Examiner: Laurent Mydlarski Associate Examiner: B. Rabi Baliga
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1 McGill University Department of Mechanical Engineering Final Exam MECH 346: Heat Transfer, Winter 2009 April 29, :00 to 17:00 (Duration: 3 hours Closed book. Closed notes. Faculty-standard calculator. A crib sheet is attached. There are 5 questions of different difficulties. Do all 5 problems. Maximum grade = 60. Write all your answers in the examination booklet(s. Examiner: Laurent Mydlarski Associate Examiner: B. Rabi Baliga Question #1 (/12 Are the following statements true or false? If false, explain why. (Do NOT simply say that the opposite of the statement is true this does not consist of an explanation. a In general, the thermal conductivity of liquids is greater than that of gases. b In conduction problems, convection at a surface corresponds to a Robin (mixed type of boundary condition. c Fins of effectiveness less than one impede heat transfer, when comparing the same surface (exposed to the same conditions without a fin. d The Fourier number can be interpreted as a non-dimensional time. e The Péclet number quantifies a flow s transport of energy by advection to its transport of energy by conduction. f In a zero-pressure-gradient, laminar flow over a flat plate, the hydrodynamic boundary layer thickness grows as x 1/2. g Turbulent internal flows of fluids of Prandtl number greater than one become hydrodynamically fully developed before they become thermally fully developed. h The volumetric thermal expansion coefficient (β of water can be approximated as 1/T. i The Rayleigh number is the product of the Grashof number and the Péclet number. j If the change in temperature of one of the fluids in a heat exchanger is negligible, then the correction factor to the log-mean temperature difference asymptotes to zero. k As the temperature of a blackbody is increased, the radiation it emits peaks at longer wavelengths. l Highly polished metals have emissivities that approach 1. 1
2 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 2 Question #2 (/12 A shell-and-tube heat exchanger is to heat 2.8 kg/s of water from 15 C to 82 C by hot engine oil flowing through the shell. The oil makes a single shell pass entering at 207 C, having a mass flow rate of 2.5 kg/s. The water flows though 10 brass tubes of 22.9 mm inside diameter, 25.4 mm outside diameter and length (per pass of 10 m. Each tube makes two passes though the shell. Assuming fully developed flow for the water, determine i the required convective heat transfer coefficient of the oil in the shell and ii the power required to pump the water through the tubes (which can be assumed to be smooth. Properties of brass: k = 11 W/(m K Properties of engine oil at 400 K: c p = 2337J/(kg K Properties of water: Temp.[ C] ν[m 2 /s] α[m 2 /s] k[w/(m K] c p [J/(kg K] ρ[kg/m 3 ] The appropriate correlations to use for the flow inside the tubes are: Nu D = 3.66 for laminar flow Nu D = 0.023Re 4/5 D Pr0.4 for turbulent flow with 0.7 Pr 160; Re D 10 4 ; L/D h 10; properties at T b For a 1-shell pass / 2 tube pass heat exchanger, the NTU relation is: NTU = (1 + Cr 2 1/2 ln( E 1 2/ǫ (1+Cr, where E = and C E+1 (1+Cr 2 r = C min /C max 1/2
