So i received a 12v 300w dc element today and it is huge and i mean Huge 50cm long with 3 u shaped element coils. The old one was also 12v 300w and 17cm long connected from the panels (series 4x [email protected])directly to a geyser. i Wonder if this one will work? i Know the amps is on the panels are less and will take longer to heat up. Tested it with 4x12v battery bank and in 10 seconds it was HOT. Does size matter?🤭

1 hour ago, Off Grid Farmer said:

So i received a 12v 300w dc element today and it is huge and i mean Huge 50cm long with 3 u shaped element coils. The old one was also 12v 300w and 17cm long connected from the panels (series 4x [email protected])directly to a geyser. i Wonder if this one will work? i Know the amps is on the panels are less and will take longer to heat up. Tested it with 4x12v battery bank and in 10 seconds it was HOT. Does size matter?🤭

The estimated time to increase the water temp by 1.71 degrees would take about 1 hour if you use 300W @ 12V for a, capacity of 150L.  You can use this time or increase for other geyser capacities. 

The current well insulated geysers have about a 75Wh heat loss per hour which is about 1.8kWh per 24hrs.

 

Size always matters.  Americans would say its cubic inches that matter, the more the better

1 hour ago, Scorp007 said:

The estimated time to increase the water temp by 1.71 degrees would take about 1 hour if you use 300W @ 12V for a, capacity of 150L.  You can use this time or increase for other geyser capacities. 

The current well insulated geysers have about a 75Wh heat loss per hour which is about 1.8kWh per 24hrs.

 

i Will install it early tomorrow and test it. 

Tested the element on 12v 800Ah battery bank for 2 hours. It consumed 50.10 Ah and heated 150L water from 10'c to 62'c. The panels and inverter kept the soc at 85%.

1 hour ago, Off Grid Farmer said:

Tested the element on 12v 800Ah battery bank for 2 hours. It consumed 50.10 Ah and heated 150L water from 10'c to 62'c. The panels and inverter kept the soc at 85%.

Very interesting to use only 600Wh to heat the water by 52 degrees on the 150L geyser. 

4 hours ago, Scorp007 said:

Very interesting to use only 600Wh to heat the water by 52 degrees on the 150L geyser. 

Must have something to do with the size and triple coils. It behaved like a electromagnetic induction coil, expansion of the conductor allows for less resistance with heat.   

Something does not add up for me.

Q = m * c * ∆T

Q = heat energy (Joules, J) m = mass of a substance (kg), c = specific heat (units J/kg∙K or J/kg∙Oc), ∆ is a symbol meaning "the change in"

∆T = change in temperature (Kelvins, K or Degree Celsius)

Water has a specific heat capacity of 4187 J/kg°C

The power of the element is known so the time can be calculated

To raise the temperature by 52 Degree Celsius

Q = m * c * ∆T

Q = 150 * 4187 * (62-10)

Q = 32658600 J

Q = 32658.6 kJ

Q= 32.658 MJ

Conversion of the energy (Q) to power (P): 1 Watt=1 J/s

t(s) = Q / P

t(s) = 32658600 / 300

t(s) = 108862 Seconds

t(m) = 108862/60

t(m) = 1814.367

t(h) = 1814.367/60

t(h) = 30.239

Time in hours minutes seconds 30h 14m 22s

 

If I take the time provided (2 hours) then I can calculate the power of the element.

P = Q / t(s)

P = 32658600 / 7200

P = 4535.916 W

P = 4.535 kW

So unless that geyser is a sun geyser I cannot see that a 300 watt element can raise the temperature of 150 liters of water by 52 °C in 2 hours time.

 

29 minutes ago, GerhardK83 said:

Something does not add up for me.

Q = m * c * ∆T

Q = heat energy (Joules, J) m = mass of a substance (kg), c = specific heat (units J/kg∙K or J/kg∙Oc), ∆ is a symbol meaning "the change in"

∆T = change in temperature (Kelvins, K or Degree Celsius)

Water has a specific heat capacity of 4187 J/kg°C

The power of the element is known so the time can be calculated

To raise the temperature by 52 Degree Celsius

Q = m * c * ∆T

Q = 150 * 4187 * (62-10)

Q = 32658600 J

Q = 32658.6 kJ

Q= 32.658 MJ

Conversion of the energy (Q) to power (P): 1 Watt=1 J/s

t(s) = Q / P

t(s) = 32658600 / 300

t(s) = 108862 Seconds

t(m) = 108862/60

t(m) = 1814.367

t(h) = 1814.367/60

t(h) = 30.239

Time in hours minutes seconds 30h 14m 22s

 

If I take the time provided (2 hours) then I can calculate the power of the element.

P = Q / t(s)

P = 32658600 / 7200

P = 4535.916 W

P = 4.535 kW

So unless that geyser is a sun geyser I cannot see that a 300 watt element can raise the temperature of 150 liters of water by 52 °C in 2 hours time.

 

It seems like we get to about the same calculated result. Normally a 3kW element takes about 2 hrs to heat a 150L geyser but that is from a higher temperature than 10 degrees. Even after running my geyser to fill a bath my temperature only gets as low as 22 degrees.

I can only suspect that the temperature was taken close to the element and no movement of water around it. It would be interesting to find what the temp of water coming out of the hot water outlet was.

On 2023/08/26 at 3:54 PM, Off Grid Farmer said:

So i received a 12v 300w dc element today and it is huge and i mean Huge 50cm long with 3 u shaped element coils. The old one was also 12v 300w and 17cm long connected from the panels (series 4x [email protected])directly to a geyser. i Wonder if this one will work? i Know the amps is on the panels are less and will take longer to heat up. Tested it with 4x12v battery bank and in 10 seconds it was HOT. Does size matter?🤭

If I may ask where did you get the Element and will it be compatible with a Kwikot and Tecron 150L Geyser?

 

If I may ask where did you get the Element and will it be compatible with a Kwikot and Tecron 150L Geyser?

Sorry for the late reply. i Bought the element from JBH solar Lichtenburg. i Don't see a problem if you fit this element in the geyser if you have enough panels or battery capacity. Check out JBH solar website for more options jbhsolar.co.za