Tanning salon service size

[Todd Z]

Ok I'm new to the forum and there is a lot of info in my question so bare with me. I looked at a tanning salon the other day that had 8 beds and a spray tan booth. The building had a 300A 208 3 phase service. The beds are on two-pole breakers and had a 208-230 buck boost trasformers. 1.5kva for the three big beds and 1kva for the five small beds. Here are the actual amprobe readings.
1-67A 2-34A 3-34A 4-15A 5-28A 6-28A
7-28A 8-66A 9-52A (rounded up)
There is also 22A of lighting load and a 50A three phase A/C. The utility demand meter shows peak demand of 57kw. The owner says there are times were all beds and the A/C are running together. My problem is the owner hired an electrical engineer to tell him if he could remove the three 28A beds and replace them with a 67A bed and two 34A beds. The electrical engineer never left the office and based all calculations off of the max demand. I think that is the wrong way to figure it being the max demand is based on a 15min window. Your thoughts?

 

[augie47]

Just an opinion, but your demand meter load is slightly over 150 amps on a 300 amp service and with the bed swap you are only adding 51 amps. The engineer is correct that the time frame on the POCO demand would not necessarily take into account the run time on the beds, but it does not seem you are running that close to me.
I would not be too concerned with doing the swap, but since the owner hired the engineer it opens the door for you to do the larger job

 

[GoldDigger]

Just an opinion, but your demand meter load is slightly over 150 amps on a 300 amp service and with the bed swap you are only adding 51 amps. The engineer is correct that the time frame on the POCO demand would not necessarily take into account the run time on the beds, but it does not seem you are running that close to me.
I would not be too concerned with doing the swap, but since the owner hired the engineer it opens the door for you to do the larger job

Todd did not say one way or the other, but my impression was that the engineer said it would be OK to do the small job, based on the meter demand and Todd is the one concerned that it might not be appropriate.

 

[scotteng]

I agree with Augie. The methodology the engineer is using is correct and the service does not appear overburdened at the moment or with the additional load. It is often better to have 1-year load history than to try and calculate the load. A 1-year load history will show actual diversity that you can't account for in an NEC 220 load calc alone. Regardless of what the owner thinks is happening, at any given point in time is it more likely that some beds will not be running as people are getting dressed/undressed before and after their tanning sessions.

 

[Npstewart]

Just an opinion, but your demand meter load is slightly over 150 amps on a 300 amp service and with the bed swap you are only adding 51 amps.

If im not mistaken, aren't you required to take the existing peak demand @ 125% and the additional load at 100%? So the existing load is actually 71.3 kW. That puts you at 249A so you are still OK.

 

[Todd Z]

I meant to expand on my question some more but had to go to a meeting. The engineer said it would work but if at any given the current exceeds 300a on any one phase then the breaker would trip. The demand meter does not show those short periods of high current draw. Not to mention the NEC allowing ocpd to only carry 80% of its rating.

 

[augie47]

If im not mistaken, aren't you required to take the existing peak demand @ 125% and the additional load at 100%? So the existing load is actually 71.3 kW. That puts you at 249A so you are still OK.

My bad You are correct, (220..87...Thanks for the "schoolin )
I would not be too concerned, but admittedly that puts it a bit closer.

 

[Npstewart]

I meant to expand on my question some more but had to go to a meeting. The engineer said it would work but if at any given the current exceeds 300a on any one phase then the breaker would trip. The demand meter does not show those short periods of high current draw. Not to mention the NEC allowing ocpd to only carry 80% of its rating.

Why would the current exceed 300A if there is less than 300A of load?


Also, there's no requirement in the NEC for OCP to only carry 80% of its load. Its 125% of continuous plus 100% of non-continuous. The only thing in that place that I would consider continuous is the lights unless someone really needs a 3 hour tan

 

[Todd Z]

I wrote the loads down in columns, L1,L2,and L3. After balancing them out each phase was over 300a with new beds installed(not adjusting for power factor). Maybe this is wrong but the way I see it, you can't use the demand meter amount. The demand meter is averaging a 15 minute period so if everything is running at the same time for 5 minutes then everything shuts off for the remaining 10min then the demand meter isn't going to show a very high demand. But if the latter happens then the breaker will trip.

 

[GoldDigger]

Why would the current exceed 300A if there is less than 300A of load?

Since the beds, etc. are fed from only two phase wires each, they are essentially single phase loads. Staying under 300A per phase requires at least a minimal amount of load balancing as well as looking at the total KVA over three phases. Not as bad as it would be with line to neutral loads, but it still requires attention given the disparity in sizes of the various loads.