3 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 3 Question #3 (/12 A conical cavity is formed by two parallel disks whose centres are aligned and separated by a distance L=0.25m. The cavity is evacuated. The upper disk is 0.5 m in diameter. The lower disk is 1 m in diameter and is very thin, such that its inside and outside surface temperatures are 350 K. The backside of the side wall is insulated. All surfaces can be approximated as being diffuse and gray with emissivities ǫ=0.8. Air at T =300 K blows over the outer surface of the lower disk with a convection coefficient of 20 W/(m 2 K. Determine (i the temperature of the inner surface of the upper disk, and (ii the temperature of the (insulated side wall. The shape factor between two concentric parallel disks is given by: F 12 = 1 2 ( S [S 2 4(D 2 /D 1 2 ] 1/2, where: S = R2 2, R 2 1 and where: R 1 = D 1 /(2L and R 2 = D 2 /(2L Though it is not required to solve this problem, the area of the side wall is given by: A 3 = π(r 2 2 r2 1 (1 + L2 (r 2 r 1 1/2, where r = D/2. 2
4 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 4 Question #4 (/12 An aluminum fin of square cross-section is attached to an integrated circuit that is maintained at a temperature of 77 C. The fin is m long and is m m in cross-section. A fan blows air at 27 C over the fin at a speed of 12 m/s and in a direction normal to a side of the fin (see figure. (i Calculate ml, where m = hp/(ka c. (ii What is the heat transfer rate from the fin? (iii Calculate the temperature at the midpoint of the fin (x = L/2. (iv How fast would the air have to blow to double the heat transfer rate? Advice: If you know how to easily do part (iv, do it. If not, move on and come back later. Properties of aluminum: k = 236 W/(m K. Properties of air: Temp.[K] ν[m 2 /s] α[m 2 /s] k[w/(m K] Use the following correlation for forced convective heat transfer over cylinders of square crosssection with air flowing normal to one face: Nu l hd k f = 0.102Re D Pr1/3. Range: < Re D < 10 5 Properties evaluated at T f = 1 2 (T sfc + T
5 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 5 Question #5 (/12 A baker bakes a loaf of bread which has dimensions 0.15m 0.15m 0.30m. When the baker removes the bread from the oven, it is at a uniform temperature T i = 450K. The baker leaves the bread on a cooling rack (which allows the bread to cool evenly on all sides in the ambient, quiescent air at T = 300K. (i What is the convective heat transfer coefficient, h, acting on the bread immediately upon its removal from the oven? (ii Assuming that h is constant at the value calculated in part (i, what is the temperature of the centre of the loaf after 20 minutes? (iii How does the assumption in (ii affect the answer you calculated in (ii? Properties of bread: ρ = 280 kg/m 3, c = 100 J/(kg K and k = W/(m K. Properties of air: Temp.[K] ν[m 2 /s] α[m 2 /s] k[w/(m K] Use the King correlation for natural convective heat transfer over isothermal rectangular blocks: Nu l hl k f = 0.55Ra 1/4 l, where l = L hl v, and where L (L h +L v v = the vertical dimension and L h = the longer of the two horizontal dimensions. Range: 10 4 < Ra l < 10 9 Properties evaluated at T f = 1 2 (T sfc + T
