Immersion thermostat keeps tripping

[Little Clanger]

12 hours ago, John Carroll said:

21 ohms @ 230V, UK standard voltage?  gives a power output of 2.519kW, it requires 9.42kWh to heat 162L from 10C to 60C so a max of 3 hrs 44min to achieve this, and just over 4 hours if stat set to 65C.

Thanks for the calculations. That's interesting. I usually leave the immersion on for the full Economy 7 overnight 7 hours, on the principle that as it is thermostatically controlled, it will switch itself off when the set temperature is reached. Now, I will reset the timer to the minimum time required and see if that makes any difference.

[Little Clanger]

12 hours ago, SteamyTea said:

Could it be the sleeve the stat goes in is bent and touching the element?

 

12 hours ago, John Carroll said:

.......  there's only one way to check it out, remove the heating element.

In answer to both these points, yes, I'm resigned to draining the tank and taking the element out

[SteamyTea]

1 minute ago, Little Clanger said:

resigned to draining the tank and taking the element out

Get the beach towels out.

[JohnMo]

13 hours ago, John Carroll said:

it requires 9.42kWh to heat 162L from 10C to 60C

But it unlikely the whole cylinder will be at 10 degs, the bottom yes, but the top of the cylinder is likely to mid 30s or above

[Little Clanger]

5 hours ago, SteamyTea said:

Get the beach towels out.

🙄

[Little Clanger]

5 hours ago, JohnMo said:

But it unlikely the whole cylinder will be at 10 degs, the bottom yes, but the top of the cylinder is likely to mid 30s or above

So immersion can be on for even shorter length of time?

[Nickfromwales]

21 hours ago, Little Clanger said:

Initially, I'm renewing the thermostat in the lower immersion heater, with a different make (a Backer to replace a Tesla).

 

A continuity test on the element gave me 21 ohms, which I understand is about right.

 

Confusingly, it was working fine last night - heated a whole 162 litre cylinder without tripping.

 

By the way, a note to NickfromWales - I haven't been able to follow your advice about returning the cylinder because I bought the cylinder without any immersion heaters fitted. I bought the immersion heaters from Screwfix, but didn't keep the receipt, so can't return them.

You will have an e-account at Screwfix which they will pull up with your postcode usually. They would take them back without a receipt iirc, my mates a serial ‘returner’ there…….

 

Worst case you’ll get a credit note. 

[Nickfromwales]

6 hours ago, Little Clanger said:

 

In answer to both these points, yes, I'm resigned to draining the tank and taking the element out

My work here is done lol 🥳

[John Carroll]

1 hour ago, Little Clanger said:

So immersion can be on for even shorter length of time?

 

You might save a few bob due to the tiny bit extra of cylinder losses but it won't fix your current problem. Assuming the top 40L or so still at 60C when the night heating cuts in then you will have to heat 120L of mains water at (my) 12C to 60C, requiring, 120*(60-12)/860, 6.7kWh, add say another kWh for cylinder losses, 6.7+1.0, 7.7kWh, requiring 7.7/2.519, 3.06hrs, say 3hrs 15 minutes and a very worst case of, assuming a whole cylinder reheat, of 160*(60-12)/860/2.519, 3.55hrs, say 4hrs, you will be well covered to programme it on for 4hrs before the end of the night rate as this will cover cold mains at 5/6C in the debths of winter.

Edited March 8 by John Carroll

[Little Clanger]

3 hours ago, Nickfromwales said:

You will have an e-account at Screwfix which they will pull up with your postcode usually. They would take them back without a receipt iirc, my mates a serial ‘returner’ there…….

 

Worst case you’ll get a credit note. 

OK, thanks 

[Little Clanger]

3 hours ago, John Carroll said:

 

You might save a few bob due to the tiny bit extra of cylinder losses but it won't fix your current problem. Assuming the top 40L or so still at 60C when the night heating cuts in then you will have to heat 120L of mains water at (my) 12C to 60C, requiring, 120*(60-12)/860, 6.7kWh, add say another kWh for cylinder losses, 6.7+1.0, 7.7kWh, requiring 7.7/2.519, 3.06hrs, say 3hrs 15 minutes and a very worst case of, assuming a whole cylinder reheat, of 160*(60-12)/860/2.519, 3.55hrs, say 4hrs, you will be well covered to programme it on for 4hrs before the end of the night rate as this will cover cold mains at 5/6C in the debths of winter.

Again, thanks for the calculations, John

[Little Clanger]

On 07/03/2025 at 21:43, John Carroll said:

21 ohms @ 230V, UK standard voltage?  gives a power output of 2.519kW, it requires 9.42kWh to heat 162L from 10C to 60C so a max of 3 hrs 44min to achieve this, and just over 4 hours if stat set to 65C.

Have you tried the bottom stat in the upper immersion?, you can easily monitor that during the day by draining off some water to get it to cut in on boost.

Can you post a photo of the wiring at/around the stat.

I seem to remember that the relationships between resistance, voltage, power and current are covered by the Ohms Law formulae, but what is the formula for calculating power required to heat a given volume of water through a given temperature change?

 

Thanks

[SteamyTea]

42 minutes ago, Little Clanger said:

but what is the formula for calculating power required to heat a given volume of water through a given temperature change

There are two.

The basic one, that is good enough for domestic usage is:

 

SHC x mass x ΔT

 

Where:

SHC = specific heat capacity of the material: unit J.kg^-1.K^-1 where J is energy in joules and K is temperature in kelvin.

Mass = kg, kilogram.

ΔT = change in temperature, T₁ - T₀

 

So for 200 lt of water, which just happens to have a density of 1000 kg.m^-3 and a SHC of 4.18 J.gram^-1 [4.18 kJ.kg^-1], to be raised in temperature from 303K to 343K [30°C to 70°C].

 

4.18 [kJ.kg^-1.K^-1] x 200 [kg] x (343 - 303) [T₁ - T₀] = 33,440 kJ

 

This can be converted to kWh by multiplying by 0.00027778

 

33,440 [kJ] x 0.00027778 = 9.289 kWh.

 

Dividing the kWh by the power of the heater gives the time to reach the desired temperature.

So if the heater was 3 kW.

 

9.289 [kWh] / 3 [kW] = 3.096 h or 186 minutes.

 

The more complicated formula involves exponential growth and the surround environment temperature.

So:

T_F = T_i -(T_i - T_a) x EXP^{(-k x t)}

 

Where:

T_F = Final Temperaure

T_i = Initial Temperature

T_a = Ambient Temperature

EXP^{(-k x t)} = Exponent of base e, where ^k is a unit less number and ^t is time.

 

This basically shows that as the material gets hotter, it looses more energy to the surroundings, and takes disproportionally longer and longer to get to a set temperature.

 

 

 

Edited March 9 by SteamyTea

[John Carroll]

I 've used a a vey simple few for decades.

Energy, kWh = litres x (T2-T1) / 860.

Time, hrs = kWh/heating power in kW.

And. To calculate power required to instantenously heat water.....

Power. kW = LPM x 60 x (T2-T1)  / 860 or any variation of this.

 

[Little Clanger]

Thank you both. Very useful