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FUNDAMENTALS OF HEAT TRANSFER .pdf
- 1. FUNDAMENTALS OF HEAT FLOW: THE ANALYTICAL SOLUTIONS TO THE HEAT EQUATION USING AN INTEGRAL METHOD By Wasswa Derrick
- 3. 2 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 INSULATED METAL RODS IN SERIES OF DIFFERENT THERMO CONDUCTIVITY FOR AN INFINITE METAL ROD COMBINATION...............11 HOW DO WE DEAL WITH CONVECTION AT THE SURFACE AREA OF THE SEMI- INFINITE METAL ROD ......................................................................................................................13 EQUAL FIXED TEMPERATURES AT THE END OF AN INSULATED METAL ROD....17 UNEQUAL FIXED TEMPERATURES AT THE END OF AN INSULATED METAL ROD. ....................................................................................................................................................................20 HOW DO WE DEAL WITH OTHER TYPES OF BOUNDARY CONDITIONS?..................23 HOW DO WE DEAL WITH NATURAL CONVECTION AT THE SURFACE AREA OF A SEMI-INFINITE METAL ROD FOR FIXED END TEMPERATURE.....................................25 HOW DO WE DEAL WITH METAL RODS OF FINITE LENGTH 𝒍 ? ...................................35 HOW DO WE DEAL WITH THE CASE WHERE THE FLUX AT THE END OF THE METAL ROD IS ZERO ........................................................................................................................41 HOW DO WE DEAL WITH CYLINDRICAL CO-ORDINATES FOR AN INFINITE RADIUS CYLINDER? ..........................................................................................................................44 HOW DO WE DEAL WITH NATURAL CONVECTION AT THE SURFACE AREA OF A SEMI-INFINITE CYLINDER FOR FIXED END TEMPERATURE ........................................48 HOW DO WE DEAL WITH METAL RODS OF FINITE RADIUS CYLINDER?.................53 HOW DO WE DEAL WITH RECTANGULAR CO-ORDINATES IN NATURAL CONVECTION?......................................................................................................................................58 LET US ANALYSE THE RATE OF COOLING IN NATURAL CONVECTION....................68 WHAT HAPPENS WHEN THE INITIAL TEMPERATURE IS A FUNCTION OF X? .......70
- 4. 3 SEMI INFINITE WALL ANALYTICAL SOLUTION TO THE HEAT EQUATION. What does semi-infinite wall mean? It means that the wall is limited at one end but extends to infinity in the other direction. For a cylindrical metal rod, semi infinite means that the rod has a finite radius but extends to infinity along its length. We shall be using these terms in the analyses to follow. 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 → ∞
- 5. 4 Equation 1 may be integrated once to get 𝑙𝑛 𝑑𝑌 𝑑𝜂 = 𝑐1 − 𝜂2 𝑑𝑌 𝑑𝜂 = 𝑐2𝑒−𝜂2 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. 5 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
- 7. 6 Since 𝑙 and 𝑇∞ are constants independent of time 𝜕(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
- 8. 7 You notice that the initial condition is satisfied by the temperature profile above i.e., 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. 8 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. 9 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. 10 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. 11 HOW DO WE DEAL WITH INSULATED METAL RODS IN SERIES OF DIFFERENT THERMO CONDUCTIVITY FOR AN INFINITE METAL ROD COMBINATION We know that for rod 1, the temperature variation at any point will be given by the infinite rod formula. 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = 𝒆 −𝒙𝟏 √𝟐𝜶𝟏𝒕 Where: 𝛼1 = 𝐾1 𝜌1𝐶1 For rod 2, the temperature variation at any point is given by; 𝑻 − 𝑻∞ 𝑻𝟏 − 𝑻∞ = 𝒆 −𝒙𝟐 √𝟐𝜶𝟐𝒕 But the temperature 𝑇1 can be got from the equation above by substituting for 𝑥 = 𝑙 as shown below: (𝑇1 − 𝑇∞) = (𝑇𝑠 − 𝑇∞)𝑒 −𝑙 √2𝛼1𝑡 Substituting in the equation above for rod 2, we get (𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆 −( 𝒍 √𝟐𝜶𝟏𝒕 + 𝒙𝟐 √𝟐𝜶𝟐𝒕 ) You notice that if the two rods are of the same material, we reduce back to the equation
- 13. 12 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = 𝒆 −𝒙 √𝟐𝜶𝒕 We can extend this analysis to finite length metal rods once we find the solution to them.
- 14. 13 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
- 15. 14 ∫ (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 )
- 16. 15 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 )
- 17. 16 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
- 18. 17 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 )
- 19. 18 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
- 20. 19 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.
- 21. 20 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 )
- 22. 21 ∫ ( 𝜕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:
- 23. 22 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 )]
- 24. 23 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 )]
- 25. 24 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.
- 26. 25 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
- 27. 26 ∫ (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
- 28. 27 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:
- 29. 28 2ℎ𝑃 𝐴𝜌𝐶 𝑡0 = 𝑎 𝑡0 = 𝑎 𝑟𝜌𝐶 4ℎ And 𝑎 = 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 𝟎. 𝟎𝟐𝟒𝟔𝟔 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 𝟒𝒉 Where: 𝛼 = 𝐾 𝜌𝐶 The equation becomes: 𝒙𝟐 = 𝟐𝜶[𝐥𝐧 ( 𝑻𝒔 − 𝑻∞ 𝑻 − 𝑻∞ )]𝟐 × 𝒕 − 𝒂𝑲𝒓 𝟐𝒉 [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.
