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<title>Renewable energy</title></titleStmt>

<publicationStmt><distributor>BASE and Oxford Text Archive</distributor>


<availability><p>The British Academic Spoken English (BASE) corpus was developed at the

Universities of Warwick and Reading, under the directorship of Hilary Nesi

(Centre for English Language Teacher Education, Warwick) and Paul Thompson

(Department of Applied Linguistics, Reading), with funding from BALEAP,

EURALEX, the British Academy and the Arts and Humanities Research Board. The

original recordings are held at the Universities of Warwick and Reading, and

at the Oxford Text Archive and may be consulted by bona fide researchers

upon written application to any of the holding bodies.

The BASE corpus is freely available to researchers who agree to the

following conditions:</p>

<p>1. The recordings and transcriptions should not be modified in any


<p>2. The recordings and transcriptions should be used for research purposes

only; they should not be reproduced in teaching materials</p>

<p>3. The recordings and transcriptions should not be reproduced in full for

a wider audience/readership, although researchers are free to quote short

passages of text (up to 200 running words from any given speech event)</p>

<p>4. The corpus developers should be informed of all presentations or

publications arising from analysis of the corpus</p><p>

Researchers should acknowledge their use of the corpus using the following

form of words:

The recordings and transcriptions used in this study come from the British

Academic Spoken English (BASE) corpus, which was developed at the

Universities of Warwick and Reading under the directorship of Hilary Nesi

(Warwick) and Paul Thompson (Reading). Corpus development was assisted by

funding from the Universities of Warwick and Reading, BALEAP, EURALEX, the

British Academy and the Arts and Humanities Research Board. </p></availability>




<recording dur="00:45:17" n="5546">


<respStmt><name>BASE team</name>



<langUsage><language id="en">English</language>



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<personGrp role="speakers" size="26"><p>number of speakers: 26</p></personGrp>





<item n="speechevent">Lecture</item>

<item n="acaddept">Engineering</item>

<item n="acaddiv">ps</item>

<item n="partlevel">UG2</item>

<item n="module">Heat transfer</item>




<u who="nf0831"> on <pause dur="0.3"/> use of language in <pause dur="0.5"/> lectures </u><u who="om0832" trans="latching"> yes </u><u who="nf0831" trans="latching"> # so # </u><u who="om0832" trans="overlap"> for the benefit of international students </u><u who="nf0831" trans="latching"> sorry </u><pause dur="0.2"/> <u who="om0832" trans="pause"> for the benefit of international students </u><u who="nf0831" trans="overlap"> right <pause dur="0.4"/> # so there's no need to sort of adjust your hair or or worry that you wore the wrong clothes today <pause dur="1.2"/> # <pause dur="0.9"/> okay what i want to do today is <pause dur="1.2"/> to <pause dur="0.5"/> # just go over again this idea of a a thermal resistance model <pause dur="0.7"/> and why it's so useful because <pause dur="0.3"/> i think your group like <pause dur="0.2"/> # <pause dur="1.3"/> # many <pause dur="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 <pause dur="1.2"/><kinesic desc="puts on transparency" iterated="n"/> # <pause dur="1.2"/> i'll go back to <pause dur="0.2"/> # an example <pause dur="0.2"/> we looked at <pause dur="0.9"/> # earlier in the course of just what a real <pause dur="0.2"/> heat transfer problem might <pause dur="0.5"/> get to <pause dur="0.8"/> this isn't <pause dur="0.2"/> a particularly complicated system <pause dur="0.5"/> it's just <pause dur="0.4"/> # a sort of a building wall with a window in it so something that you <pause dur="0.3"/> you come across every day <pause dur="0.9"/> but if you look at it in heat transfer terms <pause dur="0.4"/> there's many different sort of pathways that the heat can go <pause dur="0.8"/> so you've got a a sort of a channel through here through the brick through the cavity insulation brick <pause dur="0.4"/> with <trunc>conduc</trunc> convection and radiation <pause dur="0.4"/> to the surface from the inside <pause dur="0.3"/> from the surface from the outside <pause dur="0.5"/> you've got a direct brick path here along the window frame <pause dur="0.4"/> you've got a conduction path through the

actual # <trunc>so</trunc> sorry on the on the window surround <pause dur="0.4"/> you've got a a <pause dur="0.3"/> path through the window frame <pause dur="0.8"/> there's the path through the window the glass <pause dur="0.3"/> whatever is in the gap <pause dur="0.4"/> the glass on the other side <pause dur="0.4"/> and all the same down here <pause dur="0.6"/> so trying to analyse this in heat transfer terms <pause dur="0.4"/> is rather difficult <pause dur="1.2"/> i've drawn a resistance model out really just to the <trunc>t</trunc> of of <pause dur="0.2"/> # <pause dur="0.4"/> essentially the top <kinesic desc="changes transparency" iterated="y" dur="4"/> half of this because it's symmetrical <pause dur="0.5"/> to show all the different sort of thermal resistances that you might put on this <pause dur="0.6"/> so here is the convection radiation <pause dur="0.2"/> from the inside surface <pause dur="0.4"/> here is the convection and radiation to the outside surface <pause dur="0.5"/> and here are the different heat transfer paths in thermal resistance <pause dur="0.2"/> notation so brick insulation brick <pause dur="0.6"/> direct brick <pause dur="0.2"/> directly through the frame <pause dur="0.5"/> glass convection and radiation these will be different coefficients between the two plate panes of glass <pause dur="0.4"/> and glass <pause dur="0.2"/> on the other side <pause dur="0.9"/> okay <pause dur="0.2"/> the reason we go for that <pause dur="0.3"/> # <pause dur="0.2"/> approach <pause dur="0.4"/> rather than trying to

actually calculate individual <pause dur="0.4"/> # components so the reason why we put everything into a thermal <pause dur="0.3"/> resistance form <pause dur="0.7"/> is that once you've done that <pause dur="0.8"/> you've basically got a very simple equation if that was electrical resistances you could add them up <pause dur="0.3"/> you wouldn't need very complicated <pause dur="0.3"/> # methods to do it it isn't like <pause dur="0.4"/> # some of the # matrix circuits that you've looked at that's just a straightforward electrical resistance circuit <pause dur="0.6"/> so you can add that up <pause dur="0.7"/> you can write it as a total thermal resistance <pause dur="3.3"/> like that <pause dur="1.7"/><vocal desc="cough" iterated="n"/><pause dur="1.2"/> with the <pause dur="0.2"/> inside here <pause dur="2.3"/> and the outside here <pause dur="2.9"/> so this is the total thermal resistance <pause dur="4.0"/> and that's the <pause dur="0.2"/> driving force like voltage in an electrical circuit <pause dur="0.6"/> so the reet <trunc>h</trunc> rate of head transfer Q-dot <pause dur="1.1"/> is just the temperature difference T-in-minus-T-out <pause dur="2.9"/> over R-total <pause dur="4.6"/> and when you write out <pause dur="0.2"/> a heat transfer problem in that sort of form <pause dur="1.3"/> what it enables you to do is to see which <pause dur="0.4"/> # components in that <pause dur="0.4"/> are going to be important <pause dur="0.3"/> for the overall <pause dur="0.2"/> heat transfer <pause dur="0.8"/> so you may have

