"Derating" a geyser element

[markus_m2]

@frivanand @Modina, in another thread you mention using two capacitors in series to "derate" a geyser element...

Could you please elaborate? This could be a GREAT solution !

My solar water heater element is around 3.5kw, and I've been considering replacing with a smaller element (to stay within the bounds of my PV production), but:

1. SWH doesn't take stock standard elements (Solahart).

2. We'd loose ability to heat up the water relatively quickly if needed. 

Idea: run parallel (seperated) circuits feeding the geyser. Controlled by wifi switches in a way that they're never connected at the same time. 

One circuit straight in (no funky electronic gizmos, straight supply), the other parallel circuit using capacitors as mentioned to "derate" the element. 

What would be needed and how would one wire this 2nd "derating" circuit?

 

 

[Modina]

Firstly note that the following discussion is only valid for resistive loads such as elements.  This will not work when your load is inductive such as a motor or 50Hz transformer.

To de-rate a heater element you could do so in three way:

 1. Use a transformer to reduce the supply voltage.  You would NOT use a transformer in the conventional sense, but use the secondary wired in SERIES to either decrease or even increase you AC voltage.  (Wired in phase would add the transformer voltage, 180deg out of phase would subtract the voltage) This works well but is very expensive.

2.  Use a simple diode, mounted on a heatsink.  The diode will only pass every other half-wave.  It results in pulsating DC.  This is cheap to do.  It has the disadvantage of not reducing the peak current but only reducing the average power.  It is also 50%.  Nothing more, nothing less.

3.  By connecting a non-polarized capacitor in series with the load.

I will cover (3) here:

Let's assume you have a 230VAC 3KW element.  We can calculate it's current as  I = P/V  = 3000/230 = 13.0A.  We can also calculate the element's resistance:   R = V^2 / P  = 230^2 / 3000 = 17.63 ohms

Let us assume you only want to run the element at 50% power.  That means you would need to connect a capacitor in series that has a impedance Z (at 50Hz) that equals the element's resistance.

So let's see what capacitance will give us an impedance of 17.63 ohms at 50Hz.

Z = 1 / [ 2pi.f.c ]   where pi = 3.14159,  f = 50Hz and c= capacitance in Farads    ... (1)

We can re-write (1) as

C = 1 / [Z.2.pi.f]
    = 1 / [17.63 x 2 x 3.14159 x 50]
    = 1.805 . 10-4  F
    = 180 uF

So you would need a 180uF non-polarised capacitor rated for 250VAC.  You could obviously connect a few smaller capacitors in parallel.  If you do, you could add a switch to select more or less paralleled capacitors and thus have some pre-set power settings to select from.

A small capacitor will have a high impedance and the element will see very little current.  To increase the current, you need to reduce the series impedance and thus increase the capacitor size.

Note that no electronic component is ideal.  For instance, a capacitor is not JUST a capacitor.  It has a certain internal resistance and even an internal inductance.  We talk of a capacitor's ESR = Electrical Series Resistance.  You get so called low ESR caps that are used in power supplies, etc.  The ESR will cause the capacitor to heat up.  Now, I have absolutely no idea how much ESR a motor starter capacitor has.  Remember that motor starters only have to work for a very short time.  I can't comment if a 180uF motor start capacitor would be happy to pass many amps of AC voltage on a continues basis.  The capacitor might not be so happy and have a short life expectancy.  One would need to try it and just check that the capacitor doesn't get too warm.



 

Edited May 2, 2023 by Modina

[markus_m2]

25 minutes ago, Superfly said:

Reading @Modina posts is like being in Physics class again . Luv'n it!

