Document Sample

Heat transfer and thermal radiation modelling HEAT TRANSFER AND THERMAL MODELLING ................................................................................ 1 Thermal modelling approaches ................................................................................................................. 2 Heat transfer modes and the heat equation ............................................................................................... 2 Thermal conductivities and other thermo-physical properties of materials .......................................... 4 Thermal inertia and energy storage ....................................................................................................... 6 MODELLING THERMAL RADIATION ................................................................................................... 7 Black-body radiation ................................................................................................................................. 7 Real bodies: interface ................................................................................................................................ 9 Emissivity............................................................................................................................................ 10 Absorptance ........................................................................................................................................ 10 Reflectance .......................................................................................................................................... 10 Transmittance ...................................................................................................................................... 11 Real bodies: bulk ..................................................................................................................................... 11 Absorptance and transmittance ........................................................................................................... 11 Scattering ............................................................................................................................................ 11 Measuring thermal radiation ................................................................................................................... 12 Infrared detectors ................................................................................................................................ 12 Bolometers and micro-bolometers ...................................................................................................... 14 Measuring thermo-optical properties .................................................................................................. 15 IR windows ......................................................................................................................................... 15 Spectral and directional modelling ......................................................................................................... 19 Two-spectral-band model ................................................................................................................... 19 Radiation emerging from a surface ......................................................................................................... 20 Radiation from a small patch to another small patch. View factors ................................................... 21 Radiation between a small patch and a large plate ............................................................................. 22 Radiation between a sphere and a small patch .................................................................................... 25 Radiative coupling in general: thermo-optical and directional effects ................................................... 26 Thermo-optical effects on simple geometries ..................................................................................... 27 View factor algebra, relevant values and exercises ............................................................................ 28 HEAT TRANSFER AND THERMAL MODELLING After a previous survey on the influence of spacecraft missions and the space environment on spacecraft thermal problems, we proceed now to a general survey of heat transfer modelling, focused on spacecraft thermal control (a more general heat transfer revision can be found aside). Thermal problems are mathematically stated as a set of restrictions (implicitly assumed data and equations, the expertise, and explicitly given data, the statement) that the sought solution must verify. It must be always kept in mind that both, the implicit equations (algebraic, differential, or integral) and the pertinent boundary conditions (fixed, or inequalities) given in the statement, are subjected to uncertainties coming from the assumed geometry, assumed material properties, assumed external interactions, etc. In this respect, it is not true that numerical methods are just approximations to the exact differential equations; all models are approximations to reality, and there is neither an exact model, nor an exact solution to a physical problem; one can just claim to be accurate enough to the envisaged purpose. A science is a set of concepts and their relations. Good notation makes concepts more clear, and helps in the developments. Unfortunately, standard heat transfer notation is not universally followed, not only on symbols but in naming too; e.g. for thermo-optical concepts three different choices can be found in the literature: A. Suffix -ivity/-ance may refer to intensive / extensive properties, as for resistivity / resistance. B. Suffix -ivity/-ance may refer to own / environment-dependent properties; e.g. emissivity (own) / absorptance (depends on oncoming radiation). This is the choice followed here. C. Suffix -ivity/-ance may refer to theoretical / practical values; e.g. emissivity of pure aluminium / emittance of a given aluminium sample. Thermal modelling approaches A model (from Latin modulus, measure) is a representation of reality that retains its salient features. The first task is to identify the system under study. Modelling usually implies approximating the real geometry to an ideal geometry (assuming perfect planar, cylindrical or spherical surfaces, or a set of points, a given interpolation function, and its domain), approximating material properties (constant values, isotropic values, reference material values, extrapolated values), and approximating the heat transfer equations (neglecting some contributions, linearising some terms, assuming a continuum media, assuming a discretization, etc.). Modelling material properties introduces uncertainties because density, thermal conductivity, thermal capacity, emissivity, and so on, depend on the base materials, their impurity contents, bulk and surface treatments applied, actual temperatures, the effects of aging, etc. Most of the times, materials properties are modelled as uniform in space and constant in time for each material, but, the worthiness of this model and the right selection of the constant-property values, requires insight. Heat transfer modes and the heat equation Heat transfer is the relaxation process that tends to do away with temperature gradients in isolated systems (recall that within them T→0), but systems are often kept out of equilibrium by imposed boundary conditions. Heat transfer tends to change the local thermal state according to the energy balance, which for a closed system says that heat, Q (i.e. the flow of thermal energy from the surroundings into the system, driven by thermal non-equilibrium not related to work or the flow of matter), equals the increase in stored energy, E, minus the flow of work, W; which, for the typical case of a perfect incompressible substance (PIS, i.e. constant thermal capacity, c, and density, ) without energy dissipation („non-dis‟), it reduces to: What is heat? (≡heat flow) Q≡EW=E+pdVWdis=HVdpWdis=mcT|PIS,non-dis (1) Notice that heat implies a flow, and thus 'heat flow' is a redundancy (the same as for work flow). Further notice that heat always refers to heat transfer through an impermeable frontier, i.e. the former equation is only valid for closed systems. The First Law applied to a regular interface implies that the heat loss by a system must pass integrally to another system, and the Second Law means that the hotter system gives off heat while the colder one takes it. In Thermodynamics one refers sometimes to „heat in an isothermal process‟, but this is a formal limit for small gradients and large periods. Here in Heat Transfer the interest is not in heat flow Q (named just heat, or heat quantity), but on heat-flow-rate Q =dQ/dt (named just heat rate, because the 'flow' characteristic is inherent to the concept of heat, contrary for instance to the concept of mass, to which two possible 'speeds' can be ascribed: mass rate of change, and mass flow rate). Heat rate, thence, is energy flow rate without work, or enthalpy flow rate at constant pressure without frictional work, i.e.: What is heat flux? (≡heat flow rate) dQ mc dT Q dt dt KAT PIS,non-dis (2) where the global heat transfer coefficient K (associated to a transfer area A and to the average temperature jump T between the system and the surroundings), is defined by the former equation; the inverse of K is named global heat resistance coefficient M≡1/K. Notice that this is the recommended nomenclature under the SI, with G=KA being the global transmittance and R=1/G the global resistance, although U has been used a lot instead of K, and R instead of M. Notice that heat (related to a path integral in a closed control volume in thermodynamics) has the positive sign when it enters the system, but heat flux, related to a control area, cannot be ascribed a definite sign until we select one side. In most heat-transfer problems, it is undesirable to ascribe a single average temperature to the system, and thus a local formulation must be used, defining the heat flow-rate density (or simply heat flux) as q dQ dA . According to the corresponding physical transport phenomena, heat flux can be related to temperature difference between the system and the environment in the classical three modes of conduction, convection, and radiation: What is heat flux density (≈heat flux)? conduction q k T (3) q K T convection q h T T 4 4 radiation q T T0 These three heat-flux models can also be viewed as: heat transfer within materials (conduction, Fourier‟s law), heat transfer within fluids (convection, Newton‟s law of cooling), and heat transfer through empty space (radiation, Stefan-Boltzmann‟s law). An important point is the non-linear temperature-dependence of radiation heat transfer, what forces the use of absolute values for temperature in any equation with radiation effects. Conduction and convection problems are usually linear in temperature (if k and h are temperature-independent), that is why it is common practice to work in degrees Celsius instead of absolute temperatures when thermal radiation is not considered. Notice that, in the case of heat conduction, the continuum hypothesis has been introduced, reducing the local formulation to a differential formulation to be solved in a continuum domain with appropriate boundary conditions (conductive to other media, convective to a fluid, or radiative to vacuum or other media), plus the initial conditions. The famous heat equation (perhaps the most studied in theoretical physics) is the energy balance for heat conduction through an infinitesimal non-moving volume, which can be deduced from the energy balance applied to a system of finite volume, transforming the area-integral to the volume-integral with GaussOstrogradski theorem of vector calculus, and considering an infinitesimal volume, i.e.: dH dt Q p V c t dV q ndA dV V 0 A V T c T q k 2T (4) t where has been introduced to account for a possible energy release rate per unit volume (e.g. by electrical dissipation, nuclear or chemical reactions). As said above, in typical heat transfer problems, convection and radiation are only boundary conditions to conduction in solids, and not field equations; when a heat-transfer problem requires solving field variables in a moving fluid, it is studied under Fluid Mechanics‟ energy equation. In radiative problems like in spacecraft thermal control (STC), the local formulation is not usually pursued to differential elements but to small finite parts (lumps) which may be assumed to be at uniform temperature (the lumped network approach). Thermal conductivities and other thermo-physical properties of materials Generic thermo-physical properties of materials can be found in any Heat Transfer text (a short list is shown in Table 1), but several problems may arise, for instance: The composite material wanted is not in the generic list. Special applications like STC usually demand special materials with specific treatments that may introduce significant variations from common data (e.g. there are different carbon-carbon composites with thermal conduction