MortarThePoint

That works on the side without the Resilient bar, but what about the side with the bar? Just a whole skin of Plywood or OSB3? Without: With on RB side:

what's that?

You're probably right and I'm looking to use resilient bar RB1 for the ceiling so could use it for the wall too. The main downside though is that I don't think the wall ends up very practical when it comes to mounting anything on it (e.g. TV, shelves etc).

This is what I have been planning and I think it achieves an STC of around 53. Wasn't overly sure what to do with the coupling to blockwork but 2x2 should be strong with the anchors. Notice only 50mm of Mineral Wool insulation. This looks similar to some I have seen, but would welcome opinions on that.

There is quite a lot online if you search for "wall STC"

Wow, I was just about to start a thread on essentially the same topic. I'm considering staggered 4x2 on 600 c/c with 6x2 top and bottom plates. Insulation woven through and likely a single sheet of 15mm Soundbloc F on each side. I hadn't given metal studs due consideration. Is this a separating wall between dwellings? I'm just wanting to keep the noise of our kids at bay.

I read of a Peltier based thermal inductor, but that's a bit of a cheat probably.

Temperature (T) should be analogous to Voltage (V) as it drives a flow of heat (Q) which would be analogous to current (I). V = I*R, T = Q*R. Would be nice to have passive equivalents to an inductor and a diode.

I hadn't heard of that before and have found the website: https://sunamp.com/ Not surprised it's been thought of before and I have found a thread in here as well.

Wowser! @nod are you still brave enough? Does he do the cement based for you too? Which one do you normally go with, still liquid? If you'd be happy to pass on contact details of your Screeder that would be great (PM welcome)

That should either have been: Q = (SUM[ A.U.dT ] + SUM[air change heat loss]) Or: dT/dt = (SUM[ A.U.dT ] + SUM[air change heat loss]) / SUM[ m.c]

Another interesting part of this is that you can approximate the way the house will respond to cyclical changes in external temperature like the diurnal temperature change. To make an analogy with electronics, the insulation acts like a resistor and the thermal mass as a capacitor. That makes for an RC low pass filter and the RC value is the Time Constant calculated above. The transfer function of an RC filter depends on the frequency of the input oscillation and the magnitude is: where RC is the time constant and w = 2*PI*f is the angular frequency and f is the frequency. For a diurnal temperature change it passes through 2*PI of angle in 24hours so w = 2*PI / 24 = 0.26 [NB: period T is 24hours which is inverse of frequency] V_in can be the temperature oscillation outside the house and V_out can be taken as the temperature oscillation inside the house [bit confusing there with in and out swapped, but that's due to the transfer function mapping an input, the changing temperature outside, to an output, the changing temperature inside]. If we used the time constant calculated above RC = 200hr and take the diurnal temperature change as 20C, the internal temperature change would be: 20C / SQRT[(0.26*200)^2 + 1] ~ 20C / (0.26*200) = 0.38C For an wRC >> 1 the SQRT[(wRC)^2 + 1] = wRC. Any time constant above 16hours should be OK for that approximation. STOP This neglects an important factor though, solar gains! If your house is completely shaded then look no further. The surface of brickwork will be hotter than the outside air temperature. Let's guess it reaches 40C and goes down to 10C over night so double the temperature. Windows let in the suns warming rays and so the heat input isn't just via conduction and air changes. Not immediately clear how to factor that in. This highlights the benefits of masonry construction and shutters in hot countries. I can conceive of 5kW of peak solar gain through windows during a hot summer and this would be 5kW/3.2kW = 156% of the heat transfer due to conductivity and air changes at dT = 10C. This could be approximated by using a compensated time constant: Q = (2*SUM[ A.U.dT ] + SUM[air change heat loss] + [Solar Gain]) SGTC = (1000* (insulated mass of concrete equivalents in tons) + 2500*(insulated mass of timber equivalents in tons)) / 3.6*(500 + heat transfer coefficient in W/K) In the simplified example in the first post, Q becomes around 5.2kW + 5kW = 10.2kW at dT = 10C [NB: 6.4kW was at dT=20C and we have doubled the heat flow due to conduction]. That makes the compensated time constant: SGTC = (64kWhr/C) / (10.2kW / 10C) = 63hours That makes the internal temperature change 20C / (0.26*63) = 1.2C. That's the minimum to maximum. [the deficiencies of all this a multiple, but it's likely to be indicative. Yes, it ignores the effect of the subfloor earth etc] If our time constant was also reduced to around a third of the uncompensated value, that would make for an SGTC=40hours temperature oscillation of 1.9C min to max. I would be pretty pleased with that. The outside temperature changing from 10C to 30C and in inside temperature changing from 19C to 21C without any heat pumps involved. There is an uncomfortable level of approximation in what I've just written, so if someone else knows of a simple was of doing a similar calculation that would be great.

Thanks, I'll look in to that. I have also seen this which is for greenhouses so may be cheaper, but I notice the question in the top right corner of the website which may suggest expensive crops. Also greenhouses probably want to be warmer than 22C

One thing I wonder though is if Rafter Roll has controlled expansion to ensure it doesn't expand up and touch the roofing membrane. There's supposed to be a 50mm gap left. SHould have gone the counter batten route, but too late now ? @ProDave your FrameTherm didn't look to puff up whereas @Thedreamer yours did. Any thoughts on that?

Adding in plasterboard and timber takes that time constant to 113hours, so 5.7hours to loss 1C at dT=20C or a temperature half life of 79hours.

Interesting to see the tank temperature. The wax plasterboard won't do anything for you up until it starts to melt and then it should clamp the plasterboard, and therefore likely the room temperature too, until the wax has all melted. After that the temperature would continue to rise, but it's a lot of heat by then.

