Week 7: radiative heat transfer
Cengel Chapters 12/13, Incropera Chapters 12/13, Welty et al. Chapter 23
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COVERED IN THIS MODULE
· Basic (really basic!) introduction to radiation as a HT mode
· Radiation emitted from a surface (surface EB)
· View factors
· Radiation exchange between surfaces
· Guidelines on how to use optical data for materials (reflectivity, transmissivity, absorptivity, emissivity) under strict assumptions
NOT COVERED IN THIS MODULE – Cengel Sections 12-1 to 12-4
· Fundamentals of radiant energy – how radiative energy/intensity varies with wavelength, temperature, angles, and how this relates to:
· Origin of radiative properties (emissivity , transmissivity , absorptivity , reflectivity ) as a function of temp, wavelength, etc
Learning objectives: By the end of this module you should be able to:
· Calculate the net rate of radiative heat transfer from a black body surface and use this information in multi-modal heat transfer problems (involving a combination of conduction, convection, radiation)
· Calculate the net rate of radiative heat transfer from an opaque, diffuse, grey body surface and use this information in multi-modal heat transfer problems (involving a combination of conduction, convection, radiation)
· Use view factors to calculate the amount of radiative heat transfer between two or more objects.
WHAT IS RADIATION, AND WHAT TYPE OF RADIATION ARE WE INTERESTED IN FOR HEAT TRANSFER CALCULATIONS?
Radiation could be defined as:
Radiation is distinct from conduction and convection in the following ways:
The electromagnetic wave spectrum
Equations that help describe radiation:
In terms of heat transfer we are interested in:
HOW ISe RADIATION DIFFERENT TO OTHER MODES OF HEAT TRANSFER?
All objects emit radiation (emissive power), but only very hot objects emit enough radiation for qrad to be considered. For example, the radiation emitted by a blackbody (see below for definition) to the atmosphere is determined by the Stefan-Boltzmann equation:
The value of the Boltzmann constant () is 5.67 x 10-8 (W/(m2K4). Therefore, for to be a significant flux, must be substantial. As an example, if we compare the heat transfer by radiation to heat transfeeby convection (assuming h = 10 W/(m2K), Tinf = 290K), we can immediately see that radiation cannot be neglected.
qrad (W/m2)
qconv (W/m2)
Radiation can occur in a vacuum – i.e. does not need a fluid (convection) or a solid (conduction) to propagate
The radiative properties of an object are dependent on the wavelength of irradiation, the angle of incidence, and the temperature of the object. HOWEVER – in this unit, we do not go into this in detail (Chap 12 Cengel) but generally provide this information as constants for the problems – this is acceptable for the common assumptions of gray, diffuse surfaces.
Finally, at a surface, the net rate of radiative heat transfer is determined by an energy balance between incident, reflected, absorbed, transmitted and emissive radiation terms. This is VERY DIFFERENT to conduction/convection.
Our goal in solving problems involving radiation is usually to determine the net rate of radiative heat transfer emitted from an object, as part of a multi-modal energy balance. We can also evaluate the net rate of radiative heat transfer between objects.
For example, think about the following situations and draw diagrams and write energy balances (involving qcond, qconv, qrad) for the following:
· BBQ plate
· Saucepan on a stove containing hot water
· Ice rink in summer
CALCULATING NET RADIATIVE HEAT TRANSFER FROM A SURFACE
When radiant energy hits a surface, there is a “mini energy balance” set up on that surface. We thus need to define several new terms which identify the incident energy or irradiation (G), the emissive power of the object itself (E), and the total amount of energy leaving the object, termed the radiosity (J – a combination of reflected incident energy and emissive power).
