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GENERAL MATHEMATICAL THEORY OF HEAT CONDUCTION USING THE INTEGRAL HEAT EQUATION.pdf

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Wasswaderrick3

In this book, we look at using the integral heat equation as the general method of solving the heat equation subject to given boundary conditions. We begin first by looking at x- directional heat conduction and look at the case of the insulated metal rod first. It is known from literature that the Fourier series yield a solution to this problem for given boundary conditions. But on analyzing the solution got, we notice that it is made up of an infinite number of terms and what this means is that we shall only have an approximate solution since we can’t in practice add up all the terms to infinity. To solve this problem, we solved the heat equation by first transforming it into an integral equation and then find an exact solution as shall be shown in the text later. In solving the heat equation, the temperature profiles that satisfy the heat equation are exponential temperature profiles and hyperbolic temperature profiles as derived in literature for heat conduction in fins. For this case of insulate metal rod, we invoke L’hopital’s rule to get the steady state temperature profile. We then extend this integral equation approach to the case where there is lateral convection along the metal rod and get also both the transient and steady state solution which agrees with theory for steady state heat conduction. After that, 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 same 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 as was the case for x-directional heat flow. We conclude by saying that this same analysis can be extended to spherical co-ordinates heat conduction

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TABLE OF CONTENTS PREFACE ............................................................................................................................................3 X-DIRECTION HEAT FLOW ..........................................................................................................4 X-DIRECTIONAL HEAT FLOW WITH NO LATERAL CONVECTION (OR INSULATED METAL ROD) .........................................................................................................4 THE INTEGRAL HEAT EQUATION METHOD. ......................................................................................6 CASE 1: FIXED END TEMPERATURE...................................................................................................9 CASE 2: CONVECTION AT THE END OF A FINITE METAL ROD...........................................................13 CASE 3: NO CONVECTION AT THE FREE END...................................................................................17 CASE 4: SEMI-INFINITE ROD CASE ..................................................................................................18 HEAT CONDUCTION WITH LATERAL CONVECTION IN x DIRECTION ...................19 CASE1: SEMI-INFINITE CASE WITH LATERAL CONVECTION:.............................................................20 CASE2: CONVECTION AT THE END OF A FINITE METAL ROD ...........................................................23 CASE 3: ZERO FLUX AT THE END OF THE METAL ROD?....................................................................26 RADIAL HEAT FLOW: ...................................................................................................................28 INSULATED HOLLOW CYLINDER WITH NO LATERAL CONVECTION .....................28 CASE1: SEMI-INFINITE HOLLOW CYLINDER.....................................................................................28 CASE 2: FIXED END TEMPERATURE.................................................................................................33 CASE3: CONVECTION AT THE FREE END .........................................................................................40 CASE4: ZERO CONVECTION AT THE FREE END ................................................................................48 HEAT CONDUCTION WITH LATERAL CONVECTION IN CYLINDRICAL CO- ORDINATES..................................................................................................................................54 CASE1: SEMI-INFINITE CASE...........................................................................................................55 CASE2: CONVECTION AT THE FREE END .........................................................................................58 CONCLUSION...................................................................................................................................67 REFERENCES..................................................................................................................................67
PREFACE In this book, we look at using the integral heat equation as the general method of solving the heat equation subject to given boundary conditions. We begin first by looking at x- directional heat conduction and look at the case of the insulated metal rod first. It is known from literature that the Fourier series yield a solution to this problem for given boundary conditions. But on analyzing the solution got, we notice that it is made up of an infinite number of terms and what this means is that we shall only have an approximate solution since we can’t in practice add up all the terms to infinity. To solve this problem, we solved the heat equation by first transforming it into an integral equation and then find an exact solution as shall be shown in the text later. In solving the heat equation, the temperature profiles that satisfy the heat equation are exponential temperature profiles and hyperbolic temperature profiles as derived in literature for heat conduction in fins. For this case of insulate metal rod, we invoke L’hopital’s rule to get the steady state temperature profile. We then extend this integral equation approach to the case where there is lateral convection along the metal rod and get also both the transient and steady state solution which agrees with theory for steady state heat conduction. After that, 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 same 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 as was the case for x-directional heat flow. We conclude by saying that this same analysis can be extended to spherical co-ordinates heat conduction.
X-DIRECTION HEAT FLOW Wasswaderricktimothy7@gmail.com X-DIRECTIONAL HEAT FLOW WITH NO LATERAL CONVECTION (OR INSULATED METAL ROD) The governing equation is: 𝛼 πœ•2 𝑇 πœ•π‘₯2 = πœ•π‘‡ πœ•π‘‘ To get the steady state temperature, we simply substitute πœ•π‘‡ πœ•π‘‘ = 0 𝑖𝑛 π‘ π‘‘π‘’π‘Žπ‘¦ π‘ π‘‘π‘Žπ‘‘π‘’ And get: πœ•2 𝑇 πœ•π‘₯2 = 0 We integrate the above equation to get the variation of temperature with distance x i.e., πœ•π‘‡ πœ•π‘₯ = 𝑐1 𝑻 = π’„πŸπ’™ + π’„πŸ The condition above that the temperature is linear in distance x will help us to solve for the transient state. At π‘₯ = π‘₯1, 𝑇 = 𝑇1 ∴ 𝑇1 = 𝑐1π‘₯1 + 𝑐2 … (𝐴) At π‘₯ = π‘₯2, 𝑇 = 𝑇2 ∴ 𝑇2 = 𝑐1π‘₯2 + 𝑐2 … . (𝐡) (𝐴 βˆ’ 𝐡) π‘™π‘’π‘Žπ‘‘π‘  π‘‘π‘œ 𝑐1 = 𝑇1βˆ’π‘‡2 π‘₯1βˆ’π‘₯2 From A 𝑐2 = 𝑇1 βˆ’ 𝑐1π‘₯1 = 𝑇1 βˆ’ 𝑇1βˆ’π‘‡2 π‘₯1βˆ’π‘₯2 π‘₯1 Substituting in the general solution, we get 𝑇 = 𝑐1π‘₯ + 𝑐2
𝑇 = 𝑇1 βˆ’ 𝑇2 π‘₯1 βˆ’ π‘₯2 π‘₯ + 𝑇1 βˆ’ 𝑇1 βˆ’ 𝑇2 π‘₯1 βˆ’ π‘₯2 π‘₯1 𝑇 βˆ’ 𝑇1 = 𝑇1 βˆ’ 𝑇2 π‘₯1 βˆ’ π‘₯2 (π‘₯ βˆ’ π‘₯1) Since π‘₯2 > π‘₯1 as shown in the diagram above, we can rearrange and get 𝑇 βˆ’ 𝑇1 = βˆ’( 𝑇1 βˆ’ 𝑇2 π‘₯2 βˆ’ π‘₯1 )(π‘₯ βˆ’ π‘₯1) From 𝑄 = βˆ’π‘˜π΄ πœ•π‘‡ πœ•π‘₯ πœ•π‘‡ πœ•π‘₯ = βˆ’( 𝑇1 βˆ’ 𝑇2 π‘₯2 βˆ’ π‘₯1 ) Upon substitution, we get: 𝑄 = π‘˜π΄( 𝑇1 βˆ’ 𝑇2 π‘₯2 βˆ’ π‘₯1 ) What we have found is that in flow in the x direction, in steady state the temperature profile is negatively linear with distance x. But what about the transient state? How do we predict it? That is what we are going to work on below: For the transient state there exists the separation of variables method which yields an infinite series solution. But in this document, we are going to use the integral equation method which will yield an exact solution not an infinite series solution. The governing equation is: 𝜢 𝝏𝟐 𝑻 ππ’™πŸ = 𝝏𝑻 𝝏𝒕 We are going to deal with a variety of boundary conditions and initial condition.
