How much ETICS makes sense? Revisited!

Posted on September 22, 2015
Tags: building, green

This is a revisit to the prior post based on improved data for EPS with Graphite (aka Neopor). It also considers a thicker but cheaper installation using a filled-cavity wall.

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 = ???0.08/mm, l = 32
  • Cork: Ve = ???0.24/mm, l = 40
  • PUR: Ve = ???0.32/mm, l = 23
So:
  • EPS: t = (25 + d * 0.18) * 0.842 / (1.11 - 1/(0.9 + d/32))
  • 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 * 0.18) * 0.842 / (1.11 - 1/(0.9 + x/32))
  • 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: 63mm
  • Cork: 52mm
  • PUR: 35mm
If I substitute back into the equations above, then:
  • EPS: t = 40 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 + 63 * (0.10 + 0.08) = ???36.34
  • 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: 0.35
  • Cork: 0.45
  • PUR: 0.41

Note: the EPS prices above are based on real prices for EPS60 blocks of graphite-enhanced EPS, made in Portugal.

Of further interest - consider what happens if I build a fully-filled cavity wall. Suppose I use the same thermal block on the inner leaf, and use 15cm tijolo on the outer leaf, with a Tyvek membrane or similar to stop air infiltration and dissuade water to enter the EPS from the outer leaf.

In effect I swap:
  • base bonding layer
  • mechaniscal fixings
  • cover bonding layer
  • mesh
  • cover bonding layer
with:
  • single fixing to hold boards in place while building the outer leaf
  • tijolo
Allow:
  • Tyvek ???4
  • Two fixings (one in each board) - instead of 8
  • 15cm tijolo (???3 plus fixing at ???3) this might be some way out!
  • Simplified procedure

This makes I, the fixed cost, about ???12 and Vf about ???0.03.

Additionally the R value for the wall is raised by that of the external tijolo, about 0.4, so the baseline R value is 1.3 and the baseline U is 0.77.

That makes the equation to solve:

EPS: (12 + x * 0.11) * 0.842 / (1.3 - 1/(0.77 + x/32))

The positive root is at 52mm.

Substituting back:
  • t = 17 years.
  • cost = ???27.72
  • U = 0.34

Which seems quite palattable! What's more, the actual cost is low and the installation mechanism is straightforward - even if we wish to increase the depth somewhat. In fact the insulation material is so cheap that the limiting factor is likely to be our stomach for loosing space to a thick wall, which is already over 46cm deep. Rounding it up to a 50cm section with 90mm of insulation yields a wall with a U value of 0.24, which is certainly good enough to reach Passivhaus if we want to.