to use you almost certainly will have to use <pause dur="0.2"/> approximations <pause dur="0.3"/> to find <pause dur="0.2"/> these thermal resistances <pause dur="0.2"/> the conduction ones are quite straightforward convection and radiation as we've looked at are more complicated <pause dur="0.7"/> but once you've got them and you've got some numbers on the diagram <pause dur="0.4"/> you can decide which are the sort of dominant <pause dur="0.2"/> # issues <pause dur="0.7"/> and <pause dur="0.2"/> if <pause dur="0.2"/> something isn't going to be important in the overall heat transfer <pause dur="0.4"/> for instance in this problem if you look at the thermal resistance of the glass <pause dur="0.4"/> it's very very small so in terms of changing the heat transfer <pause dur="0.3"/> it really wouldn't matter if you made the glass a bit thicker or if you changed to a different sort of <pause dur="0.3"/> glass <pause dur="0.9"/> on the other hand when you get to modern double glazed windows where <pause dur="0.3"/> the actual glass part of it <pause dur="0.3"/> has been # optimized to reduce heat loss <pause dur="0.5"/> you may suddenly find that the dominant channel of heat transfer is going to be through the <pause dur="0.2"/> metal frame <pause dur="2.6"/> so the reason for doing this is to get <pause dur="0.2"/> what may not be a completely correct

starting point but at least an approximate starting point <pause dur="0.4"/> where you can <pause dur="0.2"/> approach <pause dur="0.2"/> a real problem <pause dur="3.4"/> the <pause dur="1.2"/><kinesic desc="changes transparency" iterated="y" dur="15"/> driving mechanisms that we looked at the mechanisms that you've got to get into <pause dur="0.4"/> heat transfer format <pause dur="0.4"/> are conduction that we've looked at <pause dur="0.3"/> # quite a bit earlier on either for flat plates or for cylinders <pause dur="0.5"/> and both of them you can just add them up in series <pause dur="0.4"/> # you can add them up in parallel as well although we haven't looked specifically at that <pause dur="1.0"/> convection <pause dur="0.5"/> where the convected coefficient which determines the <pause dur="0.3"/> # <pause dur="0.2"/> convective <pause dur="0.3"/> resistance <pause dur="0.4"/> # you need <pause dur="0.2"/> to go through the procedures we've looked at over the last two weeks <pause dur="0.4"/> either free convection or forced convection <pause dur="0.4"/> # and working out <pause dur="0.2"/> how this relates to the fundamental properties <pause dur="0.3"/> of the fluid <pause dur="0.9"/> and the difficult one and the one that doesn't really fit <pause dur="0.2"/> into this linear model <pause dur="0.3"/> radiation <pause dur="1.2"/> and the reason why we ploughed through last week looking at how we could write radiation into an approximate linear form <pause dur="0.5"/> is so that radiation can

fit into a resistance model <pause dur="0.3"/> like everything else <pause dur="0.3"/> so that instead of having to write radiation as a fourth power relationship <pause dur="0.4"/> we can write it <pause dur="0.2"/> at least approximately <pause dur="0.3"/> as a linear relationship <pause dur="0.4"/> then <pause dur="0.3"/> radiation is described by a thermal resistance <pause dur="0.2"/> and a temperature difference <pause dur="0.3"/> to a first approximation <pause dur="0.3"/> and then you can just slip it in <pause dur="0.3"/> in the model like everything else <pause dur="0.5"/> and it's exactly the same in an electrical circuit if you had a component <pause dur="0.3"/> which didn't have a linear relationship between <pause dur="0.3"/> current and voltage <pause dur="0.4"/> or i mean you can in heat transfer like in electricity you can have # # things that are out of phase you can use complex numbers to represent thermal storage in elements but <pause dur="0.3"/> fortunately i'm not going to do that <pause dur="0.5"/> if you have something that doesn't have a linear relationship between <trunc>co</trunc> # the the current and voltage <pause dur="0.3"/> then it's much more difficult to incorporate it into a <trunc>sing</trunc> a simple <pause dur="0.2"/> linear model <pause dur="0.6"/> yes </u><u who="sm0833" trans="overlap"> can i have another look at that model <gap reason="inaudible" extent="2 secs"/> </u><pause dur="1.0"/> <u who="nf0831" trans="pause"> the <pause dur="0.2"/> <trunc>th</trunc> <pause dur="0.3"/> this one <kinesic desc="changes transparency" iterated="y" dur="3"/></u><pause dur="0.7"/> <u who="sm0833" trans="pause"> yeah </u><pause dur="1.7"/> <u who="nf0831" trans="pause"> it was meant

as a sketch rather than a a sort of something to <pause dur="0.2"/> take down in detail really just to show how you can get <pause dur="0.3"/> all the bits together </u><pause dur="3.0"/> <u who="sm0834" trans="pause"> can you put that sheet back <gap reason="inaudible" extent="1 sec"/><vocal desc="laugh" iterated="n"/></u><pause dur="0.2"/> <u who="nf0831" trans="pause"> pardon </u><pause dur="0.2"/> <u who="sm0834" trans="pause"> <gap reason="inaudible" extent="1 sec"/> sheet back up <gap reason="inaudible" extent="1 sec"/></u><pause dur="2.8"/> <u who="nf0831" trans="pause"> i can't get them both on at the same time i <trunc>onl</trunc> i only wanted </u><u who="sm0835" trans="overlap"> <gap reason="inaudible" extent="2 secs"/></u><pause dur="0.6"/> <u who="nf0831" trans="pause"> # <pause dur="0.6"/> that one we have looked at several times before <pause dur="0.3"/> # i can't get them both on at the same time <pause dur="0.4"/> # so we we will have a quick sketch a a model sketch and # i'll leave that up whilst i say the next # parts of what i want to do <pause dur="0.9"/> so really what i'm saying is <pause dur="0.2"/> you can make <pause dur="0.4"/> in principle quite a complicated model <pause dur="0.5"/> but provided everything is linear <pause dur="0.3"/> it just simplifies to <pause dur="0.2"/> total resistance <pause dur="0.3"/> heat flux is total is the <trunc>temp</trunc> overall temperature difference <pause dur="0.2"/> divided by the total <pause dur="0.2"/> resistance <pause dur="0.4"/> and then when you've <trunc>est</trunc> got you've worked through to that level <pause dur="0.3"/> you then start worrying about which of these components are important and where you need to know things <pause dur="0.2"/> more accurately <pause dur="0.8"/> and once you get to real problems there's obviously some heat transfer <pause dur="0.4"/> # going

