Page 111 - Handbook of Electrical Engineering

92 HANDBOOK OF ELECTRICAL ENGINEERING

4.2.1 Worked Example

An exciter has an open-circuit curve which has the following two pairs of data points.

V a1 = 2.0 V fd1 = 1.853

V a2 = 4.0 V fd2 = 3.693

Find the constants A and B in the exponential function that describes the saturation charac-
teristic of the exciter,

V a1 − V fd1 2.0 − 1.853 0.1470
= = = 0.478827
V a2 − V fd2 4.0 − 3.693 0.3070
V fd1 − V fd2 = 1.853 − 3.693 =−1.840
log 0.478827
e
B = = 0.400226
−1.840
V a1 − V fd1 0.1470
A = = = 0.070022
e BVf d1 e 0.741618

4.2.2 Worked Example

Repeat the example of 4.2.1 but assume the data are less accurate due to visual rounding errors in
V fd . Assume the data are,

V a1 = 2.0 V fd1 = 1.85 instead of 1.853
V a2 = 4.0 V fd2 = 3.70 instead of 3.693
2.0 − 1.85 0.15
V a1 − V fd1
= = = 0.5
V a2 − V fd2 4.0 − 3.70 0.30
V fd1 − V fd2 = 1.85 − 3.70 =−1.85
log 0.5
e
B = = 0.374674
−1.85
0.15
V a1 − V fd1
A = = = 0.075
e BVf d1 e 0.6931
or
V a2 − V fd2 0.15
A = = = 0.075
e BVf d2 e 0.6931

Hence an average error in V fd of 0.176% causes an error in B of 6.38% and an error in A of
7.11%. It is therefore important to carefully extract the data from the open-circuit curves to at least
the third decimal place.