Understanding Convection in Heat Transfer

Hello and welcome to this lecture on heat transfer. Today we're going to continue talking about the three modes of heat transfer and their rate equations. Today's lecture is going to be about convection. As a review, the three modes of heat transfer are conduction, with a rate equation that can be described by Fourier's law, convection, which is what we'll be talking about today, with a rate equation that can be described by Newton's law of cooling, and the convection entails heat transfer from a surface to a moving fluid. And finally radiation, which is typically surface to surface heat transfer, which can be described by the Stefan-Boltzmann law. So just as a review, what is the driving force for heat transfer? In other words, what makes heat flow from one place to another place? And the answer there is a temperature difference. And for convection, the temperature difference that we'll use that will show up in our rate equation is going to be an algebraic temperature difference or a delta t. So convection here in the middle is heat transfer that is due to the combined influence of bulk and random motion for fluid flow over its surface. So when you have a surface that's giving off heat and going into a fluid, that is convection. The difference between convection and conduction is that convection has this moving fluid where that fluid can circulate and can carry heat away just by the bulk fluid motion. Whereas if you recall in conduction, which typically happens in a solid, all the molecules or atoms are bound. They're not going anywhere, or at least they're not going long distances, so they kind of stay in place. So they can only transfer energy by random motion, whereas convection, that energy can be transferred by this what's called a bulk fluid motion, or the currents of the fluid that help to sweep away some of that heat. So let's stop and think about convection a little bit. So let's suppose that we have a heated plate. So this is just a hot solid surface. So it's hot and the air above it is relatively cooler. So we had air flowing, so this is not a stationary fluid, this is a moving fluid. Naturally that air flowing is going to carry away heat. So which way is heat going to travel? It's going to travel from the hot to the cold. Remember it has that that delta t driving force. So we're going to get this q convection or we could also express it as a flux with q double prime convection. So heat transfer can also happen in the opposite direction. If we switched our driving force, to be the other way. Let's say we had a cool plate and a hot fluid, and we're doing the same thing. We're circulating air, we're blowing air over the top. Naturally, heat in this case is going to flow from the hot to the cold. There's more energy, relatively speaking, in the hot fluid than in the cool plate. So that driving force for heat transfer is going to go the other way, and we'd get Q convection going from the air into the plate. So this can work either way, but the rate law that we'll use is going to be called Newton's law of cooling. Alright, so this is a similar visualization, a little bit more scientific from our textbook. So this shows the hot surface and what happens physically is if you're flowing fluid at a velocity, a bulk fluid velocity called U sub infinity, that just means the bulk fluid velocity, that fluid is going to be flowing over this hot plate. What's really happening is that Very very close to the hot surface the temperature is going to be a little bit higher The velocity is actually going to be a little bit lower due to viscous effects of that fluid and you'll see what's called this this profile this velocity distribution of velocity profile form and Similarly you'll get this temperature distribution forming so close to the plate you'll get something called a boundary layer where really they're There is kind of a molecular scale phenomenon happening here, or a very small scale phenomenon happening here, where there is a more gradual change in temperature. But typically, we're going to look at the macro level, and we're going to be looking at the difference in the surface temperature as compared to the fluid temperature. So in a lot of the problems that we'll solve, we'll be looking at the difference between these two physically separated points, the surface temperature and the fluid temperature. Because we're doing that, we'll use this algebraic temperature difference as our fluid driving force. Another way of saying that is we're essentially, in a lot of the problems that we're solving, I'm going to ignore what's happening here and just look at what's happening here and here, although we do get into more of the theory about what's happening in that boundary layer. So Newton's law of cooling is a little bit simpler than Fourier's law. It's just this simple because it uses that simple algebraic temperature difference. Newton's law of cooling tells us that the flux is equal to h. h is called the convective heat transfer coefficient multiplied by the temperature difference, our surface temperature minus the bulk fluid temperature. So if we know h and we know these two temperatures, it's pretty easy to calculate our heat flux. Thanks. The convection heat transfer coefficient has units of watts per meter squared per Kelvin. So what that's saying, notice that the Kelvin shows