in the range 400..1200 W/(m·K)). The surface treatment does not coincide with those listed. Particularly concerning the thermo-optical properties, uncertainties in solar absorptance (and to a lesser extent in emissivity) may be typically ±30% from generic data to actual surface state. The working temperature is different to the reference temperature applicable to the standard data value, and all material properties vary with temperature. For instance, very pure aluminium may reach k=237 W/(m·K) at 288 K, decreases to k=220 W/(m·K) at 800 K; going down, it is k=50 W/(m·K) at 100 K, increasing to a maximum of k=25∙103 W/(m·K) at 10 K and then decreasing towards zero proportionally to T, with k=4∙103 W/(m·K) at 1 K). Duralumin (4.4%Cu, 1%Mg, 0.75%Mn, 0.4%Si) has k=174 W/(m·K), increasing to k=188 W/(m·K) at 500 K. Thermal joint conductance is heavily dependent on joint details difficult to characterise. And some joints are not fixed but rotating or sliding. However dark the problem of finding appropriate thermal data may appear, the truth is that accuracy should not be pursued locally but globally, and that there are always uncertainties in the geometry, the imposed loads, and other interactions, which render the isolated high precision quest useless and thus wasteful. Table 1. Properties of materials at 288 K. Substance Melting Melting temp. Tf K enthalpy hsl kJ/kg Density (mass) Thermal expansion (linear) Thermal capacity c J/(kg K) Thermal conductivity k W/(m K) Solar absorptance (normal) Emissivity (hemispherical, bolometric) kg/m3 .106 K-1 Alumina Aluminium Bakelite Berylliuma Brass Bronze Copper Cork Diamond Elastomer Glass (optical) Glass (pyrex) Glass (quartz) Glass (window) Glass (wool) Graphite 2300 933 1560 1200 1300 1358 4700 ~400 1070 395 877 205 1970 1400 4760 9000 3980 2710 1300 1850 8780 8800 8910 100 3500 1100 4000 2230 2650 2500 52 2250 7 24 150 11 20 18 17 1 200..300 6 3 0.5 9 3 840 896 1600 1850 400 400 390 2000 500 2000 500 840 780 820 657 750 33 220 17 190 150 50..80 393 0.05 ~2000 0.1 1.4 1.1 1.5 0.8..1.1 0.038 6 0.1..0.25 0.1..0.15 0.5..0.7 0.3..0.5 0.05 0.94 0.03..0.1 0.2..0.5 0.05 0.7 0.9 0.9 0.93 0.9 0.1 Ice 273 333 920 50 2040 2.3 0.3 0.92 Invar 1770 8100 1.7 460 12 Iron (cast-) 1800 290 7200 9..12 420 40 0.4..0.7 Magnesium 923 350 1730 26 1000 160 0.2..0.5 Methacrylate ~360 1200 60..70 1460 0.20 0.9 Nickel 1730 298 8890 13 440 90 0.2 0.05 Paper ~550 900 1500 0.07..0.13 0.3 0.95 Platinum 2040 114 21470 9 130 70 0.09 b Polyethylene ~320 920 180..400 2300 0.35 0.9 b Polystyrene ~360 30 70..130 1300 0.035 0.9 b Polyurethane ~360 25..150 150 1120 0.027 0.9 b PVC ~360 1400 90..180 960 0.10 0.9 Sand & soil 1970 ~1500 0.5 800 0.32 0.6 0.5..0.8 Silicon 1687 1650 2330 2.5 703 150 Silver 1235 110 10500 20 235 425 0.02 Steel (carbon-) 1800 7800 12 500 52 0.2 0.2 Steel (stainless-) 1750 7800 10..17 500 18 0.4 0.2..0.3 Teflon (PTFE) ~650 2150 100..200 1030 0.25 0.12 0.85 Titanium 1940 295 4530 9 610 22 0.4..0.7 0.5 Wolfram 3695 285 19400 4.5 130 200 0.45 0.09 a) Due to its stiffness (Young modulus 30% larger than steel), light weight, high melting point, high thermal and electrical conductivities, and dimensional stability over a wide temperature range, beryllium metal is used in the aerospace industry (e.g. Saturn V nozzles). Sound speed is one of the largest, 12800 m/s. It is transparent to X-rays. b) Emissivity of thin plastic films is much lower than the quoted bulk value (e.g. for 0.1 mm films, =0.15 for polyethylene and =0.60 for PVC). Unless experimentally measured on a sample, thermal conductivities from generic materials may have uncertainties of some 10%. Most metals in practice are really alloys, and thermal conductivities of alloys are usually much lower than those of their constituents, as shown in Table 2; it is good to keep in mind that conductivities for pure iron, mild steel, and stainless steel, are (80, 50, 15) W/(m·K), respectively. Besides, many common materials like graphite, wood, holed bricks, reinforced concrete, and modern composite materials, are highly anisotropic, with directional heat conductivities. And measuring k is not simple at all: in fluids, avoiding convection is difficult; in metals, minimising thermal-contact resistance is difficult; in insulators, minimising heat losses relative to the small heat flows implied is difficult; the most accurate procedures to find k are based on measuring thermal diffusivity a=k/(c) in transient experiments. Table 2. Thermal conductivities of some alloys and its elements. Alloy k [W/(m·K)] k [W/(m·K)] k [W/(m·K)] of alloy of base element of other elements Mild steel G-10400 51 (at 15 ºC) 80 (Fe) 2000 (C, diamond) (99% Fe, 0.4% C) 25 (at 800 ºC) 2000 (C, graphite, parallel) 6 (C, graphite, perpendicular) 2 (C, graphite amorphous) Stainless steel S-30400 16 (at 15 ºC) 80 (Fe) 66 (Cr) (18.20% Cr, 8..10% Ni) 21 (at 500 ºC) 90 (Ni) Unless experimentally measured on the spot, solar absorptance, , and infrared emissivity, , of a given surface can have great uncertainties, which in the case of metallic surfaces may be double or half, due to minute changes in surface finishing and weathering. Honeycomb panels Honeycomb panels (Fig. 1) are structural elements with great stiffness-to-mass ratio, widely used in aerospace vehicles. Heat transfer through honeycomb panels is non-isotropic and difficult to predict. If the effect of the cover faces is taken aside, and convection and radiation within the honeycomb cells can be neglected in comparison with conduction along the ribbons (what is the actual case in aluminiumhoneycombs), heat transfer across each of the dimensions is: T Qx kFx Ax x Lx Ty Qy kFy Ay Ly T Qz kFz Az z Lz 3 2 s with Fy s 8 with Fz 3s with Fx (5) where F is the factor modifying solid body conduction (the effective conductive area divided by the plate cross-section area), which is proportional to ribbon thickness, , divided by cell size, s (distance between opposite sides in the hexagonal cell), and depends on the direction considered (Fig. 1): x is along the ribbons (which are glued side by side), y is perpendicular to the sides, and z is perpendicular to the panel. Fig. 1. Structure of a honeycomb sandwich panel: assembled view (A), and exploded view (with the two face sheets B, and the honeycomb core C). Ribbons run along the x direction, and glued side by side in counter-phase along the y direction. Thermal inertia and energy storage A basic question on thermal control systems is to know how long the heating or cooling process takes (i.e. the thermal inertia of the system), usually with the intention to modify it, either to make the system more permeable to heat, more insulating, or more 'capacitive', to retard a periodic cooling/heating wave. When the heat flow can be imposed, the minimum time required is obtained from the energy balance, dH / dt Q , yielding t mcT Q . When a temperature gradient is imposed, an order-of-magnitude analysis of the energy balance, dH / dt Q → mcT/t=KAT, shows that the relaxation time is of the order t=mc/(KA), and, depending on the dominant heat-transfer mode in K, several extreme cases can be considered: Conduction driven case. The time it takes for the body centre to reached a mid-temperature, representative of the forcing step imposed at the surface, is t=L2/a, i.e. increases with the square of the size, decreases with thermal diffusivity, and is independent of temperature. Convection driven case. In this case, t=cL/h. Radiation driven case. In this case, t=cL/h, with h being the net thermal radiation flux; if irradiance E is dominant (e.g. solar gain with E=1370 W/m2), then h=E; if exitance M is dominant and there are only losses to the deep-space background at T0=2.7 K (0), then h=T3; in the case of heat radiation exchange with a blackbody at T0, then h=(T2+T02)(T+T0). When thermal loads are transient, with short pulses, the best way to protect equipment from large temperature excursions is to increase the thermal inertia of the system, preferably by adding some phase change material like a salt or an organic compound (within a closed container with good conductive characteristics). Exercise 1. Find the time it takes for the centre of a 1 cm glass sphere to reach a representative temperature in a heating or cooling process. Sol.: The time it takes for the centre to reach a representative temperature in a heating or cooling process (e.g. a mid-temperature between the initial and the final), is t=cL2/k=2500·800·(0.01/6)2/1=6 s, where the characteristic length of a spherical object, L=V/A=(D3/6)/(D2)=D/6, has been used. MODELLING THERMAL RADIATION Thermal radiation is the EM-radiation emitted by bodies because of its temperature, i.e. not due to radionuclear disintegration (like rays), not by stimulation with another radiation (like X-rays produced with an electron beam), not by electromagnetic resonance in macroscopic conductors (like radio waves). Although radiation with the same properties as thermal radiation can be produced by non-thermal methods (e.g. ultraviolet radiation produced by an electron beam in a rarefied gas, visible radiation produced by chemical luminescence), proper thermal radiation is emitted as a result of the thermal motions at microscopic level in atoms and molecules, increasing with temperature. Maxwell‟s equations of electromagnetism might be used to build a theoretical description of the interaction of electromagnetic radiation with matter, but it is so complicated and uncertain for real bodies (precise knowledge of material data like electrical conductivity, permittivity, and permeability, would be needed), that one has to resort to empirical data in most instances. Even so, uncertainty in surface finishing at microscopic level (<10-6 m) cannot be avoided in practice, what compromises the accuracy in extrapolating the data. We mainly consider thermal radiation exchanges in vacuum (except when planet atmospheres are considered). Black-body radiation Considering an evacuated material enclosure (of any material property, but non-interacting with the environment) at equilibrium (i.e. isothermal), and the EM radiation field created inside by the thermal vibrations of atoms at the walls, thermodynamic equilibrium between matter and radiation dictates that this radiation (named black-body radiation by Gustav Kirchhoff in 1860) must have the following properties: 1. Temperature. One may ascribe a temperature to the radiation, the temperature of the enclosure. 2. Isotropy. The radiation must be isotropic. 3. Photon gas. By quantum physics, energy is quantified, E=h=hc/ (h=6.6·10-34 J·s) and the EM waves can be viewed as EM particles, called photons. One often refers to the photon gas as an ideal gas (i.e. a set of non-interacting particles, each with an energy E=h, the main distinction being that the particles are not conservative and that they all move at the speed of light, c, but with different wavelengths). 4. Spectrum. In similarity with the fact that maximum entropy yields the Maxwell-Bolzmann distribution of molecular speeds in classical gases, maximum entropy yields the Planck distribution of photon wavelengths or frequencies: Planck‟s law in terms of spectral energy density [(J/m3)/m] is: u 8 hc hc 5 exp k T 1 (6) Although the wavelength-range extends in principle to the whole domain, 0<<, Planck‟s distribution is very peaked, particularly at lower wavelengths, and 93% of the whole energy lies in the range 0.5</Mmax<4, where Mmax=C/T and C=2.9.10-3 m·K. Human eye can only see in the range, 0.4 m<<0.7 m (the so called visible range, which can be subdivided in six 0.5 m amplitude colour bands corresponding to violet, blue, green, yellow, orange, and red, in increasing order 5. Emission. When this radiation escapes through a small hole in the enclosure (small holes appear black to the eye because they do not reflect any illuminating light), the spectral exitance [(W/m2)/m] is: Planck‟s law in spectral exitance: M c1 c exp 2 1 T 5 2 hc 2 L hc 5 exp 1 k T (7) where L is the spectral radiance [(W/m2)/sr], and c1=3.74·10-16 W·m2, c2=0.0144 m·K. Recall: h=6.626·10-34 J·s, k=1.38·10-23 J/K. Notice that exitance and emittance are referred to real surface area, whereas radiance is referred to a normal emitting area; thence, an infinitesimal emitter of area dA emits with a cosine law (projected area) but is seen with a constant radiance at all 2 steradians, and thus, M L cos d L cos 2 sin d L . Further notice that it is wrong to substitute there =c/; the correct relation is dL=Ld=Ld:, i.e.: 2hc 2 2h 3 dL d d hc h 5 2 exp c exp 1 1 k T kT (8) Planck‟s law corollaries: Wien's displacement law: Mmax=C/T, with C=2.8978·10-3 m·K=hc/(kx), where x is the root of x=5(1ex) (=4.965). Notice again the rapid spread of Planck‟s distribution with representative wavelength: at the peak, T=C, the spectral emission falls with the fifth power of Mmax. Stefan-Boltzmann‟s law: M=Md=T4, proposed by Jozef Stefan in 1879 and deduced by his student Ludwig Boltzmann in 1884, with =25k4/(15c2h3)=5.67·10-8 W/(m2·K4). Stefan used this law to find for the first time the temperature of the Sun. Planck‟s law approximations: c1 . c2 5 exp T 2 ckT In the limit of long wavelengths, it reduces to Rayleigh-Jeans law: M . 