perpendicular to the direction of the temperature differences odd edge effects around corners <pause dur="0.4"/> but at least if you've got a first stab you know <pause dur="0.2"/> which bits are important to look at <pause dur="0.3"/> and which aren't <pause dur="2.2"/> okay today i want to look at two things <pause dur="0.4"/> # the first is just to look at the second problem <pause dur="0.2"/> from the <pause dur="0.3"/> problem sheet from the tutorial last week which i didn't go over in the tutorial <pause dur="0.6"/> and then to start building up this rather open-ended problem <pause dur="0.4"/> of designing the # cooling system for <pause dur="0.3"/> # the departmental <pause dur="0.3"/> store <pause dur="0.2"/> in time for the <pause dur="0.2"/> strawberries at Wimbledon <pause dur="1.0"/> so we're going # if you want to be there's have you finished <pause dur="0.4"/> <trunc>m</trunc> <pause dur="0.2"/> resistance model sketching <pause dur="0.4"/> i'll just leave that up <pause dur="0.5"/> for the moment <pause dur="0.7"/> # so problem two in the heat transfer # sheet <pause dur="0.2"/> number four <pause dur="1.9"/> oh i'm just raise that slightly <event desc="adjusts equipment" iterated="y" dur="10"/> so i've got a bit of board <pause dur="1.0"/> at the <pause dur="0.2"/> bottom <pause dur="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 <pause dur="0.2"/> # an overhead projector but # <pause dur="0.2"/> that will have to be for the

next one <pause dur="4.4"/> okay so the second the the second problem <pause dur="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 <pause dur="0.3"/> approximations <pause dur="0.6"/> you're designing <pause dur="0.2"/> an insulation system for a furnace wall i'll do this down here and then move it up in a minute <pause dur="0.4"/> so this is # sheet four number two <pause dur="7.8"/><kinesic desc="writes on board" iterated="y" dur="6"/>

you've got a furnace wall at a thousand degrees C <pause dur="6.3"/><kinesic desc="writes on board" iterated="y" dur="3"/> and you <pause dur="0.3"/> # have got <pause dur="0.3"/> # <pause dur="0.7"/> an <pause dur="0.2"/> environment around at forty <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="2"/> degrees C <pause dur="2.8"/> a nasty hot corner of a factory <pause dur="3.8"/> you've been asked to insulate <pause dur="0.2"/> the furnace wall and you've got two <pause dur="0.4"/> different types of <pause dur="0.2"/> insulation <pause dur="1.3"/> i'm not showing these <pause dur="0.5"/> necessarily to the right relative thickness <pause dur="0.9"/> you're starting with mineral wall because that will withstand high temperatures <pause dur="0.4"/><kinesic desc="writes on board" iterated="y" dur="19"/> and this has got a thermal conductivity of seventy <pause dur="1.8"/> milliwatts <pause dur="3.1"/> per metre <pause dur="2.1"/> per degree kelvin <pause dur="1.8"/> and you're following it with fibreglass <pause dur="2.9"/> which has got a better thermal conductivity so it's <trunc>ge</trunc> better insulation material <pause dur="0.5"/> but it can't go up to such high temperatures <pause dur="0.4"/> so it's to # it's got a thermal conductivity of forty <pause dur="0.3"/> milliwatts per metre <kinesic desc="writes on board" iterated="y" dur="7"/> per degree kelvin <pause dur="10.8"/> and <pause dur="0.5"/> conveniently somebody has estimated for you the heat transfer coefficient <pause dur="0.4"/> from the surface <pause dur="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 <pause dur="0.4"/> has been done for you <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="3"/> so H <pause dur="0.2"/> for the surface <pause dur="0.5"/> has been estimated <pause dur="0.3"/> at about fifteen watts per square metre <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="5"/> per degree kelvin <pause dur="4.6"/> and that includes both convection and radiation it's the overall heat loss from the surface <pause dur="14.3"/><kinesic desc="writes on board" iterated="y" dur="9"/> and you've got two <pause dur="0.4"/> # <pause dur="0.4"/> criteria two things to satisfy <pause dur="0.3"/> on this <pause dur="0.6"/> the first thing that you must satisfy is that the outer surface <pause dur="0.3"/> mustn't be more than <pause dur="0.2"/> fifty-five degrees celsius <pause dur="0.3"/> can i take the # <pause dur="0.2"/> overhead off now </u><pause dur="0.3"/> <u who="sm0836" trans="pause"> <gap reason="inaudible" extent="1 sec"/></u><u who="nf0831" trans="latching"> 'cause it means i don't have to go around on my knees <pause dur="8.4"/><event desc="puts away screen" iterated="n"/><event desc="turns off overhead projector" iterated="n"/> so the conditions that this system has got to filfil fulfil <pause dur="0.5"/> is that the temperature at the interface between <pause dur="0.2"/> the two media <pause dur="0.4"/> has got to be no more than four-hundred <pause dur="0.2"/> degrees celsius <pause dur="8.6"/><kinesic desc="writes on board" iterated="y" dur="8"/> and the temperature at the outside surface <pause dur="0.3"/> has to be not more than fifty-five <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="10"/> degrees celsius <pause dur="8.8"/> so the surface temperature limitation is so that people don't hurt their hands when they touch it <pause dur="0.4"/> or bump into it by mistake <pause dur="0.4"/> rather high for a surface environment that people are going to be close but it's

obviously a very hot <pause dur="0.2"/> environment that it's in <pause dur="0.6"/> this interface temperature <pause dur="0.4"/> the reason there is so that the outer insulation material doesn't work at a temperature <pause dur="0.3"/> that it doesn't like to be at <pause dur="0.8"/> and so what you're asked to do is to calculate the thicknesses <pause dur="0.4"/> of the <pause dur="0.2"/> two insulation materials or the minimum thicknesses <pause dur="0.2"/> to achieve these <pause dur="0.2"/> obviously the more insulation you put on <pause dur="0.3"/> the lower these two temperatures will go the more <pause dur="0.2"/> these surfaces will approach <pause dur="0.3"/> the temperature <pause dur="0.3"/> of the surroundings <pause dur="11.0"/><kinesic desc="writes on board" iterated="y" dur="10"/> okay <pause dur="0.2"/> how do you think you can go about <pause dur="0.4"/> doing that <pause dur="1.5"/> any suggestions <pause dur="0.6"/> this is an occasion where you don't have to make any approximations any assumptions <pause dur="0.4"/> you've got all the information there to do an exact calculation <pause dur="0.5"/> if you take this <pause dur="0.3"/> # assumed value for the surface <pause dur="0.2"/> heat transfer coefficient <pause dur="4.9"/> yes </u><u who="sf0837" trans="overlap"> <gap reason="inaudible" extent="4 secs"/> </u><pause dur="0.3"/> <u who="nf0831" trans="pause"> have i indeed <pause dur="0.8"/> thank you <pause dur="1.3"/> that's a very good starting point is to actually get your data correct <pause dur="6.6"/><kinesic desc="writes on board" iterated="y" dur="2"/> in fact fifteen is a is a more sort of realistic value that you're likely to get in a <pause dur="0.4"/> #