THE INTEGRAL HEAT EQUATION METHOD. We are going to use the integral equation transform method of the heat equation to solve for the transient and steady state. The integral transform works as below: The governing equation is 𝛼 πœ•2 𝑇 πœ•π‘₯2 = πœ•π‘‡ πœ•π‘‘ Let us change this equation into an integral equation as below: 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 Postulate 1: Any temperature profile of the form 𝑇 = 𝐡𝑒 π‘Žπ‘₯ 𝛿 Where B is a constant and a can be a complex number (𝑖), or -1 or +1 can solve the heat equation. This means that even hyperbolic and trigonometric functions can solve the heat equation too. For example, take a temperature profile where π‘Ž = 1 and 𝐡 = 𝑇0 𝑇 = 𝑇0𝑒 π‘₯ 𝛿 πœ•2 𝑇 πœ•π‘₯2 = 𝑇0 𝛿2 𝑒 π‘₯ 𝛿 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = 𝑇0 𝛿 [𝑒 𝐿 𝛿 βˆ’ 1] What we have learned from the above integral is that ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = [ πœ•π‘‡ πœ•π‘₯ ] 𝐿 0 Which we shall use in later text. Anyway, to continue, let us evaluate ∫ (𝑇)𝑑π‘₯ 𝑙 0 = 𝛿[𝑒 𝐿 𝛿 βˆ’ 1]
Upon substitution in the heat equation, we get: 𝛼 𝑇0 𝛿 [𝑒 𝐿 𝛿 βˆ’ 1] = πœ• πœ•π‘‘ [𝑇0𝛿[𝑒 𝐿 𝛿 βˆ’ 1]] There’s a common term 𝑇0[𝑒 𝐿 𝛿 βˆ’ 1] which we drop out and get: 𝛼 𝛿 = 𝑑𝛿 𝑑𝑑 We finally get 𝛿 = √2𝛼𝑑 + 𝑐 We finally substitute in the temperature profile to get: 𝑇 = 𝑇0𝑒 π‘₯ √2𝛼𝑑+𝑐 Where c and 𝑇0are evaluated using the boundary and initial conditions. It can be shown that hyperbolic temperature functions like ( π‘ π‘–π‘›β„Žπ‘Žπ‘₯ π‘œπ‘Ÿ π‘π‘œπ‘ β„Žπ‘Žπ‘₯) too can solve the heat equation using the integral method above. We conclude that the general temperature profile that solves the heat equation is: 𝑇 = 𝐴𝑒 π‘₯ 𝛿 + 𝐡𝑒 βˆ’π‘₯ 𝛿 Where A and B are evaluated using the boundary conditions. From literature for heat flow in extended surfaces (fins), we are going to use the derived temperature profiles to solve for the transient state for no lateral convection in metal rods. Here the nature of m is substituted to be π‘š = 1 𝛿 Where 𝛿(𝑑) will be got by solving the heat equation. The table below will give the required temperature profiles for given boundary conditions [1]
CASE 1: 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 show that the initial condition is satisfied we postulate that π‘Žπ‘‘ 𝑑 = 0, 𝛿 = 0 and get: 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 Let us show that the above postulate implies the initial condition 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 = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )(𝑒 π‘₯ 𝛿 + 𝑒 βˆ’π‘₯ 𝛿 ) + (𝑒 πΏβˆ’π‘₯ 𝛿 + 𝑒 βˆ’(πΏβˆ’π‘₯) 𝛿 ) 𝑒 𝐿 𝛿 βˆ’ 𝑒 βˆ’πΏ 𝛿 𝑒 βˆ’(πΏβˆ’π‘₯) 𝛿 = π‘’βˆ’ (πΏβˆ’π‘₯) 0 = π‘’βˆ’βˆž(πΏβˆ’π‘₯) = 0 Similarly 𝑒 βˆ’πΏ 𝛿 = 𝑒 βˆ’πΏ 0 = π‘’βˆ’βˆžπΏ = 0 So, we are left with
𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) (𝑒 π‘₯ 𝛿) + (𝑒 πΏβˆ’π‘₯ 𝛿 ) 𝑒 𝐿 𝛿 = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )𝑒 βˆ’(πΏβˆ’π‘₯) 𝛿 + 𝑒 βˆ’π‘₯ 𝛿 = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )𝑒 βˆ’(πΏβˆ’π‘₯) 0 + 𝑒 βˆ’π‘₯ 0 = π‘’βˆ’βˆžπ‘₯ We finally get 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )π‘’βˆ’βˆž(πΏβˆ’π‘₯) + π‘’βˆ’βˆžπ‘₯ = 0 Since 𝐿 βˆ’ π‘₯ > 0 Hence at 𝑑 = 0, 𝑇 = π‘‡βˆž and hence the initial condition. Let us solve the heat equation analytically using the temperature profile above: 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = [ πœ•π‘‡ πœ•π‘₯ ] 𝐿 0 [ πœ•π‘‡ πœ•π‘₯ ] 𝐿 0 = (𝑇𝑏1 βˆ’ π‘‡βˆž) 𝛿 [( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) coth ( 𝐿 𝛿 ) βˆ’ ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1)π‘π‘œπ‘ π‘’π‘β„Ž ( 𝐿 𝛿 )] To evaluate ∫ (𝑇)𝑑π‘₯ 𝑙 0 we need to know T 𝑇 = (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 ] + π‘‡βˆž ∫ (𝑇)𝑑π‘₯ 𝑙 0 = 𝛿(𝑇𝑏1 βˆ’ π‘‡βˆž) [( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1)coth ( 𝐿 𝛿 ) βˆ’ ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) π‘π‘œπ‘ π‘’π‘β„Ž ( 𝐿 𝛿 )] + π‘‡βˆžπΏ πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 = 𝑑 𝑑𝑑 (𝛿(𝑇𝑏1 βˆ’ π‘‡βˆž) [( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) coth ( 𝐿 𝛿 ) βˆ’ ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) π‘π‘œπ‘ π‘’π‘β„Ž ( 𝐿 𝛿 )]) + 𝑑(π‘‡βˆžπΏ) 𝑑𝑑 But 𝑑(π‘‡βˆžπΏ) 𝑑𝑑 = 0 𝑠𝑖𝑛𝑐𝑒 𝐿 π‘Žπ‘›π‘‘ π‘‡βˆž π‘Žπ‘Ÿπ‘’ π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘π‘  So, we have
πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 = 𝑑 𝑑𝑑 (𝛿(𝑇𝑏1 βˆ’ π‘‡βˆž) [( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) coth ( 𝐿 𝛿 ) βˆ’ ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) π‘π‘œπ‘ π‘’π‘β„Ž ( 𝐿 𝛿 )]) Upon substitution in the heat equation, we get: 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 𝛼 𝛿 (𝑇𝑏1 βˆ’ π‘‡βˆž) [( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) coth ( 𝐿 𝛿 ) βˆ’ ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) π‘π‘œπ‘ π‘’π‘β„Ž ( 𝐿 𝛿 )] = 𝑑 𝑑𝑑 (𝛿(𝑇𝑏1 βˆ’ π‘‡βˆž) [( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) coth ( 𝐿 𝛿 ) βˆ’ ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) π‘π‘œπ‘ π‘’π‘β„Ž ( 𝐿 𝛿 )]) The factor (𝑇𝑏1 βˆ’ π‘‡βˆž) [( 𝑇𝑏2 βˆ’π‘‡βˆž 𝑇𝑏1 βˆ’π‘‡βˆž + 1) coth ( 𝐿 𝛿 ) βˆ’ ( 𝑇𝑏2 βˆ’π‘‡βˆž 𝑇𝑏1 βˆ’π‘‡βˆž + 1) π‘π‘œπ‘ π‘’π‘β„Ž ( 𝐿 𝛿 )] is common on both sides of the equation so we cross it out and get 𝛼 𝛿 = 𝑑𝛿 𝑑𝑑 Where the boundary conditions are: 𝛿 = 0 π‘Žπ‘‘ 𝑑 = 0 On solving the above we get: 𝛿 = √2𝛼𝑑 Upon substitution in the temperature profile, we get 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 OR 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) π‘ π‘–π‘›β„Ž π‘₯ √2𝛼𝑑 + π‘ π‘–π‘›β„Ž 𝐿 βˆ’ π‘₯ √2𝛼𝑑 π‘ π‘–π‘›β„Ž 𝐿 √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‘₯ ∞ + π‘ π‘–π‘›β„Ž 𝐿 βˆ’ π‘₯ ∞ π‘ π‘–π‘›β„Ž 𝐿 ∞ 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 = π‘₯ 𝛿 ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )π‘π‘œπ‘ β„Ž π‘₯ 𝛿 + ( 𝐿 βˆ’ π‘₯ 𝛿 )π‘π‘œπ‘ β„Ž 𝐿 βˆ’ π‘₯ 𝛿 𝐿 𝛿 π‘π‘œπ‘ β„Žβ„Ž 𝐿 𝛿 Upon simplification, we get 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = π‘₯ ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) π‘π‘œπ‘ β„Ž π‘₯ 𝛿 + (𝐿 βˆ’ π‘₯)π‘π‘œπ‘ β„Ž 𝐿 βˆ’ π‘₯ 𝛿 πΏπ‘π‘œπ‘ β„Žβ„Ž 𝐿 𝛿 We then substitute 𝛿 = √2𝛼𝑑 = ∞ And get 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = π‘₯ ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) + (𝐿 βˆ’ π‘₯) 𝐿 Upon substitution, we get: 𝑻 βˆ’ π‘»βˆž π‘»π’ƒπŸ βˆ’ π‘»βˆž = 𝒙 𝑳 ( π‘»π’ƒπŸ βˆ’ π‘»βˆž π‘»π’ƒπŸ βˆ’ π‘»βˆž ) + (𝟏 βˆ’ 𝒙 𝑳 ) Hence, we get a temperature profile linear in x in steady state.
