nf0831: on [0.3] use of language in [0.5] lectures 
om0832: yes 
nf0831: er so er 
om0832: for the benefit of international students 
nf0831: sorry [0.2] 
om0832: for the benefit of international students 
nf0831: right [0.4] er so there's no need to sort of adjust your hair or or 
worry that you wore the wrong clothes today [1.2] er [0.9] okay what i want to 
do today is [1.2] to [0.5] er just go over again this idea of a a thermal 
resistance model [0.7] and why it's so useful because [0.3] i think your group 
like [0.2] er [1.3] er many [0.5] sort of groups at this stage think this is 
getting all far too complicated and there are too many equations and i don't 
know what goes where [1.2] er [1.2] i'll go back to [0.2] er an example [0.2] 
we looked at [0.9] er earlier in the course of just what a real [0.2] heat 
transfer problem might [0.5] get to [0.8] this isn't [0.2] a particularly 
complicated system [0.5] it's just [0.4] er a sort of a building wall with a 
window in it so something that you [0.3] you come across every day [0.9] but if 
you look at it in heat transfer terms [0.4] there's many different sort of 
pathways that the heat can go [0.8] so you've got a a sort of a channel through 
here through the brick through the cavity insulation brick [0.4] with conduc-, 
convection and radiation [0.4] to the surface from the inside [0.3] from the 
surface from the outside [0.5] you've got a direct brick path here along the 
window frame [0.4] you've got a conduction path through the 
actual er so-, sorry on the on the window surround [0.4] you've got a a [0.3] 
path through the window frame [0.8] there's the path through the window the 
glass [0.3] whatever is in the gap [0.4] the glass on the other side [0.4] and 
all the same down here [0.6] so trying to analyse this in heat transfer terms 
[0.4] is rather difficult [1.2] i've drawn a resistance model out really just 
to the t-, of of [0.2] er [0.4] essentially the top half of this because it's 
symmetrical [0.5] to show all the different sort of thermal resistances that 
you might put on this [0.6] so here is the convection radiation [0.2] from the 
inside surface [0.4] here is the convection and radiation to the outside 
surface [0.5] and here are the different heat transfer paths in thermal 
resistance [0.2] notation so brick insulation brick [0.6] direct brick [0.2] 
directly through the frame [0.5] glass convection and radiation these will be 
different coefficients between the two plate panes of glass [0.4] and glass [0.
2] on the other side [0.9] okay [0.2] the reason we go for that [0.3] er [0.2] 
approach [0.4] rather than trying to 
actually calculate individual [0.4] er components so the reason why we put 
everything into a thermal [0.3] resistance form [0.7] is that once you've done 
that [0.8] you've basically got a very simple equation if that was electrical 
resistances you could add them up [0.3] you wouldn't need very complicated [0.
3] er methods to do it it isn't like [0.4] er some of the er matrix circuits 
that you've looked at that's just a straightforward electrical resistance 
circuit [0.6] so you can add that up [0.7] you can write it as a total thermal 
resistance [3.3] like that [1.7] [cough] [1.2] with the [0.2] inside here [2.3] 
and the outside here [2.9] so this is the total thermal resistance [4.0] and 
that's the [0.2] driving force like voltage in an electrical circuit [0.6] so 
the reet h-, rate of head transfer Q-dot [1.1] is just the temperature 
difference T-in-minus-T-out [2.9] over R-total [4.6] and when you write out [0.
2] a heat transfer problem in that sort of form [1.3] what it enables you to do 
is to see which [0.4] er components in that [0.4] are going to be important [0.
3] for the overall [0.2] heat transfer [0.8] so you may have 
to use you almost certainly will have to use [0.2] approximations [0.3] to find 
[0.2] these thermal resistances [0.2] the conduction ones are quite 
straightforward convection and radiation as we've looked at are more 
complicated [0.7] but once you've got them and you've got some numbers on the 
diagram [0.4] you can decide which are the sort of dominant [0.2] er issues [0.
