Topic: Exhaust temp of a miner (Read 673 times)

Well, after digging some more about this, I end up doing the math a little bit different and for other conditions.
I used these 3 fucntions:

Energy needed initially to heat up a space by a given amount of degrees.
Q = m * C * ΔT

Rate of heat loss for a composite wall
Qloss = A * ΔT / Rt


W = Q / T <=> T = Q / W

So, first I calculated the heat needed to initially heat up a space by a certain amount of degrees. I got a total of 1.629.818,70J + 795.429.44J = 2425248.14J
Then I converted that to [kWh] = (1.629.818,70 J + 795.429.44 J ) * 24 / 1000 = 0,6737 kWh

Then, I calculated the rate of heat loss over a day an also converted that to [kWh]
The values I got are: 797.4057 W + 398.7029 W = 1196.10W. This needs to be converted too to [kWh].
Therefore 1196.1W corresponds to 1196.1*24/1000 = 28.707 kWh.

Then I adde both values getting a totall of 0.6737 kWh+ 28.71 kWh = 29.3803 kWh

Finally I calculated the time needed to heat up the space initially (only), using the joules from the 1st step and considering a device rated at 3500W.
T = Q / W = 2.425248,14 / 3500 = ~693 s which around 12 mins!

NOTE: I did calculations for 2 different temperature differences to account for cold parts of the day an warmer parts of the day:


Do you guys think this makes sense?

Yes, I also have the feeling miners won't be able to beat boilers unless we approach it as I did initially, which was to only invest in as much miners as neeed, instead of comparing the same power for the 2 scenarios as you did it, the 400kW.
I've been working in an excel sheet to try to compute energy needs and heat losses, but I'm struggling to understand the meaning of one of the results.

Maybe you can help me understanding it.
The meaning I'm not sure I understand correctly is the result from the formula Q = m*C*ΔT. This calculates de energy needed to heat up a space, taking into account a temperature delta.
But if I want to account for heat losses, I'll calculate the heat losses over a given area and this will be given as Watts or J/s, meaning it's a rate.
So, if I want to calculate the energy needed to heat up a space taking into account the heat losses, how do I combine a value that is in Joules (the units of Q in m*C*ΔT) with the heat losses that is a rate?
Like, do I calculate Q and then calculate the heat loss rate, multiply it by 86400 seconds (a day in seconds) and finally sum it to Q and I'll have the totall energy needed to heat up the place and keep the target temperature for one day?

Exactly, because this is the only logical way to look at it, because again, you are comparing a traditional way of operating your business vs a different way, had there been no need for a boiler in the first place, this whole thing would change, since heat is a must in your business, you would treat that bill as you treat rent or anything else that you must pay every day to keep the operation ongoing.

I am interested in knowing the cost of biofuel, we know it isn't 1k a day, but I would like to know the exact cost so we can do the math with the correct figures, if I have to take a wild guess, it would be near impossible for miners to beat the boilers.

In the second year, when you say the funder funds the same amount for both, you say it just to make both of us in the same conditions? Because in the 1st year we both needed that initial investement. The 400k was needed for both of us, but in the second year, only one of us would still need the 365k€ for the biofuel. The other one also gets the same amount just to keep things even in terms of conditions?

It's rather simple, let's assume you and I are running the same business, and we are being funded by the same investor, you run boilers and I run miners, we would assume ىخ depreciation and maintenance cost is zero, the investor would pay the same amount, 35000 for the boiler and 1000 a day for the power bill, you chose to stick to the plan, I chose to find another investor who would pay me the extra money needed to run miners.

Your business for the first year:

Funding : 35000 + 365000 = 400,000
Expenses: 35000 + 365000 = 400,000
Income: Zero
Debt = 0

My business for the first year:

I would take the same funding from the same investor which is 400,000, and I would take a loan of  124,000 from the bank, I would then run my 114 miners to heat the place

Funding : 400,000
Expenses: 400,000 + 124,000 (loan)
Income: 300*365 = 109,500
Net profit:  109,500 - 124,000 = 14,500
Debt = 14,500

To clarify here, my miners make 2000 profit and pay 1700 for power, I pay my bill daily, and thus by the end of the year I make 300*365 = 109,500, but I still have a loan to pay which is 124,000, for the investor, he doesn't care, we both are heating the greenhouse equality, and the income from both businesses is the same.

at this point, I take the  109,500 and pay it back to the bank, and now I owe them 124,000 - 109,500 = -14,500

Second year, the owner/investor would still pay us both 365000 for the biofuel, assuming we both have the boiler already.

