How would the load calc break down for an 18kV Water Heater when:

Commercial Installation
Service
208y/120 3P 4W

FLA = 57A (from manufacturers spec sheet)

This w/h has 4 heating elements - 4500w each

Thanks,
Mario

[This message has been edited by Load-n-Code (edited 01-20-2005).]

[This message has been edited by Load-n-Code (edited 01-20-2005).]

Do I calculate as follows:

A) (This would be true for Single Phase)
57A * 208V = 11856VA

even though i think this is the formula for 3p
B)
x in VA/(208V*1.732)= y in amps PER phase

If B) is correct then it would seem that C) would be correct...

C)
57A * 208V * 1.732 = 20,534.59VA PER phase

I follow your math but I'm somewhat confused by the fact that the manufacturer tells you the ampere rating and the kw rating. The kw rating is all you need to size the service or the circuit. The ampere rating is going to vary depending on the voltage. So size the circuit using the kw rating and Article 422 and size the Service based on the kw rating of the water heater.

You've swapped your root 3.

3 phase VA = amps * volts * 1.732

3 phase amps = VA / volts / 1.732

The reason is that when a three phase circuit is described as carrying X amps, this means X amps per leg, and when the voltage of a three phase circuit is described, this is the line to line voltage. If you convert things to 'line to neutral' voltage, and consider 'aggregate amps' (the total current down all of the wires) then the above equations make more sense, because now VA is simply volts times amps.

In a three phase 'wye' circuit, the line to neutral voltage is the line to line voltage/1.732

P.S. I think that the manufacturer needs to supply the amp rating because with _4_ heating elements, this is probably an unbalanced load.

-Jon

Thanks George & Jon,

How is this calculated for the unbalanced load?

Thanks,
--Mario

all you need is the KW rating or FLA, if it is a storage type water heater, 120 gallons or less, 422.13 says the branch circuit supplying the water must have a rating of not less than 125% of the nameplate rating.

BEWARE that a W/H with 4 elements will NOT be balanced across three phases. Each element is actually a single phase load. Assuming they are 208 volt elements, each 4500 watt element will pull 21.65 amps. But those amps will not be distributed evenly across the phases.

Assuming 1 element is connected A-B, and 1 element is connected B-C, and 2 elements are connected C-A, you will have phase B carrying the sum of 21.65 amps at 0 degrees and 21.65 amps at 120 degrees. That adds up to 37.5 amps. But phases A and C have 21.65 amps at 0 degrees and 43.3 amps at 120 degrees. Adding that up gives 57.28 amps. So the manufacturer rating of 57A is about right (but a tad low). It's just phases A and C that will carry the 57.28 amps while phase B carries only 37.5 amps.

If the W/H had 3 elements of 6000w, then each would draw 28.82 amps (at 208 volts) and when balanced across three phases that would add up to a nice round 50.0 amps. But since there are 4 elements, it's not balanced, and the load is lopsided.

If you happen to have multiple water heaters of this type, then you'd want to rotate some of the phases to better distribute the total load. But for just one, you'll need to use the full 57.28 amps. And 125% of that is 71.6 amps.

But you'll have to figure how that fits into the code (e.g. if you need to do the 125% circuit rating in that situation). I'm not a licensed EC; I just do computer room designs and outsource the electrical work to ECs after I've done all my calculations.

[This message has been edited by pdh (edited 01-21-2005).]

OK, based on what mike d says, and you need to put it on a 125% circuit, given the 57.28 amps on 2 of the 3 phases, and 125% being 71.6 amps, you'll probably have to put it on an 80 amp circuit.

You state the service is 3 phase, but are you sure the water heater is 3 phase? I find it hard to believe a 3 phase W/H would not have 3, 6 or 9 elements.

steve

While doing a little shopping for water heaters a couple months ago, to get some specifications an options for planning a house build way in the future, I downloaded technical details on many water heaters from several manufacturers, including commercial models because they had more of the point-of-use types I was interested in. In these technical details were wiring instructions and diagrams. I looked at the 3 phase models, too, because I was curious. Some can even be rewired between 1 phase and 3 phase.

The larger commercial models did have a multiple of 3 for the number of elements. But some smaller models did have 2 or 4 elements. What caught my attention was that the replacement element model numbers for 1 phase and 3 phase versions of the same model line were the same. That's not easy to do unless the element really has 6 coils, and that would be an expensive way to manufacture them.

I did look at a 2 element model for 3 phase rather closely. It did indeed place one element on A-B and the other element on B-C. So the B phase would be drawing more current than A or C. My guess is this was done to minimize the overall costs and use the same line of elements between 1 phase and 3 phase, and make it possible to field re-wire the phase configuration.

In smaller models, 2 elements, or maybe 4 elements, makes more sense due to having upper and lower heating elements. 6 elements would make sense, too, but the cost is getting high there, and that would not be viable for smaller models. If you need 9000 watts, 2 elements of 4500 watts is much less costly than 6 elements of 1500 watts.

