Course Name: Heat Transfer Course Code: UME 720 Topic: Thermal Radiation by Dr. Madhup Kumar Mittal Associate Professor Department of Mechanical Engineering Thapar Institute of Engineering & Technology Fundamentals of Thermal Radiation 1. Radiation is the energy emitted by a body in the form of energy waves which are also called electromagnetic waves because they consist of both electric and magnetic field waves. That is why radiation emitted by a body is also called electromagnetic radiation. 2. Electromagnetic radiation can be described as a stream of photons, which are massless particles each travelling in a wave-like pattern and moving at the speed of light. Electric field wave Magnetic field wave Flashlight Photons 3. There are many types of electromagnetic radiation such as gamma ray radiation, X-ray radiation, microwave radiation, thermal radiation etc. 4. Different types of electromagnetic radiations are produced through various mechanisms. For example, gamma rays are produced by nuclear reactions, X-rays by the bombardment of metals with highenergy electrons, microwaves by special types of electron tubes such as klystrons and magnetrons. continued…. Fundamentals of Thermal Radiation 5. In 1864, James Maxwell observed an interesting mechanism which also produced electromagnetic radiations. He observed that electromagnetic radiations are produced whenever electric charges accelerate, decelerate or change the direction of their movement. accelerates or decelerate electromagnetic radiation electromagnetic radiation change of direction 6. The observations of Maxwell can be used as an explanation for emission of radiation from any hot material. β’ In a hot solid material, the molecules continuously vibrate. The continuous vibration of molecules in solids results in production of electromagnetic radiations. β’ In hot liquid or gas, the molecules collide with each other randomly, send each other off in different directions at different speeds. The collision of molecules in liquid and gases result in production of electromagnetic radiations. vibrating charge electromagnetic radiation electromagnetic radiation Solid material Liquid or gas β’The amount of radiation emitted from a hot material depends on the temperature of the material producing the radiation. The radiation emitted by a body as a result of its temperature is called Thermal Radiation. Fundamentals of Thermal Radiation Thermal Radiation • Thermal radiation is the thermal energy emitted by a body in the form of electromagnetic waves as a result of its temperature. • Electromagnetic waves can be described as a stream of massless energy particles, called photons (proposed by Max Planck), each moving at the speed of light in a wave like pattern whose wavelength is given by π = πΤπ. • Each photon is considered to have an energy of π¬ = π. πΤπ, where h = 6.62×10-34 J.s is Planck’s constant. • The strength or energy of thermal radiation emitted by a body depends upon the temperature of body. As E ∝ 1/λ, thermal radiation emitted by bodies at low temperature (or at room temperature) has high wavelength and vice-versa. Flashlight Photons Continued… Fundamentals of Thermal Radiation Thermal Radiation • All bodies at a temperature above absolute zero emit thermal radiations over a wide range of wavelength. However, most of the thermal radiation (≈ 99%) emitted by a body lies in the wavelength range of 0.1 to 100 μm, hence the thermal radiation is also defined as the portion of electromagnetic spectrum that extends from 0.1 to 100 μm. • The radiation emitted by bodies at room temperature lies in the infrared range of spectrum. • Bodies start emitting noticeable radiation (radiation