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Integral method of the Analytic solutions to the heat equation With Experimental Results.pdf
- 1. THE ANALYTICAL SOLUTION TO THE HEAT EQUATION USING AN INTEGRAL METHOD By Wasswa Derrick
- 3. TABLE OF CONTENTS SEMI INFINITE WALL ANALYTICAL SOLUTION TO THE HEAT EQUATION................3 ALTERNATIVE SOLUTION TO THE SEMI-INFINITE WALL PROBLEM ...........................5 HOW DO WE EXPLAIN THE EXISTENCE OF THE FOURIER LAW IN STEADY STATE FOR SEMI-INFINITE ROD?................................................................................................8 HOW DO WE DEAL WITH CONVECTION AT THE SURFACE AREA OF THE SEMI- INFINITE METAL ROD ......................................................................................................................11 EQUAL FIXED TEMPERATURES AT THE END OF AN INSULATED METAL ROD....15 UNEQUAL FIXED TEMPERATURES AT THE END OF AN INSULATED METAL ROD. ....................................................................................................................................................................18 HOW DO WE DEAL WITH OTHER TYPES OF BOUNDARY CONDITIONS?..................21 HOW DO WE DEAL WITH NATURAL CONVECTION AT THE SURFACE AREA OF A SEMI-INFINITE METAL ROD FOR FIXED END TEMPERATURE.....................................23 WHAT HAPPENS WHEN THE INITIAL TEMPERATURE IS A FUNCTION OF X? .......39 HOW DO WE DEAL WITH CYLINDRICAL CO-ORDINATES FOR AN INFINITE RADIUS CYLINDER? ..........................................................................................................................41
- 4. SEMI INFINITE WALL ANALYTICAL SOLUTION TO THE HEAT EQUATION. The differential equation to be solved is 𝜕𝑇 𝜕𝑡 = 𝛼 𝜕2 𝑇 𝜕𝑥2 Where the initial and boundary conditions are 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒕 𝑻 = 𝑻∞ 𝒂𝒕 𝒙 = ∞ 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒕 𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒙 We postulate: 𝑌 = 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ And 𝜂 = 𝑥 2√𝛼𝑡 We get 𝑑2 𝑌 𝑑𝜂2 + 2𝜂 𝑑𝑌 𝑑𝜂 = 0 (1) With the transformed boundary and initial conditions 𝑌 → 0 𝑎𝑠 𝜂 → ∞ And 𝑌 = 1 𝑎𝑡 𝜂 = 0 The first condition is the same as the initial condition 𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0 and the boundary condition 𝑇 → 𝑇∞ 𝑎𝑠 𝑥 → ∞ Equation 1 may be integrated once to get 𝑙𝑛 𝑑𝑌 𝑑𝜂 = 𝑐1 − 𝜂2 𝑑𝑌 𝑑𝜂 = 𝑐2𝑒−𝜂2
- 5. And integrated once more to get 𝑌 = 𝑐3 + 𝑐2 ∫ 𝑒−𝜂2 𝑑 𝜂 Applying the boundary conditions to the equation, we get 𝑌 = 1 − erf ( 𝑥 2√𝛼𝑡 ) 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = 𝟏 − 𝐞𝐫𝐟 ( 𝒙 𝟐√𝜶𝒕 ) Or 𝑻𝒔 − 𝑻 𝑻𝒔 − 𝑻∞ = 𝐞𝐫𝐟 ( 𝒙 𝟐√𝜶𝒕 )
- 6. ALTERNATIVE SOLUTION TO THE SEMI-INFINITE WALL PROBLEM The problem of the semi-infinite wall could also be solved as below: Given the boundary and initial conditions 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒕 𝑻 = 𝑻∞ 𝒂𝒕 𝒙 = ∞ 𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 Since the rod is infinite in one direction, we say 𝑙 = ∞ And the governing equation 𝜕𝑇 𝜕𝑡 = 𝛼 𝜕2 𝑇 𝜕𝑥2 We assume an exponential temperature profile that satisfies the boundary conditions: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 −𝑥 𝛿 We can satisfy the initial condition if we assume that 𝛿 will have a solution as 𝛿 = 𝑐𝑡𝑛 Where c and n are constants so that at 𝑡 = 0, 𝛿 = 0 and the initial condition is satisfied as shown below. 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 −𝑥 0 = 𝑒−∞ = 0 𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0 We then transform the heat governing equation into an integral equation as: 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = ∫ 𝜕𝑇 𝜕𝑡 𝑑𝑥 𝑙 0 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ 𝑇𝑑𝑥 𝑙 0 𝑇 = (𝑇𝑠 − 𝑇∞)𝑒 −𝑥 𝛿 + 𝑇∞ 𝜕 𝜕𝑡 ∫ 𝑇𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ ((𝑇𝑠 − 𝑇∞)𝑒 −𝑥 𝛿 + 𝑇∞)𝑑𝑥 = 𝜕 𝜕𝑡 [𝛿(𝑇𝑠 − 𝑇∞)(1 − 𝑒 −𝑙 𝛿 )] − 𝜕(𝑙𝑇∞) 𝜕𝑡 𝑙 0 Since 𝑙 and 𝑇∞ are constants independent of time
- 7. 𝜕(𝑙𝑇∞) 𝜕𝑡 = 0 So 𝜕 𝜕𝑡 ∫ 𝑇𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 [𝛿(𝑇𝑠 − 𝑇∞)(1 − 𝑒 −𝑙 𝛿 )] Since 𝑙 = ∞, we substitute for 𝑙 and get 𝜕 𝜕𝑡 ∫ 𝑇𝑑𝑥 𝑙 0 = 𝑑𝛿 𝑑𝑡 (𝑇𝑠 − 𝑇∞) We go ahead and find 𝜕2 𝑇 𝜕𝑥2 = (𝑇𝑠 − 𝑇∞) 𝛿2 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = − (𝑇𝑠 − 𝑇∞) 𝛿 (𝑒 −𝑙 𝛿 − 1) = (𝑇𝑠 − 𝑇∞) 𝛿 (1 − 𝑒 −𝑙 𝛿 ) Since 𝑙 = ∞, we substitute for 𝑙 and get ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = − (𝑇𝑠 − 𝑇∞) 𝛿 (𝑒 −𝑙 𝛿 − 1) = (𝑇𝑠 − 𝑇∞) 𝛿 Substituting into the integral equation, we get 𝛼 (𝑇𝑠 − 𝑇∞) 𝛿 = 𝑑𝛿 𝑑𝑡 (𝑇𝑠 − 𝑇∞) The boundary conditions are 𝛿 = 0 𝑎𝑡 𝑡 = 0 We find 𝛿 = √2𝛼𝑡 We substitute in the temperature profile and get 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = 𝒆 −𝒙 √𝟐𝜶𝒕 You notice that the initial condition is satisfied by the temperature profile above i.e.,
- 8. At 𝑡 = 0 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = 𝒆 −𝒙 √𝟐𝜶𝒕 Becomes 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = 𝒆 −𝒙 𝟎 = 𝒆−∞ = 𝟎 Hence 𝑻 = 𝑻∞ throughout the rod at 𝑡 = 0 Observation. The two equations 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = 𝒆 −𝒙 √𝟐𝜶𝒕 And 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = 𝟏 − 𝐞𝐫𝐟 ( 𝒙 𝟐√𝜶𝒕 ) Should give the same answer. Indeed, they give answers that are the same with a small error since the error function is got from tables after rounding off yet in the exponential temperature profile there is no rounding off.