 

[Npstewart]

I just looked back and realized they were single phase.

We do a lot of tanning salons for a huge chain in my area. Service size is ALWAYS a big deal. We can usually only put around 4-6 beds in each store cause each bed is typically 50A demand. They also use buck/boost transformers after they get their final inspection because they like to have the voltage at 240 to get the most out of the bulbs.

 

[John120/240]

They also use buck/boost transformers after they get their final inspection because they like to have the voltage at 240 to get the most out of the bulbs.

What is your reasoning to use buck-boost transformers after final inspection ? Could not an acceptable installation be accomplished using buck-boost transformers before inspection ?

 

[iwire]

I wrote the loads down in columns, L1,L2,and L3. After balancing them out each phase was over 300a with new beds installed(not adjusting for power factor).

You measured the single phase amps and added them together?

That does not work, you would have to convert all the single phase loads to watts and total that up.

The service is capable of almost 108,000 watts non-continuos.

 

[Todd Z]

You measured the single phase amps and added them together?

That does not work, you would have to convert all the single phase loads to watts and total that up.

The service is capable of almost 108,000 watts non-continuos.

I agree with your statement of the service being capable of 108kw but what if one phase conductor is loaded at say 350A and the other 2 phases are at 250A. The breaker will still trip. Converting everything to Watts or KVA assumes equal phase balance which isn't the case in the real world.

 

[iwire]

I agree with your statement of the service being capable of 108kw but what if one phase conductor is loaded at say 350A and the other 2 phases are at 250A. The breaker will still trip.

Of course it will.

Converting everything to Watts or KVA assumes equal phase balance which isn't the case in the real world.

But it is the only way to do it.

You can't just add up the single phase amps on a three phase service, it gives false and meaningless result.

 

[Todd Z]

Of course it will.

But it is the only way to do it.

You can't just add up the single phase amps on a three phase service, it gives false and meaningless result.

Ok so if I take amp readings of every load on say C-phase and then add them all together it won't be the actual amperage on the C-phase service conductor? Please explain?

 

[Npstewart]

What is your reasoning to use buck-boost transformers after final inspection ? Could not an acceptable installation be accomplished using buck-boost transformers before inspection ?

Not sure why but they dont like to show it on the drawings because they dont always install them. Sometimes they put them in, sometimes they don't.

 

[iwire]

Ok so if I take amp readings of every load on say C-phase and then add them all together it won't be the actual amperage on the C-phase service conductor?

Weird huh?

Please explain?

I can demonstrate it but it will take someone smarter than I to explain why.

First off balance is important but for the sake of discussion lets start with a fictional perfectly balanced load, no power factor etc.

Lets say you have a 3 phase panel with six 2 pole breakers, each breaker reads exactly 20 amps on your amp clamp.

Lets say the breakers are evenly distributed across the phases

Breaker 1 on A&B

Breaker 2 on B&C

Breaker 3 on C&A

Breaker 4 on A&B

Breaker 5 on B&C

Breaker 6 on C&A

We used each phase four times at 20 amps each.

4*20=80 amps

By adding everything up as you did we come out with 80 amps per phase.

Do we agree?

Now lets do it with watts.

20 amps single phase @ 208 volts = 4160 watts per load.

6 loads @ 4160 watts each results in 24,960 watts.

Now lets convert that to amps.

24,960/208/1.73=69 amps per phase.

A difference of 11 amps on just 80, with 250 the difference would be even more.



I would do what the engineer says to.

 

[growler]

I would do what the engineer says to.

So would I. I get paid to install the new tanning beds and if the engineer screws up I get paid to fix that also. It's a no lose situation.

They may need a better service, with the new larger beds they may have to increase the size of the AC units also. It's amazing how much heat those suckers put out.

 

[GoldDigger]

Ok so if I take amp readings of every load on say C-phase and then add them all together it won't be the actual amperage on the C-phase service conductor? Please explain?

Maybe easier to see, but without the mathematical rigor in a simpler, non-balanced situation:

Consider only two loads, A to C and B to C. Each 80 amps.
The current on A will be 80 amps, the current on B will be 80 amps, but because the two loads are pulling currents on the C wire which are out of phase the resultant total current on C will be the vector sum rather than the arithmetic sum. Specifically 80 x (1.73 / 2) = 69. That gives you the same numbers as growler's calculation by watts (which includes the 1.73 factor that comes into the three phase wattage calculation for the same reason!)