in visible range) at temperatures above 527°C (800 K). visible π¬ = π. πΤπ Spectrum of Electromagnetic Radiation A surface that reflects red while absorbing the remaining parts of the incident light appears red to the eye Fundamentals of Thermal Radiation Concept of Black Body • A Black body is an idealized body which is used as a standard body against which the radiative properties of real bodies are compared. • A Black body is a perfect absorber, i.e., it absorbs all the incident radiation. Black body All radiation absorbed Real body Partial radiation absorbed • A Black body is also a perfect emitter, i.e., at a specified temperature, no surface can emit more energy than a black body. Eblackbody > Erealbody Both bodies at the same temperature T Fundamentals of Thermal Radiation A body emits radiation in all directions 1012 Red 1014 Visible light region 5800 K 4000 K Locus of max. power λT = 2898 μm-k Eb (λ), W/m2-μm A body emits radiation in all directions over a wide range of wavelength. In 1901, Max Planck developed the following relation for the amount of radiation energy emitted per unit time, per unit surface area, at a wavelength λ by a black body at an absolute temperature T. Violet Planck’s law for black body radiation 1010 2000 K 1000 K 108 500 K 300 K 106 100 K 104 102 Eb(λ, T) is called monochromatic (single wavelength) emissive power of black body. 1 0.01 0.1 1 10 λ (μm) 100 1000 Fundamentals of Thermal Radiation Total emissive power of black body (Stefan-Boltzman Law) The total emissive power of a black body [Eb(T)] is defined as the total radiant energy emitted in all directions over the entire wavelength range, per unit time, per unit surface area, by a black body at an absolute temperature T. π=∞ πΈπ π = ΰΆ± πΈπ π, π ππ π=0 = ΰΆ± π=0 πΆ1 ππ π5 ππ₯π πΆ2 Τππ − 1 = 5.67 × πΈπ π = ππ π 4 10−8 π4 = ππ (W/m2) πΈπ π, π πΈπ π, π π=∞ πΈπ π π4 dλ λ where, σb = Stefan-Boltzman constant = 5.67×10−8 W/m2-K4 Fundamentals of Thermal Radiation Monochromatic Emissivity (ελ) of non-black body (real body) The ratio of the monochromatic emissive power of a real body [E(λ,T)] to the monochromatic emissive power of black body [Eb(λ,T)] at the same wavelength and temperature is called the monochromatic emissivity of real body. πΈ(π, π) ππ = πΈπ (π. π) Emissivity (ε) of non-black body (real body) The ratio of the total emissive power of a real body [E(T)] to the total emissive power of black body [Eb(T)], both being at the same temperature, is called the average emissivity or simply the emissivity of non-black body. T = const. E (λ) λ ελ 1 Blackbody, ε =1 ε = average emissivity for non-black body π=∞ β«=πΧ¬β¬0 πΈ π, π ππ πΈ(π) πΈ(π) π= = π=∞ = πΈπ (π) β«Χ¬β¬ ππ π 4 πΈ π, π ππ π π=0 0 ελ for non-black body 0 ∴ πΈ(π) = π πΈπ (π) = π ππ π 4 = Emissive power of any real body at temperature T λ Fundamentals of Thermal Radiation Absorptivity, reflectivity and transmissivity of a body When a radiant energy emitted by a body impinges upon a semi-transparent body, a portion of it is absorbed, a part of it is reflected back and the remaining portion is transmitted as shown in the Fig. below: ……(1) Absorbed radiation semi-transparent (Ga) body Transmitted radiation (Gt) • The absorptivity, reflectivity and transmissivity of real body are not constant. Rather, they vary with the wavelength of radiation impinging on it. 1 αλ 0 • The absorptivity (α), reflectivity (ρ) and transmissivity (τ) in Eq. (1) are the average values of monochromatic absorptivity (αλ), monochromatic reflectivity (ρλ) and monochromatic transmissivity (τλ) respectively, averaged over the entire wavelength range of impinging radiation on the body. 1 ρλ 0 1 τλ 0 αavg λ ρavg λ τavg λ Fundamentals of Thermal Radiation Types of radiative bodies (radiative surfaces) Incident radiation (1) Black Body (Black surface) • A black body is one which absorbs all the radiation impinging on it. • For black body: α = 1; ρ = τ = 0 • Black body is also a perfect emitter, i.e., ε = 1 for black body. It means, at a specified temperature, no surface can emit more energy than a black body. • There does not exists a perfectly black body which can absorb all the incident radiation. Hence, black body is a hypothetical ideal body which is used as a standard body against which the radiative properties of real bodies are compared. (2) Real Body (Non-black body, Non-black surface, Real surface) • A real body is a non-black body which absorbs only a certain portion of total radiation striking on it. Thus the absorptivity of a real body is blow unity (α < 1). • The absorptivity of a real body is wavelength dependent, i.e., the absorptivity of a real body does not remain constant over the entire range of wavelength of incident radiation. Thus for real body: α < 1 and α = f (λ). • The other radiation properties such as ρ, τ and ε are also less than one and wavelength dependent. α =1 πΈ(π) = π πΈπ (π) = π ππ π 4 Both bodies at the same temperature T 1 αλ 0 αavg λ 1 ρλ 0 1 τλ 0 ρavg λ τavg λ continued…. Fundamentals of Thermal Radiation (3) Gray Body (Gray surface) • For the purpose of simplification of analysis, all real bodies are assumed to have constant values of radiation properties irrespective of wavelength of incident radiation. • Thus for gray body: (α, ρ, τ & ε) < 1 and (α, ρ, τ & ε) ≠ f (λ). εreal surface 1 ε α real surface τ real surface τ 0 ρ gray surface λ 0 1 α gray surface λ ρ real surface 1 ρ • The condition of constant radiation properties too is not satisfied by the real body and as such gray body is also a hypothetical concept like a black body. λ 0 1 α 0 εgray surface τ gray surface λ (4) Opaque Body (Opaque surface) • An opaque body is one which does not allow any amount of radiation to transmit through it. Thus for opaque body: τ = 0 and α + ρ =1. • Since most of the solids and liquids encountered in engineering are almost non-transparent (opaque) to thermal radiation, all radiation thermal analysis is done by assuming bodies involved to be opaque. τ=0 Opaque body & α + ρ =1 No radiation transmitted Fundamentals of Thermal Radiation Large enclosure behave like a black body A large enclosure at temperature ‘T’ behave like a black body because radiation emitted by an interior surface will undergo multiple reflections and some amount of radiation will be absorbed by inner surface upon each reflection. Hence entire radiation will be eventually absorbed by the enclosure. Kirchoff’s Law of Radiation Kirchoff’s law of radiation states that the ratio of the emissive power ‘E’ to the absorptivity ‘α’ is same for all the bodies which are in thermal equilibrium with their surrounding enclosure. 1 Large enclosure at temp. T Small body 1 (α1) at temp. T 2 Large enclosure at temp. T Small body 2 (α2) at temp. T 3 Large enclosure at temp. T πΈ1 πΈ2 πΈ3 πΈπ = = = πΌ1 πΌ2 πΌ3 πΌπ Small body 3 (α3) at temp. T continued…. Proof of Kirchoff’s Law of Radiation Let us consider a large enclosure which encloses a small body ‘1’ of surface area A1. Let us assume both the enclosure and body are at the same temperature T. Under the steady state condition, thermal equilibrium must exist between the body and the enclosure, i.e. Radiation emitted by the body (1) = radiation absorbed by the body (1) πΈ (π) …(1) π΄1 πΈ1 π = πΌ1 π΄1 πΈπ (π) or, πΈπ (π) = 1 πΌ1 Now we remove body ‘1’ and replace it by body ‘2’ having surface area A2, absorptivity α2 