- 30. 29 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)
- 31. 30 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
- 32. 31 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
- 33. 32 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
- 34. 33 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.
- 35. 34 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.
- 36. 35 HOW DO WE DEAL WITH METAL RODS OF FINITE LENGTH 𝒍 ? 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
- 37. 36 𝑇 = 𝑇1 𝑎𝑡 𝑥 = 𝑙 𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0 Provided 𝛿 = 0 𝑎𝑡 𝑡 = 0 , then the initial condition above is satisfied 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:
- 38. 37 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 − ℎ𝑃 𝐴𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 … … . . 𝑏) 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∞) 𝛿 ( 2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 (𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿) ) ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = ∫ ((𝑇𝑠 − 𝑇∞)𝑒− 𝑥 𝛿 (1− 𝑥 𝑙 ) [ 𝑥 𝑙 (𝑇1 − 𝑇∞) (𝑇𝑠 − 𝑇∞) + (1 − 𝑥 𝑙 )])𝑑𝑥 𝑙 0 ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = [ (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
- 39. 38 2ℎ 𝑟𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = 2ℎ 𝑟𝜌𝐶 [(𝑇1 − 𝑇∞) + (𝑇𝑠 − 𝑇∞)]𝛿 substitute for (𝑇1 − 𝑇∞) and get 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∞) 𝛿 ( 2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 (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
- 40. 39 𝛿 = √ 𝐾𝐴 ℎ𝑃 (1 − 𝑒 −2ℎ𝑃 𝐴𝜌𝐶 𝑡 ) 2ℎ𝑃 𝐴𝜌𝐶 𝑡 ≪ 1 𝑒 −2ℎ𝑃 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
- 41. 40 (𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆− 𝒙 𝜹 (𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆 − 𝒙 √𝑲𝑨 𝒉𝑷 (𝟏−𝒆 −𝟐𝒉𝑷 𝑨𝝆𝑪 𝒕 ) Which is what we got before. Using the temperature profile above, we can go ahead and also derive the governing equations for rods in series as shown before for an infinite metal rod combination.
- 42. 41 HOW DO WE DEAL WITH THE CASE WHERE THE FLUX AT THE END OF THE METAL ROD IS ZERO 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝒅𝑻 𝒅𝒙 = 𝟎 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 )] As solved before, 𝛿 will be equal to: 𝛿 = √ 𝐾𝐴 ℎ𝑃 (1 − 𝑒 −2ℎ𝑃 𝐴𝜌𝐶 𝑡 ) For small times, we use 2ℎ𝑃 𝐴𝜌𝐶 𝑡 ≪ 1
- 43. 42 𝑒 −ℎ𝑃 𝐴𝜌𝐶 𝑡 = 1 − 2ℎ𝑃 𝐴𝜌𝐶 𝑡 1 − 𝑒 −2ℎ𝑃 𝐴𝜌𝐶 𝑡 = 2ℎ𝑃 𝐴𝜌𝐶 𝑡 𝛿 = √2𝛼𝑡 First of all, how do we realize the temperature profile above. The temperature profile above is observed when you stick wax particle at the end of the metal rod and heat the rod from the other end i.e., we substitute 𝑥 = 𝑙 in the temperature profile above and get: (𝑇 − 𝑇∞) = (𝑇𝑠 − 𝑇∞)( 𝛿 𝑙 + 𝛿 ) We substitute: 𝑇 = 37℃ 𝑝𝑜𝑖𝑛𝑡 𝑎𝑡 𝑤ℎ𝑖𝑐ℎ 𝑤𝑎𝑥 𝑏𝑒𝑔𝑖𝑛𝑠 𝑡𝑜 𝑚𝑒𝑙𝑡 For short times 𝛿 = √2𝛼(𝑡 − 𝑡0) Now making length 𝑙 the subject of the formula above and substituting for 𝛿, we get: 𝒍𝟐 = [ 𝑻𝒔 − 𝑻 𝑻 − 𝑻∞ ]𝟐 × 𝟐𝜶(𝒕 − 𝒕𝟎) Where:
- 44. 43 2ℎ𝑃 𝐴𝜌𝐶 𝑡0 = 0.02466 From the above equation of length, we can tell the time it will take the wax to melt. We can also verify above that the relationship of length 𝑙 against time above is obeyed
- 45. 44 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 𝒓 𝝏 𝝏𝒓 (𝒓 𝝏𝑻 𝝏𝒓 ) = 𝝏𝑻 𝝏𝒕 In steady state 𝜕𝑇 𝜕𝑡 = 0 We end up with 𝜕 𝜕𝑟 (𝑟 𝜕𝑇 𝜕𝑟 ) = 0 We can then integrate once to get ∫ ( 𝜕 𝜕𝑟 (𝑟 𝜕𝑇 𝜕𝑟 )) 𝑑𝑟 = ∫(0)𝑑𝑟 And get 𝑟 𝜕𝑇 𝜕𝑟 = 𝐶1 Therefore 𝜕𝑇 𝜕𝑟 = 𝐶1 𝑟 We can go ahead and find the temperature profile as a function of radius r. Back to the PDE, to get the transient state 𝜶 𝒓 𝝏 𝝏𝒓 (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
- 46. 45 The temperature profile that satisfies the conditions above is 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 −(𝑟−𝑟1) 𝛿 We transform the equation above into an integral equation and take integrals from 𝑟 = 𝑟1 to 𝑟 = 𝑅 = ∞. 𝛼 𝑟 𝜕 𝜕𝑟 (𝑟 𝜕𝑇 𝜕𝑟 ) = 𝜕𝑇 𝜕𝑡 We first multiply through by radius r and get 𝛼 𝜕 𝜕𝑟 (𝑟 𝜕𝑇 𝜕𝑟 ) = 𝑟 𝜕𝑇 𝜕𝑡 We then take integrals and get ∫ [𝛼 𝜕 𝜕𝑟 (𝑟 𝜕𝑇 𝜕𝑟 )]𝑑𝑟 𝑅 𝑟1 = ∫ (𝑟 𝜕 𝜕𝑡 𝑇)𝑑𝑟 𝑅 𝑟1 We get 𝛼𝑟 𝜕𝑇 𝜕𝑟 = ∫ (𝑟 𝜕 𝜕𝑡 𝑇)𝑑𝑟 𝑅 𝑟1 Divide through by r and get 𝛼 𝜕𝑇 𝜕𝑟 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑟 𝑅 𝑟1 But 𝜕𝑇 𝜕𝑟 = ∫ ( 𝜕2 𝑇 𝜕𝑟2 )𝑑𝑟 𝑅 𝑟1 So, the PDE becomes: 𝜶 ∫ ( 𝝏𝟐 𝑻 𝝏𝒓𝟐 )𝒅𝒓 𝑹 𝒓𝟏 = 𝝏 𝝏𝒕 ∫ (𝑻)𝒅𝒓 𝑹 𝒓𝟏 We then go ahead to solve and find 𝛿 as before. 𝜕𝑇 𝜕𝑟 = − 𝑇𝑠 − 𝑇∞ 𝛿 𝑒 −(𝑟−𝑟1) 𝛿 ∫ ( 𝜕2 𝑇 𝜕𝑟2 ) 𝑑𝑟 𝑅 𝑟1 = [ 𝜕𝑇 𝜕𝑟 ] 𝑅 𝑟1 = − (𝑇𝑠 − 𝑇∞) 𝛿 [𝑒 −(𝑟−𝑟1) 𝛿 ] 𝑅 = ∞ 𝑟1 = (𝑇𝑠 − 𝑇∞) 𝛿
- 47. 46 𝑇 = (𝑇𝑠 − 𝑇∞)𝑒 −(𝑟−𝑟1) 𝛿 + 𝑇∞ ∫ 𝑇𝑑𝑟 𝑅=∞ 𝑟1 = ∫ ((𝑇𝑠 − 𝑇∞)𝑒 −(𝑟−𝑟1) 𝛿 )𝑑𝑟 𝑅=∞ 𝑟1 + ∫ 𝑇∞𝑑𝑟 𝑅=∞ 𝑟1 = 𝛿(𝑇𝑠 − 𝑇∞) + 𝑇∞(𝑅 − 𝑟1) 𝜕 𝜕𝑡 ∫ 𝑇𝑑𝑟 𝑅 𝛿 = 𝑑𝛿 𝑑𝑡 (𝑇𝑠 − 𝑇∞) + 𝑑(𝑇∞(𝑅 − 𝑟1)) 𝑑𝑡 But 𝑑(𝑇∞(𝑅 − 𝑟1)) 𝑑𝑡 = 0 Since 𝑇∞, 𝑅, 𝑟1 are constants independent of time. So 𝜕 𝜕𝑡 ∫ 𝑇𝑑𝑟 𝑅 𝛿 = 𝑑𝛿 𝑑𝑡 (𝑇𝑠 − 𝑇∞) substituting all the above in the integral equation, we get 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑟2 )𝑑𝑟 𝑅 𝑟1 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑟 𝑅 𝑟1 𝛼 𝛿 (𝑇𝑠 − 𝑇∞) = 𝑑𝛿 𝑑𝑡 (𝑇𝑠 − 𝑇∞) Divide through by (𝑇𝑠 − 𝑇∞) and get 𝛼 𝛿 = 𝑑𝛿 𝑑𝑡 𝛼 𝛿 = 𝑑𝛿 𝑑𝑡 The boundary conditions are: 𝛿 = 0 𝑎𝑡 𝑡 = 0 𝛿 = √2𝛼𝑡 We substitute 𝛿 in the temperature profile and get: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 −(𝑟−𝑟1) 𝛿 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = 𝒆 −(𝒓−𝒓𝟏) √𝟐𝜶𝒕
- 48. 47 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.
