On Fri, 8 Dec 2017 03:22:18 -0800 (PST), whisky-dave
wrote:
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There were no tools that were any use unless you can account for those variables.
And you can account for them by going 'worst case'.
To get a false result you mean.
To get an approximation of what is required, yes.
So what is the result then.
I'll re-word that *again* for you:
'Then you can calculate the result'.
If you just want a number then it might as well be 42.
The result isn't something you pluck out of the air, it's something
you calculate using the information available using and the science
and maths that accompany such.
Did you use the worst case for radiators, what would you say is the worst case output for a 2KW radiator ?
WTF are you waffling on about now?
You really don't have a clue do you.
Re what you are waffling about? No, or I wouldn't have asked.
You say John Rumm has given the tools if that is the case why canl;t you use those, can you at least explain that.
Yes, I can, very easily. No. The answer is 'no' because I don't have
the measurements and U values for the construction of your lab.
Neither do I.
But *you* can find out eh?
Not for construction I can't,
Why not? Does it look like brick on the outside? Where there has been
any damage to the inside does it look like breeze blocks or
plasterboard etc?
no more so that yuo can get the construction detaila for grenfell tower or any other building.
No more that you could get you mean? I might have a chance, given
reasonable access and under the same conditions your labs are
currently under.
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If the builders and their company DON'T have the correct constructionn details then how am I suppse to aquire them. ?
And you have asked about the fabric of the building have you (not
electrical or other stuff because they are builders not electricians).
Why do you think I should have
I don't.
or even need construction details.
To work out the answer to your question of 'How many radiators'.
It has nothing to do with me.
It seems it is since you posted about the heaters here.
I'm not going to try measuring the lab using 1 metre rulers either.
Why not, can't get past 10m without taking your socks off?
Why would I need to take my socks off. ?
FFS! To count past ten you thicko!
Are you so stupid yuo can't estimate the size of a room from a picture.
Yes and not only that, it's stupid to expect any one to do so when you
are in there yourself. Are you not willing to do anything yourself,
apart from bleat and cry?
So, neither you nor the establishment have a tape measure (or could
get one)?
And even if I did give you the measurements I doubt that'd help you as yuo;d still come up with excuses.
Oh the irony (idiot).
I've given you the dimentions so what is your next excuse ?
You have given me rough dimensions of the lab, that's it.
It takes me 21 strides do walk the lengh of lab 253 so about 15 metres from door to door,
So, it that a real measurement disguised with you normal weird BS or
just another example of dave weirdness?
So another excuse is it.
The lab is 15 metres long.
The lab is 12 metres long.
It is 3 metres high.
So what excuses do you have left ?
Doors, windows, fabric, temperature requirements ...
Now see if you can work it out then.
Nope, can't.
I knew that all along.
You knew *nothing* 'all along' and you still know the same!
Of course you'll now come up with loads more excuses.
Yes, like, are there no windows?
https://www.dropbox.com/s/35ntzcnna5...1940s.jpg?dl=0
see windows.
Oh yes, windows! Who would have thought ... rolls eyes
Now, what size are they ... or if the wall is mostly window, what
percentage of the wall is window would you say?
In the vide you can so NO windows as they were replaced with whatever builders use to tempally cover up such gaps in windows.
Well we aren't really interested in anything that is temporary,
especially if it isn't fully sealed etc.
What is the construction of the walls?
Reinforced concrete like a lot of building of the 1950s-60s.
Ok, we are getting there ... but is it solid though to the inside or
is the inside lined with something (like plasterboard on a studwork of
some sort). If you knock on the inside of a plain bit of outside wall,
does it sound hollow or solid and if you look at the edge of the outer
wall (say at a window), how thick does it look?
What is the construction of the ceiling and floor?
same I assume going by the noise created.
Ok.
How many doors and what size are they?
See all this info is avaible in the video.
Nope, that gives me no idea of the sizes of the doors, just measure
them ffs.
Only goes to show how little a clue you have about all this the fact
that you fail to supply all the required information, in spite of
having it explained to you several times.
No on eworking in the premisies needs to kn ow such things you ****ing idiot.
You do, if you want to answer your own question.
How many teachers do you think know the construction details of teh schools and classrooms they work in.
Irrelevant.
How many tube drivers know the construction details of teh platforms and tunnels they use.
Irrelevant (although the chances are they might know more about it
than you do of your environment).
You an idiot for thinking that I should know these details.
