CONVENCION , INTERCAMBIO DE CALOR.ppt

Lecture Series 1Convection Heat Transfer

John Richter--Staff EngineerOxford, OHThe Thermal Management Team of AT&AKevin Parker--Staff EngineerRaleigh, NCStuart Brogden--TMT SupervisorNashville, TNTMT VisionFor each design center to have the people, skills, and tools to effectively utilize thermal analysis within the design process to optimize products and reduce design cycle time and cost. Effectively the vision is to use thermal analysis to help make Schneider 2000+ a reality!

Questions You Might Be Asking?What is convection?How do I calculate a heat transfer coefficient?Is turbulent flow better than laminar flow?What is Forced Convection? What is Natural Convection?What IN THE WORLD does an understanding of convection have to do with =SE= Design?

Lecture OutlineI. Fundamentals of ConvectionII. Forced vs. Natural Convection III. Practical convection notes for designing =SE= equipmentIV. Bus bar example problemV. Engineering calculatorsVI. Final questions

Convection Heat Transfer: Convection is at the interface between fluid mechanics and heat transfer. Convection is the rate of heat transfer due to fluid diffusion (random molecular motion) and bulk fluid motion (advection) over a surface. q is the heat flux or heat flow per unit area (W/m2).Fundamentals h is the heat transfer coefficient or film coefficient and has the units (W/m2-K) or (W/in2-C). h is a function of the temperature, geometry, velocity, and fluid properties. Tsurface is the temperature of the surface (K or oC). T is the temperature of the bulk or freestream fluid (K or oC).

FundamentalsL. Prandtl in 1904 first recognized that the flow about a solid body can be divided into two regions: a very thin layer in the neighborhood of the body (boundary layer) where friction plays an essential role and the remaining region outside this layer, where friction may be neglected.

Fundamentals Starting from the leading edge the BL grows (y increases) as you move along the plate. Therefore the heat flux (q) and the heat transfer coefficient (h) must decrease. In other words, the thicker the BL the lower the heat transfer.

FundamentalsLaminar vs. TurbulentFirst step in any convection problem is to determine whether the BL is laminar or turbulent. Turbulent flow enhances the transfer of momentum and energy and therefore increases the heat transfer coefficient. 3 Regions: laminar sublayer, buffer layer, and turbulent layer. Flat plate - Reynolds No., Rex = uxc/ = 5 x 105

FundamentalsThe fundamental problem of convection is to calculate h

Average Heat Transfer coefficient Calculate from BL Theory Calculate from empirical correlations Calculate by numerical simulation or physical experiments

Method of Calculating Heat Transfer Coefficients1. Recognize the flow geometry, determine characteristic length (L)2. Calculate Tf by Tf = (Ts + T )/ 2 3. Evaluate the fluid properties at Tf4. Calculate Pr = Cp /k, Re = uL/ or Ra = g(Ts-T)PrL3/25. Choose the appropriate correlation for h

Forced Convection: Flat Plate - Laminar (ReL < 5x105 ) : Flat Plate - Turbulent (ReL > 5x105 ): Convection of heat due to fluid motion caused by external means (fan, pump, wind) Typical Range: 15-120 W/m2K for air Characteristic Length (L) is the plate lengthFlat Plate Empirical Equations

Natural (Free) Convection: Convection of heat due to fluid motion caused by buoyant forces (hot air rises) Typical range: 1-15 W/m2K Characteristic Length (L) is plate lengthVertical Plate Empirical EquationsLaminar and Turbulent (10-1 < RaL < 1012, transition at 109):

Practical Considerations For Design1. Short enclosures produce little draft / Tall enclosures produce good draft.2. Avoid clutter around critical parts (driving force for flow is small)4. Radiation heat transfer is often critical!5. High altitudes will reduce air density and reduce heat transfer.6. Turbulent flow enhances heat transfer. Rough surfaces can help transition flow.Natural (Free) Convection: 7. Orient critical parts to maximize convection (ex. Vertical vs. Horizontal).8. Baffles can be used to effectively direct air to hot components.3. Locate exit vents at top (preferably on the top) and inlets at bottom1. Forced convection flow should be in same direction as buoyancy flow.2. A Blowing system produces more turbulence, a positive pressure inside enclosure, and lower operation temperatures for fans.3. Exhausting system allows more control of inlet air to direct to critical parts and can eliminate recirculation of hot air in the enclosure.4. High altitudes reduce fan effectiveness.5. Incoming flow direction and location can cause a short circuit flow to the exhaust. Can be fixed with baffles.Forced Convection:

Example Problem: Steady-stateAdiabatic at the endsNegligible Q at thin edgesIgnore radiationCu matl negligible temp gradient through thicknessHow much current can this bus bar carry?Does the amount of current depend on the length?NO?Surf 1Surf 2

