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RADIAL HEAT CONDUCTION SOLVED USING THE INTEGRAL EQUATION .pdf

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Wasswaderrick3

We look at the case of radial heat flow. Again, in radial heat flow, the temperature profiles that satisfy the boundary and initial conditions are the exponential and hyperbolic functions as derived in literature of conduction in fins. We use the technique of transforming the PDE into an integral equation. But in the case of radial heat flow, we have to multiply through by r the heat equation and then introduce integrals. We do this to avoid introducing integrals of the form of the exponential integral whose solutions cannot be expressed in the form of a simple mathematical function. We look at the case of a semi-infinite hollow cylinder for both insulated and non-insulated cases and then find the solution. We also look at cases of finite radius hollow cylinders subject to given boundary conditions. We notice that the solutions got for finite radius hollow cylinders do not reduce to those of semi-infinite hollow cylinders. We conclude by saying that this same analysis can be extended to spherical co-ordinates heat conduction.

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Using the integral heat equation Wasswa Derrick 4/27/24 Applied Mathematics
TABLE OF CONTENTS PREFACE ............................................................................................................................................3 RADIAL HEAT CONDUCTION WITH NO LATERAL CONVECTION...................................4 CASE1: SEMI-INFINITE HOLLOW CYLINDER...........................................................................................4 CASE 2: FIXED END TEMPERATURE ......................................................................................................9 CASE3: CONVECTION AT THE FREE END .............................................................................................16 CASE4: ZERO CONVECTION AT THE FREE END ....................................................................................24 RADIAL HEAT CONDUCTION WITH LATERAL CONVECTION........................................30 CASE1: SEMI-INFINITE CASE...............................................................................................................31 CASE2: CONVECTION AT THE FREE END .............................................................................................34 CONCLUSION...................................................................................................................................43 REFERENCES..................................................................................................................................44
PREFACE We look at the case of radial heat flow. Again, in radial heat flow, the temperature profiles that satisfy the boundary and initial conditions are the exponential and hyperbolic functions as derived in literature of conduction in fins. We use the technique of transforming the PDE into an integral equation. But in the case of radial heat flow, we have to multiply through by r the heat equation and then introduce integrals. We do this to avoid introducing integrals of the form of the exponential integral whose solutions cannot be expressed in the form of a simple mathematical function. We look at the case of a semi-infinite hollow cylinder for both insulated and non-insulated cases and then find the solution. We also look at cases of finite radius hollow cylinders subject to given boundary conditions. We notice that the solutions got for finite radius hollow cylinders do not reduce to those of semi-infinite hollow cylinders. We conclude by saying that this same analysis can be extended to spherical co- ordinates heat conduction.
RADIAL HEAT CONDUCTION WITH NO LATERAL CONVECTION CASE1: SEMI-INFINITE HOLLOW CYLINDER. The governing PDE is: 𝜢 𝝏𝟐 𝑻 ππ’“πŸ + 𝜢 𝟏 𝒓 𝝏𝑻 𝝏𝒓 = 𝝏𝑻 𝝏𝒕 The boundary conditions are 𝑇 = 𝑇𝑠 π‘Žπ‘‘ π‘Ÿ = π‘Ÿ1 𝑇 = π‘‡βˆž π‘Žπ‘‘ π‘Ÿ = ∞ The initial condition is: 𝑇 = π‘‡βˆž π‘Žπ‘‘ 𝑑 = 0 The temperature profile that satisfies the conditions above is 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 We transform the equation above into an integral equation and take integrals with limits from π‘Ÿ = π‘Ÿ1 to π‘Ÿ = 𝑅 = ∞. 𝛼 πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ = πœ•π‘‡ πœ•π‘‘ We take integrals and get
𝛼 ∫ πœ•2 𝑇 πœ•π‘Ÿ2 π‘‘π‘Ÿ 𝑅 π‘Ÿ1 + 𝛼 ∫ [ 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = πœ• πœ•π‘‘ ∫ (𝑇)π‘‘π‘Ÿ 𝑅 π‘Ÿ1 Where: 𝑅 = ∞ Let us look at the integral below: ∫ [ 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 πœ•π‘‡ πœ•π‘Ÿ = βˆ’ 𝑇𝑠 βˆ’ π‘‡βˆž 𝛿 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ∫ [ 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = βˆ’ ∫ [ 1 π‘Ÿ 𝑇𝑠 βˆ’ π‘‡βˆž 𝛿 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 Upon rearranging, we get: ∫ [ 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = βˆ’ 𝑇𝑠 βˆ’ π‘‡βˆž 𝛿 𝑒 π‘Ÿ1 𝛿 ∫ [ 1 π‘Ÿ 𝑒 βˆ’π‘Ÿ 𝛿 ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 Calling 𝑒 = π‘Ÿ 𝛿 π‘Ÿ = 𝑒𝛿 π‘‘π‘Ÿ = 𝛿𝑑𝑒 So, we have ∫ [ 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = βˆ’ 𝑇𝑠 βˆ’ π‘‡βˆž 𝛿 𝑒 π‘Ÿ1 𝛿 ∫ [ 1 𝑒 π‘’βˆ’π‘’ ]𝑑𝑒 𝑅 𝑒 π‘Ÿ1 𝛿 This integral is in the form of the Exponential integral i.e., ∫[ 1 π‘₯ π‘’βˆ’π‘₯ ]𝑑𝑒 = 𝐸𝑖(βˆ’π‘₯) + 𝑐 and is a non-elementary function. This means that the integral cannot be expressed as a simple function. To avoid this problem, we have to multiply through by r the heat equation as shown below and solve. 𝛼 ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ 𝑅 π‘Ÿ1 + 𝛼 ∫ [ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ 𝑅 π‘Ÿ1 Let us evaluate:
∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ 𝑅 π‘Ÿ1 πœ•2 𝑇 πœ•π‘Ÿ2 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿2 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ∫(π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿2 ∫ [π‘Ÿπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ] π‘‘π‘Ÿ We use ∫(𝑒𝑣)𝑑π‘₯ = 𝑒𝑣 βˆ’ ∫ 𝑣 𝑑𝑒 𝑑π‘₯ 𝑑π‘₯ Let us evaluate: ∫ [π‘Ÿπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ] π‘‘π‘Ÿ Let 𝑒 = π‘Ÿ π‘Žπ‘›π‘‘ 𝑑𝑣 π‘‘π‘Ÿ = 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 β„Žπ‘’π‘›π‘π‘’ 𝑣 = βˆ’π›Ώπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ∫ [π‘Ÿπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ] π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 + 𝛿 ∫ (𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ) π‘‘π‘Ÿ ∫ [π‘Ÿπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ] π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 βˆ’ 𝛿2 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 Now let us put the limits ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ ∞ π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿2 ([βˆ’π‘Ÿπ›Ώπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ] ∞ π‘Ÿ1 βˆ’ 𝛿2 [𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ] ∞ π‘Ÿ1 ) = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 (π‘Ÿ1 + 𝛿) ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ ∞ π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 (π‘Ÿ1 + 𝛿) Now let us evaluate the integral: ∫ [ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ ∞ π‘Ÿ1 πœ•π‘‡ πœ•π‘Ÿ = βˆ’ 𝑇𝑠 βˆ’ π‘‡βˆž 𝛿 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ∫ [ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ ∞ π‘Ÿ1 = ∫ [βˆ’ 𝑇𝑠 βˆ’ π‘‡βˆž 𝛿 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ] π‘‘π‘Ÿ ∞ π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž)[𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ] ∞ π‘Ÿ1 ∫ [ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ ∞ π‘Ÿ1 = βˆ’(𝑇𝑠 βˆ’ π‘‡βˆž)
Now let us evaluate: ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ 𝑅=∞ π‘Ÿ1 𝑇 = (𝑇𝑠 βˆ’ π‘‡βˆž)𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 + π‘‡βˆž ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ 𝑅=∞ π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž) ∫ (π‘Ÿπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 )π‘‘π‘Ÿ 𝑅=∞ π‘Ÿ1 + ∫ π‘‡βˆžπ‘‘π‘Ÿ 𝑅=∞ π‘Ÿ1 = 𝛿(𝑇𝑠 βˆ’ π‘‡βˆž) + π‘‡βˆž(𝑅 βˆ’ π‘Ÿ1) ∫ (π‘Ÿπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 )π‘‘π‘Ÿ 𝑅=∞ π‘Ÿ1 = π‘Ÿ1𝛿 + 𝛿2 ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ 𝑅=∞ π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž)(π‘Ÿ1𝛿 + 𝛿2) + π‘‡βˆž(𝑅 βˆ’ π‘Ÿ1) Upon substituting all the above in the integral heat equation, we get 𝛼 [ (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 (π‘Ÿ1 + 𝛿) βˆ’ (𝑇𝑠 βˆ’ π‘‡βˆž)] = πœ• πœ•π‘‘ [(𝑇𝑠 βˆ’ π‘‡βˆž)(π‘Ÿ1𝛿 + 𝛿2 ) + π‘‡βˆž(𝑅 βˆ’ π‘Ÿ1)] But πœ•(π‘‡βˆž(𝑅 βˆ’ π‘Ÿ1)) πœ•π‘‘ = 0 𝑠𝑖𝑛𝑐𝑒 π‘‡βˆž π‘Žπ‘›π‘‘ (𝑅 βˆ’ π‘Ÿ1) π‘Žπ‘Ÿπ‘’ π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘π‘  So, we are left with 𝛼 [ (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 (π‘Ÿ1 + 𝛿) βˆ’ (𝑇𝑠 βˆ’ π‘‡βˆž)] = πœ• πœ•π‘‘ [(𝑇𝑠 βˆ’ π‘‡βˆž)(π‘Ÿ1𝛿 + 𝛿2 )] 𝛼 [ 1 𝛿 (π‘Ÿ1 + 𝛿) βˆ’ 1] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 + 𝛿2 )] Upon substitution, we get 𝛼 π‘Ÿ1 𝛿 = (π‘Ÿ1 + 2𝛿) 𝑑𝛿 𝑑𝑑 The boundary condition is: 𝛿 = 0 π‘Žπ‘‘ 𝑑 = 0 So, we get: π›Όπ‘Ÿ1 ∫ 𝑑𝑑 𝑑 0 = ∫ (π‘Ÿ1𝛿 + 2𝛿2) 𝛿 0 𝑑𝛿 Upon simplification, we get:
π›Όπ‘Ÿ1𝑑 = π‘Ÿ1𝛿2 2 + 2𝛿3 3 We get: 6π›Όπ‘Ÿ1𝑑 = 3π‘Ÿ1𝛿2 + 4𝛿3 i.e., πŸ‘π’“πŸπœΉπŸ + πŸ’πœΉπŸ‘ βˆ’ πŸ”πœΆπ’“πŸπ’• = 𝟎 Which is a cubic function and can be solved to get 𝛿 as a function of time t. You notice that the initial condition is satisfied for the above temperature profile. We can also go ahead and look at situations where there is natural convection and other situations where the radius r is finite and not infinite.
CASE 2: FIXED END TEMPERATURE The initial and boundary conditions are: 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 𝑻 = π‘»π’ƒπŸ βˆ’ π‘»βˆž 𝒂𝒕 𝒙 = 𝟎 𝑻 = π‘»π’ƒπŸ βˆ’ π‘»βˆž 𝒂𝒕 𝒙 = 𝑳 We start with a temperature profile below: 𝑇 βˆ’ π‘‡βˆž = 𝐴𝑒 π‘₯ 𝛿 + 𝐡𝑒 βˆ’π‘₯ 𝛿 We evaluate the constants A and B using the boundary conditions and get the temperature profile below: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) π‘ π‘–π‘›β„Ž π‘₯ 𝛿 + π‘ π‘–π‘›β„Ž 𝐿 βˆ’ π‘₯ 𝛿 π‘ π‘–π‘›β„Ž 𝐿 𝛿 The temperature profile above can be referenced in textbooks for heat flow in extended surfaces like in the book [1]. To transform the above equation to cylindrical co-ordinates, we use the substitutions, 𝐿 = (π‘Ÿ2 βˆ’ π‘Ÿ1) And π‘₯ = π‘Ÿ βˆ’ π‘Ÿ1 And the temperature profile becomes: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )π‘ π‘–π‘›β„Ž π‘Ÿ βˆ’ π‘Ÿ1 𝛿 + π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 The governing PDE is: 𝜢 𝝏𝟐 𝑻 ππ’“πŸ + 𝜢 𝟏 𝒓 𝝏𝑻 𝝏𝒓 = 𝝏𝑻 𝝏𝒕 The initial and boundary conditions are: 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎
𝑻 = π‘»π’ƒπŸ βˆ’ π‘»βˆž 𝒂𝒕 𝒓 = π’“πŸ 𝑻 = π‘»π’ƒπŸ βˆ’ π‘»βˆž 𝒂𝒕 𝒓 = π’“πŸ The temperature profile that satisfies the above conditions is 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )π‘ π‘–π‘›β„Ž π‘Ÿ βˆ’ π‘Ÿ1 𝛿 + π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 To simplify matters, we can express the temperature profile as below: 𝑇 βˆ’ π‘‡βˆž = (𝑇𝑏2 βˆ’ π‘‡βˆž)π‘ π‘–π‘›β„Ž π‘Ÿ βˆ’ π‘Ÿ1 𝛿 π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 + (𝑇𝑏1 βˆ’ π‘‡βˆž)π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 Simply as: 𝑇 βˆ’ π‘‡βˆž = π΄π‘ π‘–π‘›β„Ž [ π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ] + π΅π‘ π‘–π‘›β„Ž [ π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ] Where: 𝐴 = (𝑇𝑏2 βˆ’ π‘‡βˆž) π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 And 𝐡 = (𝑇𝑏1 βˆ’ π‘‡βˆž) π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 So, the temperature profile we are going to use is: 𝑻 βˆ’ π‘»βˆž = π‘¨π’”π’Šπ’π’‰ [ 𝒓 βˆ’ π’“πŸ 𝜹 ] + π‘©π’”π’Šπ’π’‰ [ π’“πŸ βˆ’ 𝒓 𝜹 ] 𝛼 πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ = πœ•π‘‡ πœ•π‘‘ Multiplying through by r the heat equation becomes: π›Όπ‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 πœ•π‘‡ πœ•π‘Ÿ = πœ•π‘Ÿπ‘‡ πœ•π‘‘ Transforming the PDE into an integral equation, we get: 𝛼 [∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] = πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1
Let us evaluate the integral ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 πœ•2 𝑇 πœ•π‘Ÿ2 = 1 𝛿2 (π΄π‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ) + π΅π‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 = 1 𝛿2 (π΄π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ) + π΅π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] Let us evaluate the integral ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 We use ∫(𝑒𝑣)𝑑π‘₯ = 𝑒𝑣 βˆ’ ∫ 𝑣 𝑑𝑒 𝑑π‘₯ 𝑑π‘₯ Let 𝑒 = π‘Ÿ π‘Žπ‘›π‘‘ 𝑑𝑣 π‘‘π‘Ÿ = π‘ π‘–π‘›β„Ž ( π‘Ÿβˆ’π‘Ÿ1 𝛿 ) β„Žπ‘’π‘›π‘π‘’ 𝑣 = 𝛿cosh( π‘Ÿβˆ’π‘Ÿ1 𝛿 ) ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿβˆ’π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ = π‘Ÿπ›Ώ cosh ( π‘Ÿβˆ’π‘Ÿ1 𝛿 ) βˆ’ 𝛿 ∫ (cosh ( π‘Ÿβˆ’π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ = π‘Ÿπ›Ώ cosh ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 𝛿2 (sinh( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 )) ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [π‘Ÿπ›Ώ cosh ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 )] π‘Ÿ2 π‘Ÿ1 βˆ’ 𝛿2 [sinh( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 )] π‘Ÿ2 π‘Ÿ1 Simplifying, we get: ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 )) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = π‘Ÿ2𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 βˆ’ 𝛿2 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) Now let us evaluate the integral ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 Let 𝑒 = π‘Ÿ π‘Žπ‘›π‘‘ 𝑑𝑣 π‘‘π‘Ÿ = π‘ π‘–π‘›β„Ž ( π‘Ÿ2βˆ’π‘Ÿ 𝛿 ) β„Žπ‘’π‘›π‘π‘’ 𝑣 = βˆ’π›Ώcosh( π‘Ÿ2βˆ’π‘Ÿ 𝛿 )
∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ) + 𝛿 ∫ (cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ cosh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ) βˆ’ 𝛿2 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [βˆ’π‘Ÿπ›Ώ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )] π‘Ÿ2 π‘Ÿ1 βˆ’ 𝛿2 [sinh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )] π‘Ÿ2 π‘Ÿ1 Simplifying, we get: ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 (π‘Ÿ2𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 βˆ’ 𝛿2 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝐡 (π‘Ÿ1𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Simplifying we get: ∫ (𝒓 𝝏𝟐 𝑻 ππ’“πŸ )𝒅𝒓 π’“πŸ π’“πŸ = [𝐴 ( π‘Ÿ2 𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1 𝛿 βˆ’ sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝐡 ( π‘Ÿ1 𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2 𝛿 + sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Now let us evaluate the integral: ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [π΄π‘ π‘–π‘›β„Ž [ π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ] + π΅π‘ π‘–π‘›β„Ž [ π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ]] π‘Ÿ2 π‘Ÿ1 = (𝐴 βˆ’ 𝐡)sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) ∫ [ 𝝏𝑻 𝝏𝒓 ] 𝒅𝒓 π’“πŸ π’“πŸ = 𝑨 𝐬𝐒𝐧𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 ) βˆ’ 𝑩𝐬𝐒𝐧𝐑( π’“πŸ βˆ’ π’“πŸ 𝜹 ) Now let us evaluate the integral πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑇 = π΄π‘ π‘–π‘›β„Ž [ π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ] + π΅π‘ π‘–π‘›β„Ž [ π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ] + π‘‡βˆž ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 𝐴 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + π‘‡βˆž 2 (π‘Ÿ2 2 βˆ’ π‘Ÿ1 2 )