6 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 6 Equations of potential use In Cartesian co-ordinates: = î x + ĵ y + ˆk z, 2 = 2 x y z 2 In cylindrical co-ordinates: = ê r r + êθ r 2 = 2 r r In spherical co-ordinates: θ + ê z z, r r 2 θ z = 1 2 r = ê r r + êθ r θ + êφ rsinθ 2 = 1 r 2 r (r2 r + 1 r 2 sinθ φ, r (r r r 2 θ z 2 θ (sinθ θ + 1 r 2 sin 2 θ 2 φ 2 Conservation of Energy (integral form: Ė in Ėout + Ėg = dest dt Fourier s Law: q = ka T or q = k T, where q = q/a = heat flux Newton s Law of Cooling: q = ha(t s T Heat (Diffusion Equation: ρc p dt dt = (k T + q Thermal Resistances: Plane Wall Conductive Resistance: Cylindrical Shell Conductive Resistance: R tcond = L/(kA R tcond = ln(r 2 /r 1 /(2πLk Spherical Shell Conductive Resistance: R tcond = ( 1 r 1 1 r 2 /(4πk Convective Resistance: R tconv = 1/(hA Radiative Resistance: R trad = 1/(h r A Contact Resistance: R tcont = 1/(h contact A Fin Resistance: R tfin = θ o /q fin = θ o /(η f q max = 1/(η f ha f Critical Radii: Cylinder: r cr = k/h Sphere: r cr = 2k/h
7 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 7 Fin Theory: d Fin Equation: (ka dx c dt (T T dx hdas dx d Fin Equation (assuming A c = const. A s = Px: 2 θ m 2 θ = 0, where θ = T T dx 2 and m 2 = hp ka c Case 1 Very Long Fin BCs: θ(x = 0 = θ o ; lim L θ(x = L = 0 θ θ o = T(x T T o T = exp( mx q fin = M = θ o hpkac Case 2 Finite Length Fin BCs: θ(x = 0 = θ o ; k dθ θ θ o = T(x T T o T dx x=l = hθ(x = L = cosh[m(l x]+(h/(mksinh[m(l x] cosh[ml]+(h/(mksinh[ml] q fin = M sinh[ml]+(h/(mkcosh[ml] cosh[ml]+(h/(mksinh[ml] dθ Case 3 Fin with Adiabatic Tip BCs: θ(x = 0 = θ o ; dx x=l = 0 θ θ o = T(x T T o T = cosh[m(l x] cosh[ml] q fin = Mtanh[mL] Fin Efficiency = η f q fin /q max = q fin /(ha f θ o For a case 3 fin, η f = tanh[ml]/[ml] Fin Effectiveness = ǫ f q fin /q without fin = q fin /(ha c,base θ o = R t,base /R t,fin Approximations for Fins with Non-Adiabatic Tips q finnon adiabatic tip = Mtanh[mL c ] η finnon adiabatic tip = tanh[ml c ]/[ml c ] where L c = L + A tip /perimeter Conduction Shape Factors: q = ks T (i.e., R t = 1/(Sk Lumped Capacitance Analysis: valid for Bi hlc k s < 0.1; L c V/A θ(t θ i = T(t T T i T = exp[ ( ha s ρv c t] = exp[ Bi Fo] Q(t = (ρv cθ i [1 exp[ ( ha s ρv c t]] = ρv c(t i T αt (Fo L 2 c, α k ρc p
8 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 8 1-D Unsteady Conduction in a Plane Wall: θ (x = x/l, t = αt/l 2 ; Bi = T(x,t T T i T = n=1 C n exp( ζnt 2 cos(ζ n x where: C n = 4sin(ζ n 2ζ n +sin(2ζ n and ζ n is given by: ζ n tan(ζ n = Bi Q/Q o = E(t=0 E(t ρc p V (T i T = ρcp [T i T(x,t]dV ρc p V (T i T = 1 V (1 θ dv For t 0.2: θ = C 1 exp( ζ 2 1 t cos(ζ 1 x Q/Q o = 1 sin(ζ 1 ζ 1 θ o, where θ o = C 1 exp( ζ 2 1t 1-D Unsteady, Radial Conduction in an Infinite Cylinder: θ (r = r/r o, t = αt/ro 2; Bi = T(r,t T T i T = n=1 C n exp( ζn 2t J 0 (ζ n r where: C n = 2 J 1 (ζ n J ζ n J0 2(ζ n+j1 2(ζ and ζ n n is given by: ζ 1 (ζ n n J 0 (ζ n = Bi Q/Q o = E(t=0 E(t ρc p V (T i T = ρcp [T i T(x,t]dV ρc p V (T i T = 1 V (1 θ dv For t 0.2: θ = C 1 exp( ζ 2 1 t J 0 (ζ 1 r Q/Q o = 1 2θ o ζ 1 J 1 (ζ 1, where θ o = C 1 exp( ζ 2 1t 1-D Unsteady, Radial Conduction in a Sphere: θ (r = r/r o, t = αt/r 2 o ; Bi = T(r,t T T i T = n=1 C n exp( ζ 2 n t 1 where: C n = 4[sin(ζ n ζ n cos(ζ n ] 2ζ n sin(2ζ n Q/Q o = E(t=0 E(t ρc p V (T i T = ρcp [T i T(x,t]dV ρc p V (T i T ζ n r sin(ζ n r and ζ n is given by: 1 ζ n cot(ζ n = Bi (1 θ dv = 1 V For t 0.2: θ = C 1 exp( ζ1t 2 1 ζ 1 r sin(ζ 1 r Q/Q o = 1 3θ o ζ [sin(ζ ζ 1 cos(ζ 1 ], where θo = C 1exp( ζ1 2t