- 49. 48 HOW DO WE DEAL WITH NATURAL CONVECTION AT THE SURFACE AREA OF A SEMI-INFINITE CYLINDER FOR FIXED END TEMPERATURE The governing equation is: 𝜶 𝒓 𝝏 𝝏𝒓 (𝒓 𝝏𝑻 𝝏𝒓 ) − 𝒉𝑷 𝑨𝝆𝑪 (𝑻 − 𝑻∞) = 𝝏𝑻 𝝏𝒕 𝑃 = 2𝜋𝑟 𝐴 = 2𝜋𝑟𝑑 Where: 𝑑 = ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 The boundary conditions are: 𝑇 = 𝑇𝑠 𝑎𝑡 𝑟 = 𝑟1 𝑇 = 𝑇∞ 𝑎𝑡 𝑟 = 𝑅 = ∞ The initial condition is: 𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0 The temperature profile that satisfies the boundary conditions above is: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 −(𝑟−𝑟1) 𝛿 𝛼 𝑟 𝜕 𝜕𝑟 (𝑟 𝜕𝑇 𝜕𝑟 ) − ℎ𝑃 𝐴𝜌𝐶 (𝑇 − 𝑇∞) = 𝜕𝑇 𝜕𝑡 Multiply through by r to get 𝛼 𝜕 𝜕𝑟 (𝑟 𝜕𝑇 𝜕𝑟 ) − ℎ𝑃 𝐴𝜌𝐶 𝑟(𝑇 − 𝑇∞) = 𝑟 𝜕𝑇 𝜕𝑡 We then transform the PDE into an integral equation and take the limits to be from 𝑟 = 𝑟1 to 𝑟 = 𝑅 = ∞ 𝛼 ∫ [𝛼 𝜕 𝜕𝑟 (𝑟 𝜕𝑇 𝜕𝑟 )]𝑑𝑟 𝑅=∞ 𝑟1 − ℎ 𝑑𝜌𝐶 ∫ 𝑟(𝑇 − 𝑇∞)𝑑𝑟 𝑅=∞ 𝑟1 = 𝜕 𝜕𝑡 ∫ (𝑟𝑇)𝑑𝑟 𝑅=∞ 𝑟1 We get 𝛼𝑟 𝜕𝑇 𝜕𝑟 − ℎ 𝑑𝜌𝐶 ∫ 𝑟(𝑇 − 𝑇∞) 𝑅=∞ 𝑟1 𝑑𝑟 = 𝜕 𝜕𝑡 ∫ (𝑟𝑇)𝑑𝑟 𝑅=∞ 𝑟1
- 50. 49 We divide through by r to get: 𝛼 𝜕𝑇 𝜕𝑟 − ℎ 𝑑𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑟 𝑅=∞ 𝑟1 = 𝜕 𝜕𝑡 ∫ 𝑇𝑑𝑟 𝑅=∞ 𝑟1 But 𝜕𝑇 𝜕𝑟 = ∫ ( 𝜕2 𝑇 𝜕𝑟2 )𝑑𝑟 𝑅 𝑟1 So, the PDE becomes: 𝜶 ∫ ( 𝝏𝟐 𝑻 𝝏𝒓𝟐 )𝒅𝒓 𝑹 𝒓𝟏 − 𝒉 𝒅𝝆𝑪 ∫ (𝑻 − 𝑻∞) 𝑹=∞ 𝒓𝟏 𝒅𝒓 = 𝝏 𝝏𝒕 ∫ (𝑻)𝒅𝒓 𝑹 𝒓𝟏 ℎ 𝑑𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑟 𝑅=∞ 𝑟1 = 𝛿ℎ 𝑑𝜌𝐶 (𝑇𝑠 − 𝑇∞) From the derivations above, we get: ∫ ( 𝜕2 𝑇 𝜕𝑟2 ) 𝑑𝑟 𝑅 𝑟1 = (𝑇𝑠 − 𝑇∞) 𝛿 𝜕 𝜕𝑡 ∫ 𝑇𝑑𝑟 𝑅 𝑟1 = 𝑑𝛿 𝑑𝑡 (𝑇𝑠 − 𝑇∞) Upon substitution of the above expressions in the integral equation, we get: 𝛼 𝛿 − ℎ 𝑑𝜌𝐶 𝛿 = 𝑑𝛿 𝑑𝑡 𝛼 𝛿 − ℎ 𝑑𝜌𝐶 𝛿 = 𝑑𝛿 𝑑𝑡 𝛼 − ℎ 𝑑𝜌𝐶 𝛿2 = 𝛿 𝑑𝛿 𝑑𝑡 The boundary conditions are: 𝛿 = 0 𝑎𝑡 𝑡 = 0 The solution of the equation above is 𝛿 = √ 𝛼𝑑𝜌𝐶 ℎ (1 − 𝑒 − 2ℎ𝑡 𝑑𝜌𝐶)
- 51. 50 𝛼 = 𝐾 𝜌𝐶 And get 𝛿 = √ 𝐾𝑑 ℎ (1 − 𝑒 − 2ℎ𝑡 𝑑𝜌𝐶) The temperature profile becomes: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 −( 𝑟−𝑟1 𝛿 ) Substitute for 𝛿 and get 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = 𝒆 −( 𝒓−𝒓𝟏 √𝑲𝒅 𝒉 (𝟏−𝒆 − 𝟐𝒉𝒕 𝒅𝝆𝑪) ) You notice that the initial condition is satisfied by the above temperature profile. What do we observe for short time in transient state? For short time, the exponential is small and it becomes: 𝑒 − 2ℎ𝑡 𝑑𝜌𝐶 = 1 − 2ℎ𝑡 𝑑𝜌𝐶 After using binomial expansion of the exponential in the above And (1 − 𝑒 − 2ℎ𝑡 𝑑𝜌𝐶) = 2ℎ𝑡 𝑑𝜌𝐶 𝛿 = √ 𝐾𝑑 ℎ (1 − 𝑒 − 2ℎ𝑡 𝑑𝜌𝐶) Becomes 𝛿 = √2𝛼𝑡 Where: 𝛼 = 𝐾 𝜌𝐶