- 9. HOW DO WE EXPLAIN THE EXISTENCE OF THE FOURIER LAW IN STEADY STATE FOR SEMI-INFINITE ROD? The Fourier law states: 𝑄 = −𝑘𝐴 𝜕𝑇 𝜕𝑥 Under steady state. It can be stated as: 𝜕𝑇 𝜕𝑥 = − 𝑄 𝑘𝐴 Under steady state. To satisfy the Fourier law under steady state, we postulate the temperature profile to be: 𝑻 − 𝑻∞ = 𝑸 𝒌𝑨 𝜹𝒆 −𝒙 𝜹 𝛿 is a function of time 𝑡 and not distance 𝑥 We believe that after solving for 𝛿, 𝛿 will be directly proportional to time t so that 𝛿 = 𝑘𝑡𝑛 sothat at 𝑡 = ∞ , 𝛿 = ∞ And taking the first derivative of temperature with distance x at 𝑡 = ∞ , we get 𝜕𝑇 𝜕𝑥 |𝑡=∞ = − 𝑄 𝑘𝐴 𝑒 −𝑥 𝛿 = − 𝑄 𝑘𝐴 𝑒 −𝑥 ∞ = − 𝑄 𝑘𝐴 𝑒0 𝝏𝑻 𝝏𝒙 = − 𝑸 𝒌𝑨 Hence the Fourier law is satisfied. Now let us go ahead and solve for 𝛿. Recall PDE 𝜕𝑇 𝜕𝑡 = 𝛼 𝜕2 𝑇 𝜕𝑥2 The initial condition is 𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0
- 10. The boundary conditions are 𝑇 = 𝑇∞ 𝑎𝑡 𝑥 = ∞ 𝜕𝑇 𝜕𝑥 |𝑥=0 = − 𝑄 𝑘𝐴 The temperature profile that satisfies the conditions above is 𝑇 − 𝑇∞ = 𝑄 𝑘𝐴 𝛿𝑒 −𝑥 𝛿 We transform the PDE into an integral equation by integrating over the whole length of the metal rod. 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 And using the temperature profile, we get ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = [ 𝜕𝑇 𝜕𝑥 ] 𝑙 0 = 𝑄 𝑘𝐴 (1 − 𝑒 −𝑙 𝛿 ) Since 𝑙 = ∞, we substitute for 𝑙 and get ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = [ 𝜕𝑇 𝜕𝑥 ] 𝑙 0 = 𝑄 𝑘𝐴 𝑇 = 𝑄 𝑘𝐴 𝛿𝑒 −𝑥 𝛿 + 𝑇∞ 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ ( 𝑄 𝑘𝐴 𝛿𝑒 −𝑥 𝛿 + 𝑇∞)𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 [ 𝑄 𝑘𝐴 𝛿2 (1 − 𝑒 −𝑙 𝛿 )] + 𝜕(𝑙𝑇∞) 𝜕𝑡 𝜕(𝑙𝑇∞) 𝜕𝑡 = 0 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 [ 𝑄 𝑘𝐴 𝛿2 (1 − 𝑒 −𝑙 𝛿 )] Since 𝑙 = ∞, we substitute for 𝑙 and get 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 [ 𝑄 𝑘𝐴 𝛿2 ] We then substitute into the integral equation
- 11. 𝛼 𝑄 𝑘𝐴 = 2𝛿 𝑑𝛿 𝑑𝑡 ( 𝑄 𝑘𝐴 ) The boundary conditions are 𝛿 = 0 𝑎𝑡 𝑡 = 0 𝛿 = √𝛼𝑡 Substituting into the temperature profile, we get 𝑇 − 𝑇∞ = 𝑄 𝑘𝐴 𝛿𝑒 −𝑥 𝛿 𝑻 − 𝑻∞ = 𝑸 𝒌𝑨 × √𝜶𝒕 × 𝒆 −𝒙 √𝜶𝒕 You notice that the initial condition is satisfied 𝝏𝑻 𝝏𝒙 |𝒕=∞ = − 𝑸 𝒌𝑨 Hence the Fourier law So, our assumption of 𝛿 = 𝑘𝑡𝑛 is satisfied
- 12. HOW DO WE DEAL WITH CONVECTION AT THE SURFACE AREA OF THE SEMI-INFINITE METAL ROD Recall that the temperature profile that satisfies the Fourier law was 𝑇 − 𝑇∞ = 𝑄 𝑘𝐴 𝛿𝑒 −𝑥 𝛿 Recall PDE 𝛼 𝜕2 𝑇 𝜕𝑥2 − ℎ𝑃 𝐴𝜌𝐶 (𝑇 − 𝑇∞) = 𝜕𝑇 𝜕𝑡 The initial condition is 𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0 The boundary conditions are 𝑇 = 𝑇∞ 𝑎𝑡 𝑥 = ∞ 𝜕𝑇 𝜕𝑥 |𝑥=0 = − 𝑄 𝑘𝐴 Remember that for a semi-infinite rod 𝑙 = ∞ We transform the PDE into an integral equation 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 − ℎ𝑃 𝐴𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 Where: We are dealing with a cylindrical metal rod. 𝑃 = 2𝜋𝑟 𝑎𝑛𝑑 𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑚𝑒𝑡𝑎𝑙 𝑟𝑜𝑑 And using the temperature profile, we get ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = 𝑄 𝑘𝐴 (1 − 𝑒 −𝑙 𝛿 ) Substitute for 𝑙 = ∞ and get ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = 𝑄 𝑘𝐴
- 13. ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝑄 𝑘𝐴 𝛿2 (1 − 𝑒 −𝑙 𝛿 ) + 𝑇∞𝑙 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ( 𝑄 𝑘𝐴 𝛿2 (1 − 𝑒 −𝑙 𝛿 )) + 𝜕(𝑇∞𝑙) 𝜕𝑡 𝜕(𝑇∞𝑙) 𝜕𝑡 = 0 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ( 𝑄 𝑘𝐴 𝛿2 (1 − 𝑒 −𝑙 𝛿 )) Substitute for 𝑙 = ∞ and get 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ( 𝑄 𝑘𝐴 𝛿2 ) = 2𝛿( 𝑄 𝑘𝐴 ) 𝑑𝛿 𝑑𝑡 ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = 𝑄 𝑘𝐴 𝛿2 (1 − 𝑒 −𝑙 𝛿 ) Substitute for 𝑙 = ∞ and get ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = 𝑄 𝑘𝐴 𝛿2 Substituting in the integral equation above, we get 𝛼 − ℎ𝑃 𝐴𝜌𝐶 𝛿2 = 2𝛿 𝑑𝛿 𝑑𝑡 The boundary condition is 𝛿 = 0 𝑎𝑡 𝑡 = 0 We solve and get 𝛿 = √ 𝐴𝜌𝐶𝛼 ℎ𝑃 (1 − 𝑒 −ℎ𝑃𝑡 𝐴𝜌𝐶 ) Substituting in the temperature profile, we get 𝑇 − 𝑇∞ = 𝑄 𝑘𝐴 𝛿𝑒 −𝑥 𝛿 𝑻 − 𝑻∞ = 𝑸 𝒌𝑨 × √ 𝑨𝝆𝑪𝜶 𝒉𝑷 (𝟏 − 𝒆 −𝒉𝑷𝒕 𝑨𝝆𝑪 ) × 𝒆 −𝒙 √𝑨𝝆𝑪 𝒉𝑷 (𝟏−𝒆 −𝒉𝑷𝒕 𝑨𝝆𝑪 )
- 14. We notice that the initial condition and boundary conditions are satisfied. For small time the term ℎ𝑃𝑡 𝐴𝜌𝐶 ≪ 1 And using binomial approximation of the exponential, we get 𝑒 −ℎ𝑃𝑡 𝐴𝜌𝐶 = 1 − ℎ𝑃𝑡 𝐴𝜌𝐶 Then (1 − 𝑒 −ℎ𝑃𝑡 𝐴𝜌𝐶 ) = ℎ𝑃𝑡 𝐴𝜌𝐶 Upon substitution in the temperature profile, we get 𝑇 − 𝑇∞ = 𝑄 𝑘𝐴 × √𝛼𝑡 × 𝑒 −𝑥 √𝛼𝑡 Upon rearranging, we get 𝑥 √𝛼𝑡 = ln ( 𝑄 𝑘𝐴 √𝛼𝑡) − ln (𝑇 − 𝑇∞) 𝑥 √𝑡 = √𝛼ln(√𝑡) + √𝛼 [ln ( 𝑄 𝑘𝐴 √𝛼) − ln(𝑇 − 𝑇∞)] What we observe is 𝒙 √𝒕 = √𝜶𝐥𝐧(√𝒕) + √𝜶 [𝐥𝐧 ( 𝑸 𝒌𝑨√𝜶 (𝑻 − 𝑻∞) )] That is what we observe for short times. When the times become big, we observe 𝑻 − 𝑻∞ = 𝑸 𝒌𝑨 × √ 𝑨𝝆𝑪𝜶 𝒉𝑷 (𝟏 − 𝒆 −𝒉𝑷𝒕 𝑨𝝆𝑪 ) × 𝒆 −𝒙 √𝑨𝝆𝑪 𝒉𝑷 (𝟏−𝒆 −𝒉𝑷𝒕 𝑨𝝆𝑪 )
