2 Synchronous Generators in parallel connection

AI Thread Summary
Two synchronous generators, G1 and G2, are connected in parallel to supply a 36-megawatt load at a power factor of 0.90 lagging. G1 operates with an excitation voltage of 40 kilovolts line-to-line, while G2 runs at unity power factor. The discussion focuses on calculating the real power dissipated by each generator, with an emphasis on the reactive power of the load being equal to that of G1. The equations for active and reactive power are highlighted, alongside the need to consider the current through G1 to determine its complex power. The conversation underscores the importance of understanding the relationships between voltage, current, and power angles in this context.
lonelypancreas
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Homework Statement


Two 3-phase 60-hertz 35-kilovolt synchronous generators (G1 and G2) each have synchronous reactances of j9 ohms per phase with negligible armature resistance. They are connected in parallel to supply a 36-megawatt load at rated voltage and power factor of 0.90 lagging. G1 has an exci-tation voltage of 40 kilovolts line-to-line. G2 is operating at unity power factor. What is the real power dissipated by each generator

Homework Equations


Power Equations

The Attempt at a Solution


I have computed the reactive power of the load which is equal to that of g1 since g2 is in unity power factor. I originally planned to get the current through gen 1 to get its complex power ie S = VI* and V is given already. However further down the drain, I acquire a real power of purely Vars in form (it has an imaginary part only)
 
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lonelypancreas said:
...negligible armature resistance...What is the real power dissipated...
 
lonelypancreas said:
What is the real power dissipated by each generator
I believe they are asking for the real power supplied by each generator.

What is the load voltage (per phase)? What are the equations for active and reactive power in terms of voltages and power angles of both machines?
 
lonelypancreas said:
exci-tation voltage of 40 kilovolts line-to-line.

Wow! That machine has some really great insulation!
 
In my opinion, the equation E1-sqrt(3).I1.jXs=Vg and E2-sqrt(3).I2.jXs=Vg and Itot=I1+I2 it is sufficient in order to find I1=function(E1real+jE1imag) and put E1real^2+E1imag^2=40^2 you can find I1.Note: you have to take into consideration I1=I1.(cosfi-jsinfi).
 
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