up in the denominator. So the higher driving force you get, the more... watts you will transport and also the bigger your surface area is for heat transfer the more heat you'll you will dissipate from the from whatever your surface is And you can see that shows up in this heat rate version. So up here is the heat flux version, down here is the heat rate version. And all you do is you multiply the heat flux version by the surface area of your surface, and that gives you a total heat rate in terms of watts rather than a flux in terms of watts per meter squared. Notice that, as I showed in the thought experiment, that you can actually flip those temperatures, and you can have heat flowing the other way. So you can have heat flowing from the hotter fluid into the colder surface, and that is pretty easily flipped. So that's where it's really important for you as an engineer solving heat transfer problems to know which form of this you should be using. And that comes usually by a fairly simple inspection of your system. However, it's very common for students to not take that extra time to really think about what's happening conceptually, and they may just be sort of taking a stab in the dark, but I promise you... you if you follow this very simple systematic way of doing it, it's not too hard to figure out. So let's take a look with an example problem. How would we solve a convection problem? I'll just read through the problem. If my roof is at a uniform temperature of 35 degrees Celsius and the ambient temperature is 20 degrees Celsius, the surface area of my roof is 50 meters squared and the convective heat transfer coefficient is 10 watts per meter squared per Kelvin, how fast is my roof losing heat? by convection to the environment. And then I want you to include the appropriate units in your answer. So if you notice, I included the roof area, so that should imply that we're using the total heat rate because we have an area. We can quantify how much heat is being lost over the entire roof rather than how much heat is being lost per meter squared for example. Okay so just to make this simple I'm going to give my house kind of a simple one sided roof. Okay so we have our Here's our T sub s, our surface temperature. Out here in the bulk fluid away from that surface we have our bulk fluid temperature T infinity and we'll assume there's some kind of breeze blowing that is helping to remove heat from the roof out into the ambient air. So we also have our H equal to 10 watts per meter squared per Kelvin. Okay, so this is a really simple plug-and-chug problem. We could use the flux form of this. where we'd have the heat lost. So here the roof surface is hotter than the ambient, so it's going to be losing heat. So our flux could be the flux per unit area is equal to our heat transfer. or coefficient times our roof surface temperature minus T infinity. So that would give us the flux version, and then we could take that flux and multiply that by our surface area. to get the total heat rate version. It's easier probably to just jump to the total heat rate version. Q equals HA times TS minus T infinity. And again, this is a really simple problem, but it's really important to master this conceptually and get really used to using this because this will just become a very small piece of more complex problems later. So if we plug and chug, we get Q equals 10 watts per meters squared per Kelvin. And it's actually important to get familiar with the units of these different terms too. Times 50 square meters multiplied by our temperature difference. Our surface temperature is 35 minus 20, and because this is a temperature difference, a temperature difference in Kelvin is the same as a temperature difference in degrees Celsius. So I'm just going to put in a Kelvin there. So this gives us a loss of 7,500, and stop and think what units this should be. And then we're also going to double check by very carefully canceling out our units. So the meters squared cancel out, Kelvin cancels out, and we get a loss of 7,500 watts. So we defined this. We were asked, how fast is the roof losing heat? So we looked at, we assumed that the roof was going to be losing heat. So we assumed that this temperature was higher than this temperature, making this... Delta T, a positive number, and because H will always be positive and our area will always be positive, we knew that our total heat rate would also be a positive number. So we'd be losing 7,500 watts of heat. And again, it's important to just... conceptually get your head around that because you may not always know which temperature which a direction that driving force will be so you may have to just make some assumptions so let's go ahead and do another example problem very similar so the roof is at the uniform temperature of 20 the ambient temperature is at 30 the surface area of the roof is still 50 and the heat transfer coefficient is still 10 how fast is my roof losing heat by convection to the environment so again we have our roof Our surface temperature and our bulk fluid temperature. So our rate of heat loss, we apply Newton's law of cooling. Q equals HA times TS minus T infinity. So I've taken the approach that our roof is still going to be losing heat. The problem tells me my roof is losing heat. So let's plug and chug here. So that is equal to 10 watts per square meter per Kelvin times 50 square