- 15. And in steady state (𝑡 = ∞), we observe 𝑻 − 𝑻∞ = 𝑸 𝒌𝑨 × √ 𝑨𝝆𝑪𝜶 𝒉𝑷 × 𝒆 −𝒙 √𝑨𝝆𝑪𝜶 𝒉𝑷 𝛼 = 𝑘 𝜌𝐶 We finally get 𝑻 − 𝑻∞ = 𝑸 𝒌𝑨 × √ 𝒌𝑨 𝒉𝑷 × 𝒆 −𝒙 √𝒌𝑨 𝒉𝑷 the heat flow in steady state is given by: 𝜕𝑇 𝜕𝑥 = − 𝑄 𝑘𝐴 𝑒 −𝑥 √𝑘𝐴 ℎ𝑃 −𝒌𝑨 𝝏𝑻 𝝏𝒙 = 𝑸𝒆 −𝒙 √𝒌𝑨 𝒉𝑷
- 16. EQUAL FIXED TEMPERATURES AT THE END OF AN INSULATED METAL ROD. PDE 𝜕𝑇 𝜕𝑡 = 𝛼 𝜕2 𝑇 𝜕𝑥2 BCs 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝟎 < 𝒕 < ∞ 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝒍 𝟎 < 𝒕 < ∞ IC 𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 𝟎 ≤ 𝒙 ≤ 𝒍 we know a Fourier series solution exists given by 𝑻 − 𝑻𝒔 𝑻∞ − 𝑻𝒔 = 𝟒 𝝅 ∑ 𝟏 𝒏 ∞ 𝒏=𝟏 𝒔𝒊𝒏 ( 𝒏𝝅𝒙 𝒍 ) 𝒆 −( 𝒏𝝅 𝟐 ) 𝜶𝒕 ( 𝒍 𝟐 )𝟐 𝒏 = 𝟏, 𝟑, 𝟓, … You notice that this solution is not entirely deterministic since it involves summing terms up to infinity. There is an alternative solution as shown below: 𝜕𝑇 𝜕𝑡 = 𝛼 𝜕2 𝑇 𝜕𝑥2 BCs 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝟎 < 𝒕 < ∞ 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝒍 𝟎 < 𝒕 < ∞ IC 𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 𝟎 ≤ 𝒙 ≤ 𝒍 We assume an exponential temperature profile that satisfies the boundary conditions: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 −𝑥 𝛿 (1− 𝑥 𝑙 )
- 17. You notice that the temperature profile above satisfies the boundary conditions. We can satisfy the initial condition if we assume that 𝛿 will assume a solution as 𝛿 = 𝑐𝑡𝑛 Where c and n are constants so that at 𝑡 = 0, 𝛿 = 0 and the initial condition is satisfied as shown below. 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 −𝑥 0 = 𝑒−∞ = 0 𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0 We transform the PDE into an integral equation 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = [ 𝜕𝑇 𝜕𝑥 ] 𝑙 0 = (𝑇𝑠 − 𝑇∞) [ (−𝑙 + 2𝑥) 𝛿𝑙 × 𝑒 −𝑥 𝛿 (1− 𝑥 𝑙 ) ] 𝑙 0 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = (𝑇𝑠 − 𝑇∞) [ (−𝑙 + 2𝑥) 𝛿𝑙 × 𝑒 −𝑥 𝛿 (1− 𝑥 𝑙 ) ] 𝑙 0 = 2(𝑇𝑠 − 𝑇∞) 𝛿 𝑇 = (𝑇𝑠 − 𝑇∞)𝑒 −𝑥 𝛿 (1− 𝑥 𝑙 ) + 𝑇∞ ∫ (𝑇)𝑑𝑥 𝑙 0 = (𝑇𝑠 − 𝑇∞) [ 𝛿𝑙 (−𝑙 + 2𝑥) × 𝑒 −𝑥 𝛿 (1− 𝑥 𝑙 ) ] 𝑙 0 + 𝑇∞𝑙 = 2(𝑇𝑠 − 𝑇∞)𝛿 + 𝑇∞𝑙 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 2(𝑇𝑠 − 𝑇∞) 𝑑𝛿 𝑑𝑡 + 𝑑(𝑇∞𝑙) 𝑑𝑡 𝑑(𝑇∞𝑙) 𝑑𝑡 = 0 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 2(𝑇𝑠 − 𝑇∞) 𝑑𝛿 𝑑𝑡 Substituting in the integral equation above, we get: 𝛼 ( 2 𝛿 ) = 2 𝑑𝛿 𝑑𝑡 𝛿 = √2𝛼𝑡 Substituting back 𝛿 into the temperature profile, we get
- 18. 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = 𝒆 −𝒙 √𝟐𝜶𝒕 (𝟏− 𝒙 𝒍 ) Or 𝑻 − 𝑻𝒔 𝑻∞ − 𝑻𝒔 = 𝟏 − 𝒆 −𝒙 √𝟐𝜶𝒕 (𝟏− 𝒙 𝒍 ) You notice that the initial condition is satisfied. You notice that when 𝑙 = ∞ , we reduce to the temperature profile we derived before 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = 𝒆 −𝒙 √𝟐𝜶𝒕 you notice that in the temperature profile developed, we get an exact solution to the problem not an approximate as the Fourier series.
- 19. UNEQUAL FIXED TEMPERATURES AT THE END OF AN INSULATED METAL ROD. 𝜕𝑇 𝜕𝑡 = 𝛼 𝜕2 𝑇 𝜕𝑥2 The boundary conditions are: 𝑇 = 𝑇𝑠 𝑎𝑡 𝑥 = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 𝑇 = 𝑇1 𝑎𝑡 𝑥 = 𝑙 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 The initial condition is 𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0 0 ≤ 𝑥 ≤ 𝑙 The temperature profile that satisfies the boundary conditions is: 𝑻 − 𝑻∞ [ 𝒙 𝒍 (𝑻𝟏 − 𝑻∞) + (𝑻𝒔 − 𝑻∞) (𝟏 − 𝒙 𝒍)] = 𝒆− 𝒙 𝜹 (𝟏− 𝒙 𝒍 ) For now, we shall have a solution where 𝛿 is proportional to time t so that at 𝑡 = 0, 𝛿 = 0 and the initial condition will be satisfied. We then transform the heat governing equation into an integral equation as: 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = ∫ 𝜕𝑇 𝜕𝑡 𝑑𝑥 𝑙 0 Where: 𝑙 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑒𝑡𝑎𝑙 𝑟𝑜𝑑 So, the integral equation becomes: 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 We go ahead and find ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = [ 𝜕𝑇 𝜕𝑥 ] 𝑙 0 𝜕𝑇 𝜕𝑥 = (𝑇𝑠 − 𝑇∞)𝑒− 𝑥 𝛿 (1− 𝑥 𝑙 ) [ 1 𝑙 (𝑇1 − 𝑇∞) (𝑇𝑠 − 𝑇∞) − 1 𝑙 ] + (𝑇𝑠 − 𝑇∞)𝑒− 𝑥 𝛿 (1− 𝑥 𝑙 ) [ 𝑥 𝑙 (𝑇1 − 𝑇∞) (𝑇𝑠 − 𝑇∞) − (1 − 𝑥 𝑙 )] ( −𝑙 + 2𝑥 𝛿𝑙 )
