Question on sizing storage

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DavidSavill said:
I have been reading https://www.hearth.com/econtent/index.php?ACT=24&fid=21&aid=55750_cVpXAbPPZAH1zYQLHYNx&board_id=1 and am stumped on storage sizes.

In the diagram it refers to Heating load of premises in kW and then Minimum heat in kW. Minimum heat of what? The boiler output or the minium heating load from the house that we can live with?

Qmin is Minimum heat output of the heating appliance in kW. Sounds like they mean the lowest steady output the boiler can maintain efficiently. As the ratio of QN to Qmin increases, the required storage volume decreases because the boiler can spend more time just keeping up with load without filling storage when load is light. The X axis is the amount of heat per load, which is constant.
 
ewdudley said:
Qmin is Minimum heat output of the heating appliance in kW. Sounds like they mean the lowest steady output the boiler can maintain efficiently. As the ratio of QN to Qmin increases, the required storage volume decreases because the boiler can spend more time just keeping up with load without filling storage when load is light. The X axis is the amount of heat per load, which is constant.

Thanks.... perhaps this graph is not giving me what I expected.

Lets assume I have a heat load of 50kW and boiler of 60kW that can go down to 30kW.

Assuming a fire time of 6 hrs, I am trying to understand the largest volume of storage I can effectively heat with that excess 10kW to use as supply when the fire has gone out (ie overnight)

I am thinking not much, excess of 10kW over 6 hrs is 60kW into the storage... but the graph (well its forumla since these numbers are off the chart a little) tells me I need 6300 L of storage.
 
DavidSavill said:
ewdudley said:
Qmin is Minimum heat output of the heating appliance in kW. Sounds like they mean the lowest steady output the boiler can maintain efficiently. As the ratio of QN to Qmin increases, the required storage volume decreases because the boiler can spend more time just keeping up with load without filling storage when load is light. The X axis is the amount of heat per load, which is constant.

Thanks.... perhaps this graph is not giving me what I expected.

Lets assume I have a heat load of 50kW and boiler of 60kW that can go down to 30kW.

Assuming a fire time of 6 hrs, I am trying to understand the largest volume of storage I can effectively heat with that excess 10kW to use as supply when the fire has gone out (ie overnight)

I am thinking not much, excess of 10kW over 6 hrs is 60kW into the storage... but the graph (well its forumla since these numbers are off the chart a little) tells me I need 6300 L of storage.

Looks like it may be one of those deals where by the time you figure out what the chart is trying to express, you already understand it to the point where it's easier without the chart.

To start you need to estimate what the useful temperature range of storage is going to be for your situation. If you have low temperature emitters you may be able to return water to storage down around 40 degC. If you have higher temperature emitters it may be hard to get much below 55 degC. So if you heat storage to 85 degC then your storage temperature delta could range anywhere from 30 to 45 kelvins.

Storage capacity is about 0.00116 kWh per liter per kelvin. Using the lower storage temperature delta of 30 kelvins, this works out to about 35 kWh per 1000 liters. The higher temperature delta would yield 52 kWh per 1000 liters of storage.

So to store 60 kWh you'd need somewhere between 1150 and 1700 liters.

But if your load is 50 kW, you'll be using up the 60 kWh of stored heat in 72 minutes.

--ewd
 
ewdudley said:
Looks like it may be one of those deals where by the time you figure out what the chart is trying to express, you already understand it to the point where it's easier without the chart.

To start you need to estimate what the useful temperature range of storage is going to be for your situation. If you have low temperature emitters you may be able to return water to storage down around 40 degC. If you have higher temperature emitters it may be hard to get much below 55 degC. So if you heat storage to 85 degC then your storage temperature delta could range anywhere from 30 to 45 kelvins.

Storage capacity is about 0.00116 kWh per liter per kelvin. Using the lower storage temperature delta of 30 kelvins, this works out to about 35 kWh per 1000 liters. The higher temperature delta would yield 52 kWh per 1000 liters of storage.

So to store 60 kWh you'd need somewhere between 1150 and 1700 liters.

But if your load is 50 kW, you'll be using up the 60 kWh of stored heat in 72 minutes.

--ewd

Thats, thats really helpful. To spin it around the other way....

Say it I wanted 5000L of storage. To shift 5000L from 55degC to 85degC would take 627,000,000 joules (4180J * 30 degC DeltaT* 5000L).

To get watts its Joules / Time(seconds), so 627,000,000 J / (21600 sec - 6hr burn)

So 29,027 watts or 29kW.

