Water Heater Sizing At Continuous?

[ritelec]

Hello all ,

Residential water heater of 5500 watts is ~ 22.92 amps.
at 125% is 28.65

Would it be correct to put it on a 30 amp circuit and still have the breaker protected at 24 amps ( 80%) with it's true amps at 22.92,
Or because of the 125% being 28.65 does it go on a 40 amp circuit?


thank you

 

[Little Bill]

Hello all ,

Residential water heater of 5500 watts is ~ 22.92 amps.
at 125% is 28.65

Would it be correct to put it on a 30 amp circuit and still have the breaker protected at 24 amps ( 80%) with it's true amps at 22.92,
Or because of the 125% being 28.65 does it go on a 40 amp circuit?


thank you

You need at least #10 wire and a 30A breaker. The 125% is on the 22A which brings it to 28A. Next size up is 30A. That is most common here, 30A for the WH of that size. We usually even put a 4500 on a 30A.

I'm sure you know this, but you don't protect the breaker, the breaker protects the wire!

 

[ritelec]

Thanks Little Bill,

so the 80% of (to protect the breaker), is for your true running amps. and nothing to do with the 125%.
The 125% of your fla would be for the circuit ampacity (lets call it virtual or you tube amps) up to 100% of that 30? or 40 or 50?

SO:

(A) I've always thought sizing the circuit @ 125% than that number had to fall to or under the 80% of the breaker then that would give you the circuit.

so 125 % = 24 puts you on a 30.... 125% = 32 puts you at a 40 etc.. etc.. Wrong?


or

(B) Go back to the 125% of fla is the circuit. The true amps cant be more than 80% ??



thank you.

 

[ritelec]

You need at least #10 wire and a 30A breaker.

In reading about protection of the water heater... the article says oc protection can't be any more than 150% of the fla... so that would make it 34.38

Which means it's not even at least.... I couldn't put it on a 40 if I wanted...correct?

TY

 

[mbrooke]

In reading about protection of the water heater... the article says oc protection can't be any more than 150% of the fla... so that would make it 34.38

Which means it's not even at least.... I couldn't put it on a 40 if I wanted...correct?

TY

Well code wise I guess its technically possible, however I hae never seen a resi 5000 watt water heater on a 40, In fact Im not sure how you would connect the 10 guage wires to it in an already small j box.

 

[ritelec]

ok. thanks...

125% of fla is the circuit. The true amps can't be more than 80% of brk ..........


been thinking of it wrongly all these years..

;- (

 

[david luchini]

Thanks Little Bill,

so the 80% of (to protect the breaker), is for your true running amps. and nothing to do with the 125%.
The 125% of your fla would be for the circuit ampacity (lets call it virtual or you tube amps) up to 100% of that 30? or 40 or 50?

There is no "80%" rule in the Code, only a 125% rule. See 210.19 & 20 for branch circuits, and 215.2 & 3 for feeders.

SO:

(A) I've always thought sizing the circuit @ 125% than that number had to fall to or under the 80% of the breaker then that would give you the circuit.

so 125 % = 24 puts you on a 30.... 125% = 32 puts you at a 40 etc.. etc.. Wrong?

This would be applying the 125% rule TWICE.

or

(B) Go back to the 125% of fla is the circuit. The true amps cant be more than 80% ??

This would be correct.

In reading about protection of the water heater... the article says oc protection can't be any more than 150% of the fla... so that would make it 34.38

Which means it's not even at least.... I couldn't put it on a 40 if I wanted...correct?

No you cannot put this heater on a 40A c/b. You could put it on a 35A c/b, though.

 

[ritelec]

Thank you

 

[Little Bill]

OP may be confusing the "mostly unenforceable" 210.21(B)(2)/table with his "80%" reference.

BTW, FLA is used usually with motor loads. So unless a water heater has a motor, you would not use FLA.
Only thing I find in the code concerning water heaters is:
422.13 which says heaters with a tank are to be considered a continuous load,
and 220.53 & 220.82.

 

[brichter]

There is no "80%" rule in the Code, only a 125% rule. See 210.19 & 20 for branch circuits, and 215.2 & 3 for feeders.



This would be applying the 125% rule TWICE.




This would be correct.

No you cannot put this heater on a 40A c/b. You could put it on a 35A c/b, though.



No you cannot put this heater on a 40A c/b. You could put it on a 35A c/b, though.

Where did you come up with this notion for the 35 amp cb?

 

[Dennis Alwon]

No you cannot put this heater on a 40A c/b. You could put it on a 35A c/b, though.

I believe you could use a 35 amp overcurrent protective device based on 422.11(E) 150% however I don't think the wire could be #10

 

[brichter]

I believe you could use a 35 amp overcurrent protective device based on 422.11(E) 150% however I don't think the wire could be #10

Agreed.

 

[ritelec]

OP may be confusing the "mostly unenforceable" 210.21(B)(2)/table with his "80%" reference.

BTW, FLA is used usually with motor loads. So unless a water heater has a motor, you would not use FLA.
Only thing I find in the code concerning water heaters is:
422.13 which says heaters with a tank are to be considered a continuous load,
and 220.53 & 220.82.

I'm referencing the 80% of continuous load to protect a non continuous load rated breaker...

I get the fla thing...

If I said the whole thing, full load amperage would it still be wrong..??