7] and [0.2] if [0.2] something isn't going to be important in the overall heat 
transfer [0.4] for instance in this problem if you look at the thermal 
resistance of the glass [0.4] it's very very small so in terms of changing the 
heat transfer [0.3] it really wouldn't matter if you made the glass a bit 
thicker or if you changed to a different sort of [0.3] glass [0.9] on the other 
hand when you get to modern double glazed windows where [0.3] the actual glass 
part of it [0.3] has been er optimized to reduce heat loss [0.5] you may 
suddenly find that the dominant channel of heat transfer is going to be through 
the [0.2] metal frame [2.6] so the reason for doing this is to get [0.2] what 
may not be a completely correct 
starting point but at least an approximate starting point [0.4] where you can 
[0.2] approach [0.2] a real problem [3.4] the [1.2] driving mechanisms that we 
looked at the mechanisms that you've got to get into [0.4] heat transfer format 
[0.4] are conduction that we've looked at [0.3] er quite a bit earlier on 
either for flat plates or for cylinders [0.5] and both of them you can just add 
them up in series [0.4] er you can add them up in parallel as well although we 
haven't looked specifically at that [1.0] convection [0.5] where the convected 
coefficient which determines the [0.3] er [0.2] convective [0.3] resistance [0.
4] er you need [0.2] to go through the procedures we've looked at over the last 
two weeks [0.4] either free convection or forced convection [0.4] er and 
working out [0.2] how this relates to the fundamental properties [0.3] of the 
fluid [0.9] and the difficult one and the one that doesn't really fit [0.2] 
into this linear model [0.3] radiation [1.2] and the reason why we ploughed 
through last week looking at how we could write radiation into an approximate 
linear form [0.5] is so that radiation can 
fit into a resistance model [0.3] like everything else [0.3] so that instead of 
having to write radiation as a fourth power relationship [0.4] we can write it 
[0.2] at least approximately [0.3] as a linear relationship [0.4] then [0.3] 
radiation is described by a thermal resistance [0.2] and a temperature 
difference [0.3] to a first approximation [0.3] and then you can just slip it 
in [0.3] in the model like everything else [0.5] and it's exactly the same in 
an electrical circuit if you had a component [0.3] which didn't have a linear 
relationship between [0.3] current and voltage [0.4] or i mean you can in heat 
transfer like in electricity you can have er er things that are out of phase 
you can use complex numbers to represent thermal storage in elements but [0.3] 
fortunately i'm not going to do that [0.5] if you have something that doesn't 
have a linear relationship between co-, er the the current and voltage [0.3] 
then it's much more difficult to incorporate it into a sing-, a simple [0.2] 
linear model [0.6] yes 
sm0833: can i have another look at that model [1.0] 
nf0831: the [0.2] th-, [0.3] this one [0.7] 
sm0833: yeah [1.7] 
nf0831: it was meant 
as a sketch rather than a a sort of something to [0.2] take down in detail 
really just to show how you can get [0.3] all the bits together [3.0] 
sm0834: can you put that sheet back [laugh] [0.2] 
nf0831: pardon [0.2] 
sm0834: sheet back up [2.8] 
nf0831: i can't get them both on at the same time i onl-, i only wanted 
sm0835: [0.6] 
nf0831: er [0.6] that one we have looked at several times before [0.3] er i 
can't get them both on at the same time [0.4] er so we we will have a quick 
sketch a a model sketch and er i'll leave that up whilst i say the next er 
parts of what i want to do [0.9] so really what i'm saying is [0.2] you can 
make [0.4] in principle quite a complicated model [0.5] but provided everything 
is linear [0.3] it just simplifies to [0.2] total resistance [0.3] heat flux is 
total is the temp-, overall temperature difference [0.2] divided by the total 
[0.2] resistance [0.4] and then when you've est-, got you've worked through to 
that level [0.3] you then start worrying about which of these components are 
important and where you need to know things [0.2] more accurately [0.8] and 
once you get to real problems there's obviously some heat transfer [0.4] er 
going 
perpendicular to the direction of the temperature differences odd edge effects 
around corners [0.4] but at least if you've got a first stab you know [0.2] 
which bits are important to look at [0.3] and which aren't [2.2] okay today i 
want to look at two things [0.4] er the first is just to look at the second 
problem [0.2] from the [0.3] problem sheet from the tutorial last week which i 
didn't go over in the tutorial [0.6] and then to start building up this rather 
open-ended problem [0.4] of designing the er cooling system for [0.3] er the 
departmental [0.3] store [0.2] in time for the [0.2] strawberries at Wimbledon 
[1.0] so we're going er if you want to be there's have you finished [0.4] m-, 