But we have already evaluated the above so we get: ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 (π‘Ÿ2𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 βˆ’ 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝐡 (π‘Ÿ1𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] + π‘‡βˆž 2 (π‘Ÿ2 2 βˆ’ π‘Ÿ1 2 ) Substituting all the above in the integral equation, we get 𝜢( π‘¨π’“πŸ 𝜹 𝐜𝐨𝐬𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 ) βˆ’ π‘¨π’“πŸ 𝜹 βˆ’ 𝑨𝐬𝐒𝐧 𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 )) + ( π‘©π’“πŸ 𝜹 𝐜𝐨𝐬𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 ) βˆ’ π‘©π’“πŸ 𝜹 + 𝑩𝐬𝐒𝐧 𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 )) + πœΆπ‘¨ 𝐬𝐒𝐧𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 ) βˆ’ πœΆπ‘©π¬π’π§π‘( π’“πŸ βˆ’ π’“πŸ 𝜹 ) = πœ• πœ•π‘‘ [𝐴 (π‘Ÿ2𝛿cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 βˆ’ 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝐡 (π‘Ÿ1𝛿cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Since πœ• πœ•π‘‘ [ π‘‡βˆž 2 (π‘Ÿ2 2 βˆ’ π‘Ÿ1 2)] = 0 We finally get after crossing out the sinh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) terms 𝛼 ( π΄π‘Ÿ2 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΄π‘Ÿ1 𝛿 ) + ( π΅π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ) = πœ• πœ•π‘‘ [𝐴 (π‘Ÿ2𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 βˆ’ 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Collecting like terms in A and B we get: 𝛼 ( π΄π‘Ÿ2 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΄π‘Ÿ1 𝛿 ) = πœ• πœ•π‘‘ (𝐴 (π‘Ÿ2𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 βˆ’ 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )))… 𝑀 𝛼 ( π΅π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ) = πœ• πœ•π‘‘ (𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))). . 𝑁 From equation N we can get an expression for 𝛿2 sinh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) which we can substitute in equation M. From equation N, we have: ∫ 𝛼 ( π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2 𝛿 )𝑑𝑑 βˆ’ π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + π‘Ÿ2𝛿 = 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) Substituting in equation M which is 𝛼 ( π‘Ÿ2 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1 𝛿 ) = πœ• πœ•π‘‘ ((π‘Ÿ2𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 βˆ’ 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))) We get 𝛼 ( π‘Ÿ2 𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1 𝛿 ) = πœ• πœ•π‘‘ ((π‘Ÿ2𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 βˆ’ [𝛼 ∫( π‘Ÿ1 𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2 𝛿 )𝑑𝑑 βˆ’ π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + π‘Ÿ2𝛿)]) Which gives 𝛼 ( π‘Ÿ2 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1 𝛿 ) = πœ• πœ•π‘‘ ((π‘Ÿ2𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 + π‘Ÿ1𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿)) βˆ’ 𝛼 ( π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2 𝛿 )
Collecting like terms we have 𝛼 𝛿 ((π‘Ÿ2 + π‘Ÿ1)cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ (π‘Ÿ2 + π‘Ÿ1)) = 𝑑𝛿 𝑑𝑑 ((π‘Ÿ2 + π‘Ÿ1)cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ (π‘Ÿ2 + π‘Ÿ1)) The like terms of ((π‘Ÿ2 + π‘Ÿ1)cosh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) βˆ’ (π‘Ÿ2 + π‘Ÿ1)) cancel out and we remain with 𝛼 𝛿 = 𝑑𝛿 𝑑𝑑 Hence, we have 𝛿 = √2𝛼𝑑 Hence the temperature profile is: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )π‘ π‘–π‘›β„Ž π‘Ÿ βˆ’ π‘Ÿ1 𝛿 + π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 Or 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )π‘ π‘–π‘›β„Ž π‘Ÿ βˆ’ π‘Ÿ1 √2𝛼𝑑 + π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ √2𝛼𝑑 π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 √2𝛼𝑑 In steady state, 𝑑 β†’ ∞ π‘ π‘œ 𝛿 β†’ ∞ Upon substitution we get 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )π‘ π‘–π‘›β„Ž π‘Ÿ βˆ’ π‘Ÿ1 ∞ + π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ ∞ π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 ∞ 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) π‘ π‘–π‘›β„Ž0 + π‘ π‘–π‘›β„Ž0 π‘ π‘–π‘›β„Ž0 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = 0 0 L’hopital’s rule is then invoked i.e., π‘™π‘–π‘š π‘₯ β†’ 𝑐 𝑓(π‘₯) 𝑔(π‘₯) = π‘™π‘–π‘š π‘₯ β†’ 𝑐 𝑓′ (π‘₯) 𝑔′(π‘₯) We differentiate the numerate and denominator with respect to 1 𝛿 since it is the one varying and we get
𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )π‘ π‘–π‘›β„Ž π‘Ÿ βˆ’ π‘Ÿ1 𝛿 + π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) π‘π‘œπ‘ β„Ž π‘Ÿ βˆ’ π‘Ÿ1 𝛿 + ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )π‘π‘œπ‘ β„Ž π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 π‘π‘œπ‘ β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 Upon simplification, we get 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) π‘π‘œπ‘ β„Ž π‘Ÿ βˆ’ π‘Ÿ1 𝛿 + ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )π‘π‘œπ‘ β„Ž π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 π‘π‘œπ‘ β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = (π‘Ÿ βˆ’ π‘Ÿ1) ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) π‘π‘œπ‘ β„Ž π‘Ÿ βˆ’ π‘Ÿ1 𝛿 + (π‘Ÿ2 βˆ’ π‘Ÿ)π‘π‘œπ‘ β„Ž 𝐿 βˆ’ π‘₯ 𝛿 (π‘Ÿ2 βˆ’ π‘Ÿ1)π‘π‘œπ‘ β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 We then substitute 𝛿 = √2𝛼𝑑 = ∞ And get 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = (π‘Ÿ βˆ’ π‘Ÿ1) ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) + (π‘Ÿ2 βˆ’ π‘Ÿ) (π‘Ÿ2 βˆ’ π‘Ÿ1) Upon substitution, we get: 𝑻 βˆ’ π‘»βˆž π‘»π’ƒπŸ βˆ’ π‘»βˆž = (𝒓 βˆ’ π’“πŸ) ( π‘»π’ƒπŸ βˆ’ π‘»βˆž π‘»π’ƒπŸ βˆ’ π‘»βˆž ) + (π’“πŸ βˆ’ 𝒓) (π’“πŸ βˆ’ π’“πŸ) Hence, we get a temperature profile linear in radius r in steady state but this temperature profile is in contrast to the logarithmic temperature profile expected. So, we have to choose which theory we take to be true. The other thing we notice is that the solution we have got doesn’t reduce to the one of the semi-infinite solution when the radius π‘Ÿ2 tends to infinity.
CASE3: CONVECTION AT THE FREE END This same analysis can be extended to a metal rod with convection at the other end of the metal rod where the temperature profile is given by [1]: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh[π‘š(𝐿 βˆ’ π‘₯)] + ( β„ŽπΏ π‘šπ‘˜ ) sinh[π‘š(𝐿 βˆ’ π‘₯)] cosh π‘šπΏ + ( β„ŽπΏ π‘šπ‘˜ ) π‘ π‘–π‘›β„Žπ‘šπΏ Or 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (𝐿 βˆ’ π‘₯) 𝛿 ] + ( β„ŽπΏπ›Ώ π‘˜ ) sinh[( 𝐿 βˆ’ π‘₯ 𝛿 )] cosh 𝐿 𝛿 + ( β„ŽπΏπ›Ώ π‘˜ ) π‘ π‘–π‘›β„Ž 𝐿 𝛿 For cylindrical co-ordinates we make the substitutions 𝐿 = (π‘Ÿ2 βˆ’ π‘Ÿ1) And π‘₯ = π‘Ÿ βˆ’ π‘Ÿ1 And the temperature profile becomes: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh[π‘š(π‘Ÿ2 βˆ’ π‘Ÿ)] + ( β„Žπ‘Ÿ2 π‘šπ‘˜ ) sinh[π‘š(π‘Ÿ2 βˆ’ π‘Ÿ)] cosh π‘š(π‘Ÿ2 βˆ’ π‘Ÿ1) + ( β„Žπ‘Ÿ2 π‘šπ‘˜ ) π‘ π‘–π‘›β„Žπ‘š(π‘Ÿ2 βˆ’ π‘Ÿ1) The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒓 = π’“πŸ βˆ’π’Œ 𝒅𝑻 𝒅𝒓 = π’‰π’“πŸ (𝑻 βˆ’ π‘»βˆž) 𝒂𝒕 𝒓 = π’“πŸ 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 This same analysis can be extended to a metal rod with convection at the other end of the metal rod where the temperature profile is given by [1]: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh[π‘š(π‘Ÿ2 βˆ’ π‘Ÿ)] + ( β„Žπ‘Ÿ2 π‘šπ‘˜ ) sinh[π‘š(π‘Ÿ2 βˆ’ π‘Ÿ)] cosh π‘š(π‘Ÿ2 βˆ’ π‘Ÿ1) + ( β„Žπ‘Ÿ2 π‘šπ‘˜ ) π‘ π‘–π‘›β„Žπ‘š(π‘Ÿ2 βˆ’ π‘Ÿ1) Or
𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + ( β„Žπ‘Ÿ2 𝛿 π‘˜ ) sinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 To show that the initial condition is satisfied we assume from the above that π‘Žπ‘‘ 𝑑 = 0, 𝛿 = 0. 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + ( β„Žπ‘Ÿ2 𝛿 π‘˜ ) sinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 Becomes: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 = 𝑒 (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 + 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ) 𝛿 𝑒 π‘Ÿ2βˆ’π‘Ÿ1 𝛿 + 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ1) 𝛿 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ) 𝛿 = π‘’βˆ’ (π‘Ÿ2βˆ’π‘Ÿ) 0 = π‘’βˆ’βˆž(πΏβˆ’π‘₯) = 0 Similarly 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ1) 𝛿 = 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ1) 0 = π‘’βˆ’βˆž(π‘Ÿ2βˆ’π‘Ÿ1) = 0 So, we are left with 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 𝑒 π‘Ÿ2βˆ’π‘Ÿ1 𝛿 = 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 = 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 0 = π‘’βˆ’βˆž(π‘Ÿβˆ’π‘Ÿ1) = 0 Hence at 𝑑 = 0, 𝑇 = π‘‡βˆž and hence the initial condition. We use this temperature profile which satisfies the initial condition to solve the heat equation and get 𝛿 as shown below: Boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒓 = π’“πŸ βˆ’π’Œ 𝒅𝑻 𝒅𝒓 = π’‰π’“πŸ (𝑻 βˆ’ π‘»βˆž) 𝒂𝒕 𝒓 = π’“πŸ 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 The governing temperature profile is:
𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + ( β„Žπ‘Ÿ2 𝛿 π‘˜ ) sinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 Which we can express as: 𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] + 𝐁𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] Where: 𝐴 = (𝑇𝑠 βˆ’ π‘‡βˆž) cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 And 𝐡 = ( β„Žπ‘Ÿ2 𝛿 π‘˜ )(𝑇𝑠 βˆ’ π‘‡βˆž) cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 The governing equation is 𝜢 𝝏𝟐 𝑻 ππ’“πŸ + 𝜢 𝟏 𝒓 𝝏𝑻 𝝏𝒓 = 𝝏𝑻 𝝏𝒕 Multiplying through by r the heat equation becomes: π›Όπ‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 πœ•π‘‡ πœ•π‘Ÿ = πœ•π‘Ÿπ‘‡ πœ•π‘‘ Transforming the PDE into an integral equation, we get: 𝛼 [∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] = πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 Let us evaluate the integral ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] + 𝐁𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ]
πœ•2 𝑇 πœ•π‘Ÿ2 = 1 𝛿2 (𝐴 cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Bsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 = 1 𝛿2 (π΄π‘Ÿ cosh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Brsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (rsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] Let us evaluate the integral ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 We use ∫(𝑒𝑣)𝑑π‘₯ = 𝑒𝑣 βˆ’ ∫ 𝑣 𝑑𝑒 𝑑π‘₯ 𝑑π‘₯ Let 𝑒 = π‘Ÿ π‘Žπ‘›π‘‘ 𝑑𝑣 π‘‘π‘Ÿ = cosh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] β„Žπ‘’π‘›π‘π‘’ 𝑣 = βˆ’π›Ώ sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] ∫ (π‘Ÿ cosh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ])π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] + 𝛿 ∫ (sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ])π‘‘π‘Ÿ ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] βˆ’ 𝛿2 (cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 βˆ’ 𝛿2 [cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 Simplifying, we get: ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1] Now let us evaluate the integral ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 Let 𝑒 = π‘Ÿ π‘Žπ‘›π‘‘ 𝑑𝑣 π‘‘π‘Ÿ = π‘ π‘–π‘›β„Ž ( π‘Ÿ2βˆ’π‘Ÿ 𝛿 ) β„Žπ‘’π‘›π‘π‘’ 𝑣 = βˆ’π›Ώcosh( π‘Ÿ2βˆ’π‘Ÿ 𝛿 ) ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ) + 𝛿 ∫ (cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ
∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ cosh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ) βˆ’ 𝛿2 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [βˆ’π‘Ÿπ›Ώ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )] π‘Ÿ2 π‘Ÿ1 βˆ’ 𝛿2 [sinh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )] π‘Ÿ2 π‘Ÿ1 Simplifying, we get: ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 )) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 (π‘Ÿ1𝛿 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 (π‘Ÿ1𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Simplifying we get: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 ( π‘Ÿ1 𝛿 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 ( π‘Ÿ1 𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2 𝛿 + sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Now let us evaluate the integral: ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] + 𝐁𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Bsinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 = 𝐴 [1 βˆ’ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] βˆ’ B sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) ∫ [ 𝝏𝑻 𝝏𝒓 ] 𝒅𝒓 π’“πŸ π’“πŸ = 𝑨 [𝟏 βˆ’ 𝐜𝐨𝐬𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 )] βˆ’ 𝐁 𝐬𝐒𝐧𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 ) Now let us add the two integrals below: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 ( π‘Ÿ1 𝛿 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 ( π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2 𝛿 + sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] + 𝐴 [1 βˆ’ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] βˆ’ B sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) And get: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [ π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + π΅π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ] Now let us evaluate the integral
πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑇 = 𝐴 cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Bsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + π‘‡βˆž ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 𝐴 ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + π‘‡βˆž(π‘Ÿ2 βˆ’ π‘Ÿ1) But we have already evaluated the above so we get: ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] + π‘‡βˆž(π‘Ÿ2 βˆ’ π‘Ÿ1) Substituting all the above in the integral equation, we get 𝛼 [ π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + π΅π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ] = πœ• πœ•π‘‘ [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Now let us collect like terms i.e., let us collect terms in A separately and terms in B separately and get 𝛼 π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) = πœ• πœ•π‘‘ [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] Cancelling out A we get: 𝛼 [ π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] … … … 𝑴 Similarly for terms in B 𝛼 [ π΅π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ] = πœ• πœ•π‘‘ [𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Cancelling out B we get: 𝛼 [ π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))]… … 𝑡 Adding equations M and N we get: 𝛼 [ π‘Ÿ1 𝛿 (sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) βˆ’ π‘Ÿ2 𝛿 ] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 (sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝛿2 [sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] βˆ’ 𝛿(𝛿 + π‘Ÿ2))] Now let us collect terms in sinh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) and get 𝛼 [ π‘Ÿ1 𝛿 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝛿2 [sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )])]
Let us drop out the common term sinh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) and get 𝛼 [ π‘Ÿ1 𝛿 ] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 + 𝛿2)] … . . 𝑱 Let us also collect out the π‘Ÿ2 terms and get: 𝛼 [βˆ’ π‘Ÿ2 𝛿 ] = πœ• πœ•π‘‘ [(βˆ’π›Ώ(𝛿 + π‘Ÿ2))] Upon simplification we get: 𝛼 [ π‘Ÿ2 𝛿 ] = πœ• πœ•π‘‘ [(𝛿2 + π‘Ÿ2𝛿)]… . . 𝑲 Let us subtract equation K from equation J and get: 𝛼 [ π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ] = πœ• πœ•π‘‘ [(π‘Ÿ2 βˆ’ π‘Ÿ1)𝛿] Upon simplification, we get: 𝛼 𝛿 = 𝑑𝛿 𝑑𝑑 The boundary conditions are: 𝛿 = 0 π‘Žπ‘‘ 𝑑 = 0 Upon integrating, we get 𝛼 ∫ 𝑑𝑑 𝑑 0 = ∫ 𝛿𝑑𝛿 𝛿 0 We finally get 𝛿 = √2𝛼𝑑 Upon substitution in the temperature profile we get, 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + ( β„Žπ‘Ÿ2 𝛿 π‘˜ ) sinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) βˆšπŸπœΆπ’• ] + ( π’‰π’“πŸ βˆšπŸπœΆπ’• π’Œ ) 𝐬𝐒𝐧𝐑[ (π’“πŸ βˆ’ 𝒓) βˆšπŸπœΆπ’• ] 𝐜𝐨𝐬𝐑 (π’“πŸ βˆ’ π’“πŸ) βˆšπŸπœΆπ’• + ( π’‰π’“πŸ βˆšπŸπœΆπ’• π’Œ )π’”π’Šπ’π’‰ (π’“πŸ βˆ’ π’“πŸ) βˆšπŸπœΆπ’•
L’hopital’s rule can be used to find the steady state temperature when time tends to infinity. Again, we notice that the temperature profile above doesn’t reduce to the semi- infinite temperature profile when π‘Ÿ2 tends to infinity because both solutions have different values of 𝛿.