and emissivity ε2. In this case, we may write: πΈ (π) …(2) π΄2 πΈ2 π = πΌ2 π΄2 πΈπ (π) or, πΈπ (π) = 2 πΌ2 πΈ1 (π) πΈ2 (π) πΈ3 (π) = = = … … … πΈπ (π) πΌ1 πΌ2 πΌ3 From above Eq. (3), we can also write that for any radiating body (having absorptivity α) which is in thermal equilibrium with the surrounding : Thus in general we can write: …(3) πΈ(π) = πΌ …(4) πΈπ (π) Large enclosure at temp. T 1 Small body 1 (A1, α1, ε1) at temp. T 2 Large enclosure at temp. T Small body 2 (A2, α2, ε2) at temp. T We also know that the ratio of the emissive power of a real body (E) to the emissive power of black body (Eb), both being at the same temperature, is called the emissivity of body, i.e., πΈ(π) …(5) βΈ« from Eq.(4) and Eq.(5), π= πΈπ (π) we have α = ε Thus, Kirchoff’s law of radiation also states that the absorptivity of a body is equal to its emissivity when the body is in thermal equilibrium with its surroundings. Radiation Heat Transfer Between Surfaces Radiation heat transfer between surfaces depends on: Surface-3 Surface-2 • Temperatures of the surfaces • Radiation properties of the surfaces • Orientation of surfaces relative to each other Surface-1 Shape factor (View factor, geometric factor, configuration factor) • Shape factor is defined as the fraction of radiative energy that is radiated from one surface and strikes the other surface directly. F12 Direct radiation from surface "1" incident upon surface "π" = Total radiation radiated from surface"1" • Similarly F21 = Direct radiation from surface "2" incident upon surface "π" Total radiation radiated from surface"2" • In general, Fij represents the fraction of the radiation leaving surface ‘i’ that strikes surface ‘j’ directly Radiation Heat Transfer Between Surfaces Salient features of shape factor When two surfaces 1 & 2, of area A1 and A2 respectively, exchange radiant energy with each other, the shape factor follows reciprocity theorem, i.e., 2 ,A 2 A1F12 = A2F21 1 , A1 A concave surface has a shape factor with respect to itself because the radiant energy coming out from one part of the surface is intercepted by the another part of the same surface. The shape factor with respect to itself for surface ‘1’ is denoted as F11. 1 F11 ≠ 0 For a flat surface or convex surface, shape factor with respect to itself is zero as no radiation leaving from their surfaces strikes themselves. F11 = 0 F11 = 0 If a enclosure consists of many surfaces, then each surfaces of enclosure follows the following relation for shape factor. F11 + F12 + F13 +………..+F1N = 1 F21 + F22 + F23 +………..+F2N = 1 ………………………………….. N 1 2 3 This is known as conservation principle or summation rule. FN1 + FN2 + FN3 +………..+FNN = 1 Shape factor for a body inside a enclosure is 1 as all radiation leaving from the exterior surface of inside body strikes the enclosure surface. Enclosure 2 1 Inner body (sphere) F12 = 1 Radiation Heat Transfer Between Surfaces Determine the shape factor from any one surface to any other surface for the following enclosure configuration: 2 3 2 1 3 1 A tube with cross-section of an equilateral triangle Thus, for surface-1 For surface-1 F11 + F12 + F13 = 1 F11 = 0 F11 = 0 (because it is a flat surface) F12 = 0.5 βΈ« F12 + F13 = 1 F13 = 0.5 and, F12 = F13 (By symmetry) Similarly for surface-2 βΈ« F12 = F13 = 0.5 F22 = 0 F23 = 0.5 F21 = 0.5 Similarly for surface-3 F33 = 0 F31 = 0.5 F32 = 0.5 Radiation Heat Transfer Between Surfaces Determine the shape factor from any one surface to any other surface for the following enclosure configuration: 2 For plane surface-1 F11 + F12 = 1 F11 = 0 (because it is a flat surface) βΈ« F12 = 1 r 1 Plane surface For hemispherical surface-2 F22 + F21 = 1 r Hemispherical