meters. multiplied by our surface temperature is 20 and our ambient temperature is 30. And again, that temperature difference, even though the temperatures are given to us in degrees Celsius, the temperature difference can readily be switched from degrees C to Kelvin. So that gives us now minus 5,000 watts. Okay, so now we have a negative flow of heat. And man, do we need to panic and freeze? freak out here? "and the answer is no. It's kind of a trick question, but this situation could arise many times as you're solving heat transfer problems. So we assume that the roof was losing heat, but we found out that it's losing minus 5 kilowatts of heat, which actually means although we assumed that heat was going out, it turns out that heat is actually... going in. So we we're losing minus 5000 watts of heat. Another way of saying that is that we're gaining so this would be let's say Q loss and this is Q gain would actually be gaining 5000 watts. watts of heat. So when you get to be solving more complex energy balance problems, this becomes very important, how you're dealing with these little sign changes. So if you were solving an energy balance problem and you wanted to know... what the final temperature of the inside of the house was, or something like that. You could make the same assumption that it's losing heat, but as long as you carefully apply the delta T in this way, you'd find out, well, it's not really losing heat, it's gaining heat, but that negative sign would save you and would make sure that you're still getting the energy balance right. So with Newton's law of cooling, it's really important. So the way I think about it is that if I have an out term from the surface, then I say Q is equal to HA times TS minus T infinity. Or if I'm calling this an in-term, I'm just assuming that the heat is coming in, and I can say Q equals HA times T infinity minus TS. And either one of those would be correct, and it doesn't actually matter if the net rate of heat is going out or in. As long as I... think about them in this way, my energy balance will turn out okay, because if you remember, the in-term is an additive term on your energy balance, and the out-term is a subtractive term on your energy balance. So that sine flip will take care of itself as long as you are able to lay out the equations this way. So while this seems pretty simple and trivial, it's actually a fairly important concept to grasp. All right, so convection. There will be a lot of different ways that convection happens. So It can happen because of forced airflow over a surface. There can be something called buoyancy-driven flow. This is actually called natural or free convection, where you don't necessarily need something like a fan to force a fluid through. You can actually have that fluid flow happening naturally because of the uneven heating of your surface relative to the bulk. Boiling is another way of sort of inducing motion in a fluid, so boiling creates convection. And similarly... Condensation does the same thing, and we'll cover each of those in detail as we get to the convection section of this course. Some notes on convection. So the driving force in convection is this algebraic temperature difference, or a delta T. So we express that as Ts minus T infinity, or T infinity minus Ts. and typically if it's an energy balance problem you would use if the if the heat were leaving the surface and you would use this form and call that energy out or if you assume that heat were coming into the surface then you would use this as your in term and it's just it's really important to carefully think through which direction the heat is flowing or even guessing and picking a direction that you think the heat is flowing so that you use the right form of this and treat that equation appropriately as either an out or an in term. Alright so the rate equation for convective heat transfer is Newton's law of cooling, a very important one to commit to memory. So with convection problems, you're often trying to find a single temperature rather than a profile. Remember with conduction, we were dealing with graphs of temperature versus x, and we were looking at these graphs. temperature profile and solving for the flux based on that profile. And I mentioned that we will go in the reverse order often. Whereas with convection, we're typically looking at sort of a sudden jump in temperature, where this might be the surface, this might be the solid surface, and this would be the fluid. So we're treating convection as if there's just this sudden jump in temperature, delta T. All right, so and I mentioned that in my bullet point, that assumption will eventually go away as we get deeper into convection. It becomes, convection is actually extremely complicated, but we'll try and treat it as simply as we can. So convection is introduced, we're introducing that now and it's actually important even as we get into conduction to understand this basic Newton's law of cooling form because we're going to use this at the boundaries of conduction problems. So here at this solid surface we might care about the temperature profile here by conduction but then at the surface or at the boundary we'll just look at this this sudden delta T between the edge of the surface or the edge of the solid and the fluid. So there will be plenty of problems where different modes of heat transfer are combined.