- 20. ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = [ 𝜕𝑇 𝜕𝑥 ] 𝑙 0 = (𝑇𝑠 + 𝑇1 − 2𝑇∞) 𝛿 𝑇 = [ 𝑥 𝑙 (𝑇1 − 𝑇∞) + (𝑇𝑠 − 𝑇∞) (1 − 𝑥 𝑙 )]𝑒 − 𝑥 𝛿 (1− 𝑥 𝑙 ) + 𝑇∞ ∫ (𝑇)𝑑𝑥 𝑙 0 = (𝑇1 − 𝑇∞) 𝑙 ∫ 𝑥 𝑙 0 𝑒 − 𝑥 𝛿 (1− 𝑥 𝑙 ) 𝑑𝑥 + (𝑇𝑠 − 𝑇∞) ∫ 𝑒 − 𝑥 𝛿 (1− 𝑥 𝑙 ) 𝑑𝑥 𝑙 0 − (𝑇𝑠 − 𝑇∞) 𝑙 ∫ 𝑥 𝑙 0 𝑒 − 𝑥 𝛿 (1− 𝑥 𝑙 ) + 𝑇∞𝑙 ∫ 𝑥 𝑙 0 𝑒 − 𝑥 𝛿 (1− 𝑥 𝑙 ) 𝑑𝑥 = 𝑙𝛿 ∫ 𝑒 − 𝑥 𝛿 (1− 𝑥 𝑙 ) 𝑑𝑥 𝑙 0 = 2𝛿 So ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝛿(𝑇𝑠 + 𝑇1 − 2𝑇∞) + 𝑇∞𝑙 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝑑𝛿 𝑑𝑡 (𝑇𝑠 + 𝑇1 − 2𝑇∞) + 𝑑(𝑇∞𝑙) 𝑑𝑡 𝑑(𝑇∞𝑙) 𝑑𝑡 = 0 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝑑𝛿 𝑑𝑡 (𝑇𝑠 + 𝑇1 − 2𝑇∞) Substituting ∫ ( 𝜕2𝑇 𝜕𝑥2) 𝑑𝑥 𝑙 0 and 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 in the integral equation, we get 𝛼 𝛿 = 𝑑𝛿 𝑑𝑡 Where: 𝛿 = 0 𝑎𝑡 𝑡 = 0 𝛿 = √2𝛼𝑡 You notice that the initial condition is satisfied since after finding the solution as done before to the PDE, 𝛿 = √2𝛼𝑡 Therefore substituting 𝛿 in the temperature profile, we get:
- 21. 𝑻 − 𝑻∞ [ 𝒙 𝒍 (𝑻𝟏 − 𝑻∞) + (𝑻𝒔 − 𝑻∞) (𝟏 − 𝒙 𝒍)] = 𝒆 − 𝒙 √𝟐𝜶𝒕 (𝟏− 𝒙 𝒍 ) you notice that at steady state (𝑡 = ∞) 𝒆 − 𝒙 √𝟐𝜶𝒕 (𝟏− 𝒙 𝒍 ) = 𝒆− 𝒙 ∞ (𝟏− 𝒙 𝒍 ) = 𝒆−𝟎 = 𝟏 The temperature profile becomes: 𝑻 − 𝑻∞ = [ 𝒙 𝒍 (𝑻𝟏 − 𝑻∞) + (𝑻𝒔 − 𝑻∞) (𝟏 − 𝒙 𝒍 )]
- 22. HOW DO WE DEAL WITH OTHER TYPES OF BOUNDARY CONDITIONS? Consider the following types of boundary conditions and initial condition: A) 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝒅𝑻 𝒅𝒙 = 𝟎 𝒂𝒕 𝒙 = 𝒍 𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 B) 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 −𝒌 𝒅𝑻 𝒅𝒙 = 𝒉(𝑻 − 𝑻∞) 𝒂𝒕 𝒙 = 𝒍 𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 Let us go about solving for the above boundary conditions but let us deal with set A boundary conditions and then we can deal with set B later. We start with a temperature profile below: 𝑇 − 𝑇∞ = (𝑇𝑠 − 𝑇∞)𝑒− 𝑥 𝛿 (1− 𝑥 𝑙 ) [ 𝑥 𝑙 (𝑇1 − 𝑇∞) (𝑇𝑠 − 𝑇∞) + (1 − 𝑥 𝑙 )] we take the derivative 𝒅𝑻 𝒅𝒙 𝒂𝒕 𝒙 = 𝒍 and equate it to 0 and get: 𝑑𝑇 𝑑𝑥 |𝑥=𝑙 = ( (𝑇1 − 𝑇∞) 𝑙 − (𝑇𝑠 − 𝑇∞) 𝑙 + (𝑇1 − 𝑇∞) 𝛿 ) 𝑑𝑇 𝑑𝑥 |𝑥=𝑙 = ( (𝑇1 − 𝑇∞) 𝑙 − (𝑇𝑠 − 𝑇∞) 𝑙 + (𝑇1 − 𝑇∞) 𝛿 ) = 0 We finally get (𝑇1 − 𝑇∞) = (𝑇𝑠 − 𝑇∞)( 𝛿 𝑙 + 𝛿 ) We substitute 𝑇1 − 𝑇∞ into the temperature profile and get (𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆− 𝒙 𝜹 (𝟏 − 𝒙 𝒍 ) [ 𝒙 𝒍 ( 𝜹 𝜹 + 𝒍 ) + (𝟏 − 𝒙 𝒍 )]
- 23. So, the temperature profile above satisfies the set A) boundary and initial conditions and we can go ahead and solve the governing equation using the temperature profile above. For set B) boundary conditions, we again start with the temperature profile below: 𝑇 − 𝑇∞ = (𝑇𝑠 − 𝑇∞)𝑒− 𝑥 𝛿 (1 − 𝑥 𝑙 ) [ 𝑥 𝑙 (𝑇1 − 𝑇∞) (𝑇𝑠 − 𝑇∞) + (1 − 𝑥 𝑙 )] we take the derivative 𝒅𝑻 𝒅𝒙 𝒂𝒕 𝒙 = 𝒍 and equate it to: 𝑑𝑇 𝑑𝑥 |𝑥=𝑙 = ( (𝑇1 − 𝑇∞) 𝑙 − (𝑇𝑠 − 𝑇∞) 𝑙 + (𝑇1 − 𝑇∞) 𝛿 ) 𝒅𝑻 𝒅𝒙 |𝒙=𝒍 = − 𝒉 𝒌 (𝑻𝟏 − 𝑻∞) We then find the required temperature profile which we can use to solve the governing equation.
- 24. HOW DO WE DEAL WITH NATURAL CONVECTION AT THE SURFACE AREA OF A SEMI-INFINITE METAL ROD FOR FIXED END TEMPERATURE 𝛼 𝜕2 𝑇 𝜕𝑥2 − ℎ𝑃 𝐴𝜌𝐶 (𝑇 − 𝑇∞) = 𝜕𝑇 𝜕𝑡 We shall use the integral approach. The boundary and initial conditions are 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒕 𝑻 = 𝑻∞ 𝒂𝒕 𝒙 = ∞ 𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 Where: 𝑻∞ = 𝒓𝒐𝒐𝒎 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 First, we assume a temperature profile that satisfies the boundary conditions as: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 −𝑥 𝛿 where 𝛿 is to be determined and is a function of time t. The governing equation is 𝛼 𝜕2 𝑇 𝜕𝑥2 − ℎ𝑃 𝐴𝜌𝐶 (𝑇 − 𝑇∞) = 𝜕𝑇 𝜕𝑡 Let us change this equation into an integral as below: 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 − ℎ𝑃 𝐴𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 … … . . 𝑏) 𝜕2 𝑇 𝜕𝑥2 = (𝑇𝑠 − 𝑇∞) 𝛿2 𝑒 −𝑥 𝛿 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = −(𝑇𝑠 − 𝑇∞) 𝛿 (𝑒 −𝑙 𝛿 − 1) But 𝑙 = ∞, upon substitution, we get ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = (𝑇𝑠 − 𝑇∞) 𝛿
- 25. ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = −𝛿(𝑇𝑠 − 𝑇∞)(𝑒 −𝑙 𝛿 − 1) But 𝑙 = ∞, upon substitution, we get ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = 𝛿(𝑇𝑠 − 𝑇∞) ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝛿(𝑇𝑠 − 𝑇∞)(𝑒 −𝑙 𝛿 − 1) + 𝑇∞𝑙 Substitute 𝑙 = ∞ and get 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝑑𝛿 𝑑𝑡 (𝑇𝑠 − 𝑇∞) + 𝜕 𝜕𝑡 (𝑇∞𝑙) 𝜕 𝜕𝑡 (𝑇∞𝑙) = 0 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝑑𝛿 𝑑𝑡 (𝑇𝑠 − 𝑇∞) Substituting the above expressions in equation b) above, we get 𝛼 − ℎ𝑃 𝐴𝜌𝐶 𝛿2 = 𝛿 𝑑𝛿 𝑑𝑡 We solve the equation above assuming that 𝛿 = 0 𝑎𝑡 𝑡 = 0 And get 𝛿 = √ 𝛼𝐴𝜌𝐶 ℎ𝑃 (1 − 𝑒 −2ℎ𝑃 𝐴𝜌𝐶 𝑡 ) 𝛿 = √ 𝐾𝐴 ℎ𝑃 (1 − 𝑒 −2ℎ𝑃 𝐴𝜌𝐶 𝑡 ) Substituting for 𝛿 in the temperature profile, we get