CASE 2: CONVECTION AT THE END OF A FINITE METAL ROD The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 βˆ’π’Œ 𝒅𝑻 𝒅𝒙 = 𝒉𝑳(𝑻 βˆ’ π‘»βˆž) 𝒂𝒕 𝒙 = 𝒍 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 [2]: 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 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 [ (𝐿 βˆ’ π‘₯) 𝛿 ] + ( β„ŽπΏπ›Ώ π‘˜ ) sinh[( 𝐿 βˆ’ π‘₯ 𝛿 )] cosh 𝐿 𝛿 + ( β„Žπ›Ώ π‘˜ )π‘ π‘–π‘›β„Ž 𝐿 𝛿 Becomes: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (𝐿 βˆ’ π‘₯) 𝛿 ] cosh 𝐿 𝛿 = 𝑒 (πΏβˆ’π‘₯) 𝛿 + 𝑒 βˆ’(πΏβˆ’π‘₯) 𝛿 𝑒 𝐿 𝛿 + 𝑒 βˆ’πΏ 𝛿 𝑒 βˆ’(πΏβˆ’π‘₯) 𝛿 = π‘’βˆ’ (πΏβˆ’π‘₯) 0 = π‘’βˆ’βˆž(πΏβˆ’π‘₯) = 0 Similarly 𝑒 βˆ’πΏ 𝛿 = 𝑒 βˆ’πΏ 0 = π‘’βˆ’βˆžπΏ = 0 So, we are left with
𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 (πΏβˆ’π‘₯) 𝛿 𝑒 𝐿 𝛿 = 𝑒 βˆ’π‘₯ 𝛿 = 𝑒 βˆ’π‘₯ 0 = π‘’βˆ’βˆžπ‘₯ = 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 𝒙 = 𝟎 βˆ’π’Œ 𝒅𝑻 𝒅𝒙 = 𝒉𝑳(𝑻 βˆ’ π‘»βˆž) 𝒂𝒕 𝒙 = 𝒍 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 The governing temperature profile is: 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 The governing equation is 𝛼 πœ•2 𝑇 πœ•π‘₯2 = πœ•π‘‡ πœ•π‘‘ Let us change this equation into an integral equation as below: 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 … … . . 𝑏) 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 ∫ ( πœ•2 𝑇 πœ•π‘₯2 )𝑑π‘₯ 𝑙 0 = [ πœ•π‘‡ πœ•π‘₯ ] 𝑙 0 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 ( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 ) 𝑇 = 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 (𝑇𝑠 βˆ’ π‘‡βˆž) + π‘‡βˆž
πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ [𝛿(𝑇𝑠 βˆ’ π‘‡βˆž) ( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 )] + πœ•(π‘™π‘‡βˆž) πœ•π‘‘ πœ•(π‘™π‘‡βˆž) πœ•π‘‘ = 0 Upon substitution of all the above in the heat equation, we get: 𝛼 (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 ( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 ) = πœ• πœ•π‘‘ [𝛿(𝑇𝑠 βˆ’ π‘‡βˆž) ( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 )] We notice that the term(𝑇𝑠 βˆ’ π‘‡βˆž) ( βˆ’ β„ŽπΏπ›Ώ π‘˜ +(π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 ) is common and can be eliminated: 𝛼 𝛿 = 𝑑𝛿 𝑑𝑑 We go ahead and solve for 𝛿 provided 𝛿 = 0π‘Žπ‘‘ 𝑑 = 0 and get the expression 𝛿 = √2𝛼𝑑 So, the final solution for the finite metal rod with convective flux at the end of the metal rod is: 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = ( 𝐜𝐨𝐬𝐑 [ (𝑳 βˆ’ 𝒙) 𝜹 ] + ( π’‰π‘³πœΉ π’Œ ) 𝐬𝐒𝐧𝐑 [( 𝑳 βˆ’ 𝒙 𝜹 )] 𝐜𝐨𝐬𝐑 𝑳 𝜹 + ( π’‰π‘³πœΉ π’Œ ) π’”π’Šπ’π’‰ 𝑳 𝜹 ) OR 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = ( 𝐜𝐨𝐬𝐑 [ (𝑳 βˆ’ 𝒙) βˆšπŸπœΆπ’• ] + ( π’‰π‘³βˆšπŸπœΆπ’• π’Œ )𝐬𝐒𝐧𝐑 [( 𝑳 βˆ’ 𝒙 βˆšπŸπœΆπ’• )] 𝐜𝐨𝐬𝐑 𝑳 βˆšπŸπœΆπ’• + ( π’‰π‘³βˆšπŸπœΆπ’• π’Œ )π’”π’Šπ’π’‰ 𝑳 βˆšπŸπœΆπ’• ) For the steady state solution, we set 𝑑 β†’ ∞, but first let us first divide through by 𝛿 and get: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = ( 1 𝛿 cosh [ (𝐿 βˆ’ π‘₯) 𝛿 ] + ( β„ŽπΏ π‘˜ ) sinh [( 𝐿 βˆ’ π‘₯ 𝛿 )] 1 𝛿 cosh 𝐿 𝛿 + ( β„ŽπΏ π‘˜ ) π‘ π‘–π‘›β„Ž 𝐿 𝛿 )
Now set 𝑑 β†’ ∞ and get: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 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‘₯) (i.e., differentiate the denominator and numerator with respect to 1 𝛿 ) but we differentiate with respect to the varying parameter and this is 1 𝛿 and get: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = ( (𝐿 βˆ’ π‘₯) 𝛿 sinh [ (𝐿 βˆ’ π‘₯) 𝛿 ] + cosh [ (𝐿 βˆ’ π‘₯) 𝛿 ] + ( β„ŽπΏ π‘˜ ) (𝐿 βˆ’ π‘₯)cosh [( 𝐿 βˆ’ π‘₯ 𝛿 )] 𝐿 𝛿 sinh 𝐿 𝛿 + π‘π‘œπ‘ β„Ž 𝐿 𝛿 + ( β„ŽπΏ π‘˜ )πΏπ‘π‘œπ‘ β„Ž 𝐿 𝛿 ) We now substitute 𝑑 = ∞ in the formula above and get: 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = ( 𝟏 + ( 𝒉𝑳 π’Œ )(𝑳 βˆ’ 𝒙) 𝟏 + ( 𝒉𝑳 π’Œ ) 𝑳 ) The above is the steady state solution and it shows temperature is linearly negative in x.
CASE 3: NO CONVECTION AT THE FREE END The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝒅𝑻 𝒅𝒙 = 𝟎 𝒂𝒕 𝒙 = 𝒍 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[ (𝐿 βˆ’ π‘₯) 𝛿 ] cosh 𝐿 𝛿 Again, solving the heat equation using the temperature profile above yields 𝛿 = √2𝛼𝑑 For steady state conditions we set 𝑑 β†’ ∞ And get: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 1 Hence temperature will be uniformly 𝑇𝑠 throughout the metal rod.
CASE 4: SEMI-INFINITE ROD CASE The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝑻 = π‘»βˆž 𝒂𝒕 𝒙 = ∞ 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 The temperature profile that satisfies the boundary and initial conditions is: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 βˆ’π‘₯ 𝛿 Again, solving the heat equation using the temperature profile above yields 𝛿 = √2𝛼𝑑 For steady state conditions we set 𝑑 β†’ ∞ And get: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 1 Hence temperature will be uniformly 𝑇𝑠 throughout the metal rod. This analysis can be extended to cases where there is lateral convection along the metal rod and yield π‘š = √ β„Žπ‘ƒ π‘˜π΄ in steady state.
HEAT CONDUCTION WITH LATERAL CONVECTION IN x DIRECTION The governing conduction equation is: 𝛼 πœ•2 𝑇 πœ•π‘₯2 βˆ’ β„Žπ‘ƒ 𝐴𝜌𝐢 (𝑇 βˆ’ π‘‡βˆž) = πœ•π‘‡ πœ•π‘‘
CASE1: SEMI-INFINITE CASE WITH LATERAL CONVECTION: The boundary and initial conditions are 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒕 𝑻 = π‘»βˆž 𝒂𝒕 𝒙 = ∞ 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 Where: π‘»βˆž = π’“π’π’π’Ž π’•π’†π’Žπ’‘π’†π’“π’‚π’•π’–π’“π’† First, we assume a temperature profile that satisfies the boundary conditions as: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 βˆ’π‘₯ 𝛿 where 𝛿 is to be determined and is a function of time t and not x. for the initial condition, we assume 𝛿 = 0 at 𝑑 = 0 seconds so that the initial condition is satisfied i.e., Since at 𝑑 = 0, 𝛿 = 0 we get 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 βˆ’π‘₯ 0 = π‘’βˆ’βˆž = 0 Hence 𝑇 = π‘‡βˆž Which is the initial condition. The governing partial differential equation is: 𝛼 πœ•2 𝑇 πœ•π‘₯2 βˆ’ β„Žπ‘ƒ 𝐴𝜌𝐢 (𝑇 βˆ’ π‘‡βˆž) = πœ•π‘‡ πœ•π‘‘ Let us change transform the heat equation into an integral equation as below: 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 βˆ’ β„Žπ‘ƒ 𝐴𝜌𝐢 ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 … … . . 𝑏) πœ•2 𝑇 πœ•π‘₯2 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿2 𝑒 βˆ’π‘₯ 𝛿 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = βˆ’(𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 (𝑒 βˆ’π‘™ 𝛿 βˆ’ 1)
But for a semi-infinite cylindrical rod, 𝑙 = ∞, upon substitution, we get ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = βˆ’π›Ώ(𝑇𝑠 βˆ’ π‘‡βˆž)(𝑒 βˆ’π‘™ 𝛿 βˆ’ 1) But 𝑙 = ∞, upon substitution, we get ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = 𝛿(𝑇𝑠 βˆ’ π‘‡βˆž) 𝑇 = (𝑇𝑠 βˆ’ π‘‡βˆž)𝑒 βˆ’π‘₯ 𝛿 + π‘‡βˆž ∫ (𝑇)𝑑π‘₯ 𝑙 0 = βˆ’π›Ώ(𝑇𝑠 βˆ’ π‘‡βˆž)(𝑒 βˆ’π‘™ 𝛿 βˆ’ 1) + π‘‡βˆžπ‘™ Substitute 𝑙 = ∞ and get πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 = 𝑑𝛿 𝑑𝑑 (𝑇𝑠 βˆ’ π‘‡βˆž) + πœ• πœ•π‘‘ (π‘‡βˆžπ‘™) Since π‘‡βˆž π‘Žπ‘›π‘‘ 𝑙 are constants πœ• πœ•π‘‘ (π‘‡βˆžπ‘™) = 0 πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 = 𝑑𝛿 𝑑𝑑 (𝑇𝑠 βˆ’ π‘‡βˆž) Substituting the above expressions in equation b) above, we get 𝛼 βˆ’ β„Žπ‘ƒ 𝐴𝜌𝐢 𝛿2 = 𝛿 𝑑𝛿 𝑑𝑑 We solve the equation above with initial condition 𝛿 = 0 π‘Žπ‘‘ 𝑑 = 0 And get 𝛿 = √ π›Όπ΄πœŒπΆ β„Žπ‘ƒ (1 βˆ’ 𝑒 βˆ’2β„Žπ‘ƒ 𝐴𝜌𝐢 𝑑 )
