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ME 452 - Heat Transfer Radiation Heat Transfer Lesson Outline • View Factor – – – – Reciprocity Relation Summation Rule Superposition Rule Symmetry Rule • Blackbody Radiation • Graybody Radiation • Radiosity • Direct Method for Solving Radiation Problems Radiation Big Picture • Radiation can be quite complex. Most of our analyses will be greatly simplified. – Black body radiation: – Gray body correction: – Directional correction (View Factor): – Shielding correction, and/or correction for non-vacuum radiation (absorption by gases in atmosphere) Q blackbody As Ts4 T4 4 4 Qgraybody As Ts T Q12 F12As T14 T24 View Factor • Thermal radiation follows the same rules as other electromagnetic radiation • Directionality is important – i.e. radio or TV antennae • The View Factor (F) corrects for directionality – Note the view factor is different depending upon which surface is sending/receiving Q A1 A2 1 cos 1 cos 2 F12 dA1dA2 2 Q A1 A1 A2 A1 r Q A2 A1 1 F21 Q A2 A2 cos 2 cos 1 A1 A2 r 2 dA2dA1 Four Basic Rules of View Factors • • • • Reciprocity Relation Summation Rule Superposition Rule Symmetry Rule Reciprocity Relation • If the view factor of one surface is known, the view factor of the other surface can be calculated using the Reciprocity Relation: A1F12 A2 F21 Summation Rule • Because radiation is a line of sight phenomenon, the most common application involves enclosed spaces or enclosures • Summation Rule: – The sum of the view factors from a surface to all other surfaces within an enclosure must equal one N F j 1 i j 1 N number of surfaces in enclosure F view factors Superposition Rule • Superposition Rule: – The view factor from a surface “i” to a surface “j” is equal to the sum of view factors from surface “i” to the parts of surface “j” – This is useful when complex view factors must be calculated • In example to right: – Known view factors: » 1 to 2 » 1 to 2 and 3 – Allows calculation of view factor 1 to 3 Symmetry Rule • Symmetry Rule: – Two or more surfaces that have symmetry with another surface will all have the same view factor from that surface – In example to right • Surfaces 2,3,4 and 5 will all have the same view factor from surface 1 Four Basic Rules of View Factors • Reciprocity Relation A1F12 A2 F21 • Summation Rule N F j 1 i j 1 • Superposition Rule F12,3 F12 F13 • Symmetry Rule Solved View Factors • Most common view factors have been calculated and tabulated • A few common 3-D solutions can readily be found Graphical View Factors • Less accurate, but less cumbersome are the graphical representations of view factors 2D View Factors • Just as with 3D view factors, many common 2D configurations have been calculated and tabulated Crossed Strings Method • The Crossed Strings Method is used to calculate view factors for infinitely long surfaces (2 ½ D) Fi j F12 crossed strings uncrossed strings 2 string on surface i L5 L6 L3 L4 2 L1 Direct Vs indirect determination of View factors for N-sided enclosure We need N2 view factors 1 (example N=5) 2 N 5 25 2 2 5 3 From which, we need to directly find N ( N 1) 5(5 1) 10 2 2 The remaining 15 , we find them indirectly using the previous rules 4 Blackbody Radiation • Blackbody radiation should be familiar at this point • How to incorporate the view factor between two surfaces: Q12 F12As T14 T24 • An enclosure may have multiple surfaces, therefore: 4 4 Qi Qi j Fi jAi Ti T j N N j 1 j 1 Graybody • A blackbody is an idealization of the perfect opaque diffuse emitter and absorber – Opaque - nontransparent – Diffuse – independent of direction • A graybody corrects for the amount of reflected radiation – Still assumes opaque and diffuse • Radiosity (J) – The total amount of radiation coming from a surface, including the emitted and reflected amounts of radiation Radiosity • Radiosity is the total amount of radiation leaving a surface – Reflected from other sources – Emitted from this surface • A blackbody would absorb all incident radiation (and therefore reflect none) – Therefore the radiosity of a blackbody is equal to the emissive power Radiosity • A graybody radiosity is equal to the reflected radiation plus emitted radiation: emitted reflected J i radiation i radiation i J i iTi 4 iGi • Because a graybody is opaque (the transmittance equals zero) absorptivity reflectivity transmissivity 1 therefore: 1 opaque 0 1 gray J i iTi 4 1 i Gi Electrical Analogy (Surface Resistance) • Note that Surface Resistance is different than the electrical analogy applied before when we used Thermal