sphere of radius r closed by a plane surface We know that: A1F12 = A2F21 (By reciprocity theorem) ∴ F21 A1 = F A2 12 ππ 2 = × 1 = 0.5 1Τ2(4ππ 2 ) βΈ« F22 = 0.5 Radiation Heat Transfer Between Surfaces Determine the shape factor from any one surface to any other surface for the following enclosure configuration: Spherical enclosure For surface-1 of inner sphere F11 + F12 = 1 F11 = 0 (because it is a convex surface) βΈ« F12 = 1 2 For spherical enclosure surface-2 1 (sphere) F22 + F21 = 1 r 2r A1F12 = A2F21 (By reciprocity theorem) ∴ F21 = Sphere of radius r inside spherical enclosure of radius 2r A1 F12 A2 4ππ 2 = × 1 = 0.25 2 4π(2π) βΈ« F22 = 0.75 Radiation Heat Transfer Between Surfaces Net radiation heat exchange between black surfaces • Consider two black opaque surfaces of arbitrary shape of area A1 & A2, and maintained at uniform temperatures T1 and T2 respectively, as shown in Figure. T1 > T2 α=1 ε=1 ρ=0 τ=0 • Amount of radiation energy leaving surface ‘1’ per unit time per unit area is given by: Eb1 = σb T14 WΤm2 • Total amount of radiation energy leaving from entire surface ‘1’ per unit time is given by: Q1 = π΄1 σb T14 W • Amount of radiation energy leaving surface ‘1’ and striking surface ‘2’ is given by: Q12 = π΄1 πΉ12 σb T14 W • Similarly, the amount of radiation energy leaving surface ‘2’ and striking surface ‘1 ‘is given by: Q21 = π΄2 πΉ21 σb T24 W • The net rate of radiation heat transfer from surface ‘1’ to surface ‘2’ can be calculated as: Q12 πππ‘ = Q12 − Q21 = π΄1 πΉ12 σb T14 − π΄2 πΉ21 σb T24 • By applying reciprocity relation (i.e., A1F12 = A2F21), the above equation can be written as: Q12 πππ‘ = π΄1 πΉ12 σb T14 − T24 Question A small sphere (outside diameter = 60 mm) with a surface temperature of 300°C is located at the geometric centre of a large hollow sphere (inside diameter = 360 mm) with an inner surface temperature of 15°C. Calculate how much of emission from the inner surface of the large sphere is incident upon the outer surface of the small sphere; assume that both sides approach black body behaviour. Also determine the net interchange of radiation heat between the two spheres. Solution The net interchange of heat between the two sphere is: 2 ππππ‘ = π΄1 πΉ12 σb T14 − T24 r1 ππππ‘ = 4π × 0.032 × 1 × 5.67 × 10−8 300 + 273 1 r2 ππππ‘ = 0.0113 × 5.67 × 573 100 4 288 − 100 4 − 15 + 273 4 = 64.66 π (π¨ππ) 4 Radiation Heat Transfer Between Surfaces Radiation Heat Transfer Between Gray (Non-Black) Surfaces Irradiation (G) It denotes the total radiant energy incident upon unit surface area per unit time. Its unit is W/m2. Radiosity (J) It denotes the total radiant energy leaving from unit surface area per unit time. Its unit is W/m2. J = εEb + ρπΊ = εσb T 4 + ρπΊ For black surface, ε = 1 & ρ = 0 βΈ« For black surface, J = Eb, i.e., the radiosity of black surface is equal to its emissive power. Opaque gray surface with emissivity ε & at temperature T Radiation Heat Transfer Between Surfaces Net Radiation Heat Transfer from or to a Gray Surface Let us consider a opaque gray surface i of surface area Ai and at a temperature Ti. Let Gi be the total radiant energy incident upon the unit surface area per unit time. Ji For a surface i, the radiosity (Ji) can be expressed as: π½π = ππ πΈππ + ππ πΊπ (W/m2) Gi ....(1) For opaque surface i: πΌπ +ππ = 1 ππ, since, πΌπ = ππ (from Kirchoff ′ s law) ππ = 1 − πΌπ Surface i ρi G i εiEbi Ai, εi ∴ ππ = 1 − ππ βΈ« Eq (1) can be written as: π½π = ππ πΈππ + 1 − ππ πΊπ ππ, π½π − ππ πΈππ πΊπ = 1 − ππ ....(2) Continued… Radiation Heat Transfer Between Surfaces Net Radiation Heat Transfer from or to a Gray Surface π½π − ππ πΈππ πΊπ = 1 − ππ ....