Question 6.5: Consider the generator data of Example 6.1. Calculate the di......

The generator operates at a power factor of 0.85 at its rated voltage of 1.0 per unit. Therefore, \phi = 31.8°. The generator full-load current is 4183.8 A = 1.0 per unit. The terminal voltage vector can be drawn to scale, the Ir drop ( = 0.0012 per unit) is added, and the vector IX_{q} = 1.8 per unit is drawn to locate the q axis. Current I can be resolved into direct axis and quadrature axis components, and the phasor diagram is completed as shown in Figure 6.12b and the values of V_{d}, I_{d}, V_{q}, I_{q}, and E are read from it. The analytical solution is as follows:
The load current is resolved into active and reactive components I_{r} = 0.85 per unit and I_{x} = 0.527 per unit, respectively. Then, from the geometric construction shown in Figure 6.13:
(δ – β) = \tan^{-1} (\frac{X_{q}I_{r} \ + \ rI_{x}}{V_{a} \ + \ rI_{r} \ – \ X_{q}I_{x}})
= \tan^{-1}(\frac{(1.8)(0.85) \ + \ (0.0012)(0.527)}{1 \ + \ (0.0012)(0.85) \ + \ (1.8)(0.527)})  = 38.14°          (6.107)
From the above calculation, resistance can even be ignored without an appreciable error. Thus, (δ – β + \phi) = 69.93°; this is the angle of the current vector with the q axis. Therefore,
I_{q} = I_{a} \cos (δ – β – \phi) = 0.343 \ pu, \ i_{q} = 0.594 \ pu
I_{d} = -I_{a} \sin (δ – β – \phi) = -0.939 \ pu, \ i_{d} = -1.626 \ pu
V_{q} = V_{a} \cos (δ – β ) = 0.786 \ pu, \ v_{q} = 1.361 \ pu
V_{d} = -V_{a} \sin (δ – β ) = -0.618 \ pu, \ v_{d} = 1.070 \ pu
The machine generated voltage is
E = V_{q} + rI_{q} – X_{d}I_{d} = 2.66 \ pu
The infinite bus voltage is simply the machine terminal voltage less the IZ drop subtracted vectorially:
V_{\infty }= V_{a}∠0° – I_{a}∠31.8° Z∠84.3° = 0.94∠ – 4.8°
The infinite bus voltage lags the machine voltage by 4.8°. Practically, the infinite bus voltage will be held constant and the generator voltage changes, depending on the system impedance and generator output.