If you would show me how you figured this.
2007-01-02 10:16:59 · 6 answers · asked by cwhite1973 1 in Home & Garden ➔ Maintenance & Repairs
Add it up, 32kw. now go back to the last question I answered for you and apply the same math. Hint, these are inductive loads so you don't have to worry about start up loading.
2007-01-02 10:24:38 · answer #1 · answered by bearcat 4 · 0⤊ 0⤋
I assume that all this equipment isn't going into a dwelling unit. I believe for a commercial install the demand factor for 5 pieces of kitchen equipment is 70%.
Take 3kw + 3kw + 3kw + 20kw + 3kw = 32kw
32kw x .7 (demand factor) = 22.4kw
However, I think that the demand load needs to be at least as big as the 2 biggest pieces of equipment.
20kw + 3kw = 23kw
So, you would need to figure the demand load at 23kw for your calculations.
2007-01-02 20:43:11 · answer #2 · answered by the4biddendonut 2 · 0⤊ 0⤋
its a simple mathamatical equation first you need to change to Watts, 3kW=3000W
Demand(amps)= Watts divided by Volts
so in australia where volts is 240v it'd be
3000 / 240 = 12.5 amps
hope this helps
2007-01-02 18:45:27 · answer #3 · answered by plzsome1helpme 2 · 0⤊ 2⤋
add all the watts up and devide by the voltage applyed and that should give you the amp draw or just call up you electrical supplier and ask them . they will be glad to give you an answer because they will be selling you the juice!
2007-01-02 19:31:11 · answer #4 · answered by Anonymous · 0⤊ 1⤋
32kilo watts x 65% = 95 amps
2007-01-02 18:32:11 · answer #5 · answered by T C 6 · 0⤊ 1⤋
104 amps. 4 amps on every kw
2007-01-02 18:21:36 · answer #6 · answered by Jekyl and Hyde 2 · 0⤊ 1⤋