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molo
10-11-2007, 08:40 PM
Is there some way I could instal an independent usage meter on an electric water heater to isolate the usage of the heater? I am interested in doing this because I want to see how much the operating cost goes up in colder months.

This is a thread that started from a similar plumbing thread

TIA,
Molo

Bob NH
10-11-2007, 09:49 PM
You would have to buy a meter and socket.

Search e*b*a*y on (kwh, kwhr, kilowatt) meter and take your pick. You will also need a socket that will cost you less than $40 at HD. You will be all set for less than $100.

hj
10-12-2007, 05:50 AM
Is the water heater jacket warm? If not, then the insulation factor is such that there is probably not enough difference to worry about it. And unless you also have gas available, even when you know the answer you won't be able to do anything about it.

molo
10-12-2007, 06:05 AM
I'm trying to determine how much is lost not only through the water jacket, but also through the 10' of uninsulated pipes (prior to going into an insulated a heated room). I wonder if insulaing the pipes will save $1, $5, or $20 a month. I have no idea how much the cost is increased because those pipes are uninsulated, they are hot/warm to touch. Again, it is an electric hot water tank.

TIA,
Molo

Bob NH
10-12-2007, 07:28 AM
The biggest difference in hot water heating from month to month will be from changes in amount of water used and changes in the incoming water temperature. Any differences due to different leakage of heat in different seasons will be small compared to that.

You will not be able to make sense out of differences in electrical consumption unless you measure the gallons of water used and the temperature of the incoming water.

If you are concerned about the amount of heat lost through the insulation and pipe you would be better off spending a fraction of the cost of the meter on extra insulation. That will reduce the losses to a minimum and the difference will be small compared to variations caused by usage and water temperature.

jadnashua
10-12-2007, 07:54 AM
I'll reiterate...measuring the time it takes between the tank turning on to reheat the water after no useage and knowing the amount and delta between when the thermostat turns off and back on again will let you calculate the power usage and therefore the heat loss. if you do this now before the incoming water gets really cold and then again sometime say in February, you should have a good data point. If you can wait long enough, do it again in say August after everything has heat-soaked. You could probably hook something up to tell you how often the thing ran by wiring in a clock circuit for cheap, although you would need to be very careful how you did it to remain safe. There are plug-in watt meters (not applicable here), and probably a wired in version that might be less expensive than using a meter. But, other than interesting, adding insulation is by far less expensive.

Some spec sheets include info on the insulation R-factor. If not, maybe an e-mail or call to the tech support of the manufacturer might give you the info you desire without any further effort.

BrianJohn
10-12-2007, 11:20 AM
Emon makes a digital meter than can be installed with out disconnecting the branch circuit., around $800.00

http://www.electricsubmeter.com/emon/kwh-meters/class-1000.htm

jadnashua
10-12-2007, 11:28 AM
You could probably use a clamp on ammeter. then, take the time it runs to bring the water back up and factor that into how often it runs to reheat the water without use and figure your power usage. Clamp-on ammeters aren't particularly expensive and are useful for more mundane things. Well, it really comes back to how often and how long it runs. Since this is electric, you should already know how many watts it draws. Just figure out how often it runs. A current sensing relay and an old analog electric clock - use the relay to turn a normal 110vac clock on. Start it at 12, then overnight see how long the heater was running. 4500W/hr or whatever power the elements are in your tank. Do that for several nights.

Lots of ways to skin this cat...

got_nailed
10-12-2007, 06:41 PM
I donít know how many watts it takes to raise a gallon of water 10*.

So all this is going to come down to is how much isolation you can pack around your tank.

The main cost is getting the water hot. Once it is hot then it dose not take much to keep it hot.

If you go with this test then you need ever bit of info onÖ
How much water
Flow rate of the water
How much power
Temp of incoming water
Temp of room air
Temp of the temp of the water in the tank
PF and voltage

You will need to be able to change only one of the factors at a time to do this test. If you can not show that you only changed one or two factors at a time then we have a case of apples and oranges. But I would rather have a banana.

If you go back to the test of the CSV and much money you can save all it is, is an oppion because there test did not contain all the info. No two days or weeks or months you will use the water in the same amounts. Today I flushed the water closet at a different time compared to turning on the shower. This can change the amount the pump could cycle.

Iím not trying to be an ass but I think you will spend too much money on something that will just be something that someone did and have now way to back up your findings. Find the CSV test that was done on this site and pick it apart. You might see where Iím coming from.

molo
10-12-2007, 07:48 PM
Thanks,

AI'l look into the CSV test.

Molo

Furd
10-12-2007, 08:00 PM
Molo, don't bother looking for the thread on the CSV. What got_nailed is trying to state is that unless you have a laboratory full of instruments and a computer program to integrate all of the readings it is almost impossible to achieve what you desire.

Insulate the pipes and be done with it. I suggest that you use pre-formed fiberglass insulation with a wall thickness of no less than one inch.

jadnashua
10-12-2007, 08:17 PM
Oh come on...the math is simple. 1BTU=3.415179W. One BTU will raise one pound of water 1-degree F. A gallon weighs (approximately) 8 pounds. So to raise a gallon of water 10-degrees it's 8*10*3.415179 or approximately 273W.

If you know how much water is in the tank (he does). You measure the temp of the outlet water. Shut it off. Let it sit for say 8-hours. Measure the temp, and it is straight-forward, simple math (other than figuring out the conversion factors) that a 5th grader could do. Are you smarter than a 5th grader?

that should tell you the amount of heat (and therefore power) lost over time, since it will take that many watts to reheat it to the normal temp. It gets messier with a flame driven system, since you have to then figure the efficiency, but the same thing applies. If the WH is the only thing operating, and you shut it off, you can do exactly the same thing if you watch the gas meter before and after your test period.

molo
10-12-2007, 09:18 PM
I'm going to pursue the technique that jad suggests, I'll need to consider the time it takes to get the "hot " water to the tap.

Thanks for the suggestions,

got_nailed
10-13-2007, 07:29 AM
Thank you jadnashua I was looking for the math but could not find it.

jadnashua
10-13-2007, 04:36 PM
Run it out of the T&P valve...no delay, as hot as it gets.