AC MACHINES-1 (66761) Theory

Example 1. A 15 kVA, 2000/200 V transformer has an iron loss of 250 W and full-load copper loss 350 W. During the day it is loaded as follows :

Calculate the all-day efficiency.

Solution. Rating of transformer = 15 kVA

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3. Example: The transformer of example 18 operates with the following loads during a 24-hr period: 1 ½ times rated kva, power factor = 0.8, 1hr; 1 ¼ times rated kva, power factor = 0.8, 2hr; rated kva, power factor = 0.9, 3hr; ½ rated kva, power factor = 1.0, 6hr; ¼ rated kva, power factor = 0.8; no-load, 4hr. Calculate the all-day efficiency.

. Solution: Energy output, kw-hr Energy losses, kw-hr W1 = 1.5 x 5 x 0.8 x = 6.0 (1 ½)2 x 0.112 x 1 = 0.252 W2 = 1.25 x 0.8 x 2 = 10.0 (1 ½)2 x 0.112 x 2 = 0.350 W3 = 1 x 5 x 0.9 x 3 = 13.5 1 x 0.112 x 3 = 0.336 W6 = 0.5 x 5 x 1.0 x 6 = 15.0 (1/2)2 x 0.112 x 6 = 0.168 W8 = 0.25 x 5 x 1.0 x 8 = 10.0 (1/4)2 x 0.112 x 8 = 0.056 ____ Total. . . . . . . . 54.5 Iron = 0.04 x 24 = 0.960 _____ Total. . . . . . . . .. . . . 2.122 All-day Efficiency = (1 – 2.122/54.5 + 2.122) x 100 = 96.25%