Electrical calculation question

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Matt Mckenzie

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Not understanding can someone please help. four customer supplied 240 volt 2400watt base board heaters to be installed in an store. The service for the building is 120volt/208volt 3 phase. What will be the connected load of the heaters?
 
That;s one of those "separates the men from the boys" questions :D
Is this prep for a test ? It certainly incorporates some mathematical electrical knowledge.
I'm sure of few of us will accept the challenge. One key to keep in mind when you are operating 240 v resistance loads on 208 v, the "constant" is the resistance.
 
The 2400 watt base board heaters are designed to operate at 240 volts. 2400 w @ 240 volts = 10 amps.
The wattage of the heater is I^2 x resistance. 10^2 x R = 2400 w . The resistance then = 24 ohms.
Operating the heater at 208 volts, the amperage is 208 V/ 24 = 8.7 amps. wattage = 8.7^2 x 24 = 1817 watts.
Lower voltage = lower heat.
 
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The OP stated four customer-supplied heaters, as in one store. No mention was ever made of four customers IMHO.

The first answers addressed the load of each heater, figuring that someone would go on from there to add up the load of all four.

The "customer-supplied" part rationalizes why 240V heaters were being used in a 208V environment. (As if an electrician might not end up doing the same thing based on model availability.)
 
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junkhound: I took the liberty of deleting your post.
(a) I could not see where it added to the thread, only confused the issue and (b)
the answers supplied seemed to be on the right track .


Matt :from this point it is an issue as to how the heaters will be connected. As far as total load, you have a number. As far as actual load to each phase, it depends on how the single phase heaters are connected to the three phase service.
 
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