Question 7.5: Consider the machine of Example 7.4 operating at 0.8 PF lead......
Consider the machine of Example 7.4 operating at 0.8 PF leading. Construct the V_{t} versus I_{a} curve by computing selected points on the curve.
From Example 7.4 at 0.8 PF leading, we found that for the rated terminal voltage, we need an excitation voltage of
E_{f}=7560.935\ \mathrm{V}Substituting in Eq. (7.28b), the characteristic is described by
V_{t}^{2}+[2]Z_{s}||I_{a}|\cos(\psi+\phi)|V_{t}+|Z_{s}|^{2}|I_{a}|^{2}-|E_{f}|^{2}=0 (7.28b)
V_{t}^{2}-2.064V_{t}I_{a}+(3.244\,I_{a}^{2}-5.7168\times10^{7})=0The first point to find is at no load, I_{\alpha}=0, where we find that
V_{\mathrm{f}}=E_{f}=7560.935~\mathrm{V}The full-load current is 418.37 A, and we now compute {\mathit{V}}_{t} for half load as
V_{t_{1/2}}=7770.5~\mathrm{V}The other root of the equation is negative. At full load, we get
V_{t_{I}}=7967.434\,\mathrm{V}For two times full load, we get
V_{t_{2}}=8322.9~\mathrm{V}Note that the terminal voltage increases with an increase in load.
The voltage, V_{t}, will be single-valued when the quadratic has two equal roots, and this takes place at
As a result, the current at the knee (k) of the curve is
I_{a_{k}}=5122.3\ \mathrm{A}This is almost 12 times full load. The corresponding value of \mathbf{}V_{t} is obtained as
V_{t_{k}}=5286.5 VThe final point to obtain is that for which V_{t}\equiv0, and is found to be at
I_{a}=4198.2{\mathrm{A}}That is about 10 times full-load current.
Important characteristics of the synchronous machine are given by the reactive capability curves. These give the maximum reactive power loadings corresponding to various active power loadings for the rated voltage operation. Armature heating constraints govern the machine for power factors from rated to unity. Field heating represents the constraints for lower power factors. Figure 7.16 shows a typical set of curves for a large turbine generator.