I thought the exact same, coupled with the realisation that in my youth this stuff was completely uninteresting, but now I find it fascinating (especially now one can put it to such cool practical use)

Thank you @Modina, really appreciate you taking the time to explain! Will go through your post in detail when I have time, and research more on my own, before embarking on this cool project...Will post results 

[Modina]

About capacitors and the ESR.  I think it would be better to use 2 or even better, 4 or 5 smaller capacitors connected in parallel.  For instance, using 4x  45uF caps in parallel will reduce the ESR by a factor 4.  Many smaller capacitors will be able to handle high current much better than one single large capacitor.

@markus_m2 When I was a high school kid my dad was very disappointed in me for not showing any interest when he worked on his cars, doing some basic maintenance.  Years later, I completely overtook him, by overhauling a GTI engine completely, even opened parts of the K-Jetronic fuel injection components which where sealed and regarded as "unserviceable" items.  Since then, my cars never saw the inside of a service centre.  So yes, learning theoretical aspects when you do not have a direct application is boring.  I have little interest in planes or tractors or even Ferraris, because I will never own, fly or drive them.  I might be more interested in new BMW tech, because I might, in 6 years from now, be able to afford such a car, buying from the 2nd hand market.  

I know many engineers that are dreamers.  Very good with theory and they can tell you everything about various fighter-jets, performance cars, etc.  But often they don't know the difference between the radiator cap and the oil cap.  A school friend's dad was a science professor and he was literally about to pour engine oil into the radiator, luckily my friend could stop him. 

Edited May 3, 2023 by Modina

[markus_m2]

LOL, yes, I have a similar story - car broken down in the middle of Botswana, with my "physics doctorate" mate standing next to me gobsmacked (handing me the tools) while I hot-wired the failed immobilizer....He will NEVER live that story down

 

[Modina]

I was asked back channel to calculate the power that the 3KW element would see when using a 140uF series capacitor.  I will calculate it here:

A 140uF capacitor will have an impedance Z = 1/ [2pi.f.c]  = 1/[2 x 3.14159 x 50 x 140E-6] = 22.736 ohms

The impedance of the capacitor will form a voltage divider with the resistance of the element.  We can calculate the new voltage that will be seen by the element thus:
V_load = V_in x R_load / [R_load + Z_cap]          where:  V_load is the voltage over the element
                                                                                            V_in is the 230V mains voltage
                                                                                            R_load is the geyser resistance of 17.7ohms
                                                                                            Z_cap is the impedance of the capacitor @50Hz, i.e 22.7 ohms
            = 230 x 17.7 / (17.7 + 22.7)

            = 100.77 V

The power dissipated by the 3KW element is:
 P_load = V_load ^2 / R_load
             = 100.77^2 / 17.7  = 573.7 Watts

[BritishRacingGreen]

21 hours ago, Modina said:

Firstly note that the following discussion is only valid for resistive loads such as elements.  This will not work when your load is inductive such as a motor or 50Hz transformer.

To de-rate a heater element you could do so in three way:

 1. Use a transformer to reduce the supply voltage.  You would NOT use a transformer in the conventional sense, but use the secondary wired in SERIES to either decrease or even increase you AC voltage.  (Wired in phase would add the transformer voltage, 180deg out of phase would subtract the voltage) This works well but is very expensive.

2.  Use a simple diode, mounted on a heatsink.  The diode will only pass every other half-wave.  It results in pulsating DC.  This is cheap to do.  It has the disadvantage of not reducing the peak current but only reducing the average power.  It is also 50%.  Nothing more, nothing less.

3.  By connecting a non-polarized capacitor in series with the load.

I will cover (3) here:

Let's assume you have a 230VAC 3KW element.  We can calculate it's current as  I = P/V  = 3000/230 = 13.0A.  We can also calculate the element's resistance:   R = V^2 / P  = 230^2 / 3000 = 17.63 ohms

Let us assume you only want to run the element at 50% power.  That means you would need to connect a capacitor in series that has a impedance Z (at 50Hz) that equals the element's resistance.

So let's see what capacitance will give us an impedance of 17.63 ohms at 50Hz.