Calculating h: Tf = (Ts + T )/ 2= 77.5C = 350.7KRa = 3.232 x 108 havg= 5.1 W/m2KRe = 72,849 havg= 9.4 W/m2K

Example Problem: Steady-stateAdiabatic at the endsNegligible q at top and botIgnore radiationCu matl negligible temp gradient through thicknessHow much current can this bus bar carry?Does the amount of current depend on the length?Inatl conv = 631 Ahnatl conv = 5.1 W/m2Khforced conv= 9.4 W/m2KCu@105C = 2.31x10-8 -mIforced conv= 856 A

Engineering CalculatorsAdvanced Technology & Analytics Home PageAnalytics Home PageEng. Reference MaterialNatural Convection CalculatorBusbar Rating CalculatorOthers

Questions?

Hello and welcome to Lecture Series 1: Convection Heat Transfer.This presentation will be given by Kevin Parker and John Richter of the Thermal Management Team of AT&A. The purpose of the Thermal Lecture Series is to equip thermal engineers at each design center with the theoretical skills to effectively conduct thermal analysis. We believe that a theoretical understanding of the physics involved in heat transfer is the critical foundation that all thermal analysis must be built upon.At this point, you might have some questions about what you will learn in this presentation. Questions like, What is convection? What is forced convection and what is natural convection? Is turbulent flow better than laminar flow? How do I calculate a heat transfer coefficient? And what in the world does an understanding of convection have to do with Schneider Electric design? Well hopefully over the course of this presentation, these questions and maybe some others will be answered.This is an overview of what we plan to cover in this presentation. First, Kevin Parker will go over some fundamentals of convection heat transfer. Then I will take you through the differences between Forced and Natural Convection. Next we will discuss the practical applications of these ideas to the design of Schneider Electric equipment. We will look at a bus bar example problem. Then we will discuss some calculators on the intranet that can help you in your thermal analysis.Convection is at the interface between fluid mechanics and heat transfer. We have to understand both boundary layer fluid mechanics and conduction heat transfer to understand convection. Convection can be defined as the rate of heat transfer due to fluid diffusion (random molecular motion) and bulk fluid motion (advection) over a surface.

Newtons Law of cooling defined here is the fundamental equation used to calculate the rate of convective heat transfer. This law states that the heat flux from a surface is proportional to the temperature difference between the surface and the surrounding fluid.The heat transfer coefficient is a function of the temperature of the surface, the geometry of the surface, the velocity of the fluid flowing over the surface and the fluid properties.

Ludwig Prandtl in 1904 first recognized that the flow about a solid body can be divided into two regions: a very thin layer in the neighborhood of the body (boundary layer) where friction plays an essential role and the remaining region outside this layer, where friction may be neglected.

There is both a velocity boundary layer and a thermal boundary layer. In the velocity boundary layer the fluid is stuck to the wall at y=0. This is known as a no-slip condition. The velocity in this layer therefore goes from zero at the wall to nearly the freestream value at the edge of the boundary layer. Since the fluid is stuck to the wall at y=0 the tempature of the fluid is equal to the temperature of the wall and heat is transferred by conduction. There is no fluid motion in the direction of heat flow.

For a gas, i.e. air, the thermal boundary layer is approximately equal to the velocity boundary layer.

So at the wall, Fouriers law can be equated to Newtons law of cooling. From this equality the heat transfer coefficient can be determined.

Lets look at an example to get a better understanding of this equation. Here we have a copper busbar. We want to orient this bar so that we will get the highest cooling.

So examining the equation above we can see that the BL grows (y increases) as you move along the plate. Therefore the heat flux (q) and the heat transfer coefficient (h) must decrease. In other words, the thicker the BL the lower the heat transfer.

Therefore if the busbar is in the verical orientation then we will have a larger boundary layer and a smaller heat transfer coefficient then if the busbar is in the horizontal orientation.

First step in any convection problem is to determine whether the BL is laminar or turbulent.

Lets look at flow over a flat plate. Looking at the boundary layer in the vertical direction we can define 3 distinct regions. The laminar sublayer begins at the leading edge and extends the full lenght of the plate. The buffer layer is a mixture of laminar and turbulent flow that begins at the transition point and extends the remaining length of the plate. The last layer is the turbulent layer that begins after the transition point and extends the remaining length of the plate.

The flat plate critical Reynolds number occurs at 5 x 105, but the actual onset of turbulence depends on the shape of the leading edge, the roughness of the plate and the presence of upstream noise.

Turbulent flow enhances the transfer of momentum and energy and therefore increases the heat transfer coefficient.

