Calculating wire size and OCPD for an EEMAX tank-less hot water device

[harsha1972]

Given: EEMAX model EX1608T2 208V/2 phase, 16.6KW (This is 2 single phase 208V branch circuits) and the EEMAX unit and the panel board terminals are 75C rated.

Assumptions: Total distance is about 125ft from the panel to the EEMAX, 4 phase conductors in one raceway, non-continuous 100% resistive load (pf=1)

Calculations:

1. Use 4-#8AWG THHN 90C rated conductors. (Maximum terminal ampacity is 50A @ 75 deg. C rating for EEMAX and panel board).
2. Now derating the ampacity due to 4 current carrying conductors in one raceway by 80% = 55X0.8=44A
3. At about 125feet, voltage drop is about 7.78V and the circuit current is = 16.6KWX1000/(208V-7.78V)= 82.9A (Also, the voltage drop is 3.74%. No need to up-size the conductor yet)
4. Each 208V/1phase circuit will draw approximately (82.9A/2=41.45A) and the #8AWG ampacity is acceptable (44A per step 2 above)
5. NEC allows the next higher over current protective device to protect the conductor ampacity which in this case will be a 45A/2pole breaker.

So, the conclusion is that per NEC code, I can use 4-#8AWG THHN 90C rated conductors protected by 2-45A/2pole breakers. (Plus 1-#10AWG equip. ground in 3/4" EMT) Had the EEMAX or the panelboard terminal temperature rating had been 60 deg. C, then I will have to go with 4-#6AWG THHN with a set of 50A/2 pole breakers.



Can someone verify the above calculations?

 

[suemarkp]

Do the instructions give any clue as to what the ampacity should be on the two circuits or the max breaker size to protect it? You seem to be assuming the load is evenly split between the two, and it may not be (could be some control load on one of them).

Also, I don't believe you can use the actual terminal voltage to determine the amp loading. You need to use standard values (e.g. 208V) and not reduce for voltage drop. Adding the voltage drop is creating a larger amp number anyway, so I guess you can do this to be safer (but the actual amp draw will be less in a resistive load because it won't draw 16.6 KW when supplied with less than 208V).

I get a 79.8A load when assuming 16.6 KW at 208V. If the load is evenly divided, then two 40A branch circuits would be compliant. Two 40A branch circuits run with #8 THHN in the same pipe would still be compliant. That seems to be similar to the few others I've seen installed.

 

[Smart $]

The owners manual says 2 ? 39A... so that's the value you use (unless the nameplate states otherwise).

 

[david luchini]

Two 40A branch circuits run with #8 THHN in the same pipe would still be compliant.

I agree with two 40A branch circuits with #8 THHN. Two 40A branch circuits with #8 THWN would also be OK.

Had the EEMAX or the panelboard terminal temperature rating had been 60 deg. C, then I will have to go with 4-#6AWG THHN with a set of 50A/2 pole breakers.

If the terminations had a 60 deg. C rating, then two 40A branch circuits with #8 THHN or two 40A branch circuits with #8 THWN would still be acceptable. If you used TW, you would need two 40A branch circuits with #6 TW.

 

[harsha1972]

I?m still not convinced that a 40A/2pole breaker will work for a 16.6KW load on a 208V/2phase circuit at a distance of about 125ft. We all know with distance, voltage will drop, but the load still remains the same. So based on the fundamentals, the load will draw I=Watt/ (volt x pf). Current will go up with the distance as voltage drops.
I?m afraid the 40A/2P breaker will trip if the load draws anything over 39A for a long time. I can up-size the breaker to 45A on #8AWG since the terminal ratings are 75 deg. C. (max. 50A breaker for #8AWG on 75C column per NEC 310.15(B)(16), but due to 4 conductors in the same raceway, I cannot go past 45A.)

EEMAX tech support says if I go pass 100ft, I should go with 4-#6AWG?s and 50A/2Pole breakers. I think that?s an over-kill So, I?m still leaning towards 4-#8AWG THHN with 45A/2pole breakers. (Again I?m afraid the 40A/2P will trip as the distance gets involved.)

 

[infinity]

I?m still not convinced that a 40A/2pole breaker will work for a 16.6KW load on a 208V/2phase circuit at a distance of about 125ft. We all know with distance, voltage will drop, but the load still remains the same. So based on the fundamentals, the load will draw I=Watt/ (volt x pf). Current will go up with the distance as voltage drops.
I?m afraid the 40A/2P breaker will trip if the load draws anything over 39A for a long time. I can up-size the breaker to 45A on #8AWG since the terminal ratings are 75 deg. C. (max. 50A breaker for #8AWG on 75C column per NEC 310.15(B)(16), but due to 4 conductors in the same raceway, I cannot go past 45A.)

EEMAX tech support says if I go pass 100ft, I should go with 4-#6AWG?s and 50A/2Pole breakers. I think that?s an over-kill So, I?m still leaning towards 4-#8AWG THHN with 45A/2pole breakers. (Again I?m afraid the 40A/2P will trip as the distance gets involved.)

You're using the incorrect formula, this is a resistive load so by adding VD (lower voltage at the load) the load will draw less current.

 

[harsha1972]

You're using the incorrect formula, this is a resistive load so by adding VD (lower voltage at the load) the load will draw less current.

I learned back in school for resistive loads, the power factor is 1, thus the formula becomes I=Watt/Volt. For inductive and capacitive loads p.f<1, and the formula still holds true. So, lower voltage means higher current on the circuit.

 

[GoldDigger]

I learned back in school for resistive loads, the power factor is 1, thus the formula becomes I=Watt/Volt. For inductive and capacitive loads p.f<1, and the formula still holds true. So, lower voltage means higher current on the circuit.

But your problem is that, unlike a motor, the heating element will not be able to deliver its rated power (wattage) at any value of voltage other than the one it is rated at. The formula becomes I = Watts@voltage/voltage, where Watts@voltage is V²/R. That means that I = V/R in this case, fortunately for Mr. Ohm.

 

[harsha1972]

Finally got it after seeing this drawing. Like everyone said here, I see now why the current will decrease with the increase in conductor lenght.