How to read a RTU nameplate

[electricalintern]

How do I determine the size of the RTU breaker and FLA from the nameplate

Do I take the MCA and multiply it by 1.25 or take the largest load and multiply it by 1.25

I would typical use the max fuse size of 25 per the nameplate and take the MCA of 21.

But I see the Compr A and B only shows a RLA and the fan loads shows FLA.

When I add the RLA and FLA amp together I get 17.75 AMP.

Can you add RLA and FLA together?

 

[wwhitney]

If it has MCA (as it should), that's your minimum circuit ampacity. No further multipliers, just size the wire to be at least the MCA.

If there's no MCA, and there is just a compressor RLA and a fan FLA, no other loads, and RLA > FLA as I would expect is always true, then MCA = 1.25 * RLA + FLA.

Edit: your label shows two RLAs and three FLAs. The MCA is calculated for you already, but if it weren't, you'd take the largest number * 1.25 plus all the other numbers. So for it, you'd have 6.8 * 1.25 + 6.2 + 0.8 + 3.4 + 0.25 = 19.15A.

That doesn't quite match the 20A shown, but notice that with the optional 1.8A accessory, they list an MCA of 21A, which does match 19.15A + 1.8A = 20.95A. So the manufacture just rounded 19.15A up to 20A. Not sure if that's what the listing standard requires, or if they could have rounded it down to 19A.

Cheers, Wayne

 

[augie47]

As noted, for practical purposed you can pretty much ignore everything except below:


Your MCA (conductor size) Your MOCP (In this case it's fuse or breaker but use caution as sometimes it's just fuse)

With this low a MCA the SCCR should not be an issue but on larger units it can pose a major problem as most are 5ka and often the available is greater.

 

[tom baker]

This subject is #1 for confusion. In the next version of the WA State electrical rules there will be note explaining how to size conductors.

 

[electricalintern]

If I have two Rooftop Units with 21 MCA each and a maximum breaker size of 25 amps each, can I consolidate them into a single breaker? If so, what would be the size of the combined breaker?

 

[infinity]

If I have two Rooftop Units with 21 MCA each and a maximum breaker size of 25 amps each, can I consolidate them into a single breaker? If so, what would be the size of the combined breaker?

No. The MaxOCPD is 25 amps both units will not run on that circuit.

 

[wwhitney]

If I have two Rooftop Units with 21 MCA each and a maximum breaker size of 25 amps each, can I consolidate them into a single breaker? If so, what would be the size of the combined breaker?

Infinity answered your question. For bonus information (ignore if confusing):

If you wanted to run a feeder to a small panel with two 25A circuit breakers, feeding (2) of the machines with the nameplate you posted (presumably each with the extra power exhaust, since you specified 21A MCA), then the minimum ampacity of that feeder wouldn't be 21A * 2, as that would include an extra 25% of the 6.8A largest RLA twice, once for each unit. Instead it would be 40.2A (which IIRC can be rounded down to 40A), because you only have to include the extra 25% of the single largest motor on the feeder.

A small difference in this case, larger with more machines or with larger motors.

Cheers, Wayne

 

[electricalintern]

If it has MCA (as it should), that's your minimum circuit ampacity. No further multipliers, just size the wire to be at least the MCA.

If there's no MCA, and there is just a compressor RLA and a fan FLA, no other loads, and RLA > FLA as I would expect is always true, then MCA = 1.25 * RLA + FLA.

Edit: your label shows two RLAs and three FLAs. The MCA is calculated for you already, but if it weren't, you'd take the largest number * 1.25 plus all the other numbers. So for it, you'd have 6.8 * 1.25 + 6.2 + 0.8 + 3.4 + 0.25 = 19.15A.

That doesn't quite match the 20A shown, but notice that with the optional 1.8A accessory, they list an MCA of 21A, which does match 19.15A + 1.8A = 20.95A. So the manufacture just rounded 19.15A up to 20A. Not sure if that's what the listing standard requires, or if they could have rounded it down to 19A.

Cheers, Wayne

When the MCA is 75.5 AMP on the RTU nameplate, do you use 100% of the load when calculating the total load for the store?

Since if the store has 208V/3P/400 AMP service and there is a total of (4) RTU, that means (4*75.5*360)/360 = 302A

If the existing service is 400AMP, that means I am only left with 98 AMP for the entire store.

Is there a demand factor I and use to lower the RTU load

 

[wwhitney]

When the MCA is 75.5 AMP on the RTU nameplate, do you use 100% of the load when calculating the total load for the store?

Since if the store has 208V/3P/400 AMP service and there is a total of (4) RTU, that means (4*75.5*360)/360 = 302A

No, you get to apply the procedure in post #7.

So you need to look beyond the MCA on the RTU and determine what the largest RLA/FLA number contributing to the MCA is. If it were, say, 40A, then without the extra 25% that goes into MCA, the sum of the motor ratings is just 65.5A. Which would make the total load 75.5 + 3 * 65.5 = 272A. Replace 40A with the correct number and revise the calculation accordingly.

Just double checking that the units are 3 phase units, not single phase units.

Cheers, Wayne

 

[topgone]

When the MCA is 75.5 AMP on the RTU nameplate, do you use 100% of the load when calculating the total load for the store?

Since if the store has 208V/3P/400 AMP service and there is a total of (4) RTU, that means (4*75.5*360)/360 = 302A

If the existing service is 400AMP, that means I am only left with 98 AMP for the entire store.

Is there a demand factor I and use to lower the RTU load

MCA = 125% x (motor rated current + heater current). Can't just add the four MCAs.

 

[electricalintern]

So I don't have the name plate but I do have the cut sheet from the manufacturer. See attach for reference

So if I can't add the MCA up, I tried to do the following calculation
1.25*16.5+4.3+2.8=27.725 AMP.
I don't get how the MCA is 75.5A

 

[electrofelon]

So I don't have the name plate but I do have the cut sheet from the manufacturer. See attach for reference

So if I can't add the MCA up, I tried to do the following calculation
1.25*16.5+4.3+2.8=27.725 AMP.
I don't get how the MCA is 75.5A

Something is definitely missing. Likely Either there are multiple compressors not shown for some reason, or there is electric heat. If the compressor shown is the largest motor, than you can deduct the extra 25% from three of those, so (75.5*4) - (3*(16.5*.2)), so not a big reduction.

 

[topgone]

So I don't have the name plate but I do have the cut sheet from the manufacturer. See attach for reference

So if I can't add the MCA up, I tried to do the following calculation
1.25*16.5+4.3+2.8=27.725 AMP.
I don't get how the MCA is 75.5A

Did I see you posted there are 4 of them here? 1.25*16.5+ 3(16.5)+4.3+2.8 = 77 ~ 75A. Just trying to make out what you have there.

 

[Elect117]

There are two different voltages? The first image shows 460V and the third attachment (detailing the 75.5A MCA) shows 230V.

I find adding up loads easier in KVA or KW. While that might might seem complicated at first, it is easier in the long run. Especially if you are focusing in more commercial applications where you will have multiple voltages and there are single and three phase loads.