[0.2] resistance model sketching [0.4] i'll just leave that up [0.5] for the 
moment [0.7] er so problem two in the heat transfer er sheet [0.2] number four 
[1.9] oh i'm just raise that slightly so i've got a bit of board [1.0] at the 
[0.2] bottom [4.3] reason i'd hoped this would be on video is to to show people 
how difficult it is if you don't have a separate whiteboard and a [0.2] er an 
overhead projector but er [0.2] that will have to be for the 
next one [4.4] okay so the second the the second problem [0.5] is one where in 
fact you don't have to it's a it's a very straightforward problem you don't 
have to do any [0.3] approximations [0.6] you're designing [0.2] an insulation 
system for a furnace wall i'll do this down here and then move it up in a 
minute [0.4] so this is er sheet four number two [7.8] 
you've got a furnace wall at a thousand degrees C [6.3] and you [0.3] er have 
got [0.3] er [0.7] an [0.2] environment around at forty [0.2] degrees C [2.8] a 
nasty hot corner of a factory [3.8] you've been asked to insulate [0.2] the 
furnace wall and you've got two [0.4] different types of [0.2] insulation [1.3] 
i'm not showing these [0.5] necessarily to the right relative thickness [0.9] 
you're starting with mineral wall because that will withstand high temperatures 
[0.4] and this has got a thermal conductivity of seventy [1.8] milliwatts [3.1] 
per metre [2.1] per degree kelvin [1.8] and you're following it with fibreglass 
[2.9] which has got a better thermal conductivity so it's ge-, better 
insulation material [0.5] but it can't go up to such high temperatures [0.4] so 
it's to er it's got a thermal conductivity of forty [0.3] milliwatts per metre 
per degree kelvin [10.8] and [0.5] conveniently somebody has estimated for you 
the heat transfer coefficient [0.4] from the surface [0.4] so this sort of 
first estimate of you know roughly what temperature it is so what heat transfer 
coefficient 
is it going to be [0.4] has been done for you [0.6] so H [0.2] for the surface 
[0.5] has been estimated [0.3] at about fifteen watts per square metre [0.2] 
per degree kelvin [4.6] and that includes both convection and radiation it's 
the overall heat loss from the surface [14.3] and you've got two [0.4] er [0.4] 
criteria two things to satisfy [0.3] on this [0.6] the first thing that you 
must satisfy is that the outer surface [0.3] mustn't be more than [0.2] fifty-
five degrees celsius [0.3] can i take the er [0.2] overhead off now [0.3] 
sm0836: 
nf0831: 'cause it means i don't have to go around on my knees [8.4] so the 
conditions that this system has got to filfil fulfil [0.5] is that the 
temperature at the interface between [0.2] the two media [0.4] has got to be no 
more than four-hundred [0.2] degrees celsius [8.6] and the temperature at the 
outside surface [0.3] has to be not more than fifty-five [0.2] degrees celsius 
[8.8] so the surface temperature limitation is so that people don't hurt their 
hands when they touch it [0.4] or bump into it by mistake [0.4] rather high for 
a surface environment that people are going to be close but it's 
obviously a very hot [0.2] environment that it's in [0.6] this interface 
temperature [0.4] the reason there is so that the outer insulation material 
doesn't work at a temperature [0.3] that it doesn't like to be at [0.8] and so 
what you're asked to do is to calculate the thicknesses [0.4] of the [0.2] two 
insulation materials or the minimum thicknesses [0.2] to achieve these [0.2] 
obviously the more insulation you put on [0.3] the lower these two temperatures 
will go the more [0.2] these surfaces will approach [0.3] the temperature [0.3] 
of the surroundings [11.0] okay [0.2] how do you think you can go about [0.4] 
doing that [1.5] any suggestions [0.6] this is an occasion where you don't have 
to make any approximations any assumptions [0.4] you've got all the information 
there to do an exact calculation [0.5] if you take this [0.3] er assumed value 
for the surface [0.2] heat transfer coefficient [4.9] yes 
sf0837: [0.3] 
nf0831: have i indeed [0.8] thank you [1.3] that's a very good starting point 
is to actually get your data correct [6.6] in fact fifteen is a is a more sort 
of realistic value that you're likely to get in a [0.4] er 
with convection and radiation combined thank you [2.2] okay what do you need to 
know [1.2] in order to calculate [0.3] one or both of these [2.8] so the 
thicknesses here i can call it X-A [8.7] and X-B [6.1] you could [0.4] for 
instance write down the equation including one of those and then decide what 
you knew in the equation [0.3] and what you didn't know [3.2] 
sm0838: Q-N through the material [0.7] 
nf0831: sorry 
sm0838: 'cause [0.4] the conduction through the material [0.3] 
nf0831: yeah [0.6] 
sm0838: is going to be [0.8] the radiation and convection [0.6] 
nf0831: that's right so you know that the conduction through the wall [12.8] is 
equal to the [1.3] i'll do it with a Q-dot to say that it's a rate [0.9] a [0.