𝛿 = √ 𝐾𝐴 β„Žπ‘ƒ (1 βˆ’ 𝑒 βˆ’2β„Žπ‘ƒ 𝐴𝜌𝐢 𝑑 ) Substituting for 𝛿 in the temperature profile, we get 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = 𝒆 βˆ’π’™ βˆšπ‘²π‘¨ 𝒉𝑷 (πŸβˆ’π’† βˆ’πŸπ’‰π‘· 𝑨𝝆π‘ͺ 𝒕 ) From the equation above, we notice that the initial condition is satisfied i.e., 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 The equation above predicts the transient state and in steady state (𝑑 = ∞) it reduces to 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = 𝒆 βˆ’βˆš( 𝒉𝑷 𝑲𝑨 )𝒙 What are the predictions of the transient state? For transient state the governing solution is: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 βˆ’π‘₯ √𝐾𝐴 β„Žπ‘ƒ (1βˆ’π‘’ βˆ’2β„Žπ‘ƒ 𝐴𝜌𝐢 𝑑 ) Let us make π‘₯ the subject of the equation of transient state and get: π‘₯ = [ln ( 𝑇𝑠 βˆ’ π‘‡βˆž 𝑇 βˆ’ π‘‡βˆž )] Γ— √ 𝐾𝐴 β„Žπ‘ƒ (1 βˆ’ 𝑒 βˆ’2β„Žπ‘ƒ 𝐴𝜌𝐢 𝑑 )
CASE2: CONVECTION AT THE END OF A FINITE METAL ROD The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 βˆ’π’Œ 𝒅𝑻 𝒅𝒙 = 𝒉𝑳(𝑻 βˆ’ π‘»βˆž) 𝒂𝒕 𝒙 = 𝒍 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: [2] 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 To show that the initial condition is satisfied we see from the above that π‘Žπ‘‘ 𝑑 = 0, 𝛿 = 0. 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›Ώ π‘˜ )π‘ π‘–π‘›β„Ž 𝐿 𝛿 Becomes: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (𝐿 βˆ’ π‘₯) 𝛿 ] cosh 𝐿 𝛿 = 𝑒 (πΏβˆ’π‘₯) 𝛿 + 𝑒 βˆ’(πΏβˆ’π‘₯) 𝛿 𝑒 𝐿 𝛿 + 𝑒 βˆ’πΏ 𝛿 𝑒 βˆ’(πΏβˆ’π‘₯) 𝛿 = π‘’βˆ’ (πΏβˆ’π‘₯) 0 = π‘’βˆ’βˆž(πΏβˆ’π‘₯) = 0 Similarly 𝑒 βˆ’πΏ 𝛿 = 𝑒 βˆ’πΏ 0 = π‘’βˆ’βˆžπΏ = 0 So, we are left with
𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 (πΏβˆ’π‘₯) 𝛿 𝑒 𝐿 𝛿 = 𝑒 βˆ’π‘₯ 𝛿 = 𝑒 βˆ’π‘₯ 0 = π‘’βˆ’βˆžπ‘₯ = 0 Hence at 𝑑 = 0, 𝑇 = π‘‡βˆž and hence the initial condition. The governing temperature profile is: 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 The governing equation is 𝛼 πœ•2 𝑇 πœ•π‘₯2 βˆ’ β„Žπ‘ƒ 𝐴𝜌𝐢 (𝑇 βˆ’ π‘‡βˆž) = πœ•π‘‡ πœ•π‘‘ Let us change this equation into an integral equation as below: 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 βˆ’ β„Žπ‘ƒ 𝐴𝜌𝐢 ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 … … . . 𝑏) 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 βˆ’ 2β„Ž π‘ŸπœŒπΆ ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 ∫ ( πœ•2 𝑇 πœ•π‘₯2 )𝑑π‘₯ 𝑙 0 = [ πœ•π‘‡ πœ•π‘₯ ] 𝑙 0 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 ( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 ) ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = |βˆ’π›Ώ(𝑇𝑠 βˆ’ π‘‡βˆž) ( sinh [ (𝐿 βˆ’ π‘₯) 𝛿 ] + β„ŽπΏπ›Ώ π‘˜ cosh [ (𝐿 βˆ’ π‘₯) 𝛿 ] π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 )| 𝑙 0 ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = 𝛿(𝑇𝑠 βˆ’ π‘‡βˆž) ( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 ) 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
𝑇 = 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 (𝑇𝑠 βˆ’ π‘‡βˆž) + π‘‡βˆž πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ [𝛿(𝑇𝑠 βˆ’ π‘‡βˆž) ( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 )] + πœ•(π‘™π‘‡βˆž) πœ•π‘‘ πœ•(π‘™π‘‡βˆž) πœ•π‘‘ = 0 Upon substitution of all the above in the heat equation, we get: 𝛼 (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 ( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 ) βˆ’ 2β„Ž π‘ŸπœŒπΆ 𝛿(𝑇𝑠 βˆ’ π‘‡βˆž)( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 ) = πœ• πœ•π‘‘ [𝛿(𝑇𝑠 βˆ’ π‘‡βˆž)( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 )] We notice that the term(𝑇𝑠 βˆ’ π‘‡βˆž) ( βˆ’ β„ŽπΏπ›Ώ π‘˜ +(π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 ) is common and can be eliminated and what this signifies is that the nature of (𝑇𝑠 βˆ’ π‘‡βˆž) doesn’t matter and so we get: 𝛼 𝛿 βˆ’ 2β„Ž π‘ŸπœŒπΆ 𝛿 = 𝑑𝛿 𝑑𝑑 We go ahead and solve for 𝛿 provided 𝛿 = 0π‘Žπ‘‘ 𝑑 = 0 and get the expression 𝛿 = √ 𝐾𝐴 β„Žπ‘ƒ (1 βˆ’ 𝑒 βˆ’2β„Žπ‘ƒ 𝐴𝜌𝐢 𝑑 ) So, the final solution for the finite metal rod with convective flux at the end of the metal rod is: 𝑻 βˆ’ π‘»βˆž 𝑻𝑺 βˆ’ π‘»βˆž = ( 𝐜𝐨𝐬𝐑 [ (𝑳 βˆ’ 𝒙) 𝜹 ] + ( π’‰π‘³πœΉ π’Œ ) 𝐬𝐒𝐧𝐑 [( 𝑳 βˆ’ 𝒙 𝜹 )] 𝐜𝐨𝐬𝐑 𝑳 𝜹 + ( π’‰π‘³πœΉ π’Œ ) π’”π’Šπ’π’‰ 𝑳 𝜹 )
CASE 3: ZERO FLUX AT THE END OF THE METAL ROD? The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝒅𝑻 𝒅𝒙 = 𝟎 𝒂𝒕 𝒙 = 𝒍 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 The equation that satisfies the conditions is [3] 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh[π‘š(𝐿 βˆ’ π‘₯)] cosh π‘šπΏ In terms of 𝜹 we get 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh[ (𝐿 βˆ’ π‘₯) 𝛿 ] cosh 𝐿 𝛿 For the initial condition to be satisfied, we assume that at 𝑑 = 0, 𝛿 = 0 The governing equation is 𝛼 πœ•2 𝑇 πœ•π‘₯2 βˆ’ β„Žπ‘ƒ 𝐴𝜌𝐢 (𝑇 βˆ’ π‘‡βˆž) = πœ•π‘‡ πœ•π‘‘ Let us change this equation into an integral equation as below: 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 βˆ’ β„Žπ‘ƒ 𝐴𝜌𝐢 ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 … … . . 𝑏) 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 βˆ’ 2β„Ž π‘ŸπœŒπΆ ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = [ πœ•π‘‡ πœ•π‘₯ ] 𝑙 0 = (𝑇𝑠 βˆ’ π‘‡βˆž) tanh( 𝐿 𝛿 ) 𝛿 ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = (𝑇𝑠 βˆ’ π‘‡βˆž)𝛿tanh( 𝐿 𝛿 )
𝑇 = cosh [ (𝐿 βˆ’ π‘₯) 𝛿 ] cosh 𝐿 𝛿 (𝑇𝑠 βˆ’ π‘‡βˆž) + π‘‡βˆž πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ [𝛿(𝑇𝑠 βˆ’ π‘‡βˆž) tanh ( 𝐿 𝛿 )] + πœ•(π‘™π‘‡βˆž) πœ•π‘‘ πœ•(π‘™π‘‡βˆž) πœ•π‘‘ = 0 Upon substitution of all the above in the heat equation, we get: 𝛼(𝑇𝑠 βˆ’ π‘‡βˆž) tanh ( 𝐿 𝛿 ) 𝛿 βˆ’ 2β„Ž π‘ŸπœŒπΆ (𝑇𝑠 βˆ’ π‘‡βˆž)𝛿 tanh ( 𝐿 𝛿 ) = πœ• πœ•π‘‘ [𝛿(𝑇𝑠 βˆ’ π‘‡βˆž) tanh ( 𝐿 𝛿 )] We notice that the term (𝑇𝑠 βˆ’ π‘‡βˆž)tanh( 𝐿 𝛿 ) is common and can be eliminated and what this signifies is that the nature of (𝑇𝑠 βˆ’ π‘‡βˆž) doesn’t matter and so we get: 𝛼 𝛿 βˆ’ 2β„Ž π‘ŸπœŒπΆ 𝛿 = 𝑑𝛿 𝑑𝑑 We go ahead and solve for 𝛿 provided 𝛿 = 0π‘Žπ‘‘ 𝑑 = 0 and get the expression 𝛿 = √ 𝐾𝐴 β„Žπ‘ƒ (1 βˆ’ 𝑒 βˆ’2β„Žπ‘ƒ 𝐴𝜌𝐢 𝑑 ) So, the final solution for the finite metal rod with zero flux at the end of the metal rod is: 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = ( 𝐜𝐨𝐬𝐑[ (𝑳 βˆ’ 𝒙) 𝜹 ] 𝐜𝐨𝐬𝐑 𝑳 𝜹 ) It can be shown that the initial condition is satisfied.
RADIAL HEAT FLOW: INSULATED HOLLOW CYLINDER 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 bookInvalid source specified. 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 )
HEAT CONDUCTION WITH LATERAL CONVECTION IN CYLINDRICAL CO-ORDINATES 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 ) This same analysis can be extended to spherical co-ordinates.
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 EXTENDED SURFACES (FINS)," in Fundamentals Of Heat and Mass Transfer, New Delhi, NEW AGE INTERNATIONAL PUBLISHERS, 2006, p. 132. [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. [3] 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, Oregon, John Wiley & Sons, Inc., 2008, p. 237. .
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GENERAL MATHEMATICAL THEORY OF HEAT CONDUCTION USING THE INTEGRAL HEAT EQUATION.pdf

  • 1. Wasswa Derrick 4/27/24 Applied Mathematics
  • 2.
