Given a generator and a system with reactances of X_{s}=1.2 and X_{e}=0.2, both on a l00-MVA base, assume a generator terminal voltage of 0.95 p.u. The infinite-bus voltage is unknown. Find the minimum permissible output var for the p.u. output watts varying from zero to 1 in steps of 0.25.
The given parameters inserted in the static-stability limit curve equation
[Eq. (7.53)] yield
P^{2}+\left[Q-\frac{V_{t}^{2}}{2}\left(\frac{1}{X_{e}}-\frac{1}{X_{s}}\right)\right]^{2}=\left[\frac{V_{t}^{2}}{2}\left(\frac{1}{X_{s}}+\frac{1}{X_{e}}\right)\right]^{2} (7.53)
P^{2}+(Q-1.88)^{2}=6.93For P = 0, we have by simple substitution:
Q=-0.752Similarly, we get
For P = 0.25: Q = -0.740.
For P = 0.5: Q = -0.704.
For P = 0.75: Q = -0.643.
For P = 1.00: Q= -0.555.