with convection and radiation combined thank you <pause dur="2.2"/> okay what do you need to know <pause dur="1.2"/> in order to calculate <pause dur="0.3"/> one or both of these <pause dur="2.8"/> so the thicknesses here i can call it X-A <pause dur="8.7"/><kinesic desc="writes on board" iterated="y" dur="9"/> and X-B <pause dur="6.1"/> you could <pause dur="0.4"/> for instance write down the equation including one of those and then decide what you knew in the equation <pause dur="0.3"/> and what you didn't know </u><pause dur="3.2"/> <u who="sm0838" trans="pause"> Q-N through the material </u><pause dur="0.7"/> <u who="nf0831" trans="pause"> sorry </u><u who="sm0838" trans="overlap"> 'cause <pause dur="0.4"/> the conduction through the material </u><pause dur="0.3"/> <u who="nf0831" trans="pause"> yeah </u><pause dur="0.6"/> <u who="sm0838" trans="pause"> is going to be <pause dur="0.8"/> the radiation and convection </u><pause dur="0.6"/> <u who="nf0831" trans="pause"> that's right so you know that the conduction through the wall <pause dur="12.8"/><kinesic desc="writes on board" iterated="y" dur="14"/> is equal to the <pause dur="1.3"/> i'll do it with a Q-dot to say that it's a rate <pause dur="0.9"/><kinesic desc="writes on board" iterated="y" dur="1"/> a <pause dur="0.5"/> flow per unit time is equal to the heat flux <pause dur="0.3"/> # Q-dot <kinesic desc="writes on board" iterated="y" dur="18"/> loss <pause dur="3.5"/> from the surface <pause dur="13.4"/> so you're making the usual assumption that you're in steady state <pause dur="0.8"/> so that's a good starting point <pause dur="7.6"/> having made that start <pause dur="0.2"/> what can you do next <pause dur="19.2"/> there's no <pause dur="0.3"/> tricks there's no <pause dur="0.3"/> # # there's there's no <pause dur="0.2"/> assumptions in this you've just got to decide <pause dur="0.4"/> how you can find a value for this thickness <pause dur="0.6"/> and this thickness and once you've done one it'll be perfectly obvious how to do the <pause dur="0.4"/> the other one </u><pause dur="5.1"/> <u who="sm0839" trans="pause"> do you need to get a value of the <pause dur="0.9"/> each resistance </u><pause dur="1.2"/> <u who="nf0831" trans="pause"> you need to get a

value of each resistance so you could say that Q-dot-N <pause dur="4.0"/><kinesic desc="writes on board" iterated="y" dur="5"/> is the <pause dur="0.2"/> temperature difference <pause dur="0.6"/> # we've used A and B for the materials so if i call this sort of <kinesic desc="writes on board" iterated="y" dur="6"/> T-one <pause dur="1.0"/> T-two <pause dur="2.2"/> T-three <pause dur="2.1"/> we can say that Q-dot <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="16"/> is equal to T-one minus T-two <pause dur="1.4"/> over <pause dur="0.8"/> R <pause dur="0.2"/> the thermal resistance of layer A <pause dur="1.9"/> which is equal to <pause dur="0.6"/> T-two minus T-three <pause dur="0.8"/> over the thermal resistance of layer B <pause dur="1.8"/> how would you calculate the thermal resistance <pause dur="0.4"/> of each layer </u><pause dur="10.5"/> <u who="sm0840" trans="pause"> i think it's something <pause dur="0.5"/> multiplied by the <pause dur="1.4"/> K-A </u><pause dur="1.3"/> <u who="nf0831" trans="pause"> # <pause dur="0.4"/> the thermal <kinesic desc="writes on board" iterated="y" dur="3"/> resistance is the thickness <pause dur="1.5"/> any advance on multiply </u><pause dur="3.6"/> <u who="sm0841" trans="pause"> divide </u><pause dur="0.5"/> <u who="nf0831" trans="pause"> divide <pause dur="3.4"/><kinesic desc="writes on board" iterated="y" dur="7"/><vocal desc="laughter" iterated="y" dur="1"/> okay the thickness is X divided by <pause dur="0.3"/> K <pause dur="0.7"/> and one more </u><pause dur="0.6"/> <u who="sm0842" trans="pause"> A </u><pause dur="0.4"/> <u who="nf0831" trans="pause"> A that's right so <pause dur="0.2"/> the thickness is X over K-A <pause dur="4.6"/><kinesic desc="writes on board" iterated="y" dur="4"/> so are you getting any <pause dur="0.3"/> closer <pause dur="1.1"/> to <pause dur="0.5"/> # <pause dur="0.3"/> a solution on this so you know that Q-dot-N <pause dur="0.3"/> is equal to Q-dot-loss <pause dur="0.6"/> you know that the same Q-dot-N is going through each <pause dur="0.2"/> layer <pause dur="0.6"/> so that you can write to the conduction equation for the first layer <pause dur="0.3"/> and for the second layer <pause dur="0.3"/> and in each case for each layer the <pause dur="0.3"/> # the X-over-K-A is equal to the <pause dur="0.5"/> resistance <pause dur="2.3"/> and the things that you know <pause dur="0.5"/> on this

'cause they give it in the problem you know the temperatures <pause dur="0.8"/> 'cause that's what your <pause dur="0.3"/> criteria that you're trying to set <pause dur="0.8"/> you know <pause dur="1.2"/> the <pause dur="0.2"/> K value <pause dur="2.2"/> one problem with this is that nowhere in the <pause dur="0.3"/> question have you got anything <pause dur="0.4"/> to do with the area <pause dur="1.0"/> so you certainly don't know <pause dur="0.3"/> the area <pause dur="2.0"/> and there's no information given from which you can calculate <pause dur="0.4"/> the area <pause dur="1.3"/>

and if you think about it it shouldn't really matter <pause dur="0.4"/> 'cause you're <trunc>defi</trunc> you're designing something for a building wall <pause dur="0.3"/> and you know it's got to have certain temperatures <pause dur="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 </u><pause dur="0.9"/> <u who="sm0843" trans="pause"> can could you work it out <pause dur="0.6"/> per unit area </u><u who="nf0831" trans="latching"> you can work it out per unit area so that's that's the way that i would approach this <pause dur="0.3"/><kinesic desc="writes on board" iterated="y" dur="1"/> 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 <pause dur="0.3"/> so you can work <pause dur="0.2"/> per <kinesic desc="writes on board" iterated="y" dur="4"/> unit area <pause dur="0.8"/> so we've established this we've established this <pause dur="0.4"/> we've established that you need to work per unit area <pause dur="8.1"/><kinesic desc="writes on board" iterated="y" dur="14"/> 'cause there's no value <pause dur="3.6"/> for A <pause dur="0.3"/> and it shouldn't matter <pause dur="0.4"/> what the value of A is you should have the same <pause dur="0.3"/> temperatures at the interface if you make the <trunc>ar</trunc> the <pause dur="0.2"/> the wall twice as big in area or half as big <pause dur="0.3"/> in area <pause dur="1.4"/> okay so you're still left with the problem though you know <pause dur="0.3"/> you're going to work per unit area <pause dur="0.4"/> you want to know the <pause dur="0.3"/> value <pause dur="0.4"/> of X <pause dur="0.6"/> you know the temperatures <pause dur="0.3"/> we've established you don't need the area you'll work per unit area <pause dur="0.8"/> how can you find