5] flow per unit time is equal to the heat flux [0.3] er Q-dot loss [3.5] from 
the surface [13.4] so you're making the usual assumption that you're in steady 
state [0.8] so that's a good starting point [7.6] having made that start [0.2] 
what can you do next [19.2] there's no [0.3] tricks there's no [0.3] er er 
there's there's no [0.2] assumptions in this you've just got to decide [0.4] 
how you can find a value for this thickness [0.6] and this thickness and once 
you've done one it'll be perfectly obvious how to do the [0.4] the other one [5.
1] 
sm0839: do you need to get a value of the [0.9] each resistance [1.2] 
nf0831: you need to get a 
value of each resistance so you could say that Q-dot-N [4.0] is the [0.2] 
temperature difference [0.6] er we've used A and B for the materials so if i 
call this sort of T-one [1.0] T-two [2.2] T-three [2.1] we can say that Q-dot 
[0.2] is equal to T-one minus T-two [1.4] over [0.8] R [0.2] the thermal 
resistance of layer A [1.9] which is equal to [0.6] T-two minus T-three [0.8] 
over the thermal resistance of layer B [1.8] how would you calculate the 
thermal resistance [0.4] of each layer [10.5] 
sm0840: i think it's something [0.5] multiplied by the [1.4] K-A [1.3] 
nf0831: er [0.4] the thermal resistance is the thickness [1.5] any advance on 
multiply [3.6] 
sm0841: divide [0.5] 
nf0831: divide [3.4] [laughter] okay the thickness is X divided by [0.3] K [0.
7] and one more [0.6] 
sm0842: A [0.4] 
nf0831: A that's right so [0.2] the thickness is X over K-A [4.6] so are you 
getting any [0.3] closer [1.1] to [0.5] er [0.3] a solution on this so you know 
that Q-dot-N [0.3] is equal to Q-dot-loss [0.6] you know that the same Q-dot-N 
is going through each [0.2] layer [0.6] so that you can write to the conduction 
equation for the first layer [0.3] and for the second layer [0.3] and in each 
case for each layer the [0.3] er the X-over-K-A is equal to the [0.5] 
resistance [2.3] and the things that you know [0.5] on this 
'cause they give it in the problem you know the temperatures [0.8] 'cause 
that's what your [0.3] criteria that you're trying to set [0.8] you know [1.2] 
the [0.2] K value [2.2] one problem with this is that nowhere in the [0.3] 
question have you got anything [0.4] to do with the area [1.0] so you certainly 
don't know [0.3] the area [2.0] and there's no information given from which you 
can calculate [0.4] the area [1.3] 
and if you think about it it shouldn't really matter [0.4] 'cause you're defi-, 
you're designing something for a building wall [0.3] and you know it's got to 
have certain temperatures [0.3] it shouldn't matter what area it is because 
you've got to make sure it's got that temperatures whether you make it big or 
small so [0.9] 
sm0843: can could you work it out [0.6] per unit area 
nf0831: you can work it out per unit area so that's that's the way that i would 
approach this [0.3] you can say that you can't actually do the calculations 
with an area and you don't you can't put a number in for the area [0.3] so you 
can work [0.2] per unit area [0.8] so we've established this we've established 
this [0.4] we've established that you need to work per unit area [8.1] 'cause 
there's no value [3.6] for A [0.3] and it shouldn't matter [0.4] what the value 
of A is you should have the same [0.3] temperatures at the interface if you 
make the ar-, the [0.2] the wall twice as big in area or half as big [0.3] in 
area [1.4] okay so you're still left with the problem though you know [0.3] 
you're going to work per unit area [0.4] you want to know the [0.3] value [0.4] 
of X [0.6] you know the temperatures [0.3] we've established you don't need the 
area you'll work per unit area [0.8] how can you find 
though the heat flux so the only thing you're really left with in one of these 
equations [0.3] is to find the heat flux [0.3] per unit area [1.3] how can you 
do that [14.4] and [0.2] it looks like you could do it either for the heat loss 
from the surface or for the conduction through the wall [0.5] because they're 
[0.6] the same [2.0] so [0.3] can you do either of those which one can you do 
[13.9] anyone going to hazard a [0.2] a try [5.1] how do you calculate [0.8] 
heat loss from the surface you you you've established that to calculate the 
conduction [0.5] you need to know [0.7] the property that you've got to 
calculate [0.5] so you can't calculate the conduction flux direc-, directly [0.
