SlideShare a Scribd company logo

TRANSIENT AND STEADY STATE HEAT CONDUCTION WITH NO LATERAL CONVECTION SOLVED USING THE INTEGRAL EQUATION TRANSFORM.pdf

W
Wasswaderrick3

The document discusses transient and steady state heat conduction in one-dimensional rectangular coordinates, detailing the governing heat equation and integration methods for solving temperature variations. It presents findings on temperature profiles under different boundary conditions and initial conditions, exploring both direct and integral transform methods to derive exact solutions. The final solutions include models for heat flow in extended surfaces and metal rods with lateral convection, showcasing how temperature profiles evolve over time.

Science
1 of 16
Download to read offline
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
TRANSIENT STATE AND STEADY STATE HEAT CONDUCTION WITH NO LATERAL CONVECTION By Wasswa Derrick Wasswaderricktimothy7@gmail.com ONE DIMENSIONAL HEAT CONDUCTION RECTANGULAR CO-ORDINATES. The governing equation is: 𝛼 𝜕2 𝑇 𝜕𝑥2 = 𝜕𝑇 𝜕𝑡 To get the steady state temperature, we simply substitute 𝜕𝑇 𝜕𝑡 = 0 𝑖𝑛 𝑠𝑡𝑒𝑎𝑦 𝑠𝑡𝑎𝑡𝑒 And get: 𝜕2 𝑇 𝜕𝑥2 = 0 We integrate the above equation to get the variation of temperature with distance x i.e., 𝜕𝑇 𝜕𝑥 = 𝑐1 𝑻 = 𝒄𝟏𝒙 + 𝒄𝟐 The condition above that the temperature is linear in distance x will help us to solve for the transient state. At 𝑥 = 𝑥1, 𝑇 = 𝑇1 ∴ 𝑇1 = 𝑐1𝑥1 + 𝑐2 … (𝐴) At 𝑥 = 𝑥2, 𝑇 = 𝑇2 ∴ 𝑇2 = 𝑐1𝑥2 + 𝑐2 … . (𝐵) (𝐴 − 𝐵) 𝑙𝑒𝑎𝑑𝑠 𝑡𝑜 𝑐1 = 𝑇1−𝑇2 𝑥1−𝑥2 From A 𝑐2 = 𝑇1 − 𝑐1𝑥1 = 𝑇1 − 𝑇1−𝑇2 𝑥1−𝑥2 𝑥1 Substituting in the general solution, we get
𝑇 = 𝑐1𝑥 + 𝑐2 𝑇 = 𝑇1 − 𝑇2 𝑥1 − 𝑥2 𝑥 + 𝑇1 − 𝑇1 − 𝑇2 𝑥1 − 𝑥2 𝑥1 𝑇 − 𝑇1 = 𝑇1 − 𝑇2 𝑥1 − 𝑥2 (𝑥 − 𝑥1) Since 𝑥2 > 𝑥1 as shown in the diagram above, we can rearrange and get 𝑇 − 𝑇1 = −( 𝑇1 − 𝑇2 𝑥2 − 𝑥1 )(𝑥 − 𝑥1) From 𝑄 = −𝑘𝐴 𝜕𝑇 𝜕𝑥 𝜕𝑇 𝜕𝑥 = −( 𝑇1 − 𝑇2 𝑥2 − 𝑥1 ) Upon substitution, we get: 𝑄 = 𝑘𝐴( 𝑇1 − 𝑇2 𝑥2 − 𝑥1 ) 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: 𝜶 𝝏𝟐 𝑻 𝝏𝒙𝟐 = 𝝏𝑻 𝝏𝒕 We are going to deal with a variety of boundary conditions and initial condition.