Z = 1 / [ 2pi.f.c ]   where pi = 3.14159,  f = 50Hz and c= capacitance in Farads    ... (1)

We can re-write (1) as

C = 1 / [Z.2.pi.f]
    = 1 / [17.63 x 2 x 3.14159 x 50]
    = 1.805 . 10-4  F
    = 180 uF

So you would need a 180uF non-polarised capacitor rated for 250VAC.  You could obviously connect a few smaller capacitors in parallel.  If you do, you could add a switch to select more or less paralleled capacitors and thus have some pre-set power settings to select from.

A small capacitor will have a high impedance and the element will see very little current.  To increase the current, you need to reduce the series impedance and thus increase the capacitor size.

Note that no electronic component is ideal.  For instance, a capacitor is not JUST a capacitor.  It has a certain internal resistance and even an internal inductance.  We talk of a capacitor's ESR = Electrical Series Resistance.  You get so called low ESR caps that are used in power supplies, etc.  The ESR will cause the capacitor to heat up.  Now, I have absolutely no idea how much ESR a motor starter capacitor has.  Remember that motor starters only have to work for a very short time.  I can't comment if a 180uF motor start capacitor would be happy to pass many amps of AC voltage on a continues basis.  The capacitor might not be so happy and have a short life expectancy.  One would need to try it and just check that the capacitor doesn't get too warm.



 

Well explained @Modina , thanks. 

For  the ordinary mortals like me and others , what  Modina has actually done here is carefully selecting the right capacitance to affect a  particular phase angle between voltage and current , due to the combined R+C circuit , in order to affect the wanted amounted of real power across the element. Without wasting real power to do so , and without needing to alter the voltage magnitude. (well we will waste a bit of real power due to the ESR of capacitor , as he explained). EDIT : And without the horrible need to chop the sinewave with a triac or SCR in order to reduce power.

This is  not unlike the fundamentals needed  to entertain grid tie power control . The element is replaced by an inductor (reactance) , one side is fed by a generator (grid), the capacitor is replaced by an inverter  which synthesizes the required difference in  phase angle artificially in relation to the generator voltage . As a result the magnitude and direction of real power is controlled by the inverter increasing or decreasing the phase angle.  Because we can reverse power flow direction by merely adjusting the phase angle , this now becomes the bidirectional nature of the inverter. In the one direction we push power towards the grid and onto the loads , in the other direction we pull power from the grid in order to charge battery when required.

I look forward to some practical  test of the 'derated' element. The ESR might be a bigger problem than we think , I am interested to learn.

EDIT : don't use your current clamp to measure the resultant current thru the element , cause it will fool you by still showing the full 13amps for a 3kW element.

Edited May 3, 2023 by BritishRacingGreen

[Modina]

4 minutes ago, BritishRacingGreen said:

I look forward to some practical  test of the 'derated' element. The ESR might be a bigger problem than we think , I am interested to learn.

Well, @frivan reported having used these 2x 70uF caps but he didn't say for how long.  Him only getting just under 600W will keep the current on the low side.  If someone wants to run the element at substantially higher values, say over 1KW, then I would strongly recommend using 4 caps.  But this is only an educated guess.  One should run such a setup for 20 minutes and then feel how warm or not these caps run.  I wouldn't want them to run more than a few degC over ambient.

[frivan]

2 minutes ago, Modina said:

Well, @frivan reported having used these 2x 70uF caps but he didn't say for how long.  Him only getting just under 600W will keep the current on the low side.  If someone wants to run the element at substantially higher values, say over 1KW, then I would strongly recommend using 4 caps.  But this is only an educated guess.  One should run such a setup for 20 minutes and then feel how warm or not these caps run.  I wouldn't want them to run more than a few degC over ambient.

The running caps I got had a plastic outer, so not great for heat dissipation. I got the impression that the power fluctuated with voltage. So, 210V would draw noticeably less power than 230V. I actually used a Sonoff dual relay and TH16 to bring in the two caps at different times and follow the solar profile... then I took out my prepaid meter and stopped using the capacitors.