though the heat flux so the only thing you're really left with in one of these equations <pause dur="0.3"/> is to find the heat flux <pause dur="0.3"/> per unit area <pause dur="1.3"/> how can you do that <pause dur="14.4"/> and <pause dur="0.2"/> it looks like you could do it either for the heat loss from the surface or for the conduction through the wall <pause dur="0.5"/> because they're <pause dur="0.6"/> the same <pause dur="2.0"/> so <pause dur="0.3"/> can you do either of those which one can you do <pause dur="13.9"/> anyone going to hazard a <pause dur="0.2"/> a try <pause dur="5.1"/> how do you calculate <pause dur="0.8"/> heat loss from the surface you you you've established that to calculate the conduction <pause dur="0.5"/> you need to know <pause dur="0.7"/> the property that you've got to calculate <pause dur="0.5"/> so you can't calculate the conduction flux <trunc>direc</trunc> directly <pause dur="0.7"/> how do you calculate the heat loss <pause dur="0.3"/> from the surface </u><pause dur="11.8"/> <u who="sm0844" trans="pause"> <gap reason="inaudible" extent="1 sec"/> <pause dur="0.2"/> <vocal desc="clears throat" iterated="n"/> <gap reason="inaudible" extent="2 secs"/> </u><pause dur="0.8"/> <u who="nf0831" trans="pause"> yes </u><u who="sm0844" trans="latching"> <gap reason="inaudible" extent="1 sec"/></u><u who="nf0831" trans="overlap"> now you've got a <trunc>con</trunc> coefficient that's right so you've got a coefficient that <pause dur="0.5"/> tells you about both convection and radiation the sum of them <pause dur="1.0"/> so what equation can you do to write down <pause dur="0.8"/> to give you the rate of heat loss from the surface </u><u who="sm0845" trans="latching"> T-S minus T-A <pause dur="0.4"/> <gap reason="inaudible" extent="1 sec"/></u><pause dur="0.4"/> <u who="nf0831" trans="pause"> that's right so you can say that Q-<sic corr="dot">lot</sic>-loss <pause dur="3.2"/><kinesic desc="writes on board" iterated="y" dur="15"/> is the <pause dur="0.2"/> temperature

difference <pause dur="1.2"/> divided by <pause dur="0.2"/> the <pause dur="0.2"/> surface resistance <pause dur="3.2"/> where <pause dur="0.3"/> R-S <pause dur="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 <pause dur="0.5"/> in this term <pause dur="0.3"/> because you've got the surface heat transfer coefficient <pause dur="0.3"/> it's easier to say that R-S is <kinesic desc="writes on board" iterated="y" dur="3"/> one-<pause dur="0.3"/>over-A <pause dur="0.7"/> multiplied by <pause dur="0.4"/> the <pause dur="0.5"/> heat transfer coefficient for the surface <pause dur="0.7"/> so <kinesic desc="writes on board" iterated="y" dur="10"/> Q-dot-loss <pause dur="0.2"/> is <pause dur="0.4"/> A <pause dur="0.8"/> multiplied by H-S <pause dur="1.6"/> # <pause dur="0.2"/> multiplied by <pause dur="0.6"/> T-S-<pause dur="0.6"/>minus-T-A <pause dur="5.2"/> so finally you're getting somewhere that you can actually get some <pause dur="0.3"/> # numbers out <pause dur="0.3"/> you can't get a value for A there's nothing given to give you a value A <pause dur="0.4"/> but if you calculate Q-dot-loss <pause dur="0.2"/> over A <pause dur="9.5"/><kinesic desc="writes on board" iterated="y" dur="6"/> so the heat loss per unit area <pause dur="2.3"/><kinesic desc="writes on board" iterated="y" dur="6"/> is therefore H-S multiplied by T-S-minus-T-A <pause dur="0.4"/> sorry H-S has just gone off the top of the board as it always does when i need it <pause dur="0.3"/> so that's fifteen <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="6"/> watts <pause dur="0.8"/> per square metre <pause dur="1.0"/> per degree kelvin <pause dur="1.4"/> multiplied by the temperature difference fifty-five minus forty <pause dur="5.1"/><kinesic desc="writes on board" iterated="y" dur="12"/> which is therefore fifteen multiplied by <pause dur="0.4"/> fifteen <pause dur="0.3"/> which is

two-hundred-and-twenty-five </u><pause dur="2.2"/> <u who="sm0846" trans="pause"> it's not fifty-five minus forty kelvin </u><pause dur="1.4"/> <u who="nf0831" trans="pause"> # </u><u who="sm0847" trans="overlap"> <gap reason="inaudible" extent="2 secs"/></u><u who="nf0831" trans="latching"> okay it's <trunc>i</trunc> it's fifty-five it's fifty-five of of if you if you get a temperature interval <pause dur="0.6"/> i should have said it fifty-five-minus-forty <pause dur="0.6"/> measured in units of degrees kelvin or in degrees celsius 'cause the temperature interval is the same <pause dur="0.3"/> whether you're in kelvin <pause dur="0.4"/> or celsius <pause dur="3.8"/> so <pause dur="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 <pause dur="0.3"/> when it's a temperature difference because the interval is the same <pause dur="0.4"/> # <pause dur="0.2"/> the unit <pause dur="0.2"/> can be either <pause dur="1.0"/> so you've got fifteen watts per square metre per degree kelvin <pause dur="0.4"/> multiplied by a temperature interval <pause dur="0.3"/> of fifteen kelvin <pause dur="0.3"/> or two-hundred-and-twenty-five <kinesic desc="writes on board" iterated="y" dur="5"/> watts <pause dur="0.4"/> per <pause dur="0.2"/> square metre <pause dur="5.5"/> okay so you know the heat loss <pause dur="0.2"/> per unit area <pause dur="0.4"/> and if the heat <pause dur="0.4"/> loss is equal to the heat conduction then the heat loss per unit area <pause dur="0.3"/> must be equal to the heat conduction <pause dur="0.4"/> per <pause dur="0.2"/> unit area <pause dur="0.8"/> so we can therefore say <pause dur="4.8"/><kinesic desc="writes on board" iterated="y" dur="5"/> the heat conduction <pause dur="0.4"/> per unit area <pause dur="0.3"/> through either of <kinesic desc="writes on board" iterated="y" dur="8"/> of the media <pause dur="1.9"/>

is also two-hundred-and-twenty-five <pause dur="0.3"/> watts <pause dur="0.8"/> per square metre <pause dur="5.5"/> so finally we've said that the heat conduction <pause dur="0.5"/> is the temperature difference divided by <pause dur="0.2"/> the thermal resistance <pause dur="0.4"/> and we've said that the thermal resistance is X-<pause dur="0.3"/>over-<pause dur="0.2"/>K-A <pause dur="2.1"/> so the heat conduction per unit area <pause dur="2.3"/> is therefore for the first material <pause dur="0.4"/> equal to the temperature difference <pause dur="6.4"/><kinesic desc="writes on board" iterated="y" dur="14"/> divided by <pause dur="0.8"/> X <pause dur="1.7"/> and multiplied by <pause dur="0.3"/> K <pause dur="0.2"/> for that <pause dur="0.4"/> medium <pause dur="1.4"/> so we're dividing by <pause dur="0.5"/> X-over-K <pause dur="1.1"/> and we've taken the A on to this side because we're working with everything <pause dur="0.4"/> per unit area <pause dur="2.8"/> so finally we have an equation where the only unknown <pause dur="0.3"/> is a thing that you want to calculate the thickness <pause dur="0.3"/> of this <pause dur="0.2"/> material <pause dur="1.6"/> this is for medium one the first one so i'll write this as X-A <pause dur="0.4"/> and K-A <pause dur="1.3"/> so the thickness of the <pause dur="0.4"/> mineral wall <pause dur="12.7"/><kinesic desc="writes on board" iterated="y" dur="13"/> X-A just <pause dur="0.3"/> cross-multiplying that equation <pause dur="0.5"/><kinesic desc="writes on board" iterated="y" dur="3"/> is therefore equal to the temperature <pause dur="0.2"/> difference <pause dur="0.7"/> that's the temperature difference <pause dur="0.2"/> i'll just put this down and see if the diagram will reappear <pause dur="1.0"/><kinesic desc="reveals covered diagram" iterated="n"/> yeah there's a temperature difference across the mineral wall