So if the heat load in the house was 50kW, then you would need an 80kW boiler to be able to heat the house AND charge the storage.

I googled lots of this, would appreciate a sanity check - does that sound right? :)
 
DavidSavill said:
Say it I wanted 5000L of storage. To shift 5000L from 55degC to 85degC would take 627,000,000 joules (4180J * 30 degC DeltaT* 5000L).

To get watts its Joules / Time(seconds), so 627,000,000 J / (21600 sec - 6hr burn)

So 29,027 watts or 29kW.

So if the heat load in the house was 50kW, then you would need an 80kW boiler to be able to heat the house AND charge the storage.

I googled lots of this, would appreciate a sanity check - does that sound right? :)

Your numbers are -- how do you say -- spot on, dead naughts.

As for sanity, 50 kW is anywhere from four to ten times the amount of heat load I think one would expect for a 'normal' house. Where does the 50 kW number come from?
 
ewdudley said:
Your numbers are -- how do you say -- spot on, dead naughts.

As for sanity, 50 kW is anywhere from four to ten times the amount of heat load I think one would expect for a 'normal' house. Where does the 50 kW number come from?

Had the heat load sized by a a few different radiator folks. They all come up from about 45-55 kW. I came up with 49kW using http://www.heatlosscalculator.co.uk/.

Its a big house, 380m2 with insulation that poor, but the best that can be achieved. Its actually three small houses joined together so there are 9 external walls.
 
DavidSavill said:
Its a big house, 380m2 with insulation that poor, but the best that can be achieved. Its actually three small houses joined together so there are 9 external walls.

Right, the one with the courtyard? The floor plan looked like wonderful architecture for living, like an ancient Roman extended family setup.

Well 50 kW, that's a shload of heat. As they say on hearth.com, go ahead and reduce you heating load by two thirds, install low-temperature radiant panels, build yourself a Garn barn, and Bob's your uncle! All it takes is a 'metric a**load of cubic dollars'.
 
I might have missed the info -- what temp is the design heat load based on? How often/how many days each heating season do you experience this, on average? What's the average monthly low temperature you need to plan for? It may make more sense to plan for average lows and then supplement with other heat rather than plan for design heat loss at a temperature extreme.

Example: my shop design heat loss was 35,000 btuh at -37C. But the highest heat loss I measured winter of 2010-11 (very typical, cold, average winter with lows to -37C) was right around 18,000 btuh, and more typical heat loss is in the 12-14,000 btu range. That makes a big difference in sizing the amount of storage. I have 1000 gal pressurized storage, and there has been no period yet where this will not last two days between burns, and often 3 days. I have not had to use supplemental heat. At design heat loss I would be burning every day.
 
jebatty said:
I might have missed the info -- what temp is the design heat load based on? How often/how many days each heating season do you experience this, on average? What's the average monthly low temperature you need to plan for? It may make more sense to plan for average lows and then supplement with other heat rather than plan for design heat loss at a temperature extreme.

Example: my shop design heat loss was 35,000 btuh at -37C. But the highest heat loss I measured winter of 2010-11 (very typical, cold, average winter with lows to -37C) was right around 18,000 btuh, and more typical heat loss is in the 12-14,000 btu range. That makes a big difference in sizing the amount of storage. I have 1000 gal pressurized storage, and there has been no period yet where this will not last two days between burns, and often 3 days. I have not had to use supplemental heat. At design heat loss I would be burning every day.

Yeah I hear you, I look at everyones heat load here and get very envious. I am worried we if undersize the boiler and are stuffed. Seems if you oversize it and have storage to back it up then worst case is you dont start the fire as often.....

Average temps last year were
May 6.15 Jun 5.61 Jul 4.43 Aug 7.23 Sept 9.44

Lows were
May -5.72 Jun -4.94 Jul -5.17 Aug -3.06 Sep -4.22

So nothing like what your getting.

Real life heat loss - got me wondering but its hard to know. One of the houses where we have rads now we have 5.5kW of radiators in a combined kitchen/lounge. They dont heat it enough over June/Jul, can never get it above 19degC unless we turn on the electric heaters even when the boiler is going hard. If we go to bed with the boiler roaring around 5 hrs later the temp will be down below 14degC, by 8 hrs we are down to about 10degC which is where daddy is out swearing at the boiler with the kids complaining they are cold :)

My thinking was whack another 3kW in that space, so the radiators can work at lower temps and we might be on the sweet spot.
 
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