4 In the limit of short wavelengths, it reduces to Wien‟s law: M Black-body spectral fraction. Computing the fraction of black-body radiation within a spectral band is important is many applications, what can be helped by the mathematical equality (obtained by integrating by parts a series expansion of Plank‟s law): F0 1 T 4 c d 5 1 c2 0 exp T 1 e x 3 x 3x 2 6 x 6 4 4 n1 n 15 with x nc2 T (9) Two infinite black-bodies in a parallel-plate configuration exchange a heat flux of qij T j4 Ti 4 . Radiation-exchange between real bodies is modelled by introducing separate directional and spectral factors when possible (only for isothermal diffuse surfaces with only two spectral bands of interest), or by statistical ray tracing modelling in the more general case (using Monte Carlo method). Exercise 2. A manufacturer of electrical infrared heaters quotes in the applications of its products a maximum heating power of 1.2 MW/m2. What can be deduced about the operation temperature of its heaters? Ans.: Assuming the 1/4heater elements were black-bodies, from q T 4 T04 , we deduce 1/4 4 4 6 8 T T0 q 300 1.2 10 5.76 10 2100 K (1800 ºC). There are some heater elements close to black-bodies, as carbon heaters, whereas typical industrial heaters use kanthal wire (an iron-chromium alloy), which has an emissivity =0.7, and would need to be operated at 2300 K to yield that power, what is not realistic because its melting temperature is below 2000 K. There are, however, other metals withstanding higher temperatures (wolfram works above 3000 K in halogen lamps), but they are much more expensive and difficult to work with: they oxidise, they are brittle, etc. Real bodies: interface The ideal black-body model is in essence an interface model, describing the radiation entering or leaving a small hole in a cavity. The interaction of thermal radiation with real bodies departs from the black-body model in several respects: At the surface (i.e. an interface with abrupt change in refractive index). Real bodies do not absorb all the incident energy because there is some reflection and some transmission. If the transmitted energy is totally absorbed shortly within the body (say in less than 1 mm), the body is said to be opaque, and the absorption process can be ascribed to the interface, calling „absorptance‟ the fraction of the incident energy absorbed (i.e. not reflected back). As a consequence of the energy balance, a partially absorbing surface must be partially emitting, i.e. at the same temperature, real bodies emit less energy than black-bodies, what is quantified by the factor named emissivity. See thermo-optical surface properties data. At the bulk (of a constant or slowly varying refractive index media). Real bodies transmit radiation energy with some absorption (intensity decays exponentially along the path), and some scattering (re-radiation at the same or different wavelength in other directions than the path). According to the decay length, substances are grouped into two limit cases: opaque materials (if the decay length is less than the thickness of interest, and transparent materials (if the decay length is much larger than the thickness of interest); for instance, a 20 nm gold layer (deposited on a transparent substrate) is transparent enough to see trough it (it has 20% transmittance in the visible range). One should always keep in mind that ascribing physical properties to a geometrical surface is just a simplifying limit; in reality, like in a black-body cavity, radiant energy is absorbed or emitted within a sizeable thickness, not just at a geometrical surface. Emissivity Real surfaces emit less energy than the ideal black-body at the same temperature, what can be measured by an energy balance test in a non-equilibrium arrangement (e.g. within a cryogenic vacuum chamber). Surface emissivity is defined in detail by TMT/MT,bb. When emissivity does not change with direction (what is termed diffuse emission, or Lambertian emission in honour of J.H. Lambert‟s 1760 “Photometria”), the emitted power flux varies proportionally to the projected area of emission, i.e. with the cosine law, M=M0cos; in that case, a hot spherical surface is seen with a uniform flat brightness due to area compensation (this is also the case of a black-body); however, a metallic hot sphere appears darker at the centre because metal emissivity is greater towards the horizon, whereas hot non-metal spheres look brighter at the centre because emissivity of dielectrics tends to cero at levelling angles. Black-body emission verifies Lambert‟s cosine law. Polished metals have low infrared emissivities, of the order =0.1, but non-metals emit nearly as blackbodies (say =0.9) in the infrared, nearly irrespective of structure or visible colour (e.g. white paint emits nearly the same as black paint, and the same for human race skin colours). Emissivity of wolfram (tungsten) varies a lot with temperature and wavelength (and direction too, as for metals); its total hemispherical emissivity increases from =0.09 at 300 K to =0.39 at 3000 K (with a large spectral slope; at 3000 K, =0.45 at =0.5 m and =0.20 at =4 m). Notice that the radiation emitted by a lamp bulb (and quartz covered heaters) is limited by bulb transmittance (normal glass bulbs have a cut-off at 3 m, and quartz bulbs at 5 m with a dip at 2.8 m). . Absorptance Real surfaces do not absorb all the incident radiation received, but only a fraction , which can be measured by an energy balance. At the maximum level of detail, it can be shown that absorptance and emissivity are related by Kirchhoff's law (1859), T=T, since, if one considers an element of a real surface as part of a black-body cavity (i.e. in equilibrium at uniform temperature), the isotropy preservation of black-body radiation and the local energy balance implies that T=T. For opaque surfaces in general, what is not reflected is absorbed: =. Water absorbs practically all at 3 m, PVC at 3.5 m. Reflectance Real surfaces reflect part of the incident irradiation, , which can be measured with a radiometer, first measuring the irradiance (radiant flux incident on the surface by unit area, E), and thence the normal radiance (radiant flux exiting the surface by unit area and unit solid angle, L), and applying =L/E. Preservation of the isotropy in the interaction of a real surface with black-body radiation dictates that bidirectional reflectance at a given wavelength is the same if both directions (incident and reflected) are exchanged, T''=''T. Detailed reflectance measurements are computed by dividing the increment of exitance from a real surface by the irradiance used for the probing. For opaque surfaces in general, what is not absorbed is reflected: =. Reflection at real surfaces always has some scattering. Several limit cases are of most interest: Specular reflection, when there is no scattering and the reflected ray has an opposite angle to the incident ray, relative to the surface normal, 1=2. Mirrors approach this behaviour. Polished metals are good mirrors in the visible, infrared and microwave bands, although common mirrors are not first-surface mirrors but second-surface mirrors where a metal coating (silver in most cases) is behind a transparent glass sheet. Diffuse reflection, when reflectance is uniform for all outgoing direction. The reflected power flux varies proportionally to the projected area of emission, i.e. with the cosine law, M=M0cos. When a planar surface of such a perfect diffuser is illuminated by a beam in any direction, the surface appears uniformly illuminated, due to area compensation, however, in the case of an illuminated sphere it appears brighter in the region perpendicular to the shining beam. Retro-reflection, when the reflected ray goes out precisely in the same direction than the incident ray. The fact that the Moon is seen uniformly illuminated by the Sun at full Moon (instead of being brighter at the centre as for a perfect diffuser), is explained by the retroreflective properties of lunar dust. Real directional reflectance data are often simplified in two-term contributions: perfect diffuser reflectance, and specular reflectance. Transmittance Transmittance at an interface is the fraction of incident radiation energy that propagates to the rear of the interface, always with a change in direction (from the incident direction), which can be collimated (refraction), or scattered. An energy balance indicates that, at any interface, absorptance plus reflectance plus transmittance must equal unity, =1. Transmission is really a bulk effect. Real bodies: bulk Bulk effects on radiation-matter interaction are rarely considered in spacecraft thermal control, where the model of opaque surfaces is the rule; only a few cases of transparent materials are used in STC, notably second surface mirrors, and viewing windows in vehicles and space suits. Absorptance and transmittance Radiation absorption and transmission are bulk processes (ascribed to the surface when the penetration distance is very small). When considering bulk behaviour, instead of reflection (re-transmission backwards) one considers scattering, which is the re-transmission in all directions (backward and forward) except in the prolongation of the incoming ray, which is termed transmission. Hence, the energy balance establishes that absorption plus scattering plus transmission equals unity. Absorption (or better, transmission) within a medium is characterised by an attenuation or extinction coefficient, , (be careful to avoid confusion with surface absorptance with the same symbol; now has dimensions of m-1), such that radiation intensity falls exponentially along the path as I(x)=I0exp(x), what is known as Beer-Lambert's law. The extinction factor includes the effect of absorption and scattering, and is a function of wavelength. For finite thickness, the optical depth, , is defined by Iout=Iinexp(), and depends on wavelength. Clear sky has a total optical depth of 0.35 along the vertical path; aerosols increase the optical depth, making the Sun difficult to locate when >4. In engineering problems however, it is still common to talk about absorptance and transmittance factors (not coefficients) when dealing with finite transparent materials, and apply =1 globally. When transmission occurs in a collimated way, it is termed refraction, and the ray directions verify Snell‟s law, n1sin1=n2sin2, where n is the refractive index and the angle with the normal to the interface. Other times, transmission is not collimated but scattered, loosing the ability to form images (the material is then said to be translucent). Scattering In general, scattering is the process in which particles (material or electromagnetic) travelling along a given direction are deflected as a result of collision (interaction) with other particles (material particles). Electromagnetic scattering can be due to different processes, classified as elastic and non-elastic. Elastic scattering, where the wavelength is preserved. It may take place under several circumstances: o At interfaces, what gives way to diffuse reflection. o At molecular level in the medium, what is known as Rayleigh scattering. The scattered pattern is lobular symmetric (i.e. axisymmetric and symmetric to the normal plane), and the intensity is proportional to -4 (i.e. scatters more the lower the wavelength, what gives way to the bluish of our atmosphere and oceans), o At particles comparable in size to the radiation-wavelength, what is known as Mie scattering (G. Mie solved in 1908 Maxwell equations for the interaction of an EM-wave with a dielectric sphere) or Tyndall‟s effect (J. Tyndall was the first to attribute in 1859 the bluish of the sky by selective scattering, later explained by Rayleigh). The scattered pattern is lobular non-symmetric, larger forward, with intensity independent of frequency. Non-elastic, where the wavelength changes, like in Raman scattering at molecular level. Measuring thermal radiation Radiometers measure radiation coming from a field of view and falling onto a detector. The field of view (FOV) is delimited by a series of holes, or focused by refractive lenses, or mirror reflectors. The incoming radiation may be due to emission by objects in the FOV, by reflection on them from other bodies, and by transmission from the background). We only deal here with thermal radiation, and thermal detectors are described below. Detectors for shorter-wavelength radiations may be photographic films (for visible, ultraviolet, X-rays), supersaturated phase-change media (e.g. Wilson cloud chamber), gas-discharge devices (e.g. Geiger counter), etc. Detectors for longer-wavelength radiations are resonant electrical circuits known as aerials or antennas, with size proportional to wavelength, and sometimes with a reflector to concentrate the EM-field to be detected. The primary standard (the World Radiometric Reference, from the World Meteorological Organization) is based on absolute cavity radiometers. An absolute radiometer consist of a black cavity with an