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14kw solar pv, storing excess in a large heat store

Greenblood

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Sep 4, 2021
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How would I calc the ideal volume to do this? I’ve maxed put the roof space on my garage and have 3xsingle phase inverters with 3ximmersuns so the 14kw is split between the phases. I will install a heat store with 3 immersion heaters so that any excess energy is stored as heat for use during the night heating the floor of the house.
cheers
steve
 
If we can calculate the volume of water and the required temperaure rise, we can answer this question.

The following formula is used to calculate the power of heating element needed to heat a specific volume of water by a given temperature rise in 1 hour.

volume in litres x 4 x temperature rise in degrees centigrade / 3412

(4 being a factor and 3412 being a given constant)

for example 100 litres of water, to be heated from 20ºC to 50ºC, giving a temperature rise of 30ºC would give –

100 x 4 x 30 / 3412 = 3.52 kWh
 
If we can calculate the volume of water and the required temperaure rise, we can answer this question.

The following formula is used to calculate the power of heating element needed to heat a specific volume of water by a given temperature rise in 1 hour.

volume in litres x 4 x temperature rise in degrees centigrade / 3412

(4 being a factor and 3412 being a given constant)

for example 100 litres of water, to be heated from 20ºC to 50ºC, giving a temperature rise of 30ºC would give –

100 x 4 x 30 / 3412 = 3.52 kWh
Smashin thanos
 
How would I calc the ideal volume to do this? I’ve maxed put the roof space on my garage and have 3xsingle phase inverters with 3ximmersuns so the 14kw is split between the phases. I will install a heat store with 3 immersion heaters so that any excess energy is stored as heat for use during the night heating the floor of the house.
cheers
steve
So 14kw is panel sticker wattage? because actual power produced in winter is going to be less than half that as you get half the sun in the winter as you do the summer. so 7kw ... your going to have "extra" of that? to the point that its going to useful to heat water with? me thinks no
 
a. Ascertain your yield using a prediction tool, with angles/directions etc. (kWh per day generated)
b. Work out what your nominal daytime load is in kW when it's sunny. Multiply that by the number of daylight hours. (kWh per day used in house)
c. Subtract b from a (kWh spare for water heating)
d. Work out your desired water temp (unlikely to be >80 deg C)
e. use the numbers form c & d to determine how much water you can heat up

Do it for summer & winter to get a feel for things, then decide. A 200litre tank might have a standing loss of only 1kWh per day with very good pipe insulation, so there's no reason why you can't have a few days worth stored.
 
So 14kw is panel sticker wattage? because actual power produced in winter is going to be less than half that as you get half the sun in the winter as you do the summer. so 7kw ... your going to have "extra" of that? to the point that its going to useful to heat water with? me thinks no
I live in the south of France, so am hopeful of a decent output even in winter, I also plan on instaling more panels in time. A rough estimation I should get 17kwh per day production in December. Summer is three times that. Yes, I will use a fair amount during the day running the heat pump and household, but want to store excess and need an appropriately sized tank.
Looking at my smart meeter bill, i used 40kwh per daytime last January...
 
I have a 3000L tank to do that. Just to give an idea of energy stored in there, I'll expand on the calculations done by @solarsimon - maybe this gives you some ballpark figures to compare with.

The specific heat capacity for water is 4.2 kJ/(kg K). A 3000L tank, at 80C, stores in principle ~300kWh of energy, but let's assume the ambient temperature is 20C and that's what we start out with as the temperature in the tank. We want to figure out how much more we can store. Thus, the temperature delta in the tank we want to look at is 80°C - 20°C = 60°C

The calculation with all units:

E = (4.2 kJ/kg°C) ((80 °C) - (20 °C)) (3000 liter) (1 kg/liter)
= 4.2 * 60 * 3000
= 756000 kJ


Since 1 kJ = 1kWs we multiply this by 1/3600 (hours/seconds) to get kWh - this makes for around 210kWh.
 
Who says you can't make power in the winter?

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2 days ago I had a peak production of 4kw from 5.5kw of panels. I was very impressed.
 
I can't wait to see how well I produce this summer.
Then I can decide how much more to install.
 
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