ALTERNATING GENERATORS AND SYNCHRONOUS MOTORS. 137 
From Equation (26) this factor may be calculated fairly 
exact: 
To 
ae a r, 5 
0.313 
where %”%= 3 
axrxl ax tx (u+5) 
oxo o(u + 0) 
and r= i : 
; ae xf 4dxf, abxd’ 
n m m-+b 
with 
es 0.313 ae OLB. 10": 
° 8.75 1.59.5, 8.79 9.5 X (0.28+0.45) 260-+54 10°” 
2.4X0.2 °~  2.4x(0.28-+0.2) 
eal 0.313 eee 2 a6 
*  4X1.25X9 4X 7.25X9  4X6X7.25° 84-464+15 10°” 
2 ae 5.6 ah 
10 > 
then o=1+ Fe ae 1.22, 
The value 6 = 1.25 is very close to the actual factor. 
In order to determine four points of the open cireuit char- 
acteristic, proceed as in Example 23, Par. 178. 
f 
239. Calculation of Field Flux.—By means of Equation 
(33) it has been found that for the generation of 3,500 
volts between any two phases, a useful flux of 4.12 x 10° 
lines is required, the corresponding field flux being - 
ee Oo HF 1.95 4,12 k 10° — 5.15. 10° lines. The 
useful flux varies proportionally with the voltage; that 
is, for 4,200 volts, 7, = 4.95 x 10°; or, for 3,000 volts, 
F, = 3.54 X 10°, ete. 
Assuming that the leakage factor is the same for various 
degrees of saturation, the corresponding field fluxes are: 
For 4,200 volts, 6.18 X 10°, and for 3,000 volts, 4.42 x 10° 
ete. : 