CASE4: ZERO CONVECTION AT THE FREE END The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒓 = π’“πŸ 𝒅𝑻 𝒅𝒓 = 𝟎 𝒂𝒕 𝒓 = π’“πŸ 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 Recall the compact temperature profile that satisfies the boundary and initial conditions is [2]: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ) cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) 𝛼 πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ = πœ•π‘‡ πœ•π‘‘ So, the temperature profile we are going to use is: 𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 ( π’“πŸ βˆ’ 𝒓 𝜹 ) Where: 𝐴 = (𝑇𝑠 βˆ’ π‘‡βˆž) cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 Multiplying through by r the heat equation becomes: π›Όπ‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 πœ•π‘‡ πœ•π‘Ÿ = πœ•π‘Ÿπ‘‡ πœ•π‘‘ Transforming the PDE into an integral equation, we get: 𝛼 [∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] = πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 Let us evaluate the integral ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1
πœ•2 𝑇 πœ•π‘Ÿ2 = 1 𝛿2 (π΄π‘π‘œπ‘ β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 = 1 𝛿2 (π΄π‘Ÿπ‘π‘œπ‘ β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) Let us evaluate the integral ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 We use ∫(𝑒𝑣)𝑑π‘₯ = 𝑒𝑣 βˆ’ ∫ 𝑣 𝑑𝑒 𝑑π‘₯ 𝑑π‘₯ Let 𝑒 = π‘Ÿ π‘Žπ‘›π‘‘ 𝑑𝑣 π‘‘π‘Ÿ = cosh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] β„Žπ‘’π‘›π‘π‘’ 𝑣 = βˆ’π›Ώ sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] ∫ (π‘Ÿ cosh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ])π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] + 𝛿 ∫ (sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ])π‘‘π‘Ÿ ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] βˆ’ 𝛿2 (cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 βˆ’ 𝛿2 [cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 Simplifying, we get: ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1] ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] Simplifying we get: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 ( π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])]
Now let us evaluate the integral: ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 = 𝐴 [1 βˆ’ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] ∫ [ 𝝏𝑻 𝝏𝒓 ] 𝒅𝒓 π’“πŸ π’“πŸ = 𝑨 [𝟏 βˆ’ 𝐜𝐨𝐬𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 )] Now let us add the two integrals below: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 ( π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + [cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] + 𝐴 [1 βˆ’ cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] And get: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [ π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] Now let us evaluate the integral πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑇 = 𝐴 cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + π‘‡βˆž ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 𝐴 ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + π‘‡βˆž(π‘Ÿ2 βˆ’ π‘Ÿ1) But we have already evaluated the above so we get: ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] + π‘‡βˆž(π‘Ÿ2 βˆ’ π‘Ÿ1)
Substituting all the above in the integral equation, we get 𝛼 [ π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] = πœ• πœ•π‘‘ [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] Now let us collect like terms i.e., let us collect terms in A separately and get: 𝛼 π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) = πœ• πœ•π‘‘ [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] Cancelling out A we get: 𝛼 [ π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] Here is the catch lets us expand sinh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) and cosh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) into exponentials and get: Calling 𝛽 = π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 𝛼 [ π‘Ÿ1 𝛿 sinh 𝛽] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 sinh 𝛽 + 𝛿2[cosh𝛽 βˆ’ 1])] Now we have: 𝛼 [ π‘Ÿ1 𝛿 ( 𝑒𝛽 βˆ’ π‘’βˆ’π›½ 2 )] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿( 𝑒𝛽 βˆ’ π‘’βˆ’π›½ 2 ) + 𝛿2 [( 𝑒𝛽 + π‘’βˆ’π›½ 2 ) βˆ’ 1])] Collecting like terms i.e., terms in 𝑒𝛽 and π‘’βˆ’π›½ and constant terms and get: 𝛼 [ π‘Ÿ1 𝛿 ( 𝑒𝛽 2 )] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿( 𝑒𝛽 2 ) + 𝛿2 ( 𝑒𝛽 2 ))] Dropping out 𝑒𝛽 we get: 𝛼 [ π‘Ÿ1 𝛿 ] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 + 𝛿2)] … . 𝑴 Similarly collecting terms in π‘’βˆ’π›½ we get: 𝛼 [ π‘Ÿ1 𝛿 ( βˆ’π‘’βˆ’π›½ 2 )] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿( βˆ’π‘’βˆ’π›½ 2 ) + 𝛿2 [( +π‘’βˆ’π›½ 2 )])] Dropping out the π‘’βˆ’π›½ we get:
𝛼 [ π‘Ÿ1 𝛿 ] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 βˆ’ 𝛿2)] … … 𝑡 Collecting out the β€˜constant’ terms we have: 0 = πœ• πœ•π‘‘ [(βˆ’π›Ώ2)] So, what we have is πœ• πœ•π‘‘ [(𝛿2)] = 0 … 𝑷 We are going to substitute equation P in equations M and N and get From equation M, we have: 𝛼 [ π‘Ÿ1 𝛿 ] = πœ•(π‘Ÿ1𝛿) πœ•π‘‘ + πœ•(𝛿2 ) πœ•π‘‘ But from equation P, we have πœ• πœ•π‘‘ [(𝛿2)] = 0 So, we end with 𝛼 [ π‘Ÿ1 𝛿 ] = πœ•(π‘Ÿ1𝛿) πœ•π‘‘ + 0 Upon simplification, we have: 𝛼 𝛿 = 𝑑𝛿 𝑑𝑑 … . 𝑱 Similarly, from equation N, we have 𝛼 [ π‘Ÿ1 𝛿 ] = πœ•(π‘Ÿ1𝛿) πœ•π‘‘ βˆ’ πœ•(𝛿2 ) πœ•π‘‘ But from equation P, we have: πœ• πœ•π‘‘ [(𝛿2)] = 0 So, we end with: 𝛼 [ π‘Ÿ1 𝛿 ] = πœ•(π‘Ÿ1𝛿) πœ•π‘‘ βˆ’ 0 We finally end with:
𝛼 𝛿 = 𝑑𝛿 𝑑𝑑 … 𝑲 We see that equation K and J are similar and so we solve for 𝛿 to get: 𝛿 = √2𝛼𝑑 We finally substitute in the temperature profile to get: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ) cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) OR 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh ( π‘Ÿ2 βˆ’ π‘Ÿ √2𝛼𝑑 ) cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 √2𝛼𝑑 )
RADIAL HEAT CONDUCTION WITH LATERAL CONVECTION The governing equation looks is: 𝜢 𝝏𝟐 𝑻 ππ’“πŸ + 𝜢 𝟏 𝒓 𝝏𝑻 𝝏𝒓 βˆ’ 𝒉𝑷 𝑨𝝆π‘ͺ (𝑻 βˆ’ π‘»βˆž) = 𝝏𝑻 𝝏𝒕
CASE1: SEMI-INFINITE CASE The governing equation is: 𝜢 𝝏𝟐 𝑻 ππ’“πŸ + 𝜢 𝟏 𝒓 𝝏𝑻 𝝏𝒓 βˆ’ 𝒉𝑷 𝑨𝝆π‘ͺ (𝑻 βˆ’ π‘»βˆž) = 𝝏𝑻 𝝏𝒕 𝑃 = 4πœ‹π‘Ÿ 𝐴 = 2πœ‹π‘Ÿπ‘‘ Where: 𝑑 = β„Žπ‘’π‘–π‘”β„Žπ‘‘ π‘œπ‘“ π‘π‘¦π‘™π‘–π‘›π‘‘π‘’π‘Ÿ Upon substituting the above in the heat equation, we get: 𝜢 𝝏𝟐 𝑻 ππ’“πŸ + 𝜢 𝟏 𝒓 𝝏𝑻 𝝏𝒓 βˆ’ πŸπ’‰ 𝒅𝝆π‘ͺ (𝑻 βˆ’ π‘»βˆž) = 𝝏𝑻 𝝏𝒕 The boundary conditions are: 𝑇 = 𝑇𝑠 π‘Žπ‘‘ π‘Ÿ = π‘Ÿ1 𝑇 = π‘‡βˆž π‘Žπ‘‘ π‘Ÿ = 𝑅 = ∞ The initial condition is: 𝑇 = π‘‡βˆž π‘Žπ‘‘ 𝑑 = 0 The temperature profile that satisfies the boundary conditions above is: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿
𝛼 πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ βˆ’ 2β„Ž π‘‘πœŒπΆ (𝑇 βˆ’ π‘‡βˆž) = πœ•π‘‡ πœ•π‘‘ We multiply through by r the heat equation as shown below and solve. π›Όπ‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 πœ•π‘‡ πœ•π‘Ÿ βˆ’ 2β„Ž π‘‘πœŒπΆ π‘Ÿ(𝑇 βˆ’ π‘‡βˆž) = πœ•π‘Ÿπ‘‡ πœ•π‘‘ We transform the PDE into an integral equation and take the limits to be from π‘Ÿ = π‘Ÿ1 to π‘Ÿ = 𝑅 = ∞ 𝛼 ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ 𝑅 π‘Ÿ1 + 𝛼 ∫ [ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 βˆ’ 2β„Ž π‘‘πœŒπΆ ∫ (π‘Ÿ(𝑇 βˆ’ π‘‡βˆž))π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ 𝑅 π‘Ÿ1 We evaluated those integrals before to be: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ ∞ π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 (π‘Ÿ1 + 𝛿) ∫ [ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ ∞ π‘Ÿ1 = βˆ’(𝑇𝑠 βˆ’ π‘‡βˆž) 2β„Ž π‘‘πœŒπΆ ∫ (𝑇 βˆ’ π‘‡βˆž)π‘‘π‘Ÿ 𝑅=∞ π‘Ÿ1 = π›Ώβ„Ž π‘‘πœŒπΆ (𝑇𝑠 βˆ’ π‘‡βˆž) From the derivations above, we get: ∫ ( πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 πœ• πœ•π‘‘ ∫ π‘‡π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = 𝑑𝛿 𝑑𝑑 (𝑇𝑠 βˆ’ π‘‡βˆž) ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ 𝑅=∞ π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž)(π‘Ÿ1𝛿 + 𝛿2) + π‘‡βˆž(𝑅 βˆ’ π‘Ÿ1) ∫ (π‘Ÿ(𝑇 βˆ’ π‘‡βˆž))π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž)(π‘Ÿ1𝛿 + 𝛿2 ) πœ• πœ•π‘‘ ∫ (π‘Ÿ(𝑇 βˆ’ π‘‡βˆž))π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = πœ• πœ•π‘‘ [(𝑇𝑠 βˆ’ π‘‡βˆž)(π‘Ÿ1𝛿 + 𝛿2 ) + π‘‡βˆž(𝑅 βˆ’ π‘Ÿ1)] But πœ•(π‘‡βˆž(𝑅 βˆ’ π‘Ÿ1)) πœ•π‘‘ = 0 𝑠𝑖𝑛𝑐𝑒 π‘‡βˆž π‘Žπ‘›π‘‘ (𝑅 βˆ’ π‘Ÿ1) π‘Žπ‘Ÿπ‘’ π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘π‘ 
So, we are left with 𝛼 [ (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 (π‘Ÿ1 + 𝛿) βˆ’ (𝑇𝑠 βˆ’ π‘‡βˆž)] βˆ’ (𝑇𝑠 βˆ’ π‘‡βˆž) 2β„Ž π‘‘πœŒπΆ (π‘Ÿ1𝛿 + 𝛿2) = πœ• πœ•π‘‘ [(𝑇𝑠 βˆ’ π‘‡βˆž)(π‘Ÿ1𝛿 + 𝛿2)] We are left with: 𝛼 π‘Ÿ1 𝛿 βˆ’ 2β„Ž π‘‘πœŒπΆ (π‘Ÿ1𝛿 + 𝛿2) = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 + 𝛿2)] Multiplying through by 𝛿 we get: π›Όπ‘Ÿ1 βˆ’ 2β„Ž π‘‘πœŒπΆ 𝛿(π‘Ÿ1𝛿 + 𝛿2) = πœ• πœ•π‘‘ [𝛿(π‘Ÿ1𝛿 + 𝛿2)] Calling 𝛿(π‘Ÿ1𝛿 + 𝛿2) as X i.e., 𝑋 = 𝛿(π‘Ÿ1𝛿 + 𝛿2) We get: π›Όπ‘Ÿ1 βˆ’ 2β„Ž π‘‘πœŒπΆ 𝑋 = 𝑑𝑋 𝑑𝑑 We solve the above 0DE with limits that at 𝑑 = 0 𝑋 = 0 since 𝛿 = 0 π‘Žπ‘‘ 𝑑 = 0 And we get: 𝑋 = πΎπ‘‘π‘Ÿ1 2β„Ž (1 βˆ’ 𝑒 βˆ’2β„Žπ‘‘ π‘‘πœŒπΆ ) Substituting for X we get: 𝛿(π‘Ÿ1𝛿 + 𝛿2) = πΎπ‘‘π‘Ÿ1 2β„Ž (1 βˆ’ 𝑒 βˆ’2β„Žπ‘‘ π‘‘πœŒπΆ ) Upon simplifying we get: πœΉπŸ‘ + π’“πŸπœΉπŸ βˆ’ π‘²π’…π’“πŸ πŸπ’‰ (𝟏 βˆ’ 𝒆 βˆ’πŸπ’‰π’• 𝒅𝝆π‘ͺ ) = 𝟎 Which is a cubic function that can be solved to get 𝛿 as a function of time. In steady state when time t tends to infinity, we get: πœΉπŸ‘ + π’“πŸπœΉπŸ βˆ’ π‘²π’…π’“πŸ πŸπ’‰ = 𝟎
CASE2: CONVECTION AT THE FREE END The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒓 = π’“πŸ βˆ’π’Œ 𝒅𝑻 𝒅𝒓 = π’‰π’“πŸ (𝑻 βˆ’ π‘»βˆž) 𝒂𝒕 𝒓 = π’“πŸ 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 This same analysis can be extended to a metal rod with convection at the other end of the metal rod where the temperature profile is given by [1]: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh[π‘š(π‘Ÿ2 βˆ’ π‘Ÿ)] + ( β„Žπ‘Ÿ2 π‘šπ‘˜ ) sinh[π‘š(π‘Ÿ2 βˆ’ π‘Ÿ)] cosh π‘š(π‘Ÿ2 βˆ’ π‘Ÿ1) + ( β„Žπ‘Ÿ2 π‘šπ‘˜ ) π‘ π‘–π‘›β„Žπ‘š(π‘Ÿ2 βˆ’ π‘Ÿ1) Or 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + ( β„Žπ‘Ÿ2 𝛿 π‘˜ ) sinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 To show that the initial condition is satisfied we assume from the above that π‘Žπ‘‘ 𝑑 = 0, 𝛿 = 0. 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + ( β„Žπ‘Ÿ2 𝛿 π‘˜ ) sinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 Becomes: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 = 𝑒 (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 + 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ) 𝛿 𝑒 π‘Ÿ2βˆ’π‘Ÿ1 𝛿 + 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ1) 𝛿 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ) 𝛿 = π‘’βˆ’ (π‘Ÿ2βˆ’π‘Ÿ) 0 = π‘’βˆ’βˆž(πΏβˆ’π‘₯) = 0 Similarly 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ1) 𝛿 = 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ1) 0 = π‘’βˆ’βˆž(π‘Ÿ2βˆ’π‘Ÿ1) = 0
So, we are left with 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 𝑒 π‘Ÿ2βˆ’π‘Ÿ1 𝛿 = 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 = 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 0 = π‘’βˆ’βˆž(π‘Ÿβˆ’π‘Ÿ1) = 0 Hence at 𝑑 = 0, 𝑇 = π‘‡βˆž and hence the initial condition. We use this temperature profile which satisfies the initial condition to solve the heat equation and get 𝛿 as shown below: Boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒓 = π’“πŸ βˆ’π’Œ 𝒅𝑻 𝒅𝒓 = π’‰π’“πŸ (𝑻 βˆ’ π‘»βˆž) 𝒂𝒕 𝒓 = π’“πŸ 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 The governing temperature profile is: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + ( β„Žπ‘Ÿ2 𝛿 π‘˜ ) sinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 Which we can express as: 𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] + 𝐁𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] Where: 𝐴 = (𝑇𝑠 βˆ’ π‘‡βˆž) cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 And 𝐡 = ( β„Žπ‘Ÿ2 𝛿 π‘˜ )(𝑇𝑠 βˆ’ π‘‡βˆž) cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 The governing equation is
𝜢 𝝏𝟐 𝑻 ππ’“πŸ + 𝜢 𝟏 𝒓 𝝏𝑻 𝝏𝒓 βˆ’ 𝒉𝑷 𝑨𝝆π‘ͺ (𝑻 βˆ’ π‘»βˆž) = 𝝏𝑻 𝝏𝒕 Where: 𝑃 = 4πœ‹π‘Ÿ 𝐴 = 2πœ‹π‘Ÿπ‘‘ Upon substitution of the above in the heat equation, we get: 𝜢 𝝏𝟐 𝑻 ππ’“πŸ + 𝜢 𝟏 𝒓 𝝏𝑻 𝝏𝒓 βˆ’ πŸπ’‰ 𝒅𝝆π‘ͺ (𝑻 βˆ’ π‘»βˆž) = 𝝏𝑻 𝝏𝒕 Multiplying through by r the heat equation becomes: π›Όπ‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 πœ•π‘‡ πœ•π‘Ÿ βˆ’ 2β„Ž π‘‘πœŒπΆ π‘Ÿ(𝑇 βˆ’ π‘‡βˆž) = πœ•π‘Ÿπ‘‡ πœ•π‘‘ Transforming the PDE into an integral equation, we get: 𝛼 [∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] βˆ’ 2β„Ž π‘‘πœŒπΆ ∫ (π‘Ÿ(𝑇 βˆ’ π‘‡βˆž))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 Let us evaluate the integral ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1
𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] + 𝐁𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] πœ•2 𝑇 πœ•π‘Ÿ2 = 1 𝛿2 (𝐴 cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Bsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 = 1 𝛿2 (π΄π‘Ÿ cosh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Brsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (rsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] Let us evaluate the integral ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 We use ∫(𝑒𝑣)𝑑π‘₯ = 𝑒𝑣 βˆ’ ∫ 𝑣 𝑑𝑒 𝑑π‘₯ 𝑑π‘₯ Let 𝑒 = π‘Ÿ π‘Žπ‘›π‘‘ 𝑑𝑣 π‘‘π‘Ÿ = cosh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] β„Žπ‘’π‘›π‘π‘’ 𝑣 = βˆ’π›Ώ sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] ∫ (π‘Ÿ cosh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ])π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] + 𝛿 ∫ (sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ])π‘‘π‘Ÿ ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] βˆ’ 𝛿2 (cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 βˆ’ 𝛿2 [cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 Simplifying, we get: ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1] Now let us evaluate the integral ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 Let 𝑒 = π‘Ÿ π‘Žπ‘›π‘‘ 𝑑𝑣 π‘‘π‘Ÿ = π‘ π‘–π‘›β„Ž ( π‘Ÿ2βˆ’π‘Ÿ 𝛿 ) β„Žπ‘’π‘›π‘π‘’ 𝑣 = βˆ’π›Ώcosh( π‘Ÿ2βˆ’π‘Ÿ 𝛿 )
∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ) + 𝛿 ∫ (cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ cosh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ) βˆ’ 𝛿2 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [βˆ’π‘Ÿπ›Ώ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )] π‘Ÿ2 π‘Ÿ1 βˆ’ 𝛿2 [sinh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )] π‘Ÿ2 π‘Ÿ1 Simplifying, we get: ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 )) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 (π‘Ÿ1𝛿 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 (π‘Ÿ1𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Simplifying we get: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 ( π‘Ÿ1 𝛿 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 ( π‘Ÿ1 𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2 𝛿 + sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Now let us evaluate the integral: ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] + 𝐁𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Bsinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 = 𝐴 [1 βˆ’ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] βˆ’ B sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) ∫ [ 𝝏𝑻 𝝏𝒓 ] 𝒅𝒓 π’“πŸ π’“πŸ = 𝑨 [𝟏 βˆ’ 𝐜𝐨𝐬𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 )] βˆ’ 𝐁 𝐬𝐒𝐧𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 ) Now let us add the two integrals below: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 ( π‘Ÿ1 𝛿 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 ( π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2 𝛿 + sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] + 𝐴 [1 βˆ’ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] βˆ’ B sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) And get: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [ π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + π΅π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ]
Now let us evaluate: 2β„Ž π‘‘πœŒπΆ ∫ (π‘Ÿ(𝑇 βˆ’ π‘‡βˆž))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 Let us evaluate: ∫ (π‘Ÿ(𝑇 βˆ’ π‘‡βˆž))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] + 𝐁𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] π‘Ÿ(𝑇 βˆ’ π‘‡βˆž) = π΄π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Brsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] ∫ (π‘Ÿ(𝑇 βˆ’ π‘‡βˆž))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ])π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (rsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] We have already evaluated the integrals above so we get: ∫ (π‘Ÿ(𝑇 βˆ’ π‘‡βˆž))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 (π‘Ÿ1𝛿 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 (π‘Ÿ1𝛿cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Now let us evaluate the integral πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑇 = 𝐴 cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Bsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + π‘‡βˆž ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 𝐴 ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + π‘‡βˆž(π‘Ÿ2 βˆ’ π‘Ÿ1) But we have already evaluated the above so we get: ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] + π‘‡βˆž(π‘Ÿ2 βˆ’ π‘Ÿ1) Substituting all the above in the integral equation, we get
𝛼 [ π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + π΅π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ] βˆ’ 2β„Ž π‘‘πœŒπΆ [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 (π‘Ÿ1𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] = πœ• πœ•π‘‘ [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Now let us collect like terms i.e., let us collect terms in A separately and terms in B separately and get 𝛼 π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 2β„Ž π‘‘πœŒπΆ [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] = πœ• πœ•π‘‘ [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] Cancelling out A we get: 𝛼 [ π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] βˆ’ 2β„Ž π‘‘πœŒπΆ [(π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] … … … 𝑴 Similarly for terms in B 𝛼 [ π΅π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ] βˆ’ 2β„Ž π‘‘πœŒπΆ [𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] = πœ• πœ•π‘‘ [𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Cancelling out B we get: 𝛼 [ π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ] βˆ’ 2β„Ž π‘‘πœŒπΆ [𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] …… 𝑡 Adding equations M and N we get: 𝛼 [ π‘Ÿ1 𝛿 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) βˆ’ π‘Ÿ2 𝛿 ] βˆ’ 2β„Ž π‘‘πœŒπΆ [(π‘Ÿ1𝛿 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝛿2 [sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] βˆ’ 𝛿(𝛿 + π‘Ÿ2))] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝛿2 [sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] βˆ’ 𝛿(𝛿 + π‘Ÿ2))]…. 𝑃 Now let us collect terms in sinh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) and get 𝛼 [ π‘Ÿ1 𝛿 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] βˆ’ 2β„Ž π‘‘πœŒπΆ [(π‘Ÿ1𝛿 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝛿2 [sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝛿2 [sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )])] Let us drop out the common term sinh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) and get 𝛼 [ π‘Ÿ1 𝛿 ] βˆ’ 2β„Ž π‘‘πœŒπΆ [(π‘Ÿ1𝛿 + 𝛿2 ] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 + 𝛿2)]… . . 𝑱 Let us also collect out the π‘Ÿ2 terms from equation P and get: 𝛼 [βˆ’ π‘Ÿ2 𝛿 ] βˆ’ 2β„Ž π‘‘πœŒπΆ [βˆ’π›Ώ(𝛿 + π‘Ÿ2)] = πœ• πœ•π‘‘ [(βˆ’π›Ώ(𝛿 + π‘Ÿ2))] Upon simplification we get:
𝛼 [ π‘Ÿ2 𝛿 ] βˆ’ 2β„Ž π‘‘πœŒπΆ [𝛿(𝛿 + π‘Ÿ2)] = πœ• πœ•π‘‘ [(𝛿2 + π‘Ÿ2𝛿)]… . . 𝑲 Let us subtract equation K from equation J and get: 𝛼 [ π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ] βˆ’ 2β„Ž π‘‘πœŒπΆ [(π‘Ÿ2 βˆ’ π‘Ÿ1)𝛿] = πœ• πœ•π‘‘ [(π‘Ÿ2 βˆ’ π‘Ÿ1)𝛿] Upon simplification, we get: 𝛼 𝛿 βˆ’ 2β„Ž π‘‘πœŒπΆ 𝛿 = 𝑑𝛿 𝑑𝑑 𝛼 βˆ’ 2β„Ž π‘‘πœŒπΆ 𝛿2 = 𝛿 𝑑𝛿 𝑑𝑑 We solve the equation above with initial condition 𝛿 = 0 π‘Žπ‘‘ 𝑑 = 0 And get 𝛿 = √ 𝐾𝑑 2β„Ž (1 βˆ’ 𝑒 βˆ’4β„Ž π‘‘πœŒπΆ 𝑑 ) Upon substitution in the temperature profile we get, 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + ( β„Žπ‘Ÿ2 𝛿 π‘˜ ) sinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) √ 𝑲𝒅 πŸπ’‰ (𝟏 βˆ’ 𝒆 βˆ’πŸ’π’‰ 𝒅𝝆π‘ͺ 𝒕 ) ] + ( π’‰π’“πŸ √ 𝑲𝒅 πŸπ’‰ (𝟏 βˆ’ 𝒆 βˆ’πŸ’π’‰ 𝒅𝝆π‘ͺ 𝒕 ) π’Œ ) 𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) √ 𝑲𝒅 πŸπ’‰ (𝟏 βˆ’ 𝒆 βˆ’πŸ’π’‰ 𝒅𝝆π‘ͺ 𝒕 ) ] 𝐜𝐨𝐬𝐑 (π’“πŸ βˆ’ π’“πŸ) √ 𝑲𝒅 πŸπ’‰ (𝟏 βˆ’ 𝒆 βˆ’πŸ’π’‰ 𝒅𝝆π‘ͺ 𝒕 ) + ( π’‰π’“πŸ √ 𝑲𝒅 πŸπ’‰ (𝟏 βˆ’ 𝒆 βˆ’πŸ’π’‰ 𝒅𝝆π‘ͺ 𝒕 ) π’Œ ) π’”π’Šπ’π’‰ (π’“πŸ βˆ’ π’“πŸ) √ 𝑲𝒅 πŸπ’‰ (𝟏 βˆ’ 𝒆 βˆ’πŸ’π’‰ 𝒅𝝆π‘ͺ 𝒕 )
In steady state when time t tends to infinity, we get 𝛿 = √ 𝐾𝑑 2β„Ž And the temperature profile becomes: 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) βˆšπ‘²π’… πŸπ’‰ ] + ( π’‰π’“πŸ βˆšπ‘²π’… πŸπ’‰ π’Œ ) 𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) βˆšπ‘²π’… πŸπ’‰ ] 𝐜𝐨𝐬𝐑 (π’“πŸ βˆ’ π’“πŸ) βˆšπ‘²π’… πŸπ’‰ + ( π’‰π’“πŸ βˆšπ‘²π’… πŸπ’‰ π’Œ ) π’”π’Šπ’π’‰ (π’“πŸ βˆ’ π’“πŸ) βˆšπ‘²π’… πŸπ’‰ Again, we notice that the temperature profile above doesn’t reduce to the semi- infinite temperature profile when π‘Ÿ2 tends to infinity because both solutions have different values of 𝛿. It can be shown that other transient boundary value problems where the hollow cylinder is finite and not semi-infinite, the value of 𝛿 for lateral convection case will be as derived above: 𝛿 = √ 𝐾𝑑 2β„Ž (1 βˆ’ 𝑒 βˆ’4β„Ž π‘‘πœŒπΆ 𝑑 )
CONCLUSION It should be noted that these equations are the ideal solutions of heat flow and that experimental results are needed to verify the nature of some parameters like the nature of the heat transfer coefficient at the end of the metal rod β„ŽπΏ. Also, experimental results are needed to verify the transient nature of the heat flow.
REFERENCES [1] C.P.Kothandaraman, "Heat Transfer with Extendeded Surfaces(Fins)," in Fundamentals of Heat and Mass Transfer, New Delhi, NEW AGE INTERNATIONAL PUBLISHERS, 2006, p. 129. [2] C. E. W. R. E. W. G. L. R. James R. Welty, "HEAT TRANSFER FROM EXTENDED SURFACES," in Fundamentals of Momentum, Heat, and Mass Transfer 5th Edition, Corvallis, Oregon, John Wiley & Sons, Inc., 2000, pp. 236-237.

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RADIAL HEAT CONDUCTION SOLVED USING THE INTEGRAL EQUATION .pdf

  • 1. Using the integral heat equation Wasswa Derrick 4/27/24 Applied Mathematics
  • 2.
  • 3. TABLE OF CONTENTS PREFACE ............................................................................................................................................3 RADIAL HEAT CONDUCTION WITH NO LATERAL CONVECTION...................................4 CASE1: SEMI-INFINITE HOLLOW CYLINDER...........................................................................................4 CASE 2: FIXED END TEMPERATURE ......................................................................................................9 CASE3: CONVECTION AT THE FREE END .............................................................................................16 CASE4: ZERO CONVECTION AT THE FREE END ....................................................................................24 RADIAL HEAT CONDUCTION WITH LATERAL CONVECTION........................................30 CASE1: SEMI-INFINITE CASE...............................................................................................................31 CASE2: CONVECTION AT THE FREE END .............................................................................................34 CONCLUSION...................................................................................................................................43 REFERENCES..................................................................................................................................44
  • 4. PREFACE We look at the case of radial heat flow. Again, in radial heat flow, the temperature profiles that satisfy the boundary and initial conditions are the exponential and hyperbolic functions as derived in literature of conduction in fins. We use the technique of transforming the PDE into an integral equation. But in the case of radial heat flow, we have to multiply through by r the heat equation and then introduce integrals. We do this to avoid introducing integrals of the form of the exponential integral whose solutions cannot be expressed in the form of a simple mathematical function. We look at the case of a semi-infinite hollow cylinder for both insulated and non-insulated cases and then find the solution. We also look at cases of finite radius hollow cylinders subject to given boundary conditions. We notice that the solutions got for finite radius hollow cylinders do not reduce to those of semi-infinite hollow cylinders. We conclude by saying that this same analysis can be extended to spherical co- ordinates heat conduction.