Equally balanced, 18000 watts would be 50 amps on 208/120. But the stated 57 amps didn't come close to the 125% figure of 62.5 amps you get from the well balanced case. So I assumed that wasn't indicating the necessary circuit size. So I ran the numbers for the current based on balancing 4 elements the best that can be done on 3 phase, and it came up with 57.282196187 amps. Close enough to 57. Thus I have to conclude that this is a water heater with 4 elements wired as such, because I can see no other reason for listing 57 amps on the nameplate.

Anyway, I do know for a fact that water heaters are made this way with that number of elements. With the load numbers matching, what else can it be?

Thanks for replying guys.

mike d,
Is the 125% for OCP, VA (panel schedule/load calcs), wire sizing, all?

pdh,

How did you come up with 57.282196187 ?


stamcon,

Maybe the spec sheet is incorrect, or they do not list all available configurations.

The w/h is a Rheem Eclipse ME85-18.

If I am not allowed to post links to items we're dicussing let me know and I'll edit it out.

The product spec sheet is available in pdf on the right side of the page at the below link.
http://www.rheem.com/dealers/catalogComm_detail.asp?id=389

How would this be wired?

AB @ 1 element = 4500VA / 208V = 21.63A - 30A 2 POLE OCP
BC @ 1 element = 4500VA / 208V = 21.63A - 30A 2 POLE OCP
CA @ 1 element = 4500VA / 208V = 21.63A - 30A 2 POLE OCP
AB @ 1 element = 4500VA / 208V = 21.63A - 30A 2 POLE OCP

Seems like the above is Single Phase...

PH-A - in VA?
PH-B - in VA?
PH-C - in VA?

Thanks Again Guys,
--Mario

[This message has been edited by Load-n-Code (edited 01-22-2005).]

 
pdh nailed it. The first 3 elements can be treated as balanced 13.5kW 3ø load and using root3 in calcs, but the fourth 1ø can only “add” to two pahses on the 3ø load. Remember that using root3 ‘shortcut’ is only for balanced-voltage/balanced-current applications.

{Be sure that the 4500W element rating is at 208V and not 240V|

Smaller 3ø HVAC duct heaters can have the same unbalanced-load design, an will get your tit in the wringer on branch-circuit sizing if overlooked.

-

A precise way to figure per-phase loading is with excel calcs. Root3 is not needed here.

    




[This message has been edited by Bjarney (edited 01-22-2005).]

Thanks Bjarney,

would the PH-A total also be 6750?

(edited here)
i thought that element 4 would be AB.

--Mario

[This message has been edited by Load-n-Code (edited 01-22-2005).]

L-n-C, you're correct. It should be fixed now.

Load-n-code:

OK, here's my "show your work". I'll show the full precision (which is way more than actually necessary) my programs did their calculations with. Real life, of course, varies enough to make anything more then 2 or 3 digits just noise.

I use 1.73205080756887729 for the square root of 3. I derive the phase to phase voltage as 120 volts times the square root of 3 to get 207.84609690826525 volts. That voltage and 4500 watts (per element) gives 21.650635094610969 amps and 9.6 ohms (I'm assuming purely resistive, which is not necessarily the case).

One element was assumed to be wired A-B giving the current on phase A at an angle of 30 degrees. Two elements were assumed to be wired A-C giving twice the current (43.301270189221938 amps) on phase A at an angle of 90 degrees. The X and Y vector components of the A-B current on phase A are 10.825317547 and 18.75. The X and Y vector components of the A-C current on phase A are 43.301270189 and 0.0. The sums of these vectors are 54.12658773 and 18.75. X squared is 2929.6875 and Y squared is 351.5625. The sum of squares is 3281.25. The square root of 3281.25 is 57.282196186948 which is the vector sum of the two currents. The arctangent in degrees gives the final current phase angle of 70.893394649.

Note that these figures are based on current, not watts. The power factor is less than 1. But as you know, it's the current that exhibits voltage drop over conductors and heats them up. The 57.282196186948 amps on phases A and C give each a load of 6873.86354243376 volt-amps for a power factor (relative to 6750 watts and assuming the elements are purely resistive) of 0.98198050606196574, one leading and the other lagging.

Bjarney's figures are for watts, and are the ones you'd use to figure power costs and heat rise. But you need to use the amps to figure circuit load and voltage drop. And if you're power billing is based on real plus reactive power per phase, you end up paying a wee bit more to handle this imbalance. If you had three identical water heaters like this, you'd want to have them in different phase rotations to balance out the current and improve the power factor.

Thanks pdh,

Bjarney, how did you insert that spreadsheet?

Also where do I enable HTML for posting?


--Mario

[This message has been edited by Load-n-Code (edited 01-23-2005).]

In Bjarney's example

Would the total VA for this heater be equal to:
PH-A - 6873.86354243376VA
PH-B - 6873.86354243376VA
PH-C - 4582.57569495584VA (PH-A*2/3)
TOTAL: 18330.30277982336VA

There is more information now available. The manufacturer states that their 4500w 208v elements have 9.61 ohms resistance.

--Mario

[This message has been edited by Load-n-Code (edited 01-24-2005).]

The 9.61 ohms figure is 0.01 more than what I came up with. I don't know how I could have been so far off.