THE INTEGRAL 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 𝜕2 𝑇 𝜕𝑥2 = 𝜕𝑇 𝜕𝑡 Let us change this equation into an integral equation as below: 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 Postulate 1: Any temperature profile of the form 𝑇 = 𝐵𝑒 𝑎𝑥 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 𝑎 = 1 and 𝐵 = 𝑇0 𝑇 = 𝑇0𝑒 𝑥 𝛿 𝜕2 𝑇 𝜕𝑥2 = 𝑇0 𝛿2 𝑒 𝑥 𝛿 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = 𝑇0 𝛿 [𝑒 𝐿 𝛿 − 1] What we have learned from the above integral is that ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = [ 𝜕𝑇 𝜕𝑥 ] 𝐿 0 Which we shall use in later text. Anyway, to continue, let us evaluate ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝛿[𝑒 𝐿 𝛿 − 1]
Upon substitution in the heat equation, we get: 𝛼 𝑇0 𝛿 [𝑒 𝐿 𝛿 − 1] = 𝜕 𝜕𝑡 [𝑇0𝛿[𝑒 𝐿 𝛿 − 1]] There’s a common term 𝑇0[𝑒 𝐿 𝛿 − 1] which we drop out and get: 𝛼 𝛿 = 𝑑𝛿 𝑑𝑡 We finally get 𝛿 = √2𝛼𝑡 + 𝑐 We finally substitute in the temperature profile to get: 𝑇 = 𝑇0𝑒 𝑥 √2𝛼𝑡+𝑐 Where c and 𝑇0are evaluated using the boundary and initial conditions. It can be shown that hyperbolic temperature functions like ( 𝑠𝑖𝑛ℎ𝑎𝑥 𝑜𝑟 𝑐𝑜𝑠ℎ𝑎𝑥) 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 𝑥 𝛿 + 𝐵𝑒 −𝑥 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 𝑚 = 1 𝛿 Where 𝛿(𝑡) will be got by solving the heat equation. The table below will give the required temperature profiles for given boundary conditions [1]
TRANSIENT AND STEADY STATE HEAT CONDUCTION WITH NO LATERAL CONVECTION SOLVED USING THE INTEGRAL EQUATION TRANSFORM.pdf
CASE 1: FIXED END TEMPERATURE The initial and boundary conditions are: 𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 𝑻 = 𝑻𝒃𝟏 − 𝑻∞ 𝒂𝒕 𝒙 = 𝟎 𝑻 = 𝑻𝒃𝟐 − 𝑻∞ 𝒂𝒕 𝒙 = 𝑳 We start with a temperature profile below: 𝑇 − 𝑇∞ = 𝐴𝑒 𝑥 𝛿 + 𝐵𝑒 −𝑥 𝛿 We evaluate the constants A and B using the boundary conditions and get the temperature profile below: 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) 𝑠𝑖𝑛ℎ 𝑥 𝛿 + 𝑠𝑖𝑛ℎ 𝐿 − 𝑥 𝛿 𝑠𝑖𝑛ℎ 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 𝑎𝑡 𝑡 = 0, 𝛿 = 0 and get: 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) 𝑠𝑖𝑛ℎ 𝑥 𝛿 + 𝑠𝑖𝑛ℎ 𝐿 − 𝑥 𝛿 𝑠𝑖𝑛ℎ 𝐿 𝛿 Let us show that the above postulate implies the initial condition 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) 𝑠𝑖𝑛ℎ 𝑥 𝛿 + 𝑠𝑖𝑛ℎ 𝐿 − 𝑥 𝛿 𝑠𝑖𝑛ℎ 𝐿 𝛿 = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ )(𝑒 𝑥 𝛿 + 𝑒 −𝑥 𝛿 ) + (𝑒 𝐿−𝑥 𝛿 + 𝑒 −(𝐿−𝑥) 𝛿 ) 𝑒 𝐿 𝛿 − 𝑒 −𝐿 𝛿 𝑒 −(𝐿−𝑥) 𝛿 = 𝑒− (𝐿−𝑥) 0 = 𝑒−∞(𝐿−𝑥) = 0 Similarly 𝑒 −𝐿 𝛿 = 𝑒 −𝐿 0 = 𝑒−∞𝐿 = 0 So, we are left with
𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) (𝑒 𝑥 𝛿) + (𝑒 𝐿−𝑥 𝛿 ) 𝑒 𝐿 𝛿 = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ )𝑒 −(𝐿−𝑥) 𝛿 + 𝑒 −𝑥 𝛿 = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ )𝑒 −(𝐿−𝑥) 0 + 𝑒 −𝑥 0 = 𝑒−∞𝑥 We finally get 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ )𝑒−∞(𝐿−𝑥) + 𝑒−∞𝑥 = 0 Since 𝐿 − 𝑥 > 0 Hence at 𝑡 = 0, 𝑇 = 𝑇∞ and hence the initial condition. Let us solve the heat equation analytically using the temperature profile above: 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = [ 𝜕𝑇 𝜕𝑥 ] 𝐿 0 [ 𝜕𝑇 𝜕𝑥 ] 𝐿 0 = (𝑇𝑏1 − 𝑇∞) 𝛿 [( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) coth ( 𝐿 𝛿 ) − ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1)𝑐𝑜𝑠𝑒𝑐ℎ ( 𝐿 𝛿 )] To evaluate ∫ (𝑇)𝑑𝑥 𝑙 0 we need to know T 𝑇 = (𝑇𝑏1 − 𝑇∞) [ ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) 𝑠𝑖𝑛ℎ 𝑥 𝛿 + 𝑠𝑖𝑛ℎ 𝐿 − 𝑥 𝛿 𝑠𝑖𝑛ℎ 𝐿 𝛿 ] + 𝑇∞ ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝛿(𝑇𝑏1 − 𝑇∞) [( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1)coth ( 𝐿 𝛿 ) − ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) 𝑐𝑜𝑠𝑒𝑐ℎ ( 𝐿 𝛿 )] + 𝑇∞𝐿 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝑑 𝑑𝑡 (𝛿(𝑇𝑏1 − 𝑇∞) [( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) coth ( 𝐿 𝛿 ) − ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) 𝑐𝑜𝑠𝑒𝑐ℎ ( 𝐿 𝛿 )]) + 𝑑(𝑇∞𝐿) 𝑑𝑡 But 𝑑(𝑇∞𝐿) 𝑑𝑡 = 0 𝑠𝑖𝑛𝑐𝑒 𝐿 𝑎𝑛𝑑 𝑇∞ 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠 So, we have
𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝑑 𝑑𝑡 (𝛿(𝑇𝑏1 − 𝑇∞) [( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) coth ( 𝐿 𝛿 ) − ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) 𝑐𝑜𝑠𝑒𝑐ℎ ( 𝐿 𝛿 )]) Upon substitution in the heat equation, we get: 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 𝛼 𝛿 (𝑇𝑏1 − 𝑇∞) [( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) coth ( 𝐿 𝛿 ) − ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) 𝑐𝑜𝑠𝑒𝑐ℎ ( 𝐿 𝛿 )] = 𝑑 𝑑𝑡 (𝛿(𝑇𝑏1 − 𝑇∞) [( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) coth ( 𝐿 𝛿 ) − ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) 𝑐𝑜𝑠𝑒𝑐ℎ ( 𝐿 𝛿 )]) The factor (𝑇𝑏1 − 𝑇∞) [( 𝑇𝑏2 −𝑇∞ 𝑇𝑏1 −𝑇∞ + 1) coth ( 𝐿 𝛿 ) − ( 𝑇𝑏2 −𝑇∞ 𝑇𝑏1 −𝑇∞ + 1) 𝑐𝑜𝑠𝑒𝑐ℎ ( 𝐿 𝛿 )] is common on both sides of the equation so we cross it out and get 𝛼 𝛿 = 𝑑𝛿 𝑑𝑡 Where the boundary conditions are: 𝛿 = 0 𝑎𝑡 𝑡 = 0 On solving the above we get: 𝛿 = √2𝛼𝑡 Upon substitution in the temperature profile, we get 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) 𝑠𝑖𝑛ℎ 𝑥 𝛿 + 𝑠𝑖𝑛ℎ 𝐿 − 𝑥 𝛿 𝑠𝑖𝑛ℎ 𝐿 𝛿 OR 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) 𝑠𝑖𝑛ℎ 𝑥 √2𝛼𝑡 + 𝑠𝑖𝑛ℎ 𝐿 − 𝑥 √2𝛼𝑡 𝑠𝑖𝑛ℎ 𝐿 √2𝛼𝑡 In steady state, 𝑡 → ∞ 𝑠𝑜 𝛿 → ∞ Upon substitution we get
𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ )𝑠𝑖𝑛ℎ 𝑥 ∞ + 𝑠𝑖𝑛ℎ 𝐿 − 𝑥 ∞ 𝑠𝑖𝑛ℎ 𝐿 ∞ 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) 𝑠𝑖𝑛ℎ0 + 𝑠𝑖𝑛ℎ0 𝑠𝑖𝑛ℎ0 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = 0 0 L’hopital’s rule is then invoked i.e., 𝑙𝑖𝑚 𝑥 → 𝑐 𝑓(𝑥) 𝑔(𝑥) = 𝑙𝑖𝑚 𝑥 → 𝑐 𝑓′ (𝑥) 𝑔′(𝑥) We differentiate the numerate and denominator with respect to 1 𝛿 since it is the one varying and we get 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = 𝑥 𝛿 ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ )𝑐𝑜𝑠ℎ 𝑥 𝛿 + ( 𝐿 − 𝑥 𝛿 )𝑐𝑜𝑠ℎ 𝐿 − 𝑥 𝛿 𝐿 𝛿 𝑐𝑜𝑠ℎℎ 𝐿 𝛿 Upon simplification, we get 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = 𝑥 ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) 𝑐𝑜𝑠ℎ 𝑥 𝛿 + (𝐿 − 𝑥)𝑐𝑜𝑠ℎ 𝐿 − 𝑥 𝛿 𝐿𝑐𝑜𝑠ℎℎ 𝐿 𝛿 We then substitute 𝛿 = √2𝛼𝑡 = ∞ And get 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = 𝑥 ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) + (𝐿 − 𝑥) 𝐿 Upon substitution, we get: 𝑻 − 𝑻∞ 𝑻𝒃𝟏 − 𝑻∞ = 𝒙 𝑳 ( 𝑻𝒃𝟐 − 𝑻∞ 𝑻𝒃𝟏 − 𝑻∞ ) + (𝟏 − 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 −𝒌 𝒅𝑻 𝒅𝒙 = 𝒉𝑳(𝑻 − 𝑻∞) 𝒂𝒕 𝒙 = 𝒍 𝑻 = 𝑻∞ 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 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = cosh[𝑚(𝐿 − 𝑥)] + ( ℎ𝐿 𝑚𝑘 ) sinh[𝑚(𝐿 − 𝑥)] cosh 𝑚𝐿 + ( ℎ𝐿 𝑚𝑘 ) 𝑠𝑖𝑛ℎ𝑚𝐿 Or 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = cosh [ (𝐿 − 𝑥) 𝛿 ] + ( ℎ𝐿𝛿 𝑘 ) sinh[( 𝐿 − 𝑥 𝛿 )] cosh 𝐿 𝛿 + ( ℎ𝐿𝛿 𝑘 ) 𝑠𝑖𝑛ℎ 𝐿 𝛿 To show that the initial condition is satisfied we assume from the above that 𝑎𝑡 𝑡 = 0, 𝛿 = 0. 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = cosh [ (𝐿 − 𝑥) 𝛿 ] + ( ℎ𝐿𝛿 𝑘 ) sinh[( 𝐿 − 𝑥 𝛿 )] cosh 𝐿 𝛿 + ( ℎ𝛿 𝑘 )𝑠𝑖𝑛ℎ 𝐿 𝛿 Becomes: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = cosh [ (𝐿 − 𝑥) 𝛿 ] cosh 𝐿 𝛿 = 𝑒 (𝐿−𝑥) 𝛿 + 𝑒 −(𝐿−𝑥) 𝛿 𝑒 𝐿 𝛿 + 𝑒 −𝐿 𝛿 𝑒 −(𝐿−𝑥) 𝛿 = 𝑒− (𝐿−𝑥) 0 = 𝑒−∞(𝐿−𝑥) = 0 Similarly 𝑒 −𝐿 𝛿 = 𝑒 −𝐿 0 = 𝑒−∞𝐿 = 0 So, we are left with
𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 (𝐿−𝑥) 𝛿 𝑒 𝐿 𝛿 = 𝑒 −𝑥 𝛿 = 𝑒 −𝑥 0 = 𝑒−∞𝑥 = 0 Hence at 𝑡 = 0, 𝑇 = 𝑇∞ 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 −𝒌 𝒅𝑻 𝒅𝒙 = 𝒉𝑳(𝑻 − 𝑻∞) 𝒂𝒕 𝒙 = 𝒍 𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 The governing temperature profile is: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = cosh [ (𝐿 − 𝑥) 𝛿 ] + ( ℎ𝐿𝛿 𝑘 ) sinh[( 𝐿 − 𝑥 𝛿 )] cosh 𝐿 𝛿 + ( ℎ𝐿𝛿 𝑘 ) 𝑠𝑖𝑛ℎ 𝐿 𝛿 The governing equation is 𝛼 𝜕2 𝑇 𝜕𝑥2 = 𝜕𝑇 𝜕𝑡 Let us change this equation into an integral equation as below: 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 … … . . 𝑏) 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 ∫ ( 𝜕2 𝑇 𝜕𝑥2 )𝑑𝑥 𝑙 0 = [ 𝜕𝑇 𝜕𝑥 ] 𝑙 0 = (𝑇𝑠 − 𝑇∞) 𝛿 ( − ℎ𝐿𝛿 𝑘 + (𝑠𝑖𝑛ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑐𝑜𝑠ℎ 𝐿 𝛿 ) 𝑐𝑜𝑠ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑠𝑖𝑛ℎ 𝐿 𝛿 ) 𝑇 = cosh [ (𝐿 − 𝑥) 𝛿 ] + ( ℎ𝐿𝛿 𝑘 )sinh[( 𝐿 − 𝑥 𝛿 )] cosh 𝐿 𝛿 + ( ℎ𝐿𝛿 𝑘 )𝑠𝑖𝑛ℎ 𝐿 𝛿 (𝑇𝑠 − 𝑇∞) + 𝑇∞
𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 [𝛿(𝑇𝑠 − 𝑇∞) ( − ℎ𝐿𝛿 𝑘 + (𝑠𝑖𝑛ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑐𝑜𝑠ℎ 𝐿 𝛿 ) 𝑐𝑜𝑠ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑠𝑖𝑛ℎ 𝐿 𝛿 )] + 𝜕(𝑙𝑇∞) 𝜕𝑡 𝜕(𝑙𝑇∞) 𝜕𝑡 = 0 Upon substitution of all the above in the heat equation, we get: 𝛼 (𝑇𝑠 − 𝑇∞) 𝛿 ( − ℎ𝐿𝛿 𝑘 + (𝑠𝑖𝑛ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑐𝑜𝑠ℎ 𝐿 𝛿 ) 𝑐𝑜𝑠ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑠𝑖𝑛ℎ 𝐿 𝛿 ) = 𝜕 𝜕𝑡 [𝛿(𝑇𝑠 − 𝑇∞) ( − ℎ𝐿𝛿 𝑘 + (𝑠𝑖𝑛ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑐𝑜𝑠ℎ 𝐿 𝛿 ) 𝑐𝑜𝑠ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑠𝑖𝑛ℎ 𝐿 𝛿 )] We notice that the term(𝑇𝑠 − 𝑇∞) ( − ℎ𝐿𝛿 𝑘 +(𝑠𝑖𝑛ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑐𝑜𝑠ℎ 𝐿 𝛿 ) 𝑐𝑜𝑠ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑠𝑖𝑛ℎ 𝐿 𝛿 ) is common and can be eliminated: 𝛼 𝛿 = 𝑑𝛿 𝑑𝑡 We go ahead and solve for 𝛿 provided 𝛿 = 0𝑎𝑡 𝑡 = 0 and get the expression 𝛿 = √2𝛼𝑡 So, the final solution for the finite metal rod with convective