Re: Fastened in place appliances
Bryan, I agree that the most common way to interpret this it that you apply the demand and then add 25%. But that still only gets us to 100%
Figure it this way and see what i mean...
4500 va Water Heater X 75% = 3375
1500 va Dishwasher X 75% = 1125
1000 va Disposal X 75% = 750 - Largest Rated Motor
800 va Compactor X 75% = 600
TOTAL VA demand = 5850 VA
1000 va largest rated motor X 25% = 250 va
if you add the disposal demand to the extra 25% we added for 430.24, we only equal 1000 va.

That is why in my original post i suggested adding 50% to achieve a full 125%
if i were doing an actual calculation for an actual installation, err on the high side. we are only talking about a few amps, which most likely will not affect wire and service size, (200 amps is minimum in our jurisdiction anyway). Then it is only a very isolated incident, as the AC is generally going to be the largest rated motor load, making the discussion moot. however, when teaching apprentices how to do the calc, and preparing them for the journeyman exam, how is it going to be interpreted on the test? is the most widely accepted method going to fly? to me, the code does not address the specifics of 'what if' in this situation, and someone ages ago, said that by applying a demand factor of 75% IF there are 4 or more fastened in place, and IF one of them is also your highest rated motor, that you add 25% of the FLC of that motor, thinking that 75% demand is the same as 100% continuous duty, and adding 25% is equal to 125% of the motor under a demand factor.
[ April 19, 2004, 09:40 AM: Message edited by: bradlittle ]