And you are an idiot for thinking I would think you would *know* any
of these details. The task is for you to *find out*.
Do you think MPs know the construction details of the houses of parliament ?
Grow up.
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They are apparently argon filled.
So, slowly we are getting there ...
why they breezeblock the last one off I don't know and wasn't expecting them to.
Could they see you from anywhere via that window?
What the **** are you talking about.
OMG, you are so s l o w ... It was in case they didn't want to watch
you stumbling about in your confusion all the time! ;-)
As I am on the 2nd floor I don't think any student is tall enough unless they use a drone.
Quite possible these days (but see above).
As studetns aren't allowed on the roof of the next building I don;t see how ANYONE other than the builders and maintaince will be abnle to see in.
;-)
And the windows were all coated withb a gold film and NO ONE coulds see in.
So you throw that one in at the end do you, like it's the least
important.
What differnce does this make anyway ?
None. It was a joke.
Your next excuse will be how many frames and what wood and thic=kness are the frames.
How many windows (facing the outside) yes.
what do you mean by facing outwards ?
Not going into another room or internal area of the same temperature.
If you arnelt going to lok at pictures
I did, didn't really help.
how will I describe the number of the windowws are you asking for
Right and actual question at last.
Take the height and with of the glass of one frame on one wall and
assuming they are roughly the same size, work out the total area of
glass for 'that wall'. Then subtract that from the already calculated
for that wall and that should give you the area of wall (and one u
value) and the area of glass (and u value) for that wall. You
basically do the same for each of the walls (and ceilings / floors
etc), along with the temperature differential for each (max inside
temp over minimum outside temp) and add them all up.
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would it really make a difernce if so how much because the heaters were order BEFORE the windows were even installed.
Yes, because this is all to do with the heat loss from the building,,
*nothing* to do with heaters (yet)..
So you claim to know hiow to do these calcualtions so do them.
'Do I claim', no, I have never 'claimed' as I've never actually tried
but I don't see any reason why I couldn't (certainly to the level we
are considering here) and if I couldn't I'm sure others here would
(which is all I suggested in the first place). The point is, no others
are going to bother with you, till you have bothered yourself yo come
up with a decent list of stuff.
eg, You say the lab is 15m long by 3m high and if one wall was
therefore 15 x 3 that would equal 45 sq/m and if that was solid
concrete that would have a u value of (guess, without more detail), .4
then it would be:
http://www.combustionresearch.com/U-...materials.html
So very simply for just one wall
Ht = A U (ti - to)
where
Ht = transmission heat loss (W)
A = area of exposed surface (m2)
U = overall heat transmission coefficient (W/m2K)
ti = inside air temperature (oC)
to= outside air temperature (oC)
So, Ht would be for that plain wall = 45 x .5 x (20 - 0) = 450 (W/hr)?
(Seems a bit low to me)?
https://www.engineeringtoolbox.com/h...ion-d_748.html
http://www.sensiblehouse.org/nrg_heatloss.htm
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So how can I possible predict what heating would be required without knowing the details
Exactly, you can't. But you can get a *good idea* with some details,
even if they aren't precise.
I didn't know what sort of windows what size or what type until they had been installed I didnlt even know they had been istalled until; they took down the hording that covered the gaps where the windows were.
And now you do ... and doing any calcs before that would be pretty
pointless other than for that specific situation.
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Nope, it's always been the case (unless you actually are God, rather
than just thinking you are)?
But you claim I should know these things.
Nope ... I have suggested there was no reason you couldn't find out.
Step 1. Work out how much heat the building loses.
I;ve no idea how to work that out
We have given you loafs of ideas how you do that. What you want is us
to do it for you but we can't because we have no tangible information.
and I doubt yuo do either you'll just come out with your usual BS.
Looks like I might stand a better chance than you eh.
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Full input setting but possibly not on their highest output.
I've no idea what their highest output is do you ?
Yes, less than could be possible when on setting III and could be more
than the heat output on setting III when on setting II.
Tape measure, notepad, pencil, common sense ....
Wall 1.
15 x 3 m = 45 sq meter (u value for solid concrete x mm thick)
6 x windows 2.5 x 1.5m = 22.5m (u value for triple glazed glass)
Calculate the total thermal value for that wall.
Work that against the desired internal temperature for the worst
external temperature and that will give you the energy required to
bring the inside of that wall up to your desired temperature / hour.
You then add in the values calculated for the other (3?) walls, floor
and ceiling and there you have it.
Cheers, T i m