- 26. 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = 𝒆 −𝒙 √𝑲𝑨 𝒉𝑷 (𝟏−𝒆 −𝟐𝒉𝑷 𝑨𝝆𝑪 𝒕 ) From the equation above, we notice that the initial condition is satisfied i.e., 𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 The equation above predicts the transient state and in steady state (𝑡 = ∞) it reduces to 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = 𝒆 −√( 𝒉𝑷 𝑲𝑨 )𝒙 What are the predictions of the transient state? Let us make 𝑥 the subject of the equation of transient state and get: 𝑥2 = [ln ( 𝑇𝑠 − 𝑇∞ 𝑇 − 𝑇∞ )]2 × 𝐾𝐴 ℎ𝑃 (1 − 𝑒 −2ℎ𝑃 𝐴𝜌𝐶 𝑡 ) When the time duration is small and 2ℎ𝑃 𝐴𝜌𝐶 𝑡 ≪ 1 We use the binomial expansion approximation 𝑒 −2ℎ𝑃 𝐴𝜌𝐶 𝑡 = 1 − 2ℎ𝑃 𝐴𝜌𝐶 𝑡 Substituting in the equation of 𝑥2 as the subject, we get 𝑥2 = 2𝛼[ln ( 𝑇𝑠 − 𝑇∞ 𝑇 − 𝑇∞ )]2 × 𝑡 Where: 𝛼 = 𝐾 𝜌𝐶 We can include an intercept term 𝑡0 which is observed experimentally i.e., 𝑥2 = 2𝛼[ln ( 𝑇𝑠 − 𝑇∞ 𝑇 − 𝑇∞ )]2 × (𝑡 − 𝑡0) Where:
- 27. 𝑡0 = 𝑎 𝑟𝜌𝐶 2ℎ And 𝑎 = 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 2.5926 × 10−6 The above implies that 𝛿 = 0 𝑎𝑡 𝑡 = 𝑡0 That there is a lag in the motion of the heat boundary layer by a time 𝑡0. The equation becomes 𝒙𝟐 = 𝟐𝜶[𝐥𝐧 ( 𝑻𝒔 − 𝑻∞ 𝑻 − 𝑻∞ )]𝟐 × 𝒕 − 𝟐𝜶[𝐥𝐧 ( 𝑻𝒔 − 𝑻∞ 𝑻 − 𝑻∞ )]𝟐 × 𝒂 𝒓𝝆𝑪 𝟐𝒉 Where: 𝛼 = 𝐾 𝜌𝐶 The equation becomes: 𝒙𝟐 = 𝟐𝜶[𝐥𝐧 ( 𝑻𝒔 − 𝑻∞ 𝑻 − 𝑻∞ )]𝟐 × 𝒕 − 𝒂𝑲𝒓 𝒉 [𝐥𝐧 ( 𝑻𝒔 − 𝑻∞ 𝑻 − 𝑻∞ )]𝟐 What that equation says is that when you stick wax particles on a long metal rod (𝑙 = ∞) at distances x from the hot end of the rod and note the time t it takes the wax particles to melt, then a graph of 𝑥2 against 𝑡 is a straight-line graph with an intercept as stated by the equation above when the times are small. The equation is true because that is what is observed experimentally. The intercept above leads to an increase in time of flow of a boundary layer.
- 28. Since the graph above is a straight-line graph, it shows that 𝑇𝑠 IS NOT a function of time. In the equation above we substitute 𝑇 = 37℃ which is the temperature at which wax begins to melt. You notice that by varying the radius of the rod and plotting a graph of 𝑥2 against time t for melting wax at the sides of the rod, from the intercept, the constant ‘a’ above can be measured and from the gradient, 𝑇𝑠 can be measured since 𝑇 = 37℃ in the equation. For an aluminium rod of radius 2mm, 𝑇𝑠 was found to be 57℃. NB • The temperature at which wax begins to melt is 37℃ • From experiment, it was found that 𝑇𝑠 is not the temperature of the flame at the beginning of the metal rod. To get 𝑇𝑠 we plot the graph of 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 −√( ℎ𝑃 𝐾𝐴 )𝑥 𝐥𝐧(𝑻 − 𝑻∞) = 𝒍𝒏(𝑻𝒔 − 𝑻∞) − √( 𝒉𝑷 𝑲𝑨 )𝒙 y = 0.0002x 0 0.01 0.02 0.03 0.04 0.05 0.06 0 50 100 150 200 250 300 350 x^2(m^2) t(seconds) A Graph of x^2 against time t(sec)
- 29. A graph of ln(𝑇 − 𝑇∞) against x gives an intercept 𝑙𝑛(𝑇𝑠 − 𝑇∞) from which 𝑇𝑠 can be measured. From experiment, using an aluminium rod of radius 2mm and using a thermoconductivity value of 𝟐𝟑𝟖 𝑾 𝒎𝑲 ⁄ , The heat transfer coefficient h of aluminium was found to be 𝟑. 𝟎𝟓𝟓𝟐𝟓 𝑾 𝒎𝟐𝑲 ⁄ . From experiment a graph of temperature (℃) against distance 𝑥 looks as below for an aluminium rod of radius 2mm in steady state: The value of 𝑇𝑠 is lower than the value of the flame 𝑇𝑓 because of heat convection at the beginning of the rod. i.e. From the equation 0 50 100 150 200 250 300 350 400 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 Temperature (C) Distance (x) A graph of temperature against distance x
- 30. 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 −√( ℎ𝑃 𝐾𝐴 )𝑥 Plotting a graph of ln(𝑇 − 𝑇∞) against x (excluding temperature at x=0) gives an intercept 𝑙𝑛(𝑇𝑠 − 𝑇∞) from which 𝑇𝑠 can be measured but the value of 𝑇𝑠 got is not the value of the temperature of the flame at 𝑥 = 0(𝑻𝒇). THEORY There is a relationship between 𝑇𝑠 and temperature of the flame(𝑻𝒇) at 𝑥 = 0. First of all, we can postulate an existence of a flux at the beginning of the rod independent of time i.e., 𝑞̇|𝑥=0 = ℎ𝑓(𝑇𝑓 − 𝑇𝑠) Where: ℎ𝑓 = ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 𝑎𝑡 𝑏𝑒𝑔𝑖𝑛𝑛𝑖𝑛𝑔 𝑜𝑓 𝑟𝑜𝑑 But we CAN’T equate this flux to 𝑞̇ = 𝑘 𝜕𝑇 𝜕𝑥 |𝑥=0 Because 𝑘 𝜕𝑇 𝜕𝑥 |𝑥=0 = 𝑘 (𝑇𝑠 − 𝑇∞) 𝛿