  • 5. RADIAL HEAT CONDUCTION WITH NO LATERAL CONVECTION CASE1: SEMI-INFINITE HOLLOW CYLINDER. The governing PDE is: 𝜢 𝝏𝟐 𝑻 ππ’“πŸ + 𝜢 𝟏 𝒓 𝝏𝑻 𝝏𝒓 = 𝝏𝑻 𝝏𝒕 The boundary conditions are 𝑇 = 𝑇𝑠 π‘Žπ‘‘ π‘Ÿ = π‘Ÿ1 𝑇 = π‘‡βˆž π‘Žπ‘‘ π‘Ÿ = ∞ The initial condition is: 𝑇 = π‘‡βˆž π‘Žπ‘‘ 𝑑 = 0 The temperature profile that satisfies the conditions above is 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 We transform the equation above into an integral equation and take integrals with limits from π‘Ÿ = π‘Ÿ1 to π‘Ÿ = 𝑅 = ∞. 𝛼 πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ = πœ•π‘‡ πœ•π‘‘ We take integrals and get
  • 6. 𝛼 ∫ πœ•2 𝑇 πœ•π‘Ÿ2 π‘‘π‘Ÿ 𝑅 π‘Ÿ1 + 𝛼 ∫ [ 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = πœ• πœ•π‘‘ ∫ (𝑇)π‘‘π‘Ÿ 𝑅 π‘Ÿ1 Where: 𝑅 = ∞ Let us look at the integral below: ∫ [ 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 πœ•π‘‡ πœ•π‘Ÿ = βˆ’ 𝑇𝑠 βˆ’ π‘‡βˆž 𝛿 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ∫ [ 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = βˆ’ ∫ [ 1 π‘Ÿ 𝑇𝑠 βˆ’ π‘‡βˆž 𝛿 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 Upon rearranging, we get: ∫ [ 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = βˆ’ 𝑇𝑠 βˆ’ π‘‡βˆž 𝛿 𝑒 π‘Ÿ1 𝛿 ∫ [ 1 π‘Ÿ 𝑒 βˆ’π‘Ÿ 𝛿 ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 Calling 𝑒 = π‘Ÿ 𝛿 π‘Ÿ = 𝑒𝛿 π‘‘π‘Ÿ = 𝛿𝑑𝑒 So, we have ∫ [ 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = βˆ’ 𝑇𝑠 βˆ’ π‘‡βˆž 𝛿 𝑒 π‘Ÿ1 𝛿 ∫ [ 1 𝑒 π‘’βˆ’π‘’ ]𝑑𝑒 𝑅 𝑒 π‘Ÿ1 𝛿 This integral is in the form of the Exponential integral i.e., ∫[ 1 π‘₯ π‘’βˆ’π‘₯ ]𝑑𝑒 = 𝐸𝑖(βˆ’π‘₯) + 𝑐 and is a non-elementary function. This means that the integral cannot be expressed as a simple function. To avoid this problem, we have to multiply through by r the heat equation as shown below and solve. 𝛼 ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ 𝑅 π‘Ÿ1 + 𝛼 ∫ [ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ 𝑅 π‘Ÿ1 Let us evaluate:
  • 7. ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ 𝑅 π‘Ÿ1 πœ•2 𝑇 πœ•π‘Ÿ2 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿2 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ∫(π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿2 ∫ [π‘Ÿπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ] π‘‘π‘Ÿ We use ∫(𝑒𝑣)𝑑π‘₯ = 𝑒𝑣 βˆ’ ∫ 𝑣 𝑑𝑒 𝑑π‘₯ 𝑑π‘₯ Let us evaluate: ∫ [π‘Ÿπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ] π‘‘π‘Ÿ Let 𝑒 = π‘Ÿ π‘Žπ‘›π‘‘ 𝑑𝑣 π‘‘π‘Ÿ = 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 β„Žπ‘’π‘›π‘π‘’ 𝑣 = βˆ’π›Ώπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ∫ [π‘Ÿπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ] π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 + 𝛿 ∫ (𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ) π‘‘π‘Ÿ ∫ [π‘Ÿπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ] π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 βˆ’ 𝛿2 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 Now let us put the limits ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ ∞ π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿2 ([βˆ’π‘Ÿπ›Ώπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ] ∞ π‘Ÿ1 βˆ’ 𝛿2 [𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ] ∞ π‘Ÿ1 ) = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 (π‘Ÿ1 + 𝛿) ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ ∞ π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 (π‘Ÿ1 + 𝛿) Now let us evaluate the integral: ∫ [ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ ∞ π‘Ÿ1 πœ•π‘‡ πœ•π‘Ÿ = βˆ’ 𝑇𝑠 βˆ’ π‘‡βˆž 𝛿 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ∫ [ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ ∞ π‘Ÿ1 = ∫ [βˆ’ 𝑇𝑠 βˆ’ π‘‡βˆž 𝛿 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ] π‘‘π‘Ÿ ∞ π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž)[𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 ] ∞ π‘Ÿ1 ∫ [ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ ∞ π‘Ÿ1 = βˆ’(𝑇𝑠 βˆ’ π‘‡βˆž)
  • 8. Now let us evaluate: ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ 𝑅=∞ π‘Ÿ1 𝑇 = (𝑇𝑠 βˆ’ π‘‡βˆž)𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 + π‘‡βˆž ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ 𝑅=∞ π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž) ∫ (π‘Ÿπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 )π‘‘π‘Ÿ 𝑅=∞ π‘Ÿ1 + ∫ π‘‡βˆžπ‘‘π‘Ÿ 𝑅=∞ π‘Ÿ1 = 𝛿(𝑇𝑠 βˆ’ π‘‡βˆž) + π‘‡βˆž(𝑅 βˆ’ π‘Ÿ1) ∫ (π‘Ÿπ‘’ βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 )π‘‘π‘Ÿ 𝑅=∞ π‘Ÿ1 = π‘Ÿ1𝛿 + 𝛿2 ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ 𝑅=∞ π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž)(π‘Ÿ1𝛿 + 𝛿2) + π‘‡βˆž(𝑅 βˆ’ π‘Ÿ1) Upon substituting all the above in the integral heat equation, we get 𝛼 [ (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 (π‘Ÿ1 + 𝛿) βˆ’ (𝑇𝑠 βˆ’ π‘‡βˆž)] = πœ• πœ•π‘‘ [(𝑇𝑠 βˆ’ π‘‡βˆž)(π‘Ÿ1𝛿 + 𝛿2 ) + π‘‡βˆž(𝑅 βˆ’ π‘Ÿ1)] But πœ•(π‘‡βˆž(𝑅 βˆ’ π‘Ÿ1)) πœ•π‘‘ = 0 𝑠𝑖𝑛𝑐𝑒 π‘‡βˆž π‘Žπ‘›π‘‘ (𝑅 βˆ’ π‘Ÿ1) π‘Žπ‘Ÿπ‘’ π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘π‘  So, we are left with 𝛼 [ (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 (π‘Ÿ1 + 𝛿) βˆ’ (𝑇𝑠 βˆ’ π‘‡βˆž)] = πœ• πœ•π‘‘ [(𝑇𝑠 βˆ’ π‘‡βˆž)(π‘Ÿ1𝛿 + 𝛿2 )] 𝛼 [ 1 𝛿 (π‘Ÿ1 + 𝛿) βˆ’ 1] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 + 𝛿2 )] Upon substitution, we get 𝛼 π‘Ÿ1 𝛿 = (π‘Ÿ1 + 2𝛿) 𝑑𝛿 𝑑𝑑 The boundary condition is: 𝛿 = 0 π‘Žπ‘‘ 𝑑 = 0 So, we get: π›Όπ‘Ÿ1 ∫ 𝑑𝑑 𝑑 0 = ∫ (π‘Ÿ1𝛿 + 2𝛿2) 𝛿 0 𝑑𝛿 Upon simplification, we get:
  • 9. π›Όπ‘Ÿ1𝑑 = π‘Ÿ1𝛿2 2 + 2𝛿3 3 We get: 6π›Όπ‘Ÿ1𝑑 = 3π‘Ÿ1𝛿2 + 4𝛿3 i.e., πŸ‘π’“πŸπœΉπŸ + πŸ’πœΉπŸ‘ βˆ’ πŸ”πœΆπ’“πŸπ’• = 𝟎 Which is a cubic function and can be solved to get 𝛿 as a function of time t. You notice that the initial condition is satisfied for the above temperature profile. We can also go ahead and look at situations where there is natural convection and other situations where the radius r is finite and not infinite.
  • 10. CASE 2: FIXED END TEMPERATURE The initial and boundary conditions are: 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 𝑻 = π‘»π’ƒπŸ βˆ’ π‘»βˆž 𝒂𝒕 𝒙 = 𝟎 𝑻 = π‘»π’ƒπŸ βˆ’ π‘»βˆž 𝒂𝒕 𝒙 = 𝑳 We start with a temperature profile below: 𝑇 βˆ’ π‘‡βˆž = 𝐴𝑒 π‘₯ 𝛿 + 𝐡𝑒 βˆ’π‘₯ 𝛿 We evaluate the constants A and B using the boundary conditions and get the temperature profile below: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) π‘ π‘–π‘›β„Ž π‘₯ 𝛿 + π‘ π‘–π‘›β„Ž 𝐿 βˆ’ π‘₯ 𝛿 π‘ π‘–π‘›β„Ž 𝐿 𝛿 The temperature profile above can be referenced in textbooks for heat flow in extended surfaces like in the book [1]. To transform the above equation to cylindrical co-ordinates, we use the substitutions, 𝐿 = (π‘Ÿ2 βˆ’ π‘Ÿ1) And π‘₯ = π‘Ÿ βˆ’ π‘Ÿ1 And the temperature profile becomes: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )π‘ π‘–π‘›β„Ž π‘Ÿ βˆ’ π‘Ÿ1 𝛿 + π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 The governing PDE is: 𝜢 𝝏𝟐 𝑻 ππ’“πŸ + 𝜢 𝟏 𝒓 𝝏𝑻 𝝏𝒓 = 𝝏𝑻 𝝏𝒕 The initial and boundary conditions are: 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎
  • 11. 𝑻 = π‘»π’ƒπŸ βˆ’ π‘»βˆž 𝒂𝒕 𝒓 = π’“πŸ 𝑻 = π‘»π’ƒπŸ βˆ’ π‘»βˆž 𝒂𝒕 𝒓 = π’“πŸ The temperature profile that satisfies the above conditions is 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )π‘ π‘–π‘›β„Ž π‘Ÿ βˆ’ π‘Ÿ1 𝛿 + π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 To simplify matters, we can express the temperature profile as below: 𝑇 βˆ’ π‘‡βˆž = (𝑇𝑏2 βˆ’ π‘‡βˆž)π‘ π‘–π‘›β„Ž π‘Ÿ βˆ’ π‘Ÿ1 𝛿 π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 + (𝑇𝑏1 βˆ’ π‘‡βˆž)π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 Simply as: 𝑇 βˆ’ π‘‡βˆž = π΄π‘ π‘–π‘›β„Ž [ π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ] + π΅π‘ π‘–π‘›β„Ž [ π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ] Where: 𝐴 = (𝑇𝑏2 βˆ’ π‘‡βˆž) π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 And 𝐡 = (𝑇𝑏1 βˆ’ π‘‡βˆž) π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 So, the temperature profile we are going to use is: 𝑻 βˆ’ π‘»βˆž = π‘¨π’”π’Šπ’π’‰ [ 𝒓 βˆ’ π’“πŸ 𝜹 ] + π‘©π’”π’Šπ’π’‰ [ π’“πŸ βˆ’ 𝒓 𝜹 ] 𝛼 πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ = πœ•π‘‡ πœ•π‘‘ Multiplying through by r the heat equation becomes: π›Όπ‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 πœ•π‘‡ πœ•π‘Ÿ = πœ•π‘Ÿπ‘‡ πœ•π‘‘ Transforming the PDE into an integral equation, we get: 𝛼 [∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] = πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1
  • 12. Let us evaluate the integral ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 πœ•2 𝑇 πœ•π‘Ÿ2 = 1 𝛿2 (π΄π‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ) + π΅π‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 = 1 𝛿2 (π΄π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ) + π΅π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] Let us evaluate the integral ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 We use ∫(𝑒𝑣)𝑑π‘₯ = 𝑒𝑣 βˆ’ ∫ 𝑣 𝑑𝑒 𝑑π‘₯ 𝑑π‘₯ Let 𝑒 = π‘Ÿ π‘Žπ‘›π‘‘ 𝑑𝑣 π‘‘π‘Ÿ = π‘ π‘–π‘›β„Ž ( π‘Ÿβˆ’π‘Ÿ1 𝛿 ) β„Žπ‘’π‘›π‘π‘’ 𝑣 = 𝛿cosh( π‘Ÿβˆ’π‘Ÿ1 𝛿 ) ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿβˆ’π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ = π‘Ÿπ›Ώ cosh ( π‘Ÿβˆ’π‘Ÿ1 𝛿 ) βˆ’ 𝛿 ∫ (cosh ( π‘Ÿβˆ’π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ = π‘Ÿπ›Ώ cosh ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 𝛿2 (sinh( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 )) ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [π‘Ÿπ›Ώ cosh ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 )] π‘Ÿ2 π‘Ÿ1 βˆ’ 𝛿2 [sinh( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 )] π‘Ÿ2 π‘Ÿ1 Simplifying, we get: ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 )) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = π‘Ÿ2𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 βˆ’ 𝛿2 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) Now let us evaluate the integral ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 Let 𝑒 = π‘Ÿ π‘Žπ‘›π‘‘ 𝑑𝑣 π‘‘π‘Ÿ = π‘ π‘–π‘›β„Ž ( π‘Ÿ2βˆ’π‘Ÿ 𝛿 ) β„Žπ‘’π‘›π‘π‘’ 𝑣 = βˆ’π›Ώcosh( π‘Ÿ2βˆ’π‘Ÿ 𝛿 )
  • 13. ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ) + 𝛿 ∫ (cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ cosh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ) βˆ’ 𝛿2 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [βˆ’π‘Ÿπ›Ώ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )] π‘Ÿ2 π‘Ÿ1 βˆ’ 𝛿2 [sinh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )] π‘Ÿ2 π‘Ÿ1 Simplifying, we get: ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 (π‘Ÿ2𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 βˆ’ 𝛿2 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝐡 (π‘Ÿ1𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Simplifying we get: ∫ (𝒓 𝝏𝟐 𝑻 ππ’“πŸ )𝒅𝒓 π’“πŸ π’“πŸ = [𝐴 ( π‘Ÿ2 𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1 𝛿 βˆ’ sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝐡 ( π‘Ÿ1 𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2 𝛿 + sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Now let us evaluate the integral: ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [π΄π‘ π‘–π‘›β„Ž [ π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ] + π΅π‘ π‘–π‘›β„Ž [ π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ]] π‘Ÿ2 π‘Ÿ1 = (𝐴 βˆ’ 𝐡)sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) ∫ [ 𝝏𝑻 𝝏𝒓 ] 𝒅𝒓 π’“πŸ π’“πŸ = 𝑨 𝐬𝐒𝐧𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 ) βˆ’ 𝑩𝐬𝐒𝐧𝐑( π’“πŸ βˆ’ π’“πŸ 𝜹 ) Now let us evaluate the integral πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑇 = π΄π‘ π‘–π‘›β„Ž [ π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ] + π΅π‘ π‘–π‘›β„Ž [ π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ] + π‘‡βˆž ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 𝐴 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + π‘‡βˆž 2 (π‘Ÿ2 2 βˆ’ π‘Ÿ1 2 )
  • 14. But we have already evaluated the above so we get: ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 (π‘Ÿ2𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 βˆ’ 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝐡 (π‘Ÿ1𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] + π‘‡βˆž 2 (π‘Ÿ2 2 βˆ’ π‘Ÿ1 2 ) Substituting all the above in the integral equation, we get 𝜢( π‘¨π’“πŸ 𝜹 𝐜𝐨𝐬𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 ) βˆ’ π‘¨π’“πŸ 𝜹 βˆ’ 𝑨𝐬𝐒𝐧 𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 )) + ( π‘©π’“πŸ 𝜹 𝐜𝐨𝐬𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 ) βˆ’ π‘©π’“πŸ 𝜹 + 𝑩𝐬𝐒𝐧 𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 )) + πœΆπ‘¨ 𝐬𝐒𝐧𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 ) βˆ’ πœΆπ‘©π¬π’π§π‘( π’“πŸ βˆ’ π’“πŸ 𝜹 ) = πœ• πœ•π‘‘ [𝐴 (π‘Ÿ2𝛿cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 βˆ’ 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝐡 (π‘Ÿ1𝛿cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Since πœ• πœ•π‘‘ [ π‘‡βˆž 2 (π‘Ÿ2 2 βˆ’ π‘Ÿ1 2)] = 0 We finally get after crossing out the sinh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) terms 𝛼 ( π΄π‘Ÿ2 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΄π‘Ÿ1 𝛿 ) + ( π΅π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ) = πœ• πœ•π‘‘ [𝐴 (π‘Ÿ2𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 βˆ’ 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Collecting like terms in A and B we get: 𝛼 ( π΄π‘Ÿ2 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΄π‘Ÿ1 𝛿 ) = πœ• πœ•π‘‘ (𝐴 (π‘Ÿ2𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 βˆ’ 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )))… 𝑀 𝛼 ( π΅π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ) = πœ• πœ•π‘‘ (𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))). . 𝑁 From equation N we can get an expression for 𝛿2 sinh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) which we can substitute in equation M. From equation N, we have: ∫ 𝛼 ( π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2 𝛿 )𝑑𝑑 βˆ’ π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + π‘Ÿ2𝛿 = 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) Substituting in equation M which is 𝛼 ( π‘Ÿ2 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1 𝛿 ) = πœ• πœ•π‘‘ ((π‘Ÿ2𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 βˆ’ 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))) We get 𝛼 ( π‘Ÿ2 𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1 𝛿 ) = πœ• πœ•π‘‘ ((π‘Ÿ2𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 βˆ’ [𝛼 ∫( π‘Ÿ1 𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2 𝛿 )𝑑𝑑 βˆ’ π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + π‘Ÿ2𝛿)]) Which gives 𝛼 ( π‘Ÿ2 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1 𝛿 ) = πœ• πœ•π‘‘ ((π‘Ÿ2𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ1𝛿 + π‘Ÿ1𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿)) βˆ’ 𝛼 ( π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2 𝛿 )
