welcome to today's heat transfer lecture we're going to continue talking about the three modes of heat transfer and their rate equations just at a very introductory level today's topic is radiation and this is the final mode of heat transfer so you've seen the slide before if you've watched the other videos but just as a refresher we have the three modes of heat transfer being conduction convection and radiation so radiative heat transfer is heat transfer that typically occurs from one surface to another surface and the rate equation that corresponds with it is called the stefan boltzmann law so just a review question if this is the third time with this question because it's important so what is the driving force for heat transfer and that is a temperature difference so for conduction it was a temperature gradient because in conduction you're typically tracking how heat is is moving through a solid with convection it's a just a delta t it's an algebraic t1 minus t infinity and then for radiation it's going to be a nonlinear term it's going from surface to surface and it's actually going to look something like this the temperature of surface one to the fourth power minus the temperature of surface 2 to the 4th power so the dependence on temperature is certainly very different than in conduction and convection however it's still important to know that the heat is going to flow from the high higher temperature to the lower temperature surface all right so just to review again heat transfer rated heat transfer is energy that is emitted by matter due to changes in the electron configurations of its atoms or molecules and is transported as electromagnetic waves or photons so with radiation you're actually having a surface because of its temperature it's going to be shooting out photons out into space and then for that heat that heat or that thermal energy to enter another surface it's got to be intercepted by that other surface so the there is an important distinction between radiation and conduction and convection whereas radiation doesn't require matter in between these two surfaces there isn't that can be a total vacuum and you can still get heat coming from here to here because it's by radiation or by these electromagnetic waves that pulls out from a surface to another surface so let's just try and put some practicality and experience to help you get your head around how radiation works so i'm sure most of you have stood by a fire before so if you stand by a fire you'll notice that you get hot obviously but as you you get hot but that heat doesn't come from the air around you the air around you isn't necessarily getting overly hot you feel that heat coming to you directly from the flame so if you were to block that if you put something in between you you would actually stop feeling the heat so radiation or thermal energy emitted by radiation works a lot like light it just happens to be it's i mean just like light thermal radiation is part of the electromagnetic spectrum it's just coming from a different wavelength than visible light so if you just like if you blocked your view from the fire you wouldn't be able to see the fire if you were to block your view from the fire you also wouldn't be able to feel it because you'd be blocking that thermal radiation so this is a surface to surface phenomenon if we thought of this fire as a surface and you as a surface some of that electrons sorry not electrons photons and electromagnetic energy is going to be emitted from that fire and you are going to intercept some of it which would cause your body to absorb some of that heat to the point that you could physically feel it and even though there would have to be air for the fire to be burning the heat transfer doesn't require that error in between so and largely air is mostly transparent to radiation so even if there are a vacuum between you and the fire you'd still be able to feel that heat and this is just the same way that earth gets heat from the sun it's 93 93 million miles of just emptiness vacuum and yet heat can still be transferred very effectively from the sun across that 93 million miles all the way out to earth okay so let's introduce the stefan boltzmann law as i've mentioned the stephen boltzmann law will tell us about the rate at which heat is transferred from one surface to another so the surface that's emitting thermal energy is going to do so in proportion to its temperature to the fourth power i showed you that temperature difference driving force this is where it originates from so the stefan boltzmann law tells us that the emissive power or the total rate at which energy is being emitted from a surface is proportional to epsilon epsilon is a is a property of the surface called the emissivity of a surface so a white painted wall is going to have a different emissivity than a rock or and both of us are going to have a different emissivity than a steel surface so this is a property of the particular surface will determine largely how much heat is able to be admitted from that surface so the emissive power is proportional to the emissivity multiplied by this variable sigma sigma is just a constant it's always the same that's called the stefan boltzmann constant and it happens to be equal to 5.67 times 10 to the minus 8 with units of watts per meter squared per kelvin to the fourth and you'll notice those units come in really handy when you're actually applying this law so we have the emissive powers proportional to epsilon times sigma times the temperature of the surface to the fur to the fourth power this quantity sigma times the surface temperature to the fourth power is also known as the emissive power of a black body so a black body one of the assumptions of a black body is that it emits perfectly so there is nothing else hindering heat transfer except for the temperature itself we use this epsilon as a nice convenient term to as a relative ratio of how much does that surface emit relative to that of a black body so this is going to be um a decimal number between 0 and 1 that tells us okay i emit half or three-fourths of what a black body would emit so epsilon is always going to be between zero and one with one being again the emissivity of a black body so a perfect emitter the best you could do would be to have a black body or the highest rate of heat transfer you could get by radiation is if you have the surface being a black body okay so let's talk a little bit more about the stefan boltzmann law so the surface is going to emit thermal energy so here's a surface it's going to be emitting this thermal energy with its emissive power of e the surface may also be losing heat by convection that's not exactly what we're talking about right now and again you