The Shop > Electronics & IC Programing
Water Heater Monitoring
DavidA:
It is fairly straightforward to work out the best route to take.
It takes 1 Watt to raise 1 CC of water by 1 degree Celcius.
So 1 Kw will raise 1 Litre by 1 degree in 1 second
So you need to know the volume of the cylinder, the starting temperature of the water and the power rating of the heat coils.
From this you can calculate the time it will take to raise the water from ambient to the required temperature.
The better the insulation the longer it will take to fall back to ambient and thus the less time the heater will be on bringing it back up to heat.
It would also make sense to only have the tank as big as needed. Less water to heat up.
Dave.
DavidA:
Back to the original problem.
I was wondering if you could leave out the toroid and wind the turns directly around one of the power leads. Obviously not both as it wouldn't work.
If you then place a resistor between the ends of the coil you may be able to measure the voltage generated across the winding by the current induced into it from the current passing through the cable.
As the current through the cable varies so would the current in the coil and from this you would get a change in the voltage across the resistor.
Dave.
Manxmodder:
Hi David, where you say "So 1 Kw will raise 1 Litre by 1 degree in 1 second"
Is that heat transfer rate constant throughout the range from ambient to boiling point,or does the transfer rate slow down as the water temp vs element temp differential becomes less?......OZ.
vtsteam:
There are 4 variable classes in your experiment and 3 data points. No good conclusions can be drawn about what the data means. If you want to know if it graphs as straight line (the original assumption) or a curve, or discontinuous straight lines, or discontinuous curves, you need more data.
Variable classes:
110 V
220V
light bulb (non-linear resistance)
water heater element (non-linear resistance)
Available Data Points:
4.9s ma x 110 V ......60 watt bulb
8.2ma x 110V ..........100 watt bulb
102.8 ma x 220 V ....3700 watt water heater (single element assumed)
Do a 3rd bulb test at 110v to characterize that curve,
And do a 2 element 220V water heater test to determine if those points appear to lie on the same curve as the 110V points. You still need another data point @220V to determine if that curve is a straight line or not. Which, unfortunately you can't get (at least with the water heater.) But you should be able to tell if those points lie on the extended 110V curve.
If they do, you just need to write your formula, and you're good to go with your new induced wattmeter.
David Jupp:
--- Quote from: DavidA on January 06, 2015, 08:16:27 AM ---
It takes 1 Watt to raise 1 CC of water by 1 degree Celcius.
So 1 Kw will raise 1 Litre by 1 degree in 1 second
--- End quote ---
I think there is a factor of 4.2 missing - specific heat capacity of water = 4.2 J/K/g
4.2 kJ required to heat 1kg (1 ltr) of water by 1K (or C) -> to raise 1 ltr of water by 1 K in one second would require power of 4.2kW.
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