Sorry for the lateness of the response, I have been out and about.

The PSI value is essentially the reduction in U-Value at the junction.

As for the pretty colours on the isotherms, red and white is good (i.e. warm) blue is bad as it is cold. The colour inbetween are the temps inbetween. The colours on the heat flux are heat flows - red and white are the largest heat losses.

I don't see that insulating the party wall will help a huge amount, but I could run the numbers again tomorrow with 50mm of EPS on the party wall (very simple model) to get an idea.

Obviously adding EWI will help over 95% or more of the building, but you guys asked what the thermal bridge was, so there it is!

Hi Dave,

I'm not understanding this either. If you multiply W/mK x length you still don't get a measure of heat loss The height would need to be factored in as well wouldn't it to get W/m2K - and hence the actual heat loss by calculation?

As I asked above, I'snt the PSI value meaningless if you cannot put a number on the actual loss?

What djh wrote supports my somewhat vague understanding of what psi (linear conductance) values are. Let's see if I can a) check my understanding and b) explain it with an example:

Suppose you have a wall with a U-value of 0.1 W/m²·K with a studding or something with a psi value of 0.2 W/m·K. If the wall is 2.5 metres high and 4 metres wide with studs on 1 metre centres (yes, I know, rather implausible in the UK but let's keep the arithmetic simple). There are three studs, each 2.5 metres long for a total length of 7.5 metres.

The total heat loss co-efficient of the wall would be area × U-value + stud-length × psi-value = (2.5 × 4 × 0.1) + (7.5 × 0.2) = 1 + 1.5 = 2.5 W/K.

If the temperature differential is 20 °C (i.e., 20 K) then the heat loss would be 20 × 2.5 = 50 W.

The bit I'm rather vague on is the idea that the psi value is in addition to the basic U-value; I think it's the incremental loss over the basic U-value. It doesn't have a width as such as that's already taken into account in its evaluation. E.g., a particular I-beam in particular insulation would have some psi value. If you doubled the thickness of the I-beam or changed the insulation you'd have to find a new psi value. On the other hand, if you decided to go from 600 mm centres to 400 mm then you'd just have to increase the total length you multiply the psi value by in order to calculate the total heat-loss coefficient.

Posted By: fostertomIs the point of psi-value, that once calcd, you can apply it 'per m' all over the place?

Yes.

And to calc it in the first place you model 1m x 1m of the junction in Therm.

I think you'd model 1 metre of the junction in Therm; 1 m × 1 m wouldn't make sense for a linear feature.

The therm model has no length. The 1m is how far to either side of the junction the model must extend to ensure that the heat flux and isotherms can stabalise. Also the 1m is not set in stone. The model should extend to three times the thickness of the element (wall) or at least 1m, whichever is smaller. Although this is not often followed as 1m tends to do the trick in most cases.

Apparently, the way this works is that the PSI value is used to calculate a Y-value, which is measured in the units we all love and understand - W/m2K. This can then be used to determine the actual heat loss.

All is explained here http://www.pinewood-structures.co.uk/content/1/95/psi-values--y-values-and-thermal-bridging.html

Yes, best to ignore the blag and just get to the explanation of what's what

Though I'd still like to see it all put into context. IE.

Overall predicted heat loss and the contribution a thermal bridge makes to it.

Yes, of course. There's actually a worked example on the link

Well I have just re-done some calcs.

I actually got the first ones slightly wrong - in-so-far-as I tweeked the model to the one shown in the pics, but forgot (DUR!) to re-run the calculations. The PSI value of the justion that I showed is actaully 0.198 W/mK. Not as bad as 0.3 that I quoted previously but still poor.

Here is the interesting thing though...... insulating the party wall with 50 mm of EPS makes almost NO difference to the PSI value. It goes down to 0.193 W/mK. Hardly worth the effort from a thermal point of view.

So, there it is, insulating the party wall doesn't offer much (if any) improvement in this sort of situation. And, infact, it makes things worse for your neighbour. Their corner between the party wall and external walls gets slightly colder, as the insulation on 'your' side, reduces the amount of heat you contribute to keeping the party wall warm.

@tom, @mike, I'm having trouble seeing what you find difficult about psi-values. Let me try an example.

Suppose I have a 2.5 m x 2.5 m wall with a U-value of 0.1. So the heat loss is 0.1 x 2.5 x 2.5 W/K
Suppose I have a 1 m x 1.5 m window with a U-value of 1. So the heat loss is 1 x 1 x 1.5 W/K

So if I put the window in the wall, the total heat loss H is 0.1 x (2.5 x 2.5 - 1 x 1.5) + 1 x 1 x 1.5 W/K = 1.975 W/K

But that's not quite right, because the heat doesn't flow out at right angles where the window meets the wall. It sneakily curves to find the path of least resistance. So the psi-value adjusts for that. Somebody makes a cross-sectional picture of the edge of the window and the wall and the clever program calculates how heat actually flows through it, and what the difference per metre is.

Suppose it is 0.08. Then you work out the length to apply it over - the perimeter of the window; in this case 1 + 1.5 + 1 + 1.5 = 5, multiply them together 5 x 0.08 = 0.4 and add it to the heat loss H' = 1.975 + 0.4 = 2.375 W/K

Of course, the psi-value for the sill and jambs and head could all be different, to complicate things further.

Posted By: djhIt sneakily curves to find the path of least resistance

Yes, it is sneaky. It goes from hot to cold, as the internal surfaces (at the very first molecule level) are at the same temperature the heat path is perpendicular, at the next molecule level the windows is slightly colder than the wall (worse U-Value), so some of the heat that is rocketing through the window (ten times faster than the wall in the above example if assume identical thickness for both) goes forwards toward outside, but some will go to heat the wall, this goes on all the way though. Kind of two steps forwards and one to the side and half a one forwards (bit like a knight in chess, or me dancing in the 1970's).
So it should be possible to do it by vectors if you know the thermal inertia of the wall and window (and any other materials involved), the thickness's/profiles and the temperature gradients, probably end up drawing an ellipse (looks like one on the pictures).
In 3D it will look like an egg.

It seems to me, that's very well reasoned.