As a review, the three modes of heat transfer are conduction, with a rate equation that can be described by Fourier's law, convection, which is what we'll be talking about today, with a rate equation that can be described by Newton's law of cooling, and the convection entails heat transfer from a surface to a moving fluid. And finally radiation, which is typically surface to surface heat transfer, which can be described by the Stefan-Boltzmann law. So just as a review, what is the driving force for heat transfer?

In other words, what makes heat flow from one place to another place? And the answer there is a temperature difference. And for convection, the temperature difference that we'll use that will show up in our rate equation is going to be an algebraic temperature difference or a delta t. So convection here in the middle is heat transfer that is due to the combined influence of bulk and random motion for fluid flow over its surface.

So when you have a surface that's giving off heat and going into a fluid, that is convection. The difference between convection and conduction is that convection has this moving fluid where that fluid can circulate and can carry heat away just by the bulk fluid motion. Whereas if you recall in conduction, which typically happens in a solid, all the molecules or atoms are bound.

They're not going anywhere, or at least they're not going long distances, so they kind of stay in place. So they can only transfer energy by random motion, whereas convection, that energy can be transferred by this what's called a bulk fluid motion, or the currents of the fluid that help to sweep away some of that heat. So let's stop and think about convection a little bit.

So let's suppose that we have a heated plate. So this is just a hot solid surface. So it's hot and the air above it is relatively cooler.

So we had air flowing, so this is not a stationary fluid, this is a moving fluid. Naturally that air flowing is going to carry away heat. So which way is heat going to travel?

It's going to travel from the hot to the cold. Remember it has that that delta t driving force. So we're going to get this q convection or we could also express it as a flux with q double prime convection. So heat transfer can also happen in the opposite direction.

If we switched our driving force, to be the other way. Let's say we had a cool plate and a hot fluid, and we're doing the same thing. We're circulating air, we're blowing air over the top. Naturally, heat in this case is going to flow from the hot to the cold. There's more energy, relatively speaking, in the hot fluid than in the cool plate.

So that driving force for heat transfer is going to go the other way, and we'd get Q convection going from the air into the plate. So this can work either way, but the rate law that we'll use is going to be called Newton's law of cooling. Alright, so this is a similar visualization, a little bit more scientific from our textbook.

So this shows the hot surface and what happens physically is if you're flowing fluid at a velocity, a bulk fluid velocity called U sub infinity, that just means the bulk fluid velocity, that fluid is going to be flowing over this hot plate. What's really happening is that Very very close to the hot surface the temperature is going to be a little bit higher The velocity is actually going to be a little bit lower due to viscous effects of that fluid and you'll see what's called this this profile this velocity distribution of velocity profile form and Similarly you'll get this temperature distribution forming so close to the plate you'll get something called a boundary layer where really they're There is kind of a molecular scale phenomenon happening here, or a very small scale phenomenon happening here, where there is a more gradual change in temperature. But typically, we're going to look at the macro level, and we're going to be looking at the difference in the surface temperature as compared to the fluid temperature.

So in a lot of the problems that we'll solve, we'll be looking at the difference between these two physically separated points, the surface temperature and the fluid temperature. Because we're doing that, we'll use this algebraic temperature difference as our fluid driving force. Another way of saying that is we're essentially, in a lot of the problems that we're solving, I'm going to ignore what's happening here and just look at what's happening here and here, although we do get into more of the theory about what's happening in that boundary layer.

So Newton's law of cooling is a little bit simpler than Fourier's law. It's just this simple because it uses that simple algebraic temperature difference. Newton's law of cooling tells us that the flux is equal to h. h is called the convective heat transfer coefficient multiplied by the temperature difference, our surface temperature minus the bulk fluid temperature. So if we know h and we know these two temperatures, it's pretty easy to calculate our heat flux.

Thanks. The convection heat transfer coefficient has units of watts per meter squared per Kelvin. So what that's saying, notice that the Kelvin shows up in the denominator. So the higher driving force you get, the more... watts you will transport and also the bigger your surface area is for heat transfer the more heat you'll you will dissipate from the from whatever your surface is And you can see that shows up in this heat rate version.