The fundamental problem of convection is to find the heat transfer coefficient h. To do this we can calculate h from boundary layer theory, calculate h from empirical correlations, find h by performing numerical simulation or find h by doing physical experiments.

In engineering, we often use the average heat transfer coefficient as apposed to the local heat transfer coefficient. The average heat transfer coefficient can be found by integrating the local heat transfer cofficient over the entire surface area and then dividing by the surface area.

Now that Kevin has shown you the fundamentals of convection, I want to give you a clear methodology for calculating the heat transfer coefficient which is so critical for a convection heat transfer problems. 1. The first step is to recognize the flow geometry. Is the flow over a cylinder or a flat plate or a sphere? Is the surface at a constant temperature or a constant heat flux? The flow geometry determines the characteristic length which is designated as L in the equations for calculating the heat transfer coefficient. 2. The second step is to calculate the Film Temperature. The Film Temperature is the average temperature between the Surface Temperature and the Bulk Fluid Temperature.3. Then this Film Temperature is used to determine the fluid properties to use in the calculation. These properties are the specific heat, viscosity, kinematic viscosity, and the density (note that the Prandtl number can often be obtained also when getting these values)4. After obtaining these properties, you can calculate the dimensionless parameters important for determining h. For forced convection these parameters are the Prandtl number and the Reynolds number and for natural convection they are the Prandtl number and the Rayleigh number.5. Then the last step is use these parameters to determine whether to use the laminar or turbulent correlation. The flow is determined to be laminar or turbulent based on the Reynolds number (forced convection) or the Rayleigh number (natural convection).Now that we have looked at a general method for calculating the heat transfer coefficient, lets look at different flow conditions that are common to convection heat transfer.First of all lets consider the Forced Convection application over a flat plate. For hand calculations, a flat plate assumption is often good enough to get approximate results. If the geometry is much more complicated than this, then I would use a numerical simulation tool like Icepak. Forced Convection is defined as the transfer of heat due to fluid motion over a surface caused by some external source. This source could be a fan or a pump or wind. For air, the typical range of the heat transfer coefficient for forced convection is 15 to 120 Watts per meter squared degrees Kelvin. Knowing the approximate range is important for catching calculation errors. If you would calculate an h of say 1000 for forced air, then you should definitely question your results. As I mentioned before, the characteristic length is an important parameter for determining h. For a flat plate, this is the length of the plate in the direction of flow. Here is the correlation for laminar flow. Notice that this is for Reynolds numbers less than 500,000. Also notice once again that this is for calculating the average heat transfer coefficient. The dimensionless parameter on the left is referred to as the Nusselt number which is equal to the heat transfer coefficient times the characteristic length divided by the fluid conductivity. This is a function of two dimensionless parameters, the Reynolds number and the Prandtl number. The Reynolds number is equal to the fluid density times the bulk fluid velocity times the characteristic length divided by the viscosity. The Prandtl number can usually be obtained in a heat transfer reference book when looking up the fluid properties at the film temperature. Now this is the Turbulent correlation which is good for Reynolds numbers greater than 500,000. You can observe from these equations that you will calculate higher heat transfer coefficients for turbulent flow which means better heat transfer. This is due to the mixing effect involved in turbulent flow.Now lets look at natural convection which is also called free convection. We will look at a common situation of natural convection along a vertical flat plate. Natural convection is defined as the transfer of heat due to fluid motion caused by buoyancy forces. As everyone knows, hot air rises. As the air heats up, the density of the air decreases which in turn causes an upward movement of air. The heat transfer due to natural convection is typically much less than forced convection with typical values of around 1-15 Watts per meter squared degree Kelvin.The common dimensionless parameter used to describe this flow is the Rayleigh number which is equal to this equation. g is the gravitational acceleration. Beta is the expansion coefficient of the fluid and is equal to the inverse of the temperature in degrees Kelvin for ideal gases. The temperature difference is between the bulk fluid and the surface temperature. There is also the Prandtl number, the characteristic length cubed, divided by the kinematic viscosity. The kinematic viscosity is different from the viscosity and is equal to the viscosity, mu, divided by the fluid density. Here is the natural convection correlation for a vertical flat plate. It is good for laminar and turbulent conditions and for Rayleigh numbers between 1 and 10 trillion.Kevin has given us some of the fundamentals of convection heat transfer, and we have looked at calculating the heat transfer coefficient for some basic situations. Now lets look at some practical applications of what we have learned for designing Schneider Electric equipment. First lets look at natural convection applications:1. When designing enclosures, remember to try and take advantage of the chimney effect. This is done by making the enclosure taller and narrower.2. Try to avoid clutter below or above your critical hot spots. This tends to block the flow and increase the part temperature. With natural convection, the driving force is very small therefore any type of blockage greatly reduces the amount of heat transfer.3. As a general rule of thumb, try to locate exit vents as near to the top as possible. Preferably on the top surface of your enclosure. If the air has to turn to exit the enclosure than this will increase the flow resistance of your system and therefore increase your temperatures. Also try to locate the inlet vents as low as possible to bring in the coolest air at the bottom and allow the air to accelerate through the system.4. Remember that radiation effects are very critical for these types of problems and cannot be ignored. Increasing the radiation from your hot spots can dramatically reduce your temperature. 