  • 15. Collecting like terms we have 𝛼 𝛿 ((π‘Ÿ2 + π‘Ÿ1)cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ (π‘Ÿ2 + π‘Ÿ1)) = 𝑑𝛿 𝑑𝑑 ((π‘Ÿ2 + π‘Ÿ1)cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ (π‘Ÿ2 + π‘Ÿ1)) The like terms of ((π‘Ÿ2 + π‘Ÿ1)cosh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) βˆ’ (π‘Ÿ2 + π‘Ÿ1)) cancel out and we remain with 𝛼 𝛿 = 𝑑𝛿 𝑑𝑑 Hence, we have 𝛿 = √2𝛼𝑑 Hence the temperature profile is: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )π‘ π‘–π‘›β„Ž π‘Ÿ βˆ’ π‘Ÿ1 𝛿 + π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 Or 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )π‘ π‘–π‘›β„Ž π‘Ÿ βˆ’ π‘Ÿ1 √2𝛼𝑑 + π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ √2𝛼𝑑 π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 √2𝛼𝑑 In steady state, 𝑑 β†’ ∞ π‘ π‘œ 𝛿 β†’ ∞ Upon substitution we get 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )π‘ π‘–π‘›β„Ž π‘Ÿ βˆ’ π‘Ÿ1 ∞ + π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ ∞ π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 ∞ 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) π‘ π‘–π‘›β„Ž0 + π‘ π‘–π‘›β„Ž0 π‘ π‘–π‘›β„Ž0 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = 0 0 L’hopital’s rule is then invoked i.e., π‘™π‘–π‘š π‘₯ β†’ 𝑐 𝑓(π‘₯) 𝑔(π‘₯) = π‘™π‘–π‘š π‘₯ β†’ 𝑐 𝑓′ (π‘₯) 𝑔′(π‘₯) We differentiate the numerate and denominator with respect to 1 𝛿 since it is the one varying and we get
  • 16. 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )π‘ π‘–π‘›β„Ž π‘Ÿ βˆ’ π‘Ÿ1 𝛿 + π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 π‘ π‘–π‘›β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) π‘π‘œπ‘ β„Ž π‘Ÿ βˆ’ π‘Ÿ1 𝛿 + ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )π‘π‘œπ‘ β„Ž π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 π‘π‘œπ‘ β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 Upon simplification, we get 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) π‘π‘œπ‘ β„Ž π‘Ÿ βˆ’ π‘Ÿ1 𝛿 + ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )π‘π‘œπ‘ β„Ž π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 π‘π‘œπ‘ β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = (π‘Ÿ βˆ’ π‘Ÿ1) ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) π‘π‘œπ‘ β„Ž π‘Ÿ βˆ’ π‘Ÿ1 𝛿 + (π‘Ÿ2 βˆ’ π‘Ÿ)π‘π‘œπ‘ β„Ž 𝐿 βˆ’ π‘₯ 𝛿 (π‘Ÿ2 βˆ’ π‘Ÿ1)π‘π‘œπ‘ β„Ž π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 We then substitute 𝛿 = √2𝛼𝑑 = ∞ And get 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = (π‘Ÿ βˆ’ π‘Ÿ1) ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) + (π‘Ÿ2 βˆ’ π‘Ÿ) (π‘Ÿ2 βˆ’ π‘Ÿ1) Upon substitution, we get: 𝑻 βˆ’ π‘»βˆž π‘»π’ƒπŸ βˆ’ π‘»βˆž = (𝒓 βˆ’ π’“πŸ) ( π‘»π’ƒπŸ βˆ’ π‘»βˆž π‘»π’ƒπŸ βˆ’ π‘»βˆž ) + (π’“πŸ βˆ’ 𝒓) (π’“πŸ βˆ’ π’“πŸ) Hence, we get a temperature profile linear in radius r in steady state but this temperature profile is in contrast to the logarithmic temperature profile expected. So, we have to choose which theory we take to be true. The other thing we notice is that the solution we have got doesn’t reduce to the one of the semi-infinite solution when the radius π‘Ÿ2 tends to infinity.
  • 17. CASE3: CONVECTION AT THE FREE END This same analysis can be extended to a metal rod with convection at the other end of the metal rod where the temperature profile is given by [1]: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh[π‘š(𝐿 βˆ’ π‘₯)] + ( β„ŽπΏ π‘šπ‘˜ ) sinh[π‘š(𝐿 βˆ’ π‘₯)] cosh π‘šπΏ + ( β„ŽπΏ π‘šπ‘˜ ) π‘ π‘–π‘›β„Žπ‘šπΏ Or 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (𝐿 βˆ’ π‘₯) 𝛿 ] + ( β„ŽπΏπ›Ώ π‘˜ ) sinh[( 𝐿 βˆ’ π‘₯ 𝛿 )] cosh 𝐿 𝛿 + ( β„ŽπΏπ›Ώ π‘˜ ) π‘ π‘–π‘›β„Ž 𝐿 𝛿 For cylindrical co-ordinates we make the substitutions 𝐿 = (π‘Ÿ2 βˆ’ π‘Ÿ1) And π‘₯ = π‘Ÿ βˆ’ π‘Ÿ1 And the temperature profile becomes: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh[π‘š(π‘Ÿ2 βˆ’ π‘Ÿ)] + ( β„Žπ‘Ÿ2 π‘šπ‘˜ ) sinh[π‘š(π‘Ÿ2 βˆ’ π‘Ÿ)] cosh π‘š(π‘Ÿ2 βˆ’ π‘Ÿ1) + ( β„Žπ‘Ÿ2 π‘šπ‘˜ ) π‘ π‘–π‘›β„Žπ‘š(π‘Ÿ2 βˆ’ π‘Ÿ1) The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒓 = π’“πŸ βˆ’π’Œ 𝒅𝑻 𝒅𝒓 = π’‰π’“πŸ (𝑻 βˆ’ π‘»βˆž) 𝒂𝒕 𝒓 = π’“πŸ 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 This same analysis can be extended to a metal rod with convection at the other end of the metal rod where the temperature profile is given by [1]: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh[π‘š(π‘Ÿ2 βˆ’ π‘Ÿ)] + ( β„Žπ‘Ÿ2 π‘šπ‘˜ ) sinh[π‘š(π‘Ÿ2 βˆ’ π‘Ÿ)] cosh π‘š(π‘Ÿ2 βˆ’ π‘Ÿ1) + ( β„Žπ‘Ÿ2 π‘šπ‘˜ ) π‘ π‘–π‘›β„Žπ‘š(π‘Ÿ2 βˆ’ π‘Ÿ1) Or
  • 18. 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + ( β„Žπ‘Ÿ2 𝛿 π‘˜ ) sinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 To show that the initial condition is satisfied we assume from the above that π‘Žπ‘‘ 𝑑 = 0, 𝛿 = 0. 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + ( β„Žπ‘Ÿ2 𝛿 π‘˜ ) sinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 Becomes: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 = 𝑒 (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 + 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ) 𝛿 𝑒 π‘Ÿ2βˆ’π‘Ÿ1 𝛿 + 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ1) 𝛿 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ) 𝛿 = π‘’βˆ’ (π‘Ÿ2βˆ’π‘Ÿ) 0 = π‘’βˆ’βˆž(πΏβˆ’π‘₯) = 0 Similarly 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ1) 𝛿 = 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ1) 0 = π‘’βˆ’βˆž(π‘Ÿ2βˆ’π‘Ÿ1) = 0 So, we are left with 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 𝑒 π‘Ÿ2βˆ’π‘Ÿ1 𝛿 = 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 = 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 0 = π‘’βˆ’βˆž(π‘Ÿβˆ’π‘Ÿ1) = 0 Hence at 𝑑 = 0, 𝑇 = π‘‡βˆž and hence the initial condition. We use this temperature profile which satisfies the initial condition to solve the heat equation and get 𝛿 as shown below: Boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒓 = π’“πŸ βˆ’π’Œ 𝒅𝑻 𝒅𝒓 = π’‰π’“πŸ (𝑻 βˆ’ π‘»βˆž) 𝒂𝒕 𝒓 = π’“πŸ 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 The governing temperature profile is:
  • 19. 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + ( β„Žπ‘Ÿ2 𝛿 π‘˜ ) sinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 Which we can express as: 𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] + 𝐁𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] Where: 𝐴 = (𝑇𝑠 βˆ’ π‘‡βˆž) cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 And 𝐡 = ( β„Žπ‘Ÿ2 𝛿 π‘˜ )(𝑇𝑠 βˆ’ π‘‡βˆž) cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 The governing equation is 𝜢 𝝏𝟐 𝑻 ππ’“πŸ + 𝜢 𝟏 𝒓 𝝏𝑻 𝝏𝒓 = 𝝏𝑻 𝝏𝒕 Multiplying through by r the heat equation becomes: π›Όπ‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 πœ•π‘‡ πœ•π‘Ÿ = πœ•π‘Ÿπ‘‡ πœ•π‘‘ Transforming the PDE into an integral equation, we get: 𝛼 [∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] = πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 Let us evaluate the integral ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] + 𝐁𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ]
  • 20. πœ•2 𝑇 πœ•π‘Ÿ2 = 1 𝛿2 (𝐴 cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Bsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 = 1 𝛿2 (π΄π‘Ÿ cosh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Brsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (rsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] Let us evaluate the integral ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 We use ∫(𝑒𝑣)𝑑π‘₯ = 𝑒𝑣 βˆ’ ∫ 𝑣 𝑑𝑒 𝑑π‘₯ 𝑑π‘₯ Let 𝑒 = π‘Ÿ π‘Žπ‘›π‘‘ 𝑑𝑣 π‘‘π‘Ÿ = cosh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] β„Žπ‘’π‘›π‘π‘’ 𝑣 = βˆ’π›Ώ sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] ∫ (π‘Ÿ cosh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ])π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] + 𝛿 ∫ (sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ])π‘‘π‘Ÿ ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] βˆ’ 𝛿2 (cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 βˆ’ 𝛿2 [cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 Simplifying, we get: ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1] Now let us evaluate the integral ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 Let 𝑒 = π‘Ÿ π‘Žπ‘›π‘‘ 𝑑𝑣 π‘‘π‘Ÿ = π‘ π‘–π‘›β„Ž ( π‘Ÿ2βˆ’π‘Ÿ 𝛿 ) β„Žπ‘’π‘›π‘π‘’ 𝑣 = βˆ’π›Ώcosh( π‘Ÿ2βˆ’π‘Ÿ 𝛿 ) ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ) + 𝛿 ∫ (cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ
  • 21. ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ cosh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ) βˆ’ 𝛿2 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [βˆ’π‘Ÿπ›Ώ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )] π‘Ÿ2 π‘Ÿ1 βˆ’ 𝛿2 [sinh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )] π‘Ÿ2 π‘Ÿ1 Simplifying, we get: ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 )) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 (π‘Ÿ1𝛿 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 (π‘Ÿ1𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Simplifying we get: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 ( π‘Ÿ1 𝛿 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 ( π‘Ÿ1 𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2 𝛿 + sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Now let us evaluate the integral: ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] + 𝐁𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Bsinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 = 𝐴 [1 βˆ’ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] βˆ’ B sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) ∫ [ 𝝏𝑻 𝝏𝒓 ] 𝒅𝒓 π’“πŸ π’“πŸ = 𝑨 [𝟏 βˆ’ 𝐜𝐨𝐬𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 )] βˆ’ 𝐁 𝐬𝐒𝐧𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 ) Now let us add the two integrals below: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 ( π‘Ÿ1 𝛿 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 ( π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2 𝛿 + sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] + 𝐴 [1 βˆ’ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] βˆ’ B sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) And get: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [ π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + π΅π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ] Now let us evaluate the integral
  • 22. πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑇 = 𝐴 cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Bsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + π‘‡βˆž ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 𝐴 ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + π‘‡βˆž(π‘Ÿ2 βˆ’ π‘Ÿ1) But we have already evaluated the above so we get: ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] + π‘‡βˆž(π‘Ÿ2 βˆ’ π‘Ÿ1) Substituting all the above in the integral equation, we get 𝛼 [ π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + π΅π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ] = πœ• πœ•π‘‘ [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Now let us collect like terms i.e., let us collect terms in A separately and terms in B separately and get 𝛼 π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) = πœ• πœ•π‘‘ [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] Cancelling out A we get: 𝛼 [ π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] … … … 𝑴 Similarly for terms in B 𝛼 [ π΅π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ] = πœ• πœ•π‘‘ [𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Cancelling out B we get: 𝛼 [ π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))]… … 𝑡 Adding equations M and N we get: 𝛼 [ π‘Ÿ1 𝛿 (sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) βˆ’ π‘Ÿ2 𝛿 ] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 (sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝛿2 [sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] βˆ’ 𝛿(𝛿 + π‘Ÿ2))] Now let us collect terms in sinh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) and get 𝛼 [ π‘Ÿ1 𝛿 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝛿2 [sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )])]
  • 23. Let us drop out the common term sinh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) and get 𝛼 [ π‘Ÿ1 𝛿 ] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 + 𝛿2)] … . . 𝑱 Let us also collect out the π‘Ÿ2 terms and get: 𝛼 [βˆ’ π‘Ÿ2 𝛿 ] = πœ• πœ•π‘‘ [(βˆ’π›Ώ(𝛿 + π‘Ÿ2))] Upon simplification we get: 𝛼 [ π‘Ÿ2 𝛿 ] = πœ• πœ•π‘‘ [(𝛿2 + π‘Ÿ2𝛿)]… . . 𝑲 Let us subtract equation K from equation J and get: 𝛼 [ π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ] = πœ• πœ•π‘‘ [(π‘Ÿ2 βˆ’ π‘Ÿ1)𝛿] Upon simplification, we get: 𝛼 𝛿 = 𝑑𝛿 𝑑𝑑 The boundary conditions are: 𝛿 = 0 π‘Žπ‘‘ 𝑑 = 0 Upon integrating, we get 𝛼 ∫ 𝑑𝑑 𝑑 0 = ∫ 𝛿𝑑𝛿 𝛿 0 We finally get 𝛿 = √2𝛼𝑑 Upon substitution in the temperature profile we get, 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + ( β„Žπ‘Ÿ2 𝛿 π‘˜ ) sinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) βˆšπŸπœΆπ’• ] + ( π’‰π’“πŸ βˆšπŸπœΆπ’• π’Œ ) 𝐬𝐒𝐧𝐑[ (π’“πŸ βˆ’ 𝒓) βˆšπŸπœΆπ’• ] 𝐜𝐨𝐬𝐑 (π’“πŸ βˆ’ π’“πŸ) βˆšπŸπœΆπ’• + ( π’‰π’“πŸ βˆšπŸπœΆπ’• π’Œ )π’”π’Šπ’π’‰ (π’“πŸ βˆ’ π’“πŸ) βˆšπŸπœΆπ’•
  • 24. L’hopital’s rule can be used to find the steady state temperature when time tends to infinity. Again, we notice that the temperature profile above doesn’t reduce to the semi- infinite temperature profile when π‘Ÿ2 tends to infinity because both solutions have different values of 𝛿.
