How much ETICS makes sense?

Let us suppose that I plan to build a house with the Artebel block system using pillar formwork blocks 250mm deep, and the BTE25 blocks. The U value is given as 1.07 for the blocks and we have some thermal bridging from the pillars and a bit of extra resistance at boundaries, so let's assume a composite U value of 1.11, or R0.9.

Now I need to decide how much ETICS to install.

I know or assume that:

• the thermal resistance, R, increases linearly with depth
• the U value is the reciprocal of the summed resistances from the blocks and the insulation
• there is a cost of installing ETICS relating to the glue, mechanical fixings, mesh reinforcement and so on - and labour to fit.
• the installation cost is largely independant of the depth of the insulation (on a new build) except for the cost of mechanical fittings.
• the cost of insulation is broadly proportional to its depth

Let us assume that there is an equivalence between the inflation rate of heating fuel and the discount rate I use for future values.

The metric I will use is to find the shortest repayment duration, from a simplsitic division of the insulation cost by the savings in a year.

I need to know the number of 'degree hours' of heating (or cooling) that I have in a year.

Tomar seems a little warmer than you might expect given the location between Porto and Lisboa, so I assumed:

• 2 months at 3 degrees
• 1 month at 6 degrees
• 1 month at 9 degrees
• 1 month at 12 degrees

This is 33 'month degrees', or 23760 'hour degrees'.

Assume that heating costs are based on the cost of pellets - about €0.05 per kWHr. Heat pump air conditioning units run with quite a high COP in Central Portugal and have similar running costs.

Definitions:

• d, the depth of insulation, in mm
• l, the conductivity (lambda) of the insulation, normalised to mm
• Rw, the resistance of the basic wall, 0.9
• Re, the insulation of the ETICS, d/l
• U, the insulated U value, 1/(Rw+Re)
• I, the fixed cost of installing ETICS
• Vf, the variable cost of the fixings per mm
• Ve, the variable cost of the insulation per mm
• Ci, the cost of the insulation, I + d * (Vf + Ve)
• U', the change in U value by insulating, 1/Rw - U
• H, the number of 'hour degrees' annually
• F, the amount of fuel saved per year, U' * H / 1000 in kWHr
• Cf, the cost of the fuel as a savings per year, F * 0.05, where 0.05 is the cost per kWHr
• t, the number of years to repay, Ci/Cf

Now:

• I is the estimated cost of fixing and is €25
• Vf is the cost of the fixings and is €0.10/mm

Thus:

t = Ci/Cf

Substitute:

t = (I + d * (Vf + Ve)) / Cf

Again, Cf (and then F):

t = (I + d * (Vf + Ve)) / (U' * H/1000 * 0.05)

Simplify:

t = (I + d * (Vf + Ve)) / (U' * H / 20000)

Simplify, H is 23760:

t = (I + d * (Vf + Ve)) / (U' * 1.19)

Move that 1.19 term so we can expand U':

t = (I + d * (Vf + Ve)) * 0.842 / U'

Substitute for U':

t = (I + d * (Vf + Ve)) * 0.842 / (1/Rw - 1/(Rw + Re))

Substitute Rw and Re:

t = (I + d * (Vf + Ve)) * 0.842 / (1.11 - 1/(0.9 + d/l))

Substitute I and Vf which are known:

t = (25 + d * (0.1 + Ve)) * 0.842 / (1.11 - 1/(0.9 + d/l))

Consider:

• EPS: Ve = €0.10/mm, l = 37
• Cork: Ve = €0.24/mm, l = 40
• PUR: Ve = €0.32/mm, l = 23

So for EPS:

• EPS: t = (25 + d * 0.2) * 0.842 / (1.11 - 1/(0.9 + d/37))
• Cork: t = (25 + d * 0.34) * 0.842 / (1.11 - 1/(0.9 + d/40))
• PUR: t = (25 + d * 0.42) * 0.842 / (1.11 - 1/(0.9 + d/23))

I want to select the most cost effective depth in each case, so I want to minimise t as a function of d.

My calculus is rusty, so I cheat at this point. Step up the Online Derivative Calculator.

To make it easy, I rephrase as a function of x (so x means 'the depth'):

• EPS: (25 + x * 0.2) * 0.842 / (1.11 - 1/(0.9 + x/37))
• Cork: (25 + x * 0.34) * 0.842 / (1.11 - 1/(0.9 + x/40))
• PUR: (25 + x * 0.42) * 0.842 / (1.11 - 1/(0.9 + x/23))

You can:

• paste these functions into the box at the top of the derivative calculator page
• press Go
• press Roots/zeros for the First Derivative (F'(x))
• Look for the positive root

And the answers are:

• EPS: 65mm
• Cork: 52mm
• PUR: 35mm

If I substitute back into the equations above, then:

• EPS: t = 49 years
• Cork: t = 61 years
• PUR: t = 54 years

Note that the actual costs of the insulation are estimated as (I + d * (Vf + Ve)):

• EPS: 25 + 65 * (0.10 + 0.10) = €38.00
• Cork: 25 + 52 * (0.10 + 0.24) = €42.70
• PUR: 25 + 35 * (0.10 + 0.32) = €39.70

So - this is not a result of inflated estimates for ETICS costs. These costs are low and assume that in a new build we do not have huge costs resulting from rearranging door and window stones, changing the roof overhangs etc.

The U values implied are (1/(0.9 + d/l)):

• EPS: 0.38
• Cork: 0.45
• PUR: 0.41

Surprised? I was - I had expected an optimised depth to be greater.