remember all those terms from before so you'll notice on this figure we also have energy can also be intercepted by this surface from elsewhere it could be coming from another surface out up here which is also emitting let's call that e1 and so whatever from there is emitted from this surface whatever portion of that hits this surface is referred to as g so g is the energy absorption due to irradiation actually i misspoke g is the actual irradiation hitting the surface so that's a given in terms of flux in watts per meter squared so that's what's coming in and hitting our surface but we have to multiply that by this alpha term so alpha is referred referred to as the surface absorptivity and the product of those two is what is actually absorbed so some of that energy is going to hit our surface but not all of it will necessarily be absorbed by the surface some of it's going to bounce off and be reflected and we'll get much more into this in chapters 12 and 13 when we get really in depth into radiation so for now we're just trying to define these basic terms and help you get a feel for how radiation works basically and there may be some simplistic radiation problems that you'll be solving even before we get to the actual radiation chapters of the book all right let's go back to our thought experiment now so we're standing next to the fire so as i mentioned if you got blocked then that radiation wouldn't necessarily make its way to your actual body and you wouldn't feel it the same way because there's something blocking you so only what your body is able to actually what's actually incident on your body is what your body would have the chance of absorbing that irradiation would hit your body and then your body is going to have a certain absorptivity again a number between 0 and 1 suggesting that your body won't absorb 100 percent of what hits it but it'll only absorb a portion of that and that will be a function of the absorptivity of your skin and of your clothes and of your hair so radiation if you haven't had this thought already radiation from this fire is actually going to be emitted out in all kinds of directions so that your friends standing next to you they would also absorb some of that radiation and a lot of people could take advantage of the warmth from this fire because it's going to be emitting out in all kinds of different directions so radiation is going to have this as we're studying radiation heat transfer problem is going to have this extra complicating factor where that radiation isn't necessarily being absorbed and emitted just to one surface it could be it would be hitting multiple surfaces so the person would another thing about radiation is that if you noticed the form of that emissive power equation from the stefan boltzmann law where e is equal to the emissivity times the stefan boltzmann constant times t to the fourth this suggests that any surface that has a measurable temperature is going to be emitting energy so while i've drawn these heat transfer lines going from this flame out to the person actually the person is going to be emitting some energy back and the person themselves will also be emitting energy out in all directions it's just that the person will have a much lower temperature than the fire so its effect won't be nearly as noticeable as that of actual flame but any surface that actually has a measurable non-absolute zero temperature would be emitting radiative energy pretty much at all times so radiation does get very complicated where you get into something called a view factor which says how much of that radiation is actually going from the fire to this person so it gets really complicated there's a lot of geometry involved but first we're going to introduce a little bit of a simplified form of the stefan boltzmann law that'll help us start to solve some simpler problems so if we have radiation to the surroundings so we have our surface here it's going to be emitting radiation this q double prime rad is the net radiative heat flux coming from the surface so that takes into account both e what the surface is emitting and g what is hitting the surface and if we assume that our surface is going out uniformly to all of its surroundings that will make dealing with all that what i called the view factor a lot easier to deal with so when we get this particular circumstance when we have a small object that is completely surrounded by other surfaces at uniform temperature so if we could approximate our surroundings all occurring at this variable at this temperature t subscript surroundings then we can get this much simpler form of this radiation heat heat transfer relationship so when we have the special case that our surface is exposed to large surroundings and those surroundings that are at this uniform temperature t surroundings then we have the incident irradiation is coming in at g we're going to assume that's coming in uniformly we can treat our surroundings because they're very large as a black body so we can assume that the that the that g what's hitting our surface is going to be coming as if it were emitted by a black body sigma times t surroundings to the fourth we can also make another assumption that if alpha or the absorptivity of our surface is equal to epsilon or the emissivity of our surface then we can greatly simplify this equation to be so the net radiative heat transfer from our surface to the surroundings would be what it's emitting minus what it is absorbing and if we were to plug in so this ts in parenthesis means the emissivity of a black body at as a function of the surroundings temperature this so this term would actually be epsilon sigma times ts to the fourth if we substituted this sigma ts to the fourth being the emissivity or the emissive power of a black body so if we do some algebra we end up with this much simpler equation the net rate of radiation exchange between the surface and its surroundings is equal to epsilon or the emissivity of our surface which we've made the assumption that it's equal to the absorptivity of our surface times sigma which is the stefan boltzmann constant times the temperature of the surface of the fourth power minus the temperature of the surroundings to the fourth power so this again is given in terms of flux notice the double prime there so if we wanted to get the absolute rate of heat we would just multiply this flux by the total surface area of our surface and we'd get this total heat rate form of the equation radiation is is complex and i assure you that we'll go into this in much more detail in chapters 12 and 13 so the key takeaway for now is to remember this basic form when you have a surface that is exchanging heat via radiation to surroundings that are at uniform temperature