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The most insulating grade TLA110 has a thermal conductivity of 0.043W/m.K compared to PIR of 0.023W/m.K so you will need 0.043/0.23 or 1.87 times the thickness to compensate.

Have you tried the Energy Saving Trust, they should have lists of installers, 0800 444 202 should get you through to their local to you office. As to the U-value some confusion here I think. The thermal conductivity of silver or grey EPS is likely to be about 0.033W/m.K. A 175mm cavity of a brick/block wall when filled with this is likely to have a U-value of about 0.16W/m2.K

Air tightness relates to bulk air movement, vapour permeability relates to a physical/chemical process which allows water vapour to transfer through a material which has no bulk air permeability You might also find the attached document useful. It has a section at the end on Perm conversions Vapour_Resistances includes perm conversions.pdf

Yes there is a difference. Air tightness relates to bulk air movement, vapour permeability relates to a physical/chemical process which allows water vapour to transfer through a material which has no bulk air permeability

@SteamyTea , section 2.3, page 11, here https://www.bre.co.uk/filelibrary/SAP/2012/SAP-2012_9-92.pdf

SAP assumes that the natural unpressurised infiltration is 1/20th of the 50Pa value.

An ordinary 25mm cavity has a thermal resistance of 0.18m2.K/W, a cavity with a low emissivity surface 0.44m2.K/W. (possibly the foil layer of the PIR) a cavity with 25mm of rockwool, thermal conductivity 0.035W/m.K will have a thermal resistance of 25/35 m2.K/W or 0.71m2.K/W Solid walls with a core of rubble fill do have a better than expected thermal resistance, typically 1.0m2.K/W When finished with plaster, pageV Mortar has a thermal conductivity of about 0.9W/m.K whilde granite is about 2.8W/m.K so I do not think so. hstp102011-u-values-and-traditional-buildings.pdf

IMHO, yes. The problem will be balancing and measuring the airflow. Any condensing in the heat exchanger would be an added complication. If you do not have it you may find information on your MVHR in this database https://www.ncm-pcdb.org.uk/sap/pcdbsearch.jsp?pid=34

You would need the exhaust and intake temperatures, but there are at least two additional problems, 1) The volumes of the two flows will probably not be the exactly the same, and 2) The heat from the fan motors will be added to the airstreams.

No, I'll rephrase - Only if there is an air path from the warm side of the stud around or through the insulated studs. The foil should be nearest the brick . It acts as a vapour check to reduce water vapour moving from warm to cold.

Only if there is an air path from the heated space around or through the insulated studs.

The m3 is a volume element, the volume 'leaking' per hour. I am not sure what you are asking, perhaps an example will help. With an infiltration of 407.5m3 / hr (1.0ACH) and an internal surface area of 377m2 then 407.5/377 = 1.08m3/m2/hr

hello and welcome, Attached are the BRE conventions for most situations, specifically pages 19-21 for what you are looking for. BR443-October-2019_consult.pdf

hello and welcome, not quite, if the extract air is humid enough and the incoming air fairly low temp the water vapour in the extract air can condenses to liquid in the mvhr heat exchanger, In the process giving heat to the incoming air. The liquid water is drained from the heat exchanger via a condensate drain. as to the PIV will there be trickle vents in the humid rooms?

Agree with @SteamyTea. Given good but not quite Passivhaus standards, walls/ceiling/floor to about 0.12W/m2.K, windows below 1.0W/m2.K, infiltration less than 1.0ach (with MVHR) annual space heating not more than 1000kWh, hot water 2500kWh (a lot less if WWHR installed on showers). Appliance electricity should be below 3650kWh/yr, current electricity use could give a guide here. So total of no more than 7150kWh/yr unless you are including an electric car. With further conservation measures perhaps the total could be reduced by 2500kWh/yr.

@howplum, I get U=0.28, given central England and a typical heating regime this will reduce heat loss by about 40kWh/m2/year

They are two different parameters. The second is the thermal conductivity of a single material and is independent of material thickness. A two brick solid wall has a U-value of approximately 2.03. This means it has a thermal resistance of 1/2.03 or 0.49m2K/W Where you have a thermal conductivity the thermal resistance of a layer is thickness in metres/thermal conductivity. You can then add all the thermal resistances together and take the reciprocal to get the new U-value. 75mm of your insulation will be needed to get below the requirement for renovations to achieve not more than U=0.3 (Part L1B) It will give U=0.234.

You will generate about that amount but to make a saving you will have to use it to substitute for electricity from the grid. If you have a 9 to 5 lifestyle you may only use 20-30% with the rest going out to the grid.

Yes.

Its easy, just select your location and follow your nose! For Forfar it gives about 890kWh/yr per kWp with about 4kWh/day per kWp in summer and about 0.7kWh per day per kWp in Jan/Dec

The lowest thermal conductivity for TLA I have found is 0.043W/m.K. Typical values for EPS for floors is around 0.035W/m.K. So you would need 43mm of TLA to give the same insulating effect as 35mm of EPS. You would need 215mm of TLA to provide the same thermal resistance as 175mm of EPS

English regs, N.B. Part L1A not L1B

100mm of 0.035W/m.K EPS would for a 8m x 15m slab give a U-value of 0.21 which is better than the Part L1A limiting value of 0.25, but CO2 emissions are calculated on a basis of 0.13 and you would have to add extra insulation elsewhere to compensate. 100mmm of PIR would give U = 0.15 would require less if any extra insulation to compensate.

And much safer in regard of interstitial condensation with the higher vapour resistance layer inside the lower. 175mm of 0.035 mineral wool between timbers and 75mm of PIR below will get 0.13W/m.K

In your situation you can expect at least 1700kWh/yr with up to 14kWh on a cold cloudless day in April/May or Sept/Oct. A 250l DHW cylinder can absorb at least 18kWh from near cold. So with a solar diverter you could export very little. .