A question of heat
Discussion
Considering fitting a form of watery storage heating to our current central heating which is wood fired. Which is great as it chucks out huge amounts of heat but is very peaky. Great when you are in and feeding it but in the morning it does nothing as the fire has gone out over night.
So how much heat energy would i be able to store in 100litres of water saying I could heat it up to 75C and then drop it to 35C heating the house so a 40C drop.
Also would this be enough to heat a two bedroomed house for an hour or two?
So how much heat energy would i be able to store in 100litres of water saying I could heat it up to 75C and then drop it to 35C heating the house so a 40C drop.
Also would this be enough to heat a two bedroomed house for an hour or two?
I've no idea if what you are actually suggesting will work, but for fag packet work, let's start with simple O-level Physics:
Stored Energy
The formula is:
Q=MTS
M=Mass of water = 100 litres = 100kg
T=Temperature difference = 40 degrees
S=Specific Heat Capacity of water = 4.187kJ per kilo per degree
So Q = 16,748kJ
Energy requirement & time
I don't know the recommended central heating figure per room. I would guess at 2kW per room (a decent start in terms of electrical bar heaters), and let's assume four significant rooms.
Therefore, you would want 8kW
(Memo: The Worcester-Bosch site suggests the 25Si combi boiler, with central heating capacity of 7.2-24kW)
Theoretically, if you could draw down the energy at a constant power rating (which you can't, as the water in your storage unit will be cooling), that equates to 2,093.5 seconds of power = 35 minutes. Using the 25Si figures, that would be more like 10-15 minutes.
Already, it sounds to me like you're gonna need a bigger tank.
You could develop a more complex and accurate model, working out the energy output of your radiators at the beginning (75 degrees), then for each subsequent minute (work out how much energy/temp has been lost in one minute, then recalculate for the next minute, and so on).
It would need a plumber or someone better qualified. But I think the numbers are pointing to your system delivering a useful kick-start in the morning, but even in the morning another system would need to kick in. But these are very rough, theoretical numbers and I'd suggest you get someone better qualified.
Stored Energy
The formula is:
Q=MTS
M=Mass of water = 100 litres = 100kg
T=Temperature difference = 40 degrees
S=Specific Heat Capacity of water = 4.187kJ per kilo per degree
So Q = 16,748kJ
Energy requirement & time
I don't know the recommended central heating figure per room. I would guess at 2kW per room (a decent start in terms of electrical bar heaters), and let's assume four significant rooms.
Therefore, you would want 8kW
(Memo: The Worcester-Bosch site suggests the 25Si combi boiler, with central heating capacity of 7.2-24kW)
Theoretically, if you could draw down the energy at a constant power rating (which you can't, as the water in your storage unit will be cooling), that equates to 2,093.5 seconds of power = 35 minutes. Using the 25Si figures, that would be more like 10-15 minutes.
Already, it sounds to me like you're gonna need a bigger tank.
You could develop a more complex and accurate model, working out the energy output of your radiators at the beginning (75 degrees), then for each subsequent minute (work out how much energy/temp has been lost in one minute, then recalculate for the next minute, and so on).
It would need a plumber or someone better qualified. But I think the numbers are pointing to your system delivering a useful kick-start in the morning, but even in the morning another system would need to kick in. But these are very rough, theoretical numbers and I'd suggest you get someone better qualified.
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