[Modina]

Yes, these sort of motor start capacitors are normally plastic these days.  A capacitor is not a component that should get warm at all.  Voltage fluctuations will definitely cause power variations, 20V fluctuations would indeed be noticeable.

[Scorp007]

Thanks for the time to explain and calculate. 

Due to the Xc being quite a bit more than R do you ignore the PF that @BritishRacingGreen

Also mentioned this relationship. 

 

[sjp100]

Awesome thread, exactly what I needed. @Modina would you mind double checking my math please.

I have a 2kw element that I need to run at 1kw. My element has a resistance of 25 ohm, thus:

V=(P*R)^0.5 ... V=(1000*25)^0.5 ... V=158v

Thus the capacitance impedance needed is Z_cap=((V_in*R_load)/V_load)-R_load    V_load is the voltage over the element
                                                                                                                                            V_in is the 230V mains voltage
                                                                                                                                            R_load is the geyser resistance of 25ohms
                                                                                                                                            Z_cap is the impedance of the capacitor @50Hz

Thus Z_cap=((230*25)/158)-25 ... Z_cap=11.392 ohms

Thus the capacitance needed is c=1/(2*pi*f*Z_cap) ... c=1/(2*3.14*50*11.392 ... c=279uF

 

Now from what I remember from Varsity, and according to monsieur Google, the capacitance is added together for capacitors in parallel, so 6*45uF capacitors like this one should be close enough, I can remove or add one if needed.

 

[Scorp007]

7 hours ago, sjp100 said:

Awesome thread, exactly what I needed. @Modina would you mind double checking my math please.

I have a 2kw element that I need to run at 1kw. My element has a resistance of 25 ohm, thus:

V=(P*R)^0.5 ... V=(1000*25)^0.5 ... V=158v

Thus the capacitance impedance needed is Z_cap=((V_in*R_load)/V_load)-R_load    V_load is the voltage over the element
                                                                                                                                            V_in is the 230V mains voltage
                                                                                                                                            R_load is the geyser resistance of 25ohms
                                                                                                                                            Z_cap is the impedance of the capacitor @50Hz

Thus Z_cap=((230*25)/158)-25 ... Z_cap=11.392 ohms

Thus the capacitance needed is c=1/(2*pi*f*Z_cap) ... c=1/(2*3.14*50*11.392 ... c=279uF

 

Now from what I remember from Varsity, and according to monsieur Google, the capacitance is added together for capacitors in parallel, so 6*45uF capacitors like this one should be close enough, I can remove or add one if needed.

 

A post from Modina on 31 July. Your calcs seem to be correct. 

I don't read or follow any publications anymore.  I have divorced myself from technical stuff.  I refuse to listen to this post-Covid bullshit. 

[sjp100]

34 minutes ago, Scorp007 said:

A post from Modina on 31 July. Your calcs seem to be correct. 

I don't read or follow any publications anymore.  I have divorced myself from technical stuff.  I refuse to listen to this post-Covid bullshit. 

Thank you. Do you think the capacitor will work, or do you maybe have a different suggestion? 

https://www.glolighting.co.za/vossloh-schwabe-528555-89-capacitor-type-a-45mf-250v-with-250mm-leads-stud.html

[Scorp007]

49 minutes ago, sjp100 said:

Thank you. Do you think the capacitor will work, or do you maybe have a different suggestion? 

https://www.glolighting.co.za/vossloh-schwabe-528555-89-capacitor-type-a-45mf-250v-with-250mm-leads-stud.html

@frivan used it. See earlier in this topic. It should work but not sure if one should not try to get caps rated at 400V AC. Also without trying it I am not sure how reliable if in circuit for a few hours at a time. With only 1kW the geyser can take long to heat up.

I would 1st try it with 2 caps and test and measure before buying 6 caps. 

Use a fine over current MCB like a 10A MCB. 