absorber connected to a heat sink through a precision heat flux transducer (a thermopile), upon which two beams can be directed using appropriate shutters: the sample solar irradiance, and a calibrated beam from a radiant electrical heater, controlled to maintain the same heat flux with and without the sample beam. In other versions, two opposite cavities are used, connected through the heat-flux assembly; if Pshut is the heater power with the solar beam shut, and Popen the power when sampling, the total beam irradiance is found from E=k(PshutPopen), with k obtained from Eele=kPshut, with Eele and Pshut measured. As for any electronic sensor, periodic calibration is needed (against a controlled black-body cavity). Thermal radiometers can be classified on different basis: type of detector, spectral range, directional range, array size, etc. Infrared detectors According to detector type, measuring thermal radiation can be based on different effects: Thermal effects. Incoming radiation is focused on a thermal detector (a tiny blackened electrical thermometer), whose temperature variation is measured. Two thermometric effects can be used: electric resistance (with a tiny thermistor called „bolometer‟), and thermoelectric voltage (a series array of thermocouples called „thermopile‟). For a given irradiation, the response is the same for any spectral distribution, but as emissive power falls rapidly with temperature, thermal detectors are not suitable for low temperatures. Thermal detectors are the most common for total thermal radiation, but used to have lower sensitivity and response time than quantum detectors; nowadays, thermistors and thermopiles made by metal deposition, are bridging the gap. The response of thermal detectors depends on the detector-body temperature, which must be controlled. Quantum or photon effect. Incoming radiation causes an electric charge release that is measured by photovoltaic, photoelectric, or photoconductive effects. Quantum effect detectors have an upper bound in wavelength response, and work best in a narrow waveband just below that cut- off wavelength, were sensibility is much greater than for thermal detectors. For a given irradiation power, the response is proportional to the number of photons, and thus to , since E=Nh=Nhc/. Quantum-effect detectors are based on electron-transitions in semiconductors, notably in the valence band of CdTe-HgTe alloys (known as HgCdTe), which requires cryogenic cooling for good signal-to-noise ratio, and more recently in the conduction well in GaAs. Optical effects. Incoming radiation is visually compared with radiation emitted by a calibrated source (optical pyrometer). Chemical effects. Incoming radiation cause a chemical reaction. Since thermal radiation is not very energetic, it only applies to visible and near infrared detection in special photographic film. According to spectral range: Total radiometers. They measure total radiation (i.e. the integral effect of all wavelengths, always limited by the optics). Sometimes, detectors with a narrow-band sensitivity are used to infer total radiation. Spectro-radiometers. They measure in a narrow spectral band, selected by appropriate spectral filters, or a polychromators (dispersion in a prism or in a fibre optic, or diffraction in a diffraction grating) and special filters (resulting in a monochromator device). o In the near IR band (say 0.7..1.4 m. Silicon and photographic detectors are used (IR-CCD from 1978). In this range, SiO2 has high transmittance (used in fibre optics), and water has low absorption. Used for night vision with CCD image intensifiers, and for spectroscopic analysis. Notice that sometimes near-IR lighting is used as an active means to enhance night vision. o In the short or middle IR band (say 3..5 m, centred around the first atmospheric window), indium antimonide (InSb) and mercury cadmium telluride (HgCdTe) detectors are used. With these kinds of detectors, IR-guided missiles follow the thermal signature left by aircraft (the exhaust nozzle and plume are at some 1000 K). o In the long or far IR band (say 8..14 m, centred around the main atmospheric window), HgCdTe bolometers are used. According to field of view (directional range): Normal radiometers. They measure radiation coming from a narrow field of view. In the case of solar radiation they are known as pyroheliometers. Hemispherical radiometers. Only used to measure total solar radiation at ground level, for meteorological or solar-energy applications. They are known as pyranometers (see below). According to the temperature range: Pyrometers, if specially suited to high temperature measurement. Radiometers. In general. According to the spatial scanning: Point radiometers. They use one single sensor (belonging to one of the mentioned types) to yield a single spatial measurement of radiation or the associated temperature. Thermal cameras (or thermo-cameras, or infrared cameras). They yield a two-dimensional measurement. Old devices were based on a mechanical 2-D scanner and a point radiometer; others used a linear array of sensors and 1-D mechanical scanning, while modern ones (since 1980s) use a 2-D array of sensors electrically scanned. The most accurate and quick-response IR sensors use HgCdTe detectors at cryogenic temperature, but they are very expensive and difficult to maintain. Un-cooled IR detectors, using micro-bolometers (see below) are cheaper, smaller, consume less power, and need no cooling time (they must be temperature-stabilised for proper accuracy). Micro-bolometer cameras are used for accurate temperature measurement, but their resolution is currently limited to 0.5 mega-pixel (640480); cheaper pyroelectric materials (which have better spatial resolution and response time, but lack accuracy, and need periodic chopping) can be used for more qualitative work (e.g. night vision). Thermography is synonymous of IR imaging. Modern thermal cameras (of less than 0.5 mega-pixel) cost an order of magnitude more than corresponding visual digital cameras of more than 10 mega-pixel (by the way, it helps a lot taking visual images at the same time as infrared images). Bolometers and micro-bolometers A bolometer (from Gr. bolo, thrown) is a thermal-radiation sensor based on the electric resistance change with temperature. The first bolometer, made by the Am. astronomer Samuel Langley in 1878, consisted of two platinum strips covered with lampblack, one strip was shielded from the radiation and one exposed to it, forming two branches of a Wheatstone bridge, using a galvanometer as indicator. Micro-bolometers are tiny bolometers (micro-machined in a CMOS silicon wafer, see Fig. 2) used in detector arrays in modern un-cooled thermal cameras. It is a grid of vanadium oxide or amorphous silicon heat sensors atop a corresponding grid of silicon. Infrared radiation from a specific range of wavelengths strikes the vanadium oxide and changes its electrical resistance. Fig. 2. Sketch of a micro-bolometer structure (and a design by Fluke-Infrared Solutions). The word bolometric is sometimes used as synonymous of total (i.e. spectral integral), but what a bolometer detects depend on the filters used. Pyranometers and helliometers A pyranometer (from Gr. pyr, fire, ano, upwards), sometimes named solarimeter, is a thermopile-sensor radiometer (Fig. 3) that measures all incoming solar radiation (hemispherical, i.e. 2 stereo-radians, and total, i.e. from 0.3 m to 3 m in practice). It is the typical device used in meteorology and solar energy applications; it is un-powered, and typical sensitivity is 10 V/(W/m2). Fig. 3. Sketch of a pyranometer: 1, glass dome; 2, thermopile sensor (an array of thermocouples arranged in series and wrap around a dielectric film, as detailed in the insert); 3, thermal block; 4, radiation protector. A pyranometer with a shadow band or shading disk blocking the direct solar beam (0.49 rad of arc), measures the hemispherical total diffuse sky radiation. Solar beam power can be deduced by difference, although another kind of instrument, the pyroheliometer, which has a narrow field of view (some 5º) is used for that purpose. Measuring thermo-optical properties Two basic thermo-optical quantities are measured at a material surface: emissivity and reflectance, and the others are computed from them. Emissivity is measured by detecting incoming radiation from an opaque body at temperature T, under a cryogenic vacuum, and dividing the result by the corresponding Planck‟s equation value. Either spectral or total measurements are carried out. Reflectance is measured by dividing the increase in irradiation detected from an opaque body (i.e. to discount emission and transmission), by the increase in irradiation shining on the object when illuminated (i.e. to discount other reflections). Albedo can be measured using two opposite pyranometers aligned with solar radiation. Absorptance in opaque bodies is computed from the energy balance =1. The equality between absorptance and emissivity only applies in general to the detailed balance: T=T. Absorptance in transparent media is measured in terms of the exiting radiation, used to compute an extinction coefficient (includes scattering). On photovoltaic cells (e.g. in solar arrays) not all the absorbed energy goes to thermal energy; for solar cells of electrical efficiency (VI)max/(EA), thermal absorption is th=Fp, where Fp is the packaging factor for the cells (cell area divided by panel area A, around 0.8), E is a standard normal irradiance (1370 W/m2 for space cells, but 1000 W/m2 for terrestrial cells), and (VI)max the maximum electrical power delivered. Transmittance in transparent bodies is computed in terms of the extinction coefficient and the reflectance at interfaces. Maxwell theory shows that reflectance at a dioptric interface is =(n1n2)2/(n1n2)2; e.g. from air to glass or vice versa, =(0.33/2.33)2=0.02 IR windows The spectral range of most narrow band radiation thermometers is typically determined by the optical filter. Filters are used to restrict response to selected wavelengths to meet the need of a particular application. For example, the 5±0.2 μm band is used to measure glass surface temperature because glass surface emits strongly in this region, but poorly below or immediately above this band. Next, the 3.43±0.2 μm band is often used for temperature measurement of thin films or polyethylene-type plastics etc. Atmospheric filter. The atmospheric filter depends a lot on actual water content in the atmosphere, and aerosols content in general. Clean air has two main windows in the IR, besides the visual and radio windows: Visible window. Incoming solar radiation energy is 95% in the =0.3..3 m range (10% in the UV, 40% in the visible, and 50% in the infrared). Short IR window, in the range =3..5 m. Complete absorption by CO2 in the range 4.2..4.5 m, what is used in remote sensing to detect mean air temperature in the tropospherestratosphere. The high absorption band in 58 m is used in remote sensing to measure water content in the air. Long IR window, in the range =8..14 m. This is the main atmospheric window (see Fig. 4), being highly transparent to water vapour, carbon dioxide, smoke, and dust, although there is a small absorption band by ozone at 9.510 m, what can be used to measure ozone abundance. This long-IR (or far-infrared) band is used in remote sensing to measure surface temperature from satellites. The most used material for long-IR optics is n-doped polycrystalline germanium. Since the optical refractive index of germanium is high, the reflectance from each surface is high and the net transmittance through the germanium is relatively low. The refractive index of germanium is 4.0, resulting in 36% reflectance per surface. The transmittance of uncoated germanium is only 47% through a 1 mm thick piece. In order to improve the IR transmittance of the window, a suitable antireflection coating is applied (some 2 m thick low refracting index material like thorium fluoride, ThF), reducing window reflectance to <1%, thereby raising the transmittance to >95%. Radio window, in the range =0.01..1 m (corresponding to 30 MH..30 GHz), i.e. including the microwave range up to the Ku band. Fig. 4. The Earth‟s atmospheric filter (clear sky): a) general electromagnetic opacity (NASA/IPAC), and b) detailed transmittance. The three main atmospheric windows are: the solar band (0.3..3 m) that gives us the Sun light and heat (and allows our seeing the stars), the long infrared band (8..14 m) that allows for some Earth cooling, and the radio band (10-2..10 m, including microwaves) that allows for space radio-communications. Fig. 5. The Earth‟s atmospheric filter for clear-sky conditions. Planet emittance as looking from outside. Earth emission energy is over 90% in the =3..30 m range, with the peak around Mmax=10 m. To this peak in the spectrum corresponds a black-body temperature around 300 K (in any case, close to 288 K average surface temperature). However, from the total emittance as seeing from outside, a black-body temperature of 253 K would be deduced. A good simple model is then to take T0=288 K