{R}-\tan^{-1}\frac{\omega RC}{1-\omega^2 LC} =cos^{-1}\; 0.9$$

Solving this equation we can obtain a $C$. As now we have

$$\tan^{-1}\frac{\omega L}{R}>\tan^{-1}\frac{\omega RC}{1-\omega^2 LC}$$

i.e.,

$$\frac{\omega L}{R}>\frac{\omega RC}{1-\omega^2 LC}$$

we get an even smaller capacitance

$$C<\frac{L}{R^2+\omega^2 L^2}$$

more practically implementable.