  • 3. TABLE OF CONTENTS PREFACE ............................................................................................................................................3 X-DIRECTION HEAT FLOW ..........................................................................................................4 X-DIRECTIONAL HEAT FLOW WITH NO LATERAL CONVECTION (OR INSULATED METAL ROD) .........................................................................................................4 THE INTEGRAL HEAT EQUATION METHOD. ......................................................................................6 CASE 1: FIXED END TEMPERATURE...................................................................................................9 CASE 2: CONVECTION AT THE END OF A FINITE METAL ROD...........................................................13 CASE 3: NO CONVECTION AT THE FREE END...................................................................................17 CASE 4: SEMI-INFINITE ROD CASE ..................................................................................................18 HEAT CONDUCTION WITH LATERAL CONVECTION IN x DIRECTION ...................19 CASE1: SEMI-INFINITE CASE WITH LATERAL CONVECTION:.............................................................20 CASE2: CONVECTION AT THE END OF A FINITE METAL ROD ...........................................................23 CASE 3: ZERO FLUX AT THE END OF THE METAL ROD?....................................................................26 RADIAL HEAT FLOW: ...................................................................................................................28 INSULATED HOLLOW CYLINDER WITH NO LATERAL CONVECTION .....................28 CASE1: SEMI-INFINITE HOLLOW CYLINDER.....................................................................................28 CASE 2: FIXED END TEMPERATURE.................................................................................................33 CASE3: CONVECTION AT THE FREE END .........................................................................................40 CASE4: ZERO CONVECTION AT THE FREE END ................................................................................48 HEAT CONDUCTION WITH LATERAL CONVECTION IN CYLINDRICAL CO- ORDINATES..................................................................................................................................54 CASE1: SEMI-INFINITE CASE...........................................................................................................55 CASE2: CONVECTION AT THE FREE END .........................................................................................58 CONCLUSION...................................................................................................................................67 REFERENCES..................................................................................................................................67
  • 4. PREFACE In this book, we look at using the integral heat equation as the general method of solving the heat equation subject to given boundary conditions. We begin first by looking at x- directional heat conduction and look at the case of the insulated metal rod first. It is known from literature that the Fourier series yield a solution to this problem for given boundary conditions. But on analyzing the solution got, we notice that it is made up of an infinite number of terms and what this means is that we shall only have an approximate solution since we can’t in practice add up all the terms to infinity. To solve this problem, we solved the heat equation by first transforming it into an integral equation and then find an exact solution as shall be shown in the text later. In solving the heat equation, the temperature profiles that satisfy the heat equation are exponential temperature profiles and hyperbolic temperature profiles as derived in literature for heat conduction in fins. For this case of insulate metal rod, we invoke L’hopital’s rule to get the steady state temperature profile. We then extend this integral equation approach to the case where there is lateral convection along the metal rod and get also both the transient and steady state solution which agrees with theory for steady state heat conduction. After that, 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 same 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 as was the case for x-directional heat flow. We conclude by saying that this same analysis can be extended to spherical co-ordinates heat conduction.
  • 5. X-DIRECTION HEAT FLOW Wasswaderricktimothy7@gmail.com X-DIRECTIONAL HEAT FLOW WITH NO LATERAL CONVECTION (OR INSULATED METAL ROD) The governing equation is: 𝛼 πœ•2 𝑇 πœ•π‘₯2 = πœ•π‘‡ πœ•π‘‘ To get the steady state temperature, we simply substitute πœ•π‘‡ πœ•π‘‘ = 0 𝑖𝑛 π‘ π‘‘π‘’π‘Žπ‘¦ π‘ π‘‘π‘Žπ‘‘π‘’ And get: πœ•2 𝑇 πœ•π‘₯2 = 0 We integrate the above equation to get the variation of temperature with distance x i.e., πœ•π‘‡ πœ•π‘₯ = 𝑐1 𝑻 = π’„πŸπ’™ + π’„πŸ The condition above that the temperature is linear in distance x will help us to solve for the transient state. At π‘₯ = π‘₯1, 𝑇 = 𝑇1 ∴ 𝑇1 = 𝑐1π‘₯1 + 𝑐2 … (𝐴) At π‘₯ = π‘₯2, 𝑇 = 𝑇2 ∴ 𝑇2 = 𝑐1π‘₯2 + 𝑐2 … . (𝐡) (𝐴 βˆ’ 𝐡) π‘™π‘’π‘Žπ‘‘π‘  π‘‘π‘œ 𝑐1 = 𝑇1βˆ’π‘‡2 π‘₯1βˆ’π‘₯2 From A 𝑐2 = 𝑇1 βˆ’ 𝑐1π‘₯1 = 𝑇1 βˆ’ 𝑇1βˆ’π‘‡2 π‘₯1βˆ’π‘₯2 π‘₯1 Substituting in the general solution, we get 𝑇 = 𝑐1π‘₯ + 𝑐2
  • 6. 𝑇 = 𝑇1 βˆ’ 𝑇2 π‘₯1 βˆ’ π‘₯2 π‘₯ + 𝑇1 βˆ’ 𝑇1 βˆ’ 𝑇2 π‘₯1 βˆ’ π‘₯2 π‘₯1 𝑇 βˆ’ 𝑇1 = 𝑇1 βˆ’ 𝑇2 π‘₯1 βˆ’ π‘₯2 (π‘₯ βˆ’ π‘₯1) Since π‘₯2 > π‘₯1 as shown in the diagram above, we can rearrange and get 𝑇 βˆ’ 𝑇1 = βˆ’( 𝑇1 βˆ’ 𝑇2 π‘₯2 βˆ’ π‘₯1 )(π‘₯ βˆ’ π‘₯1) From 𝑄 = βˆ’π‘˜π΄ πœ•π‘‡ πœ•π‘₯ πœ•π‘‡ πœ•π‘₯ = βˆ’( 𝑇1 βˆ’ 𝑇2 π‘₯2 βˆ’ π‘₯1 ) Upon substitution, we get: 𝑄 = π‘˜π΄( 𝑇1 βˆ’ 𝑇2 π‘₯2 βˆ’ π‘₯1 ) What we have found is that in flow in the x direction, in steady state the temperature profile is negatively linear with distance x. But what about the transient state? How do we predict it? That is what we are going to work on below: For the transient state there exists the separation of variables method which yields an infinite series solution. But in this document, we are going to use the integral equation method which will yield an exact solution not an infinite series solution. The governing equation is: 𝜢 𝝏𝟐 𝑻 ππ’™πŸ = 𝝏𝑻 𝝏𝒕 We are going to deal with a variety of boundary conditions and initial condition.
  • 7. THE INTEGRAL HEAT EQUATION METHOD. We are going to use the integral equation transform method of the heat equation to solve for the transient and steady state. The integral transform works as below: The governing equation is 𝛼 πœ•2 𝑇 πœ•π‘₯2 = πœ•π‘‡ πœ•π‘‘ Let us change this equation into an integral equation as below: 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 Postulate 1: Any temperature profile of the form 𝑇 = 𝐡𝑒 π‘Žπ‘₯ 𝛿 Where B is a constant and a can be a complex number (𝑖), or -1 or +1 can solve the heat equation. This means that even hyperbolic and trigonometric functions can solve the heat equation too. For example, take a temperature profile where π‘Ž = 1 and 𝐡 = 𝑇0 𝑇 = 𝑇0𝑒 π‘₯ 𝛿 πœ•2 𝑇 πœ•π‘₯2 = 𝑇0 𝛿2 𝑒 π‘₯ 𝛿 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = 𝑇0 𝛿 [𝑒 𝐿 𝛿 βˆ’ 1] What we have learned from the above integral is that ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = [ πœ•π‘‡ πœ•π‘₯ ] 𝐿 0 Which we shall use in later text. Anyway, to continue, let us evaluate ∫ (𝑇)𝑑π‘₯ 𝑙 0 = 𝛿[𝑒 𝐿 𝛿 βˆ’ 1]
  • 8. Upon substitution in the heat equation, we get: 𝛼 𝑇0 𝛿 [𝑒 𝐿 𝛿 βˆ’ 1] = πœ• πœ•π‘‘ [𝑇0𝛿[𝑒 𝐿 𝛿 βˆ’ 1]] There’s a common term 𝑇0[𝑒 𝐿 𝛿 βˆ’ 1] which we drop out and get: 𝛼 𝛿 = 𝑑𝛿 𝑑𝑑 We finally get 𝛿 = √2𝛼𝑑 + 𝑐 We finally substitute in the temperature profile to get: 𝑇 = 𝑇0𝑒 π‘₯ √2𝛼𝑑+𝑐 Where c and 𝑇0are evaluated using the boundary and initial conditions. It can be shown that hyperbolic temperature functions like ( π‘ π‘–π‘›β„Žπ‘Žπ‘₯ π‘œπ‘Ÿ π‘π‘œπ‘ β„Žπ‘Žπ‘₯) too can solve the heat equation using the integral method above. We conclude that the general temperature profile that solves the heat equation is: 𝑇 = 𝐴𝑒 π‘₯ 𝛿 + 𝐡𝑒 βˆ’π‘₯ 𝛿 Where A and B are evaluated using the boundary conditions. From literature for heat flow in extended surfaces (fins), we are going to use the derived temperature profiles to solve for the transient state for no lateral convection in metal rods. Here the nature of m is substituted to be π‘š = 1 𝛿 Where 𝛿(𝑑) will be got by solving the heat equation. The table below will give the required temperature profiles for given boundary conditions [1]
  • 9.