so it's a thousand minus four-hundred <pause dur="7.6"/><kinesic desc="writes on board" iterated="y" dur="4"/> again it's an interval so we can write it as <kinesic desc="writes on board" iterated="y" dur="3"/> kelvin or degrees celsius <pause dur="3.7"/> multiplied by the <kinesic desc="writes on board" iterated="y" dur="15"/> thermal conductivity of that medium <pause dur="0.4"/> seventy milliwatts per metre per degree kelvin so <pause dur="0.4"/> point <trunc>ne</trunc> zero-seven <pause dur="1.7"/> watts per metre <pause dur="0.9"/> per degree kelvin <pause dur="3.0"/> and divided by <pause dur="0.4"/> the <pause dur="0.2"/> heat flux per unit area <pause dur="0.7"/><kinesic desc="writes on board" iterated="y" dur="6"/> two-hundred-and-twenty-<pause dur="0.3"/>five <pause dur="1.8"/> watts <pause dur="0.2"/> per square metre <pause dur="3.6"/> okay with a bit of luck that's going to end up with the right dimensions <pause dur="0.4"/> # although it takes a long time i always put dimensions in equations 'cause then you can check <pause dur="0.3"/> if you've lost a term <pause dur="0.4"/> so the <pause dur="0.4"/> K cancel out with the <pause dur="0.5"/> K-to-the-minus-one <pause dur="0.6"/> watts cancel out <pause dur="1.2"/> and you're left with metres-to-the-minus-one divided by metres-to-the-minus-two <pause dur="0.3"/> or therefore metres <pause dur="0.2"/> which is what you would want for a <pause dur="0.4"/> thickness <pause dur="0.4"/> so if you can just <pause dur="0.3"/> # do the calculation on that <pause dur="43.2"/> a number </u><pause dur="1.0"/> <u who="sm0847" trans="pause"> point-one-eight-seven </u><pause dur="0.3"/> <u who="nf0831" trans="pause"> sorry </u><pause dur="0.3"/> <u who="sm0847" trans="pause"> point-one-eight-seven </u><u who="nf0831" trans="latching"> point-one-eight-seven that's that's all right i <pause dur="0.2"/> put left behind where i did mine i calculated <trunc>point-one-n</trunc>-nine my number <pause dur="0.5"/> so that's <kinesic desc="writes on board" iterated="y" dur="2"/>

point-one-eight-seven about point-one-nine <pause dur="1.0"/> metres <pause dur="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 <pause dur="0.3"/> of a wall <pause dur="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 <pause dur="0.5"/> # equally if you get sort of ten metres then <pause dur="0.3"/> you think this probably isn't realistic so you go back again and calculate but <pause dur="0.2"/> certainly that looks like <pause dur="0.3"/> a reasonable number <pause dur="1.1"/> so just doing the same for the other <pause dur="0.2"/> # <pause dur="0.2"/> # insulation material <pause dur="0.3"/> the fibreglass <pause dur="3.3"/><kinesic desc="writes on board" iterated="y" dur="15"/> exactly the same calculation <pause dur="0.2"/> but X-B <pause dur="1.8"/> is therefore equal to the temperature difference across the fibreglass four-hundred minus fifty-five <pause dur="1.3"/> degrees kelvin <pause dur="0.2"/> or celsius <pause dur="0.4"/> multiplied by the thermal conductivity of the fibreglass <pause dur="0.3"/> point-zero-four <pause dur="2.1"/><kinesic desc="writes on board" iterated="y" dur="12"/> watts per metre per degree kelvin <pause dur="8.7"/> so a better insulation material <pause dur="0.3"/> and again divided by two-hundred-and-twenty-five <pause dur="8.5"/><kinesic desc="writes on board" iterated="y" dur="8"/>

which is equal to </u><pause dur="17.0"/> <u who="sm0848" trans="pause"> zero-six <gap reason="inaudible" extent="1 sec"/></u><u who="nf0831" trans="overlap"> <trunc>part</trunc> point </u><pause dur="0.4"/> <u who="sm0848" trans="pause"> zero-six </u><u who="nf0831" trans="overlap"> that sounds about right <pause dur="0.7"/> i was going to say was it's something like <pause dur="0.2"/> one-and-a-half times <pause dur="0.4"/> point-zero-four so that's about point-zero-six <pause dur="3.1"/><kinesic desc="writes on board" iterated="y" dur="2"/> metres so again sort of numbers that look as if you could build <pause dur="0.3"/> an insulation system out of it </u><u who="sm0849" trans="latching"> what are they <gap reason="inaudible" extent="2 secs"/></u><pause dur="0.4"/> <u who="nf0831" trans="pause"> sorry </u><pause dur="0.2"/> <u who="sm0849" trans="pause"> what are they <gap reason="inaudible" extent="1 sec"/> <pause dur="0.5"/> blocks <pause dur="1.4"/> <gap reason="inaudible" extent="1 sec"/> wall blocks </u><u who="nf0831" trans="overlap"> wall blocks <pause dur="0.4"/> they're sort of <trunc>s</trunc> # compressed <pause dur="0.2"/> # i mean they're they're solid insulation <pause dur="0.5"/> blocks <pause dur="0.4"/> that you use for high temperature <pause dur="0.2"/> # sort of a first <pause dur="0.3"/> <trunc>li</trunc> lining layer you you <pause dur="0.3"/> you would actually have a steel layer <pause dur="0.8"/> or you you would have <pause dur="0.4"/> a steel layer or <unclear>somewhere</unclear> a firebrick layer on the very <pause dur="0.3"/> inside <pause dur="0.5"/> and then they're they're sort of lightweight fibre <pause dur="0.4"/> compressed fibre blocks <pause dur="0.8"/> that you can <pause dur="0.6"/> build as a sort of an insulation wall <pause dur="2.0"/>