  • 25. CASE4: ZERO CONVECTION AT THE FREE END The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒓 = π’“πŸ 𝒅𝑻 𝒅𝒓 = 𝟎 𝒂𝒕 𝒓 = π’“πŸ 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 Recall the compact temperature profile that satisfies the boundary and initial conditions is [2]: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ) cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) 𝛼 πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ = πœ•π‘‡ πœ•π‘‘ So, the temperature profile we are going to use is: 𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 ( π’“πŸ βˆ’ 𝒓 𝜹 ) Where: 𝐴 = (𝑇𝑠 βˆ’ π‘‡βˆž) cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 Multiplying through by r the heat equation becomes: π›Όπ‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 πœ•π‘‡ πœ•π‘Ÿ = πœ•π‘Ÿπ‘‡ πœ•π‘‘ Transforming the PDE into an integral equation, we get: 𝛼 [∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] = πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 Let us evaluate the integral ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1
  • 26. πœ•2 𝑇 πœ•π‘Ÿ2 = 1 𝛿2 (π΄π‘π‘œπ‘ β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 = 1 𝛿2 (π΄π‘Ÿπ‘π‘œπ‘ β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) Let us evaluate the integral ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 We use ∫(𝑒𝑣)𝑑π‘₯ = 𝑒𝑣 βˆ’ ∫ 𝑣 𝑑𝑒 𝑑π‘₯ 𝑑π‘₯ Let 𝑒 = π‘Ÿ π‘Žπ‘›π‘‘ 𝑑𝑣 π‘‘π‘Ÿ = cosh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] β„Žπ‘’π‘›π‘π‘’ 𝑣 = βˆ’π›Ώ sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] ∫ (π‘Ÿ cosh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ])π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] + 𝛿 ∫ (sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ])π‘‘π‘Ÿ ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] βˆ’ 𝛿2 (cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 βˆ’ 𝛿2 [cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 Simplifying, we get: ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1] ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] Simplifying we get: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 ( π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])]
  • 27. Now let us evaluate the integral: ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 = 𝐴 [1 βˆ’ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] ∫ [ 𝝏𝑻 𝝏𝒓 ] 𝒅𝒓 π’“πŸ π’“πŸ = 𝑨 [𝟏 βˆ’ 𝐜𝐨𝐬𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 )] Now let us add the two integrals below: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 ( π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + [cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] + 𝐴 [1 βˆ’ cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] And get: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [ π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] Now let us evaluate the integral πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑇 = 𝐴 cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + π‘‡βˆž ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 𝐴 ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + π‘‡βˆž(π‘Ÿ2 βˆ’ π‘Ÿ1) But we have already evaluated the above so we get: ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] + π‘‡βˆž(π‘Ÿ2 βˆ’ π‘Ÿ1)
  • 28. Substituting all the above in the integral equation, we get 𝛼 [ π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] = πœ• πœ•π‘‘ [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] Now let us collect like terms i.e., let us collect terms in A separately and get: 𝛼 π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) = πœ• πœ•π‘‘ [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] Cancelling out A we get: 𝛼 [ π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] Here is the catch lets us expand sinh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) and cosh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) into exponentials and get: Calling 𝛽 = π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 𝛼 [ π‘Ÿ1 𝛿 sinh 𝛽] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 sinh 𝛽 + 𝛿2[cosh𝛽 βˆ’ 1])] Now we have: 𝛼 [ π‘Ÿ1 𝛿 ( 𝑒𝛽 βˆ’ π‘’βˆ’π›½ 2 )] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿( 𝑒𝛽 βˆ’ π‘’βˆ’π›½ 2 ) + 𝛿2 [( 𝑒𝛽 + π‘’βˆ’π›½ 2 ) βˆ’ 1])] Collecting like terms i.e., terms in 𝑒𝛽 and π‘’βˆ’π›½ and constant terms and get: 𝛼 [ π‘Ÿ1 𝛿 ( 𝑒𝛽 2 )] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿( 𝑒𝛽 2 ) + 𝛿2 ( 𝑒𝛽 2 ))] Dropping out 𝑒𝛽 we get: 𝛼 [ π‘Ÿ1 𝛿 ] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 + 𝛿2)] … . 𝑴 Similarly collecting terms in π‘’βˆ’π›½ we get: 𝛼 [ π‘Ÿ1 𝛿 ( βˆ’π‘’βˆ’π›½ 2 )] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿( βˆ’π‘’βˆ’π›½ 2 ) + 𝛿2 [( +π‘’βˆ’π›½ 2 )])] Dropping out the π‘’βˆ’π›½ we get:
  • 29. 𝛼 [ π‘Ÿ1 𝛿 ] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 βˆ’ 𝛿2)] … … 𝑡 Collecting out the β€˜constant’ terms we have: 0 = πœ• πœ•π‘‘ [(βˆ’π›Ώ2)] So, what we have is πœ• πœ•π‘‘ [(𝛿2)] = 0 … 𝑷 We are going to substitute equation P in equations M and N and get From equation M, we have: 𝛼 [ π‘Ÿ1 𝛿 ] = πœ•(π‘Ÿ1𝛿) πœ•π‘‘ + πœ•(𝛿2 ) πœ•π‘‘ But from equation P, we have πœ• πœ•π‘‘ [(𝛿2)] = 0 So, we end with 𝛼 [ π‘Ÿ1 𝛿 ] = πœ•(π‘Ÿ1𝛿) πœ•π‘‘ + 0 Upon simplification, we have: 𝛼 𝛿 = 𝑑𝛿 𝑑𝑑 … . 𝑱 Similarly, from equation N, we have 𝛼 [ π‘Ÿ1 𝛿 ] = πœ•(π‘Ÿ1𝛿) πœ•π‘‘ βˆ’ πœ•(𝛿2 ) πœ•π‘‘ But from equation P, we have: πœ• πœ•π‘‘ [(𝛿2)] = 0 So, we end with: 𝛼 [ π‘Ÿ1 𝛿 ] = πœ•(π‘Ÿ1𝛿) πœ•π‘‘ βˆ’ 0 We finally end with:
  • 30. 𝛼 𝛿 = 𝑑𝛿 𝑑𝑑 … 𝑲 We see that equation K and J are similar and so we solve for 𝛿 to get: 𝛿 = √2𝛼𝑑 We finally substitute in the temperature profile to get: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ) cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) OR 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh ( π‘Ÿ2 βˆ’ π‘Ÿ √2𝛼𝑑 ) cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 √2𝛼𝑑 )
  • 31. RADIAL HEAT CONDUCTION WITH LATERAL CONVECTION The governing equation looks is: 𝜢 𝝏𝟐 𝑻 ππ’“πŸ + 𝜢 𝟏 𝒓 𝝏𝑻 𝝏𝒓 βˆ’ 𝒉𝑷 𝑨𝝆π‘ͺ (𝑻 βˆ’ π‘»βˆž) = 𝝏𝑻 𝝏𝒕
  • 32. CASE1: SEMI-INFINITE CASE The governing equation is: 𝜢 𝝏𝟐 𝑻 ππ’“πŸ + 𝜢 𝟏 𝒓 𝝏𝑻 𝝏𝒓 βˆ’ 𝒉𝑷 𝑨𝝆π‘ͺ (𝑻 βˆ’ π‘»βˆž) = 𝝏𝑻 𝝏𝒕 𝑃 = 4πœ‹π‘Ÿ 𝐴 = 2πœ‹π‘Ÿπ‘‘ Where: 𝑑 = β„Žπ‘’π‘–π‘”β„Žπ‘‘ π‘œπ‘“ π‘π‘¦π‘™π‘–π‘›π‘‘π‘’π‘Ÿ Upon substituting the above in the heat equation, we get: 𝜢 𝝏𝟐 𝑻 ππ’“πŸ + 𝜢 𝟏 𝒓 𝝏𝑻 𝝏𝒓 βˆ’ πŸπ’‰ 𝒅𝝆π‘ͺ (𝑻 βˆ’ π‘»βˆž) = 𝝏𝑻 𝝏𝒕 The boundary conditions are: 𝑇 = 𝑇𝑠 π‘Žπ‘‘ π‘Ÿ = π‘Ÿ1 𝑇 = π‘‡βˆž π‘Žπ‘‘ π‘Ÿ = 𝑅 = ∞ The initial condition is: 𝑇 = π‘‡βˆž π‘Žπ‘‘ 𝑑 = 0 The temperature profile that satisfies the boundary conditions above is: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿
  • 33. 𝛼 πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 1 π‘Ÿ πœ•π‘‡ πœ•π‘Ÿ βˆ’ 2β„Ž π‘‘πœŒπΆ (𝑇 βˆ’ π‘‡βˆž) = πœ•π‘‡ πœ•π‘‘ We multiply through by r the heat equation as shown below and solve. π›Όπ‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 πœ•π‘‡ πœ•π‘Ÿ βˆ’ 2β„Ž π‘‘πœŒπΆ π‘Ÿ(𝑇 βˆ’ π‘‡βˆž) = πœ•π‘Ÿπ‘‡ πœ•π‘‘ We transform the PDE into an integral equation and take the limits to be from π‘Ÿ = π‘Ÿ1 to π‘Ÿ = 𝑅 = ∞ 𝛼 ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ 𝑅 π‘Ÿ1 + 𝛼 ∫ [ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ 𝑅 π‘Ÿ1 βˆ’ 2β„Ž π‘‘πœŒπΆ ∫ (π‘Ÿ(𝑇 βˆ’ π‘‡βˆž))π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ 𝑅 π‘Ÿ1 We evaluated those integrals before to be: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ ∞ π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 (π‘Ÿ1 + 𝛿) ∫ [ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ ∞ π‘Ÿ1 = βˆ’(𝑇𝑠 βˆ’ π‘‡βˆž) 2β„Ž π‘‘πœŒπΆ ∫ (𝑇 βˆ’ π‘‡βˆž)π‘‘π‘Ÿ 𝑅=∞ π‘Ÿ1 = π›Ώβ„Ž π‘‘πœŒπΆ (𝑇𝑠 βˆ’ π‘‡βˆž) From the derivations above, we get: ∫ ( πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 πœ• πœ•π‘‘ ∫ π‘‡π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = 𝑑𝛿 𝑑𝑑 (𝑇𝑠 βˆ’ π‘‡βˆž) ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ 𝑅=∞ π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž)(π‘Ÿ1𝛿 + 𝛿2) + π‘‡βˆž(𝑅 βˆ’ π‘Ÿ1) ∫ (π‘Ÿ(𝑇 βˆ’ π‘‡βˆž))π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = (𝑇𝑠 βˆ’ π‘‡βˆž)(π‘Ÿ1𝛿 + 𝛿2 ) πœ• πœ•π‘‘ ∫ (π‘Ÿ(𝑇 βˆ’ π‘‡βˆž))π‘‘π‘Ÿ 𝑅 π‘Ÿ1 = πœ• πœ•π‘‘ [(𝑇𝑠 βˆ’ π‘‡βˆž)(π‘Ÿ1𝛿 + 𝛿2 ) + π‘‡βˆž(𝑅 βˆ’ π‘Ÿ1)] But πœ•(π‘‡βˆž(𝑅 βˆ’ π‘Ÿ1)) πœ•π‘‘ = 0 𝑠𝑖𝑛𝑐𝑒 π‘‡βˆž π‘Žπ‘›π‘‘ (𝑅 βˆ’ π‘Ÿ1) π‘Žπ‘Ÿπ‘’ π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘π‘ 
  • 34. So, we are left with 𝛼 [ (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 (π‘Ÿ1 + 𝛿) βˆ’ (𝑇𝑠 βˆ’ π‘‡βˆž)] βˆ’ (𝑇𝑠 βˆ’ π‘‡βˆž) 2β„Ž π‘‘πœŒπΆ (π‘Ÿ1𝛿 + 𝛿2) = πœ• πœ•π‘‘ [(𝑇𝑠 βˆ’ π‘‡βˆž)(π‘Ÿ1𝛿 + 𝛿2)] We are left with: 𝛼 π‘Ÿ1 𝛿 βˆ’ 2β„Ž π‘‘πœŒπΆ (π‘Ÿ1𝛿 + 𝛿2) = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 + 𝛿2)] Multiplying through by 𝛿 we get: π›Όπ‘Ÿ1 βˆ’ 2β„Ž π‘‘πœŒπΆ 𝛿(π‘Ÿ1𝛿 + 𝛿2) = πœ• πœ•π‘‘ [𝛿(π‘Ÿ1𝛿 + 𝛿2)] Calling 𝛿(π‘Ÿ1𝛿 + 𝛿2) as X i.e., 𝑋 = 𝛿(π‘Ÿ1𝛿 + 𝛿2) We get: π›Όπ‘Ÿ1 βˆ’ 2β„Ž π‘‘πœŒπΆ 𝑋 = 𝑑𝑋 𝑑𝑑 We solve the above 0DE with limits that at 𝑑 = 0 𝑋 = 0 since 𝛿 = 0 π‘Žπ‘‘ 𝑑 = 0 And we get: 𝑋 = πΎπ‘‘π‘Ÿ1 2β„Ž (1 βˆ’ 𝑒 βˆ’2β„Žπ‘‘ π‘‘πœŒπΆ ) Substituting for X we get: 𝛿(π‘Ÿ1𝛿 + 𝛿2) = πΎπ‘‘π‘Ÿ1 2β„Ž (1 βˆ’ 𝑒 βˆ’2β„Žπ‘‘ π‘‘πœŒπΆ ) Upon simplifying we get: πœΉπŸ‘ + π’“πŸπœΉπŸ βˆ’ π‘²π’…π’“πŸ πŸπ’‰ (𝟏 βˆ’ 𝒆 βˆ’πŸπ’‰π’• 𝒅𝝆π‘ͺ ) = 𝟎 Which is a cubic function that can be solved to get 𝛿 as a function of time. In steady state when time t tends to infinity, we get: πœΉπŸ‘ + π’“πŸπœΉπŸ βˆ’ π‘²π’…π’“πŸ πŸπ’‰ = 𝟎
  • 35. CASE2: CONVECTION AT THE FREE END The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒓 = π’“πŸ βˆ’π’Œ 𝒅𝑻 𝒅𝒓 = π’‰π’“πŸ (𝑻 βˆ’ π‘»βˆž) 𝒂𝒕 𝒓 = π’“πŸ 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 This same analysis can be extended to a metal rod with convection at the other end of the metal rod where the temperature profile is given by [1]: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh[π‘š(π‘Ÿ2 βˆ’ π‘Ÿ)] + ( β„Žπ‘Ÿ2 π‘šπ‘˜ ) sinh[π‘š(π‘Ÿ2 βˆ’ π‘Ÿ)] cosh π‘š(π‘Ÿ2 βˆ’ π‘Ÿ1) + ( β„Žπ‘Ÿ2 π‘šπ‘˜ ) π‘ π‘–π‘›β„Žπ‘š(π‘Ÿ2 βˆ’ π‘Ÿ1) Or 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + ( β„Žπ‘Ÿ2 𝛿 π‘˜ ) sinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 To show that the initial condition is satisfied we assume from the above that π‘Žπ‘‘ 𝑑 = 0, 𝛿 = 0. 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + ( β„Žπ‘Ÿ2 𝛿 π‘˜ ) sinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 Becomes: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 = 𝑒 (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 + 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ) 𝛿 𝑒 π‘Ÿ2βˆ’π‘Ÿ1 𝛿 + 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ1) 𝛿 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ) 𝛿 = π‘’βˆ’ (π‘Ÿ2βˆ’π‘Ÿ) 0 = π‘’βˆ’βˆž(πΏβˆ’π‘₯) = 0 Similarly 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ1) 𝛿 = 𝑒 βˆ’(π‘Ÿ2βˆ’π‘Ÿ1) 0 = π‘’βˆ’βˆž(π‘Ÿ2βˆ’π‘Ÿ1) = 0
  • 36. So, we are left with 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 𝑒 π‘Ÿ2βˆ’π‘Ÿ1 𝛿 = 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 𝛿 = 𝑒 βˆ’(π‘Ÿβˆ’π‘Ÿ1) 0 = π‘’βˆ’βˆž(π‘Ÿβˆ’π‘Ÿ1) = 0 Hence at 𝑑 = 0, 𝑇 = π‘‡βˆž and hence the initial condition. We use this temperature profile which satisfies the initial condition to solve the heat equation and get 𝛿 as shown below: Boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒓 = π’“πŸ βˆ’π’Œ 𝒅𝑻 𝒅𝒓 = π’‰π’“πŸ (𝑻 βˆ’ π‘»βˆž) 𝒂𝒕 𝒓 = π’“πŸ 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 The governing temperature profile is: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + ( β„Žπ‘Ÿ2 𝛿 π‘˜ ) sinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 Which we can express as: 𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] + 𝐁𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] Where: 𝐴 = (𝑇𝑠 βˆ’ π‘‡βˆž) cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 And 𝐡 = ( β„Žπ‘Ÿ2 𝛿 π‘˜ )(𝑇𝑠 βˆ’ π‘‡βˆž) cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 The governing equation is
