CWI - how to measure effectiveness?

[David J]

I have been wondering how to measure the 'effectiveness' of adding CWI to a domestic house, apart from seeing the resultant change in energy costs at a much later date.

It is clearly not good enough to simply measure the surface temperatures of the inside and outside of the exterior walls, as one is the source and the other is the sink of the heat transfer through different materials. And the exterior wall temperature will always be that of the ambient climate temperature.

I have read about R and U factors, but that does not help. I have also read that thermal transfer (in buildings) is analagous to electrical flow through conductors, so what really is required is to obtain the thermal drops across the wall sections.

So, taking a typical modern exterior wall construction, 4 temperatures are needed as follows.

T1. - surface temp of inside inner wall (plaster) T2. - surface temp of outside inner wall (thermal block) T3. - surface temp of inner outer wall (brick) T4. - surface temp on outer outer wall (brick)

With the CWI installed and thermal transfer reduced, under steady state there will be no change to T1 and T4, but an increase in T2 and a decrease in T3 with the majority of the thermal drop now occuring across the cavity insulation.

Does the above reasoning make sense?

David

[Tim W]

David J wibbled on Friday 27 November 2009 15:42

That's about it (ignoring the thickness of the plaster and assuming T1 is the temperature of the inner face of the thermal block).

Do it like a resistors in series calculation:

Current is analagous to power transmission Potential difference is analagous to temperature difference.

I can never remember which is U/R/K/whatever is analagous to resistance (in this case more like resistance per unit area), but a look on Wikipedia:

is the R value. U values are the inverse.

[Andrew Gabriel]

That's not the case. The outside wall must be warmer than the ambient temperature in order to lose heat to it. Similarly, the internal side must be cooler than the air temperature in order for the room to lose heat to the wall. The difference in temperature between the air and the wall is (roughly) proportional to the energy loss at the inside. This is also true at the outside but factors such as wind-chill and dampness will affect the figures and make comparison readings impossible, so you'll need to measure this on the inside where wind-chill and evaporation are hopefully a constant (zero).

It should be more like this...

T0. - ambient indoor temperature T1. - surface temp of inside inner wall (plaster) T2. - surface temp of outside inner wall (thermal block) T3. - surface temp of inner outer wall (brick) T4. - surface temp on outer outer wall (brick) T5. - ambient outdoor temperature

Assuming T0 and T5 remain the same... There will be an increase in T2 and a decrease in T3 as you say. However, there will also be a (smaller) increase in T1 and decrease in T4, because less energy is being drawn though the wall.

Measuring the difference between T0 and T1 (or T0 and T2, or T1 and T2, as none of these thermal elements change) will give you a figure which is proportional to the temperature loss through the wall.

[The Natural Philosopher]

I just look it up in a manual that tells me that the heatloss with CWI is about a third as without...

Since mots people have dine lofts and double glazed, that means about a

60% saving in fuel bills, as the walls are the dominant loss now.

[Andrew Gabriel]

^^^^^^^^^^^ Sorry, should say _energy_ loss.

[David J]

Andrew - thanks, I find your correction above is very convincing.

Now to figure out a good way to measure T2 & T3, before the CWI is installed!

David

[Andrew Gabriel]

Quite apart from being difficult to measure, I don't think that measurement will tell you anything useful quantitatively.

Take several temperature reading sets of T0, T1, and T5 (I'd include T4 too, but it's less useful). Each temperature set is all of them at the same time.

Do this again after the CWI. If you get a reading set with the same value for (T0 - T5), you will hopefully find (T0 - T1) is lower. To a first approximation, if old (T0 - T1) is 3 times the value of new (T0 - T1), then your heat loss was 3 times higher before.

If you don't get a reading set with the same value for (T0 - T5) before and afterwards, then you can still work it out, but you'll have to include a correction factor for the difference.

An infrared thermometer is excellent for measuring wall temperatures. Ideally, you want to use the same thermometer for measuring the air temperature, which an IR thermometer can't do. What you can do is to measure the temperature of something such as a small sheet of paper hanging in the air, at least a metre from the wall (but same place every time), and make sure it doesn't have the sun on it, or it won't be at air temperature.

[YAPH]

You could get a reasonable estimate using the whole house boiler sizing method to estimate the total heatloss of your house and do the calculation with and without CWI.

There are links at

to some online calculators, and a spreadsheet with which you can easily do what-if calculations.

[NT]

A simple option is to plug a heater of known power in, let temp stabilise, and measure temp diff between indoors and out. This takes account of the whole picture.

NT

[dennis]

You van make the maths easier by using an angle grinder to cut a 1 m square section out of the wall. Mount this ay one end of a well insulated box containing the heater. Let it run until it stabilises and then measure the temps.