and we make the assumption that the absorptivity of the surface is equal to its emissivity this is called the gray surface assumption then we get that the flux by radiation is equal to epsilon times sigma times our surface temperature to the fourth minus the surroundings temperature to the fourth so just store this equation away this is the thing that you'll need to remember and then just keep in mind that radiation will get a lot more complicated with the geometry of very complex systems and and a lot of different things especially when we take away these different black body assumptions all right so let's practice applying the stephen boltzmann law to a very simple radiative heat transfer problem and this problem says if my roof is at a uniform temperature of 35 degrees c the effective temperature of the surroundings is 20 degrees c and the surface area of my roof is 50 meters squared the emissivity and absorptivity is 0.85 that's the surface the roof how fast is my roof losing heat by radiation to the environment we want to make sure we do this with the right units so let's spell this problem out let's look at our same house with a this nice simple flat roof okay so we're making the assumption that all the surroundings in the environment are going to be at a uniform temperature t surroundings so by making that assumption this roof by comparison is a very small object when you consider the whole sky and all of its surroundings so we can apply this assumption where we use the rate law that the flux the net radiative heat flux from my roof out to the surroundings is equal to the emissivity of my roof multiplied by the stefan boltzmann constant and my roof temperature to the fourth minus the surrounding temperature to the fourth so this is a fairly simple plug and chug problem where we want to identify what is q rad here and notice that we want the total rate of heat loss and not just the rate of not just the flux not the rate of heat loss per unit area of my roof so let's go ahead and plug and chug in here so we have qrad is equal to epsilon times sigma times the surface temperature to the fourth minus the surroundings temperature to the fourth which would give us units of watts per meter squared so in order to get that into the total rate of heat loss form or the total heat rate form we would just multiply this equation by the area so our total rate of radiative heat exchange is going to be epsilon times sigma times the area times ts to the fourth minus t surroundings to the fourth and that would give us units of watts so let's go ahead and plug in chug so this is equal to zero point eight which is our emissivity multiplied by the stefan boltzmann's constant which is 5.67 times 10 to the minus 8 and it has units of watts per meter squared per kelvin to the fourth now my roof temperature is 35 but if you recall that is 35 degrees c so we need to add 273 to convert from degree c to kelvin and we need to convert to absolute temperature before taking this guy to the fourth power minus all right the surroundings temperature was 20 again that was degree c so we need to add 273 here and we need to make sure we do that before raising it to the fourth power so that's how our equation looks and that comes out let's double check our units so the temperature is going to be come out to be kelvin to the fourth so it's always a good idea to specify the equation first and then use units just as a check so we get this kelvin to the fourth canceling out here oh and i'm glad we are doing this unit check because that tells me that i forgot to multiply by our surface area which was 50 meters squared so i caught myself from making a a big error just by doing this unit check so that 50 meters squared cancels out here and we do indeed end up with units of watts so i'm going to spare you the plugging and chugging you can type this into your calculator or into an excel sheet but we end up with the roof losing 3 900 and 26 watts so my roof would be losing that much heat by radiation to its surroundings just as a point of comparison and maybe as a little bit of a sanity check we did a convection problem with a very similar problem with even the same temperatures and i believe we got a rate of heat loss of 7 500 watts from convection so it's good to know that these are at least on the same order of magnitude which is an identity check that we probably are getting these about right okay so just some notes on radiation the driving force is the difference in temperatures to the fourth power for each surface that driving force will basically be the same even as we get into more complex geometries and even as we get away from that black body assumption radiation and convection can happen in parallel so you could have the roof certainly losing heat you could have a radiative term here and a convective term here and actually you could have it gaining heat by radiation and losing heat by convection because the driving forces are a little bit different convection is going to happen to the based on the air temperature and radiation is going to happen based on the temperature of all the surroundings so those could actually be going against each other you could also get radiation directly from the sun hitting the roof which would cause it to absorb heat and probably the reason that it got hotter than its ambient in the first place so radiation becomes increasingly important at high temp at high surface temperatures so back to that person standing by the fire example there probably is convection happening where that the fire does heat up the air and then that air would in turn heat you up but really what you notice is the radiation you notice that energy being directly absorbed by your body from the fire so if someone blocked you you from the fire from your view of the fire you would immediately notice that you're not absorbing as much heat whereas if someone blocked you you might still be getting a tiny bit of heat by convection but it wouldn't be as noticeable and that's because that fire is at such a high temperature that radiation is going to dominate as t1 gets really really high this rate these rates become very large and you could even neglect convection because it's so small by comparison so for now we're just going to consider radiation to surroundings at uniform temperature so we're just going to be using that qrad double prime the flux is equal to epsilon sigma times t1 to the fourth minus t2 to the fourth which applies when you have a an object that is exchanging heat by radiation with its surroundings and that the object is relatively small compared to the big size of its surroundings so we'll use this form of the stephen boltzmann law so for now this is the form that i want you to remember