2007-03-20 12:45:17 · 5 answers · asked by ? 2 in Science & Mathematics ➔ Engineering
A typical midwest home (2200 ft²) with central air conditioning and gas heat and hot water has a peak load of around 6 kVA.
75 kVA / 6 kVA per home = 12.5 homes.
Assuming this is a liquid filled distribution transformer, most utilties will load the transformer beyond nameplate during peak conditions. A conservative overload rating might be 120% of nameplate. This means the 75 kVA transformer could serve 75 x 1.20 = 90 kVA under peak conditions and still achieve a normal life.
90 kVA / 6 kVA per home = 15 homes.
Larger homes will have a higher load. For example a 4000 ft² home will have a peak load of around 9 kVA.
Notice that these numbers are substantially llower than the values you would expect for homes with a 200 or 400 amp service panel. This is because there is a HUGE difference between connected load and the actual coincident loads as measured at the electric meter and seen by the transformer.
Note: If this is a dry-type transformer, the allowable loading is closer to the nameplate rating.
As a side note, the proper designation is kVA (small k, capital VA):
k is the prefix for 1000
V is capitalized to pay homage to Alessandro Volta
A is capitalized to pay homage to André-Marie Ampère
2007-03-20 15:20:25 · answer #1 · answered by Thomas C 6 · 2⤊ 0⤋
75 Kva Transformer
2016-10-01 10:46:47 · answer #2 · answered by ? 4 · 0⤊ 0⤋
It depends on the amount of amperage each item is drawing.
If you divide 75000 VA by 120V you get 625A.
By code, only load the transformer to 80% so .8 * 625A = 500A.
So 500A worth of load.
2007-03-20 12:54:46 · answer #3 · answered by Nick J 2 · 0⤊ 1⤋
Call it 2 houses
2007-03-20 12:59:59 · answer #4 · answered by Anonymous · 0⤊ 1⤋
750 100 watt light bulbs.
2007-03-20 12:53:22 · answer #5 · answered by John S 6 · 0⤊ 1⤋