(2) ρiGi Gi If, Ji > Gi The net rate of heat transfer from surface i of surface area Ai is denoted by Qi and is expressed as: ππ = π΄π (W) π½π − ππ πΈππ π½π − 1 − ππ εiEbi ππ = π΄π ( πΊπ − π½π ) Ai, εi Surface i ππ = π΄π ( π½π − πΊπ ) If, Gi > Ji The net rate of heat transfer to surface i can be expressed as: Ji ⇒ ππ = Eq (3) can be represented in the form of electrical network as given below: I V1 Ji V2 R e π π1 − 2 πΌ= π π π΄π ππ πΈππ − π½π πΈππ − π½π πΈππ − π½π ....(3) ⇒ ππ = = = 1 − ππ 1 − ππ π π π΄π ππ π½π − πΈππ π½π − πΈππ = 1 − ππ π π π΄π ππ Surface i Ji 1−ε = Aε Qi Ebi The factor π π = (1 − ππ )Τπ΄π ππ is called surface resistance because this resistance is related to surface properties of the radiating body. ....(4) Eq (4) can be represented in the form of electrical network as given below: 1−ε = Aε Qi (W) Surface i Ebi Radiation Heat Transfer Between Surfaces Radiation Heat Exchange Between Any Two Gray Surface Surface j Let us consider two opaque gray surfaces i & j of surface area Ai & Aj, and maintained at uniform temperatures Ti & Tj respectively. Jj Radiation heat exchange between surface i and surface j can be expressed as: Ji Surface i Eq (1) can be represented in the form of electrical network as given below: πππ = π΄π π½π πΉππ − π΄π π½π πΉππ Surface j By using reciprocity relation, i.e., Ai Fij = Aj Fji, above Eq. becomes πππ = π΄π πΉππ π½π − π½π ππ, π½π − π½π π½π − π½π πππ = = 1 π ππ π΄π πΉππ Jj ....(1) The factor π ππ = 1Τπ΄π πΉππ is called space resistance, because this resistance is related to the gap and orientation between the radiating surfaces . Ji Surface i Radiation Heat Transfer in Two-Surface Enclosures Let us consider an enclosure consisting of two opaque gray surfaces 1 & 2 of surface area A1 & A2, and maintained at uniform temperatures T1 & T2 respectively. Surfaces 1 & 2 have emissivities ε1 & ε2 respectively. Ji T1 > T2 Gi Q12 Q1 1 Q2 ρiGi εiEbi 1 Q 1 Surface i 2 ππ = π΄π ( π½π − πΊπ ) Ai, εi Eb1 1 − ε1 A1 ε1 (W) As the surface 1 and surface 2 making an enclosure, so we can write: Net rate of radiation heat transfer from surface 1 (Q1) = Net radiation heat exchange between surface 1 and surface 2 (Q12) = Net rate of radiation heat transfer to surface 2 (Q2) 1 Q1 J1 Q12 J1 J2 1 π΄1 πΉ12 βΈ« we can also write for an enclosure consisting of surfaces 1 & 2 that: Net rate of radiation heat transfer from surface 1 to surface 2, (Q12)net = Q1 = Q12 = Q2 π12 πππ‘ πΈπ1 − π½1 π½1 − π½2 π½2 − πΈπ2 = = = 1 − π1 1 1 − π2 π΄1 π1 π΄1 πΉ12 π΄2 π2 2 Q2 Q2 J2 2 Eb2 ...(1) continued…. Radiation Heat Transfer in Two-Surface Enclosures π12 πππ‘ πΈπ1 − π½1 π½1 − π½2 π½2 − πΈπ2 = = = 1 − π1 1 − π2 1 π΄1 π1 π΄1 πΉ12 π΄2 π2 ...(1) Q12 Q1 1 From above Eq. (1), we can write following three separate equations: 1 1 − π1 ..(2) π½1 − π½2 = π12 πππ‘ πΈπ1 − π½1 = π12 πππ‘ π΄1 πΉ12 π΄1 π1 ..(3) Q2 2 π½2 − πΈπ2 = π12 1 − π2 π΄2 π2 πππ‘ ..(4) Adding Eqs. (2), (3) & (4), we get: πΈπ1 − πΈπ2 = π12 ⇒ π12 ⇒ π12 πππ‘ πππ‘ 1 − π1 1 1 − π2 + + π΄1 π1 π΄1 πΉ12 π΄2 π2 πΈπ1 − πΈπ2 πΈπ1 − πΈπ2 = = 1 − π1 1 − π2 π 1 + π 12 + π 2 1 π΄1 π1 + π΄1 πΉ12 + π΄2 π2 πππ‘ ππ π14 − π24 = 1 − π1 1 − π2 1 + + π΄1 π1 π΄1 πΉ12 π΄2 π2 ..(5) Eq (5) can be represented in the form of electrical E b1 network as shown in the fig. Q1 J1 Q12 R12 R1 1 J2 Q2 R2 2 If both the surfaces are black, i.e., ε1 = ε2 = 1, then this Eq. is reduced to π12 πππ‘ = π΄1 πΉ12 ππ π14 − π24 Eb2 Radiation Heat Transfer in Two-Surface Enclosures The net radiation heat flow between two gray surfaces, forming an enclosure, can be represented by an electric circuit as shown in Figure : π12 πππ‘ πΈπ1 − πΈπ2 π΄1 ππ π14 − π24 = = 1 − π1 1 − π2 1 − π1 1 − π2 π΄1 1 1 + + + + π΄1 π1 π΄1 πΉ12 π΄2 π2 π1 πΉ12 π2 . π΄2 Net radiation heat