5. High altitudes reduce the ability for heat transfer because the air density is less. This will result in higher temperatures.6. As I mentioned earlier, turbulent flow enhances heat transfer. Having rough surfaces can help to cause the flow to become turbulent.7. Keep in mind the orientation of your parts in your system. Kevin mentioned this earlier how the orientation can improve your heat transfer.8. Finally remember that you can use baffles to direct flow to hot spots. 1. For forced convection, try to design your flow system to be in the same direction as the buoyancy flow. For example, you would want your fans to blow up through a heat sink and not down. 2. There are two types of flow systems--A Blowing System and An Exhausting System. A blowing system is one where the fans blow into the enclosure. An exhausting system is one where the fans blow out of the enclosure. Which one is best, depends on the application. A blowing system produces more turbulence inside your enclosure for better heat transfer. It produces a positive pressure inside the enclosure which is important for keeping dust out, and has cooler air passing through the fans for longer fan life.3. Exhausting systems allow you to have more control of the incoming air so that you can direct it to critical locations, and can help to eliminate recirculation of hot air in the system.4. High altitudes reduce fan effectiveness. 5. Finally the location of the incoming air vent to the fan and the direction of that air coming in can cause a flow short circuit. This is when the incoming air comes into the enclosure and flows directly out without touching the hot parts. Using baffles on the inlet is a cost effective way to eliminate this problem.Now lets look at an example problem. I think this will help you understand what we have looked at so far. We will look at designing a bus bar with the parameters as shown.There are two basic design questions that we want to look at. First, how much current can this bar carry? And does the amount of current depend on the length? Here is a list of our assumptions for this problem: steady-state conditions, no heat flow across the ends, we will ignore any heat flow from the thin edges and also radiation effects. We will also assume that there will be a negligible temperature gradient across the thickness which means that the two sides are at the same temperature. Now we will begin with the energy rate equation. Because it is steady-state, the change in stored energy is zero. There is no energy coming in. So our equation reduces to this. The energy being generated is from the current which is I-squared R. And the energy leaving the bus bar is by convection from surface one and two. Now we can input the relationship for convection heat transfer and simplify because surface 1 and 2 are the same. With that we can input a value for the resistance as shown in the bottom left corner. This is a relationship for calculating the resistivity, rho, as a function of temperature. Rho not is the resistivity at 20 degrees Celsius and alpha is the temperature coefficient of resistivity for copper. These values can be looked up in reference material. Now we can solve for the current. One of the things that I have found very helpful is to not input values until the very end. This helps you to be able to see relationships between variables. For instance, we can see from this equation that the current is directly proportional to the bus bar width. On the other hand the current is proportional to the square root of the thickness. This means that increasing the bus bar width will have a greater effect on the current amount than increasing the thickness. Back to our question, is the amount of current dependent on the length of the bar? Well lets continue and see, but it appears at this point that it does not.Now we can calculate the film temperature so that we can look up the air properties and plug them into the natural convection correlation. When we do this we calculate a Rayleigh number of 3.2 X 10^8 and a heat transfer coefficient of about 5. Now just for comparison lets calculate a laminar forced convection heat transfer coefficient. We will assume a fluid speed fo 3 meters per second. We get a Reynolds number of 73,000 and a heat transfer coefficient of about 9 watts per meter K. One thing to notice is that the length of the bus bar can be found in these equations for the heat transfer coefficient. There for the length still has an effect on the current.Now we can input these heat transfer coefficients into our original equation. For the natural convection situation, we get a current of 631 amps. And for the forced convection situation we get a current of 856 amps. Notice that even though the forced convection heat transfer coefficient was almost twice the natural convection, you did not get twice the current. You can see this in the equation above that the current is proportional to the square root of the heat transfer coefficient.The TMT intranet site contains a lot of information will be helpful to you for conducting thermal analysis. There is special section called on-line calculators that has thermal analysis calculators. One of these calculators is for natural convection heat transfer coefficients along a vertical wall. Another calculator is a bus bar rating calculator. I recommend that you visit this site and review the information that is available.This concludes the Lecture Series 1 on Convection Heat Transfer. If you have any questions please call the Thermal Management Team of the AT&A group. We would like to spend some time discussing this presentation with you and answering any questions that you might have.