Your business for the second year:

Funding :  365000
Expenses: 365000
Income: Zero
Debt = 0
Total expenses to this point = 765,000

My business for the second year:

Funding : 365000
Expenses: 365000
Income: 300*365 = 109,500
Net profit:  365000 + 109,500 - 14,500 = 460,000
Debt = 0
Total expenses to this point = 765,000

By the end of the second year, I would have made 460,000 in profit while still having a larger capital than yours, I got 114 miners you got an old boiler,  also I am pretty close to hitting total ROI  765,000 - 460,000 = 305,000 while your business has spent a total of 765,000 with 0 return and your ROI is never.

In my initial post, I assumed the initial cost is boiler vs miner, which is nearly half a million difference, with 1000 for the power bill being fixed because you need the boiler either way, it's pretty much the same logic.

Obviously, this example favored miners simply because I used 1000/day for the boiler bill, which can't be true, it was just a random number that shows how "under some strange circumstances" using miners is the better option if the actual power bill is 100/ day, everything changes.

I understand your approach and it makes sense in a specific scenario. I see your approach as when you use what you don't spend on one side and add it to the earnings of the other side.
But in practical terms, I can't say that if a boiler spends 1000€ a day in biofuel, then, using miners, I'll have a 14 month ROI because that's not going to happen. I see it only as a theoretical approach that beneficts miners. Unless I'm missing how you would translate the 1300€ a day in practical terms.

Your logic is different from mine, you assume only 300 profit i assume 300 + 1000, the 1000 is the amount of money you save every day for not using boilers.

This is not (using miners vs no miners at all) it is (using miners vs boilers).

If it still not clear, let me know I will explain it in detail.

@mikeywith, I was reviewing your math from post #15 and trying to do it on my way but with your numbers and I find different numbers.

Costs of a boiler over a year
Purchase: 35000€
fuel / day: 1000€.

This would be 35.000€ + (1.000€ * 365) = 400.000€ per year.

------------------------------------

Costs of 400.000W of miners / 3500W per miner = 114 miners

114 miners 3.5kW * 24h * 0,18€/kW = 1.724€ per day. Meaning 630.000€ per year + 524.000€ = 1.154.000€ per year

Now, if I assume your 2000€ estimate of Bitcoin per day, means 524.000€ + 630.000€ - 730.000€ = 424.000€ still missing to break even

Then, assuming 300€ per day after electricity bill, I still would need 424.000€ / (300€ * 30 days) = 47 more months, meaning around 5 years in total to break even.
Am I wrong?

In the meantime, while I wait for some email replies (I sent a couple of other emails asking for technical data of burner / boiling systems for greenhouses), I start reading about heat loss calculation and with the help of some AI I end up using this formula to calculate heat loss, depending on several factors, such as greenhouse cover material, thickness and other parameters:

Q_loss = (k.A.ΔT) / d

where:
Q_loss is the heat lost in Watts
k is the thermal conductivity of the insulating material (the greenhouse cover in this case) in W/m.C
A is the area through which the heat is being lost in m²
ΔT is the difference between inside and outside temperature in Kelvin or ºC
d is the thickness of the insulating material

So, for my specific case, I'm considering a tunnel-shaped greenhouse covered covered with a twin-wall polycarbonate material of the same dimensions for the 900m³ volume (30m x 10m x 3m, L, W, H).

k, according to AI can be considered 0.22 W/m.C (it will be lower because this value is for solid polycarbonate sheet, but for the sake of this calculation I will use this value).
A is the area of the 2 tops of the greenhouse plus the circular shape cover of the greenhouse. I excluded the ground area and will ignore the heat loss from the ground for now! I calculated this as 549.78m².
ΔT will be 14ºC to go along with previous assumptions from 3ºC up to 17ºC.
d will be 4mm in this case which is the worst case scenario I found here in the following table:

Plycarbonate	R-Value	K-Value
10mm twin-wall Polycarbonate	1.89	4.55
8mm twin-wall Polycarbonate	1.60	4.76
6mm twin-wall Polycarbonate	1.54	5
4mm twin-wall Polycarbonate	1.43	5.27

Source: https://greenhouseemporium.com/greenhouse-insulation/

But the values in this table doesn't seem to respect the SI unit system because from what is said in the site where the table is, the K-Value:

For instance that value of 5.27 (4mm twin-wall Polycarbonate K-Value) is the BTUs that escapes from a 1 ft² of 1 inch thick insulation material. But in the table they have a 4 mm thick material, not 1 inch (which is around 25.4 mm), so I don't even understand the relationship between the 5.27 and the 4mm thick material.
I even tried to convert this value to Watts to see if I could find any similarities between the value from AI and the value I got after converting but not even the units could be compared.
As 1ft² = 0.0929m², therefore, 5.27/0.0929 = 56.73BTU would escape from 1m² of 25.4mm thick insulation material.
Knowing that 1W = 3.41BTU/h, then 56.73BTU/h would be 16.64W. Putting all together, we would get 16.64W/m² of heat loss according to the values of that table (unless I did something wrong). But as I said, SI units doesn't match. So I have no idea how to interpret the 5.27BTU/h or how to relate it with the values from SI units.

Anyway, I considered the 0.22W/m.C value AI gave me.
Plugging these values in theQ_loss formula I would have:
Q_loss = (0.22W/m.K * 549.78m²*14ºC) / 0.004m = 423330W or 424kW
which seems to be a huge value, so I'm not sure if I did anything wrong!

Ok, yesterday I was supposed to do that simple math but I end up having an health issue and I had to go back home.

Ok, I think, given the numbers, I'll have to include some heat loss no matter what. But when time comes, I'll be back to this with the heat loss.
But before I need to find a way of comparing these values between the 2 systems.
I'm going to try to find another boiler / burner seller in my country that can provide me some data about those systems so that I can run the same numbers and see if I can get to some conculsion!

yeah the amazon electric heater 240 volt 2500 watts for 66 pounds is about the exact same heat as 1 s19k pro 2700 watts

Also if he adds braiins software he can turn it down

Fine so you want to assume that heat loss is 0, and delta T is 14, here is the math.

air density at 3c = 1.223kg/m
1.223*900*1005 = 1,106,203 * delta t of

15,486,849 joules > the heat required to raise the temps inside 900m3  by 14 degrees Celsius.

T = Q/W  

Example 1:

for 1*antminer S21 =

T = 15,486,849J/3500W/60s = 73 mins

Example 2:

for 2*antminers

T = 15,486,849/3500*2/60 = 146 mins

If you really want to assume no heat loss at all and up to 1 day to reach that delta t of 14, then you just use Q and T and solve for W
Since  T = Q/W then W = Q / T

Q = 15,486,849J > We got it from the above equation
 T = 24*60*60 = 86400 > Converting 1 day to seconds
W = 15,486,849/86400 = 179W

That means even 179W is good enough to heat a 900m3 area in 24 hours with 0% heat loss and perfect conditions, obviously, since there is no way that heat won't escape from whatever walls you are building, this never will never do it.

But if you really are okay with waiting a whole day to bring that temp up, then I think with good insulation, you can probably do well with just 5-10KW, your main issue now would be to discover how much heat loss you are going to encounter, if the place was already built, you could make a cheap test like Phil suggested, you get a 3KW heater put it there for 24hours while monitoring the temps, whatever that gets you is what 1*S19 would do, doesn't matter the source of heat, you just focus on the KW ratings and get an idea of what 3kw, 6kw, 9kw would do to your greenhouse if you get 9KW heater and it fits just fine, then 9KW worth of miners would do the exact same thing, air doesn't care how you generate that heat, heat is energy after all.

If the average person outputs 100w of heat, you put 90 lazy people inside the greenhouse and they would raise the room temp exactly the same as 3kw miners or 3kw heaters would.

you know if he has access to this product and he could return it he could simply  set some up and see how they do. then return them.

https://www.sylvane.com/delonghi-trrs0715e-oil-filled-radiator-heater.html?

two of the above on high give a constant 10,000 btus. about the same as one of the s19k pro

here are some on u k amazon I think they are 240
volt

https://www.amazon.co.uk/RWFlame-Radiator-Protection-Adjustable-Thermostat/dp/B0C9GXFYTD/ref=sr_1_2?

bigger model below 2500 watt so that is about 8500 btu

https://www.amazon.co.uk/Pro-Breeze®-Filled-Radiator-Built/dp/B095142QWJ/ref=sr_1_4?