- 52. 51 Let us make r the subject of the temperature profile above: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 −( 𝑟−𝑟1 𝛿 ) 𝑟 − 𝑟1 𝛿 = [ln ( 𝑇𝑠 − 𝑇∞ 𝑇 − 𝑇∞ )] (𝑟 − 𝑟1) = 𝛿 [ln ( 𝑇𝑠 − 𝑇∞ 𝑇 − 𝑇∞ )] Let us square both sides: (𝑟 − 𝑟1)2 = 𝛿2 ([ln ( 𝑇𝑠 − 𝑇∞ 𝑇 − 𝑇∞ )])2 Substitute for 𝛿 and get: 𝛿 = √2𝛼𝑡 (𝒓 − 𝒓𝟏)𝟐 = 𝟐𝜶([𝐥𝐧 ( 𝑻𝒔 − 𝑻∞ 𝑻 − 𝑻∞ )])𝟐 × 𝒕 That is what is observed. The above graph of (𝑟 − 𝑟1)2 against time t in seconds gives a straight-line graph. To get the above graph, we fix wax particles at distance (𝑟 − 𝑟1) on an infinite radius cylinder and note the time it takes them to melt and then plot a graph of (𝑟 − 𝑟1)2 against time t which is a straight-line graph We can include a time delay 𝑡0. From 2ℎ𝑡0 𝑑𝜌𝐶 = 𝑎 Where: 𝑎 = 0.02466 𝑡0 = 𝑎 𝑑𝜌𝐶 2ℎ (𝑟 − 𝑟1)2 = 2𝛼([ln ( 𝑇𝑠 − 𝑇∞ 𝑇 − 𝑇∞ )])2 × (𝑡 − 𝑡0)
- 53. 52 (𝑟 − 𝑟1)2 = 2𝛼([ln ( 𝑇𝑠 − 𝑇∞ 𝑇 − 𝑇∞ )])2 × (𝑡 − 𝑎 𝑑𝜌𝐶 2ℎ ) (𝑟 − 𝑟1)2 = 2𝛼([ln ( 𝑇𝑠 − 𝑇∞ 𝑇 − 𝑇∞ )])2 × 𝑡 − 2𝛼([ln ( 𝑇𝑠 − 𝑇∞ 𝑇 − 𝑇∞ )])2 × 𝑎𝑑𝜌𝐶 2ℎ (𝒓 − 𝒓𝟏)𝟐 = 𝟐𝜶([𝐥𝐧 ( 𝑻𝒔 − 𝑻∞ 𝑻 − 𝑻∞ )])𝟐 × 𝒕 − 𝑲([𝐥𝐧 ( 𝑻𝒔 − 𝑻∞ 𝑻 − 𝑻∞ )])𝟐 × 𝒂𝒅 𝒉 What the equation above says that if you fix wax particles at distance r and note the time taken for each to melt, then a graph of (𝑟 − 𝑟1)2 against t is a straight-line graph with an intercept as above. In using the equation above, it should be remembered that the expressions for temperature 𝑇𝑠 developed earlier on are still valid since this temperature is not necessarily the temperature at 𝑥 = 0.
- 54. 53 HOW DO WE DEAL WITH METAL RODS OF FINITE RADIUS CYLINDER? The boundary and initial conditions are 𝑻 = 𝑻𝒔 𝒂𝒕 𝒓 = 𝒓𝟏 −𝒌 𝒅𝑻 𝒅𝒓 = 𝒉(𝑻 − 𝑻∞) 𝒂𝒕 𝒓 = 𝒓𝟐 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 − (𝑟−𝑟1) 𝛿 (1− (𝑟−𝑟1) (𝑟2−𝑟1) ) [ (𝑟 − 𝑟1) (𝑟2 − 𝑟1) (𝑇1 − 𝑇∞) (𝑇𝑠 − 𝑇∞) + (1 − (𝑟 − 𝑟1) (𝑟2 − 𝑟1) )] Let 𝑟2 − 𝑟1 = 𝑙 𝑻 − 𝑻∞ = (𝑻𝒔 − 𝑻∞)𝒆 − (𝒓−𝒓𝟏) 𝜹 (𝟏− (𝒓−𝒓𝟏) 𝒍 ) [ (𝒓 − 𝒓𝟏) 𝒍 (𝑻𝟏 − 𝑻∞) (𝑻𝒔 − 𝑻∞) + (𝟏 − (𝒓 − 𝒓𝟏) 𝒍 )] Which says 𝑇 = 𝑇𝑠 𝑎𝑡 𝑟 = 𝑟1 𝑇 = 𝑇1 𝑎𝑡 𝑟 = 𝑟2 𝑇 = 𝑇∞ 𝑎𝑡 𝑡 = 0 Provided 𝛿 = 0 𝑎𝑡 𝑡 = 0 , then the initial condition above is satisfied we take the derivative 𝒅𝑻 𝒅𝒓 𝒂𝒕 𝒓 = 𝒓𝟐 and equate it to − ℎ 𝑘 (𝑇1 − 𝑇∞) and get: 𝑑𝑇 𝑑𝑟 |𝑟=𝑟2 = ( (𝑇1 − 𝑇∞) 𝑙 − (𝑇𝑠 − 𝑇∞) 𝑙 + (𝑇1 − 𝑇∞) 𝛿 ) 𝑑𝑇 𝑑𝑟 |𝑟=𝑟2 = − ℎ 𝑘 (𝑇1 − 𝑇∞) We equate the two and get
- 55. 54 ( (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 − (𝒓−𝒓𝟏) 𝜹 (𝟏 − (𝒓−𝒓𝟏) 𝒍 ) [ (𝒓 − 𝒓𝟏) 𝒍 ( 𝜹𝒌 𝜹𝒌 + 𝒍𝒌 + 𝒉𝒍𝜹 ) + (𝟏 − (𝒓 − 𝒓𝟏) 𝒍 )] This the temperature profile that satisfies the boundary and initial conditions below: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒓 = 𝒓𝟏 −𝒌 𝒅𝑻 𝒅𝒓 = 𝒉(𝑻 − 𝑻∞) 𝒂𝒕 𝒓 = 𝒓𝟐 𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 Let us go ahead and solve for 𝛿 The governing equation is 𝜶 𝒓 𝝏 𝝏𝒓 (𝒓 𝝏𝑻 𝝏𝒓 ) − 𝒉𝑷 𝑨𝝆𝑪 (𝑻 − 𝑻∞) = 𝝏𝑻 𝝏𝒕 We multiply through by r to get: 𝜶 𝝏 𝝏𝒓 (𝒓 𝝏𝑻 𝝏𝒓 ) − 𝒉𝑷 𝑨𝝆𝑪 𝒓(𝑻 − 𝑻∞) = 𝒓 𝝏𝑻 𝝏𝒕 We then transform the PDE into an integral equation 𝛼 ∫ (𝛼 𝜕 𝜕𝑟 (𝑟 𝜕𝑇 𝜕𝑟 )) 𝑑𝑟 𝑟2 𝑟1 − ℎ 𝑑𝜌𝐶 ∫ 𝑟(𝑇 − 𝑇∞) 𝑟2 𝑟1 𝑑𝑟 = 𝜕 𝜕𝑡 ∫ (𝑟𝑇)𝑑𝑟 𝑟2 𝑟1 And get 𝛼𝑟 𝜕𝑇 𝜕𝑟 − ℎ 𝑑𝜌𝐶 ∫ 𝑟(𝑇 − 𝑇∞) 𝑟2 𝑟1 𝑑𝑟 = 𝑟 𝜕 𝜕𝑡 ∫ 𝑇𝑑𝑟 𝑟2 𝑟1 We then divide through by r to get
