Water Heater Monitoring

[lordedmond]

Can I butt in on this

Excuse the safety warning but be extremely careful with CT,s

They must always be 1 connected to the ammeter or 2 fitted with a short across the out put terminals

This is the case In circuit or out

The short is the same as the meter and no harm will come to the CT

Reason they are capable of producing a high voltage on the output terminals

It's the turns ratio one turn on the primary and many on the secondary eg it a step up transformer
please be careful this does not apply to VT or PT both are 110vac and a short on these will kill them

Stuart

[awemawson]

Now if the original poster measures current using his device, and sources that current from 110 volts (as he did) and gets a correlation between wattages (VxA) and indication of 12.2 watts per milliamp, what I am saying is if he sources his power from 220 volts each milliamp no long represents 12.2 Watts, it now represents 4 x 12.2 = 48.8 Watts per milliamp.

His reading of 102.8 mA on the actual heater equates to 48.8 x 102.8 = 5kW which is rather high as the heater is rated at 3.8 kW but is probably within the (in)accuracy of the equipment being used to measure it.

  The above is of course on reflection a load of old balony 

He is measuring current, which he has previously related to power at 110 at 12.2 watts per  milliamp.

So measuring the heater running off 240 his 1 mA reading is now (240 / 110) * 12.2 or 26.62 watts per milliamp. So 102.8 mA represents 2.7 kW

Again probably within the measurement accuracy 

[vtsteam]

Total agreement with you on that last, Andrew

Or a manufacturer rated an  2500 watt element as a 3700 watt.

Nah, impossible.......

[vtsteam]

Seriously, though....

element ages, resistance wire thins, resistance increases, wire heats hotter
hotter wire increases resistance
element ages, scale builds up on shield
scale builds up on shield, element is insulated, temp builds up inside
temp builds up, resistance wire thins further
wire thins, resistance increases, wire heats hotter
hotter wire increases resistance
etc.

[BillTodd]

There is no saving in letting the water cool then reheating, unless the water is left to cool to ambient for some while.

I'm not so sure about that - if the water cools, then the heat loss will be reduced and because less heat has been lost the energy required to bring it back up to temperature will be less than would have been required to maintain the temperature.


Russell

OK not absolutely 'no saving' , but the saving are vanishingly small if the quantity of water is large and the insulation is good.

Essentially the money is better spent on insulation than on trying to match the use to the household . Unless the water drops to ambient for sometime, there is no serious power saving (don't forget to include losses in the heating process as you're trying to get it hot again in a hurry).

[RussellT]

I agree that you will make best savings by adding insulation, but a timeswitch will also produce significant savings.  In a lot of households water only needs to be heated for a couple of hours a day if the tank is well enough insulated.

saving are vanishingly small

I'm not sure I'd agree with that either.  I was making the point about a tank with no water drawn off to clarify that there must be a saving but in practice things are more complicated.  In most domestic hot water systems with a storage tank the hot water sits on top of the incoming cold, so it's quite possible to have the bottom of the tank warming up as the incoming water is cooler than the air surrounding the tank while the top half is losing heat.  Interestingly in that situation it's possible to reduce the water temperature at the tap by putting the heater on because the heater starts convection currents and destratifies and mixes the contents of the tank.

There are lots more factors at work here too - for example where is the thermostat?  Most gas heating systems have a thermostat about two thirds of the way down the tank and so won't heat more water until a third of the hot water has been drawn.  In most electric systems the thermostat is combined with the heater so won't come on until most of the hot water has been used.

As far as losses in the heating process go they are practically zero for electric heating as all the electricity is turned into heat and is surrounded by water - the heat has to go into the water. (I'm assuming cables are correctly sized etc).

In the case of gas heating if you are heating a cold tank the boiler efficiency will be improved compared to a tank that is already hot because the water returned to the boiler will be cooler and heat transfer in the boiler will be more efficient.

Ok. I'll get off my hobby horse now.

Russell

[Manxmodder]

Cheers Russell,that last part answers my earlier query about heat transfer efficiency between the heat source and the heated liquid becoming less effective as the 2 temperatures become closer.

That is how I have always understood it to be......OZ.

[awemawson]

"Rate of heat transfer is proportional to the absolute temperature difference"

That was the answer to the ONLY technical question I was asked in the interview for University  applied physics course

There were lots of other none technical ones and after all they had my exam results but I was rather shocked to say the least - perhaps they were surprised that I got it right 

[Manxmodder]

Cheers Andrew,this is evidenced by simple observation of how the cooling system on a car engine works more effectively in cold weather conditions vs heatwave conditions. The only thing that has changed is the heat differential between the engine coolant and ambient air temperature.

It therefore is also clear that a good deal of energy can be saved with a water heating device by having the upper limit on the thermostat set at a temperature that isn't heating the water to a higher temp than needed.
.....OZ. 

[sparky961]

Who knew such a mundane topic could elicit such spirited response? I'm thrilled.

