Ok. I'm doing a design for installation for quite a bit of PTAC units into an existing building (hotel). The PTAC unit has a 19.6 MCA (208V single phase). So for example I want to put 15 of these PTAC units on a three phase panel. If I balance the loads (theoretically, of course they all could be on or only 1 could be on) I have 196A on each leg. In regards to 215.2 (1) do I have to size the conductors for 125 percent of this load or not? Basically I don't really know whether the PTAC units will be a continous load or not. Sure, it's safer to size the feeder/OCP for this panel at 196*1.25 = 245A or 250A but is it really required? Can I use a 225A instead? Or even 200A? I know a continuous load is defined as something that runs for 3 hours straight, but as a new EE I've come across many situations where I'm not sure if something runs for 3 hours straight or not. Like I said, the safe thing would be to assume everything might run for 3 hours straight, but that wouldn't be the most efficient design. Also, is there continous factor already figured in the 19.6 MCA as calculated in accordance with the NEC?
PTAC panel feeder/ocp sizing
- Thread starter new_ee
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petersonra
Senior Member
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- Northern illinois
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- Semi-retired engineer
Is anything that is controlled by a thermostat a continuous load?
It might run for more than three hours now and then, but not on any regular basis.
It might run for more than three hours now and then, but not on any regular basis.
jtester
Senior Member
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- Las Cruces N.M.
new_ee said:
There is no violation if you feel it is necessary to design these loads a continuous, but I agree with Bob, it is not required. You correctly noted that the MCA is going to be larger than the load.
Jim T
- Location
- New Jersey
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- Journeyman Electrician
Yes, the MCA or minimum circuit ampacity already has a 125% factor added in. For example an AC unit with a compressor and evaporator fan would use this calculation: Comp FLA(125%)+E fan FLA = MCA. The manufacturer of your PTAC would use a similar calculation to determine the MCA. The actual running load ampacity of the unit would be up to 20% lower than the MCA.
I'm curious as to how you arrived at the 196 amps per leg on the three phase panelboard?
infinity said:
I usually put the MCA as the load on my panelboard. 208V single phase PTAC units with 19.6 MCA and there are 15 of them. First 10 i put on the left side of the panel and then the other 5 i put on the right side skipping the first 2 spaces. Each leg has 10 connections at 19.6A so i get 196A per leg.
suemarkp said:
I agree you have the math right, but since the MCA of 19.6 already has an extra 25% of the largest motor current added in, the actual current will only be about 133 amps. A 200A panel should work fine.
New_EE:
You made the very common mistake of just adding all the currents connected to the A phase (just using A as an example). But some of the loads are connected from A to B, and some are connected from A to C. These currents aren't in phase, so they don't simply add.
To find the input current, you want to add all the loads in KVA', and then divide by the line-line voltage, and divide by sqrt(3). (Assuming a balanced panel).
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steve066 said:
This is the reason I asked how he did his ampacity calculation. There seems to be some key components missing from his formula.
steve066 said:
So if I fill out a panel schedule for a 3 phase panel where I simply add all the currents on each phase A, B, C (and total them up) does that number have any significance? I've seen that done on many panelboard schedules. If it was all single phase loads or all three phase loads would it make any difference? Also when I do my KVA calculation this is what I come up with: Each PTAC at 208V single phase so 19.6 * 208 = 4076.8 VA. 4076.8VA * 15 = 61152 VA. 61152 VA / (208V * sqrt(3)) = 169.74A.
sceepe
Senior Member
new_ee said:
New EE, here is some advice that is hopefully worth more than you paid for it. Always Always Always add your loads in RLP/P (Running load power / Pole). So your loads were such that you could add them up and get the correct answer this time. If you had single phase loads, 2 phase loads, and 3 phase loads in the same panel it might not work out that way. Get in the habit of converting your loads to VA per pole. Then add power (never amps) and convert the total to amps. If you make a spreadsheet that does this for you, you will eliminate a lot of potential math errors in your calcs.
new_ee said:
Yes, I think your 170A is correct. And that already includes an extra 25% (the difference between the actual current drawn by the units, and their minimum circuit ampacity).
Simply adding the currents gives correct answers if all the loads were line to neutral loads (since all the currents on one phase would be "in phase"). It would also work for a single phase panel (say 120/240V).
If you just add all the currents on a typical three phase panel, then you get a answer that is larger than the actual current. In most cases, that probably wouldn't be a big deal - it leaves some spare capacity for future. So I'm guessing that some people may do this just to keep the math simple. That's fine, but when you get a result that's a little over your panel capacity, then its probably time to do a more exact calc.
Steve
sceepe said:
Not sure how to get the loads in RLP/P as you put it. If its 208 single phase load then how do i get the right VA per pole. Obviously if I just take 120V * 19.6 VA per pole thats not going to work out as I have already demonstrated. I already figured (although I dont quite understand the reasoning i kind of just figured it out) if I take (sqrt(3)/2) * 19.6 = 16.97 A and if I put that number on the panel schedule I come up with the 170 A thats correct per phase. So I'm trying to come up with a better panel schedule that will always allow me to get the real answer with a combination of 120V single phase, 208V single phase, and three phase on a panel.
For a single phase line-line load, the standard method is to divide the total VA (or watts) by 2, and put half on each phase.
We had a long discussion a short time ago about this very thing. But to make a long story short, I think we found that this is also an approximation. It actually gives the correct answer if the panel board is balanced. At any rate, it is a better approximation than adding the currents. (For your example, it would actually give the correct answer, since your panel is balanced.)
For three phase loads, divide the total VA by 3, and put one third on each phase.
Steve
We had a long discussion a short time ago about this very thing. But to make a long story short, I think we found that this is also an approximation. It actually gives the correct answer if the panel board is balanced. At any rate, it is a better approximation than adding the currents. (For your example, it would actually give the correct answer, since your panel is balanced.)
For three phase loads, divide the total VA by 3, and put one third on each phase.
Steve
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