I've got some grand plans in this area, but would use mains voltage compliant wiring etc everywhere so that it could be swapped back to mains if needed (e.g. future house buyer).

I'd accept 300K on a warm day I'm seriously considering having a 48V lighting system in the house because I think I should be able to design and install it myself all outside of part P and have better efficiency than OTS led lights, as well as use a 48V DC battery charged off Economy7 or solar.

As a comparison, just a panel of plasterboard behind massless U=0.2 insulation would heat up at a rate of dT/dt = 4W / (10kg * 1090J/kgK) = 3.6e-4 C/s gives ~1hour per degree C.

64kWhr per degree C is a lot. That's a Tesla per 2C. If you could effectively retrieve electricity from thermal heat, we'd have some pretty cheap solar batteries.

I never thought I'd spend so much time considering wax as a constituent building material. It's all to do with the latent heat of fusion being so high in comparison with heat capacity and that waxes can be chosen to melt at useful temperatures. There are two places it has occurred to me to use it: In a ASHP buffer tank - during off peak electricity hours you could heat a buffer tank to just above the melting point of the wax within it. For example, if you had a total of 100kg of wax in capsules inside a buffer tank, melting that wax would take about 2MJ of heat or 5.6kWhr. A wax that melts at the temperature you want to run your UFH water at and so as the wax solidifies it will give a constant temperature. You wouldn't have to heat the floor itself whilst you are running the ASHP or run the ASHP when you are heating the floors. Has anyone done this? Shouldn't be particularly expensive. Underneath rafters in a loft space - lofts can get too hot. If you had wax under the rafters below the insulation it would stop the plasterboard getting hot. As an idea, if the under tile void is 20C higher than the wax's melting point and you had U=0.2 of insulation, 4W/m2 would pass through the insulation into the wax. If you had a 1.1mm layer of wax it would weigh 1kg/m2 and absorb 200kJ in the process of melting. That would take 14 hours so never actually melt unless the average inside temperature went above the wax's melting point, averaged across about a day. You would choose a wax at a comfortable upper limit to the room temperature. I have seen that Knauf produce a plasterboard (ComfortBoard) with beads of wax in on this principle. I think it is very expensive though. Has anyone used that or done anything similar? Useful information of paraffin wax as a phase change material (PCM): https://www.intechopen.com/books/paraffin-an-overview/paraffin-as-phase-change-material

Yes effectively I have also found this paper which has more detail and shows my thinking wasn't entirely crazy. https://www.osti.gov/etdeweb/servlets/purl/20203120

To factor in timber, pine has a heat capacity of around 2500kJ/kg, much higher than concrete which surprised me. Time Constant in hours =(1000* (insulated mass of concrete equivalents in tons) + 2500*(insulated mass of timber equivalents in tons)) / 3.6*(heat transfer coefficient in W/K) https://www.engineeringtoolbox.com/specific-heat-solids-d_154.html

I've been thinking about rafter insulation and diurnal temperature considerations. This has lead to to consider what the time constant of the whole house is. Calculate the sum of insulated mass heat capacity and divide it by the total heat loss rate. Q = (SUM[ A.U.dT ] + SUM[air change heat loss]) / SUM[ m.c] dT / dt = Q / c.m dt = (c.m / Q) * dT Considering a simplified example system of: (I know the proportions are silly, but its helpful to keep the numbers simple and the aspect ration about right) Cuboid house 20m(l) x 10m(w) x 10m(h) --> total A = 1000m2, A_walls = 600m2, A_floors = 400m2 (ignore heat capacity of ceilings) Average U of 0.20 W/m2K --> 4kW @ dt=20C (note thermal bridges rolled into this) PIV or MEV dominated air leakage of 100l/s --> 2.4kW @ dt = 20C Total heat loss at dT = 20C is 6.4kW Floors: Assumed concrete floors (inc screed) 300kg/m2 --> m_floors = 400m2 * 300kg/m2 = 120t Walls: Assumed blockwork inner leaf (inc plaster) 140kg/m2 --> m_walls = 130% * 600m2 * 140kg/m2 = 109t (-10% windows, +40% for internal walls) Concrete, c is around 1kJ/kgC = 1MJ/tC SUM[ m.c ] = 229t * 1MJ/tC = 229 MJ/C = 64 kWhr/C, so takes 57 kWhr to change the temperature by 1C [3.6MJ = is 1kWhr] The time to loss 1C at dT = 20C is 64kWhr / 6.4kW ~ 10hours. In exponential terms, the time constant is 200 hours since dT/dt = T/200. This yields a 'temperature half life' is 200*LN[2] = 200*0.693 = 140 hours. This is going to have a large impact on how the temperature changes during the day and night as the outside temperature and the heating change. For example, if you wanted to only heat your house using economy 7 electricity and only have a 1C temperature change in the dead of winter you'd want the thermal time constant per 1C at dT=20C to be about 17hours. You can approximate it quite easily from your SAP and knowledge of floor and wall materials (e.g. screed, blocks and plasterboard order). The SAP has a figure in it called average "Heat transfer coefficient, W/K" which is (39) in a STORMA 2012 Version generated SAP. Time Constant in hours = 273 * (insulated mass of concrete equivalents in tons) / (heat transfer coefficient in W/K) [(1,000,000J/MJ) / (3600s/h) = 273] Our design looks to be around a time constant of 90hours, so 4.5hours to loss 1C at dT=20C or a temperature half life of 62hours. This is obviously a rather simplistic approach, but I think it is useful. Would be better to include mass of timber as well.

Did you mean to add a £ figure to that? Oh that's interesting, I didn't know there were issues with annhydrite and floor coverings? Is that the laitance?