flow (Q12) for the common cases of two-surface enclosures of practical interest are given below: Concentric spheres 2 1 ∴ π12 π12 πππ‘ = π΄1 ππ π14 π24 − 1 − π1 1 − π2 π΄1 1 + + π1 πΉ12 π2 . π΄2 π΄1 4ππ12 π12 Here, F12 = 1 and = = π΄2 4ππ22 π22 π΄1 ππ π14 − π24 = πππ‘ = 1 − π1 1 − π2 π12 1 1 − π2 π12 π1 + 1 + π2 . π22 π1 + π2 . π22 Very long Concentric cylinders π12 1 πππ‘ π΄1 ππ π14 − π24 = 1 − π1 1 − π2 π΄1 1 + + . π1 πΉ12 π2 π΄2 2 Here, F12 = 1 and π΄1 ππ π14 − π24 ∴ π12 π΄1 2ππ1 πΏ π1 = = π΄2 2ππ2 πΏ π2 π΄1 ππ π14 − π24 π΄1 ππ π14 − π24 = πππ‘ = 1 − π 1 − π2 π1 1 1 − π2 π1 1 + 1 + . π1 π2 π2 π1 + π2 . π2 continued…. Radiation Heat Transfer in Two-Surface Enclosures Very large parallel plates or surfaces close to each other π12 πππ‘ π΄1 ππ π14 − π24 = 1 − π1 1 1 − π2 π΄1 + + . π1 πΉ12 π2 π΄2 Here, F12 = 1 and A1 = A2 = A ∴ π12 π΄ ππ π14 − π24 π΄ ππ π14 − π24 = πππ‘ = 1 − π 1 − π 1 1 1 2 +1+ + −1 π1 π2 π1 π2 Small object in a large enclosure π12 πππ‘ π΄1 ππ π14 − π24 = 1 − π1 1 1 − π2 π΄1 + + . π1 πΉ12 π2 π΄2 Here, F12 = 1 and ∴ π12 π΄1 →0 π΄2 π΄1 ππ π14 − π24 = π1 π΄1 ππ π14 − π24 πππ‘ = 1 − π1 +1 π1 Question Determine the heat lost by radiation per meter length of 80 mm diameter pipe at 300°C, if pipe is Case-1: located in a large room with red brick walls at a temperature of 27°C; Case-2: enclosed in a 160 mm diameter red brick conduit at a temperature of 27°C. Take ε for pipe and brick conduit = 0.79 and 0.73 respectively. Solution Case-1 Given: • emissivity of pipe = ε1 = 0.79; • radius of pipe = r1 = 80/2 mm = 40 mm = 0.04 m • temperature of pipe = T1 = 300 + 273 = 573 K • temperature of room = T2 = 27 + 273 = 300 K Pipe (1) Large room (2) Radiation heat flow between the pipe and large room is given by: π12 πΈπ1 − πΈπ2 π΄1 ππ π14 − π24 = = 1 − π1 1 1 − π2 1 − π1 1 1 − π2 π΄1 + + + + π΄1 π1 π΄1 πΉ12 π΄2 π2 π1 πΉ12 π2 . π΄2 In this case, F12 = 1 and ….(1) π΄1 →0 π΄2 π΄1 ππ π14 − π24 π΄1 ππ π14 − π24 ∴ π12 = = = π1 π΄1 ππ π14 − π24 1 − π1 1 +1 −1+1 π1 π1 ∴ π12 = 0.79 × 2π × 0.04 × 1 × 5.67 × 10−8 573 4 − 300 4 = 1122.6 π Τπ (π¨ππ) continued Question Determine the heat lost by radiation per meter length of 80 mm diameter pipe at 300°C, if pipe is Case-1: located in a large room with red brick walls at a temperature of 27°C; Case-2: enclosed in a 160 mm diameter red brick conduit at a temperature of 27°C. Take ε for pipe and brick conduit = 0.79 and 0.73 respectively. Solution Case-2 Given: • emissivity of pipe and brick conduit = ε1 & ε2 = 0.79 & 0.73 • radius of pipe = r1 = 80/2 mm = 40 mm = 0.04 m • radius of brick conduit = r2 = 160/2 mm = 80 mm = 0.08 m • temperature of pipe = T1 = 300 + 273 = 573 K • temperature of brick conduit = T2 = 27 + 273 = 300 K Radiation heat flow between the pipe and brick conduit is given by: π12 πΈπ1 − πΈπ2 π΄1 ππ π14 − π24 = = 1 − π1 1 1 − π2 1 − π1 1 1 − π2 π΄1 + + + + . π΄1 π1 π΄1 πΉ12 π΄2 π2 π1 πΉ12 π2 π΄2 In this case: πΉ12 = 1 and ….(1) π΄1 2ππ1 πΏ π1 0.04 = = = = 0.5 π΄2 2ππ2 πΏ π2 0.08 π΄1 ππ π14 − π24 π΄1 ππ π14 − π24 ∴ π12 = = 1 − π1 1 − π2 π1 1 1 − π2 π1 +1+ . + . π1 π2 π2 π1 π2 π2 2π × 0.04 × 1 × 5.67 × 10−8 573 ∴ π12 = 1 1 − 0.73 + × 0.5 0.79 0.73 4 − 300 4 = 979.05 π Τπ (π¨ππ) Pipe (1) Brick conduit (2) Radiation Shield Radiation heat transfer between two surfaces can be reduced greatly by inserting a thin, opaque and highly reflective sheet between the two surfaces. Such highly reflective thin sheet is called radiation shield. π12 Eb1 Eb2 π€ππ‘βππ’π‘ π βππππ = πΈπ1 − πΈπ2 1 − π1 1 1 − π2 + + π΄1 π1 π΄1 πΉ12 π΄2 π2 ∴ π12 π΄ ππ π14 − π24 π€ππ‘βππ’π‘ π βππππ = 1 − π 1 − π2 1 +1+ π1 π2 ∴ π12 π΄ ππ π14 − π24 π€ππ‘βππ’π‘ π βππππ = 1 1 + −1 π1 π2 π12 π€ππ‘β πππ π βππππ = We know that, for very large parallel surfaces: F12 = 1 and A1 = A2 = A ….