EMISSIVE POWER (E (W)) – radiation emitted by all objects (people, furnaces, paper clips). For “blackbody” objects that absorb ALL incident radiation (no reflection or transmission; = = 0, and = 1) and whose emissivity (; dependent on wavelength of light, ) is equal to 1, the emissive power (Eb) is described by the Stefan-Boltzmann Law:
Therefore, for a “real” object (E), whose emissivity () is <1 where Eb is the emissive power of a black body, is the Boltzmann constant (5.67x10-8 W/(m2K4)), and Ts is the surface temperature. SIMPLIFYING RADIATIVE PROPERTIES The radiative properties of an object (, , , ) are actually quite complex, in that they are functions of the wavelength () and angle () of the incident light. For each type of material, there exists graphs that show plots of each radiative property as a function of and . As you can imagine, this can get quite tricky!! Luckily, we can make some assumptions about some objects that (a) remove the dependency of the property on either and/or ; and (b) set the value of some properties to 0 or 1. Assumption about an object Properties Real object (,) constant “Blackbody” (,) = 1 (therefore = = 0) (e.g. wood, bricks, metals) “Diffuse” reflector (as opposed to spectral (or mirror-like) reflector); ie independent of ) () = constant, not dependent on angle Note that this is typical of rough objects, but not for smooth/polished surfaces. () = constant, not dependent on wavelength “Diffuse and gray” (,) = constant “Opaque, diffuse and gray” (,) = constant Another important simplification is known as “Kirschoff’s Law”, which states that at a constant atmospheric temperature, a small body surrounded by a large isothermal enclosure (e.g. the atmosphere!!) satisfies the following logic. The irradiance incident on any part of the object equals the radiation emitted by a blackbody of temperature T: and the radiation absorbed by the object per unit surface area is: The radiation emitted by the body is: Therefore, the radiation energy balance comes down to: which leaves us with the extremely satisfying conclusion: Therefore, for an opaque, diffuse, gray surface, it now follows that: and therefore Therefore, knowing only 1 of the radiative properties of the object, these assumptions allow us to solve for qrad IRRADIATION OR IRRADIANCE OR INCIDENT RADIATION (G (W)): this is the radiation that hits a surface from an external source, be it from the atmosphere, another object, the sun, a fire, etc. This incoming radiation must either be absorbed, reflected or transmitted, satisfying the equations (noting of course that G can equal 0!): The absorbance (), reflectance (), and transmittance (), are dimensionless properties unique to each and every object, which depend on the wavelength of the light (in this unit we will consider them “constant” but this is not true; for more information please see Incropera Sections 12.3-12.5). In a calculation, you can gather data about the properties, or you can make assumptions (for e.g. “no transmission or absorbance; therefore =1”) RADIOSITY (J (W)): is the total energy leaving the surface, which is: Our goal in radiation problems involving one object is to determine the net rate of radiant heat transfer, qrad (W), which is the difference between G and J, or: Therefore, in order to determine qrad, we need to know the irradiance (G), the properties of the object ( and ), and the surface temperature. If there is no specific irradiance source aside form the surroundings, it is reasonable to approximate the surroundings as a blackbody with equivalent emissivity as the object, . If we re-arrange the equation above we can see: because for an opaque object. If the object is gray and diffuse (see section below on simplifications), we know that , so This is a very common equation used to determine the net rate of radiative heat transfer from an object to the surroundings in the absence of any specific irradiance. However it is also very common, and possibly more correct, to think about the net radiation in terms of the difference between the emissive power of the object, and the radiative energy absorbed by the sun and the atmosphere separately. In terms of the solar irradiance: where GD is the “direct” solar irradiance, and Gd is the “diffuse” solar irradiance (both need to be specified for a certain location, time of day, season) and is the angle of incidence (measured normal to the surface). For the atmosphere, the irradiance is calculated as: where Tsky varies from ~230K for cold, clear-sky conditions through to ~285K for warm, cloudy-sky conditions. So then if we want to determine the net rate of radiative transfer to a surface exposed to solar and atmospheric radiation: Introducing the solar absorptivity, s, and applying Kirschoff’s Law such that at the object surface we