First comes establishing of the heat load, incl DHW, then comes the decision how to provide that heat (you say a WBS in this case), then comes a decision as to how much heat you want to store, then and only then comes the decision as to how big the TS is - though there is a typical range of TS size that is normal for any given WBS heat output. But it doesn't always follow that even with free wood a WBS is the way to go. Fabric first, really get that heat load down (remembering that it can even be zero) - certainly if the heat load is low enough the outlay for a WBS, TS and all the associated pipework can often mean it is not the best way to go, an air to air heat pump or just gas will often be the sensible choice.

The amount of energy required to heat the house and provide DHW. You will know this from the up front design, the wall floor joinery and ceiling U values and ventilated losses.

One thing to consider is that standing losses are high so it might be inappropriate to have a large TS in a very well built house.

Posted By: Michael1What is heat load

Heat load is the total amount of energy you need to heat the store for your needs.
It does not really matter how that heat (heat is the old word for energy) is generated, it is basically the same (this is probably not the place to talk about heat pumps).

Heat (or energy really) is measured in joules, which are usually stated a kWh over a time period.
So imagine your kettle, which will be about 2 litres in volume and have a 3 kW electric element.
You put in tap water at about 10°C and switch it on, after a few minutes it boils at 100°C.
So you have raised it 90°C.
Now water has a Specific Heat Capacity of 4.18 J/(kg.K) (a °C and K, Kelvin, can be treated equally and a litre of water has a mass of 1 kg).
So your 2 litres of water has a mass of 2 kg and the temperature rise is 90K

J = 4.18 x 2000 x 90
J = 752400

To convert this to Wh you divide by the number of seconds in an hour (3600)

752400 / 3600 = 209 Wh

And then to convert to kWh you divide by 1000

209 / 1000 = 0.209 kWh
That is the amount of energy you need (about 3p at 15p/kWh)

That is only part of the story though, there is Power.
This is not the same as energy.
It is really just how fast you are using that energy.
You can think of it as how quickly you use fuel in your car, generally, pootling about at low speed gives you better milage than driving at full speed (not the greatest analogy as there is an optimum speed, but you know what I mean).

So taking that kettle, and now knowing that it needs 0.209 kWh and it has a power of 3 kW, you can work out how long it takes to boil.

Time = Energy / Power

Time = 0.209 / 3

Time = 0.0667

(in this case the time is in hours (because we are using kWh, the h means hours)

So that is about 4 minutes.


Now you are thinking of getting a 500 lt, or 500 kg thermal store. This store is to supply both the hot water (DHW) and the house heating (Space Heating).
Without getting into too much detail about temperature gradient (top of tank hotter than the bottom, often called stratification) you will need a minimum temperature in it that is hot enough for bathing.
For hot water, somewhere between 50°C and 65°C is normal (not going to get into detail about legionella here).
You will also need a minimum temperature for your space heating, so somewhere between 35°C (for underfloor) and 70°C (for traditional wall mounted radiators).
So somewhere between 42.5°C and 67.5°C would seem optimal (these are mean store temperatures, the top will be hotter).

Now depending on your bathing needs, they can be somewhere between 30 lt for a quick shower and 400 lt for 4 large baths.
But let us initially work with 200 lt of water at 55°C.
Water enters the store at about 12°C from the mains (measured mine today at 11.8°)
So using the Specific Heat Capacity (SHC) calculation above (with a slight modification to make the sums easier), we have:

Energy = SHC of Water = 4.2 kJ/(kg.K)

So

kJ = 4.2 x 200 x (55-12)

kJ = 36120 kJ

kJ / 3,600s = kWh

kWh = 36,120 / 3,600s

kWh = 10

Now that is the easy bit.
Space heating is much more complicated as it is more variable, hot days you need no heating, cold days you need lots.
And so make it more complicated sometimes you need to change the power during the day as the temperature changes.
Thankfully we have a thermostat to do this and it is not a case of 'adding more logs to the fire' anymore (though you fancy a log burner so you will ).

So to work out Space Heating loads you need to know a few things.
Mainly the heat losses though the floor, ceiling/roof, walls, doors and windows. Then you need to add in air loss.
Any barrier between inside and outside i.e. a wall, will have a thermal resistance that can be expressed as a U-Value.
All a U-Value says is that for any given temperature difference, this is the expected power transmission (not going to get into Thermal Inertia here).
So a U-Value of 0.1 W/(m^2.K) is saying that for every 1°C, or 1 K) temperature difference, you can expect to need 0.1 W of power (not energy) per square meter of area.
So imaging a box with a total surface area of 100 m^2 and the temperature difference is 8°C i.e. outside temperature 12°C and inside temperature 20°C), you will need a heater of:

Power (W) = 0.1 x 100 x 8
80 W or 0.08 kW

Then say you wish to change the air in that box every 2 hours (0.5 Air Changes/Hour) and the volume is 10 m^3.
Air has a SHC of 1 kJ/(kg.K) and a density of about 1.3 kg/m^3
So you have 13 kg of air that needs heating by 8°C once every 2 hours.
Energy = 1 x 13 x 8
Energy = 104 kJ or 0.03 kWh

Allowing for 0.5 ACH
0.03 / 2 = 0.015 kWh

Now you can see above that we have a mixture of 2 units, Power (W) and Energy (kJ or kWh). This can sometimes cause confusion, but the easy way is to initially convert the power figure to an energy figure.

