Simple Ampacity Calculation Question

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kwired

kwired

Electron manager
Location
NE Nebraska
Besoeker said:
For what am I being cut slack?
We use kW too............
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I guess maybe I should have looked a little harder at the OP. He did say he can calculate the demand load (which would be in kW when using the table) but I guess was asking how to determine ampacity. "but am super at how to figure the ampacity" is a little confusing but is probably a typing error or an auto-correct thing changed what he wanted to say. You are right that first step to answering that question is knowing what voltage each appliance operates at, and possibly even having to figure out maximum unbalance across three phases in some instances.

Determining allowable demand factors for cooking appliances for single phase equipment distributed across three phases can get a little tricky
 
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TrayLo1201

Member
Location
Lubbock, TX, United States
Thanks for the input everyone

Thanks for the input everyone

It is comforting in a way to know that others find this question confusing, especially me since this is my first electrical calculations class. This review is crap in my opinion, and the teaching was not much better. My first electrical class (electrical theory) I passed with flying colors with a 94 or something. This class ill be surprised if I or anyone for that matter passes. My math has always been great, hence why I pursued an electrical degree. But this teaching has been horrendous. And whomever caught my typo you are correct, it was supposed to be stumped, not super..lol. Anyway, I figured it was a 240V since its cooking equipment, but even when I divide my calculated load by 240 it does not match. Here what I have so far:

5500Wx65%=3575W for the E-ranges
6500Wx65%=4225W for the Oven
3500Wx65%=2275W for the 'Broilers'
Add together I get 10075W, if I divide by 240V I get 42 amps. Now if I were to say he meant 6500W each, 5500W each, and 3500W each that just doubles my answer to 84A. His answer key says it is 55.54 amps??? Any more input anyone? Lol
 
J

JFletcher

Senior Member
Location
Williamsburg, VA
TrayLo1201 said:
ah, so instead of doing 3 calculations with two appliances, I need to do one calculation with six appliances at the 43% demand factor?
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In this case, it seems so. Having the answer and being able to work the math backward did help. I think the question is still terrible, but all 6 appliances, as far as 220.55 is concerned, are the same type, so yeah, the 43% demand factor (for 6 appliances) of the total wattage (31000W) is what is used, divided by standard/usual voltage of those appliances, which is 240V.
 
Dennis Alwon

Dennis Alwon

Moderator
Staff member
Location
Chapel Hill, NC
Occupation
Retired Electrical Contractor
kwired said:
Now put those same loads on a three phase service/feeder and see how much your brain hertz when calculating the demand load for that service/feeder.;)
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It is not that hard. There is a good example in annex D that shows how this is done.

max number of ranges between and 2 phase 6 units / 3 = 2

2x 2= 4 Ranges @ 50% (col B)= 31000 X 50% = 15500 kw Demand

15,500/2 = 7750 kw per phase

7750 X 3 = 23,250 Kw for 3 phases-- I think this is correct
 
kwired

kwired

Electron manager
Location
NE Nebraska
Dennis Alwon said:
It is not that hard. There is a good example in annex D that shows how this is done.

max number of ranges between and 2 phase 6 units / 3 = 2

2x 2= 4 Ranges @ 50% (col B)= 31000 X 50% = 15500 kw Demand

15,500/2 = 7750 kw per phase

7750 X 3 = 23,250 Kw for 3 phases-- I think this is correct
Click to expand...
Now throw in the fact they weren't all the same kW rating.

Using the highest value for all units is the easy way out, and won't really add much overall cost to an application with only six appliances. It isn't "wrong" to calculate a larger demand then the minimum allowed, you just haven't taken full advantage of what is possible.
 
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