know = , Noting again, that values such as s, GD, Gd, Tsky, are all available as tables/charts under specific conditions. EXAMPLE: Consider an opaque horizontal plate that is well insulated on the edges and lower surface. The plate is uniformly irradiated from above while air at T = 300K flows over the surface providing steady convection with h = 40 W/(m2K). Under steady state conditions, the surface has a radiosity of 4000 W/m2, and the plate temperature is maintained at 350K. If the total absorptivity is 0.40 of the plate, determine: (a) the irradiation on the plate (b) the total reflectivity of the plate (c) the emissive power of the plate (d) the total emissivity of the plate (Answers: 6000 W/m2, 0.6, 400 W/m2, 0.47) RADIATIVE EXCHANGE BETWEEN OBJECTS – INTRODUCING VIEW FACTORS (Fij) When considering the net rate of transfer leaving a single body, we were not interested in the specific direction of the heat transfer (except that it was AWAY from the object), nor the fraction of the energy that is intercepted by another object. Consider 2 objects separated in space Q: How do we determine qij? There are several rules that allow us to use view factors relatively easily for a range of different objects: 1) Reciprocity: 2) Summation: 3) Superposition: Finally, given the definition of the view factor, you can also have radiant heat transfer from one side of object i to another side of the SAME object i, therefore: Fii and Fjj are NOT necessarily equal to 0!!! There are tables of view factors for common geometries (e.g. 2 parallel plates, 2 perpendicular plates meeting to form an edge) and also graphs that plot view factor as a function of the specific properties of these geometries (e.g. Fij as a function of the size of the parallel plates to the distance between them, etc). See Incropera Section 13.1 for tables, graphs, and examples. Now we can use view factors in combination with net radiative transfer equations to determine the rate of heat transferred from one object to another. RADIANT HEAT TRANSFER – BETWEEN BLACK SURFACES Applying the reciprocity rule () And regardless of the number of objects we have, the net rate of heat transfer from object i: Therefore, if we can determine the view factors, we can work out the net rate of heat transfer from one object to another, or the net rate from one object in total. RADIANT HEAT TRANSFER FROM DIFFUSE, GRAY SURFACES For the black body case, we could simply work with the emissive power leaving a surface (E, or more specifically, Ebi), but now we deal with the radiosity, which is the combination of emissive power AND reflected incident radiation. And if we apply Kirschoff’s Law OK, so now we can set up a similar equation for rate of heat transfer: If we then substitute the for Gi, we get: We can re-arrange this into a similar form to Ohm’s law, to identify the surface resistance (Ri) to radiation: So now, the direction of radiation is TO the surface if Ji > Ebi and FROM the surface if Ji < Ebi. Note, for the temperature to remain constant, there must be some other heat transfer mode operating (conduction, convection etc). Furthermore, the surface resistance to radiation for a black body is 0 since , and hence, and , because the surface LOSES as much energy as it GAINS. So we can solve problems related to radiative transfer between 2 black bodies as resistance in series type problems. RADIATIVE TRANSFER BETWEEN ANY TWO SURFACES Applying reciprocity rule: Again, in analogy to Ohm’s Law, we can state this as: So again, we can solve using a resistances in series analogy. Also and CRUCIALLY METHODS OF SOLVING RADIATION PROBLEMS BETWEEN OBJECTS So now there are really 2 approaches to solving radiation problems involving two objects, depending on whether or not it is the q or the surface temp Ti that is specified in the problem. For surfaces with specific q: For surfaces with specified Ti: EXAMPLE: A furnace cavity, in the form of a cylinder of 75-mm diameter and 150-mm length, is open at one end to the surroundings that are at 27C. The sides and bottom may be approximated as black bodies, are heated electrically, are well insulated, and are maintained at 1350C and 1650C respectively. How much power is required to maintain the furnace conditions? EXAMPLE: In manufacturing, the special coating on a curved solar absorber surface of area A = 15 m2 is cured by exposing it to an infrared heater of width W = 1 m. The absorber and heater are each of length L = 10 m and are separated by a distance of H = 1 m. The heater is at T = 1000K and has an emissivity of 0.9, while the absorber is at T = 600K and has an emissivity of 0.5. The system is in a large room whose walls are at 300K. What is the net rate of heat transfer to the absorber surface? 程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com