I can tell you the s19 k pro does 2700 watts
I could sell you my 3000 dollar in coupons for 600 dollars

you would save 2400 buying s19k pros you need 5 at 9600 usd - 3000 in coupons = 6600 for 5 2700 watt miners which is 42000 btu's


to see if 5 miners are enough order the heaters from amazon.

 cost for the test is 330 pounds you would be really accurate by running the 5 heaters. best way to test is that way.

my belief is 5 heaters may be enough

and 5 s19k pros would then be enough

Guys, I just want to add that you completely lost me!
Like, I'm reading and reading and I start see numbers dancing on my screen and now moving to the walls while I hear some kindergarten song!

So, not making any math just adding a few things



One thing would be to drop that 9000m³ and just do it for 10 000m³. It would make a ton of things way simpler.
Second, I would drop that recommendation, I don't know where it came from but it makes little sense.
Raising in a perfect sealed environment the temperature of 10 000 m³ of air by 10C in an hour requires 35KW.

500KW system is madness, again we have 50KW boilers running on due fuel for stables, of course we don't have to keep the same temperature there as in an office (cows don't like over 20C) but still, 500kw is insane!!!!

And I don't know why from air you guys went to water!
A boiler that heats water then heats though pipes the building is far more expensive, yes, it's better distributed, it's more comfortable as it avoid having a few not so nice zones, it loses less heat to the outside since the pipes are indoor inside the building before the insulated walls, but it's costlier! Way more, it might be different from county to country but here we look at tens of hounds of euros, a
A Purmo radiator (so upper brand) rated for 2000W of maxim heat transfer is about 100euros, for 35KW you already have 2000 euros just in radiators, then pipes then a lot more things. HVAC ducts are way cheaper and way easier to fix.

Anyhow, as I said , you guys lost me!


Hmm, only outside insulation in a cold climate is also a bit bad, take for example my basement, if it weren't for a sheet of drywall and 2cm of hardwool before the PSS barier the basement will stay cold for a while as you need to heat up also a 60cm think wall made out of rock concrete. It's a common cause for arguing here in blocks of flats as some tenants lose heat by heating the concrete walls of empty apartments below, and we still have -20C windy nights!


I think this would be the biggest disadvantage of overcapacity here, I don't understand why would you need a system that would rise the termpature from scratch to what you want in 10 minutes only for it to stand then on standby for hours and not earning through mining even one penny!


Single sheet glass, 25*100 meters height 4 meters, outside -20 inside 20C gets me 240KW.
500KW is overkill!

Yes, I understand, however, even if it was 9000m³, I think the math would still hold correct for comparisons.


I need to correct myself. I didn't mean m². I men't simply 'm'. What I mean is that they measure 1m³ of firewood but they don't say m³. They simpy mention 'm'.
For instance:
I ask him: For hoow much you sell your firewood???
He replies: 50€ per meter!

But they measure 1m³ despite the fact that, while speaking, then only mention "per meter". They don't mention "per cubic meter".


Ok, I see that we need more specific context, so I'll give it here and is a way for me to be able to come back here in the future and see exactly the asusmptions I did to get to the numbers.
So, I'm considering a greenhouse. Period. No heating homes nor any other enclosed spaces. Let's talk only about greenhouses. Just to try to narrow the discussion. And this greenhouse is one of those with hard walls, not made of plastic or whatever  it is. I'm also not sure what is the exact material of the greenhouse.
Also, I'm going to consider the actual temperatures of my local/national weather report site for this day - 06-03-2024, around 11:00 am.

Let's consider 3ºC of outside temperature, and target temperature of 17ºC inside the greenhouse. So, delta of 14ºC, for the worst case scenario. And even if we have a variation of 4ºC or 5ºC around the day, let's ignore it! Let's pretend the plants or whatever we are growing inside doesn't car for such variation, beacuse in nature, all plants in the world will suffer temperature variations of probably more than this!

Another note is that yes, I'm aware that the miner wants to run constantly without power outtages, that's why I mentioned I was thinking to build the valve system to redirect the air flow in/out of the greenhouse based on temperature and humidity inside the greenhouse. Turning the miner off is out of the question, for the sake of math. heheh.

About the amount of time to heat up the space, yeah, I want to check that too, just to learn if I should include it in the report or not. I mean, if there is a gigantic difference, then maybe I should change something to make that difference smaller, otherwise, maybe I don't even need to mention it in the report!

Regarding the amount of time I need to recover from a power outtage, I don't ahve any specific requirement. Let's just assume that if we can recover the temp in less than a day, we will be fine.


For now let's assume no heat losses nor even heat distribution is needed.