9 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 9
10 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 10
11 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 11
12 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 12
13 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 13
14 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 14
15 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 15
16 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 16
17 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p D Unsteady Conduction in a Semi-Infinite Solid: Case 1 Constant Surface Temperature (T s T(x,t T s x T i T s = erf[ 2 ] αt q s (t = k(t s T i παt Case 2 Constant Surface Heat Flux (q o T(x, t T i = 2q o k (αt π 1/2 exp[ x2 q s = q o Case 3 Surface Convection T(x,t T i T T i q o x 4αt ] q k erfc[ x = erfc[ x 2 αt ] ( exp[ hx k + h2 αt s(t = h[t T(x = 0, t] 2-D and 3-D Unsteady Conduction : Heat transfer for the intersection of 2 bodies: θcombined solid = ( θ = ( θ ( combined solid Q Qo = ( Q + ( [ ( ] Q total Q o 1 Q o 1 Q 2 Qo 1 Heat transfer for the intersection of 3 bodies: θ i ( k ] 2 θ i intersection solid 1 ( θ θ i θcombined solid = ( θ = ( θ ( combined solid Q Qo = ( Q + ( [ ( ( Q total Q o 1 Q o 1 Q 2 Qo + Q 1] Q o θ i θ i intersection solid 1 ( θ θ i 3 2 αt ] erf c[ x 2 αt + h intersection solid 2 [ 1 ( Q Qo 1] [ 1 ( Q Q o ( θ intersection solid ] 2 θ i 2 αt k ] intersection solid 3 Review of Fluid Mechanics : Conservation of Mass: ρ t + ( u ρ + ρ( u = 0 Conservation of Momentum for a constant property (ρ, µ flow: ρ D u Dt = p + ρ g + µ 2 u Conservation of Energy: ρ De Dt = (k T p ( u + µφ + q 2 [ ( u where (in 2-D: Φ = + 2 ( u y + v x x 2 + ( u x ] ( u x + v 2 y
18 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 18 Non-Dimensional Quantities: Re L = Reynolds number = UL ν Pr = Prandtl number = ν (α k α ρc p Ec = Eckert number = U 2 c p(t w T τw 1 2 ρu2 C f = Skin friction coefficient = Nu L = Nusselt number = hl St = Stanton number = Pe L = Péclet number = UL α k f h = Nu ρu cp Re Pr = Re Pr Gr L = Grashof number = gβ(tw T L3 ν 2 Ra L = Rayleigh number = gβ(tw T L3 να Momentum / Heat Transfer Analogy: (where β 1 ρ = Gr L Pr If Pr = 1: 1 2 C f = St If Pr 1 : 1 2 C f = St Pr 2/3 (for 0.6 < Pr < 50 ( ρ = 1 T P T for an ideal gas Solutions to the Laminar Boundary Layer Equations for an Isothermal, Heated Flat Plate: δ = x 5.0Re 1/2 x C fx = τwx 1 2 ρu2 C fl = τw L 1 2 ρu2 = 0.664Re 1/2 x = 1.328Re 1/2 L Nu x = hxx k f = 0.332Re 1/2 (Local (Average x Pr1/3 (Local for 0.6 < Pr < 50 Nu L = h LL k f = 0.664Re 1/2 L Pr 1/3 (Average for 0.6 < Pr < 50 δ/δ t Pr 1/3 Nu x = 0.565Pe 1/2 x (Local forpr 0.05, Pe x 100 All properties evaluated at T f Tw+T 2 Empirical Relations for Turbulent Flow over an Isothermal, Heated Flat Plate: δ = x 0.37Re 1/5 x C fx = τwx 1 2 ρu2 C fl = τw L 1 2 ρu2 = Re 1/5 x = 0.074Re 1/5 L Nu x = hxx k f = Re 4/5 (Local (Average x Pr1/3 (Local for 0.6 < Pr < 50 Nu L = h LL k f = 0.037Re 4/5 L Pr 1/3 (Average for 0.6 < Pr < 50 All properties evaluated at T f Tw+T 2 Correlations apply for < Re x < 10 8
