Equipment drawing WAY more kwh than it is supposed to...

kateddApril 7, 2010

We 'inherited' one of the 'swim-in-place' pool systems. It is an above ground vinyl pool with an hydrolic swim-current generator.

The meter was spinning so fast I could barely find the black part of the wheel.

I did a worst case scenario calculation assuming both the water quality filtration unit pump and heater were running 24 hours a day (which they were not) and used the (actual) use time of the swim system ... 3 hours total for the 9 days: (please check my calculations... I am a biologist, not a mathematician/physicist... heh)

Heater: 5.5 kw heater running 24 hours for 9 days = 1188 kwh/9 days

Filtration pump: 120w = 0.120 kwh x24x9 = 25.92 kwh/9 days

Current unit: (5 hp x 746 watts)=3.73 kw x 3 hr = 11.19 kwh/9 days

A high estimation of household electrical use: 225 kwh/9 days

Total estimation for the 9 days, assuming 24 hour operation of pump and heater: 1450.11 kwh/9days.

At a projected cost of $435 for the month, that is bad enough, but the actual reading for the 9 days was 2337 kwh/9 days (estimated cost for the month $701) and that was with the pump and heater NOT running 24 hours a day. Obviously, the system has been shut down...

I am going to assume that the electrician who ran the 220V/30A line that this beast plugs into did everything right (mostly because I can not think of a way they could have screwed it up that would account for this) and start troubleshooting the three components of the system. Filter pump, heater and hydrolic pump unit.

I am going to start with the heater because it apparently froze and the solder joint of the tube that holds the heating element busted, spewing water all over the insides. I think I can figure out where to put the amp-meter probes on that one (although hints would be welcome). The one that I am really not sure where to start with is the hydrolic unit. Any suggestions?

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katedd

Sorry... I left out the most important part:

Thank you for taking the time to read my mini-PHd thesis and thank you in advance for any help you can give.

    Bookmark   April 7, 2010 at 7:17PM
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katedd

Update: We have narrowed it down to the heater:

System unplugged: Wheel on meter takes 24 counts to turn
System plugged in, but heater turned off: 24 counts
System plugged in, heater on: 4.5 counts

Tomorrow, we will start checking to make sure the electrician wired it all correctly and then start looking for where the issue is... any ideas are still very appreciated.

Kate

    Bookmark   April 7, 2010 at 8:40PM
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hrajotte

It seems you answered your own question. Nothing appears to be wrong. That heater sucks up power. It probably cycles on & off (I hope!) and therefore will not be "on" at least some of the time.
I have an electric HW heater with a 5500 watt lower element. It's on a separate meter, which really flies when the lower element is on!

    Bookmark   April 7, 2010 at 10:43PM
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tim45z10

Make sure the line voltage is coming from 2 seperate legs. If the entire power is only on one leg it can have an affect on the meter.

    Bookmark   April 7, 2010 at 11:47PM
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brickeyee

"Make sure the line voltage is coming from 2 seperate legs. If the entire power is only on one leg it can have an affect on the meter."

Say what?

The power for the heater would be 1/4th if it was connected across 120 V.

Meters are not affected by load imbalance between the two 120 V legs.

    Bookmark   April 8, 2010 at 7:37AM
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katedd

"It seems you answered your own question. Nothing appears to be wrong. That heater sucks up power. It probably cycles on & off (I hope!) and therefore will not be "on" at least some of the time.
I have an electric HW heater with a 5500 watt lower element. It's on a separate meter, which really flies when the lower element is on!"
-----

I know that the heater draws a lot, compared to other things, but the calculations I did that show that we are using almost twice what we 'should' be using assumed that the heater was on 24 hours a day, everyday (which it is not).

You indicate that you have a 5500 W heater also... I am assuming you do not have over a $700 electric bill each month, which is what we will have if I keep it on. lol

We are going to try changing the element first.

    Bookmark   April 9, 2010 at 12:27AM
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tim45z10

If the system is a 220 and the both breakers are on the same 110 vac input it can have an effect on the meter.

    Bookmark   April 9, 2010 at 1:23AM
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DavidR

If the load is 240v and you try to power it from two breakers on the same leg of the mains, it most certainly will "affect" the meter - because the load won't receive any power!

Kate, you can't really read the heat

    Bookmark   April 9, 2010 at 2:51AM
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DavidR

Oops, sorry, let's try that again.

If the load is 240v and you try to power it from two breakers on the same leg of the mains, it most certainly will "affect" the meter - because the load won't receive any power!

Kate, you can't really read the heat er's actual load from the hoome's kWh meter readings. There are too many other loads mucking up the numbers.

I think you need to put an ammeter right on that heeter. You could also measure the element's resistance with an ohmmeter and calculate the current with Ohm's Law. Ohm's Law : power = voltage squared, divided by resistance in Ohms.

Don't just go changing parts willy-nilly; do some diagnostic work first. A cheap multimeter from the local harware store will serve you well and probably svave you money in the long run.

All that said, I'm concered about your installation there. You say this machine is connected to a 30 amp supply. For sustained loads that supply should be limited to 24 amps, or 5760 watts. The 5.5kW heatr alone, even at rated power, would just about max out that circuit.

Motor loads are different and I'm not an expert on them, so I'll defer to others if they say it's OK. But it seems to me that if the nameplate rating on that motor is really 5hp, it shouldn't be connected to a 30 amp supply along with a 5.5kW heatr. Anybody else here want to correct or amplify this?

    Bookmark   April 9, 2010 at 2:55AM
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brickeyee

"You could also measure the element's resistance with an ohmmeter and calculate the current with Ohm's Law. Ohm's Law : power = voltage squared, divided by resistance in Ohms."

Except that heaters, like light bulbs, change there resistance between cold and hot.

As the element heats up its resistance rises, and by enough to affect long term power dissipation.

The heater will NOT stay on all the time either.
It has a thermostat controlling the water temperature.

    Bookmark   April 9, 2010 at 3:50PM
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DavidR

> As the element heats up its resistance rises, and
> by enough to affect long term power dissipation.

True - but maybe not very significant for this situation. For an immersion element, the temperature of the nichrome wire buried inside is going to stay pretty close to the water temp, no?

From 20 deg C to 90 deg C, the resistance of nichrome wire increases 1.7%. Suppose the rated power of the element is 4800 watts and it operates at 240v. The resistance of the element at operating temperature is 12 ohms. Let's assume the operating temp is 90 deg C, a little high for water you're going to swiom in, but it makes things easy. The resistance at 20 deg C is 12/1.017, or 11.8 ohms. Close enough. (Actually, that 1.7% is probably pretty close to the inaccuracy of a cheap hardwware store multimeter.)

So, in this case I think a resistance measurement will probably serve as an acceptable "go / no-go" test.

That said, a cheap clip-on ammeter would probably be even less accurate, and would cost more, but might be easier to use.

    Bookmark   April 11, 2010 at 7:37PM
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weedmeister

My question is, what else do you have in this house that is sucking up power? Heated floors?

    Bookmark   April 11, 2010 at 9:33PM
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brickeyee

"For an immersion element, the temperature of the nichrome wire buried inside is going to stay pretty close to the water temp, no?"

Not even close.

The element is packed with ceramic to spread the heat out for a reason.

    Bookmark   April 12, 2010 at 8:49AM
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