Line Supply (Voltage)
208 Volt
Load (Desired Voltage)
Approximately 240 Volt
Load
20 Amps
Phase
single
Using table III of the Buck-Boost Section Tables, we find a .75KVA unit can boost a line voltage of 208V to a load voltage of 236V with an available load of 23.44 amps (5.53 KVA) using Diagram D.

By using Diagram D, the Voltage Ratio looks like this:
Line Voltage: 240V (Two 120VWindings in Series; 120V +120V)
Load Voltage: 272V (All Windings in Series; 120V + 120V + 16V +16V)
We can find the Actual Load Voltage when using this connection and the available line voltage of 208V by solving:
240 = 208 272 X
240(X)=272(208) X=(272 X 208) /240 X=236 Volts

Looking at the Load (Available) KVA we find:
In this example, the current flow is through the two 16-Volt Windings connected in Series (16V+16V=32V). Therefore the available Load Current of the Connection is found by:
KVA = .75KVA = 23.44 Amps
KV .032KV

Using System Voltages we can find the Available KVA by:
KVA=KV (Amps) .236KV (23.44 Amps) 5.53KVA
5.53KVA / .75KV=7.73 Times the Nameplate KVA of Selected Unit
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