as reference surface data, and deduce an average emissivity value, , such that M=T04, what yields =0.60. Further spectral details of the Earth emission are (Fig. 5): Nearly half of Earth‟s emitted energy is in the long-IR atmospheric window (8..14 m), at an apparent temperature of 288 K. Without this window, the Earth would become much too warm to support life, and possibly so warm that it would lose its water as Venus did early in solar system history. Outside the long-IR atmospheric window, i.e. when the atmosphere is opaque (5..8 m and >14 m), emission is perceived as coming from a black-body atmosphere at 218 K, with a total average (spectrum integral) of 258 K, what correlates well with ground measurements of sky temperature, which are (TambTsky)bolo30 K and (TambTsky)espectral70 K. Measuring the equivalent sky temperature from ground on a clear night, one gets consistent values: for total radiation, (Tamb-Tsky)bolo30 K, and, in the atmospheric window (TambTsky)IRwindow70 K. When clouds are present, the visible and infrared windows disappear, leaving just the radio window, because liquid water, even in finely dispersed aerosols like in clouds, have a much higher absorption and scattering than water vapour (some 104 times higher in the main 8..14 m infrared window). Glass filter. There are several types of glass: Window glass (common glass, comprising >90% of all glass production), also known as sodalime or crown glass (SiO2 75%, Na2O 15%, CaO 10%), is a low-melting-temperature glass used for windows and containers. Transmission has a window in 0.3<<2.5 m, with a heap transmittance profile that drops from 0.9 at 0.5 m to 0.6 at 1 m for a 10 mm thick glass, with near-IR transmittance falling rapidly with thickness (Fig. 6). An ordinary second-surface mirror has a solar absorptance of =0.14 (aluminized; =0.07 if silvered). Notice that a glass window do not let ultraviolet and infrared radiation pass, what explains the green-house effect, and why filament-emission heats up bulbs of incandescent lamps (average absorptance from a 3000 K source is around 0.7). Quartz glass is pure silica (>99.5% SiO2, also known as fused silica). Quartz windows (a few mm to a few cm thick) have some 90% radiation transmittance in the range 0.2<<2.5 m. An interesting application of selective transmission is the transparent mirror furnace, used for observing crystal growth up to 1300 K; in this furnace, a window (or the whole furnace) is made of pyrex glass (transparent in the visible) with an internal gold deposition (some 20 nm) specular in the infrared and with some transparency (=0.2) in the visible (pyrex is a thermal and chemical resistant glass, used for laboratory and oven work, with SiO2 80%, B2O3 13%, also known as borosilicate glass). Fig. 6. Glass transmittance (lamp bulb, normal pane, thick pane). Water filter. Water is transparent in the visible (lowest extinction coefficient is 0.02 m-1 at 0.55 m, growing to 1 m-1 at the practical cut-offs of 0.3 m and 0.7 m, becoming progressively opaque in the infrared, with high absorption near 3 m (what is used in IR heaters). Total transmission of solar radiation through 1 m of water is 35%. Infrared windows. Table 3 presents data for some infrared transparent materials. Table 3. IR window materials (ordered by spectral band). Material Formula transmission band Notes Sapphire Al2O3 0.15..5.5 m Very hard Calcium fluoride CaF2 0.15.. 10 m Soluble Barium fluoride BaF2 0.15.. 12 m Soluble, fragile IR polymer 0.15.. 22 m Soft Zinc selenide ZnSe 0.5.. 22 m Soft. Transmit 99% Germanium Ge 1.8.. 22 m Hard Thermochromic infrared shutters. Vanadium oxides change their crystalline network at a certain temperature from an IR-transparent semiconductor to a IR-opaque metal in the short IR band (3..5 m). The most used is VO2, which has the transition temperature around 67 ºC. The activation is performed by pulse heating a very thin gold deposition layer (in some 15 ms), changing the transmittance from 55% to <1%. Solar collector filter. In some applications like solar energy collectors, high absorptance with little emission is wanted, but, for a given material, high usually implies high ; a selective coating, however, may accomplish that, e.g. by deposition of a thin layer of a dielectric over a metal substrate, such that the coating be transparent to long wavelength (and the metal substrate is a low emitter), but opaque and absorbent at short wavelengths (e.g. SiO2 deposition on aluminium shows an abrupt cut-off wavelength at 1.5 m). Care should be paid to distinguish a solar shade filter, designed to prevent solar radiation to let through (as in sun glasses, snow goggles, and space suits), from a solar collector filter, whose objective is to absorb solar energy without letting long IR radiation to escape (like in a greenhouse as a whole, or in the oxide coated metal here described. Spectral and directional modelling Radiation and matter interaction is difficult to model because of its inherent directional and spectral properties: Directionality. Rays have direction of propagation. That direction may change by reflection, refraction and dispersion in general. And propagation can be halted by opaque bodies. Spectrality. EM-radiation has a spectrum of propagating wavelengths (or frequencies, or energies), and matter is very selective to absorption-transmission-emission of different wavelengths. As said before, the basis thermal radiation model is the so called black-body, which is a body that shows no particular directional or spectral characteristics: Directionally, a blackbody absorbs all incident radiation, and emits radiation, just proportional to the effective frontal area, i.e. with the cosine law: M=M0cos or E=E0cos. A Lambertian surface (be it blackbody or not) is a „perfect diffuser‟ and its exitance (due to own emission or reflected scattering from others) does not depends on direction (i.e. a sphere will be viewed uniformly brilliant like a frontal disc), because although the emitted power from a given area element is reduced by the cosine of the emission angle, the size of the observed area is increased by a corresponding amount. Spectrally, a blackbody absorbs all incident radiation (independently of wavelength), and emits radiation in accordance with Planck‟s law. Two-spectral-band model For most thermal problems, both on ground and in space, it is good enough to consider only two types of thermal radiation, corresponding to non-overlapping regions of the spectrum, and only spectral-averaged property values in each band: Solar radiation, corresponding to a quasi-blackbody at 5800 K, which peaks in the visible range (at about 0.5 m), and has some 10% of its energy within the ultraviolet range (0.3..0.4 m), 40% in the visible range (0.4..0.7 m), and 50% in the near-IR range (0.7..3 m). For the interaction of solar radiation with matter, both for direct sunshine and for albedo, the following properties are defined: o Solar absorptance, s (usually without sub-index, since there is no possible confusion because by Kirchhoff‟s law it should be equal to the emissivity of a body at around 6000 K, what is of no practical interest because materials cannot withstand such high temperatures). o If only the solar absorptance is given for a surface, it should be understood that the surface is opaque and that the rest of the incoming energy is diffusively reflected, =1. In more detailed numerical simulations (e.g. in ESARAD), four parameters are given to model the interaction of solar radiation with matter: solar absorptance , solar transmittance , solar diffuse reflectance diff, and solar specular reflectance spec, such that ++diff+spec=1. Infrared radiation, corresponding to a quasi-blackbody at 300 K, which peaks in the far-infrared range (at about 10 m), and has near 90% of its energy within the far-infrared range (3..30 m). The properties averaged for this spectral band are applied to radiation emitted, absorbed, transmitted, or reflected by a given material at whatever the real temperature, from 100 K to 1000 K. The following properties are defined: o Infrared emissivity, IR (usually without sub-index, since there is no possible confusion). infrared absorptance is never explicitly given since, by Kirchhoff‟s law, it is equal to the emissivity of the body, IR=IR. o If only the infrared emissivity is given for a surface, it should be understood that the surface is opaque, that it absorbs infrared radiation with IR=, and that the rest of the incoming energy is diffusively reflected, IR=1IR. In more detailed analysis (e.g. in ESARAD), four parameters are given to model the interaction of infrared radiation with matter: infrared emissivity , infrared transmittance , infrared diffuse reflectance diff, and infrared specular reflectance spec, such that ++diff+spec=1. Thermo-optic properties based on this two-band model can be found in any Heat Transfer book; vales for typical thermal control surfaces used in spacecraft thermal design are presented aside. Even when radiation comes from intermediate-temperature sources, as red-hot materials (at 1000 K or more), and incandescent lamps (up to 3000 K), the splitting of radiation effects in just the two bands described above, may be a good approximation, much simpler than taking care of all spectral details. However, for thermal radiation of very hot objects (in the range 1500..3000 K), the change with temperature of IR must be considered, but visible radiation can still be neglected in the energy balance (only some 5% of the energy consumed in an incandescent lamp goes to visible light). Radiation emerging from a surface Radiation emerging from a surface is called exitance, M (sometimes inappropriately called emittance, and formerly known as radiosity, j). Radiation energy going out of an interface per unit area and time, M [W/m2], can be due to own emission of the body, reflections from other sources in the front, or transmission from other sources in the rear; M=Mbb+E+Erear. Exitance measures the power density emerging from a surface, and it coincides with the irradiance received by a close-up facing surface (e.g. a surface close to the Sun will get E=M=TS4=5.67·10-8·57804=63·106 W/m2, decreasing with distance RSp as E=M(RS/RSp)2, so that at 1 AU E=M(RS/RSp)2=63·106·(0.7·109)/(150·109))2=1370 W/m2. Recall that irradiance is the incident energy per unit area and time, E [W/m2], and measures the power density received (from which, a fraction is absorbed, according to its absorptance). If the surface is not specified, the maximum flux is assumed, i.e. a unit surface normal to the incident radiation; e.g. extra-terrestrial solar irradiation at 1 AU, E=1370 W/m2. Irradiance is measured by the effect of the incoming radiation (focused or not) on a detector (thermal effects, or quantum effects). For point-like radiation emitting sources, total power [W] through any concentric sphere is conserved along the radius in non-absorbing media, and the same is true for radiant intensity Id/d [W/sr], i.e. power per unit solid angle. For extended surfaces (i.e. those that subtend a finite solid angle from the viewer, radiance is defined as the energy exiting the surface by unit normal area to the viewer, unit solid angle, and unit time, L [W/(m2·sr)], which coincides with the radiation falling on a detector per unit detector area and solid angle in non-dissipative optical systems (really, radiance divided by the index of refraction squared is invariant in geometric optics). Radiance is a useful magnitude because it indicates how much of the power issuing from an emitting or reflecting surface will be received by an optical system looking at the surface from some angle of view (the solid angle subtended by the optical system's entrance pupil, like in our eye). Radiance of a non-uniform source, like a half moon, depends on the viewing point (direction and distance), whereas radiance of a uniform source like the Sun, does not. Looking from the Sun to the Earth, a small patch of 1 m2 at the Sun surface emits L=M/=TS4/=63·106 W/m2/=20·106 W/(m2·sr), i.e. 20·106 W per unit solid angle towards its frontal direction (in other directions, this patch emits with the cosine law; e.g. zero in the tangential direction). At the Sun-Earth distance, RSE=150·109 m, a 1 m2 frontal patch subtends a solid angle from the Sun of =(1 m2)/RSE2=1/(150·109)2=44·10-24 sr (the whole Earth subtends =RE2/RSE2=5.7·10-9 sr, or 2RE/RSE=85·10-6 rad, and the Sun from the Earth =RS2/RSE2=68·10-6 sr, or 2RS/RSE=0.01 rad), so that the 1 m2 frontal patch at the Earth gets L=20·106·44·10-24=0.9·10-15 W from the 1 m2 frontal patch at the Sun. If we add up the contribution from the whole solar disc, we get LRS2=20·106·44·10-24(0.7·109)2=1370 W/m2 for the irradiation on a 1 m2 facing panel at the Earth (outside the atmosphere). Radiation from a small patch to another small patch. View factors Consider a differential surface patch dA1 a hemisphere centred on it (Fig. 7), and a radiant power d 1 [W] exiting from dA1, either because of its own emission (d1=dA1T4) or due to reflection from an incident flux (d1=dA1E, as for celestial albedo). We