[GerhardK83]

I would use this capacitor available from Tecsa https://www.tecsareco.co.za/product/capacitor-45mfd--450v/506

There are 2 different capacitors available for single phase induction motors because there are different types of single phase induction motors namely Capacitor Start Induction Run and Capacitor Start Capacitor Run induction motors, the capacitor linked above is used as a run capacitor for an induction motor that is why the enclosure is aluminium so that the heat can be radiated to the environment, the capacitor in the plastic housing is used as a starting cap.

[sjp100]

8 hours ago, GerhardK83 said:

I would use this capacitor available from Tecsa https://www.tecsareco.co.za/product/capacitor-45mfd--450v/506

There are 2 different capacitors available for single phase induction motors because there are different types of single phase induction motors namely Capacitor Start Induction Run and Capacitor Start Capacitor Run induction motors, the capacitor linked above is used as a run capacitor for an induction motor that is why the enclosure is aluminium so that the heat can be radiated to the environment, the capacitor in the plastic housing is used as a starting cap.

Thank you. There is a Tecsa right around the corner from me so it suits me perfectly 

Quick update. Got 3 capacitors from Tecsa, all they had in stock. More available at other branches so will do a trip next week. Unfortunately they only have the start capacitors and don't stock the run capacitors. Kinda forced to use them but will keep searching for run capacitors.

Edited September 2, 2023 by sjp100
Added info

[JustinSchoeman]

You also need to take power factor into account.  The power factor of a capacitive dropper is roughly equal to the output voltage to the input votage.

So, if you halve the power, the power factor will be around 0.7.  Inverters are designed and rated for a PF of 1, and generally derate from there.

So, while you halve the heating capacity, you only reduce the effective inverter load by 30%.  So the win may not be as big as you expect.

[frivan]

These were the capacitors I used. The aircon shop said they were 70uF running caps. It operated for a year without issues.

[Scorp007]

30 minutes ago, frivan said:

These were the capacitors I used. The aircon shop said they were 70uF running caps. It operated for a year without issues.

I must say I haven't seen any new alu cased caps in the last perhaps 10yrs. All the caps we sell for AC gate motors that are start and run are the plastic cased.

[TaliaB]

38 minutes ago, Scorp007 said:

I must say I haven't seen any new alu cased caps in the last perhaps 10yrs. All the caps we sell for AC gate motors that are start and run are the plastic cased.

Digikey sell aluminum caps (Nichicon) but they are expensive R380 for 220 microfarad 400v

[sjp100]

So wiring isn't my strong suite, am I wiring them up correctly? All 3 capacitors in parallel. They don't have any markings or so can't see if there is a designated + or - side, but since they are AC, I doubt it.

[frivan]

16 minutes ago, sjp100 said:

So wiring isn't my strong suite, am I wiring them up correctly? All 3 capacitors in parallel. They don't have any markings or so can't see if there is a designated + or - side, but since they are AC, I doubt it.

You got it. With the double terminals on each pole of the capacitor it is easy to jump to the next. Just watch out to discharge before working on them. I had a circuit to bring in my second capacitor and I had a contactor to bypass the capacitors. It depends on what you need.

[sjp100]

2 minutes ago, frivan said:

You got it. With the double terminals on each pole of the capacitor it is easy to jump to the next. Just watch out to discharge before working on them. I had a circuit to bring in my second capacitor and I had a contactor to bypass the capacitors. It depends on what you need.

Thank you. Will I only see a proper voltage drop once I connect a large load on it? Measuring with no load doesn't change the voltage. 230v in and 230v out.

[frivan]

5 minutes ago, sjp100 said:

Thank you. Will I only see a proper voltage drop once I connect a large load on it? Measuring with no load doesn't change the voltage. 230v in and 230v out.

Jip. I think I got 180V and 70V but can't remember which way around. I did notice that fluctuating utility voltage had a big impact on the power. Probably because of V*V/R.