19 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 19 Empirical Relations for Flow over an Isothermal, Heated Flat Plate with Mixed Boundary Layer Conditions: C fl = 0.074Re 1/5 L 1742Re 1 L (Average Nu L = (0.037Re 4/5 L 871Pr 1/3 (Average for 0.6 < Pr < 50 All properties evaluated at T f Tw+T 2 Correlations apply for < Re L < 10 8 and assume Re xcrit = Empirical Relations for Flow over an Isothermal, Heated Flat Plate with Unheated Starting Lengths: For laminar flow with an unheated starting length ξ and x > ξ: Nu x = Nux ξ=0, where Nu [1 (ξ/x 3/4 ] 1/3 x ξ=0 = 0.332Re 1/2 x Pr1/3 (for 0.6 < Pr < 50 For turbulent flow with an unheated starting length ξ and x > ξ: Nu Nu x = x ξ=0, where Nu [1 (ξ/x 9/10 ] 1/9 x ξ=0 = Re 4/5 x Pr1/3 (for 0.6 < Pr < 50 Empirical Relations for Flow over a Flat Plate with Uniform Surface Heat Flux BC: For laminar flow: Nu x = 0.453Rex 1/2Pr1/3 For turbulent flow: Nu x = Rex 4/5 Pr 1/3 and: T s (x = T + q w /h x Empirical Relations for Forced Convection over Other Geometries: Correlations will be given with the question Flow through Tube Banks: (S n = pitch normal to the mean flow and S p = pitch parallel to the mean flow ( For inline tubes: u max = u Sn S n D ( For staggered tubes: u max = max of: u Sn S n D or ( S u n/2 [(S n/2 2 +(S p 2 ] 1/2 D ( P = 2f G 2 max N µw 0.14 ρ µ b where G max = ρu max and N = number of transverse rows of tubes f = [ ] [(S n D/D] Re max for staggered tubes [ ] f 0.08S = p/d Re 0.15 [(S n D/D] D/Sp max for inline tubes where: Re max = umaxd ν q = h(nπdl T LM where: T LM = (Tw T i (T w T o ln( Tw T i and N is the total number of tubes Tw To
20 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 20 (Forced Internal Flow: ṁ = ρa cs U AV Re DH = U AV D H, where D ν H = 4Acs P For laminar flow: ( x F Dh D 0.06ReDH For turbulent flow: ( x F Dh D 4.4Re 1/6 D H For the purposes of this course, we will say FD flow occurs for ( x F Dh D For pipeflow in general: Darcy friction factor = f 8τw ; τ ρuav 2 w = ro 2 For laminar pipeflow: U/U max = 1 (r/r o 2 ; U = U AV = U max /2; U max = r2 o 4µ f = 64 Re DH = (4C f For turbulent pipeflow: dp dx [ ] dp dx > 10 f = (1.82log 10 Re DH ; for smooth tubes (otherwise use the Moody Diagram For laminar flow: ( x F Dt D 0.06ReDH Pr For turbulent flow: ( ( x F Dt xf D Dh D Acs Bulk Temperature = T b ρucptdacs Acs ρucpdacs Newton s Law of Cooling: q w = h(t w T b h For thermally fully-developed flow: = 0 x Average Heat Flux: q w = h[ T] mean Energy balance on an internal flow: dq conv = q w Pdx = ṁc pdt b or q conv = x 2 x 1 q w Pdx = ṁc p(t b2 T b1 dt b w P dx ṁc p = P ṁc p h(t w T b For fully-developed pipeflow in with a constant surface heat flux: T b = T bi + q w P ṁc p x [ T] mean = T w T b Nu D = 4.36 (for laminar flow in circular pipes For fully-developed pipeflow in with a constant surface temperature: ln ( T o T i = PL ṁc p h [ T] mean = T LM = To T ( i Nu D = 3.66 ln To T i (for laminar flow in circular pipes Overall Heat Transfer Coefficient, U, is defined by: UA = U i A i = U o A o 1 ΣR t