So up here is the heat flux version, down here is the heat rate version. And all you do is you multiply the heat flux version by the surface area of your surface, and that gives you a total heat rate in terms of watts rather than a flux in terms of watts per meter squared. Notice that, as I showed in the thought experiment, that you can actually flip those temperatures, and you can have heat flowing the other way.

So you can have heat flowing from the hotter fluid into the colder surface, and that is pretty easily flipped. So that's where it's really important for you as an engineer solving heat transfer problems to know which form of this you should be using. And that comes usually by a fairly simple inspection of your system. However, it's very common for students to not take that extra time to really think about what's happening conceptually, and they may just be sort of taking a stab in the dark, but I promise you...

you if you follow this very simple systematic way of doing it, it's not too hard to figure out. So let's take a look with an example problem. How would we solve a convection problem? I'll just read through the problem. If my roof is at a uniform temperature of 35 degrees Celsius and the ambient temperature is 20 degrees Celsius, the surface area of my roof is 50 meters squared and the convective heat transfer coefficient is 10 watts per meter squared per Kelvin, how fast is my roof losing heat?

by convection to the environment. And then I want you to include the appropriate units in your answer. So if you notice, I included the roof area, so that should imply that we're using the total heat rate because we have an area. We can quantify how much heat is being lost over the entire roof rather than how much heat is being lost per meter squared for example.

Okay so just to make this simple I'm going to give my house kind of a simple one sided roof. Okay so we have our Here's our T sub s, our surface temperature. Out here in the bulk fluid away from that surface we have our bulk fluid temperature T infinity and we'll assume there's some kind of breeze blowing that is helping to remove heat from the roof out into the ambient air.

So we also have our H equal to 10 watts per meter squared per Kelvin. Okay, so this is a really simple plug-and-chug problem. We could use the flux form of this.

where we'd have the heat lost. So here the roof surface is hotter than the ambient, so it's going to be losing heat. So our flux could be the flux per unit area is equal to our heat transfer. or coefficient times our roof surface temperature minus T infinity.

So that would give us the flux version, and then we could take that flux and multiply that by our surface area. to get the total heat rate version. It's easier probably to just jump to the total heat rate version. Q equals HA times TS minus T infinity. And again, this is a really simple problem, but it's really important to master this conceptually and get really used to using this because this will just become a very small piece of more complex problems later.

So if we plug and chug, we get Q equals 10 watts per meters squared per Kelvin. And it's actually important to get familiar with the units of these different terms too. Times 50 square meters multiplied by our temperature difference. Our surface temperature is 35 minus 20, and because this is a temperature difference, a temperature difference in Kelvin is the same as a temperature difference in degrees Celsius.

So I'm just going to put in a Kelvin there. So this gives us a loss of 7,500, and stop and think what units this should be. And then we're also going to double check by very carefully canceling out our units.

So the meters squared cancel out, Kelvin cancels out, and we get a loss of 7,500 watts. So we defined this. We were asked, how fast is the roof losing heat? So we looked at, we assumed that the roof was going to be losing heat. So we assumed that this temperature was higher than this temperature, making this...

Delta T, a positive number, and because H will always be positive and our area will always be positive, we knew that our total heat rate would also be a positive number. So we'd be losing 7,500 watts of heat. And again, it's important to just...

conceptually get your head around that because you may not always know which temperature which a direction that driving force will be so you may have to just make some assumptions so let's go ahead and do another example problem very similar so the roof is at the uniform temperature of 20 the ambient temperature is at 30 the surface area of the roof is still 50 and the heat transfer coefficient is still 10 how fast is my roof losing heat by convection to the environment so again we have our roof Our surface temperature and our bulk fluid temperature. So our rate of heat loss, we apply Newton's law of cooling. Q equals HA times TS minus T infinity. So I've taken the approach that our roof is still going to be losing heat. The problem tells me my roof is losing heat.