okay <pause dur="0.2"/> so <pause dur="0.2"/> that's <pause dur="0.2"/> big scale high temperature conventional engineering <pause dur="0.3"/> # sort of calculation <pause dur="1.1"/> # <pause dur="0.2"/> and the key thing is to break it down into the you know look <trunc>a</trunc> look at the problem look at the bits where you've got the information <pause dur="0.4"/> decide <pause dur="0.3"/> where <pause dur="0.2"/> you need to sort of go first to find the information <pause dur="0.3"/> you don't have <pause dur="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 <pause dur="0.8"/> is to go back to this problem we've started to look at in the last lecture <pause dur="0.6"/> on designing whether you can design a <pause dur="0.2"/> little portable <pause dur="0.4"/> # chilling system <pause dur="0.4"/> # so that the department store can <pause dur="0.5"/> present flowers or strawberries <pause dur="0.4"/> in its <pause dur="0.2"/> # # appropriate place to encourage people to <pause dur="0.3"/> buy them <pause dur="0.5"/> so this is example five-six in your notes <pause dur="5.2"/><kinesic desc="writes on board" iterated="y" dur="3"/> and this is completely open-ended # i <trunc>ha</trunc> i deliberately hadn't <pause dur="0.3"/> done the calculations before we do this we may end up that

it's completely <pause dur="0.2"/> not a good idea you can't it won't work <pause dur="0.4"/> # and so <pause dur="0.3"/> # you go back and think about something else <pause dur="0.6"/> so what we decided <pause dur="0.5"/> is that this sort of <kinesic desc="writes on board" iterated="y" dur="6"/> chiller unit <pause dur="6.2"/> is going to be made it's going to be <pause dur="0.2"/> cylindrical so that <pause dur="0.2"/> people can sort of reach in easily <pause dur="8.9"/><kinesic desc="writes on board" iterated="y" dur="6"/> and it's going to have <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="5"/> insulation on the outside we've <trunc>in</trunc> initially trying <pause dur="1.5"/> something like the insulation you'd put on a refrigerator <pause dur="0.3"/> so about sixty <kinesic desc="writes on board" iterated="y" dur="16"/> millimetres <pause dur="14.7"/> we'd taken a total height <pause dur="0.2"/> in the lecture last week of one metre <pause dur="6.8"/><kinesic desc="writes on board" iterated="y" dur="4"/> # a total external <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="6"/> diameter <pause dur="0.5"/> of six-hundred millimetres <pause dur="5.3"/> and an insulation <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="17"/> thickness <pause dur="0.6"/> of <pause dur="0.2"/> sixty <pause dur="0.4"/> millimetres </u><pause dur="9.3"/> <u who="sm0850" trans="pause"> does that bit need to be sixty as well <pause dur="0.7"/> that last one you've just drawn </u><pause dur="0.3"/> <u who="nf0831" trans="pause"> sorry </u><pause dur="0.4"/> <u who="sm0850" trans="pause"> that last one you've just drawn </u><pause dur="0.6"/> <u who="nf0831" trans="pause"> this last line i've not drawn doesn't matter this last line i have drawn doesn't need to be sixty <pause dur="0.5"/><kinesic desc="writes on board" iterated="y" dur="15"/> because this thickness here there's ice in here <pause dur="2.6"/> and there's the fresh produce <pause dur="0.2"/> sitting on top <pause dur="1.3"/><event desc="changes pen" iterated="n"/> think i've got a a fresh produce colour that i can <pause dur="0.3"/> draw in <pause dur="9.9"/><kinesic desc="writes on board" iterated="y" dur="3"/> don't think i can draw anything like a rose so we'll have a sort of a <pause dur="0.9"/> # <pause dur="0.2"/> # <pause dur="0.2"/> nursery flower <pause dur="0.7"/> and the the condition here <pause dur="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 <pause dur="0.4"/> so we've said that the condition <kinesic desc="indicates point on board" iterated="n"/> here <pause dur="0.8"/> is <pause dur="0.3"/> that <pause dur="0.2"/> the <pause dur="0.6"/> temperature <pause dur="0.6"/> at this point <pause dur="0.9"/> should be <pause dur="0.5"/> four <pause dur="0.2"/> degrees </u><pause dur="0.8"/> <u who="sm0851" trans="pause"> so we've got one work that out </u><pause dur="0.3"/> <u who="nf0831" trans="pause"> so </u><pause dur="0.8"/> <u who="sm0852" trans="pause"> that's too high </u><pause dur="0.4"/> <u who="nf0831" trans="pause"> that's <pause dur="1.6"/> the first <pause dur="0.2"/> calculation to do is to find the thickness <kinesic desc="indicates point on board" iterated="n"/><pause dur="0.3"/> of <pause dur="0.2"/> that bit <pause dur="1.3"/> the ice is in here <pause dur="0.8"/> and again first approximation we'll assume that it stays exactly at zero degrees as it melts <pause dur="0.5"/> which if you look at real melting ice is <pause dur="0.2"/> almost true <pause dur="0.3"/> so we've got ice in <kinesic desc="indicates point on board" iterated="n"/> here that we hope keeps this system cold <pause dur="0.3"/> throughout the working day <pause dur="2.0"/> we've got sixty millimetres of insulation round all the other <pause dur="0.2"/> surfaces <pause dur="0.4"/><kinesic desc="writes on board" iterated="y" dur="4"/> of the ice <pause dur="6.4"/> and <pause dur="0.3"/> we'd established that to <pause dur="0.5"/> put the produce in <pause dur="0.4"/> we need <kinesic desc="writes on board" iterated="y" dur="11"/> probably a height of about three-hundred millimetres <pause dur="0.5"/> here <pause dur="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 <pause dur="0.4"/> # and # <pause dur="0.2"/> so <pause dur="0.3"/> <trunc>y</trunc> you have to look at it and decide what you think are going to be the <pause dur="0.5"/> key <pause dur="0.3"/> # <pause dur="0.2"/> factors that determine the rate of

heat transfer <pause dur="0.3"/> so that you can do your first calculation <pause dur="0.5"/> 'cause your first calculation says that you're going to need ten tons of ice <pause dur="0.3"/> to achieve this <pause dur="0.3"/> then you know it's not going to work <pause dur="0.2"/> whatever you do <pause dur="0.3"/> in your calculations <pause dur="0.2"/> and so you go away and think about something else <pause dur="0.3"/> if your first calculation says yes this looks quite a good idea <pause dur="0.4"/> then you think about doing the calculation better <pause dur="0.4"/> to understand the physical system <pause dur="0.3"/> better <pause dur="1.3"/> so <pause dur="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 <pause dur="0.4"/> through these sides <pause dur="0.3"/> got heat transfer from the ice across the <pause dur="0.5"/> probably thin layer of insulation here <pause dur="0.3"/> heat transfer from the ice <pause dur="0.5"/> across to the air round here <pause dur="0.7"/> and also <pause dur="0.2"/> this is probably sitting on the <pause dur="1.1"/> nice plush carpet in the <pause dur="0.5"/> soft <pause dur="0.2"/> furnishings part of the <pause dur="0.2"/> store <pause dur="0.3"/> so we've got some heat transfer by conduction <pause dur="0.4"/> down <pause dur="0.2"/> to the <pause dur="0.3"/> ground or sorry conduction actually <pause dur="0.2"/> in positive heat transfer