Just to make sure things don't get confused, the calculations I made earlier assumed the phases where two elements were connected are A and C. That leaves phase B as the phase with the lower current. Bjarney's spreadsheet has the 4th element on A and B, leaving C as the phase with the lower current. This is how Load-n-Code's 01/24 post has it. So for here I'll go with that phase connection.

Phase C is not 2/3 of A or B. It's going to have 4500 VA and watts, and unity power factor (assuming resistive elements). It's current will be 37.5 amps. If you were sizing that conductor alone, 125% would 46.875 and it could go on a 50 amp circuit. But you can't split three phases up that way, so you have to size the whole circuit at whatever the highest conductor current is, which are the two (A and B) which have 3 elements attached. So the lower current on phase C doesn't really matter.

Thanks pdh,
Is this correct based upon your last post:

PH-A - 6873.86354243376 VA
PH-B - 6873.86354243376 VA
PH-C - 4500 VA
TOTAL: 18247.72708486752 VA

What about OCP:

The manufacturer recommended OCP for...
208V 3PH for the 18kW Model is 80A
(in their enhanced spec sheets)

Does that imply that there will be only one 80A (or next standard size up) 3-pole ckt. breaker with unbalanced loads as described above?

Thanks,
--Mario

[This message has been edited by Load-n-Code (edited 01-25-2005).]

That looks correct for the 3 VAs.

Since 125% of the current on phases A and B is 57.282196186948 times 1.25 equals 71.602745233685 then a 70 amp breaker won't meet the 125% requirement. So that's an 80 amp breaker needed.

Had the wattage been equally balanced over 3 phases, it could have been protected at 70 amps. The imbalance kicks the requirement up to 80 amps. So it's obviously necessary to know the situation. The manufacturer's specifications were right all along.

pdh,

Thanks,
Mario

[This message has been edited by Load-n-Code (edited 01-26-2005).]

The electrical plan reviewer at the AHJ wanted the loads balanced on all phases (A,B,C) for the water heater on the panel schedule.

As follows:
57A * 208V * 1.732 = 20534.592 / 3 = 6844.864

PH-A: 6845VA
PH-B: 6845VA
PH-C: 6845VA


This got me thinking about the following quote regarding PH-C VA for the water heater by pdh.

Quote: pdh - (posted 01-25-2005 08:02 AM)

"It's going to have 4500 VA and watts, and unity power factor (assuming resistive elements). It's current will be 37.5 amps."


Regarding the above quote...

Is the unity power factor for the PH-C load a result of having a balanced load for 3 of the elements in water heater?

Therfore is it correct to say that the 4500w can be divided by a PF of 1.0 (because of unity power factor) for a result of a 4500VA load on PH-C?

Thanks,
Mario

The power factor is only unity assuming the elements are purely resistive. They're probably close enough that it doesn't matter, but you'd have to consult the specifications to be sure (or test it).

As to why one phase has a unity power factor, that's really better answered as why the other two don't.

If there were 3 (or 6 or 9, etc) elements in the heater, each on a different phase, either P-N or P-P (all the same way, of course), then the overall power factor would be the same, and virtually unity (to the extent the elements are not reactive). In the case of P-P, the currents on each element are phase shifted from the voltages on each phase wire (as in any P-P connection), but when the currents meet at the junction, one leads and the other lags, and they cancel out (some current just flows between them) to no phase shift.

In the case of 4 elements (this water heater) wired P-P where A-B gets 2 of them, the above description of cancelling out only applies to phase C because it has the same number of elements wires in each direction (one to A and one to B). So it has the same current leading as lagging. The other 2 phase wires (A and B) have the angular imbalance because there is one element in one direction, and two elements (more current) in the other direction.

If you had 3 such unbalanced 4 element water heaters, you could stagger them 3 ways to get a net total lower load than if you just wired them up all the same, making it equivalent to a 12 element water heater system. If you did have them, doing this would be good at least to get the better power factor as seen beyond the point where their circuits meet.

You might be tempted to favor phase C for some other loads a bit more due to the lower loads from this water heater on that phase. But that wouldn't be a good idea, and your AHJ is probably enforcing that. Consider the water heater fails some day in the near or far future. You're not around, and the next guy to come along to do the wiring has no idea how you balanced other loads to match this unbalanced load (even if he does understand it is an unbalanced load). He might end up wiring the phases differently (he's got a 1 in 3 chance of getting supply phase C connected to the terminal with only 2 elements). If he gets it wrong, your balance with other loads is all thrown off.

So basically, for the purpose of calculating building loads, figure the water with each phase having the current equal to the worst cases. Only if you were putting 3 such heaters right next to each other, and leaving everything well documented, might you consider staggering the phases between them to get a slightly lower load. And even then the AHJ might not allow you to calculate everything that way (and I can see why not for the safety reasons).

FYI, I'm not an electrician. I design/manage computer/network data centers, and do have to deal with things somewhat like this surrounding UPSes and such, sometimes. Then I hand off the work to an EC who, if I did my end right, won't have to come back and tell me the code, or face a red tag. So I learn a lot of this stuff so I can get it right the first time and not have to re-design things all over.

pdh,
Thank you, again.
Mario