want to know how much radiation will impinge on another infinitesimal patch dA2 (in the hemisphere or not), i.e. the irradiation E [W/m2] it gets, and how a „viewer‟ at dA2 will „see‟ dA1 (i.e. how much energy per unit time, per unit area at the source, and per unit solid angle of the optical system, will get a detector at dA2 from dA1). Fig. 7. Notation for studying radiation from a differential patch to its viewing hemisphere (sometimes is used for the polar angle or co-latitude, instead of ). Radiant power, 1, and emittance (exitance, really), Md1/dA1, are not distributed uniformly in all directions. The most-used model is of diffuse exitance, valid for black-body emission, real-body emission with direction-independent emissivity, and ideal diffuse reflectors. For this diffuse-surface model, the emission is still non-isotropic, but varies with the cosine of the polar angle, cos, what implies that diffuse exitance is isotropic per unit normal area, what adds sense to the definition of radiance, since, for a diffuse surface, radiance (the „brightness seen by an observer‟ in the appropriate waveband) does not depends on the viewing direction. A black-body is a perfect diffuser, as can be demonstrated by considering a black-body enclosure at equilibrium, consisting of dA1 and the hemisphere above; black-body radiation being isotropic implies d212=d221 (i.e. output=input), and thus: 2 2 d 12 d 21 L 1 , 1 L 0, 0 d cos d 221 d2 cos 2 L 0, 0 1 2 1 r12 d d 212 d1 cos 1 L 1 , 1 22 r12 (10) i.e. the normal radiance of dA1 seen from dA2 (which cannot depend on the chosen patch), coincides with the radiance of dA2 seen from dA1 at any angle. The irradiance received by the patch dA2 from dA1, Ed2/dA2 [W/m2], depends on distance and orientation of both patches; with the diffuse assumption, it is E(r,1,2)=E(1,0,0)cos(1)cos(2)/r2. The relation between M (the source strength) and E(1,0,0) (the sink strength from both patches parallel at unit distance), is found in terms of the radiance of dA1, with the result E(1,0,0)=L=M/. The demonstration is based on the assumption of diffuse surface dA1 (i.e. directionally uniform radiance, Ld21/(ddA1cos(1)) [W/(m2·sr)]), and the energy balance of the whole hemisphere: d1 Md1 2 Ld d cos L 2 sin d d cos 1 1 1 1 1 1 1 0 2 M L (11) which shows that the power emitted by a surface dA1 per unit real area, M [W/m2], equal -times the power received by unit area of a surface dA2 parallel to the former and at unit distance; i.e. E(1,0,0)=L=M/..[W/m2]. View factor definition. Consider two infinitesimal surface patches, dA1 and dA2 (Fig. 8), in arbitrary position and orientation, defined by their separation distance r12, and their respective tilting relative to the line of centres, 1 and 2, with 01/2 and 02/2 (i.e. seeing each other). The radiation intercepted by surface dA2 coming from a diffuse surface dA1 will be its radiance, times its perpendicular area dA1, times the solid angle subtended by dA2, d12; i.e. d212=L1dA1d12=L1(dA1cos(1))dA2cos(2)/r2. The view factor, F12, is defined as the fraction intercepted from the total emitted energy, F12d212/(MdA1), i.e.: dF12 L d d cos 1 cos 1 cos 1 d2 cos 2 d 212 1 12 1 d12 2 M1d1 M1d1 d1 d1 r12 cos 1 cos 2 2 r12 (12) d2 Fig. 8. Geometry for view-factor definition. The view factor concept can be extended to finite surface areas which are both isothermal and diffuse; the problem is just of integration (although not a trivial one): dFij cos i cos j dAj Fij 1 Ai r 2 ij Ai A j cos i cos j rij2 dAi dAj (13) A general view on view factor expressions (algebra, values, and exercises) is presented further below, but some cases of especial relevance are covered in detail first. Radiation between a small patch and a large plate We want to analyse the radiation from a small source patch of area dA1, to a large planar surface, but we start by considering the simpler case of a plane irradiated by a point power source (Fig. 9) at a distance H and strength 1 (in watts for total radiation, or in lumens for visual radiation; by the way, this is basic for the design of artificial light-appliance distribution). The point source has an isotropic intensity I1=1/(4) (in W/sr for total radiation, or in lm/sr=cd for visual radiation), and a differential patch, dA2, on the plane, at a distance R from the sub-source point (a distance d12 H 2 R2 to the source); the radiation falling on dA2 is d12=I1d12, where d12 is the solid angle subtended by dA2 from the source, namely d12=dA2/d122=dA2cos(2)/d122, 2 being the angle between the viewing direction and the perpendicular to the plane; finally, the radiation per unit area, E2 (irradiance, in W/m2, for total radiation, or illuminance, in lm/m2=lx, for visual radiation), is E2=d12/dA2=(1/(4))H/(H2+R2)3=(1/(4H2))cos3(2), the famous cosine-cube law of illumination, represented in Fig. 9. It can be checked that the whole plane gets half of the source power, i.e. E2 2 RdR 1 2 . 0 Fig. 9. Geometric sketch and irradiance distribution, E2(R), on a plane due to an isotropic point source of power 1 at a distance H. Coming back to the radiation from a small source of area dA1 to a planar surface problem, the difference with the point-source problem is that radiation from the source patch is not isotropic in the 4 steradians, but goes as cos1 in the hemisphere facing dA1. If the emitting patch is parallel to the plane, at a distance H, and emits a power d1=M1dA1 diffusively (e.g. as a black-body, of emittance M1=T14), the radiation intensity in the direction 1 is dI1=(M1/)dA1=(M1/)dA1cos(1) (it would be uniform if per unit projected area in that direction), and the irradiation on the plane dE2=dI1cos2/d122=(M1dA1/(H2)cos4(1), since 2=1=arcos(H/d12) in this case. Notice that this is a cosine-to-the-fourth law, instead of the cosinecube for the point source. It can be checked that now the full radiated energy impinges on the plane, dE2 2 RdR d1 , instead of the half, for the point source case. 0 If the emitting patch is not parallel to the plane, but tilted an angle so that the normal to the patch intersects the plane at the point (xn,0) in the plane coordinates (x,y) centred at the sub-patch point (tan()=xn/H), then the distance from the patch to a generic point is d12=(H2+x2+y2)1/2, the distance from the patch to the point (xn,0) is d0=(H2+xn2)1/2, and the distance between the two points rn=((xxn)2+y2)1/2. The angular departure of a point (x,y) to the normal direction from the patch, 1, is given by the cosine law of triangles, cos1=(d122+d02rn2)/(2d12d0). The radiation intensity in the direction 1 is again dI1=(M1/)dA1cos1, and the irradiation at a generic point on the plane dE2=dI1cos2/d122=(M1dA1Hcos1/)/d123, since cos2=H/d12. Substitution yields the explicit form: dE 2 2 H 2 xn H 2 x 2 y 2 HM 1dA1 H 2 xxn 2 (14) which is represented in Fig. 10 for the case xn=H (45º tilting of the patch), and the reference case xn=0 (which corresponds to the parallel-patch case, solved before). A contour map of irradiance levels on the plane (x,y) is shown, to point out the fact that it is no longer axi-symmetric. Notice that the maximum irradiance is in between the closest point (the sub-patch point, x=0) and the perpendicular point (the intersection with the normal to the patch, x=xn). Finally notice that the source patch only shines on the semi-plane x>H2/xn (the other half, x<H2/xn, would be irradiated by the rear side of the patch. Fig. 10. Geometric sketch, and irradiance distribution on a plane, E2, due to a small radiation patch dA1 of emittance M1 at a distance H; a contour plot for the latter is plotted for the case xn=H (45º tilted patch), and a profile of the irradiances at the line y=0 for both xn=H (45º tilted patch) and .xn=0 (parallel patch). Let us now consider the radiation from an infinite planar surface to a small tilted patch of area dA1, with tan()=xn/H as before. The distance from a generic point (x,y) to the patch is d12=(H2+x2+y2)1/2, as before, and the same for the distance from the point to the patch, d0=(H2+xn2)1/2, the distance between the generic point and the normal point, rn=((xxn)2+y2)1/2, and the angular departure of a point (x,y) to the normal direction from the patch, cos1=(d122+d02rn2)/(2d12d0). The radiation intensity emitted by dA2=dxdy in the direction 2 is dI2=(M2/)dA2cos2, and its contribution to the irradiation at the patch dA1 is dE1=dI2cos1/d122=(M2dA2Hcos2/)/d123, since cos1=H/d12. Substitution yields the explicit form: d 2 E1 2 H 2 xn H 2 x 2 y 2 HM 2 dxdy H 2 xxn 2 (15) which is the same as (14) but with subindices 1 and 2 exchanged, because the second-order differential used here is just an artifice to match differentials, since dxdydA2. This equivalence means that the irradiation E on a given small patch (1), due to the emittance M of another fixed small patch (2), is the same whichever is considered the source, provided both radiate diffusively. If we know the emittance at every point in the plane, M2(x,y), integration of (15) would gives us the total irradiance on dA1 due to a planar distribution of sources. The simplest case is when the plane has uniform emittance (i.e. an isothermal plane), in which case, performing the integration (the change y 1 x 2 tan makes it easier) yields the result: E1 x xm dx 2 H 2 xn H 2 x 2 y 2 HM 2 H 2 xxn 1 dy M 2 2 H 2 H 2 xn 2 M2 1 cos 2 (16) The view factor from the plane (recall that it must be isothermal) to the patch, is the power received on the patch, E1dA1, divided by the power emitted by the plane, M2A2, i.e. dF21E1dA1/M2A2, which tends to zero for an infinite plate; but the view factor from the tilted patch to the infinite plate, F12, is finite, and can be calculated from the previous one by the reciprocity relation, dA1F12=A2dF21 (see the view factor algebra, below), to yield F12=E1/M2=(1+cos())/2, as can be found in the compilation on Table 5. Of course, this view factor can be computed directly from the definition, F12=(cos()cos())dA2/(r2), with 1 being the angular position of a patch in 2 from the normal to patch 1, computed by the cosine-rule as cos(1)=(r02+r2d02)(2rr0), with r0 1 tan 2 being the distance from patch 1 to the normal intersecting point in the plane (in units of patch-plane separation), r 1 x2 y 2 being the distance 2 from patch 1 to patch 2 (in units of patch-plane separation), and d0 x2 y tan being the distance from patch 3 to the normal intersecting point of patch 1 in the plane (in units of patch-plane separation); cos(2)=1/r is built from the angular position of patch 1 from the normal to patch 2, and dA2=dxdy; namely: r02 r 2 d 02 , cos 1 cos 1 cos 2 2rr0 F12 dxdy with r2 1 cos 1 , tan 2 r F12 1 tan cos y sin 2 1 y2 3 dy 1 cos 2 2 2 r 1 x y 2 d 0 x 2 y tan r0 1 tan 2 (17) 2 Radiation between a sphere and a small patch We want first to analyse the radiation from a small source patch of area dA1, to a large spherical surface, but we start by considering the simpler case of a sphere of radius R being irradiated from a point source of radiant power 1 at a distance H. The point source has an isotropic intensity I1=1/(4), and a differential patch on the sphere,d2A2=Rsin()dd, at a distance Rsin() from the axis (a distance 2 d12 H R 1 cos R 2 sin 2 to the source); the radiation falling on d2A2 is d212=I1d212, where d212 is the solid angle subtended by d2A2 from the source, namely d212=d2A2/d122=dA22cos2/d122, 2 being the angle between the viewing direction and the perpendicular to the sphere, given by the cosine law of triangles, cos(2)=(d122+R2(R+H)2)/(2Rd12); finally, the radiation per unit area is E2=d212/d2A2=(1/(4))cos(2)/d122; when E2() is plotted for given data (R,H), a bell-shape irradiance distribution is obtained, with a maximum at the closest point, E2(0)=(1/(4))/H2, falling to zero axisymmetrically at the tangential point t=arcos(R/(R+H)), and nil on the sphere shadow. When a central facing patch is considered instead of the point source, the result is similar: d1=M1dA1, dI1=(M1/)dA1=(M1/)dA1cos(1), dE2=dI1cos(2)/d122=(M1dA1)cos(1)cos(2)/d122; again, when dE2() is plotted for given data (R,H), a bell-shape irradiance distribution is obtained, with a maximum at the closest point, dE2(0)=(M1dA1/)/H2, falling to zero at the tangential point more quickly than in the case of the point source. If the emitting patch is not parallel to the sphere, but tilted an angle , two cases must be considered, depending on the relative position of the sphere and the plane passing by the patch, what is delimited by the semi-angle subtended by the tangent to the sphere from the patch centre, t=arcsin(R/(R+H)): if </2t, the whole projected sphere is seen by the patch; if >/2t, no part of the projected sphere is seen by the patch; and if /2t<</2t, only a part of the projected sphere is seen by the patch. Let us solve the first case, i.e. 0<</2arcsin(R/(R+H)). The irradiation on a small spherical patch centred at the point (x,y,z), with z R2 x2 y 2 , is again dE2=dI1cos(2)/d122=(M1dA1)cos(1)cos(2)/d122, but now the parameters are: cos(1)=(d122+d02d12)/(2d12d0), d12 x2 y 2 ( R H z )2 , 2 d0=(R+H)/cos(), d1 x H R tan y 2 z 2 , and 2 sgn( x)arcsin x2 y 2 R 1 ; to better grasp this cumbersome (although explicit) solution, Fig. 11 presents a contour plot of the irradiance distribution in the sphere, for the limit case where =/2t when H=R. Fig. 11. Geometric sketch, and irradiance distribution, dE2, on a sphere of radius R, due to a small radiation patch dA1 of emittance M1 at a distance H; a contour plot for the latter is plotted for the limit case =/2t, with t=arcsin(R/(R+H)) (i.e. 60º tilted patch, with 30º tangent angle, for H=R), and a profile of the irradiances at the central line y=0 for both =1.05 rad (60º tilted patch) and .