- 31. We have already derived 𝛿 and it is a function of time and using it to get 𝑇𝑠 will cause 𝑇𝑠 to be a function of time yet a graph of 𝑥2 against time showed this is not true since 𝑇𝑠 is constant independent of time. We also can’t equate the above power (i.e., flux times area) to this power ℎ𝑓𝜋𝑟2 (𝑇𝑓 − 𝑇𝑠) ≠ ℎ2𝜋𝑟 ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙=∞ 0 Because ℎ2𝜋𝑟 ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙=∞ 0 = ℎ2𝜋𝑟𝛿(𝑇𝑠 − 𝑇∞) The above would also bring back 𝛿 which is a function of time and this would mean 𝑇𝑠 is a function of time which would contradict the observation of the graph 𝑥2 against time to be a straight-line graph. So, the only option we are left with is equating the fluxes at 𝒙 = 𝟎 below: 𝒉𝒇(𝑻𝒇 − 𝑻𝒔) = 𝒉(𝑻𝒔 − 𝑻∞) The expression above will give us a temperature 𝑇𝑠 independent of time since all the above factors don’t depend on time. We can make 𝑇𝑠 the subject of the formula and get: 𝑻𝒔 = 𝒉𝒇𝑻𝒇 + 𝒉𝑻∞ 𝒉 + 𝒉𝒇 From the above, it can be seen that 𝑇𝑠 is independent of time. From experiment, it was found that 𝒉𝒇 = 𝟎. 𝟑𝟏𝟖𝟐 𝑾/(𝒎. 𝑲) independent of radius of the metal rod. The temperature of the flame used was measured to be 379.5℃. Therefore, the equation becomes: 𝑥2 = 2𝛼[ln ( 𝑇𝑠 − 𝑇∞ 𝑇 − 𝑇∞ )]2 × 𝑡 − 𝑎𝐾𝑟 ℎ [ln ( 𝑇𝑠 − 𝑇∞ 𝑇 − 𝑇∞ )]2 𝒙𝟐 = 𝟐𝜶[𝐥𝐧 ( ( 𝒉𝒇𝑻𝒇 + 𝒉𝑻∞ 𝒉 + 𝒉𝒇 ) − 𝑻∞ 𝑻 − 𝑻∞ )]𝟐 × 𝒕 − 𝒂𝑲𝒓 𝒉 [𝐥𝐧 ( ( 𝒉𝒇𝑻𝒇 + 𝒉𝑻∞ 𝒉 + 𝒉𝒇 ) − 𝑻∞ 𝑻 − 𝑻∞ )]𝟐
- 32. How do we measure the heat transfer coefficient? From experiment, using an aluminium rod of radius 2mm and using a thermoconductivity value of 𝟐𝟑𝟖 𝑾 𝒎𝑲 ⁄ , the heat transfer coefficient h of aluminium was found to be 𝟑. 𝟎𝟓𝟓𝟐𝟓 𝑾 𝒎𝟐𝑲 ⁄ . From, 𝐥𝐧(𝑻 − 𝑻∞) = 𝒍𝒏(𝑻𝒔 − 𝑻∞) − √( 𝒉𝑷 𝑲𝑨 )𝒙 The gradient of A graph of ln(𝑇 − 𝑇∞) against x (excluding temperature at x=0) gives √( ℎ𝑃 𝐾𝐴 ) as the gradient from which h can be measured. h can also be got from Stefan’s law of cooling in natural convection that reduces to the Newton’s law of cooling. Stefan’s law of cooling in natural convection in a non-vacuum environment states: 𝒅𝑸 𝒅𝒕 = (𝟏 + 𝑮)𝑨𝝈𝜺[𝑻𝟒 − 𝑻∞ 𝟒 ] Where: 𝑮 = 𝒌𝑷𝒓𝒏 = 𝒆𝒙𝒑𝒆𝒓𝒊𝒎𝒆𝒏𝒕𝒂𝒍 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒏 = 𝒆𝒙𝒑𝒆𝒓𝒊𝒎𝒆𝒏𝒕𝒂𝒍 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝜺 = 𝒆𝒎𝒊𝒔𝒔𝒊𝒗𝒊𝒕𝒚 𝝈 = 𝑺𝒕𝒆𝒇𝒂𝒏 𝑩𝒐𝒍𝒕𝒛𝒎𝒂𝒏𝒏 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝑻 = 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝒊𝒏 𝒌𝒆𝒍𝒗𝒊𝒏 𝑻∞ = 𝒓𝒐𝒐𝒎 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝒊𝒏 𝒌𝒆𝒍𝒗𝒊𝒏 Where: 𝑃 𝑟 = 𝑃𝑟𝑎𝑛𝑑𝑡𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔 𝑚𝑒𝑑𝑖𝑢𝑚 𝑎𝑡 𝑟𝑜𝑜𝑚 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 The Prandtl number above is independent of temperature of the cooling body.
- 33. Considering 𝑇 = 𝑇∞ + ∆𝑇 𝑑𝑄 𝑑𝑡 = (1 + 𝐺)𝐴𝜎𝜀[(𝑇∞ + ∆𝑇)4 − 𝑇∞ 4 ] Factorizing out 𝑇∞, we get 𝑑𝑄 𝑑𝑡 = (1 + 𝐺)𝐴𝜎𝜀[𝑇∞ 4 (1 + (𝑇 − 𝑇∞) 𝑇∞ )4 − 𝑇∞ 4 ] It is known from Binomial expansion that: (1 + 𝑥)𝑛 ≈ 1 + 𝑛𝑥 𝑓𝑜𝑟 𝑥 ≪ 1 So: (1 + (𝑇 − 𝑇∞) 𝑇∞ )4 ≈ 1 + 4 (𝑇 − 𝑇∞) 𝑇∞ = 1 + 4 ∆𝑇 𝑇∞ = 𝑓𝑜𝑟 ∆𝑇 𝑇∞ ≪ 1 Simplifying, we get Newton’s law of cooling i.e. 𝒅𝑸 𝒅𝒕 = 𝟒(𝟏 + 𝑮)𝑨𝝈𝜺𝑻∞ 𝟑 (𝑻 − 𝑻∞) 𝒅𝑸 𝒅𝒕 = 𝒉𝑨(𝑻 − 𝑻∞) Where: 𝒉 = 𝟒(𝟏 + 𝑮)𝝈𝜺𝑻∞ 𝟑 Substitute for the above parameters of aluminium and get h theoretically and compare as got experimentally. How do we deal with metal rods of finite length 𝒍 ?
- 34. The boundary and initial conditions are 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 −𝒌 𝒅𝑻 𝒅𝒙 = 𝒉(𝑻 − 𝑻∞) 𝒂𝒕 𝒙 = 𝒍 𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 Let us go about solving for the above boundary conditions We start with a temperature profile below: 𝑇 − 𝑇∞ = (𝑇𝑠 − 𝑇∞)𝑒− 𝑥 𝛿 (1− 𝑥 𝑙 ) [ 𝑥 𝑙 (𝑇1 − 𝑇∞) (𝑇𝑠 − 𝑇∞) + (1 − 𝑥 𝑙 )] Which says 𝑇 = 𝑇𝑠 𝑎𝑡 𝑥 = 0 𝑇 = 𝑇1 𝑎𝑡 𝑥 = 𝑙 𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0 Provided 𝛿 = 0 𝑎𝑡 𝑡 = 0 , then the initial condition above is satisfied
- 35. we take the derivative 𝒅𝑻 𝒅𝒙 𝒂𝒕 𝒙 = 𝒍 and equate it to − ℎ 𝑘 (𝑇1 − 𝑇∞) and get: 𝑑𝑇 𝑑𝑥 |𝑥=𝑙 = ( (𝑇1 − 𝑇∞) 𝑙 − (𝑇𝑠 − 𝑇∞) 𝑙 + (𝑇1 − 𝑇∞) 𝛿 ) 𝑑𝑇 𝑑𝑥 |𝑥=𝑙 = − ℎ 𝑘 (𝑇1 − 𝑇∞) We equate the two and get ( (𝑇1 − 𝑇∞) 𝑙 − (𝑇𝑠 − 𝑇∞) 𝑙 + (𝑇1 − 𝑇∞) 𝛿 ) = − ℎ 𝑘 (𝑇1 − 𝑇∞) We finally get (𝑇1 − 𝑇∞) = (𝑇𝑠 − 𝑇∞)( 𝛿𝑘 𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 ) We substitute 𝑇1 − 𝑇∞ into the temperature profile and get (𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆− 𝒙 𝜹 (𝟏 − 𝒙 𝒍 ) [ 𝒙 𝒍 ( 𝜹𝒌 𝜹𝒌 + 𝒍𝒌 + 𝒉𝒍𝜹 ) + (𝟏 − 𝒙 𝒍 )] This the temperature profile that satisfies the boundary and initial conditions below 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 −𝒌 𝒅𝑻 𝒅𝒙 = 𝒉(𝑻 − 𝑻∞) 𝒂𝒕 𝒙 = 𝒍 𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 Let us go ahead and solve for 𝛿 The governing equation is 𝛼 𝜕2 𝑇 𝜕𝑥2 − ℎ𝑃 𝐴𝜌𝐶 (𝑇 − 𝑇∞) = 𝜕𝑇 𝜕𝑡 Let us change this equation into an integral as below: 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 − ℎ𝑃 𝐴𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 … … . . 𝑏)
