Domestic water heater, 240V, underpowered...

Thread Starter

Externet

Joined Nov 29, 2005
2,201
Hi all.
A plain 40 gallon domestic water heater, meant to run at 240V, usually has two heating elements, one for the upper section, one for the lower section each about 3500 Watts, and their thermostats switches one or the other into operation, as far as I know.

That amount of power allows the incoming water to heat up to about 150F in a 'reasonable short' amount of time. Guessing 1 hour.

With a limited amount of current capability from mains and available breakers in the distribution panel, am asking these scenarios:

1. Powering with 120V instead. Will the temperature eventually reach 150F, after a much longer time ? The heating elements themselves can still get waaay hotter than 150F.

2. Powering the rewired heating elements in series with 240V. Will the temperature eventually reach 150F, after a much longer time ? The heating elements themselves can still get waaay hotter than 150F

3. Powering the rewired heating elements in series with 120V. Will the temperature eventually reach 150F, after a much longer time ? The heating elements themselves can still get waaay hotter than 150F

4. Replacing the heating elements with ones that can be powered by the limited source, series or normal.

Fast heating is not needed at all. I believe that with time, guessing 6+ hours, water should reach its intended temperature. For a single daily shower, I believe it should work.
Do you agree ?
 
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GopherT

Joined Nov 23, 2012
8,009
This is a question of ohms law and converting the current (amps) to power (watts = volts x amps).

You also need to know that a joule = watts x seconds

And, the heat capacity of water = 4.184 joules/gram/degree C.
That means, it takes 4.184 kilojoules to heat one kilograme (liter) of water by 1-degree C.
 

crutschow

Joined Mar 14, 2008
34,281
Using 120V on the 240v heater elements or placing two in series at 240V will reduce the power by a factor of 4 since power is proportional to the square of the voltage.
The water will still heat to 150F but it will take 4 times as long.
 

Brownout

Joined Jan 10, 2012
2,390
In order to answer the question correctly, you must consider the amount of heat lost through the walls of the water heater. In the normal scenario, there would be enough power supplied to keep up with the heat loss, but by reducing the power by a factor of 4, the point at which supplied power = heat loss, or equilibrium, might not be your target. In that case, more insulation would be necessary.
 

dl324

Joined Mar 30, 2015
16,839
With a limited amount of current capability from mains and available breakers in the distribution panel
Could you elaborate on the problem you're trying to solve?

Is this an existing water heater installation? Or new? Are breakers tripping? Are lights dimming? Addressing the power distribution issue seems like the most obvious solution.

Standard water heater thermostats have no provision for wiring the elements in series. They're configured to heat water in the top of the tank and then the bottom, or to operate both elements in parallel. If you wire in series, your hacking a high current circuit; plus there isn't much room where the thermostat is installed.
 

tcmtech

Joined Nov 4, 2013
2,867
Most any modern electric water heater will work just fine on half its rated input voltage but as others said it will take four times longer to warm up when doing it.

Years ago I lived in a old rental house that had a ancient water heater in it. It was a 40 - 50 gallon unit that used a single 120 volt 1200 watt element in the bottom. It worked great but if you ran it out of hot water it took the better part of the day/night to catch up. :oops:

As for reconfiguring your present unit to run both elements in series it should be pretty easy and all the wiring you would need is already there between the upper and lower thermostats and related heating elements so it would be a simple matter of just moving a few connections around on the two thermostats to put both elements in series.

The second option would be to simply go into the main circuit breaker/fuse box and move one of the hot lines that feed it 240 volts off of the breaker/fuse and connect it to the common line buss bar instead thus giving the unit half the voltage from its power source.
 

Thread Starter

Externet

Joined Nov 29, 2005
2,201
Could you elaborate on the problem you're trying to solve?

Is this an existing water heater installation? Or new? Are breakers tripping? Are lights dimming? Addressing the power distribution issue seems like the most obvious solution.
Thanks, DL234. Current heater runs on gas, and spits dirty rusty water, so I want it gone. I have been given a new modern electric heater 240V 4KW but house has only 100 Ampere service, and a spare 120V slot for a breaker. I could make it a 240V spare slot redistributing loads. Other high current loads are range and clothesdryer. No dimming lights, no power issues, no wiring issues, all solid healthy and running fine.

...Years ago I lived in a old rental house that had a ancient water heater in it. It was a 40 - 50 gallon unit that used a single 120 volt 1200 watt element in the bottom. It worked great but if you ran it out of hot water it took the better part of the day/night to catch up. :oops:

As for reconfiguring your present unit to run both elements in series it should be pretty easy ...
Thanks, tcmtech. That is what I suspected. You experienced what I believed possible. Heating time delay is not important at all; and rewiring the heater elements is not a hurdle.
 

dl324

Joined Mar 30, 2015
16,839
a spare 120V slot for a breaker. I could make it a 240V spare slot redistributing loads.
They sell a dual breaker that fits in a single slot. Don't know if it takes power from both sides of the split phase or is available in the current you need. I have one for 2 15A circuits.

Don't think they come ganged, but I had two separate 120V breakers for the water heater in my barn.
 

BillB3857

Joined Feb 28, 2009
2,570
That double breaker must require a special box to allow 240V. My home box (Cuttler Hammer) has connection tabs that are almost the full with of a standard breaker. A 240V dual breaker for my box has to be made of two full width breakers or they can't connect to the two hot buss bars.

I would also agree with BR-549 that if gas is available, use it! Much higher recovery rate and, in the event of power failure (we were without power for 7 days due to an ice storm), hot showers are still available. We also have a gas furnace which we operated from a small gasoline generator to keep nice and toasty warm.
 

Lestraveled

Joined May 19, 2014
1,946
...... I have been given a new modern electric heater 240V 4KW but house has only 100 Ampere service, .............
4000 watts at 240 volts equals 16.6 amps. You have 100 amp service which shouldn't be a problem.

I suggest that you change the breakers on your 120 volt circuits to half sized breakers and free up the space for a ganged 240 volt 20 amp breaker. (It is perfectly OK to do this.)
 

#12

Joined Nov 30, 2010
18,224
you must consider the amount of heat lost through the walls of the water heater.
I did the math on this long before, "Energy Star" was invented. I calculated $75 per year of heat loss through the standard tank insulation and $25 per year if you added a roll of R13 to the tank. That boils down to 1.9 KWH per day for what is now an antique water heater and 0.63 KWH per day with an extra blanket of Pink Panther.

Where are you now? Contemplating a 3500 watt heater running at half voltage? So .875 KW trying to make up 0.63 KWH? That's less than an extra hour of run time. 43 minutes if you have an antique electric water heater with a roll of R-13 wrapped around the outside of it.
 

Brownout

Joined Jan 10, 2012
2,390
I'm thinking more of my setup. 50-gl water heater in a cold basement ~50 degees F in the winter.

approximating loss through the sidewall:

Diameter ~2ft.
height ~4ft
Insulation ~R15
delta T ~100 degrees F

Power loss(sidewall) = 100*2pi*4/15 Btu/hr = 166Btu/hr = 48W
Power loss(end caps) = 100*pi^2*/15(2) Btu/hr = 131Btu/hr = 38W
Power(total) = 48+38 = 86W.

and Power(In) = 3500/4 = 875W.

Yeah, he's good to go.
 
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