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