flux at the end of the metal rod is: 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = ( 𝐜𝐨𝐬𝐡 [ (𝑳 − 𝒙) 𝜹 ] + ( 𝒉𝑳𝜹 𝒌 ) 𝐬𝐢𝐧𝐡 [( 𝑳 − 𝒙 𝜹 )] 𝐜𝐨𝐬𝐡 𝑳 𝜹 + ( 𝒉𝑳𝜹 𝒌 ) 𝒔𝒊𝒏𝒉 𝑳 𝜹 ) OR 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = ( 𝐜𝐨𝐬𝐡 [ (𝑳 − 𝒙) √𝟐𝜶𝒕 ] + ( 𝒉𝑳√𝟐𝜶𝒕 𝒌 )𝐬𝐢𝐧𝐡 [( 𝑳 − 𝒙 √𝟐𝜶𝒕 )] 𝐜𝐨𝐬𝐡 𝑳 √𝟐𝜶𝒕 + ( 𝒉𝑳√𝟐𝜶𝒕 𝒌 )𝒔𝒊𝒏𝒉 𝑳 √𝟐𝜶𝒕 ) For the steady state solution, we set 𝑡 → ∞, but first let us first divide through by 𝛿 and get: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = ( 1 𝛿 cosh [ (𝐿 − 𝑥) 𝛿 ] + ( ℎ𝐿 𝑘 ) sinh [( 𝐿 − 𝑥 𝛿 )] 1 𝛿 cosh 𝐿 𝛿 + ( ℎ𝐿 𝑘 ) 𝑠𝑖𝑛ℎ 𝐿 𝛿 )
Now set 𝑡 → ∞ and get: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = ( 0 0 ) L’hopital’s rule is then invoked i.e., 𝑙𝑖𝑚 𝑥 → 𝑐 𝑓(𝑥) 𝑔(𝑥) = 𝑙𝑖𝑚 𝑥 → 𝑐 𝑓′ (𝑥) 𝑔′(𝑥) (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 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = ( (𝐿 − 𝑥) 𝛿 sinh [ (𝐿 − 𝑥) 𝛿 ] + cosh [ (𝐿 − 𝑥) 𝛿 ] + ( ℎ𝐿 𝑘 ) (𝐿 − 𝑥)cosh [( 𝐿 − 𝑥 𝛿 )] 𝐿 𝛿 sinh 𝐿 𝛿 + 𝑐𝑜𝑠ℎ 𝐿 𝛿 + ( ℎ𝐿 𝑘 )𝐿𝑐𝑜𝑠ℎ 𝐿 𝛿 ) We now substitute 𝑡 = ∞ in the formula above and get: 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = ( 𝟏 + ( 𝒉𝑳 𝒌 )(𝑳 − 𝒙) 𝟏 + ( 𝒉𝑳 𝒌 ) 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 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 Recall the compact temperature profile that satisfies the boundary and initial conditions is [2]: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = cosh[ (𝐿 − 𝑥) 𝛿 ] cosh 𝐿 𝛿 Again, solving the heat equation using the temperature profile above yields 𝛿 = √2𝛼𝑡 For steady state conditions we set 𝑡 → ∞ And get: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 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 = 𝑻∞ 𝒂𝒕 𝒙 = ∞ 𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 The temperature profile that satisfies the boundary and initial conditions is: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 −𝑥 𝛿 Again, solving the heat equation using the temperature profile above yields 𝛿 = √2𝛼𝑡 For steady state conditions we set 𝑡 → ∞ And get: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 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 𝑚 = √ ℎ𝑃 𝑘𝐴 in steady state.
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. .