So let's plug and chug here. So that is equal to 10 watts per square meter per Kelvin times 50 square meters. multiplied by our surface temperature is 20 and our ambient temperature is 30. And again, that temperature difference, even though the temperatures are given to us in degrees Celsius, the temperature difference can readily be switched from degrees C to Kelvin.

So that gives us now minus 5,000 watts. Okay, so now we have a negative flow of heat. And man, do we need to panic and freeze? freak out here? "and the answer is no. It's kind of a trick question, but this situation could arise many times as you're solving heat transfer problems. So we assume that the roof was losing heat, but we found out that it's losing minus 5 kilowatts of heat, which actually means although we assumed that heat was going out, it turns out that heat is actually... going in. So we we're losing minus 5000 watts of heat. Another way of saying that is that we're gaining so this would be let's say Q loss and this is Q gain would actually be gaining 5000 watts. watts of heat. So when you get to be solving more complex energy balance problems, this becomes very important, how you're dealing with these little sign changes. So if you were solving an energy balance problem and you wanted to know... what the final temperature of the inside of the house was, or something like that. You could make the same assumption that it's losing heat, but as long as you carefully apply the delta T in this way, you'd find out, well, it's not really losing heat, it's gaining heat, but that negative sign would save you and would make sure that you're still getting the energy balance right. So with Newton's law of cooling, it's really important. So the way I think about it is that if I have an out term from the surface, then I say Q is equal to HA times TS minus T infinity. Or if I'm calling this an in-term, I'm just assuming that the heat is coming in, and I can say Q equals HA times T infinity minus TS. And either one of those would be correct, and it doesn't actually matter if the net rate of heat is going out or in. As long as I... think about them in this way, my energy balance will turn out okay, because if you remember, the in-term is an additive term on your energy balance, and the out-term is a subtractive term on your energy balance. So that sine flip will take care of itself as long as you are able to lay out the equations this way. So while this seems pretty simple and trivial, it's actually a fairly important concept to grasp. All right, so convection. There will be a lot of different ways that convection happens. So It can happen because of forced airflow over a surface. There can be something called buoyancy-driven flow. This is actually called natural or free convection, where you don't necessarily need something like a fan to force a fluid through. You can actually have that fluid flow happening naturally because of the uneven heating of your surface relative to the bulk. Boiling is another way of sort of inducing motion in a fluid, so boiling creates convection. And similarly... Condensation does the same thing, and we'll cover each of those in detail as we get to the convection section of this course. Some notes on convection. So the driving force in convection is this algebraic temperature difference, or a delta T. So we express that as Ts minus T infinity, or T infinity minus Ts. and typically if it's an energy balance problem you would use if the if the heat were leaving the surface and you would use this form and call that energy out or if you assume that heat were coming into the surface then you would use this as your in term and it's just it's really important to carefully think through which direction the heat is flowing or even guessing and picking a direction that you think the heat is flowing so that you use the right form of this and treat that equation appropriately as either an out or an in term. Alright so the rate equation for convective heat transfer is Newton's law of cooling, a very important one to commit to memory. So with convection problems, you're often trying to find a single temperature rather than a profile. Remember with conduction, we were dealing with graphs of temperature versus x, and we were looking at these graphs. temperature profile and solving for the flux based on that profile. And I mentioned that we will go in the reverse order often. Whereas with convection, we're typically looking at sort of a sudden jump in temperature, where this might be the surface, this might be the solid surface, and this would be the fluid. So we're treating convection as if there's just this sudden jump in temperature, delta T. All right, so and I mentioned that in my bullet point, that assumption will eventually go away as we get deeper into convection. It becomes, convection is actually extremely complicated, but we'll try and treat it as simply as we can. So convection is introduced, we're introducing that now and it's actually important even as we get into conduction to understand this basic Newton's law of cooling form because we're going to use this at the boundaries of conduction problems. So here at this solid surface we might care about the temperature profile here by conduction but then at the surface or at the boundary we'll just look at this this sudden delta T between the edge of the surface or the edge of the solid and the fluid. So there will be plenty of problems where different modes of heat transfer are combined.

Transcript for:Understanding Convection in Heat Transfer

Transcript for:
Understanding Convection in Heat Transfer