Resistance – Thermal resistance was used to calculate the total heat transfer rate based on a temperature difference – Surface resistance is used to calculate the net radiation from a surface • I personally do not like this method because of the ease of confusion – Therefore, we will use what is called the direct method or matrix method to solve radiation problems – The author discusses this method in detail Direct Method for Solving Radiation Problems • Procedure: – Set up a series of linear equations – Solve for applicable unknowns Surface with known Qi Ai Fi j J i J j j 1 heat transfer rate Qi Surface with known 1 i N 4 Ti J i Fi j J i J j i j 1 temperature Ti N Specific Concern – Radiation Effect on Temperature Measurement • Thermocouples are designed to read via convection • If there is a significant radiation contribution, thermocouple readings must be corrected: T f Tth th Tth4 Tw4 h T f actual fluid temperature Tth measured temperature value Tw wall temperature h convection heat transfer coefficien t emissivity of the thermocouple sensor Radiation with an Absorbing Medium • • • Air or vacuum can typically be ignored for radiation calculations (very low interaction with thermal radiation) With asymetric molecules such as H2O or CO2 there can be significant interaction This can be very complex – Gases require a volumetric phenomenon (as opposed to the surface analysis for opaque bodies) – Gas emissivity, transmissivity and absorptivity are all wavelength dependent • We will not go into detail on this subject Lesson Outline • View Factor – – – – Reciprocity Relation Summation Rule Superposition Rule Symmetry Rule • Blackbody Radiation • Graybody Radiation • Radiosity • Direct Method for Solving Radiation Problems 12–44 Two long parallel 16-cm-diameter cylinders are located 50 cm apart from each other. Both cylinders are black, and are maintained at temperatures 425 K and 275 K. The surroundings can be treated as a blackbody at 300 K. For a 1-m-long section of the cylinders, determine the rates of radiation heat transfer between the cylinders and between the hot cylinder and the surroundings. F1 2 F1 2 Crossed strings Uncrossed strings 2 2 String on surface 1 s2 D2 2s 2(D / 2) 2 s 2 D 2 s 2 0.52 0.16 2 0.5 0.099 D (0.16) F13 1 F12 1 0.099 0.901 A DL / 2 (0.16 m)(1 m) / 2 0.2513 m 2 Q 12 AF12 (T1 4 T2 4 ) (0.2513 m 2 )(0.099 )(5.67 10 8 W/m 2 .C)( 425 4 275 4 )K 4 38.0 W heat transfer between the hot cylinder and the surroundings per meter length Q 13 A1 F13(T1 4 T3 4 ) (0.5027 m 2 )(0.901)(5.67 10 8 W/m 2 .C)( 425 4 300 4 )K 4 629.8 W A1 DL (0.16 m)(1 m) 0.5027 m 2 12–43 Consider two rectangular surfaces perpendicular to each other with a common edge which is 1.6 m long. The horizontal surface is 0.8 m wide and the vertical surface is 1.2 m high. The horizontal surface has an emissivity of 0.75 and is maintained at 400 K. The vertical surface is black and is maintained at 550 K. The back sides of the surfaces are insulated. The surrounding surfaces are at 290 K, and can be considered to have an emissivity of 0.85. Determine the net rate of radiation heat transfers between the two surfaces, and between the horizontal surface and the surroundings. L1 0.8 0.5 W 1.6 F12 0.27 L2 1.2 0.75 W 1.6 A1 (0.8 m)(1.6 m) 1.28 m2 A2 (1.2 m)(1.6 m) 1.92 m2 A3 2 1.2 0.8 0.82 1.22 1.6 3.268 m2 2 A1F12 A2 F21 (1.28 )(0.27 ) (1.92 ) F21 F21 0.18 F11 F12 F13 1 0 0.27 F13 1 F13 0.73 F21 F22 F23 1 0.18 0 F23 1 F23 0.82 A1 F13 A3 F31 (1.28 )(0.73) (3.268 ) F31 F31 0.29 A2 F23 A3F32 (1.92 )(0.82 ) (3.268 ) F32 F32 0.48 J 1 1587 W/m2 , J 2 5188 W/m2 , J 3 811.5 W/m2 T1 4 J 1 (5.67 10 8 W/m 2 .K 4 )( 400 K ) 4 J 1 1 1 1 F12 ( J 1 J 2 ) F13 ( J 1 J 3 ) 1 0.75 0.27 ( J 1 J 2 ) 0.73( J 1 J 3 ) 0.75 T2 4 J 2 (5.67 10 8 W/m 2 .K 4 )(550 K ) 4 J 2 (5.67 10 8 1 3 F31 ( J 3 J 1 ) F32 ( J 3 J 2 ) 3 1 0.85 0.29 ( J 1 J 2 ) 0.48( J 1 J 3 ) W/m 2 .K 4 )( 290 K ) 4 J 3 0.85 T3 4 J 3 T1 4 J 1 (5.67 10 8 W/m 2 .K 4 )( 400 K ) 4 J 1 1 1 1 F12 ( J 1 J 2 ) F13 ( J 1 J 3 ) 1 0.75 0.27 ( J 1 J 2 ) 0.73( J 1 J 3 ) 0.75 T2 4 J 2 (5.67 10 8 W/m 2 .K 4 )(550 K ) 4 J 2 (5.67 10 8 1 3 F31 ( J 3 J 1 ) F32 ( J 3 J 2 ) 3 1 0.85 0.29 ( J 1 J 2 ) 0.48( J 1 J 3 ) W/m 2 .K 4 )( 290 K ) 4 J 3 0.85 T3 4 J 3 J 1 1587 W/m2 , J 2 5188 W/m2 , J 3 811.5 W/m2 Q 21 Q 12 A1 F12 ( J 1 J 2 ) (1.28 m 2 )(0.27)(1587 5188)W/m2 1245 W Q 13 A1 F13 ( J 1 J 3 ) (1.28 m 2 )(0.73)(1587 811.5)W/m2 725 W Q net As ( gTg4 gTs4 ) (1 m 2 )(5.67 10 8 W/m2 K 4 )[ 0.1325 (1200 K ) 4 0.1928 (600 K ) 4 ] 1.42 10 4 W HomeWork 13-24 13-37 13-41 13-43 13-44 13-32 13-41 131313-