  • 10. CASE 1: 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 show that the initial condition is satisfied we postulate that π‘Žπ‘‘ 𝑑 = 0, 𝛿 = 0 and get: 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 Let us show that the above postulate implies the initial condition 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 = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )(𝑒 π‘₯ 𝛿 + 𝑒 βˆ’π‘₯ 𝛿 ) + (𝑒 πΏβˆ’π‘₯ 𝛿 + 𝑒 βˆ’(πΏβˆ’π‘₯) 𝛿 ) 𝑒 𝐿 𝛿 βˆ’ 𝑒 βˆ’πΏ 𝛿 𝑒 βˆ’(πΏβˆ’π‘₯) 𝛿 = π‘’βˆ’ (πΏβˆ’π‘₯) 0 = π‘’βˆ’βˆž(πΏβˆ’π‘₯) = 0 Similarly 𝑒 βˆ’πΏ 𝛿 = 𝑒 βˆ’πΏ 0 = π‘’βˆ’βˆžπΏ = 0 So, we are left with
  • 11. 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) (𝑒 π‘₯ 𝛿) + (𝑒 πΏβˆ’π‘₯ 𝛿 ) 𝑒 𝐿 𝛿 = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )𝑒 βˆ’(πΏβˆ’π‘₯) 𝛿 + 𝑒 βˆ’π‘₯ 𝛿 = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )𝑒 βˆ’(πΏβˆ’π‘₯) 0 + 𝑒 βˆ’π‘₯ 0 = π‘’βˆ’βˆžπ‘₯ We finally get 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )π‘’βˆ’βˆž(πΏβˆ’π‘₯) + π‘’βˆ’βˆžπ‘₯ = 0 Since 𝐿 βˆ’ π‘₯ > 0 Hence at 𝑑 = 0, 𝑇 = π‘‡βˆž and hence the initial condition. Let us solve the heat equation analytically using the temperature profile above: 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = [ πœ•π‘‡ πœ•π‘₯ ] 𝐿 0 [ πœ•π‘‡ πœ•π‘₯ ] 𝐿 0 = (𝑇𝑏1 βˆ’ π‘‡βˆž) 𝛿 [( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) coth ( 𝐿 𝛿 ) βˆ’ ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1)π‘π‘œπ‘ π‘’π‘β„Ž ( 𝐿 𝛿 )] To evaluate ∫ (𝑇)𝑑π‘₯ 𝑙 0 we need to know T 𝑇 = (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 ] + π‘‡βˆž ∫ (𝑇)𝑑π‘₯ 𝑙 0 = 𝛿(𝑇𝑏1 βˆ’ π‘‡βˆž) [( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1)coth ( 𝐿 𝛿 ) βˆ’ ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) π‘π‘œπ‘ π‘’π‘β„Ž ( 𝐿 𝛿 )] + π‘‡βˆžπΏ πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 = 𝑑 𝑑𝑑 (𝛿(𝑇𝑏1 βˆ’ π‘‡βˆž) [( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) coth ( 𝐿 𝛿 ) βˆ’ ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) π‘π‘œπ‘ π‘’π‘β„Ž ( 𝐿 𝛿 )]) + 𝑑(π‘‡βˆžπΏ) 𝑑𝑑 But 𝑑(π‘‡βˆžπΏ) 𝑑𝑑 = 0 𝑠𝑖𝑛𝑐𝑒 𝐿 π‘Žπ‘›π‘‘ π‘‡βˆž π‘Žπ‘Ÿπ‘’ π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘π‘  So, we have
  • 12. πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 = 𝑑 𝑑𝑑 (𝛿(𝑇𝑏1 βˆ’ π‘‡βˆž) [( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) coth ( 𝐿 𝛿 ) βˆ’ ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) π‘π‘œπ‘ π‘’π‘β„Ž ( 𝐿 𝛿 )]) Upon substitution in the heat equation, we get: 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 𝛼 𝛿 (𝑇𝑏1 βˆ’ π‘‡βˆž) [( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) coth ( 𝐿 𝛿 ) βˆ’ ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) π‘π‘œπ‘ π‘’π‘β„Ž ( 𝐿 𝛿 )] = 𝑑 𝑑𝑑 (𝛿(𝑇𝑏1 βˆ’ π‘‡βˆž) [( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) coth ( 𝐿 𝛿 ) βˆ’ ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž + 1) π‘π‘œπ‘ π‘’π‘β„Ž ( 𝐿 𝛿 )]) The factor (𝑇𝑏1 βˆ’ π‘‡βˆž) [( 𝑇𝑏2 βˆ’π‘‡βˆž 𝑇𝑏1 βˆ’π‘‡βˆž + 1) coth ( 𝐿 𝛿 ) βˆ’ ( 𝑇𝑏2 βˆ’π‘‡βˆž 𝑇𝑏1 βˆ’π‘‡βˆž + 1) π‘π‘œπ‘ π‘’π‘β„Ž ( 𝐿 𝛿 )] is common on both sides of the equation so we cross it out and get 𝛼 𝛿 = 𝑑𝛿 𝑑𝑑 Where the boundary conditions are: 𝛿 = 0 π‘Žπ‘‘ 𝑑 = 0 On solving the above we get: 𝛿 = √2𝛼𝑑 Upon substitution in the temperature profile, we get 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 OR 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) π‘ π‘–π‘›β„Ž π‘₯ √2𝛼𝑑 + π‘ π‘–π‘›β„Ž 𝐿 βˆ’ π‘₯ √2𝛼𝑑 π‘ π‘–π‘›β„Ž 𝐿 √2𝛼𝑑 In steady state, 𝑑 β†’ ∞ π‘ π‘œ 𝛿 β†’ ∞ Upon substitution we get
  • 13. 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )π‘ π‘–π‘›β„Ž π‘₯ ∞ + π‘ π‘–π‘›β„Ž 𝐿 βˆ’ π‘₯ ∞ π‘ π‘–π‘›β„Ž 𝐿 ∞ 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 = π‘₯ 𝛿 ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž )π‘π‘œπ‘ β„Ž π‘₯ 𝛿 + ( 𝐿 βˆ’ π‘₯ 𝛿 )π‘π‘œπ‘ β„Ž 𝐿 βˆ’ π‘₯ 𝛿 𝐿 𝛿 π‘π‘œπ‘ β„Žβ„Ž 𝐿 𝛿 Upon simplification, we get 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = π‘₯ ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) π‘π‘œπ‘ β„Ž π‘₯ 𝛿 + (𝐿 βˆ’ π‘₯)π‘π‘œπ‘ β„Ž 𝐿 βˆ’ π‘₯ 𝛿 πΏπ‘π‘œπ‘ β„Žβ„Ž 𝐿 𝛿 We then substitute 𝛿 = √2𝛼𝑑 = ∞ And get 𝑇 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž = π‘₯ ( 𝑇𝑏2 βˆ’ π‘‡βˆž 𝑇𝑏1 βˆ’ π‘‡βˆž ) + (𝐿 βˆ’ π‘₯) 𝐿 Upon substitution, we get: 𝑻 βˆ’ π‘»βˆž π‘»π’ƒπŸ βˆ’ π‘»βˆž = 𝒙 𝑳 ( π‘»π’ƒπŸ βˆ’ π‘»βˆž π‘»π’ƒπŸ βˆ’ π‘»βˆž ) + (𝟏 βˆ’ 𝒙 𝑳 ) Hence, we get a temperature profile linear in x in steady state.
  • 14. CASE 2: CONVECTION AT THE END OF A FINITE METAL ROD The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 βˆ’π’Œ 𝒅𝑻 𝒅𝒙 = 𝒉𝑳(𝑻 βˆ’ π‘»βˆž) 𝒂𝒕 𝒙 = 𝒍 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 [2]: 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 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 [ (𝐿 βˆ’ π‘₯) 𝛿 ] + ( β„ŽπΏπ›Ώ π‘˜ ) sinh[( 𝐿 βˆ’ π‘₯ 𝛿 )] cosh 𝐿 𝛿 + ( β„Žπ›Ώ π‘˜ )π‘ π‘–π‘›β„Ž 𝐿 𝛿 Becomes: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (𝐿 βˆ’ π‘₯) 𝛿 ] cosh 𝐿 𝛿 = 𝑒 (πΏβˆ’π‘₯) 𝛿 + 𝑒 βˆ’(πΏβˆ’π‘₯) 𝛿 𝑒 𝐿 𝛿 + 𝑒 βˆ’πΏ 𝛿 𝑒 βˆ’(πΏβˆ’π‘₯) 𝛿 = π‘’βˆ’ (πΏβˆ’π‘₯) 0 = π‘’βˆ’βˆž(πΏβˆ’π‘₯) = 0 Similarly 𝑒 βˆ’πΏ 𝛿 = 𝑒 βˆ’πΏ 0 = π‘’βˆ’βˆžπΏ = 0 So, we are left with
  • 15. 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 (πΏβˆ’π‘₯) 𝛿 𝑒 𝐿 𝛿 = 𝑒 βˆ’π‘₯ 𝛿 = 𝑒 βˆ’π‘₯ 0 = π‘’βˆ’βˆžπ‘₯ = 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 𝒙 = 𝟎 βˆ’π’Œ 𝒅𝑻 𝒅𝒙 = 𝒉𝑳(𝑻 βˆ’ π‘»βˆž) 𝒂𝒕 𝒙 = 𝒍 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 The governing temperature profile is: 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 The governing equation is 𝛼 πœ•2 𝑇 πœ•π‘₯2 = πœ•π‘‡ πœ•π‘‘ Let us change this equation into an integral equation as below: 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 … … . . 𝑏) 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 ∫ ( πœ•2 𝑇 πœ•π‘₯2 )𝑑π‘₯ 𝑙 0 = [ πœ•π‘‡ πœ•π‘₯ ] 𝑙 0 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 ( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 ) 𝑇 = 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 (𝑇𝑠 βˆ’ π‘‡βˆž) + π‘‡βˆž
  • 16. πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ [𝛿(𝑇𝑠 βˆ’ π‘‡βˆž) ( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 )] + πœ•(π‘™π‘‡βˆž) πœ•π‘‘ πœ•(π‘™π‘‡βˆž) πœ•π‘‘ = 0 Upon substitution of all the above in the heat equation, we get: 𝛼 (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 ( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 ) = πœ• πœ•π‘‘ [𝛿(𝑇𝑠 βˆ’ π‘‡βˆž) ( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 )] We notice that the term(𝑇𝑠 βˆ’ π‘‡βˆž) ( βˆ’ β„ŽπΏπ›Ώ π‘˜ +(π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 ) is common and can be eliminated: 𝛼 𝛿 = 𝑑𝛿 𝑑𝑑 We go ahead and solve for 𝛿 provided 𝛿 = 0π‘Žπ‘‘ 𝑑 = 0 and get the expression 𝛿 = √2𝛼𝑑 So, the final solution for the finite metal rod with convective flux at the end of the metal rod is: 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = ( 𝐜𝐨𝐬𝐑 [ (𝑳 βˆ’ 𝒙) 𝜹 ] + ( π’‰π‘³πœΉ π’Œ ) 𝐬𝐒𝐧𝐑 [( 𝑳 βˆ’ 𝒙 𝜹 )] 𝐜𝐨𝐬𝐑 𝑳 𝜹 + ( π’‰π‘³πœΉ π’Œ ) π’”π’Šπ’π’‰ 𝑳 𝜹 ) OR 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = ( 𝐜𝐨𝐬𝐑 [ (𝑳 βˆ’ 𝒙) βˆšπŸπœΆπ’• ] + ( π’‰π‘³βˆšπŸπœΆπ’• π’Œ )𝐬𝐒𝐧𝐑 [( 𝑳 βˆ’ 𝒙 βˆšπŸπœΆπ’• )] 𝐜𝐨𝐬𝐑 𝑳 βˆšπŸπœΆπ’• + ( π’‰π‘³βˆšπŸπœΆπ’• π’Œ )π’”π’Šπ’π’‰ 𝑳 βˆšπŸπœΆπ’• ) For the steady state solution, we set 𝑑 β†’ ∞, but first let us first divide through by 𝛿 and get: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = ( 1 𝛿 cosh [ (𝐿 βˆ’ π‘₯) 𝛿 ] + ( β„ŽπΏ π‘˜ ) sinh [( 𝐿 βˆ’ π‘₯ 𝛿 )] 1 𝛿 cosh 𝐿 𝛿 + ( β„ŽπΏ π‘˜ ) π‘ π‘–π‘›β„Ž 𝐿 𝛿 )
  • 17. Now set 𝑑 β†’ ∞ and get: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 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‘₯) (i.e., differentiate the denominator and numerator with respect to 1 𝛿 ) but we differentiate with respect to the varying parameter and this is 1 𝛿 and get: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = ( (𝐿 βˆ’ π‘₯) 𝛿 sinh [ (𝐿 βˆ’ π‘₯) 𝛿 ] + cosh [ (𝐿 βˆ’ π‘₯) 𝛿 ] + ( β„ŽπΏ π‘˜ ) (𝐿 βˆ’ π‘₯)cosh [( 𝐿 βˆ’ π‘₯ 𝛿 )] 𝐿 𝛿 sinh 𝐿 𝛿 + π‘π‘œπ‘ β„Ž 𝐿 𝛿 + ( β„ŽπΏ π‘˜ )πΏπ‘π‘œπ‘ β„Ž 𝐿 𝛿 ) We now substitute 𝑑 = ∞ in the formula above and get: 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = ( 𝟏 + ( 𝒉𝑳 π’Œ )(𝑳 βˆ’ 𝒙) 𝟏 + ( 𝒉𝑳 π’Œ ) 𝑳 ) The above is the steady state solution and it shows temperature is linearly negative in x.