the heat is all coming in to the ice which is gradually <pause dur="0.2"/> melting </u><u who="sm0852" trans="latching"> can you not ignore that if it's on wheels </u><pause dur="0.2"/> <u who="nf0831" trans="pause"> sorry </u><pause dur="0.2"/> <u who="sm0852" trans="pause"> can you not ignore that if it's on wheels </u><pause dur="0.5"/> <u who="nf0831" trans="pause"> that's a good point so if we if we put it on wheels <pause dur="0.4"/> in fact it makes <trunc>sligh</trunc> a calculation slightly more <pause dur="0.2"/> more you're you're # you're putting in an extra calculation so we'll put it on wheels <pause dur="2.9"/><kinesic desc="writes on board" iterated="y" dur="2"/></u><pause dur="0.6"/> <u who="sm0853" trans="pause"> <gap reason="inaudible" extent="2 secs"/></u><pause dur="0.4"/> <u who="nf0831" trans="pause"> ah but do you need the sort of wheels that you're going to be able to sort of sit it down </u><u who="sm0853" trans="overlap"> no </u><u who="nf0831" trans="overlap"> so that people don't your customers don't start pushing it around so <pause dur="0.3"/> <vocal desc="laughter" n="sm0853" iterated="y" dur="2"/> perhaps you need some sort of retractable <kinesic desc="writes on board" iterated="y" dur="2"/> castors that <pause dur="0.7"/> will sit in here so that your customers don't wheel the strawberries out of the store with their <pause dur="0.3"/><vocal desc="laughter" n="ss" iterated="y" dur="2"/> trollies <pause dur="1.4"/> so we'll we'll put it back on the carpet i think <pause dur="1.2"/> that makes it easier 'cause if you've got a thick layer of insulation <pause dur="0.4"/> and a thick layer of carpet <pause dur="0.3"/> and some flooring here <pause dur="0.4"/> my first sort of approximation was going to say let's forget about conduction to the floor <pause dur="0.3"/> to begin with <pause dur="0.3"/> because <pause dur="0.3"/> probably <pause dur="0.4"/> the <pause dur="0.2"/> transfer <pause dur="0.3"/> from the surface is going to be much more rapid than the <pause dur="0.2"/> transfer

through sort of whatever is down here <pause dur="0.3"/> and that could be the next stage that you go back and say well was that important <pause dur="0.3"/> or <pause dur="0.3"/> not <pause dur="0.5"/> so my sort of first approach on this would <trunc>a</trunc> actually be to <pause dur="0.3"/> # quickly get the trolley off its wheels <pause dur="0.2"/> sitting on a a nice thick carpet and say <pause dur="0.3"/> # # sort of our first approximations <pause dur="4.9"/><kinesic desc="writes on board" iterated="y" dur="10"/> is to ignore the base <pause dur="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 <pause dur="0.3"/> # very important </u><pause dur="5.0"/> <u who="sm0854" trans="pause"> the sides above our ice </u><pause dur="0.4"/> <u who="nf0831" trans="pause"> yes i thought those will have <trunc>tho</trunc> those started to look very # <pause dur="0.3"/> difficult because there's going to be a sort of a temperature <kinesic desc="indicates point on board" iterated="n"/> gradient up here and we're not quite sure <pause dur="0.4"/> # at all what's going to be in there <pause dur="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 <pause dur="0.3"/> and do it later <pause dur="0.9"/><kinesic desc="writes on board" iterated="y" dur="3"/> so if we now ignore the sides above <pause dur="0.2"/> the level <pause dur="0.3"/> of the insulation <pause dur="9.3"/><kinesic desc="writes on board" iterated="y" dur="8"/> # <pause dur="0.2"/> i need a word to describe this if <trunc>w</trunc> if we <pause dur="0.4"/> call this the <pause dur="0.3"/> a <kinesic desc="writes on board" iterated="y" dur="7"/> shelf <pause dur="0.3"/> if you like <pause dur="7.1"/> then in

fact you end up with <pause dur="0.3"/> a much simpler problem <pause dur="0.5"/> to deal with <pause dur="0.7"/><kinesic desc="writes on board" iterated="y" dur="30"/> you've now got something that looks like this <pause dur="2.9"/> this is the shelf with the thickness that <pause dur="0.2"/> keeps the <pause dur="0.3"/> produce so it doesn't get too cold <pause dur="10.4"/> and this in cross-section <pause dur="0.3"/> is a cross-section through a much shorter <pause dur="0.5"/> cylinder <pause dur="2.5"/> and this is the ice <pause dur="1.8"/> at <pause dur="0.4"/> zero degrees <pause dur="0.6"/> celsius here <pause dur="0.9"/> so you've made the problem <pause dur="0.6"/> something <pause dur="0.3"/> that you've only now got <pause dur="0.2"/> two different surfaces to worry about <pause dur="0.5"/> there's the shelf <pause dur="5.6"/><kinesic desc="writes on board" iterated="y" dur="4"/> and then there's the cylindrical <pause dur="0.5"/> sides the short length of them <pause dur="10.1"/><kinesic desc="writes on board" iterated="y" dur="7"/> so that looks like the sort of thing that you've got the <pause dur="0.2"/> # tools to <pause dur="0.4"/> work with <pause dur="0.6"/> # and in fact the shelf bit is very much like the problem <pause dur="0.3"/> that we've looked just looked at <pause dur="0.5"/> so just getting down the basic <pause dur="0.2"/> information that you've got on this shelf <pause dur="0.5"/> you know it's six-hundred millimetres <pause dur="0.3"/> in external diameter and a wall thickness of sixty <pause dur="0.6"/> so that makes it <pause dur="0.2"/> # four-hundred-and-<pause dur="0.2"/>eighty thank you <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="13"/> millimetres <pause dur="0.7"/> diameter here <pause dur="5.0"/> we don't know the thickness <pause dur="0.4"/> that's the <pause dur="0.6"/> next thing to <pause dur="1.0"/>

calculate so that's the first bit of the calculation <pause dur="0.2"/> that we need to do next week <pause dur="2.4"/> because we don't know the thickness there we don't know <pause dur="0.2"/> the length down here <pause dur="0.7"/> although you would <pause dur="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 <pause dur="0.4"/> # sorry <pause dur="0.4"/> seven-hundred millimetres less the <pause dur="0.4"/> because of the insulation there it's not going to be <pause dur="0.3"/> very large so we've got a sort of feel for <pause dur="0.3"/> what that height will be <pause dur="1.2"/> and so what we'll do in the next lecture <pause dur="0.8"/> is to think about how we can describe convection and radiation from the surface

here <pause dur="0.6"/> # <pause dur="0.3"/> initially thinking about this just as being a sort of a bare plate which is the worst case <pause dur="0.6"/> # <pause dur="0.2"/> and then look at the <pause dur="0.2"/> convection and radiation <pause dur="0.3"/> and conduction through <pause dur="0.2"/> the <pause dur="0.2"/> sides <pause dur="0.5"/> i'll then finish off next week with the final calculation <pause dur="0.3"/> on <pause dur="0.2"/> the # # a sort of a a complicated system of a double glazed window <pause dur="1.1"/> okay <pause dur="0.7"/> Friday is test day <pause dur="0.4"/> please don't forget that <pause dur="0.8"/> # <pause dur="0.2"/> so come with writing implements <pause dur="0.3"/> # pen not red pen please <pause dur="0.3"/> # <pause dur="0.2"/> diagrams in pencil but writing in pen <pause dur="0.5"/> and also your university approved calculator <pause dur="0.6"/> okay thank you