- 56. 55 𝛼 𝜕𝑇 𝜕𝑟 − ℎ 𝑑𝜌𝐶 ∫ (𝑇 − 𝑇∞) 𝑟2 𝑟1 𝑑𝑟 = 𝜕 𝜕𝑡 ∫ 𝑇𝑑𝑟 𝑟2 𝑟1 But 𝜕𝑇 𝜕𝑟 = ∫ ( 𝜕2 𝑇 𝜕𝑟2 )𝑑𝑟 𝑟2 𝑟1 So, the PDE becomes: 𝜶 ∫ ( 𝝏𝟐 𝑻 𝝏𝒓𝟐 )𝒅𝒓 𝒓𝟐 𝒓𝟏 − 𝒉 𝒅𝝆𝑪 ∫ (𝑻 − 𝑻∞) 𝒓𝟐 𝒓𝟏 𝒅𝒓 = 𝝏 𝝏𝒕 ∫ (𝑻)𝒅𝒓 𝒓𝟐 𝒓𝟏 To make the following calculations not tedious, we can take 𝑟1 to be at the origin i.e., 𝑟1 = 0 and get (𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆− 𝒓 𝜹 (𝟏 − 𝒓 𝒍 ) [ 𝒓 𝒍 ( 𝜹𝒌 𝜹𝒌 + 𝒍𝒌 + 𝒉𝒍𝜹 ) + (𝟏 − 𝒓 𝒍 )] The integral equation becomes: 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑟2 ) 𝑑𝑟 𝑙 0 − ℎ 𝑑𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑟 𝑙 0 = 𝜕 𝜕𝑡 ∫ 𝑇𝑑𝑟 𝑙 0 𝜕𝑇 𝜕𝑟 = (𝑇𝑠 − 𝑇∞)𝑒− 𝑟 𝛿 (1− 𝑟 𝑙 ) [ 1 𝑙 (𝑇1 − 𝑇∞) (𝑇𝑠 − 𝑇∞) − 1 𝑙 ] + (𝑇𝑠 − 𝑇∞)𝑒− 𝑟(1− 𝑟 𝑙 ) [ 𝑟 𝑙 (𝑇1 − 𝑇∞) (𝑇𝑠 − 𝑇∞) + (1 − 𝑟 𝑙 )] ( −𝑙 + 2𝑟 𝛿𝑙 ) ∫ ( 𝜕2 𝑇 𝜕𝑟2 ) 𝑑𝑟 𝑙 0 = [ 𝜕𝑇 𝜕𝑟 ] 𝑙 0 = (𝑇𝑠 − 𝑇∞) 𝛿 + (𝑇1 − 𝑇∞) 𝛿 Substitute for (𝑇1 − 𝑇∞) = (𝑇𝑠 − 𝑇∞)( 𝛿𝑘 𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 ) ∫ ( 𝜕2 𝑇 𝜕𝑟2 ) 𝑑𝑟 𝑙 0 = [ 𝜕𝑇 𝜕𝑟 ] 𝑙 0 = (𝑇𝑠 − 𝑇∞) 𝛿 ( 2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 (𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿) ) ℎ 𝑑𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑟 𝑙 0 = ℎ 𝑑𝜌𝐶 ∫ ((𝑇𝑠 − 𝑇∞)𝑒− 𝑟 𝛿 (1− 𝑟 𝑙 ) [ 𝑟 𝑙 (𝑇1 − 𝑇∞) (𝑇𝑠 − 𝑇∞) + (1 − 𝑟 𝑙 )])𝑑𝑟 𝑙 0
- 57. 56 ∫ (𝑇 − 𝑇∞)𝑑𝑟 𝑙 0 = [ (𝑇1 − 𝑇∞) 𝑙 ∫ 𝑟𝑒− 𝑟 𝛿 (1− 𝑟 𝑙 ) 𝑑𝑟 𝑙 0 + (𝑇𝑠 − 𝑇∞) ∫ 𝑒− 𝑟 𝛿 (1− 𝑟 𝑙 ) 𝑑𝑟 𝑙 0 − (𝑇𝑠 − 𝑇∞) 𝑙 ∫ 𝑟𝑒− 𝑟 𝛿 (1− 𝑟 𝑙 ) 𝑑𝑥 𝑙 0 ] Integrating by parts shows that ∫ 𝑒 − 𝑟 𝛿 (1− 𝑟 𝑙 ) 𝑑𝑟 𝑙 0 = [( 𝑙𝛿 −𝑙 + 2𝑟 )𝑒 − 𝑟 𝛿 (1− 𝑟 𝑙 ) ] 𝑙 0 = 2𝛿 ∫ 𝑟𝑒 − 𝑟 𝛿 (1− 𝑟 𝑙 ) 𝑑𝑟 𝑙 0 = [( 𝑥𝑙𝛿 −𝑙 + 2𝑟 − 𝑙2 𝛿2 (−𝑙 + 2𝑟)2[1 − 2𝑙𝛿 (−𝑙 + 2𝑟)2] ) 𝑒 − 𝑟 𝛿 (1− 𝑟 𝑙 ) ] 𝑙 0 = 𝑙𝛿 Substituting back into the heat loss equation we get ℎ 𝑑𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑟 𝑙 0 = ℎ 𝑑𝜌𝐶 [(𝑇1 − 𝑇∞) + (𝑇𝑠 − 𝑇∞)]𝛿 substitute for (𝑇1 − 𝑇∞) and get ℎ 𝑑𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑟 𝑙 0 = ℎ 𝑑𝜌𝐶 (𝑇𝑠 − 𝑇∞)( 2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 )𝛿 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑟 𝑙 0 = (𝑇𝑠 − 𝑇∞) ( 2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 ) 𝑑𝛿 𝑑𝑡 + 𝜕(𝑙𝑇∞) 𝜕𝑡 𝜕(𝑙𝑇∞) 𝜕𝑡 = 0 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑟 𝑙 0 = (𝑇𝑠 − 𝑇∞) ( 2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 (𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿) ) 𝑑𝛿 𝑑𝑡 Substituting into the integral equation we get 𝛼 (𝑇𝑠 − 𝑇∞) 𝛿 ( 2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 (𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿) ) − ℎ 𝑑𝜌𝐶 (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