7] how do you calculate the heat loss [0.3] from the surface [11.8] 
sm0844: [0.2] [0.8] 
nf0831: yes 
sm0844: 
nf0831: now you've got a con-, coefficient that's right so you've got a 
coefficient that [0.5] tells you about both convection and radiation the sum of 
them [1.0] so what equation can you do to write down [0.8] to give you the rate 
of heat loss from the surface 
sm0845: T-S minus T-A [0.4] [0.4] 
nf0831: that's right so you can say that Q-lot-loss [3.2] is the [0.2] 
temperature 
difference [1.2] divided by [0.2] the [0.2] surface resistance [3.2] where [0.
3] R-S [3.0] you could either you can either write it as a surface resistance 
or the other way up as a surface heat transfer coefficient [0.5] in this term 
[0.3] because you've got the surface heat transfer coefficient [0.3] it's 
easier to say that R-S is one- [0.3] over-A [0.7] multiplied by [0.4] the [0.5] 
heat transfer coefficient for the surface [0.7] so Q-dot-loss [0.2] is [0.4] A 
[0.8] multiplied by H-S [1.6] er [0.2] multiplied by [0.6] T-S- [0.6] minus-T-A 
[5.2] so finally you're getting somewhere that you can actually get some [0.3] 
er numbers out [0.3] you can't get a value for A there's nothing given to give 
you a value A [0.4] but if you calculate Q-dot-loss [0.2] over A [9.5] so the 
heat loss per unit area [2.3] is therefore H-S multiplied by T-S-minus-T-A [0.
4] sorry H-S has just gone off the top of the board as it always does when i 
need it [0.3] so that's fifteen [0.2] watts [0.8] per square metre [1.0] per 
degree kelvin [1.4] multiplied by the temperature difference fifty-five minus 
forty [5.1] which is therefore fifteen multiplied by [0.4] fifteen [0.3] which 
is 
two-hundred-and-twenty-five [2.2] 
sm0846: it's not fifty-five minus forty kelvin [1.4] 
nf0831: er 
sm0847: 
nf0831: okay it's i-, it's fifty-five it's fifty-five of of if you if you get a 
temperature interval [0.6] i should have said it fifty-five-minus-forty [0.6] 
measured in units of degrees kelvin or in degrees celsius 'cause the 
temperature interval is the same [0.3] whether you're in kelvin [0.4] or 
celsius [3.8] so [1.0] when it's an absolute temperature you're you're 
completely right you've got to be certain whether you're in kelvin or celsius 
[0.3] when it's a temperature difference because the interval is the same [0.4] 
er [0.2] the unit [0.2] can be either [1.0] so you've got fifteen watts per 
square metre per degree kelvin [0.4] multiplied by a temperature interval [0.3] 
of fifteen kelvin [0.3] or two-hundred-and-twenty-five watts [0.4] per [0.2] 
square metre [5.5] okay so you know the heat loss [0.2] per unit area [0.4] and 
if the heat [0.4] loss is equal to the heat conduction then the heat loss per 
unit area [0.3] must be equal to the heat conduction [0.4] per [0.2] unit area 
[0.8] so we can therefore say [4.8] the heat conduction [0.4] per unit area [0.
3] through either of of the media [1.9] 
is also two-hundred-and-twenty-five [0.3] watts [0.8] per square metre [5.5] so 
finally we've said that the heat conduction [0.5] is the temperature difference 
divided by [0.2] the thermal resistance [0.4] and we've said that the thermal 
resistance is X- [0.3] over- [0.2] K-A [2.1] so the heat conduction per unit 
area [2.3] is therefore for the first material [0.4] equal to the temperature 
difference [6.4] divided by [0.8] X [1.7] and multiplied by [0.3] K [0.2] for 
that [0.4] medium [1.4] so we're dividing by [0.5] X-over-K [1.1] and we've 
taken the A on to this side because we're working with everything [0.4] per 
unit area [2.8] so finally we have an equation where the only unknown [0.3] is 
a thing that you want to calculate the thickness [0.3] of this [0.2] material 
[1.6] this is for medium one the first one so i'll write this as X-A [0.4] and 
K-A [1.3] so the thickness of the [0.4] mineral wall [12.7] X-A just [0.3] 
cross-multiplying that equation [0.5] is therefore equal to the temperature [0.