- 36. 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 − 2ℎ 𝑟𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = [ 𝜕𝑇 𝜕𝑥 ] 𝑙 0 𝜕𝑇 𝜕𝑥 = (𝑇𝑠 − 𝑇∞)𝑒− 𝑥 𝛿 (1− 𝑥 𝑙 ) [ 1 𝑙 (𝑇1 − 𝑇∞) (𝑇𝑠 − 𝑇∞) − 1 𝑙 ] + (𝑇𝑠 − 𝑇∞)𝑒− 𝑥 𝛿 (1− 𝑥 𝑙 ) [ 𝑥 𝑙 (𝑇1 − 𝑇∞) (𝑇𝑠 − 𝑇∞) − (1 − 𝑥 𝑙 )] ( −𝑙 + 2𝑥 𝛿𝑙 ) ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = [ 𝜕𝑇 𝜕𝑥 ] 𝑙 0 = (𝑇𝑠 − 𝑇∞) 𝛿 + (𝑇1 − 𝑇∞) 𝛿 Substitute for (𝑇1 − 𝑇∞) = (𝑇𝑠 − 𝑇∞)( 𝛿𝑘 𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 ) ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = [ 𝜕𝑇 𝜕𝑥 ] 𝑙 0 = (𝑇𝑠 − 𝑇∞)( 2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 𝛿(𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿) ) 2ℎ 𝑟𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = ℎ2𝜋𝑟 ∫ ((𝑇𝑠 − 𝑇∞)𝑒− 𝑥 𝛿 (1− 𝑥 𝑙 ) [ 𝑥 𝑙 (𝑇1 − 𝑇∞) (𝑇𝑠 − 𝑇∞) + (1 − 𝑥 𝑙 )])𝑑𝑥 𝑙 0 2ℎ 𝑟𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = ℎ2𝜋𝑟[ (𝑇1 − 𝑇∞) 𝑙 ∫ 𝑥𝑒− 𝑥 𝛿 (1− 𝑥 𝑙 ) 𝑑𝑥 𝑙 0 + (𝑇𝑠 − 𝑇∞) ∫ 𝑒− 𝑥 𝛿 (1− 𝑥 𝑙 ) 𝑑𝑥 𝑙 0 − (𝑇𝑠 − 𝑇∞) 𝑙 ∫ 𝑥𝑒− 𝑥 𝛿 (1− 𝑥 𝑙 ) 𝑑𝑥 𝑙 0 ] Integrating by parts shows that ∫ 𝑒 − 𝑥 𝛿 (1− 𝑥 𝑙 ) 𝑑𝑥 𝑙 0 = [( 𝑙𝛿 −𝑙 + 2𝑥 )𝑒 − 𝑥 𝛿 (1− 𝑥 𝑙 ) ] 𝑙 0 = 2𝛿 ∫ 𝑥𝑒 − 𝑥 𝛿 (1− 𝑥 𝑙 ) 𝑑𝑥 𝑙 0 = [( 𝑥𝑙𝛿 −𝑙 + 2𝑥 − 𝑙2 𝛿2 (−𝑙 + 2𝑥)2[1 − 2𝑙𝛿 (−𝑙 + 2𝑥)2] ) 𝑒 − 𝑥 𝛿 (1− 𝑥 𝑙 ) ] 𝑙 0 = 𝑙𝛿 Substituting back into the heat loss equation we get 2ℎ 𝑟𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = 2ℎ 𝑟𝜌𝐶 [(𝑇1 − 𝑇∞) + (𝑇𝑠 − 𝑇∞)]𝛿 substitute for (𝑇1 − 𝑇∞) and get
- 37. 2ℎ 𝑟𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = 2ℎ 𝑟𝜌𝐶 (𝑇𝑠 − 𝑇∞)( 2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 (𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿) )𝛿 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = (𝑇𝑠 − 𝑇∞) ( 2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 (𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿) ) 𝑑𝛿 𝑑𝑡 + 𝜕(𝑙𝑇∞) 𝜕𝑡 𝜕(𝑙𝑇∞) 𝜕𝑡 = 0 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = (𝑇𝑠 − 𝑇∞) ( 2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 (𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿) ) 𝑑𝛿 𝑑𝑡 Substituting into the integral equation we get 𝛼(𝑇𝑠 − 𝑇∞) ( 2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 𝛿(𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿) ) − 2ℎ 𝑟𝜌𝐶 (𝑇𝑠 − 𝑇∞) ( 2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 (𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿) ) 𝛿 = (𝑇𝑠 − 𝑇∞)( 2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 (𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿) ) 𝑑𝛿 𝑑𝑡 𝛼 − ℎ𝑃 𝐴𝜌𝐶 𝛿2 = 𝛿 𝑑𝛿 𝑑𝑡 We solve the equation above assuming that 𝛿 = 0 𝑎𝑡 𝑡 = 0 And get 𝛿 = √ 𝛼𝐴𝜌𝐶 ℎ𝑃 (1 − 𝑒 −2ℎ𝑃 𝐴𝜌𝐶 𝑡 ) 𝛿 = √ 𝐾𝐴 ℎ𝑃 (1 − 𝑒 −2ℎ𝑃 𝐴𝜌𝐶 𝑡 ) We go ahead and substitute for 𝛿 in the temperature profile below (𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆− 𝒙 𝜹 (𝟏 − 𝒙 𝒍 ) [ 𝒙 𝒍 ( 𝜹𝒌 𝜹𝒌 + 𝒍𝒌 + 𝒉𝒍𝜹 ) + (𝟏 − 𝒙 𝒍 )] When the time is small, 𝛿 using binomial approximation becomes 𝛿 = √ 𝐾𝐴 ℎ𝑃 (1 − 𝑒 −2ℎ𝑃 𝐴𝜌𝐶 𝑡 )
- 38. 2ℎ𝑃 𝐴𝜌𝐶 𝑡 ≪ 1 𝑒 −ℎ𝑃 𝐴𝜌𝐶 𝑡 = 1 − 2ℎ𝑃 𝐴𝜌𝐶 𝑡 1 − 𝑒 −2ℎ𝑃 𝐴𝜌𝐶 𝑡 = 2ℎ𝑃 𝐴𝜌𝐶 𝑡 𝛿 = √2𝛼𝑡 We substitute for 𝛿 in the temperature profile. (𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆 − 𝒙 √𝟐𝜶𝒕 (𝟏 − 𝒙 𝒍 ) [ 𝒙 𝒍 ( 𝒌√𝟐𝜶𝒕 𝒌√𝟐𝜶𝒕 + 𝒍𝒌 + 𝒉𝒍√𝟐𝜶𝒕 ) + (𝟏 − 𝒙 𝒍 )] The above equation is observed for small times. What is the flux at 𝑥 = 𝑙 From −𝒌 𝒅𝑻 𝒅𝒙 = 𝒉(𝑻𝟏 − 𝑻∞) 𝒂𝒕 𝒙 = 𝒍 Substitute for (𝑻𝟏 − 𝑻∞) and get 𝑞̇|=𝑙 = ℎ(𝑇𝑠 − 𝑇∞)( 𝛿𝑘 𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 ) You notice that at 𝑙 = 0 𝑞̇|𝑙=0 = ℎ(𝑇𝑠 − 𝑇∞) And at 𝑙 = ∞ 𝑞̇|𝑙=∞ = 0 Which is true. What happens when the length is big or tends to infinity? (𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆− 𝒙 𝜹 (𝟏 − 𝒙 𝒍 ) [ 𝒙 𝒍 ( 𝜹𝒌 𝜹𝒌 + 𝒍𝒌 + 𝒉𝑳𝒍𝜹 ) + (𝟏 − 𝒙 𝒍 )] Becomes (𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆− 𝒙 𝜹
- 39. (𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆 − 𝒙 √𝑲𝑨 𝒉𝑷 (𝟏−𝒆 −𝟐𝒉𝑷 𝑨𝝆𝑪 𝒕 ) Which is what we got before.
- 40. WHAT HAPPENS WHEN THE INITIAL TEMPERATURE IS A FUNCTION OF X? The governing equation is 𝛼 𝜕2 𝑇 𝜕𝑥2 = 𝜕𝑇 𝜕𝑡 BCs 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝟎 < 𝒕 < ∞ 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝒍 𝟎 < 𝒕 < ∞ IC 𝑻 = ∅(𝒙) 𝒂𝒕 𝒕 = 𝟎 𝟎 ≤ 𝒙 ≤ 𝒍 We assume an exponential temperature profile that satisfies the boundary conditions: 𝑇 − ∅ 𝑇𝑠 − ∅ = 𝑒 −𝑥 𝛿 (1− 𝑥 𝑙 ) The PDE becomes an integral equation given by: 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ 𝑇𝑑𝑥 𝑙 0 Let us give an example say ∅ = 𝑥 We make T the subject of the formula and get 𝑇 = ∅ + 𝑇𝑠𝑒 −𝑥 𝛿 (1− 𝑥 𝑙 ) − ∅𝑒 −𝑥 𝛿 (1− 𝑥 𝑙 ) ∅ = 𝑥 𝑇 = 𝑥 + 𝑇𝑠𝑒 −𝑥 𝛿 (1− 𝑥 𝑙 ) − 𝑥𝑒 −𝑥 𝛿 (1− 𝑥 𝑙 ) ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = [ 𝜕𝑇 𝜕𝑥 ] 𝑙 0 We go ahead and solve for 𝛿.
