Power Calculation of Y Connected Balanced Load

Electrical Circuit Analysis > Polyphase Systems

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Please refer to Fig. 1 for the following discussion

Fig. 1: Y-connected balanced load.

Average Power The average power delivered to each phase can be determined by,

$$P_{\phi}=V_{\phi} I_{\phi} \cos \theta_{I_{\phi}}^{V_{\phi}}=I_{\phi}^{2} R_{\phi}=\frac{V_{R}^{2}}{R_{\phi}}$$

where $ \theta_{I_{\phi}}^{V_{\phi}} $ indicates that $ \theta $ is the phase angle between $ V_{\phi} $ and $ I_{\phi} $. The total power to the balanced load is

$$P_{T}=3 P_{\phi}$$

or, since

$$V_{\phi}=\frac{E_{L}}{\sqrt{3}} $ \text{ and } $ I_{\phi}=I_{L} $$

then

$$P_{T}=3 \frac{E_{L}}{\sqrt{3}} I_{L} \cos \theta_{I_{\phi}}^{V_{\phi}}$$

But

$$ \left(\frac{3}{\sqrt{3}}\right)(1)=\left(\frac{3}{\sqrt{3}}\right)\left(\frac{\sqrt{3}}{\sqrt{3}}\right)=\frac{3 \sqrt{3}}{3}=\sqrt{3} $$

Therefore,

$$P_{T}=\sqrt{3} E_{L} I_{L} \cos \theta_{\Phi_{\phi}}^{V_{\phi}}=3 I_{L}^{2} R_{\phi}$$

Reactive Power The reactive power of each phase (in volt-amperes reactive) is

$$Q_{\phi}=V_{\phi} I_{\phi} \sin \theta_{I_{\phi}}^{V_{\phi}}=I_{\phi}^{2} X_{\phi}=\frac{V_{X}^{2}}{X_{\phi}}$$

The total reactive power of the load is

$$Q_{T}=3 Q_{\phi} \quad \text { (VAR) }$$

or, proceeding in the same manner as above, we have

$$Q_{T}=\sqrt{3} E_{L} I_{L} \sin \theta_{I_{\phi}}^{V}=3 I_{L}^{2} X_{\phi}$$

Apparent Power The apparent power of each phase is

$$S_{\phi}=V_{\phi} I_{\phi}$$

The total apparent power of the load is

$$S_{T}=3 S_{\phi}$$

or, as before,

$$S_{T}=\sqrt{3} E_{L} I_{L}$$

Power Factor The Power factor of th system is given by

$$ \bbox[10px,border:1px solid grey]{F_p = { P_T \over S_T} = \cos \theta_{I \phi}^{V\phi} \, \text{(leading or lagging)}}$$

Example 1: For the Y-connected load of Fig. 2:

Fig. 2: Example 1.

a. Find the average power to each phase and the total load.
b. Determine the reactive power to each phase and the total reactive power.
c. Find the apparent power to each phase and the total apparent power.
d. Find the power factor of the load.
Solution:
a. The average power is

$$ \begin{aligned} P_{\phi} &=V_{\phi} I_{\phi} \cos \theta_{I_{\phi}}^{V_{\phi}}=(100 \mathrm{~V})(20 \mathrm{~A}) \cos 53.13^{\circ}=(2000)(0.6) \\ &=1200 \mathrm{~W} \\ P_{\phi} &=I_{\phi}^{2} R_{\phi}=(20 \mathrm{~A})^{2}(3 \Omega)=(400)(3)=1200 \mathrm{~W} \\ P_{\phi} &=\frac{V_{R}^{2}}{R_{\phi}}=\frac{(60 \mathrm{~V})^{2}}{3 \Omega}=\frac{3600}{3}=1200 \mathrm{~W} \\ P_{T} &=3 P_{\phi}=(3)(1200 \mathrm{~W})=\mathbf{3 6 0 0} \mathrm{W} \\ \text { or } \\ P_{T} &=\sqrt{3} E_{L} I_{L} \cos \theta_{I_{\phi}}^{V}=(1.732)(173.2 \mathrm{~V})(20 \mathrm{~A})(0.6)=\mathbf{3 6 0 0} \mathrm{W} \end{aligned} $$

b. The reactive power is

$$\begin{aligned}Q_{\phi} &=V_{\phi} I_{\phi} \sin \theta_{I_{\phi}}^{V_{\phi}}=(100 \mathrm{~V})(20 \mathrm{~A}) \sin 53.13^{\circ}=(2000)(0.8) \\&=1600 \mathrm{VAR} \\\text { or } \quad & Q_{\phi}=I_{\phi}^{2} X_{\phi}=(20 \mathrm{~A})^{2}(4 \Omega)=(400)(4)=1600 \mathrm{VAR} \\& Q_{T}=3 Q_{\phi}=(3)(1600 \mathrm{VAR})=\mathbf{4 8 0 0 \mathrm { VAR }}\end{aligned}$$

$$Q_{T}=\sqrt{3} E_{L} I_{L} \sin \theta_{I_{\phi}}^{V_{\phi}}=(1.732)(173.2 \mathrm{~V})(20 \mathrm{~A})(0.8)=\mathbf{4 8 0 0} \mathrm{VAR}$$

c. The apparent power is

$$ S_{\phi}=V_{\phi} I_{\phi}=(100 \mathrm{~V})(20 \mathrm{~A})=\mathbf{2 0 0 0} \mathrm{VA} $$

$$S_{T}=3 S_{\phi}=(3)(2000 \mathrm{VA})=\mathbf{6 0 0 0} \mathrm{VA} $$

$$ S_{T}=\sqrt{3} E_{L} I_{L}=(1.732)(173.2 \mathrm{~V})(20 \mathrm{~A})=\mathbf{6 0 0 0} \mathrm{VA} $$

d. The power factor is

$$F_{p}=\frac{P_{T}}{S_{T}}=\frac{3600 \mathrm{~W}}{6000 \mathrm{VA}}=\mathbf{0 . 6} \text { lagging }$$

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