So, I stuck it with a meter tonight just to make sure what I was working with.  I currently get 121V and 243V, which I'm sure will vary by a few volts depending on the current grid load.  If I go with Andrew's updated math it still gets me back to my original number (which I had yet to double).  I'm favouring the hypotheses that the element isn't accurately rated or my makeshift setup isn't accurate enough, but I still have to wonder if it worked out ok for 60W (I tried 3 different bulbs) and 100W why it wouldn't for 3800W.  True that incandescent bulbs offer non-linear resistance, but the current is inversely proportional to temperature and certainly stable once warmed up.  I was indirectly measuring the current through whatever effective "on" resistance was there at 120V.  It would be better to characterize my setup using a known resistance closer to that which I'd like to measure, and some points in between but I don't have that many 100W light bulbs to connect in parallel** to get a few KW....... or do I....?

(** Boy, just about put "series" there.  That would have caused a RIOT!)

The side debate that's been going on regarding whether putting a timer on it can make it run more efficiently (to really distill it down) is essentially what I'm trying to prove or disprove to myself.  The most accurate way I can come up with is to actually measure the power it uses and log the data over time, giving me kWH to compare timer vs. no timer.  True, there are commercial meters that do just this but they cost money (why not buy a steam engine kit instead of making all from stratch? - age old question, no good answer) and more importantly I don't like the interface.  A simple display doesn't let me process the data as I will inevitably want to.

Bottom line?  Geez, I don't know. ;)

[sparky961]

Excuse the safety warning but be extremely careful with CT,s

The short is the same as the meter and no harm will come to the CT

Reason they are capable of producing a high voltage on the output terminals

Thanks, I did see that when doing some preliminary research.  You've reminded me of the importance.

[sparky961]

He is measuring current, which he has previously related to power at 110 at 12.2 watts per  milliamp.

So measuring the heater running off 240 his 1 mA reading is now (240 / 110) * 12.2 or 26.62 watts per milliamp. So 102.8 mA represents 2.7 kW

Again probably within the measurement accuracy 

I did notice that on different DVOM ranges that the reading changed fairly significantly (10's of mA).  My assumption (that bad word again) was that it had to do with the meter's internal resistance so I thought I could negate that effect by using that range for all my measurements.  Do you think I could improve accuracy by using an external resistance in the CT loop?  It did nag at me a bit that the measurement was inconsistent on those different ranges.  I suppose the way to answer the question is by collecting more data from known sources.

[vtsteam]

Sparky, all theories about why, and projections from them aside, as I said earlier, you can simply characterize your meter by a few more data points, and that's all that is really needed.

1.) do a reading for a 200 watt bulb (or so)
2.) do a reading for the water tank with both elements on (probably when it is run out of hot water completly)

That will allow you to do a graph of 5 points instead of 3. Much more meaningful. Then we can see what's going on rather than guess what's going on.

[vtsteam]

And yes, you are depending on meter resistance, which is itself variable, not fixed. It's usually reckoned in ohms per volt.

But this is again just one more possible theory, not a replacement for data, which would show you what is going on before attempting to explain why it is going on, So forget about meter resistance.

Get the data first, and then you will likely find the answer out of all the theories, because it will point to one or the other. The curve(s) will suggest the answer.

[lordedmond]

I am getting confused here not an unusual thing

Are we talking about measuring AC current still
The term ohms per volt refers to the meter when used in volts , current is measured by referencing the volt drop across a shunt ,which is very low and the meter resistance will have very little impact being in parallel with the said shunt

As to using lamps are they 100 watt ? Have you proved that because it seems like they are being used as a standard

If you want to be sure get a clamp on amp meter and a decent volt meter and go from there , then if you need to remotely measure use the figure produced to calibrate you gizmo

Normal V I calc will be ok as we are dealing with a resitive load and power factor is unity so has no effect on the result

Stuart

[vtsteam]

Okay, my turn to be confused (or perhaps, confusing).

Stuart, isn't he actually showing a few miliamps across a shorted secondary of a very, very inefficient 110 V step up transformer?

Inefficient as the transformer is, meter resistance must be the big factor in limiting current. A true short would theoretically pass infinite current no matter how low the voltage, unless there was some resistance in the circuit. That resistance is significant no matter how low, if the voltage is correspondingly low.

And I don't think the meter resistance in this case is likely to be a constant for two different primary voltages.

[BaronJ]

Hi Guys,

A number of posts have touched on, but not actually said that a burden resistor is needed on the output of a current transformer used to measure AC current.  The burden resistor is needed to firstly prevent dangerous voltages being produced by the large turns ratio and secondly to provide a reference for the current being measured.  The meter should be set to read voltage and should have, ideally, infinite impedance.  Once these conditions are met the calculations for power are straight forward.

[DavidA]

As promised above,  here is the result of my water raising experiment.

Equipment.

2,200 Watt electric kettle.
Workzone watt meter.
Therma 3 thermocouple thermometer.

Procedure.

1 litre of tap water added to kettle and left to stand for a while to allow temperature to settle.
Thermo couple placed in kettle.
Kettle plugged into powermeter.

The powermeter also displays voltage and current as well as wattage. It has a second counter on it.

Power supply to meter/kettle turned on. turned off after 120 second.

water in kettle stirred around.
Temperature taken

Result.

2,200 cc of water raised from 10 C to 37 C in 120 second.  = 27 degree raise.

Dave.