- 41. Using this integral analytical method, we can also go ahead and solve PDES of the form below: 𝜕𝑇 𝜕𝑡 = 𝛼 𝜕2 𝑇 𝜕𝑥2 + 𝑓(𝑥)
- 42. HOW DO WE DEAL WITH CYLINDRICAL CO-ORDINATES FOR AN INFINITE RADIUS CYLINDER? We know that for an insulated cylinder where there is no heat loss by convection from the sides, the governing PDE equation is 𝛼 [ 𝜕2 𝑇 𝜕𝑟2 + 1 𝑟 𝜕𝑇 𝜕𝑟 ] = 𝜕𝑇 𝜕𝑡 The boundary conditions are 𝑇 = 𝑇𝑠 𝑎𝑡 𝑟 = 𝑟1 𝑇 = 𝑇∞ 𝑎𝑡 𝑟 = ∞ The initial condition is: 𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0 The temperature profile that satisfies the conditions above is 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 −(𝑟−𝑟1) 𝛿 We transform the PDE into an integral equation 𝛼 [ 𝜕2 𝑇 𝜕𝑟2 + 1 𝑟 𝜕𝑇 𝜕𝑟 ] = 𝜕𝑇 𝜕𝑡 We take integrals from 𝑟1 to 𝑟 = 𝑅 = ∞ 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑟2 ) 𝑑𝑟 𝑅 𝑟1 + 𝛼 ∫ [ 1 𝑟 ( 𝜕𝑇 𝜕𝑟 )]𝑑𝑟 𝑅 𝑟1 = 𝜕 𝜕𝑡 ∫ 𝑇𝑑𝑟 𝑅 𝑟1 We then go ahead to solve and find 𝛿 as before. ∫ ( 𝜕2 𝑇 𝜕𝑟2 ) 𝑑𝑟 𝑅 𝑟1 = [ 𝜕𝑇 𝜕𝑟 ] 𝑅 𝑟1 = − 𝑇𝑠 − 𝑇∞ 𝛿 [𝑒 −(𝑟−𝑟1) 𝛿 ] 𝑅 = ∞ 𝑟1 = (𝑇𝑠 − 𝑇∞) 𝛿 𝜕𝑇 𝜕𝑟 = − 𝑇𝑠 − 𝑇∞ 𝛿 𝑒 −(𝑟−𝑟1) 𝛿 ∫ [ 1 𝑟 ( 𝜕𝑇 𝜕𝑟 )] 𝑑𝑟 𝑅 𝑟1 = − (𝑇𝑠 − 𝑇∞) 𝛿 ∫ 1 𝑟 𝑒 −(𝑟−𝑟1) 𝛿 𝑑𝑟 𝑅 𝑟1 ∫ 1 𝑟 𝑒 −(𝑟−𝑟1) 𝛿 𝑑𝑟 𝑅 𝑟1 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 𝑑𝑟 𝑑𝑟
- 43. 𝑢 = 1 𝑟 𝑑𝑣 𝑑𝑟 = 𝑒 −(𝑟−𝑟1) 𝛿 𝑣 = −𝛿𝑒 −(𝑟−𝑟1) 𝛿 ∫ 1 𝑟 𝑒 −(𝑟−𝑟1) 𝛿 𝑑𝑟 𝑅 𝑟1 = −𝛿𝑒 −(𝑟−𝑟1) 𝛿 𝑟 − 𝛿 𝑟 ∫ 1 𝑟 𝑒 −(𝑟−𝑟1) 𝛿 𝑑𝑟 𝑅 𝑟1 [1 + 𝛿 𝑟 ] ∫ 1 𝑟 𝑒 −(𝑟−𝑟1) 𝛿 𝑑𝑟 𝑅 𝑟1 = −𝛿𝑒 −(𝑟−𝑟1) 𝛿 𝑟 ∫ 1 𝑟 𝑒 −(𝑟−𝑟1) 𝛿 𝑑𝑟 𝑅 𝑟1 = [ −𝛿 𝑟 + 𝛿 𝑒 −(𝑟−𝑟1) 𝛿 ] 𝑅 = ∞ 𝑟1 = 𝛿 𝛿 + 𝑟1 Substituting in ∫ [ 1 𝑟 ( 𝜕𝑇 𝜕𝑟 )]𝑑𝑟 𝑅 𝑟1 = − (𝑇𝑠 − 𝑇∞) 𝛿 ∫ 1 𝑟 𝑒 −(𝑟−𝑟1) 𝛿 𝑑𝑟 𝑅 𝑟1 = − (𝑇𝑠 − 𝑇∞) 𝛿 ( 𝛿 𝛿 + 𝑟1 ) We get ∫ [ 1 𝑟 ( 𝜕𝑇 𝜕𝑟 )]𝑑𝑟 𝑅 𝑟1 = − (𝑇𝑠 − 𝑇∞) 𝛿 ( 𝛿 𝛿 + 𝑟1 ) = − (𝑇𝑠 − 𝑇∞) 𝛿 + 𝑟1 𝑇 = (𝑇𝑠 − 𝑇∞)𝑒 −(𝑟−𝑟1) 𝛿 + 𝑇∞ ∫ 𝑇𝑑𝑟 𝑅 𝑟1 = ∫ ((𝑇𝑠 − 𝑇∞)𝑒 −(𝑟−𝑟1) 𝛿 + 𝑇∞)𝑑𝑟 𝑅 𝑟1 = 𝛿((𝑇𝑠 − 𝑇∞) (1 − 𝑒 −(𝑅−𝑟1) 𝛿 ) + 𝑇∞(𝑅 − 𝑟1)) 𝜕 𝜕𝑡 ∫ 𝑇𝑑𝑟 𝑅 𝑟1 = 𝑑 𝑑𝑡 (𝛿((𝑇𝑠 − 𝑇∞) (1 − 𝑒 −(𝑅−𝑟1) 𝛿 )) + 𝜕(𝑇∞(𝑅 − 𝑟1)) 𝜕𝑡 𝜕(𝑇∞(𝑅 − 𝑟1)) 𝜕𝑡 = 0 𝜕 𝜕𝑡 ∫ 𝑇𝑑𝑟 𝑅 𝑟1 = 𝑑 𝑑𝑡 (𝛿((𝑇𝑠 − 𝑇∞) (1 − 𝑒 −(𝑅−𝑟1) 𝛿 )) Substitute for 𝑅 = ∞ and get
- 44. 𝜕 𝜕𝑡 ∫ 𝑇𝑑𝑟 𝑅 𝑟1 = 𝑑𝛿 𝑑𝑡 (𝑇𝑠 − 𝑇∞) substituting all the above in the integral equation, we get 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑟2 ) 𝑑𝑟 𝑅 𝑟1 + 𝛼 ∫ [ 1 𝑟 ( 𝜕𝑇 𝜕𝑟 )]𝑑𝑟 𝑅 𝑟1 = 𝜕 𝜕𝑡 ∫ 𝑇𝑑𝑟 𝑅 𝑟1 𝛼 (𝑇𝑠 − 𝑇∞) 𝛿 − 𝛼(𝑇𝑠 − 𝑇∞) 𝛿 + 𝑟1 = 𝑑𝛿 𝑑𝑡 (𝑇𝑠 − 𝑇∞) We go ahead and solve for 𝛿 𝛼 𝛿 − 𝛼 𝛿 + 𝑟1 = 𝑑𝛿 𝑑𝑡 The boundary conditions are: 𝛿 = 0 𝑎𝑡 𝑡 = 0 We solve for 𝛿 and get an algebraic cubic equation 𝟐𝜹𝟑 + 𝟑𝒓𝟏𝜹𝟐 − 𝟔𝒓𝟏𝜶𝒕 = 𝟎 From which 𝛿 can be got. 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.