It is a load.. it is the full, and is amperage.

Thank you.. :- )

 

[brichter]

I'm referencing the 80% of continuous load to protect a non continuous load rated breaker...

I get the fla thing...

If I said the whole thing, full load amperage would it still be wrong..??

It is a load.. it is the full, and is amperage.

Thank you.. :- )

125% and 80% are reciprocals. The ampacity of a continuously loaded circuit must be rated at 125% of the ampacity of the load. Therefore 125% of 24 amps is 30 amps. Or looking at it another way, a 30 amp circuit is limited to 24 amps continuous or 80% of 30 amps.

 

[kwired]

125% and 80% are reciprocals. The ampacity of a continuously loaded circuit must be rated at 125% of the ampacity of the load. Therefore 125% of 24 amps is 30 amps. Or looking at it another way, a 30 amp circuit is limited to 24 amps continuous or 80% of 30 amps.

That is the nuts and bolts of it. NEC doesn't use the 80% anywhere I am aware of for this kind of application, but because it is the reciprocal of 125 it can be used to figure things backward so to speak.

I often do use reciprocal methods to do ampacity adjustments. Instead of selecting a minimum size conductor then derating it for ambient temp, number of conductors in the raceway, and then look at the tables and find out it is too small I instead take minimum ampacity the load needs multiply by the reciprocal of the ampacity adjustments (80% adjustment needs to multiply by 1.25, 70% adjustment needs to multiply by 1.43) you can find the minimum size conductor needed the first time through calculations. It is not right, it is not wrong just an alternate way to do it.

 

[david luchini]

Where did you come up with this notion for the 35 amp cb?

422.11(E)(3).

 

[ritelec]

Ok.... guess I should pay better attention... did,t know they dropped the 80%.. or sec. 384-16(c) if that was it...

in reading..something isn't making sense.


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CB Sizing Examples

The following are examples of sizing rules.

Example 1: 50A continuous load and 125A noncontinuous load.

OCPD = 100% noncontinuous load + 125% continuous load = (1.00 x 125A) + (1.25 x 50A) = 187.5A

Therefore, a 200A OCPD is needed. If a 100%-rated CB is chosen, a 175A rating (125A + 50A) is acceptable.

Example 2: 300A noncontinuous load.

A 300A device is acceptable; a 100%-rated device is not needed since the load is noncontinuous.

Example 3: 200A continuous load.

OCPD = 100% noncontinuous load + 125% continuous load = (1.00 x 0A) + (1.25 x 200A) = 250A

Therefore, a 250A device is needed. If a 100%-rated CB is selected, a 200A rating is permitted.

Example 4: 16A continuous and 30A noncontinuous.

OCPD = 100% noncontinuous load + 125% continuous load = (1.00 x 30A) + (1.25 x 16A) = 50A

Therefore, a 50A device can be selected. Although 100%-rated devices typically are not available in sizes this small, the permitted rating would still be 50A (16A + 30A = 46A; rounded up to 50A).


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In the above examples... or in general..if you put a 50 amp continuous load on a 50 amp breaker (continuously rated) wouldn't it trip? or do they trip at like 50.0000000000000000001 ?


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http://forums.mikeholt.com/showthread.php?t=63261

I'll work on it

 

[GoldDigger]

There is a defined tolerance band above the rated current where the breaker is allowed but not required to trip. This is true whether the breaker is 80% or 100% rated.

Tapatalk!

 

[suemarkp]

Maybe, maybe not. It depends on the ambient temperature and how long you run that continuous load. A breaker will allow current of 10% to 25% over its rating for quite some time. It may allow 100% of its rating indefinitely. A large wire on that breaker can also act as a heat sink pulling heat away from it. So if you use the minimum size wire (e.g. high temp rated wire) and have the panel in a hot Arizona garage, it would probably eventually trip if it was loaded to its rated load for many many hours.

If you have a 50A load that is continuous, it should be on a 50*1.25=62.5A breaker (would need to use a 70A, as 62.5 is not a standard size).

A 40A load that is continuous would need a 40*1.25 = 50A breaker.

What may be tripping you up is which number is the continuous load value -- 40 or 50. If the nameplate is 40A, then correction factor for continuous results in 50A. Most of the equations have you using the 125% factor and you just use that result. You don't apply 80% to the breaker too, as you've already corrected for the continuous load with the 125% factor. You could work the problem backwards say I have a 50A breaker, so what is the largest continuous load nameplate I could connect, and that would be a 40A nameplated device that is considered a continuous use device.

Sometimes, the nameplate puts the 125% factor in for you, such as the MCA value on HVAC equipment. It would perhaps be less confusing if a water heater did the same thing instead of just using watts. A 240V 5500W heater would normally have an true amp rating of 22.9 amps. Since heater branch circuits are to be treated as a continuous load, then that 5500W heater should have a 28.6A MCA rating which would be fine on a 30A breaker.

 

[ritelec]

Thank you for the feed back..

Just that I remember when the 80% and temp. ratings of wire became a concern.. and a "big deal"..

I walked away from keeping my head in the game when I joined my Onion brothers and basically just had to run pipe and pull wire..

continued with the continuing education courses to keep lic. and permit.... but well..... guess I wasn't paying attention..

Thank you.

reciprocal................... my word for the day ;- )