[PPC]

Hello

Thank you to TKRussell and all the other folks who helped me with my sub panel problem.
It is installed and working great. What a pleasure to have a place to hook up new circuits!

My issue at hand now is as follows.

My friend just around the corner is hooking up is water heater and having problems.

It is a Giant Model 130E single element water heater rated 1500w @ 120V on the panel outside the heater and on the spec sheet. The element markings are marked something like 1125w @ 208 v and 1500w @ 120v

The spec sheet for that model states it needs a #12 wire on 120V or a #14 wire on 240V
The breaker is recommended @ 20 Amp for 120 and @ 10 Amp for 240. The tech guy from Giant seemed to think it would work with 240V.

Since there was a #14 2-conductor wire unused located near the heater we decided to try running the heater on 240v using that wire. At the Square-D panel we used a 240v 20amp breaker (10 amp was recommended but we only had 2-20 amp 240 breakers to test.

The first 20 amp breaker we tried seemed to flick the red indicator window as soon as breaker was flipped on but didn't snap breaker off. We thought maybe just a sensitive used breaker. So we tried the second breaker. It flipped on without showing any red in the indicator window and we could hear the heater element working.

We waited about an hour while doing some other things and then checked for hot water. The water was barely warm. We check the breaker again and the red indicator window was showing again but breaker still in on position. We flipped in on again and the element started heating again. This time it finished a complete heat cycle and got the water hot. But when it starts to cycle the element again it flips the red indicator window on the breaker and the element cuts out.

This water heater has never had the element in use since new but has been used for a year as a holding tank for warm water in an infloor boiler system. I think they used the thermostat but the element was disconnected.

I am wondering what is the problem with the heater.

Is the element only usable with 120v? Should we try it @ 120V?
Are the breakers faulty?
Did the element get all covered in build up or corrosion sitting in the heater for a year with the element ever in use?

We checked over the wire hook up and ground and things seem OK.

Any help would be greatly appreciated.

Thanks

[EPMiller]

Sounds to me like it is a 120 volt element. That business of a lower wattage on 208 volts doesn't make any sense. If you run a 120v element on 240 volts you could get the trip symptoms you describe. IF the element survives.

As to the wire sizing; at 120v 1500 watts you would be pulling 12.5 amps, which is not over the current rating for #14 wire. Of course that would require a 15A single pole breaker. Maybe their wattage rating is off and it is higher. Hook it up to 120 volts, measure the current and if it is above 13 amps go to #12 wire and a 20 amp circuit for a safety margin. Especially if the wire run is long.

EPM

[PPC]

Hi EPMiller

I thought a continuous load can only be 80% of the circuit rating? So that would work out to around 15.6 Amps at 100% and that is why I think they indicated #12 for 120v.

As for the lower wattage, maybe that is because that is 208v. At 240v it would work out to 1500W.

Wouldn't it trip the breaker right from the get go if it was a 120v only element?

Thanks for you comments

[donf]

No, a continuous load is rated at 125% or 1.25 as a multiplier.

For example: 12 (amps) X 1.25 (rate) = 15 (max amps). Personally, I would move to 12/2 AWG w/ground and use a 20 amp breaker.

[PPC]

Hello donf

Thanks for your response. I think we are getting off topic just a bit as my main concern is the question why is the heater having problems running 240v? If the single element is 240v capable then the #14 wire is well within limits as it is drawing half the amps as the 120v hookup.

As for the continuous load theory I think we are both talking about the same thing. Our code book up here in the snow country (Canada) words it a bit different.

Rule 8-302(2). The supply circuit must not be loaded to more than 80% of the rating of the breaker or fuse. So if an element was drawing 12.5 Amps at 120v you would require 12.5 x 10/8 = 15.6 Amp. This means we need a 20 Amp breaker or fuse and #12 copper cable.

Thanks

Dean

[donf]

Fascinating, give me a chance to learn something here. I do not believe you and I are on the same page.

A Continuous load is defined as a load that will be active for 3 or more continuous hours. In that case, you must increase the supply service rating.

The 80% load you are talking about is called derating. For example, If you are working with three or more current carrying conductors within a conduit, then the load must be derated to 80% of the full load.


