Water Heater as a Load

Hi, I am doing a project to control the temperature of a water heater for a etching tank. However, I am not sure about the representation of the load in the circuit simulation.
After doing some research, i have found that a heater is normally a resistive load instead of an inductive load, and the inductance is usually negligible.

However, after measuring the resistance across the live and neutral terminal of the water heater, the measure resistance is only 533Ω.
The water heater is rated at 240V , 100W. The region that I live in is using 240V, 50Hz AC mains.
Thus, using R = V^2/P, the Resistance is supposed to be 576Ω.

Does this mean that there is a partly inductive element in the load. If so, what value of resistor and inductor do you think that I can use to represent the load.

attached image shows the heating element of the water heater.

Swsean said:

i have found that a heater is normally a resistive load instead of an inductive load, and the inductance is usually negligible.

Click to expand...

Correct.
Power is only dissipated ie converted to heat, in a resistance.
Power is not dissipated in a reactance, inductive or capacitive.

Swsean said:

However, after measuring the resistance across the live and neutral terminal of the water heater, the measure resistance is only 533Ω.
... the Resistance is supposed to be 576Ω.

Click to expand...

Simply the difference between theory an practice.
In the real world, most measured values are slightly different from the theoretical predictions.

As for the 100w heater, there is probably a tolerance on the supply voltage, and the designer ensured that it would dissipate 100w at the lower limit of its supply voltage tolerance.
If you do the calculation you will see that the heater will dissipate 100w when the supply is only 230v.

Swsean said:

Does this mean that there is a partly inductive element in the load.

Click to expand...

No.
There will be some inductance, but its effects will be negligible at 50Hz.

Swsean said:

If so, what value of resistor and inductor do you think that I can use to represent the load.

Click to expand...

Try Z = 533 +j0 Ohms.

JimB

This is the circuit that I will use for the phase firing.


Thus, with negligible inductance, does it mean that I can use the circuit form figure 6 instead of figure 7?

The other effect you will typically see with heating elements is due to the temperature coefficient of the material.

The resistance increases as the temperature increases, in other words.

Filament lamps show a rather more extreme change as the temperature change is much higher; the "off" resistance with the element cold may be only around a third the operating resistance.

If you still have incandescent lamps lying around, perform a little experiment and measure its cold resistance.

I've a 65 watt, 120 volt floodlight. The measured resistance is 15.3 ohms. That would mean 941 watts!!

However as the filament increases in temperature all the way to white hot, its resistance increases dramatically, and the actual power is far less.

rjenkinsgb said:

The resistance increases as the temperature increases,

Click to expand...

Yes.
The element resistance will likely rise to be close to your calculated value at a typical water heater temperature.

Swsean said:

This is the circuit that I will use for the phase firing.

Click to expand...

it's best NOT to connect your control circuits directly to the power line. an optoisolator or a Solid State Relay (SSR) are the best approach. there's enough thermal inertia in the heater that a simple temperature controlled on-off system could be used. on the other hand, a very inexpensive way of controlling it would to be the use of a standard light dimmer (200Watt or higher). btw, what is the heater made of? if it's aluminum, you might not want to immerse it directly in etchant, as the ferric chloride attacks aluminum at a much higher rate than copper. perhaps a sealed steel tube around the heater and the empty space filled with water would work as a buffer.