  • 18. CASE 3: NO CONVECTION AT THE FREE END The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝒅𝑻 𝒅𝒙 = 𝟎 𝒂𝒕 𝒙 = 𝒍 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[ (𝐿 βˆ’ π‘₯) 𝛿 ] cosh 𝐿 𝛿 Again, solving the heat equation using the temperature profile above yields 𝛿 = √2𝛼𝑑 For steady state conditions we set 𝑑 β†’ ∞ And get: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 1 Hence temperature will be uniformly 𝑇𝑠 throughout the metal rod.
  • 19. CASE 4: SEMI-INFINITE ROD CASE The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝑻 = π‘»βˆž 𝒂𝒕 𝒙 = ∞ 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 The temperature profile that satisfies the boundary and initial conditions is: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 βˆ’π‘₯ 𝛿 Again, solving the heat equation using the temperature profile above yields 𝛿 = √2𝛼𝑑 For steady state conditions we set 𝑑 β†’ ∞ And get: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 1 Hence temperature will be uniformly 𝑇𝑠 throughout the metal rod. This analysis can be extended to cases where there is lateral convection along the metal rod and yield π‘š = √ β„Žπ‘ƒ π‘˜π΄ in steady state.
  • 20. HEAT CONDUCTION WITH LATERAL CONVECTION IN x DIRECTION The governing conduction equation is: 𝛼 πœ•2 𝑇 πœ•π‘₯2 βˆ’ β„Žπ‘ƒ 𝐴𝜌𝐢 (𝑇 βˆ’ π‘‡βˆž) = πœ•π‘‡ πœ•π‘‘
  • 21. CASE1: SEMI-INFINITE CASE WITH LATERAL CONVECTION: The boundary and initial conditions are 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒕 𝑻 = π‘»βˆž 𝒂𝒕 𝒙 = ∞ 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 Where: π‘»βˆž = π’“π’π’π’Ž π’•π’†π’Žπ’‘π’†π’“π’‚π’•π’–π’“π’† First, we assume a temperature profile that satisfies the boundary conditions as: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 βˆ’π‘₯ 𝛿 where 𝛿 is to be determined and is a function of time t and not x. for the initial condition, we assume 𝛿 = 0 at 𝑑 = 0 seconds so that the initial condition is satisfied i.e., Since at 𝑑 = 0, 𝛿 = 0 we get 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 βˆ’π‘₯ 0 = π‘’βˆ’βˆž = 0 Hence 𝑇 = π‘‡βˆž Which is the initial condition. The governing partial differential equation is: 𝛼 πœ•2 𝑇 πœ•π‘₯2 βˆ’ β„Žπ‘ƒ 𝐴𝜌𝐢 (𝑇 βˆ’ π‘‡βˆž) = πœ•π‘‡ πœ•π‘‘ Let us change transform the heat equation into an integral equation as below: 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 βˆ’ β„Žπ‘ƒ 𝐴𝜌𝐢 ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 … … . . 𝑏) πœ•2 𝑇 πœ•π‘₯2 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿2 𝑒 βˆ’π‘₯ 𝛿 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = βˆ’(𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 (𝑒 βˆ’π‘™ 𝛿 βˆ’ 1)
  • 22. But for a semi-infinite cylindrical rod, 𝑙 = ∞, upon substitution, we get ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = βˆ’π›Ώ(𝑇𝑠 βˆ’ π‘‡βˆž)(𝑒 βˆ’π‘™ 𝛿 βˆ’ 1) But 𝑙 = ∞, upon substitution, we get ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = 𝛿(𝑇𝑠 βˆ’ π‘‡βˆž) 𝑇 = (𝑇𝑠 βˆ’ π‘‡βˆž)𝑒 βˆ’π‘₯ 𝛿 + π‘‡βˆž ∫ (𝑇)𝑑π‘₯ 𝑙 0 = βˆ’π›Ώ(𝑇𝑠 βˆ’ π‘‡βˆž)(𝑒 βˆ’π‘™ 𝛿 βˆ’ 1) + π‘‡βˆžπ‘™ Substitute 𝑙 = ∞ and get πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 = 𝑑𝛿 𝑑𝑑 (𝑇𝑠 βˆ’ π‘‡βˆž) + πœ• πœ•π‘‘ (π‘‡βˆžπ‘™) Since π‘‡βˆž π‘Žπ‘›π‘‘ 𝑙 are constants πœ• πœ•π‘‘ (π‘‡βˆžπ‘™) = 0 πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 = 𝑑𝛿 𝑑𝑑 (𝑇𝑠 βˆ’ π‘‡βˆž) Substituting the above expressions in equation b) above, we get 𝛼 βˆ’ β„Žπ‘ƒ 𝐴𝜌𝐢 𝛿2 = 𝛿 𝑑𝛿 𝑑𝑑 We solve the equation above with initial condition 𝛿 = 0 π‘Žπ‘‘ 𝑑 = 0 And get 𝛿 = √ π›Όπ΄πœŒπΆ β„Žπ‘ƒ (1 βˆ’ 𝑒 βˆ’2β„Žπ‘ƒ 𝐴𝜌𝐢 𝑑 )
  • 23. 𝛿 = √ 𝐾𝐴 β„Žπ‘ƒ (1 βˆ’ 𝑒 βˆ’2β„Žπ‘ƒ 𝐴𝜌𝐢 𝑑 ) Substituting for 𝛿 in the temperature profile, we get 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = 𝒆 βˆ’π’™ βˆšπ‘²π‘¨ 𝒉𝑷 (πŸβˆ’π’† βˆ’πŸπ’‰π‘· 𝑨𝝆π‘ͺ 𝒕 ) From the equation above, we notice that the initial condition is satisfied i.e., 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 The equation above predicts the transient state and in steady state (𝑑 = ∞) it reduces to 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = 𝒆 βˆ’βˆš( 𝒉𝑷 𝑲𝑨 )𝒙 What are the predictions of the transient state? For transient state the governing solution is: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 βˆ’π‘₯ √𝐾𝐴 β„Žπ‘ƒ (1βˆ’π‘’ βˆ’2β„Žπ‘ƒ 𝐴𝜌𝐢 𝑑 ) Let us make π‘₯ the subject of the equation of transient state and get: π‘₯ = [ln ( 𝑇𝑠 βˆ’ π‘‡βˆž 𝑇 βˆ’ π‘‡βˆž )] Γ— √ 𝐾𝐴 β„Žπ‘ƒ (1 βˆ’ 𝑒 βˆ’2β„Žπ‘ƒ 𝐴𝜌𝐢 𝑑 )
  • 24. CASE2: CONVECTION AT THE END OF A FINITE METAL ROD The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 βˆ’π’Œ 𝒅𝑻 𝒅𝒙 = 𝒉𝑳(𝑻 βˆ’ π‘»βˆž) 𝒂𝒕 𝒙 = 𝒍 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: [2] 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 To show that the initial condition is satisfied we see from the above that π‘Žπ‘‘ 𝑑 = 0, 𝛿 = 0. 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›Ώ π‘˜ )π‘ π‘–π‘›β„Ž 𝐿 𝛿 Becomes: 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh [ (𝐿 βˆ’ π‘₯) 𝛿 ] cosh 𝐿 𝛿 = 𝑒 (πΏβˆ’π‘₯) 𝛿 + 𝑒 βˆ’(πΏβˆ’π‘₯) 𝛿 𝑒 𝐿 𝛿 + 𝑒 βˆ’πΏ 𝛿 𝑒 βˆ’(πΏβˆ’π‘₯) 𝛿 = π‘’βˆ’ (πΏβˆ’π‘₯) 0 = π‘’βˆ’βˆž(πΏβˆ’π‘₯) = 0 Similarly 𝑒 βˆ’πΏ 𝛿 = 𝑒 βˆ’πΏ 0 = π‘’βˆ’βˆžπΏ = 0 So, we are left with
  • 25. 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = 𝑒 (πΏβˆ’π‘₯) 𝛿 𝑒 𝐿 𝛿 = 𝑒 βˆ’π‘₯ 𝛿 = 𝑒 βˆ’π‘₯ 0 = π‘’βˆ’βˆžπ‘₯ = 0 Hence at 𝑑 = 0, 𝑇 = π‘‡βˆž and hence the initial condition. The governing temperature profile is: 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 The governing equation is 𝛼 πœ•2 𝑇 πœ•π‘₯2 βˆ’ β„Žπ‘ƒ 𝐴𝜌𝐢 (𝑇 βˆ’ π‘‡βˆž) = πœ•π‘‡ πœ•π‘‘ Let us change this equation into an integral equation as below: 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 βˆ’ β„Žπ‘ƒ 𝐴𝜌𝐢 ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 … … . . 𝑏) 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 βˆ’ 2β„Ž π‘ŸπœŒπΆ ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 ∫ ( πœ•2 𝑇 πœ•π‘₯2 )𝑑π‘₯ 𝑙 0 = [ πœ•π‘‡ πœ•π‘₯ ] 𝑙 0 = (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 ( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 ) ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = |βˆ’π›Ώ(𝑇𝑠 βˆ’ π‘‡βˆž) ( sinh [ (𝐿 βˆ’ π‘₯) 𝛿 ] + β„ŽπΏπ›Ώ π‘˜ cosh [ (𝐿 βˆ’ π‘₯) 𝛿 ] π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 )| 𝑙 0 ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = 𝛿(𝑇𝑠 βˆ’ π‘‡βˆž) ( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 ) 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