What you are describing is more like circuit loading calculations. It says to me that if you have a 15 amp line, then the maximum load you can place on that circuit is 12 amps (20 X .8 = 12).

However the continuous load rule says that if you have a 15 amp line and your load is going to be 12 amps then you must multiply the load by 1.25 and then size the circuit. For example 12 (amps) X 1.25 (Continuous Load multiplier) = Sized Load. (12 X 1.25 = 15)

Here that means that it would be wise to move to a 20 amp breaker and 12 AWG conductors. But supposed the load is 14 amps. Then the formula would yield (14 X 1.25 = 17.5) this would mean that it is mandatory to move to a 20 amp circuit.



17.5 amps would be higher then your mandatory 80% of the load, would you then have to move to a 25 or 30 amp circuit? I'm really curious.

[PPC]

Hi Donf

As I mentioned my main concern at hand is not the continuous load calculations but why is there a problem running on the 240v as opposed to the 120v ?

The element draws 12.5 amp at 120v. So this means the single hot conductor will of the 120v cable will be using 12.5 amp. With the 240v each hot conductor will be using only 6.25 amps thus #14 wire should be well within limits.

As for the us not being on the same page I would hope you could look at the math and see that we indeed are on the same page.

Using your calculation of 125% or 1.25 as the multiplier would be as follows for 12 amps.

12 (amps) X 1.25 (rate) = 15 (max amps).

Using our code calculation would be as follows.

12 (amps) X 10/8 = 15 (max amps). As 10/8 = 1.25 or the same multiplier you are using.

So when you stated "Personally, I would move to 12/2 AWG w/ground and use a 20 amp breaker" I assume you mean when running the single element 1500w heater on 120v. If that is the case I am in agreement with that fully using our calculations. But the main concern is why is the unit popping the breaker on 240v using the #14 as recommended in the chart above for a single element 1500w heater.

If you are still in question with regards to us being on the same page I will cut & paste right from the code book and I will let you make your conclusions from that.

In the mean time not getting any solid response back on the 240v issue other than the reply from EPMiller we decided to test unit using 120v on #12 wire. So far so good and will update the final outcome.

In meantime if anyone has any more comments on the issue with running the heater on 240v your opinions are welcome.

Thanks again

[Tev]

Think of the heating element as a resistor because that's basically what it is. With a resistive load current does not change inversely proportional to voltage, it changes directly proportional to it. This means that if you double voltage then current also doubles since the resistance is a constant and does not change. Here is some math.

1500W @ 120V is drawing 12.5 A

Resistance is volts divided by amps, in this case it's 9.6 Ohms.

At 240V and 9.6 Ohms it would draw 25 amps. 240v times 25 amps is 6000 watts

That's 6000W on an element rated at 1500.

So my guess is that the heat could not be dissipated quickly enough causing the wiring to get hot. After awhile the wires transferred the heat to the breaker causing it to trip. Also, if a nameplate says 120V then use 120V. Also also, I would expect that element to have a severely shortened lifespan. I'm a little surprised it didn't die during the experiment.

[EPMiller]

<snip>As for the continuous load theory I think we are both talking about the same thing. Our code book up here in the snow country (Canada) words it a bit different.

Rule 8-302(2). The supply circuit must not be loaded to more than 80% of the rating of the breaker or fuse. So if an element was drawing 12.5 Amps at 120v you would require 12.5 x 10/8 = 15.6 Amp. This means we need a 20 Amp breaker or fuse and #12 copper cable.
<snip>

Hah. Got to love it when the question comes from somewhere with different rule books. I guess you have to go with the #12 wire. I can get away with the #14 /15A because it is not a continuous load.

As to the correct voltage, Tev has a good explanation of why it most likely is a 120v element. It must be a pretty good one if is isn't already fried by feeding it 240v although I recall something about a 208v rating. Hope it doesn't quit early because of the high voltage experiment.

EPM

[Stratmando]

I feel it may be 120 volt element, may opened up with 240 volts applied and at same time it tripped first breaker.(forgive if I missread something)
I would check for voltage at element and an amprobe around would show it is drawing current(not open).
If it has voltage and doesn't draw current, it is open.
Never ran #14 to a Water Heater.