=0 (parallel patch). Let us now consider the radiation from a large spherical surface of radius R to a small tilted patch of area dA1, separated a distance H from the surface. Again, we only consider the case of an isothermal sphere (i.e. uniform emittance), and proceed directly to compute the power received by the patch dA1. Again, two cases must be considered, delimited by the semi-angle subtended by the tangent to the sphere from the patch centre,t=arcsin(R/(R+H)): Case A, the plane containing the patch dA1 does not cut the sphere, i.e. the patch tilting, , is small, with </2t. In this case, the view factor can be found geometrically by the projections method (it is the area of the projection on the patch-plane of the projection of the radiating sphere on the unit hemisphere centred at dA1, and divided by ): F12 cos 1 cos 2 dA2 cos r 2 1 h 2 (18) Case B, the plane containing the patch dA1 cuts the sphere, i.e. the patch tilting, , is large, with /2t<</2t. In this case, the view factor is (Chun and Naragui, 1981): F12 cos 1 y arccos x xy tan arctan x cos 2 1 h 2 (19) with x 2h h 2 tan and y 1 x2 . A compilation and plots are presented in Table 5, below. Radiative coupling in general: thermo-optical and directional effects The heat flux between two infinite parallel black-body surfaces in vacuum is given by Stefan‟s law, q12 T14 T24 , which may be recast as q12 T14 T24 T14 T24 q12 q21 M1 M 2 , with the meaning that the net heat received by surface 2 (per unit area) from surface 1, q12 (in the other direction the heat flux will have the opposite sign), is equal to the heat emitted by surface 1 that falls into surface 2, q12 M 1 E2 (i.e. the emittance of A1, or the irradiance on A2), minus the heat emitted by surface 2 that falls into surface 1 and is absorbed, q21 M 2 E1 (i.e. the emittance of A2, or the irradiance on A1). For other geometries and other thermo-optical behaviour, we want to model the heat absorbed by surface Aj that comes from Ai by Qi j Ri j Ai Ti 4 , where Rij is the so-called radiative exchange factor from surface i to surface j (i.e. the fraction of the energy exiting surface i that is eventually absorbed by face j, accounting for emission, reflection and transmission (usually restricted to the two-spectral-band model discussed above). Computing radiative exchange factors, Rij, cannot be done analytically in general, and has to be done numerically, as explained below, but some analytical expressions can be developed in the case of very simple geometries, or for more complex geometries with black-body behaviour. Thermo-optical effects on simple geometries For two infinite parallel diffusively-reflecting surfaces in vacuum, with reflectance (related to absorptance by the opaque energy balance +=1, and to emissivity by Kirchhoff‟s law =), the heat flux q12 q12 q21 can be computed by considering the interior of surface A2 (i.e. as a balance between absorption and emission, q12 2 E2 2 M 2,bb ), or by considering the exterior of surface A2 (i.e. as a balance between irradiance and exitance, q12 E2 M 2 ), as can be demonstrated by substitution of M=MbbE and =1. Another expression for the heat flux can be obtained by substitution of E from M=MbbE into q12 E2 M 2 M 1 M 2 , in the form: q12 E2 M 2 M 2 2 M 2,bb 2 M2 M 2 2 M 2,bb 1 2 M2 M 2 M 2,bb 1 2 (20) 2 which can be interpreted in terms of flux=force/resistance, a convenient way to think in terms of an electrical analogy and Ohm‟s law (I=V/R) to finally yield an expression for the heat flux similar to the black-body case ( q12 T14 T24 ): q12 M 2 M 2,bb M M1 T14 T24 M1 M 2 1,bb 1 2 1 1 1 1 1 2 1 (21) 2 1 1 2 Table 4 presents a compilation for simple geometries of this type of heat flux, and their equivalent electrical circuit. Table 4. Heat transfer per unit area in some radiative cases, and their electrical analogy. Case Electrical analogy 1. From infinite planar black-body surface at T1, to infinite planar black-body surface at T2: q12 T14 T24 2. From infinite planar surface at T1 with emissivity 1, to infinite planar surface at T2 with emissivity 2: T14 T24 q12 1 1 1 2 1 1 2 3. From infinite planar surface at T1 with emissivity 1, to infinite planar surface at T2 with emissivity 2, with an intermediate isothermal opaque plate with emissivity at its faces 3 and 4: T14 T24 q12 1 3 1 4 1 1 1 2 1 1 1 3 4 2 4. From infinite planar surface at T1 with emissivity 1, to infinite planar surface at T2 with emissivity 2, with an intermediate isothermal non-reflecting semi-transparent plate with transmittance 3 and absorptance 3 (emissivity 3=3): T14 T24 q12 1 1 1 2 1 1 2 3 3 2 View factor algebra, relevant values and exercises As said above, computing radiative exchange factors, Rij, in general cannot be done analytically, and one is forced to make use of the statistical procedure known as Monte Carlo ray-extinction method. However, in the restricted case where surfaces are opaque and diffusively radiating, and particularly in the case of black-body surfaces, some analytical solutions can be worked out using the view factor concept introduced above (Fig. 8). When considering all the surfaces under sight from a given one (enclosure theory), between the many view factors that can be considered (for a N-surfaces enclosure, N2 view factors, although N(N1)/2 can be obtained from the reciprocity relations) several relations can be established, what is known as view factor algebra: Bounding. View factors are bounded to 0Fij≤1 by definition (the view factor is a fraction). Closeness. Summing up all view factors from a given surface in an enclosure, including the possible self-view factor for concave surfaces, Fij 1 , because the same amount of radiation j emitted by a surface must be absorbed. Reciprocity. Noticing from the above equation that dAidFij=dAjdFji=(cosicosj/(rij2))dAidAj, it is deduced that Ai Fij A j F ji . Distribution. When two target surfaces are considered at once, Fi , j k Fij Fik , based on area additivity in the definition. Composition. Based on reciprocity and distribution, when two source areas are considered together, Fi j ,k Ai Fik Aj Fjk Ai Aj . Large compilations of view factors for different geometries exist, but only a few of them admit an analytical expression. Table 5 presents some of the most useful results for spacecraft thermal control. Table 5. View factors expressions. View factor 1 Frontal (=0): F12 2 1 h Case 1. From a small planar plate at an altitude H tilted to a sphere of Profile (=/2): radius R, with hH/R; the tilting angle is between Plot the normal and the line of centres. F12 1 1 x arctan with x 1 h 2 x 2h h2 Tilted: -if </2arcsin(1/(1+h)), cos F12 2 1 h -if not, cos F12 arccos x xy tan 2 2 1 h with y arctan cos x 1 y 1 x2 x 2h h2 tan , 2. From a finite planar plate at an altitude H, tilted an angle , to an infinite plane (coincides with the limit of case 1 for a close-up to the sphere). 3. From a small sphere at an altitude H=hR to a large sphere of radius R. F12 1 cos 2 1 2h h 2 F12 1 2 1 h 4. From a small cylinder (external lateral area only), at an altitude H=hR and tilted an angle , to a large sphere of radius R, is between the cylinder axis and the line of centres). Coaxial (=0): arcsin s 1 s F12 2 1 h 2h h 2 1 h Perpendicular (=/2): with s 1 x2 with elliptic integrals E(x). 0 F12 4 1 1 h 2 xE x dx Tilted cylinder: F12 1 arcsin 1 h 2 0 0 sin 1 z 2 d d 2 with z cos cos sin sin cos 5. Between two identical parallel square plates of side L and separation H, with a=L/H. F12 1 x4 ln 4ay 2 2 a 1 2a with x 1 a2 and a y x arctan arctan a x (e.g. for a=1, F12=0.2) 6. Between two identical parallel strips of width W and separation H, with h=H/W. F12 1 h2 h 7. From a disc of radius R1 to a coaxial parallel disc of radius R2 at separation H, with r1=R1/H and r2=R2/H. x y 2 with x 1 1 r12 r22 r12 and F12 y x 2 4r22 r12 8. From a sphere a radius R‟ to a frontal disc of radius R a distance D=dR apart (it must be R‟<D, but does not depend on R‟). 9. Between concentric spheres (n=2) or infinite coaxial cylinders (n=1) of radii R1 and R2>R1. 1 1 1 F12 2 1 1 2 d F12=1 F21=rn F22=1rn with rR1/R2<1 10. From a small infinite long cylinder to an infinite long parallel cylinder of radius R, with a distance H=hR between axes. F12 arcsin 1 h 10. Between two sides, 1 and 2, of an infinite long triangular prism of sides L1, L2 and L3 ( is the angle between sides 1 and 2). F12 L1 L2 L3 2 L1 1 h 1 h 2 2h cos 2 L1 L with h 2 L1 Once the view factors are known, the energy balance for any given surface Aj, which can be set in terms of its emissivity as explained in the previous point, can computed in terms of the view factors from all other surfaces in an enclosure by establishing the energy balance for each participating surface: N N M M M j M j ,bb i j Q j ,net Aj E j M j Aj E j Aj M j Ai Fij M i Aj M j 1 j 1 i 1 i 1 Ai Fij Aj j (22) This is the set of equations that would yield all the exitances Mj for known temperatures (Mj,bb=Tj4), emissivities (j) and geometry (Aj and Fij). In the case of black-body surfaces, this reduces to: Q j ,net Ai Fij Ti 4 T j4 i 1 N (23) Another case that admits a simple analytical solution is the heat exchange between two isothermal diffuse surfaces that form an enclosure. The heat flow from surface 1 to surface 2 is in general (compare with case 2 in Table 4): Q12 1 1 1 2 1 1 A1 A1F12 2 A2 T14 T24 (24) and, in the particular case where surface 1 is convex at all points, i.e. a convex closed body within a container, F12=1, and the above simplifies to: Q12 A1 T14 T24 1 A 1 1 1 1 A2 2 (25) It is interesting to realise that when the container is large (A1<<A2), the heat exchanged by radiation becomes independent of the thermo-optical properties of the enclosure (assumed opaque, of course), and the radiation inside tends to black-body radiation independently of geometry and properties. A few simple manual exercises may clarify the procedure. Exercise 3. Find the view factor from a small area dA1 normal and centred with respect to a circular disc of radius R a distance H=hR apart. Sol.: Consider the sketch in Fig. E3.1, representing two equivalent configurations for A2, the said planar disc, and the projected spherical cap. Fig. E3.1. Profile and front view for the two equivalent configurations to compute F12: the said planar disc (in bold), and the projected spherical cap. 2 We must integrate dF12 cos 1 cos 2dA2 r12 in any case, with 01arctan(R/H). In the case of the real disc, and choosing as independent variable 0rR, we have cos1=H/r12, r12 H 2 r 2 , 2=1, and dA2=2rdr, what yields: R H r12 2 rdr R cos 1 cos 2 H2 F12 dA2 2 2 2 2 r12 r12 A2 r 0 r 0 H r 2 2 d r 2 H 2 1 2 2 2 H r r 0 1 h r R Of course, we might have used 1 as independent variable to do the above integration, instead of r, but we prefer to follow another approach: to compute F12 by projecting the real disc area against the sphere centred in dA1 and bordering the disc (Fig. E3.1). In this case, and choosing as independent variable 01arctan(R/H), we have 2=0 for any differential spherical patch, r12 H 2 R 2 , and dA2=2r12sin1r12d1, what yields: cos 1 cos 2 F12 dA2 2 r12 A2 sin 2 1 arcsin 0 R H 2 R2 arctan R H 1 0 cos 1 2 r12 sin 1r12d1 2 r12 arctan R H sin 21 d1 1 0 1 1 h2 Although it is a matter of choice, this second method of view-factor computation has the advantage that the radial distance is constant and the spherical patch is always normal, simplifying the integration. Further, notice that, in terms of the angle =arctan(R/H), the area of a spherical cap of radius R+H is Acap=2(R+H)2(1cos); the solid angle subtended from dA1 is =Acap/(4(R+H)2)=(1cos)/2, and the view factor F12=sin2. Exercise 4. Consider a hemispherical shell of 1 m in diameter, at 500 K, a circular disc of 0.1 m in diameter, concentric, in the base plane and at 300 K, and the circular corona at the base that completes the closure of the hemisphere, also at 500 K. Assume that there is only heat transfer by radiation (no convection and no conduction through the contacts). Find the heat transfer received by the disc in the following cases: a) Assuming that all surfaces are black-bodies. b) Assuming that all surfaces are grey-bodies with =0.8. Sol.: The sketch and notation is presented in Fig. E4.1. Fig. E4.1. Disc radiatively heated by a hemispherical dome. a) Assuming that all surfaces are black-bodies. In this case that there are no reflections, the disc only sees the dome, and we have Q21 A2 F21 T24 T14 ; using the reciprocity relation A2F21=A1F12 simplifies the problem, since F12=1 because all the energy radiated by surface 1 falls on surface 2. It follows then Q21 A1F12 T24 T14 D12 / 4 T24 T14 =(0.12/4)·5.67·10-8·(50043004)=24 W. Notice that the heating of the disc does not depend on the temperature of the corona 3 (neither on its thermo-optical properties), if surface 2 is a black-body at a fixed temperature (but the state of 3 would have an influence on the energy balance of 2). b) Assuming that all surfaces are grey-bodies with =0.8. In this case we have a two-surface enclosure (the disc 1 of area D12/4, and the rest 2 plus 3 of area D22+D22D12)/4), with F1,2+3=1, obtaining for Q23,1 Q1,23 : Q1,23 A1 T14 T24 1 1 0.12 / 4 0.8 5.67 108 5004 3004 0.12 1 1 0.8 5 12 0.12 A1 1 1 A23 2 D 4 T 2 1 4 1 T24 1 D 4 1 1 2 2 D D2 D1 4 2 1 2 2 D 4 T 2 1 4 1 T24 1 D 1 5D D12 2 2 2 1 0.12 0.8 5.67 108 5004 3004 19 W 4 i.e. the disc receives 19 W from the rest (the hemisphere and the corona). Notice that, in this case of small receiving area, the solution is quasi-linearly proportional to 1, and independent of the particular geometry of the enclosure. Exercise 5. Consider a cubic enclosure 0.5 m in side, with its upper face a black-body at 400 K and its lower face a black-body at 300 K. Assume that there is only heat transfer by radiation (no convection and no conduction through the joins). Find the heat transfer from top to bottom under the following cases: a) Assuming all lateral surfaces to be perfect mirrors. b) Assuming all lateral surfaces to be perfect re-radiators (black-bodies externally insulated). c) Assuming all lateral surfaces to be at 300 K as the bottom. Sol.: This problem can be solved by using case 4 results in Table 4. Each surface has a face area A=0.25 m2. The view factor from the top surface to the bottom surface corresponds to case 4: two identical parallel square plates of side L=0.25 m and separation H=0.25 m, with a=L/H=1, resulting in a value F12=0.2. The four lateral surfaces can be jointly considered as surface 3 (let surface 1 be the hot one at the top), with A3=4A=1 m2 and F13=1F12=10.2=0.8. a) Assuming all lateral surfaces to be perfect mirrors. If the lateral surfaces are perfect mirrors, the bottom surface sees the whole hemisphere uniformly radiating at 400 K, and the heat received by surface 2 is Q2 A2 Th4 Tc4 =0.25·5.67·10-8·(40043004)=248 W. b) Assuming all lateral surfaces to be perfect re-radiators (black-bodies externally insulated).. If the lateral surface A3 is a perfect re-radiator, it must be at its steady state, balancing the input from the top with the output to the bottom, i.e. verifying 0 A1F13 Th4 Ta4 A3 F32 Ta4 Tc4 , which, based on the reciprocity relation A3F32=A2F23 and the symmetry of the problem (A2=A1, F23=F13), yields the relation for the adiabatic temperature Ta4 Th4 Tc4 2 . With that, the heat received by surface 2 is Q2 A1F12 Th4 Tc4 A3 F32 Ta4 Tc4 A1F12 Th4 Tc4 A3 F32 Th4 Tc4 2 = 0.25·0.2·5.67·10-8·(40043004)+1·0.2·5.67·10-8·(40043004)/2=149 W. c) Assuming all lateral surfaces to be at 300 K as the bottom. If the lateral surfaces are at the same temperature than the bottom one, there is no heat transfer with them, and the heat received by surface 2 is just Q2 A1F12 Th4 Tc4 =0.25·0.2·5.67·108 4 4 ·(400 300 )=50 W. Notice the important effect of lateral-wall conditions on the top-down heat flow. Exercise 6. Consider a cubic box 1 m in side, made of 2 mm thick aluminium plates, initially at equilibrium with the environment at 15 ºC. Assume that one of the six faces is suddenly brought to 85 ºC and maintained at that temperature, that there is only heat transfer by radiation (no convection and no conduction through the joins), and that all faces are blackbodies. Find: a) All the view factors implied. b) The net heat fluxes at each plate, just after the heating starts. c) The steady state. Sol.: This problem can be solved by using case 5 results in Table 5. Each plate has a face area A=1 m2 and emits M=T4 from each of its two sides (i.e. 931 W/m2 the hotter one, at Th=358 K, and 390 W/m2 the others when still at 288 K). The environment can always be assumed a black-body at fixed temperature, here Te=288 K. Just after the plate is brought to 358 K, all other bodies being at 288 K, there is a net heat input to the system of Q0 2 A Th4 Te4 =2·1·5.67·10-8·(35842884)=1083 W, which will tend to heat up the other 5 plates until a steady state is reached, with the plates at an intermediate temperature between the hotter one (358 K) and the environment (288 K). a) All the view factors implied. The view factor from the hot plate to the one in front corresponds to case 4: two identical parallel square plates of side L=1 m and separation H=1 m, with a=L/H=1, resulting in a value F12=0.2. The view factor from the hot plate to the four adjacent plates are equal by symmetry, and together with the former view factor must add to unity, thence, all 5 faces are seen from the hot one with the same view factor, F1j=0.2. b) The net heat fluxes at each plate, just after the heating starts. The inner hot face constantly emits Q =ATh4=5.67·10-8·1·3584=931 W, and a fifth of it, 186 W, is absorbed by each of the other faces. But, at the beginning, all the other plates are radiating Q =AT4=5.67·10-8·1·2884=390 W from each face, so that the net heat flux at each of the other 5 plates is a gain of 186+4·(390/5)390=18678=108 W, since there is no heat exchange with the environment for them. As for the heated plate itself, the net heat flux is 2·9315·78390=1082 W, since it emits 931 W by each face, and gets five equal contributions from the inside plates, plus the irradiation from the environment (this same result was already calculated by other means at the beginning). The energy balance equations for each plate are: Heated plate: 0 W 5Qf Qe W 5 AF1 j Th4 T 4 AF1e Th4 Te4 Each other plate: mc dT Qf Qe AF1 j Th4 T 4 AF1e T 4 Te4 dt where W is the necessary heat input (unknown) to maintain the heated plate at constant Th, e.g. a controlled electrical resistance, Q f the heat received by one of the 5 faces from the heated one, with a view factor F1j=0.2, and Qe the heat lost to the environment by each plate (depends on its temperature), with a view factor F1e=1. d) The steady state. Due to the symmetry of the problem, at the steady state all plate temperatures are equal, Tf (except for the heated one, of course), and are calculated from: 0 Qh Qe AF1 j T Tf AF1e Tf T 4 h 4 4 4 e F1 jTh4 F1eTe4 Tf F1 j F1e 1 4 i.e., Tf=[(0.2·3584+1·28884)/1.2]1/4=303 K (30 ºC). Notice that now, the necessary power to keep the heated plate at 85 ºC is smaller (now the interior is warmer), as we can check globally (from the exterior) or locally (focusing on the heated plate and looking at the interior). Globally, the total heat losses are to the environment from one external face at 358 K plus 5 other external faces at 303 K, totalising A(Th4Te4)+5A(Tf4Te4)= 5.67·108 ·1·(35842884)+5·5.67·10-8·1·(30342884)=992 W (against the initial 1083 W). Considering just the heated plate, its emission is the same, Q =2ATh4=2·1·5.67·10-8·3584=2·931=1862 W, 4 -8 but now it absorbs Q =ATe =5.67·10 ·1·2884=390 W from the environment and 4 -8 Q =ATf =5.67·10 ·1·3034=478 W from the plates at 303 K, with a net result of 1862390480=992 W. Exercise 7. Find the heat transfer by radiation from a long cylindrical shell 10 cm in internal diameter, at 400 K, to a concentric duct 1 cm in external diameter, at 250 K, assuming opaque surfaces with an emissivity =0.8 Sol.: Direct application of the heat exchange equation between two isothermal diffuse surfaces that form an enclosure, surface 1 being convex, yields: Q12 0.01 L 0.8 5.67 108 4004 2504 1 0.01 1 0.8 0.1 D1L T14 T24 D1L T14 T24 D1 1 D1L 1 1 A1 1 1 1 1 1 D2 1 A2 2 D2 L A1 T14 T24 W 303 L m i.e. the duct gets around 300 watts per metre of length. Exercise 8. Consider a cylindrical evacuated enclosure 0.5 m in diameter and 0.5 m in length. One of the bases is a thin plate exposed on the outside to the 2.7 K of outer-space background radiation, while the other surfaces are kept at constant temperature. Find: a) All the view factors implied. b) All the heat flows at the steady state, assuming black-body surfaces, with 400 K for the temperature-regulated surfaces. c) Same as before, but considering an emissivity =0.5 for the surfaces at 400 K. d) The same, but releasing the fixed temperature condition at the base, which should be considered now as an insulated re-radiating black-body surface Sol.: The sketch and notation is presented in Fig. E8.1. Fig. E8.1. Cylindrical enclosure with notation (two faces are considered for the circular plate exposed to deep outer space, named node 5). This problem can be solved by using case 6 results in Table 5. Each base plate has a face area A1=A2=D12/4=0.52/4=0.196 m2, and the lateral area is A3=DL=0.5·0.5=0.875 m2. a) All the view factors implied. We must assume that all surfaces are opaque isothermal and diffuse. The view factor between the circular plates is F12=(xy)/2 with x 1 1 r12 r22 r12 and y x 2 4r22 r12 ; in our case, x 1 1 r12 r22 r12 =1+4+1=6 r1=r2=R/H=(D/2)/L=0.25/0.5=1/2 and thus and 2 2 2 2 y x 4r2 r1 6 4 4 2 =5.66, what finally yields F12=(xy)/2=(6-5.66)/2=0.17. By the closeness condition, F11+F12+F13=1, with F11=0, we get F13=F23=1F12=10.17=0.83. By the reciprocity condition, A1F13=A3F31 we get F31=F32=A1F13/A3=0.196·0.83/0.875=0.186. By the closeness condition, F31+F32+F33=1, F33=1F31F32=10.1860.186=0.63, what completes all the view factor calculations (the view factor from the external face of the plate seen outer space is of course unity). b) All the heat flows at the steady state, assuming black-body surfaces, with 400 K for the temperature-regulated surfaces. Surfaces 1 and 3 are maintained at 400 K, and the exposed plate only sees a black-body at 400 K from its face 2, and a black-body at 2.7 K from its face 4, so that its thermal balance at the steady state is: 4 0 Q13,2 Q4,5 A2 T13 T24 A2 T44 T54 T244 4 T13 T54 2 i.e. the plate takes a steady temperature of T2+4=(4004+2.74)/2)1/4=336 K, receiving 4 Q13,2 A2 T13 T24 =0.196·5.67·10-8·(40043364)=142 W of heat from the other surfaces in the enclosure, and radiating the same quantity to deep space, Q4,5 =142 W. The split of Q13,2 and Q is proportional to the view factors, Q Q F =142·0.17=24 W and between Q1,2 1,2 13,2 2,1 3,2 Q3,2 Q13,2 F2,3 =142·0.83=118 W. There is no heat transfer between surfaces 1 and 3, of course. c) The same as before, but considering an emissivity =0.5 for the surfaces at 400 K. This case corresponds to an enclosure with only two isothermal surfaces: A1+3=0.196+0.875=1.07 m2, and A2=0.196 m2, with one of them non-concave, F2,1+3=1, and thence: Q2,13 A2 T T 4 2 4 1 3 D2 1 4 4 T24 T13 2 1 2 A2 1 1 A13 13 1 D 4 1 2 DL D 4 13 0.196 5.67 108 T24 4004 1 0.196 1 1 1.07 0.196 0.5 with the same type of energy balance: 0 Q13,2 Q4,5 0.196 5.67 108 T24 4004 0.196 1 1 1 1.07 0.196 0.5 0.196 5.67 10 8 T 4 2 2.7 4 that leads to the new values T2=330 K and Q13,2 132 W (against the 336 K and 142 W obtained with the black-body hypothesis, showing again that when two surfaces widely different in size exchange radiation, the thermo-optical properties of the larger are irrelevant. d) The same, but releasing the fixed temperature condition at the base, which should be considered now as an insulated re-radiating black-body surface. This is a general case with opaque diffuse surfaces, which must be solved in terms of the general radiative balance at each surface: Mi M j Q j ,net Aj E j M j 1 i 1 Ai Fij N For surface 1, the energy balance is (re-radiator): M M1 M 3 M1 Q1,net 2 0 1 1 A2 F21 A3 F31 For surface 2, the energy balance is (input from interior equals output to deep space): M M2 M3 M2 Q2,net 1 A2 T44 T54 1 1 A1 F12 A3 F32 For surface 3, the energy balance is (fixed temperature): M M 3 M 2 M 3 M 3 M 3bb Q3,net 1 1 1 1 3 A1 F13 A2 F23 A3 3 The system with 3 equations, after all the known data are substituted (areas, view factors, emissivity, M1=T14, M2=T24 and M3bb=T34), only have the three unknowns T1, T2 and M3, with the solution: T1=381 K, T2=326 K and M3=1307 W/m2. Substituting those values, the energy balances show all the heat flows implied: Q1,net 18.3 18.3 0 W Q2,net 18.3 108 126 W Q 18.3 108 126 W 1, net i.e. there must be an energy source of 126 W at surface 3 (the lateral wall kept at 400 K), to compensate the net radiative exchange with surface 1, which gets 18 W, and surface 2, which gets 108 W, being colder. The energy balance of the upper plate in Fig. E8.1 shows that it gets 18 W from the bottom base, 108 W from the lateral wall, and delivers the total 126 W to deep space. Back to Spacecraft Thermal Control

DOCUMENT INFO

Shared By:

Categories:

Tags:
Heat Transfer, thermal conductivity, heat flux, Heat Transfer Coefficient, Thermal Management, thermal transfer, Thermal Engineering, energy transfer, transfer heat, heat sinks

Stats:

views: | 120 |

posted: | 11/29/2009 |

language: | English |

pages: | 38 |

OTHER DOCS BY pptfiles

How are you planning on using Docstoc?
BUSINESS
PERSONAL

By registering with docstoc.com you agree to our
privacy policy and
terms of service, and to receive content and offer notifications.

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.