  • 26. 𝑇 = 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 (𝑇𝑠 βˆ’ π‘‡βˆž) + π‘‡βˆž πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ [𝛿(𝑇𝑠 βˆ’ π‘‡βˆž) ( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 )] + πœ•(π‘™π‘‡βˆž) πœ•π‘‘ πœ•(π‘™π‘‡βˆž) πœ•π‘‘ = 0 Upon substitution of all the above in the heat equation, we get: 𝛼 (𝑇𝑠 βˆ’ π‘‡βˆž) 𝛿 ( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 ) βˆ’ 2β„Ž π‘ŸπœŒπΆ 𝛿(𝑇𝑠 βˆ’ π‘‡βˆž)( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 ) = πœ• πœ•π‘‘ [𝛿(𝑇𝑠 βˆ’ π‘‡βˆž)( βˆ’ β„ŽπΏπ›Ώ π‘˜ + (π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 )] We notice that the term(𝑇𝑠 βˆ’ π‘‡βˆž) ( βˆ’ β„ŽπΏπ›Ώ π‘˜ +(π‘ π‘–π‘›β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘π‘œπ‘ β„Ž 𝐿 𝛿 ) π‘π‘œπ‘ β„Ž 𝐿 𝛿 + β„ŽπΏπ›Ώ π‘˜ π‘ π‘–π‘›β„Ž 𝐿 𝛿 ) is common and can be eliminated and what this signifies is that the nature of (𝑇𝑠 βˆ’ π‘‡βˆž) doesn’t matter and so we get: 𝛼 𝛿 βˆ’ 2β„Ž π‘ŸπœŒπΆ 𝛿 = 𝑑𝛿 𝑑𝑑 We go ahead and solve for 𝛿 provided 𝛿 = 0π‘Žπ‘‘ 𝑑 = 0 and get the expression 𝛿 = √ 𝐾𝐴 β„Žπ‘ƒ (1 βˆ’ 𝑒 βˆ’2β„Žπ‘ƒ 𝐴𝜌𝐢 𝑑 ) So, the final solution for the finite metal rod with convective flux at the end of the metal rod is: 𝑻 βˆ’ π‘»βˆž 𝑻𝑺 βˆ’ π‘»βˆž = ( 𝐜𝐨𝐬𝐑 [ (𝑳 βˆ’ 𝒙) 𝜹 ] + ( π’‰π‘³πœΉ π’Œ ) 𝐬𝐒𝐧𝐑 [( 𝑳 βˆ’ 𝒙 𝜹 )] 𝐜𝐨𝐬𝐑 𝑳 𝜹 + ( π’‰π‘³πœΉ π’Œ ) π’”π’Šπ’π’‰ 𝑳 𝜹 )
  • 27. CASE 3: ZERO FLUX AT THE END OF THE METAL ROD? The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝒅𝑻 𝒅𝒙 = 𝟎 𝒂𝒕 𝒙 = 𝒍 𝑻 = π‘»βˆž 𝒂𝒕 𝒕 = 𝟎 The equation that satisfies the conditions is [3] 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh[π‘š(𝐿 βˆ’ π‘₯)] cosh π‘šπΏ In terms of 𝜹 we get 𝑇 βˆ’ π‘‡βˆž 𝑇𝑠 βˆ’ π‘‡βˆž = cosh[ (𝐿 βˆ’ π‘₯) 𝛿 ] cosh 𝐿 𝛿 For the initial condition to be satisfied, we assume that at 𝑑 = 0, 𝛿 = 0 The governing equation is 𝛼 πœ•2 𝑇 πœ•π‘₯2 βˆ’ β„Žπ‘ƒ 𝐴𝜌𝐢 (𝑇 βˆ’ π‘‡βˆž) = πœ•π‘‡ πœ•π‘‘ Let us change this equation into an integral equation as below: 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 βˆ’ β„Žπ‘ƒ 𝐴𝜌𝐢 ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 … … . . 𝑏) 𝛼 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 βˆ’ 2β„Ž π‘ŸπœŒπΆ ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 ∫ ( πœ•2 𝑇 πœ•π‘₯2 ) 𝑑π‘₯ 𝑙 0 = [ πœ•π‘‡ πœ•π‘₯ ] 𝑙 0 = (𝑇𝑠 βˆ’ π‘‡βˆž) tanh( 𝐿 𝛿 ) 𝛿 ∫ (𝑇 βˆ’ π‘‡βˆž)𝑑π‘₯ 𝑙 0 = (𝑇𝑠 βˆ’ π‘‡βˆž)𝛿tanh( 𝐿 𝛿 )
  • 28. 𝑇 = cosh [ (𝐿 βˆ’ π‘₯) 𝛿 ] cosh 𝐿 𝛿 (𝑇𝑠 βˆ’ π‘‡βˆž) + π‘‡βˆž πœ• πœ•π‘‘ ∫ (𝑇)𝑑π‘₯ 𝑙 0 = πœ• πœ•π‘‘ [𝛿(𝑇𝑠 βˆ’ π‘‡βˆž) tanh ( 𝐿 𝛿 )] + πœ•(π‘™π‘‡βˆž) πœ•π‘‘ πœ•(π‘™π‘‡βˆž) πœ•π‘‘ = 0 Upon substitution of all the above in the heat equation, we get: 𝛼(𝑇𝑠 βˆ’ π‘‡βˆž) tanh ( 𝐿 𝛿 ) 𝛿 βˆ’ 2β„Ž π‘ŸπœŒπΆ (𝑇𝑠 βˆ’ π‘‡βˆž)𝛿 tanh ( 𝐿 𝛿 ) = πœ• πœ•π‘‘ [𝛿(𝑇𝑠 βˆ’ π‘‡βˆž) tanh ( 𝐿 𝛿 )] We notice that the term (𝑇𝑠 βˆ’ π‘‡βˆž)tanh( 𝐿 𝛿 ) is common and can be eliminated and what this signifies is that the nature of (𝑇𝑠 βˆ’ π‘‡βˆž) doesn’t matter and so we get: 𝛼 𝛿 βˆ’ 2β„Ž π‘ŸπœŒπΆ 𝛿 = 𝑑𝛿 𝑑𝑑 We go ahead and solve for 𝛿 provided 𝛿 = 0π‘Žπ‘‘ 𝑑 = 0 and get the expression 𝛿 = √ 𝐾𝐴 β„Žπ‘ƒ (1 βˆ’ 𝑒 βˆ’2β„Žπ‘ƒ 𝐴𝜌𝐢 𝑑 ) So, the final solution for the finite metal rod with zero flux at the end of the metal rod is: 𝑻 βˆ’ π‘»βˆž 𝑻𝒔 βˆ’ π‘»βˆž = ( 𝐜𝐨𝐬𝐑[ (𝑳 βˆ’ 𝒙) 𝜹 ] 𝐜𝐨𝐬𝐑 𝑳 𝜹 ) It can be shown that the initial condition is satisfied.
  • 29. RADIAL HEAT FLOW: INSULATED HOLLOW CYLINDER 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
  • 30. ∫ [ 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‘₯
  • 31. 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
  • 32. ∫ (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.
  • 33. 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.
  • 34. 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 bookInvalid source specified. 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
  • 35. 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
  • 36. 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 )
  • 37. ∫ (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 )
  • 38. 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 )
  • 39. 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‘₯)
  • 40. 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.
  • 41. 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
  • 42. 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:
  • 43. 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 ]
  • 44. 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
  • 45. ∫ (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
  • 46. 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 )])]
  • 47. 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’•
  • 48. 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.
  • 49. 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
  • 50. 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])]
  • 51. 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)
  • 52. 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:
  • 53. 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:
  • 54. 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 )
  • 55. HEAT CONDUCTION WITH LATERAL CONVECTION IN CYLINDRICAL CO-ORDINATES 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
  • 56. 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
  • 57. 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‘ 
  • 58. 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
  • 59. 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
  • 60. 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
  • 61. 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
  • 62. 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 )
  • 63. ∫ (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 ]
  • 64. 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
  • 65. 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:
  • 66. 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 )
  • 67. 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 ) This same analysis can be extended to spherical co-ordinates.
  • 68. 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 EXTENDED SURFACES (FINS)," in Fundamentals Of Heat and Mass Transfer, New Delhi, NEW AGE INTERNATIONAL PUBLISHERS, 2006, p. 132. [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. [3] 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, Oregon, John Wiley & Sons, Inc., 2008, p. 237. .