Electrical – How to Find Angle of Armature Current

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Two identical three-phase, Y-connected, synchronous generators are connected in parallel to equally share a load of 900 kW at 11 kV and 0.8 pf lagging. The synchronous impedance of each generator is 0.5 + jl0 R/phase. The field current of one generator is adjusted so that its armature current is 25 A at a lagging power factor. Determine (a)the armature current of the other generator, (b) the power factor of each generator, (c) the per-phase generated voltage, and (d) the power angle of each generator. What is the circulating current under no load?

My question is, how do I find the complex phasor angle of the phasor current with magnitude 25 A? I tried expressing the unknown current angle as theta and then I did a KCL to express the other generator current in terms of the 25 A phasor current and then 2 KVL equations involving the generator phasor voltages plus a power balance equation to complete the number of equations needed to solve all the unknowns. My reference phasor is the terminal voltage which is $$
\frac{11000}{\sqrt{3}}V$$.

Best Answer

a-) Each generator share the load equally. Each generator's load is 450 kW.

First of all, calculate the total load current:

$$I_{load} =\frac{900000}{\sqrt{3}\times11000\times 0.8}=59\ A $$ $$\vec{I_{load}}=59\ \angle{-36.87^\circ}\ A $$

If first generator's load is 450 kW, the power factor of this generator easily calculated as follow:

$$\cos \phi_1=\frac{450000}{\sqrt{3}\times11000\times25}=0.9447$$

and the angle of power is $$\phi_1=19.13^\circ$$

The total load current is the vectorel sum of two generator's current, i.e:

$$\vec{I_{load}}=\vec{I_1}+\vec{I_2}$$

So the current of second generator is

$$\vec{I_2}=\vec{I_{load}}-\vec{I_1}=59\ \angle{-36.87^\circ}-25\angle-19.13^\circ$$

$$\vec{I_2}=36\angle{-49^\circ}\ A$$

The remaining parts of question are easily solve from here.

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