More Related Content

PDF
TRANSIENT RADIAL HEAT CONDUCTION WITH BESSEL FUNCTIONS AND INTEGRAL TRANSFOR...
Wasswaderrick3
 
PDF
TRANSIENT STATE HEAT CONDUCTION IN NATURAL CONVECTION EXPLAINED.pdf
Wasswaderrick3
 
PDF
TRANSIENT STATE HEAT CONDUCTION FOR A SEMI-INFINITE ROD IN NATURAL CONVECTION...
Wasswaderrick3
 
PDF
GENERAL MATHEMATICAL THEORY OF HEAT CONDUCTION.pdf
Wasswaderrick3
 
PDF
THE RATE OF COOLING IN NATURAL CONVECTION EXPLAINED.pdf
Wasswaderrick3
 
PDF
DERIVATION OF THE LAW OF COOLING IN NATURAL CONVECTION.pdf
Wasswaderrick3
 
PDF
FUNDAMENTALS OF FLUID FLOW 5th Edition.pdf
Wasswaderrick3
 
PDF
FLUID MECHANICS DEMYSTIFIED 5th Edition.pdf
Wasswaderrick3
 
TRANSIENT RADIAL HEAT CONDUCTION WITH BESSEL FUNCTIONS AND INTEGRAL TRANSFOR...
Wasswaderrick3
 
TRANSIENT STATE HEAT CONDUCTION IN NATURAL CONVECTION EXPLAINED.pdf
Wasswaderrick3
 
TRANSIENT STATE HEAT CONDUCTION FOR A SEMI-INFINITE ROD IN NATURAL CONVECTION...
Wasswaderrick3
 
GENERAL MATHEMATICAL THEORY OF HEAT CONDUCTION.pdf
Wasswaderrick3
 
THE RATE OF COOLING IN NATURAL CONVECTION EXPLAINED.pdf
Wasswaderrick3
 
DERIVATION OF THE LAW OF COOLING IN NATURAL CONVECTION.pdf
Wasswaderrick3
 
FUNDAMENTALS OF FLUID FLOW 5th Edition.pdf
Wasswaderrick3
 
FLUID MECHANICS DEMYSTIFIED 5th Edition.pdf
Wasswaderrick3
 

Recently uploaded (20)

PDF
AlphaEarth Foundations and the Satellite Embedding dataset
gdgforscience
 
PPTX
EPIDURAL ANESTHESIA ANATOMY AND PHYSIOLOGY.pptx
PrabakaranMP1
 
PPTX
2. Earth - The Living Planet Module 2ELS
markjustinebarolobau
 
PPTX
Cell Membrane: Structure, Composition & Functions
Muhammad Sajid Afridi
 
DOCX
Viruses (History, structure and composition, classification, Bacteriophage Re...
Educator
 
PPTX
The KM-GBF monitoring framework – status & key messages.pptx
pensoftservices
 
PPTX
DRUG THERAPY FOR SHOCK gjjjgfhhhhh.pptx.
vipvivek704
 
PPTX
ECG_Course_Presentation د.محمد صقران ppt
lelouchml1995
 
PPTX
Taita Taveta Laboratory Technician Workshop Presentation.pptx
CALVINCEOWINO
 
PDF
An interstellar mission to test astrophysical black holes
Sérgio Sacani
 
PDF
CAPERS-LRD-z9:AGas-enshroudedLittleRedDotHostingaBroad-lineActive GalacticNuc...
Sérgio Sacani
 
PPTX
neck nodes and dissection types and lymph nodes levels
DipuSarma
 
PPTX
Introduction to Fisheries Biotechnology_Lesson 1.pptx
Esther Pongcol
 
PPTX
Protein & Amino Acid Structures Levels of protein structure (primary, seconda...
Muhammad Sajid Afridi
 
PPTX
G5Q1W8 PPT SCIENCE.pptx 2025-2026 GRADE 5
Rodel Gordzkie
 
PPTX
BIOMOLECULES PPT........................
vachieagrawal1221
 
PDF
Biophysics 2.pdffffffffffffffffffffffffff
MiraMusleh
 
PDF
ELS_Q1_Module-11_Formation-of-Rock-Layers_v2.pdf
savannaesar07
 
PPTX
ognitive-behavioral therapy, mindfulness-based approaches, coping skills trai...
OliveJolly1
 
PDF
. Radiology Case Scenariosssssssssssssss
MiraMusleh
 
AlphaEarth Foundations and the Satellite Embedding dataset
gdgforscience
 
EPIDURAL ANESTHESIA ANATOMY AND PHYSIOLOGY.pptx
PrabakaranMP1
 
2. Earth - The Living Planet Module 2ELS
markjustinebarolobau
 
Cell Membrane: Structure, Composition & Functions
Muhammad Sajid Afridi
 
Viruses (History, structure and composition, classification, Bacteriophage Re...
Educator
 
