ECN Forum Forums Code Related Discussion NEC & other Code issues Load Calc for 18kV W/H

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.