- 58. 57 And get 𝛿 = √ 𝐾𝑑 ℎ (1 − 𝑒 − 2ℎ𝑡 𝑑𝜌𝐶) We go ahead and substitute for 𝛿 in the temperature profile below (𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆 − (𝒓−𝒓𝟏) 𝜹 (𝟏 − (𝒓−𝒓𝟏) 𝒍 ) [ (𝒓 − 𝒓𝟏) 𝒍 ( 𝜹𝒌 𝜹𝒌 + 𝒍𝒌 + 𝒉𝒍𝜹 ) + (𝟏 − (𝒓 − 𝒓𝟏) 𝒍 )] When the time is small, 𝛿 using binomial approximation becomes 𝛿 = √ 𝐾𝑑 ℎ (1 − 𝑒 − 2ℎ𝑡 𝑑𝜌𝐶) When the time is small, 𝛿 using binomial approximation becomes 𝛿 = √ 𝐾𝑑 ℎ (1 − 𝑒 − 2ℎ𝑡 𝑑𝜌𝐶) 2ℎ𝑡 𝑑𝜌𝐶 ≪ 1 𝑒 − 2ℎ𝑡 𝑑𝜌𝐶 = 1 − 2ℎ𝑡 𝑑𝜌𝐶 𝑡 1 − 𝑒 − 2ℎ𝑡 𝑑𝜌𝐶 = 2ℎ𝑡 𝑑𝜌𝐶 𝑡 𝛿 = √2𝛼𝑡 We substitute for 𝛿 in the temperature profile. (𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆 − (𝒓−𝒓𝟏) √𝟐𝜶𝒕 (𝟏 − (𝒓−𝒓𝟏) 𝒍 ) [ (𝒓 − 𝒓𝟏) 𝒍 ( 𝑲√𝟐𝜶𝒕 √𝟐𝜶𝒕𝒌 + 𝒍𝒌 + 𝒉𝒍√𝟐𝜶𝒕 ) + (𝟏 − (𝒓 − 𝒓𝟏) 𝒍 )]
- 59. 58 HOW DO WE DEAL WITH RECTANGULAR CO- ORDINATES IN NATURAL CONVECTION? Consider a semi-infinite rectangular slab The governing equation is: 𝛼 𝜕2 𝑇 𝜕𝑥2 − ℎ𝑃 𝐴𝜌𝐶 (𝑇 − 𝑇∞) = 𝜕𝑇 𝜕𝑡 Where: 𝑃 = 2(𝑤 + 𝑑) 𝐴 = 𝑤 × 𝑑 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
- 60. 59 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 ∫ (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∞)
- 61. 60 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 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: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 −𝑥 √𝐾𝐴 ℎ𝑃 (1−𝑒 −2ℎ𝑃 𝐴𝜌𝐶 𝑡 )
- 62. 61 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: 2ℎ𝑃 𝐴𝜌𝐶 𝑡0 = 𝑎 = 0.02466 𝑡0 = 𝑎 𝐴𝜌𝐶 2ℎ𝑃 And 𝑎 = 0.02466 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
- 63. 62 𝒙𝟐 = 𝟐𝜶[𝐥𝐧 ( 𝑻𝒔 − 𝑻∞ 𝑻 − 𝑻∞ )]𝟐 × 𝒕 − 𝟐𝜶[𝐥𝐧 ( 𝑻𝒔 − 𝑻∞ 𝑻 − 𝑻∞ )]𝟐 × 𝒂( 𝑨𝝆𝑪 𝟐𝒉𝑷 ) Where: 𝛼 = 𝐾 𝜌𝐶 The equation becomes: 𝒙𝟐 = 𝟐𝜶[𝐥𝐧 ( 𝑻𝒔 − 𝑻∞ 𝑻 − 𝑻∞ )]𝟐 × 𝒕 − 𝒂𝑲𝑨 𝒉𝑷 [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. We follow all the rules above derived for cylindrical metal rods. 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. How do we deal with rectangular metal rods of finite length 𝒍 ? 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 )]
- 64. 63 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 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
