a) Calculate the average power per phase delivered to the Y-connected load of Example 11.1.
b) Calculate the total average power delivered to the load.
c) Calculate the total average power lost in the line.
d) Calculate the total average power lost in the generator.
e) Calculate the total number of magnetizing vars absorbed by the load.
f) Calculate the total complex power delivered by the source.
a) From Example 11.1,V_{\phi} = 115.22 V, I_{\phi} = 2.4 A, and\theta_{\phi}= -1.19 – (-36.87)= 35.68^{\circ} .Therefore
P_{\phi}= (115.22)(2.4) \cos 35.68^{\circ}= 224.64 W.
The power per phase may also be calculated from I^{2}_{\phi},R_{\phi} or
P_{\phi}= (2.4)^{2}(39)= 224.64 W.
b) The total average power delivered to the load is P_{T}= 3P_{\phi}= 673.92 W. We calculated the line voltage in Example 11.1, so we may also use Eq. 11.36 \sqrt{3}V_{L}I_{L}\cos \theta _{\phi}:
P_{T}= \sqrt{3}(199.58)(2.4) \cos 35.68^{\circ}= 673.92 W.
c) The total power lost in the line is
P_{line}= 3(2.4)^{2}(0.8) = 13.824 W.
d) The total internal power lost in the generator is
P_{gen}= 3(2.4)^{2}(0.2) = 3.456 W.
e) The total number of magnetizing vars absorbed by the load is
Q_{T} = \sqrt{3}(199.58)(2.4) \sin35.68^{\circ} .
= 483.84 VAR.
f) The total complex power associated with the source is
S_{T} = 3S_{\phi} = -3(120)(2.4) \angle36.87^{\circ} .
= -691.20 – j518.40 VA.
The minus sign indicates that the internal power and magnetizing reactive power are being delivered to the circuit. We check this result by calculating the total and reactive power absorbed by the circuit:
P = 673.92 + 13.824 + 3.456
= 691.20 W (check),
Q = 483.84 + 3(2.4)^{2}(1.5) + 3(2.4)^{2}(0.5)= 483.84 + 25.92 + 8.64
= 518.40 VAR(check).