[EPMiller]

<snip>Never ran #14 to a Water Heater.

Strat,

I have. 1000w 120v POU unit.

EPM

[Stratmando]

I have always Ran #6's for Ranges, even though overkill in many cases, if an upgrade in the future occurs, it will be ready and just the breaker needs to be changed.
If I was wiring a house with a 20 Amp Dryer, I would still pull #10, in case in the future someone wanted to use a regular? Dryer?
Down here there is a supplement to the NEC,
And says No #14 allowed, and for 30 years or more down here aluminum is not allowed.
Trailers made up north, with aluminum wiring would be brought down here, finally that wasn't allowed, Many Problems, fires, we are surrounded by Salt air, corrosion.

[Tev]

I recall something about a 208v rating.

Yeah, apparently the element itself said 1125W @208. If two elements were wired in series at 208 then each would be 1125W for 2250W total. This water heater only has a single element. Probably why the nameplate only listed 120V.

[EPMiller]

Strat,

I agree, I do the planning for upgrade thing too, but a 120v 2-1/2 gallon squirt under a vanity? If you have to go bigger you still can double the heating capacity by going to 240v.

Tev,

Good detective work. That series element idea didn't cross my mind, but the math works out. Of course 2 elements in series could also be run on 240v. Never saw elements in series.

EPM

[donf]

Just for grins, I poked around the Giant website today. Lo and behold there seems to be an interpretive error.

One statement says, "For 120 or 240 VAC" installations. However, that statement is misleading. It does not mean that any heater can be supplied by either 120 or 240 VAC feed. It actually means that for a 120 VAC, 12.5 amp circuit you should purchase a 1500 watt unit. If you are going to install the unit on a 240 VAC, 12.5 Amp circuit, you need to purchase a 3000 watt unit

This is supported by the spec. sheet that was provided by Giant. On that sheet, it lists the 120 VAC, 1500 Watt unit as needing a 20 amp line using 12/2 cable. (Continuous use device)

There is also a 3000 Watt model of the 130E product line. This is for use with the 240 VAC, 12.5 Amp installation.


FYI - extracted from the Giant Water Heater model 130 - Parts List.

#6 03G15/80 Screw-in element all 120V (-1R7N) ["1" I believe stands for 1500 watt]
04G30/80 Screw-in element all 240V (-3R7N) ["3" I believe stands for 3000 watt]

PPC: I'm concerned (or confused) with the math behind your service calculations. If you are to create a circuit that is only loaded to 80% of its maximum you are derating the circuit. On a 10 amp breaker this would mean (10 X .8 = 8 amps). The maximum load you should put on this circuit is 8 amps. (At least the way I see the formulas working, am I incorrect?)

In your situation, Giant tells you that your required amperage is 12.5 amps. Water heaters, here in the U.S. are considered to be "Continuous Use" devices. According to the NEC, that means you have to size the circuit to 125% of the required amperage. Math wise, that is 12.5 X 1.25 = 15.6 amps. To me, this would mean that the required circuit should be raised to 20 Amps. A 20 amp breaker, 12/2 w ground conductors (Supported by the spec sheet from Giant - - 120 VAC, 20 amp, 12 AWG conductors, not configured on a 15 amp line). The circuit restiance would be Voltage / Amperage = Ohms > 120 / 12.5 = 9.6
Of course if Canada does not consider this device to be a continuous use device, then yes a 15 amp line would be sufficient, but the spec sheet does call for 12 AWG conductors on a 20 amp circuit.

Calculations for the 240 VAC circuit - - Watts/ Voltage = Amperage > 3000/240 = 300/24 = 12.5.
Amperage x Continuous use = required conductor load > 12.5 X 1.25 = 15.6
Voltage / Amperage = Ohms > 240 / 12.5 = 19.2

Personally I would go with a two pole, single throw 20 Amp breaker and 12/2 AWG Conductors. While 15.6 amps is within the normal +/- 10% of allowed current, I would rather be on the safe side and have conductors that I know can handle the load. (My opinion only)