2] difference [0.7] that's the temperature difference [0.2] i'll just put this 
down and see if the diagram will reappear [1.0] yeah there's a temperature 
difference across the mineral wall 
so it's a thousand minus four-hundred [7.6] again it's an interval so we can 
write it as kelvin or degrees celsius [3.7] multiplied by the thermal 
conductivity of that medium [0.4] seventy milliwatts per metre per degree 
kelvin so [0.4] point ne-, zero-seven [1.7] watts per metre [0.9] per degree 
kelvin [3.0] and divided by [0.4] the [0.2] heat flux per unit area [0.7] two-
hundred-and-twenty- [0.3] five [1.8] watts [0.2] per square metre [3.6] okay 
with a bit of luck that's going to end up with the right dimensions [0.4] er 
although it takes a long time i always put dimensions in equations 'cause then 
you can check [0.3] if you've lost a term [0.4] so the [0.4] K cancel out with 
the [0.5] K-to-the-minus-one [0.6] watts cancel out [1.2] and you're left with 
metres-to-the-minus-one divided by metres-to-the-minus-two [0.3] or therefore 
metres [0.2] which is what you would want for a [0.4] thickness [0.4] so if you 
can just [0.3] er do the calculation on that [43.2] a number [1.0] 
sm0847: point-one-eight-seven [0.3] 
nf0831: sorry [0.3] 
sm0847: point-one-eight-seven 
nf0831: point-one-eight-seven that's that's all right i [0.2] put left behind 
where i did mine i calculated point-one-n-nine my number [0.5] so that's 
point-one-eight-seven about point-one-nine [1.0] metres [0.3] always when you 
get to this stage think about whether that seems reasonable or sort of point-
one-nine metres twenty centimetres that kind of looks like a thickness [0.3] of 
a wall [0.4] if you know you're insulating a high temperature furnace and you 
get a wall thickness of a few millimetres then you suspect you've gone wrong [0.
5] er equally if you get sort of ten metres then [0.3] you think this probably 
isn't realistic so you go back again and calculate but [0.2] certainly that 
looks like [0.3] a reasonable number [1.1] so just doing the same for the other 
[0.2] er [0.2] er insulation material [0.3] the fibreglass [3.3] exactly the 
same calculation [0.2] but X-B [1.8] is therefore equal to the temperature 
difference across the fibreglass four-hundred minus fifty-five [1.3] degrees 
kelvin [0.2] or celsius [0.4] multiplied by the thermal conductivity of the 
fibreglass [0.3] point-zero-four [2.1] watts per metre per degree kelvin [8.7] 
so a better insulation material [0.3] and again divided by two-hundred-and-
twenty-five [8.5] 
which is equal to [17.0] 
sm0848: zero-six 
nf0831: part-, point [0.4] 
sm0848: zero-six 
nf0831: that sounds about right [0.7] i was going to say was it's something 
like [0.2] one-and-a-half times [0.4] point-zero-four so that's about point-
zero-six [3.1] metres so again sort of numbers that look as if you could build 
[0.3] an insulation system out of it 
sm0849: what are they [0.4] 
nf0831: sorry [0.2] 
sm0849: what are they [0.5] blocks [1.4] wall blocks 
nf0831: wall blocks [0.4] they're sort of s-, er compressed [0.2] er i mean 
they're they're solid insulation [0.5] blocks [0.4] that you use for high 
temperature [0.2] er sort of a first [0.3] li-, lining layer you you [0.3] you 
would actually have a steel layer [0.8] or you you would have [0.4] a steel 
layer or somewhere a firebrick layer on the very [0.3] inside [0.5] and then 
they're they're sort of lightweight fibre [0.4] compressed fibre blocks [0.8] 
that you can [0.6] build as a sort of an insulation wall [2.0] 
okay [0.2] so [0.2] that's [0.2] big scale high temperature conventional 
engineering [0.3] er sort of calculation [1.1] er [0.2] and the key thing is to 
break it down into the you know look a-, look at the problem look at the bits 
where you've got the information [0.4] decide [0.3] where [0.2] you need to 
sort of go first to find the information [0.3] you don't have [1.6] the next 
thing i want to do and we won't probably complete it in the we'll complete it 
in the lecture next week [0.8] is to go back to this problem we've started to 
look at in the last lecture [0.6] on designing whether you can design a [0.2] 
little portable [0.4] er chilling system [0.4] er so that the department store 
can [0.5] present flowers or strawberries [0.4] in its [0.2] er er appropriate 
place to encourage people to [0.3] buy them [0.5] so this is example five-six 
in your notes [5.2] and this is completely open-ended er i ha-, i deliberately 
hadn't [0.3] done the calculations before we do this we may end up that 
it's completely [0.2] not a good idea you can't it won't work [0.4] er and so 
[0.3] er you go back and think about something else [0.6] so what we decided [0.
