# Heating my workshop calculations

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### Bjorn Teani

Apr 28, 2021, 8:44:28 AMApr 28
to electrodacus
Hello all !

I need your help validating my calculation before I press the "buy all the stuff" button

I'll try to keep it concise :
Two years ago I moved my workshop to an old stone building next to my house. Between my activity (bike frame building) and arranging my workshop on the spare time, I'm finally at the point where I can assemble all the solar parts an heating parts.
The plan is to push solar heated water through my concrete floor and that would be what drives my calculation. In a later part I'll use the excess energy to directly power part of my business activity adding batteries to the mix (hence my presence here).

One thing I was sure as I started this project two year ago, is the use of Dacians geniuosisty (yes that term probably does not exist) so I bought an SBMSO.

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So, here are my calculations. Please correct me if I'm wrong :

Note I consider ideal situations of solar conditions and elements efficiency. The plan is a first order of approximation (...) calculations so I can start somewhere.

The Workshop :
Volume to heat : 260m3 (105m2 x 2,5m)
Building Energy loss : 1,2W/(m3.K) (for a "not bad" isolated workshop)
Temp Delta : 15°C (its a workshop...)

Total energy loss :
260x1,2x15 = 4,6kW.
Lets consider that I need 5kW of continuous power to maintain a delta of 15°C in my 260m3 workshop.

Considering I will only be able to gather solar energy for 4 hours / day, I will have to store 24/4x5= 30kW in my water tanks during that 4 hours.
So 7,5kWh power to generate with solar and dissipated as heat in my water tanks.

Resistive Elements :
By using 36V 1200W resistive elements, I will need 6,25 of them. Lets take 6.

Solar panels :
On the solar side, I plan to buy "330W" 60 cells pannels. To reach the 7,5kWh power I would need 22 of them.

Water tank :
Now for the water reserve : Hot would be at 38°C (heating floor entry) and cold would be 28° (heating floor return), so 10° delta (maybe a bit much for the concrete also considering this would be an average). And I need 30kW in that.
Mass of water needed = 2600 l (Liter). I will take 3 x 1m3 water tanks.

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All of this is not my field of expertise and I except I could be completely wrong in my calculations. This is why I'm here... :-)

Does it all make sense ? Is it coherent ?

Last note : To start, I only plan to buy 12 panels, 3 resistive elements and 1m3 water tank and keep my stove for this winter ^^

Thank you all for the time you took to read me.

### Dacian Todea

Apr 28, 2021, 12:42:39 PMApr 28
to electrodacus
Hi Bjorn,

There are a lot of assumptions but say you need 5kW to maintain a 15°C delata to outside (seems a bit much so not that great of an insulation).
Then if you want that over a 24h period that will be 5kWh x 24h = 120kWh not sure why you divided by 4.
Each of those 1200W 36V elements is good for 4 panels two panels for each 600W element as those are made from two parallel 600W elements.
It is a good idea to start with a smaller system first. It will be good if you could measure more exactly how much energy your shop needs to maintain 15C above outside temperature if that is needed.
The 12 panels may produce in average 16kWh/day (depending on season, location and amount of sun and may be much less in winter based on location).
To increase by 1°C the water temperature in a 1000 liter tank you will need 1.16kWh of energy so in an average day with those 16kWh it will be a bit less than 14°C water temperature increase assuming zero losses from water tank to ambient.
Also air is about 1000x less dense than water so again assuming ideal isolated space with zero losses to outside then to heat the 260m3 of air by 4°C  the 1000 liter of water will drop by 1°C.  What I want to say by this is that if space is not well sealed then in a windy day you may need quite a bit more energy than in a non windy day as in top of the thermal insulation loss you also lose energy by warm air escaping outside.

### Bjorn Teani

Apr 28, 2021, 3:07:05 PMApr 28
to electrodacus
Hello Dacian !
Well right... Big mistake !  The 30kW is the power that I would need from my solar panels working for 4 hours... And store 120kWh.
So I have multiply by 4 all my needs !! This i huge ! xD

I do not really know how good my shop is insulated and I took 1.2W/(m3.K) because its the coefficient given for a "house build between 1983 and 1989 in France" and I was assuming my shop was not better.

I will try and calculate the real coefficient base and my insulation material and thickness, and windows area but even if I reach 0.8W/(m3.K) loss and lower my comfort to 12° inside when it's freezing outside, I'll need 2.5KWh continuous heating power to cover the losses, so 60kWh/day witch still doubles my needs according to my initial (wrong) calculations. So 44 panels and 6m3 of water. This is still way more than I anticipated and seems enormous.

So where am I wrong ?

### Dacian Todea

Apr 28, 2021, 3:44:27 PMApr 28
to electrodacus
My house is just 65m2 and in January (coldest month here) I need a 36C delta as outside average temperature is -16 to -17C while inside I will prefer around 20C.   There are of course warmer and much colder periods so this is just average and there are a few days where delta is 50C even 55C
In average January my house needs about 1000kWh that will translate to around 33kWh/day so an average loss of 1375W for the 36C average delta but this is an above average insulated house with very few small windows.
I never heard of that W/(m3.K) unit being used before usually is W/(m2.K) and is related to building surface area as all heat loss is through the exposed building sourface the internal volume has nothing to do with this.
Just imagine a simple rectangular building that is 2m x 5m x 2.5m with floor surface of 5 x 2 = 10m2 and a volume of 25m3 but the surface area will be made of the 6 sides (floor, roof and 4 walls) and will be in this case 55m2 and that will be the one losing the heat.
If you double the house to say 20m2 floor area 4m x 5m x 2.5m the volume will also double at 50m3 but the outside surface area will be just 85m2 so thermal losses for this building will not be double assuming same wall insulation.
The 1.2W/(m2.K) may be for a typical dual glaze low-E window and if say you have a 10cm EPS insulated wall the value will be 0.3W/(m2.K)

### Bjorn Teani

Apr 29, 2021, 3:42:00 AMApr 29
to electrodacus
Thanks again for taking the time to answer me.
You are right, losses are measured in W/(m2.K). The use of m3 is a simplification and the coefficient is a statistical value given to speed up calculations (it factors in the air renewal rate, hence the use of "volume" instead of area).

Well in my case and when you hunt for every W, it was the wrong approach.

I calculated again the energy loss of my workshop but this time going through every material and exposed surface and with a DeltaT of 15°.

The only surface where I struggled was for the floor as I have my concrete slab with the water circulating in it witch I considered being inside. And beneath it I have 4cm of λ=0.022W/m.K.so R=1,82 and U=0.55W/m2.K witch I consider to be in contact with the exterior : the soil. So I was not sure of what Delta T to take as the concrete will be heated and the soil is at constant temperature below 10m. So I (wild) guessed that the slab would be at 20° and the soil at 5°... So again DeltaT=15°

And this time I found 1,5kW (without taking in account the air renewal). And this is more in line with your own example...

So I went from 5 to 1.5 using more precise calculation. Only experience will tell if this is right.

In any case thanks for your help. This was very interesting ! And I'll report back in a few years with the real life results :-)

Have a great day !