- 65. 64 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) 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 − ℎ𝑃 𝐴𝜌𝐶 ∫ (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∞) 𝛿 ( 2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 (𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿) ) ℎ𝑃 𝐴𝜌𝐶 ∫ (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 ℎ𝑃 𝐴𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = ℎ𝑃 𝐴𝜌𝐶 [ (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
- 66. 65 Substituting back into the heat loss equation we get ℎ𝑃 𝐴𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = ℎ𝑃 𝐴𝜌𝐶 [(𝑇1 − 𝑇∞) + (𝑇𝑠 − 𝑇∞)]𝛿 substitute for (𝑇1 − 𝑇∞) and get ℎ𝑃 𝐴𝜌𝐶 ∫ (𝑇 − 𝑇∞)𝑑𝑥 𝑙 0 = ℎ𝑃 𝐴𝜌𝐶 (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∞) 𝛿 ( 2𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿 (𝛿𝑘 + 𝑙𝑘 + ℎ𝑙𝛿) ) − ℎ𝑃 𝐴𝜌𝐶 (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
- 67. 66 (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 ) 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?
- 68. 67 (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 (𝑻 − 𝑻∞) = (𝑻𝒔 − 𝑻∞)𝒆 − 𝒙 √𝑲𝑨 𝒉𝑷 (𝟏−𝒆 −𝟐𝒉𝑷 𝑨𝝆𝑪 𝒕 ) Which is what we got before.
- 69. 68 LET US ANALYSE THE RATE OF COOLING IN NATURAL CONVECTION. Consider a liquid in a cube cooling by natural convection. The heat flow rate at one side of the cube is given by: 𝑄̇|=𝑙 = ℎ𝐴(𝑇𝑠 − 𝑇∞)( 𝛿𝐾 𝛿𝐾 + 𝑙𝐾 + ℎ𝑙𝛿 ) Where: 𝑙 = 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑢𝑏𝑒 The total heat flow rate from all the six faces is 𝑄̇ = 6ℎ𝐴(𝑇𝑠 − 𝑇∞)( 𝛿𝐾 𝛿𝐾 + 𝑙𝐾 + ℎ𝑙𝛿 ) When we introduce a hot liquid in a container, heat will be conducted to through all the sides of the container until steady state is reached and 𝛿 becomes 𝛿(𝑡) = √ 𝐾𝐴 ℎ𝑃 (1 − 𝑒 −2ℎ𝑃 𝐴𝜌𝐶 𝑡 ) 𝛿 = √ 𝐾𝐴 ℎ𝑃 So, the total heat flow rate from all the sides is given by: 𝑄̇ = ℎ𝐴𝑇(𝑇𝑠 − 𝑇∞)( 𝛿𝐾 𝛿𝐾 + 𝑙𝐾 + ℎ𝑙𝛿 ) Substituting for 𝛿 , we get: 𝑄̇ = ℎ𝐴𝑇(𝑇𝑠 − 𝑇∞)( 𝐾√𝐾𝐴 ℎ𝑃 𝐾√𝐾𝐴 ℎ𝑃 + 𝑙𝐾 + ℎ𝑙√𝐾𝐴 ℎ𝑃 ) Where: 𝐴𝑇 = 6𝐴 To get the rate of change of temperature with time, we say:
- 70. 69 𝒎𝟏𝑪𝟏 𝒅𝑻 𝒅𝒕 = −𝒉𝑨𝑻(𝑻 − 𝑻∞)( 𝑲√𝑲𝑨 𝒉𝑷 𝑲√𝑲𝑨 𝒉𝑷 + 𝒍𝑲 + 𝒉𝒍√𝑲𝑨 𝒉𝑷 ) Or 𝑚1𝐶1 𝑑𝑇 𝑑𝑡 = −𝛾ℎ𝐴𝑇(𝑇 − 𝑇∞) Where: 𝛾 = 𝐾√𝐾𝐴 ℎ𝑃 𝐾√𝐾𝐴 ℎ𝑃 + 𝑙𝐾 + ℎ𝑙√𝐾𝐴 ℎ𝑃 𝛾 explains how the rate of cooling depends on the thermal conductivity and thickness and dimensions of the container of cooling body and how they affect the rate of cooling. We can extend the analysis above to cylinders holding hot liquids and get an expression for 𝛾 using the equations derived before for cylindrical coordinates.
- 71. 70 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. Using this integral analytical method, we can also go ahead and solve PDES of the form below:
- 72. 71 𝜕𝑇 𝜕𝑡 = 𝛼 𝜕2 𝑇 𝜕𝑥2 + 𝑓(𝑥) Final note: We can extend the above analysis to cylindrical co-ordinates too.