5] is that this sort of chiller unit [6.2] is going to be made it's going to be 
[0.2] cylindrical so that [0.2] people can sort of reach in easily [8.9] and 
it's going to have [0.6] insulation on the outside we've in-, initially trying 
[1.5] something like the insulation you'd put on a refrigerator [0.3] so about 
sixty millimetres [14.7] we'd taken a total height [0.2] in the lecture last 
week of one metre [6.8] er a total external [0.2] diameter [0.5] of six-hundred 
millimetres [5.3] and an insulation [0.2] thickness [0.6] of [0.2] sixty [0.4] 
millimetres [9.3] 
sm0850: does that bit need to be sixty as well [0.7] that last one you've just 
drawn [0.3] 
nf0831: sorry [0.4] 
sm0850: that last one you've just drawn [0.6] 
nf0831: this last line i've not drawn doesn't matter this last line i have 
drawn doesn't need to be sixty [0.5] because this thickness here there's ice in 
here [2.6] and there's the fresh produce [0.2] sitting on top [1.3] think i've 
got a a fresh produce colour that i can [0.3] draw in [9.9] don't think i can 
draw anything like a rose so we'll have a sort of a [0.9] er [0.2] er [0.2] 
nursery flower [0.7] and the the condition here [0.4] 
is that you want you don't want it to be so cold here that the bottoms of the 
flowers are going to freeze [0.4] so we've said that the condition here [0.8] 
is [0.3] that [0.2] the [0.6] temperature [0.6] at this point [0.9] should be 
[0.5] four [0.2] degrees [0.8] 
sm0851: so we've got one work that out [0.3] 
nf0831: so [0.8] 
sm0852: that's too high [0.4] 
nf0831: that's [1.6] the first [0.2] calculation to do is to find the thickness 
[0.3] of [0.2] that bit [1.3] the ice is in here [0.8] and again first 
approximation we'll assume that it stays exactly at zero degrees as it melts [0.
5] which if you look at real melting ice is [0.2] almost true [0.3] so we've 
got ice in here that we hope keeps this system cold [0.3] throughout the 
working day [2.0] we've got sixty millimetres of insulation round all the other 
[0.2] surfaces [0.4] of the ice [6.4] and [0.3] we'd established that to [0.5] 
put the produce in [0.4] we need probably a height of about three-hundred 
millimetres [0.5] here [9.2] okay like all real problems you look at that and 
you say that's far too difficult to do i can't do that [0.4] er and er [0.2] so 
[0.3] y-, you have to look at it and decide what you think are going to be the 
[0.5] key [0.3] er [0.2] factors that determine the rate of 
heat transfer [0.3] so that you can do your first calculation [0.5] 'cause your 
first calculation says that you're going to need ten tons of ice [0.3] to 
achieve this [0.3] then you know it's not going to work [0.2] whatever you do 
[0.3] in your calculations [0.2] and so you go away and think about something 
else [0.3] if your first calculation says yes this looks quite a good idea [0.
4] then you think about doing the calculation better [0.4] to understand the 
physical system [0.3] better [1.3] so [0.5] can you suggest any bit of this 
that we can ignore 'cause in principle we've got heat transfer here we've got 
some complicated heat transfer [0.4] through these sides [0.3] got heat 
transfer from the ice across the [0.5] probably thin layer of insulation here 
[0.3] heat transfer from the ice [0.5] across to the air round here [0.7] and 
also [0.2] this is probably sitting on the [1.1] nice plush carpet in the [0.5] 
soft [0.2] furnishings part of the [0.2] store [0.3] so we've got some heat 
transfer by conduction [0.4] down [0.2] to the [0.3] ground or sorry conduction 
actually [0.2] in positive heat transfer 
the heat is all coming in to the ice which is gradually [0.2] melting 
sm0852: can you not ignore that if it's on wheels [0.2] 
nf0831: sorry [0.2] 
sm0852: can you not ignore that if it's on wheels [0.5] 
nf0831: that's a good point so if we if we put it on wheels [0.4] in fact it 
makes sligh-, a calculation slightly more [0.2] more you're you're er you're 
putting in an extra calculation so we'll put it on wheels [2.9] [0.6] 
sm0853: [0.4] 
nf0831: ah but do you need the sort of wheels that you're going to be able to 
sort of sit it down 
sm0853: no 
nf0831: so that people don't your customers don't start pushing it around so [0.