21 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 21 Empirical Relations for Internal Forced Turbulent Convection: Correlations will be given with the question Natural (Free Convection: Natural convection dominates over forced convection when Gr L /Re 2 L 1 Empirical Relations for Natural Convection: Correlations will be given with the question Mixed Convection: Nu n mixed = Nun forced + Nun natural n 3 for assisting flow where: n 3.5 for transverse flow over a vertical plate n 4 for transverse flow over a cylinder or sphere Nu n mixed = Nu n forced Nu n natural n 3 for opposing flow Heat Exchangers: where: Overall Heat Transfer Coefficient, U, is defined by: UA = U i A i = U o A o 1 ΣR t Fouling Factor = R f 1 U dirty 1 U clean Log-Mean Temperature Difference (LMTD Method: Three governing equations: q = ṁ h c ph (T hi T ho = C h (T hi T ho (C h = ṁ h c ph q = ṁ c c pc (T co T ci = C c (T co T ci (C c = ṁ c c pc q = (UAF T LMTD ; where: T LMTD T 2 T 1 ln( T 2 / T 1 For a parallel flow HX: F = 1 T 1 = T h1 T c1 = T hi T ci T 2 = T h2 T c2 = T ho T co For a counter flow HX: F = 1 T 1 = T h1 T c1 = T hi T co T 2 = T h2 T c2 = T ho T ci For a multi-pass and cross-flow HX, F=F(P,R (where P and R are non-dimensional temperature differences that depend on the type of HX and T = T LMTDcf. For C min C max : F = 1
22 Final Exam MECH 346: Heat Transfer April 29, :00 to 17:00 p. 22 Heat Exchangers (continued: ǫ NTU Method: Three governing equations: q = ṁ h c ph (T hi T ho = C h (T hi T ho (C h = ṁ h c ph q = ṁ c c pc (T co T ci = C c (T co T ci (C c = ṁ c c pc q = ǫq max, where q max C min (T hi T ci NTU UA C min For any heat exchanger, ǫ = f (NTU, C min /C max. Analytical or graphical expressions of these relations depend on the type of heat exchanger. For C min C max : Radiative Heat Transfer: ǫ = 1 exp( NTU c = λν = m/s 2πhc E λ,b (λ, T 2 o, where: λ 5 [exp(hc o/(kλt 1] h = J s = Planck s constant k = J/K = Boltzmann s constant c o = m/s = speed of light in a vacuum Wien s Displacement Law: λ max T = µm K Stefan-Boltzmann Law: E b (T = σt 4 where σ = W/(m 2 K 4 In general: ǫ λ,θ (λ, θ, T E λ,θ (λ, θ, T/E λ,b (λ, T = spectral, directional emissivity Integrating over all directions and wavelengths, one ( obtains the 0 total, hemispherical emissivity: ǫ E/E b = G i = irradiation ǫ λ E λ,b dλ E b In general: α λ,θ (λ, θ G λ,θ,abs (λ, θ/g λ,θ (λ, θ = spectral, directional absorptivity Integrating over all directions and wavelengths, one obtains the total, hemispherical absorptivity : α G abs /G Similarly for the reflectivity (ρ and the transmisivity (τ. Note α + ρ + τ = 1 Kirchoff s Law: ǫ λ,θ = α λ,θ. If i the irradiation corresponds to emission from a blackbody at the same temperature as the surface, or ii the surface is gray and diffuse, then: ǫ = α. F ij the fraction of energy leaving surface i that intercepts surface j F ij 1 cosθ i cosθ j A i A i A j da i da j πr 2
23 Radiative Heat Transfer (continued: Reciprocity Relation: A i F ij = A j F ji Summation Rule: N j=1 F ij = 1 For a subdivided surface, A j, such that A j = N k=1 A k : F ij = N k=1 F ik and F ji = N k=1 A k F ki / N k=1 A k J i = radiosity = ǫe bi + ρg i For radiative heat transfer between diffuse, gray surfaces: q i = A i (J i G i = ( E b i J. i 1 ǫ i ǫ i A i For radiative heat transfer between multiple diffuse, gray surfaces: (J i J j q i = N j=1 q ij = ( E b i J i = N 1 ǫ i j=1 ( 1 ǫ i A i A i F ij F 0 λ = the fraction of total radiation emitted by a blackbody that falls between 0 and λ λ 0 = E λ λ,bdλ = E 0 λ,bdλ = f(λt E λ,b dλ σt
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