I was just trying to stick with the same numbers. What is valid for 9000m³ woul also be valid for 900m³ in terms of comparison, I think. But ok, let's go for the 900m³.

About my way of thinking regarding the biofuel boiler / burner, I'm not sure I understand how I can compare the system rated power with the one of a miner!
As I mentioned somehwere in this thread, a S21 rated power output is 3500W, however I think this is rather 3500Wh. Meaning it uses 3500W of power every hour. Regardless, I will consider all the input power will be converted into heat, so the output power for heating purposes, will be the same 3500Wh.

For the boiler / burner system, with the data me and the guy exchanged was for 9000m³ of greenhouse volume, so we did the math for that volume and he got to the conclusion that a 400kW or 500kW system would be needed. But now we are considering 900m³ of volume, so the boiler / burner system will be much "smaller" both in wattage and in cost. That's why I was sticking to the 9000m³ of volume, because he told me te price of these systems. For 900m³, I have no idea! And in their site, they don't show prices for any of their products!


Yes, I see that that is a mistake of me. However, if I understand, these boilers / burners, consume biofuel to heat up water. And in turn, this water will travel pipes along the greenhouse to heat up air. I hope I'm correct here! Or maybe not, I'm not sure here! Maybe it heats air that is blown into the greenhouse by ducts too! I have to make sure of this!
also why the hell would you use 9000m3 of water in your calcs, that's a lot of water to be boiling. Cheesy, you want to stick to 900m3 of air since that is what you need to heat.

The difference between 900m3 and 9000m3 is 900%, and given that the topic is heat/cooling every meter matters.



You are converting seconds to days without going to minutes

187.941.096.000J / 500000W = 375882 seconds that would be 6,264.7 minutes or  104 hours which is just about 4 days not 261 days.

Water as you already know has a higher heat capacity than air it's 4186 J/kg°C vs 1005J/kg°C, but I think you are reading the labels wrong, I don't know what a water boiler labels mean, we don't use them hear, but I suppose whatever the rating is -- it's for the output capacity of the boiler, i.e it gives you 500KW worth of heat, you use it to heat air or water is a different subject, so you shouldn't be looking at things this way, also why the hell would you use 9000m3 of water in your calcs, that's a lot of water to be boiling. , you want to stick to 900m3 of air since that is what you need to heat.


Yes that's right, it's the correct formula, obviously, this assumes no heat loss and even distribution of heat, which both don't happen in a real world sncerio, there will always be heat escaping the room depending on the insulation and the outside temp, and even heat distribution in such volume is difficult, so if the room temp was even at 20c across the whole room, and assuming no heat loss, it would be maybe only 1 degree hotter on one side of the room, and 10 degrees hotter on another spot, but ya, the average would be 5c.


So now that you say yours is 900m3 and not 9000m3, then using your friend math of 34w/m3 it means you will need 30,600KW worth of heat, the number seems pretty reasonable to me, say your outside temp at winter is 5c and you need to be 25c, with delta t of 20c and 30KW, it would take roughly half an hour to heat the place by 20c, with heat loss and all the imperfection and the fact that you don't want to be running the boiler or whatever the source of heat 24/7 then I see 30KW is pretty reasonable from your guy assuming it's really cold outside (which you have not mentioned anything about up to this point)

With miners, since they are designed to run 24/7 you may get away with less power, assuming no power outages and no heat loss alongside some good insulation, I honestly think 10KW would do you just fine for 900m3 area, probably even less.


 

@mikeywith, so that time you got of 49 minutes would be the time it takes for 5 miners to heat up the 9000m³ of 5ºC, right? did I get it right?

So, now for the boiler / burner system, with the data I have, can we do the same?
I mean, if I take the data the guy gave me, he says that considerin 35W per m³, we need 365kW, and for that he uses a 400kW or 500kW system. Knowing these systems heat water that in turn will heat up the air, I presume, then, I'll have to do the same math but with C for water, no? With miners we consider transferring heat from air to air. With these boiler systems, we consider to transfer heat from water to air! Is this the way of thinking?

If so, I'll go ahed and run the numbers.