3] [laughter] perhaps you need some sort of retractable castors that [0.7] will 
sit in here so that your customers don't wheel the strawberries out of the 
store with their [0.3] [laughter] trollies [1.4] so we'll we'll put it back on 
the carpet i think [1.2] that makes it easier 'cause if you've got a thick 
layer of insulation [0.4] and a thick layer of carpet [0.3] and some flooring 
here [0.4] my first sort of approximation was going to say let's forget about 
conduction to the floor [0.3] to begin with [0.3] because [0.3] probably [0.4] 
the [0.2] transfer [0.3] from the surface is going to be much more rapid than 
the [0.2] transfer 
through sort of whatever is down here [0.3] and that could be the next stage 
that you go back and say well was that important [0.3] or [0.3] not [0.5] so my 
sort of first approach on this would a-, actually be to [0.3] er quickly get 
the trolley off its wheels [0.2] sitting on a a nice thick carpet and say [0.3] 
er er sort of our first approximations [4.9] is to ignore the base [5.6] 
anything else you think is going to be too difficult to worry about to begin 
with and probably isn't going to be [0.3] er very important [5.0] 
sm0854: the sides above our ice [0.4] 
nf0831: yes i thought those will have tho-, those started to look very er [0.3] 
difficult because there's going to be a sort of a temperature gradient up here 
and we're not quite sure [0.4] er at all what's going to be in there [0.4] so i 
think another very good approximation is to forget about these sides as a first 
approximation you might want to come back [0.3] and do it later [0.9] so if we 
now ignore the sides above [0.2] the level [0.3] of the insulation [9.3] er [0.
2] i need a word to describe this if w-, if we [0.4] call this the [0.3] a 
shelf [0.3] if you like [7.1] then in 
fact you end up with [0.3] a much simpler problem [0.5] to deal with [0.7] 
you've now got something that looks like this [2.9] this is the shelf with the 
thickness that [0.2] keeps the [0.3] produce so it doesn't get too cold [10.4] 
and this in cross-section [0.3] is a cross-section through a much shorter [0.5] 
cylinder [2.5] and this is the ice [1.8] at [0.4] zero degrees [0.6] celsius 
here [0.9] so you've made the problem [0.6] something [0.3] that you've only 
now got [0.2] two different surfaces to worry about [0.5] there's the shelf [5.
6] and then there's the cylindrical [0.5] sides the short length of them [10.1] 
so that looks like the sort of thing that you've got the [0.2] er tools to [0.
4] work with [0.6] er and in fact the shelf bit is very much like the problem 
[0.3] that we've looked just looked at [0.5] so just getting down the basic [0.
2] information that you've got on this shelf [0.5] you know it's six-hundred 
millimetres [0.3] in external diameter and a wall thickness of sixty [0.6] so 
that makes it [0.2] er four-hundred-and- [0.2] eighty thank you [0.6] 
millimetres [0.7] diameter here [5.0] we don't know the thickness [0.4] that's 
the [0.6] next thing to [1.0] 
calculate so that's the first bit of the calculation [0.2] that we need to do 
next week [2.4] because we don't know the thickness there we don't know [0.2] 
the length down here [0.7] although you would [0.6] expect that that thickness 
is going to be quite small so we're going to be looking probably at something 
like seven-hundred [0.4] er sorry [0.4] seven-hundred millimetres less the [0.
4] because of the insulation there it's not going to be [0.3] very large so 
we've got a sort of feel for [0.3] what that height will be [1.2] and so what 
we'll do in the next lecture [0.8] is to think about how we can describe 
convection and radiation from the surface 
here [0.6] er [0.3] initially thinking about this just as being a sort of a 
bare plate which is the worst case [0.6] er [0.2] and then look at the [0.2] 
convection and radiation [0.3] and conduction through [0.2] the [0.2] sides [0.
5] i'll then finish off next week with the final calculation [0.3] on [0.2] the 
er er a sort of a a complicated system of a double glazed window [1.1] okay [0.
7] Friday is test day [0.4] please don't forget that [0.8] er [0.2] so come 
with writing implements [0.3] er pen not red pen please [0.3] er [0.2] diagrams 
in pencil but writing in pen [0.5] and also your university approved calculator 
[0.6] okay thank you