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Posted

I was a bit bored on yesterday's car journey,  so I started to let my mind wonder onto important things.

So I have taken a stab at making a DHW calculator based on cylinder volume and insulation levels.

The main points are in the pop out comment boxes.

The variables are internal cylinder height and diameter, insulation k-value, insulation thickness, top of cylinder water temperature, base of cylinder water temperature, minimum useful delivery temperature and the room that the cylinder is in ambient temperature (the cylinder base temperature is the same).

 

I am not sure if it is correct, but if anyone wants to check, there is a password in the comments, then you can see the inner workings.

Feel free to suggest any changes.

 

 

DHW Cylinder.ods

Posted

Hello @SteamyTea

You should be using the 1D heat equation derived for a cylindrical coordinate system for the heat loss calcs.
It's not a night and day impact (depends on the geometry), but I get just under 10% higher heat loss using this approach.

Posted

Then, the second point is a bit more interesting and something I'm not sure is being factored in to a lot of house heat loss calculations (not suggesting it's a big impact, haven't looked at it in the context of houses).

 

The external wall of the insulation will not be fixed at the ambient room temperature (18°C as you have it in your example). This should be a convective boundary condition. The formal definition is a Neumann boundary condition, as opposed to a Dirichlet (imposed temperature).

Where I think this could be interesting, is that the whole house heat loss calculations I've come across, assume the surface of the internal walls are at the internal temperature and vice versa for the external walls. In reality, you have a heat balance for the wall, with a convective boundary condition on either side.

 

The top "slice" of the tank in the original example lost 4.04W

Adopting the 1D heat equation in cylindrical coordinates reduced this to 3.39W

Changing the external boundary condition at the wall of the cylinder from a fixed 18°C to an ambient temperature of 18°C with a convective heat transfer coefficient of 5W/m2K reduced the loss to 2.98W (already includes impact of different formulation)

 

Would be interesting to apply this approach to the heat loss of a whole house (obviously not including the cylindrical coordinate system, unless it's a very unusual shape!)

Posted

@Owain1602

 

I did think about using an equation based on the cooling laws, but ran out of time.

The idea of a boundary layer of air is an interesting one, it is used in building heat loss calculations.

Not sure how much difference that will make in an airing cupboard, but may have a look when forced to stay inside for a day (usually when car is being serviced).

Thanks for the feed back.

Posted

Do you know which building heat loss calculations use convective boundary conditions on the internal and external walls? The simplified spreadsheets that I've seen most use just imposes the temperature on the surfaces of the internal and external walls.

 

You can see in the example values above, accounting for the convective boundary on a single slice of the tank reduced the heat loss by ~10%. When I look at the results of this, the external surface of the tank for the top slice was actually at 22°C (for an ambient temperature of 18°C).

 

I will have a look at the relative magnitude of this effect on a section of wall later this evening, to get an idea if it's significant.

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