(1) πΈπ1 − πΈπ2 1 − π1 1 1 − π3 1 − π3 1 1 − π2 + + + + + π΄1 π1 π΄1 πΉ13 π΄3 π3 π΄3 π3 π΄3 πΉ32 π΄2 π2 For very large parallel surfaces: F13 = F32 = 1 and A1 = A2 = A3 = A ∴ π12 Eb1 Eb2 Eb3 Under steady state condition Q13 = Q32 = Q12 ∴ π12 π΄ ππ π14 − π24 π€ππ‘β πππ π βππππ = 1 − π 1 − π3 1 − π3 1 − π2 1 +1+ + +1+ π1 π3 π3 π2 π΄ ππ π14 − π24 π€ππ‘β πππ π βππππ = 1 1 1 1 + −1 + + −1 π1 π3 π3 π2 ….(2) Radiation Shield Let us consider a special case when emissivities of radiating surfaces and shield are same, i.e., ε1 = ε2 = ε3 = ε 2 1 T1 T2 ε1 ε2 Q12 1 π12 shield 2 π12 π€ππ‘βππ’π‘ π βππππ π΄ ππ π14 − π24 π΄ ππ π14 − π24 = = 1 1 2 + −1 −1 π1 π2 π π€ππ‘β πππ π βππππ π΄ ππ π14 − π24 = 1 1 1 1 + −1 + + −1 π1 π3 π3 π2 T1 T3 T3 T2 ε1 ε3 ε3 ε2 Q12 1 shield shield T1 T3 T3 T4 T4 T2 ε1 ε3 ε3 ε4 ε4 ε2 Q12 π΄ ππ π14 − π24 1 π΄ ππ π14 − π24 1 = = . = π12 2 2 2 2 2 −1 + −1 −1 π π π 2 π12 π€ππ‘β π‘π€π π βπππππ If there are N shields with same surface emissivities as that of radiating surfaces, then: π€ππ‘βππ’π‘ π βππππ π΄ ππ π14 − π24 = 1 1 1 1 1 1 + −1 + + −1 + + −1 π1 π3 π3 π4 π4 π2 π΄ ππ π14 − π24 1 π΄ ππ π14 − π24 1 = = . = π12 2 2 2 2 3 3 −1 + −1 + −1 −1 π π π π π12 π€ππ‘β π π βπππππ 1 π΄ ππ π14 − π24 1 = . = π12 2 π+1 π + 1 π−1 π€ππ‘βππ’π‘ π βππππ π€ππ‘βππ’π‘ π βππππ Radiation Shield Temperature of shield in steady state condition Eb1 Eb2 Eb3 Under steady state condition Net heat transfer between surfaces 1 & 3 (Q13) = Net heat transfer between surfaces 3 & 2 (Q32), i.e., π΄ ππ π14 − π34 π΄ ππ π34 − π24 = 1 1 1 1 + − 1 + π1 π3 π3 π2 − 1 ⇒ π14 − π34 π34 − π24 = 1 1 1 1 + − 1 + π1 π3 π3 π2 − 1 ….(1) Above Eq. (1) is used to determine temperature of shield under steady state condition. 1 4 4 For special case when ε1 = ε2= ε3= ε, Eq. (1) is reduced to: π3 = π1 + π24 2 Radiation Heat Transfer in Three-Surface Enclosures On applying heat balance for each node, we get: Eb2 For node J1: 1 − π2 π΄2 π2 Q2 2 J2 ε2, A2, T2 1 π΄2 πΉ23 Q23 1 π΄1 πΉ12 Q13 Q12 3 1 π΄1 πΉ13 J1 ε1, A1, T1 1 Q1 1 − π1 π΄1 π1 Eb1 ε3, A3, T3 πΈπ1 − π½1 π½1 − π½2 π½1 − π½3 = + 1 − π1 1 1 π΄1 π1 π΄1 πΉ12 π΄1 πΉ13 or, πΈπ1 − π½1 π½1 − π½2 π½1 − π½3 − − =0 1 − π1 1 1 π΄1 π1 π΄1 πΉ12 π΄1 πΉ13 ……Eq. (1) Similarly, for node J2: πΈπ2 − π½2 π½1 − π½2 π½2 − π½3 + − =0 1 − π2 1 1 π΄2 π2 π΄1 πΉ12 π΄2 πΉ23 …Eq. (2) and, for node J3: πΈπ3 − π½3 π½2 − π½3 π½1 − π½3 + + =0 1 − π3 1 1 π΄2 πΉ23 π΄1 πΉ13 π΄3 π3 …Eq. (3) Question Consider two large parallel plates one at temperature of 727°C with emissivity 0.8 and other at temperature of 227°C with emissivity 0.4. An aluminium radiation shield with an emissivity 0.05 on both side is placed between the plates. Calculate: (1) percentage reduction in heat transfer rate between the plates. (2) Equilibrium temperature of the shield Solution plate 1 Given: • temperature of plate 1 = T1 = 727 + 273 = 1000 K • emissivity of plate 1= ε1 = 0.8 • temperature of plate 2 = T2 = 227 + 273 = 500 K • emissivity of plate 2 = ε2 = 0.4 • emissivity of shield = ε3 = 0.05 shield 3 plate 2 T1 T3 T3 T2 ε1 ε3 ε3 ε2 (1) π12 π΄ ππ π14 − π24 πΆ = = = 0.363 πΆ π€ππ‘βππ’π‘ π βππππ 1 1 1 1 π1 + π2 − 1 0.8 + 0.4 − 1 where, πΆ = π΄ ππ π14 − π24 π΄ ππ π14 − π24 π12 π€ππ‘β πππ π βππππ = 1 1 1 1 + − 1 + + π1 π3 π3 π2 − 1 π΄ ππ π14 − π24 = = 0.024 πΆ 1 1 1 1 0.8 + 0.05 − 1 + 0.05 + 0.4 − 1 βΈ« percentage reduction in heat 0.363 πΆ − 0.024 πΆ = × 100 = 93.4 % π¨ππ transfer rate between the plates 0.363 πΆ (2) Under steady state condition: Q13 = Q32 π΄ ππ π14 − π34 π΄ ππ π34 − π24 = 1 1 1 1 + − 1 + π1 π3 π3 π2 − 1 ⇒ 10004 − π34 π34 − 5004 = 1 1 1 1 + −1 + −1 0.8 0.05 0.05 0.4 ⇒ π34 = 5452.84 × 108 ∴ π3 = 859.32 πΎ (π¨ππ) END