  • 37. 𝜢 𝝏𝟐 𝑻 ππ’“πŸ + 𝜢 𝟏 𝒓 𝝏𝑻 𝝏𝒓 βˆ’ 𝒉𝑷 𝑨𝝆π‘ͺ (𝑻 βˆ’ π‘»βˆž) = 𝝏𝑻 𝝏𝒕 Where: 𝑃 = 4πœ‹π‘Ÿ 𝐴 = 2πœ‹π‘Ÿπ‘‘ Upon substitution of the above in the heat equation, we get: 𝜢 𝝏𝟐 𝑻 ππ’“πŸ + 𝜢 𝟏 𝒓 𝝏𝑻 𝝏𝒓 βˆ’ πŸπ’‰ 𝒅𝝆π‘ͺ (𝑻 βˆ’ π‘»βˆž) = 𝝏𝑻 𝝏𝒕 Multiplying through by r the heat equation becomes: π›Όπ‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 + 𝛼 πœ•π‘‡ πœ•π‘Ÿ βˆ’ 2β„Ž π‘‘πœŒπΆ π‘Ÿ(𝑇 βˆ’ π‘‡βˆž) = πœ•π‘Ÿπ‘‡ πœ•π‘‘ Transforming the PDE into an integral equation, we get: 𝛼 [∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] βˆ’ 2β„Ž π‘‘πœŒπΆ ∫ (π‘Ÿ(𝑇 βˆ’ π‘‡βˆž))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 Let us evaluate the integral ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1
  • 38. 𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] + 𝐁𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] πœ•2 𝑇 πœ•π‘Ÿ2 = 1 𝛿2 (𝐴 cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Bsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 = 1 𝛿2 (π΄π‘Ÿ cosh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Brsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (rsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] Let us evaluate the integral ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 We use ∫(𝑒𝑣)𝑑π‘₯ = 𝑒𝑣 βˆ’ ∫ 𝑣 𝑑𝑒 𝑑π‘₯ 𝑑π‘₯ Let 𝑒 = π‘Ÿ π‘Žπ‘›π‘‘ 𝑑𝑣 π‘‘π‘Ÿ = cosh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] β„Žπ‘’π‘›π‘π‘’ 𝑣 = βˆ’π›Ώ sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] ∫ (π‘Ÿ cosh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ])π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ] + 𝛿 ∫ (sinh [ (π‘Ÿ2βˆ’π‘Ÿ) 𝛿 ])π‘‘π‘Ÿ ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] βˆ’ 𝛿2 (cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [βˆ’π‘Ÿπ›Ώ sinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 βˆ’ 𝛿2 [cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 Simplifying, we get: ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1] Now let us evaluate the integral ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 Let 𝑒 = π‘Ÿ π‘Žπ‘›π‘‘ 𝑑𝑣 π‘‘π‘Ÿ = π‘ π‘–π‘›β„Ž ( π‘Ÿ2βˆ’π‘Ÿ 𝛿 ) β„Žπ‘’π‘›π‘π‘’ 𝑣 = βˆ’π›Ώcosh( π‘Ÿ2βˆ’π‘Ÿ 𝛿 )
  • 39. ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ) + 𝛿 ∫ (cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ = βˆ’π‘Ÿπ›Ώ cosh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ) βˆ’ 𝛿2 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [βˆ’π‘Ÿπ›Ώ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )] π‘Ÿ2 π‘Ÿ1 βˆ’ 𝛿2 [sinh( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )] π‘Ÿ2 π‘Ÿ1 Simplifying, we get: ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 )) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ βˆ’ π‘Ÿ1 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 ))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 1 𝛿2 [𝐴 (π‘Ÿ1𝛿 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 (π‘Ÿ1𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Simplifying we get: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 ( π‘Ÿ1 𝛿 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 ( π‘Ÿ1 𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2 𝛿 + sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Now let us evaluate the integral: ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] + 𝐁𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Bsinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]] π‘Ÿ2 π‘Ÿ1 = 𝐴 [1 βˆ’ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] βˆ’ B sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) ∫ [ 𝝏𝑻 𝝏𝒓 ] 𝒅𝒓 π’“πŸ π’“πŸ = 𝑨 [𝟏 βˆ’ 𝐜𝐨𝐬𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 )] βˆ’ 𝐁 𝐬𝐒𝐧𝐑 ( π’“πŸ βˆ’ π’“πŸ 𝜹 ) Now let us add the two integrals below: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 )π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ]π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 ( π‘Ÿ1 𝛿 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 ( π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2 𝛿 + sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] + 𝐴 [1 βˆ’ cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] βˆ’ B sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) And get: ∫ (π‘Ÿ πœ•2 𝑇 πœ•π‘Ÿ2 ) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + ∫ [ πœ•π‘‡ πœ•π‘Ÿ ] π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [ π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + π΅π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ]
  • 40. Now let us evaluate: 2β„Ž π‘‘πœŒπΆ ∫ (π‘Ÿ(𝑇 βˆ’ π‘‡βˆž))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 Let us evaluate: ∫ (π‘Ÿ(𝑇 βˆ’ π‘‡βˆž))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑻 βˆ’ π‘»βˆž = 𝑨 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] + 𝐁𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) 𝜹 ] π‘Ÿ(𝑇 βˆ’ π‘‡βˆž) = π΄π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Brsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] ∫ (π‘Ÿ(𝑇 βˆ’ π‘‡βˆž))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ])π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (rsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ] We have already evaluated the integrals above so we get: ∫ (π‘Ÿ(𝑇 βˆ’ π‘‡βˆž))π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 (π‘Ÿ1𝛿 sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 (π‘Ÿ1𝛿cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Now let us evaluate the integral πœ• πœ•π‘‘ ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 𝑇 = 𝐴 cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + Bsinh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + π‘‡βˆž ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = 𝐴 ∫ (π‘Ÿ cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ]) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + 𝐡 ∫ (π‘Ÿπ‘ π‘–π‘›β„Ž ( π‘Ÿ2 βˆ’ π‘Ÿ 𝛿 )) π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 + π‘‡βˆž(π‘Ÿ2 βˆ’ π‘Ÿ1) But we have already evaluated the above so we get: ∫ (π‘Ÿπ‘‡)π‘‘π‘Ÿ π‘Ÿ2 π‘Ÿ1 = [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] + π‘‡βˆž(π‘Ÿ2 βˆ’ π‘Ÿ1) Substituting all the above in the integral equation, we get
  • 41. 𝛼 [ π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + π΅π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ] βˆ’ 2β„Ž π‘‘πœŒπΆ [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 (π‘Ÿ1𝛿 cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] = πœ• πœ•π‘‘ [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1]) + 𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Now let us collect like terms i.e., let us collect terms in A separately and terms in B separately and get 𝛼 π΄π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 2β„Ž π‘‘πœŒπΆ [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] = πœ• πœ•π‘‘ [𝐴 (π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] Cancelling out A we get: 𝛼 [ π‘Ÿ1 𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] βˆ’ 2β„Ž π‘‘πœŒπΆ [(π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + 𝛿2 [cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ 1])] … … … 𝑴 Similarly for terms in B 𝛼 [ π΅π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ] βˆ’ 2β„Ž π‘‘πœŒπΆ [𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] = πœ• πœ•π‘‘ [𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] Cancelling out B we get: 𝛼 [ π‘Ÿ1 𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π΅π‘Ÿ2 𝛿 ] βˆ’ 2β„Ž π‘‘πœŒπΆ [𝐡 (π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) βˆ’ π‘Ÿ2𝛿 + 𝛿2 sinh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] …… 𝑡 Adding equations M and N we get: 𝛼 [ π‘Ÿ1 𝛿 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) βˆ’ π‘Ÿ2 𝛿 ] βˆ’ 2β„Ž π‘‘πœŒπΆ [(π‘Ÿ1𝛿 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝛿2 [sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] βˆ’ 𝛿(𝛿 + π‘Ÿ2))] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝛿2 [sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] βˆ’ 𝛿(𝛿 + π‘Ÿ2))]…. 𝑃 Now let us collect terms in sinh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) and get 𝛼 [ π‘Ÿ1 𝛿 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ))] βˆ’ 2β„Ž π‘‘πœŒπΆ [(π‘Ÿ1𝛿 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝛿2 [sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 (sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )) + 𝛿2 [sinh( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 )])] Let us drop out the common term sinh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) + cosh ( π‘Ÿ2βˆ’π‘Ÿ1 𝛿 ) and get 𝛼 [ π‘Ÿ1 𝛿 ] βˆ’ 2β„Ž π‘‘πœŒπΆ [(π‘Ÿ1𝛿 + 𝛿2 ] = πœ• πœ•π‘‘ [(π‘Ÿ1𝛿 + 𝛿2)]… . . 𝑱 Let us also collect out the π‘Ÿ2 terms from equation P and get: 𝛼 [βˆ’ π‘Ÿ2 𝛿 ] βˆ’ 2β„Ž π‘‘πœŒπΆ [βˆ’π›Ώ(𝛿 + π‘Ÿ2)] = πœ• πœ•π‘‘ [(βˆ’π›Ώ(𝛿 + π‘Ÿ2))] Upon simplification we get:
  • 42. 𝛼 [ π‘Ÿ2 𝛿 ] βˆ’ 2β„Ž π‘‘πœŒπΆ [𝛿(𝛿 + π‘Ÿ2)] = πœ• πœ•π‘‘ [(𝛿2 + π‘Ÿ2𝛿)]… . . 𝑲 Let us subtract equation K from equation J and get: 𝛼 [ π‘Ÿ2 βˆ’ π‘Ÿ1 𝛿 ] βˆ’ 2β„Ž π‘‘πœŒπΆ [(π‘Ÿ2 βˆ’ π‘Ÿ1)𝛿] = πœ• πœ•π‘‘ [(π‘Ÿ2 βˆ’ π‘Ÿ1)𝛿] Upon simplification, we get: 𝛼 𝛿 βˆ’ 2β„Ž π‘‘πœŒπΆ 𝛿 = 𝑑𝛿 𝑑𝑑 𝛼 βˆ’ 2β„Ž π‘‘πœŒπΆ 𝛿2 = 𝛿 𝑑𝛿 𝑑𝑑 We solve the equation above with initial condition 𝛿 = 0 π‘Žπ‘‘ 𝑑 = 0 And get 𝛿 = √ 𝐾𝑑 2β„Ž (1 βˆ’ 𝑒 βˆ’4β„Ž π‘‘πœŒπΆ 𝑑 ) Upon substitution in the temperature profile we get, 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] + ( β„Žπ‘Ÿ2 𝛿 π‘˜ ) sinh[ (π‘Ÿ2 βˆ’ π‘Ÿ) 𝛿 ] cosh (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 + ( β„Žπ‘Ÿ2 𝛿 π‘˜ )π‘ π‘–π‘›β„Ž (π‘Ÿ2 βˆ’ π‘Ÿ1) 𝛿 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) √ 𝑲𝒅 πŸπ’‰ (𝟏 βˆ’ 𝒆 βˆ’πŸ’π’‰ 𝒅𝝆π‘ͺ 𝒕 ) ] + ( π’‰π’“πŸ √ 𝑲𝒅 πŸπ’‰ (𝟏 βˆ’ 𝒆 βˆ’πŸ’π’‰ 𝒅𝝆π‘ͺ 𝒕 ) π’Œ ) 𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) √ 𝑲𝒅 πŸπ’‰ (𝟏 βˆ’ 𝒆 βˆ’πŸ’π’‰ 𝒅𝝆π‘ͺ 𝒕 ) ] 𝐜𝐨𝐬𝐑 (π’“πŸ βˆ’ π’“πŸ) √ 𝑲𝒅 πŸπ’‰ (𝟏 βˆ’ 𝒆 βˆ’πŸ’π’‰ 𝒅𝝆π‘ͺ 𝒕 ) + ( π’‰π’“πŸ √ 𝑲𝒅 πŸπ’‰ (𝟏 βˆ’ 𝒆 βˆ’πŸ’π’‰ 𝒅𝝆π‘ͺ 𝒕 ) π’Œ ) π’”π’Šπ’π’‰ (π’“πŸ βˆ’ π’“πŸ) √ 𝑲𝒅 πŸπ’‰ (𝟏 βˆ’ 𝒆 βˆ’πŸ’π’‰ 𝒅𝝆π‘ͺ 𝒕 )
  • 43. In steady state when time t tends to infinity, we get 𝛿 = √ 𝐾𝑑 2β„Ž And the temperature profile becomes: 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = 𝐜𝐨𝐬𝐑 [ (π’“πŸ βˆ’ 𝒓) βˆšπ‘²π’… πŸπ’‰ ] + ( π’‰π’“πŸ βˆšπ‘²π’… πŸπ’‰ π’Œ ) 𝐬𝐒𝐧𝐑 [ (π’“πŸ βˆ’ 𝒓) βˆšπ‘²π’… πŸπ’‰ ] 𝐜𝐨𝐬𝐑 (π’“πŸ βˆ’ π’“πŸ) βˆšπ‘²π’… πŸπ’‰ + ( π’‰π’“πŸ βˆšπ‘²π’… πŸπ’‰ π’Œ ) π’”π’Šπ’π’‰ (π’“πŸ βˆ’ π’“πŸ) βˆšπ‘²π’… πŸπ’‰ Again, we notice that the temperature profile above doesn’t reduce to the semi- infinite temperature profile when π‘Ÿ2 tends to infinity because both solutions have different values of 𝛿. It can be shown that other transient boundary value problems where the hollow cylinder is finite and not semi-infinite, the value of 𝛿 for lateral convection case will be as derived above: 𝛿 = √ 𝐾𝑑 2β„Ž (1 βˆ’ 𝑒 βˆ’4β„Ž π‘‘πœŒπΆ 𝑑 )
  • 44. CONCLUSION It should be noted that these equations are the ideal solutions of heat flow and that experimental results are needed to verify the nature of some parameters like the nature of the heat transfer coefficient at the end of the metal rod β„ŽπΏ. Also, experimental results are needed to verify the transient nature of the heat flow.
  • 45. REFERENCES [1] C.P.Kothandaraman, "Heat Transfer with Extendeded Surfaces(Fins)," in Fundamentals of Heat and Mass Transfer, New Delhi, NEW AGE INTERNATIONAL PUBLISHERS, 2006, p. 129. [2] C. E. W. R. E. W. G. L. R. James R. Welty, "HEAT TRANSFER FROM EXTENDED SURFACES," in Fundamentals of Momentum, Heat, and Mass Transfer 5th Edition, Corvallis, Oregon, John Wiley & Sons, Inc., 2000, pp. 236-237.