how much heat does a cylinder hold?

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One for the physics boffs out there...
How much heat (kwh) does a 300 litre cylinder hold..
Ta
 
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Your question does not make sense!

Could you please amend it to one that does!

Tony.
 
I'll try. I want to use the cylinder as a thermal store so I need to know how much heat is stored within it.
That is, will it hold enough heat to keep a small building warm on economy electicity tarrifs without needing back up.
Heat load of building will be around 10kw.
Water holds around 4.12 kj per kg. i need to convert this to kwh so i can understand it.
 
You are going to need a very big store depending on how long you want the store to last! Thats why solar cannot give heating.

Let me explain in simple terms. A standard 3 kW immersion will take about 90 min to heat up a cylinder from cold to say 80°C.

So the same cylinder will give out 3 kW for 90 minutes! Although in practice the capacity will be much less! This arises because you only have the capacity between the stored 80°C and say 50°C as you cannot transfer much useful heat for heating with a source of less than 50°C.

So in practice you might get about 3 kW for about 30 minutes !!! Not so much heat is it!

Thats about 10 kw for just 8 minutes!!!

Tony
 
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At what temperature is the water?.... Why not just buy a heat store... you are trying to reinvent the wheel
 
Why would i buy a heatstore?
Do they have some magical properties that a normal cylinder does not have?
That might explain why they cost so much, i believe wizards are on £70k a year...
Might retrain...
 
@ Tony. So how big would the cylinder need to be, to give 10 kw for a couple of hours?
Ta
 
Water holds around 4.12 kj per kg. i need to convert this to kwh so i can understand it.

No! Water holds 4.18 (not arguing about the number ;) ) kj per kg per degree Celsius. So raising the temperature of the water by 20°C will increase the energy contained by double that of raising the temp by 10°C.

A Watt is not a just quantity of energy; time comes into it. It is defined as one joule for one second. So if you use up 1000 joules in one second you have used up one kilowatt. The same would be true if you used up 1 joule each second for 1000 seconds.

So if you want 10kW for 2 hours that is 10 x 2 x 60 x60 kJ = 72,000 kj

Assuming you raised the water temperature by 20°C, you would need a cylinder holding 72000 ÷ 4.18 ÷20 Kg water = 861 kg = 861 litres

If you only have a 300 litre cylinder, you will have to raise the water temperature by 20 x 861 ÷ 300 = 58°C.

Don't forget that the heat output from a radiator depends on: Flow Temp, Return Temp and Room Temp.
 
Thanks for that Mr Hailsham, that was exactly the kind of information i was looking for. So if I heat the cylinder to 80 and use a blending valve for the cold water it looks as though I could be in business???
Yeehah!
thanks again,
cowboy
 
So if I heat the cylinder to 80 and use a blending valve for the cold water it looks as though I could be in business???

What you are trying to do is something like a storage heater! Put the heat in overnight, when it is cheap. The problem is that, once you have used up the stored heat, that's it; you have to wait to recharge the heater. Adjusting the heater output to match the need can be a nightmare.

You say that have estimated the requirement as 10kW, presumably to maintain the temperature at 20°C with the outside temp -1°C. Have you allowed for heating up from cold, usually 10-15% extra is necessary.

Have you thought about circulation? Hot water will leave the cylinder, initially at 80°C but when it returns it will be at a lower temp, so the heat output will drop exponentially. The greater the drop through the radiators, the more heat given out by the rad; but the quicker the heat will be used up.

In a thermal store system the heat is replaced when the temperature has dropped to say 50°C, by a boiler; though you could use electricity - but that is defeating the object of the exercise.

Unfortunately you are trying to re-invent the wheel; so don't give up the day job ;)
 
Just as a thought exercise. could you have cp's thermal store of hot water but make the tank massive then hooked up to solar panels? Heat up store at night, top it up with solar going through the day? Now super insulate the house and try to cut down of wasted heat.

I believe you could do it
 
So if I heat the cylinder to 80 and use a blending valve for the cold water it looks as though I could be in business???

What you are trying to do is something like a storage heater! Put the heat in overnight, when it is cheap. The problem is that, once you have used up the stored heat, that's it; you have to wait to recharge the heater. Adjusting the heater output to match the need can be a nightmare.

You size the thermal store to suit, so you don't run out.

In a thermal store system the heat is replaced when the temperature has dropped to say 50°C, by a boiler; though you could use electricity - but that is defeating the object of the exercise.

He is using electricity.

Unfortunately you are trying to re-invent the wheel; so don't give up the day job ;)

He is not. He is wanting to know the calcs to size up a cylinder to retain enough heat, to heat a house all day in winter. Nothing new, lots of them around. Average size 1,500 to 2,000 litres.

Solar panels help, and superinsultion and air-tightness is the best help of all.
 
As you will need a store of about 4000 litres it will take up much of the space within the house and to be reheated up it will need about 40 kW for six hours if using an overnight off peak electrical source.

All in all not very feasible in most cases!

Tony
 
all the people i deal with who have wet storage heating systems complain about the cost of the electric and the cost of running them, and they are in buildings with many of these units so the mean temp hardly drops below 10 degrees in any of the flats because there are always so many units heating up and thus losing heat to other flats/surrounding air.

i'm interested to know your reasons behind attemping to use such a expensive form of heating?
 

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