So we have a total wall loss of 0.08 kW and we know that there are 24 hours in a day.
0.08 kW x 24 h = 1.92 kWh (shall call this 2 kWh)

So the daily energy usage for DHW (10 kWh), losses though wall 2 kWh and air losses 0.36 kWh (the 0.015 kWh x 24 h), this gives a total of 12.4 kWh/day.

If that was used every day, without deviation, then your annual heat load (to answer the original question) would be 4,526 kWh (though the house would be an off size).

So you need to work out these numbers first, then you can work out what size store you need and the best way to get the energy into it.
Things to remember about your idea of combining a wood burner, solar thermal and PV is controlling the inputs at the right time.

4,526 kWh of energy is about 1 tonne of timber before burner efficiency is taken into account (don't be fooled by efficiency claims), or about the output of a 5 kWp PV system or about 4 m^ of solar thermal.
Except the output from the PV and the ST is highly variable by the hour and by the season. So you will not get much in Winter, and way too much in summer.
But before you worry about that, work out your heat loads.

Just out of interest, is this a new house/build or your existing house you are changing?
If you currently live in it, working out current heat load is pretty simple, just read the meters every day for a few weeks.

Have you ever lived with wood burner as you main source of CH/DHW? The constant cycle of bringing in wood, regular feeding and emptying, cleaning the glass etc. Lot of work.

Start with doing a drawing with all the major dimensions on it. Then note down what each component is made from i.e walls, floors, doors, windows and roof.
Work out the square meterage of each component.
Then try and find the thermal properties of each component.
No need to get too bogged down in detail.

So say you have a window that is single glazed, and it has an area of 0.6m^2
Don't worry about the frame, treat that as glass.
Glass has a thermal conductivity of 1.05 W/(m.K)
To convert thermal conductivity to the R-Value (which is part way to the U-Value), you just divide the thickness (in metres) of the material by the thermal conductivity.
So if the glass is 4mm thick, that will be

R = Length (m) / Thermal Conductivity (W/(m.K)

R-Value = 0.04 / 1.05 = 0.038

Then the U-Value is the inverse of the R-Value

U = 1 / R

U = 1 / 0.038 = 26 W/(m^2.K)

Multiply by the area

U = 26 x 0.6 = 15.6 W/K

So using an 8°C temp difference, that window allow 125 W of power to pass though to keep the temperature the same.
Over a day that is

Wh = 125 x 24 = 3000
Or 3 kWh

You can see why we use double and triple glazing.

You can get some thermal conductivity figures here (and a whole lot more)

http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html

Then they are using the terminology wrong and just adding to the confusion

Heat is work and work is heat.

There is a song about it.
https://www.youtube.com/watch?v=VnbiVw_1FNs

maybe have a play around with these online calculators !

https://www.chromalox.com/resources-and-support/calculators/comfort-heater.aspx

or

http://www.builditsolar.com/References/Calculators/HeatLoss/HeatLoss.htm

You may have to fiddle around with units and conversions !

Still, great for getting your hand in !

gg

JonG,
Ground floor is 135 as obviously the upstairs when I do it.

Posted By: SteamyTeaThen they are using the terminology wrong

Who do you think is 'they'?

Cite a reference for your definition of the term 'heat load' if you want to persist with it.

Posted By: Viking HouseIf the house is cold just put on the heating.

Did you read the OP?

So which heating do you suggest he puts on?

Posted By: Michael1This might seem a daft question but is there a danger of too much insulation and then not being able to heat the TS with a woodfired stove as it would be to hot in the rest of the house.

No real danger of too much insulation, you just have to size the stove to the needs. OK I will admit that it is not always easy to find a stove with low heat to the room and high to the store, often they are the other way around. Are you wedded to the idea of a wood stove in the lounge with flickering flames to look at - because it might be better to have an appropriately sized boiler in the utility room to run the DHW and the CH. and a v.small stove in the lounge for appearance and to heat that room only.

Heat load. Ground floor 135m2 assume 2.5m high and if the house is square you have a wall lengths of say 12m, 4 walls = 48m of wall and assume 2.5m high then a wall area of say 120m2
Form the link I gave above if the walls are 400mm thick (plus render) then you have a U value of 1.8
heat load = 1.8 x 120m2 x 20deg difference = 4320 =4.32kW

Stick on 100mm of EWI you get a U value of 0.31
heat load = 0.31 x 120m2 x 20deg difference = 744 = 0.744kW

Stick on 200mm of EWI you get a U value of 0.17
heat load = 0.17 x 120m2 x 20deg difference = 408 = 0.408kW

It is not difficult to work out the heat loads and OK that is just the walls with no account of the windows or doors so you do the sums for each element, walls, roof, windows floor etc. choose your insulation level and decide on your heating. (and heat storing)

FWIW My main house (family house) has a 40kW wood a gasification boiler with 2000 TS and at the moment I am giving the boiler one load every other day to provide DHW for 2 houses and heating for the main house.

It may be worth thinking about splitting the DHW and the Space heating systems.
They do different things, at different temperatures and at different time. Call it the 3Ds of heating if you like.

Peter, I didn't give any thought to having a stand alone wood stove purely for function and I have the room for one. I can still have a smaller pretty one in the lounge, what do you do in the summer, do you still fire the wood boiler up?