SteamyTea

The ones that wear pants under their kilts, and can't toss a caber.

So you remembered why you disliked it.

Do you use the remaining hot water to make porridge, with salt in it. Then put the horse hair shirt on.

By default, all DNOs do, to 16 A/phase. I am sure of you applied for a larger system, as long as you met the critter, it would be granted. Are we going back to the 1980s 'yuppies' with their conspicuous consumption and debt envy.

Welcome. A few of us have an interest in narrowboats. Where is it moored, and watch out for the many negative comments about WBS, I start most of them.

Do they want a little project as I have just about given up getting my Python scripts to read my MLX90614s. (can post the kits if needed)

An old mate of mine (was also once my student) looked into it about 15 years ago. I never found out what the outcome was, I think the MVHR recovered more energy in his modelling.

Not to worried about the odd litre or so, but I think I get what is happening now. Ignoring that at the moment as it just muddies the water. So 2.27 [kWh] - 0.65 [kWh] = 1.62 kWh recovered. That would imply that somewhere there is a ∆T of 23.2 K. 4.18 [kJ.kg-1.K-1] x 60 [kg] x 23.2 [∆T] = 5,818.56 kJ or 1.616 kWh I can't work out how that ∆T is created, the closest is the ∆T between the 'from shower tray" 37.1°C and the 'soil stack" 11.4°C, ∆T 25.7 K. Using that, and it seems to make sense. 4.18 [kJ.kg-1.K-1] x 60 [kg] x 25.7 [∆T] = 6,445.56 kJ or 1.79 kWh which is in the same ball park as the 1.62 kWh. I do find the 88% heat recovery a bit high and would that number double checked. (I often get muddled with numbers, it is very easy to pick up the incorrect number when working out efficiency)

Take this at face value. The energy required to heat that much water is: 4.18 [kJ.kg-1.K-1] x 60 [litres or kg] x (40 - 7.6) [∆T] = 8125.92 kJ, or 2.26 kWh. Your claim is that it uses 0.65 kWh, 2.26 times less, a 71.2% reduction. Now this is where I am getting a bit confused. Taking just the water going down the soil stack, 6.8 lt/min and comparing that to the water in from the main cold feed, 5.9 lt/min plus shower cold feed 0 lt/min and the shower hot feed 6 lt/min. 6.8 [soil stack lt/min] = 5.9 [main cold feed lt/min] + 6 [shower hot feed lt/min] + 0 [shower cold feed lt/min] 6.8 = 11.9 What am I missing?

Stop that right now, just asking to have an unattended shower. I flooded my house by popping downstairs, getting distracted and forgetting I had it running.

I seem to remember it is paintable as well. @Pocster is the CT1 man, we all told him that it will fix everything. As I drove on the M25 yesterday, between the M40 and M4, there is a very large electronic sign advertising CT1. Must be there to remind all the Boing engineers it is used to stick the doors on.

Right, as pointed out above, how is your heater more efficient. Shower waste water heat recovery has been around decades, you even get your SAP/EPC adjusted if fitted. What makes your WWHR better? Going to be a hard sell, 6lt/minute may be okay for a man, I suspect not many women will like a dribbling shower with that flow rate. If you really want to make an innovative product, think up a way to recover the energy when a bath is drained. More energy to recover from 100lt of 30⁰C bath water than 60lt of 26⁰C shower drain water. Have you got some actual usage data? We may sound a bit negative on here, but we like like a robust debate based on real data, not Hopium.

Cornwall. Like Lewis, except in every single way.

Nor here. Radioactivity is though. A really deep borehole into ancient granite should show a decent temperature gradient. As you have soft water, how about one of those water boiling taps. Am sure they can be got quite cheap. https://www.diy.com/departments/nes-home-3-in-1-instant-boiling-hot-water-kitchen-tap-chrome-tap-only-with-cool-touch/3133312774455_BQ.prd

You can also use heating degree days. Where the line intersects the y-axis is your non spacing usage. https://www.degreedays.net/

They tend to be very efficient as well. Nearly all the energy goes into the water, and as it is used immediately, there are no standing losses. If you are in a hard water area, when it scales up, buy a new ten quid kettle.

Most caulks are a water based acrylic. They have different amounts of inert fillers, the more filler, or the larger the particles, the less shrinkage. If you really want to fill something permanently, see about using an epoxy, because they generally have very small quantities of volatile (at normal temperatures), and the polymer cross linking is more thorough, they tend to stay the same volume. If you want some gap fill, then polyurethane is the material of choice, but they will, after time, shrink (just look at a car dashboard or furniture cushion).

You can take the cosine of the incident angle and multiply that by an arbitrary length to get the new arbitrary length. Then divide the power by the length to see how much it reduces by. I may have to write it up proper sometimes instead of a quick reply. Or just go to PVGIS.

Is that for that Touching the Void moment.

No idea about how good it is. Is there anything that makes you think it will be better than the £1/tube ones?

The problem here is that there is no price to compare it to, made worse by the suppliers not understanding their own tariff structures and customer needs. Just yesterday on the BBC's You and Yours they had tales of woe about billing problems.

@ToughButterCup I made a bit of an error, should have said that both the sun altitude And azimuth angles are the same. If only one angle changes, then you get greater power.

Trigonometry, Power and Specific Heat Capacity. To keep things simple, assume the same light beam, with a power of 200 W.m-2, and using the sketch above. When the sun is at 90° to a surface, the area will be 100 mm by 100 mm, so 10,000 mm2, or in proper language 0.01 m2. The power per square meter is therefore 200 [W.m-2] x 0.01 [m2] = 2 W Angle the beam over by 30°, the area changes to 115.47 mm by 115.47 mm, so 0.0133 m2, 33% greater area, at 60° it is 200 mm by 200 mm, 0.04 m2, 4 times greater area. So it goes from 2 W of power, though 1.5 W at 30° to 0.5 W at 60°. If your wall was granite (or most stone/brick/cement) it has a specific heat capacity of around 0.8 kJ.kg-1.K-1. Now a W is a J.s-1, a small amount really. The density is around 2500 kg.m-3. Taking an area of 1 m2, but just the first 10mm of wall depth that is facing the sun, this will have a mass of 2500 [kg] / (1 [m] x 0.01 [m]) = 25 kg. So to heat up 25 kg of wall, by 1 K will take: 0.8 [kJ.kg-1.K-1] x 25 [kg] = 20 kJ of energy. At a power input of 200 W.m-2 (200 J.s-1), and not taking any losses into account (which will be quite a large fraction) to increase the temperature by 1 K will take: 20,000 [J] / 200 [J.s-1] = 100 s. At 30° light beam angle, 133 s, and at 60°, 400 s.

Some of them have a small marker on them that shows up on the screen, or if yours does not, stick a little bit of tape on the bottom.

Is that all it takes, a smelly partner. Should have told my BiL that, would have saved my sister robbing him. I fitted my shower pump under the bath, it pumps to the bath mixer tap, and therefore the shower as well. Can you do something similar as a temporary measure.