Three-Phase Electric Power

Electric Power

Electric power is defined as the rate of doing work in an electric circuit. In other words, the electrical power is the energy expanded per unit time in an electric circuit, i.e.

Electric Power,

$$\mathrm{p=\frac{Work\:Done}{Time}\:\:\:\:...(1)}$$

The electrical power is measured in watts (W).

Three Phase Electric Power

As the single phase instantaneous power (for lagging load) is given by,

$$\mathrm{p=\frac{V_{m}I_{m}}{2}\cos\varphi-\frac{V_{m}I_{m}}{2}\cos(2\omega\:t-\varphi)}$$

If the RMS values of voltage and currents are V and I respectively, then

$$\mathrm{p=VI\cos\varphi-VI\cos(2\omega\:t-\varphi)\:\:\:\:...(2)}$$

Now, consider a balanced three phase system, then the above expression can be applied for each phase of the three phases of the three-phase system. Therefore, the instantaneous power in three phases can be written as,

$$\mathrm{p_{R}=V_{ph}I_{ph}\cos\varphi-V_{ph}I_{ph}\cos(2\omega\:t-\varphi)\:\:\:...(3)}$$

$$\mathrm{p_{Y}=V_{ph}I_{ph}\cos\varphi-V_{ph}I_{ph}\cos(2\omega\:t-\varphi-120^{\circ})\:\:\:...(3)\:\:\:...(4)}$$

$$\mathrm{p_{H}=V_{ph}I_{ph}\cos\varphi-V_{ph}I_{ph}\cos(2\omega\:t-\varphi-240^{\circ})\:\:\:...(3)\:\:\:...(5)}$$

Where, Vph and Iph being the RMS values of phase voltage and current respectively.

Therefore, the total three-phase instantaneous power being,

$$\mathrm{p=P_{R}+P_{Y}+P_{H}}$$

$$\mathrm{\Rightarrow\:p=[V_{ph}I_{ph}\cos\varphi-V_{ph}I_{ph}\cos(2\omega\:t-\varphi)]+[V_{ph}I_{ph}\cos\varphi-V_{ph}I_{ph}\cos(2\omega\:t-\varphi-120^{\circ})]+[V_{ph}I_{ph}\cos\varphi-V_{ph}I_{ph}\cos(2\omega\:t-\varphi-240^{\circ})]}$$

$$\mathrm{\Rightarrow\:p=3V_{ph}I_{ph}\cos\varphi-V_{ph}I_{ph}\begin{bmatrix}\cos(2\omega\:t-\varphi) \+\cos(2\omega\:t-\varphi-120^{\circ}\+\cos(2\omega\:t-\varphi-240^{\circ} \end{bmatrix}}$$

$$\mathrm{\because\begin{bmatrix}\cos(2\omega\:t-\varphi) \+\cos(2\omega\:t-\varphi-120^{\circ}\+\cos(2\omega\:t-\varphi-240^{\circ} \end{bmatrix}=0}$$

$$\mathrm{\therefore\:p=3V_{ph}I_{ph}\cos\varphi\:\:\:...(6)}$$

$$\Rightarrow\:p=\sqrt{3}V_{L}I_{L}\cos\varphi\:\:\:...(7)$$

Where, VL and IL are the line voltage and line current respectively.

From eqns. (6) and (7), it is clear that the three phase instantaneous power is constant and does not vary with the supply frequency.

Therefore, the average three phase power will be,

$$\mathrm{P=\frac{1}{2\pi}\int_{0}^{2\pi}p\:d\omega\:t=\frac{1}{2\pi}\int_{0}^{2\pi}\sqrt{3}V_{L}I_{L}\cos\varphi\:d\omega\:t}$$

$$\mathrm{\Rightarrow\:P=\sqrt{3}V_{L}I_{L}\cos\varphi\:\:\:\:...(8)}$$

Numerical Example

Determine the power drawn by a three phase load of power factor 0.85 when it is connected to a three phase supply of line voltage 415 V, 50 Hz. The current in each line is 4 A.

Solution

The three phase power is given by,

$$\mathrm{P=\sqrt{3}V_{L}I_{L}\cos\varphi}$$

$$\mathrm{P=\sqrt{3}\times\:415\times\:4\times\:0.85=2443.85\:W}$$

Updated on: 05-Jul-2021

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