Water density at 20ºC = 998.2Kg/m³ (not sure this temperature make sense here, since if the water is cold and I start the boiler, the temperature of the water will be much lower, but I took the same temp as we did for air, 20ºC)
V = 9000m³
m = 998.2Kg/m³ x 9000m³ = 8.983.800Kg
C = 4.184 J/(Kg.ºC)
ΔT = 5ºC

Q = m x C x ΔT = 8.983.800 x 4.184 x 5 = 187.941,096MJ

Then to calculate time, I'm not sure how to run the numbers. The boiler systems they have range from 400kW to 500kW, considering the 35W per m³ the guy said and the total power needed to heat up the 9000m³.
I'll just consider the 400kW just to see what numbers tells me and see if it makes sense in terms of time!

W = Q / T = 187.941.096.000J / 500000W = 261 days. Doesn't make sense! So, I'm not thinking well.

I'm overwhelmed with the amount of calcs we can do around this.
As most of you seem a lot more confortable with the subject I would kindly ask to first come up with a plan so that we don't be here round and round around numbers and methods to calculate tons of things.

First of all, I might have made a very stupid mistake. I was thinking about a greenhouse 30m long, 10m wide and 3m height. So, this is 900m³ and not 9000m³. But this is just a side note. I will need this only after chosing/deciding (with your help)  how I am going to compare the 2 systems!


Oh yess.. lol. The guy told me that the wood chip is probably the cheapest biofuel and that, for instance, aviaries use exclusively wood chip. But it's not pellets. Pellets is probably the same but with some processing already. The woo chip I'm talking is the waste remains of wood proccessing inustry. Not sure this is the correct term. But I mean different stuff with wood chip and pellets.

And in my country, even firewood is sold in square meters although I think what they sell is cubic meters. But then simply use square meters.

So, as I was saying, please allow me to take a step back and ask your (of all people) help to first decide the method of comparison that you guys think is most suitable for this scenario.
What should I be comparing, knowing that I have 2 systems to heat up / mantain the temp of the same volume? And with the data I have, what should be the parameters I should be using? I think it's here where I'm kina lost because I'm actually overwhelmed with the amount of possibilities!

Well I have a 2000 square foot home.

the furnace uses gas and is 105,000 btu.  It can heat my house to 80f in the middle of winter cheaply and quickly.  But I have really really really good insulation.

So good I have to do an air exchanger to get fresh air into the home.

Now my house is usa based so it is 120 volts but I put in 3 x

240 volt lines.

I can run 15kwatts of miners nonstop in the winter. Right now I am running 5kwatts of miners.

My furnace did not turn on since yesterday.

And the mining reduced the heating bill from $250 to around $150 most of that 150 is showers and the stove.

In the summer our gas bill is $80.

So the miners do drop my heating bill.

But at 14 cents a kwatt. a 3000 watt miner costs me about $11 a day

it earns about $14.00 so I make $3 and drop the heat bill $2.

each day .  that means $5 a day to run an s19xp profit.

never mind that I spent $3200 to buy it.

As. an aside the space you run miners can become a heatsink or heat sponge.

ie my entire garage floor is concrete six inches thick. I can get the floor fully warm 80f to 85 f by mining in the winter.

Seems pretty close to 1231j/3600seconds, so the energy for heating up 1 cubic meter of air by 1 degree Celsius, but that would be 0.342 , so 34...? I don't know obviously the starting point for that guy's math


Oh no, please no!  
While I'll probably be the closest person on this forum to that field, as we do raise shrub willow for biofuel, wood is a pain in the ass to make math on heating with it, first nobody uses cubic meters when you deal with wood, you either go for kuub or stere, and then there is the whole thing of moisture content and ash for chips(pellets or how you call them in your country) and then the type of wood. No, if you don't want headaches and you have a business plan, go with gasoline, ethanol, bunker oil, gas, anything.

But if you want to stick with it, they are rated 5 kWh/kg for soft wood (evergreens), grab the price from the first pack that is sold at your local store and that's it!

So, if it's simply heating and you're not doing your graduation paper on it, this is all you need.
Packed for consumption it's 30cents (euro) (central europe) /kilo and goes for 5kwh of energy, so 6 cents per kwh. Of course, this is packaged consumption, for a space of 9000cubic meters, so 40 by 50 by 4.5 meters (almost larger than our livestock pen) you will buy by the truck. Again, not eveything gets converted that easily but for assumptions, let's ignore a bit that!

I'll come back on this after the weekend, the math I do during Saturday and Sunday is not even kinder garden material as I'm usually typing from the sofa with beer in my hand, but I'm anyhow going to drive to my parents so I'll pick the data from out dual boilers, they are rated 50kw so that would be enough for your model while accounting for efficiency, unfortunately I can't find the manual online anymore (10 years old machines).