The KM-GBF monitoring framework – status & key messages.pptx
pensoftservices
 
DRUG THERAPY FOR SHOCK gjjjgfhhhhh.pptx.
vipvivek704
 
ECG_Course_Presentation د.محمد صقران ppt
lelouchml1995
 
Taita Taveta Laboratory Technician Workshop Presentation.pptx
CALVINCEOWINO
 
An interstellar mission to test astrophysical black holes
Sérgio Sacani
 
CAPERS-LRD-z9:AGas-enshroudedLittleRedDotHostingaBroad-lineActive GalacticNuc...
Sérgio Sacani
 
neck nodes and dissection types and lymph nodes levels
DipuSarma
 
Introduction to Fisheries Biotechnology_Lesson 1.pptx
Esther Pongcol
 
Protein & Amino Acid Structures Levels of protein structure (primary, seconda...
Muhammad Sajid Afridi
 
G5Q1W8 PPT SCIENCE.pptx 2025-2026 GRADE 5
Rodel Gordzkie
 
BIOMOLECULES PPT........................
vachieagrawal1221
 
Biophysics 2.pdffffffffffffffffffffffffff
MiraMusleh
 
ELS_Q1_Module-11_Formation-of-Rock-Layers_v2.pdf
savannaesar07
 
ognitive-behavioral therapy, mindfulness-based approaches, coping skills trai...
OliveJolly1
 
. Radiology Case Scenariosssssssssssssss
MiraMusleh
 
Ad

Featured (20)

PDF
2024 Trend Updates: What Really Works In SEO & Content Marketing
Search Engine Journal
 
PDF
Storytelling For The Web: Integrate Storytelling in your Design Process
Chiara Aliotta
 
PDF
Artificial Intelligence, Data and Competition – SCHREPEL – June 2024 OECD dis...
OECD Directorate for Financial and Enterprise Affairs
 
PDF
How to Leverage AI to Boost Employee Wellness - Lydia Di Francesco - SocialHR...
SocialHRCamp
 
PDF
2024 State of Marketing Report – by Hubspot
Marius Sescu
 
PDF
Everything You Need To Know About ChatGPT
Expeed Software
 
PDF
Product Design Trends in 2024 | Teenage Engineerings
Pixeldarts
 
PDF
How Race, Age and Gender Shape Attitudes Towards Mental Health
ThinkNow
 
PDF
AI Trends in Creative Operations 2024 by Artwork Flow.pdf
marketingartwork
 
PDF
Skeleton Culture Code
Skeleton Technologies
 
PDF
PEPSICO Presentation to CAGNY Conference Feb 2024
Neil Kimberley
 
PDF
Content Methodology: A Best Practices Report (Webinar)
contently
 
PPTX
How to Prepare For a Successful Job Search for 2024
Albert Qian
 
PDF
Social Media Marketing Trends 2024 // The Global Indie Insights
Kurio // The Social Media Age(ncy)
 
PDF
Trends In Paid Search: Navigating The Digital Landscape In 2024
Search Engine Journal
 
PDF
5 Public speaking tips from TED - Visualized summary
SpeakerHub
 
PDF
ChatGPT and the Future of Work - Clark Boyd
Clark Boyd
 
PDF
Getting into the tech field. what next
Tessa Mero
 
PDF
Google's Just Not That Into You: Understanding Core Updates & Search Intent
Lily Ray
 
PDF
How to have difficult conversations
Rajiv Jayarajah, MAppComm, ACC
 
2024 Trend Updates: What Really Works In SEO & Content Marketing
Search Engine Journal
 
Storytelling For The Web: Integrate Storytelling in your Design Process
Chiara Aliotta
 
Artificial Intelligence, Data and Competition – SCHREPEL – June 2024 OECD dis...
OECD Directorate for Financial and Enterprise Affairs
 
How to Leverage AI to Boost Employee Wellness - Lydia Di Francesco - SocialHR...
SocialHRCamp
 
2024 State of Marketing Report – by Hubspot
Marius Sescu
 
Everything You Need To Know About ChatGPT
Expeed Software
 
Product Design Trends in 2024 | Teenage Engineerings
Pixeldarts
 
How Race, Age and Gender Shape Attitudes Towards Mental Health
ThinkNow
 
AI Trends in Creative Operations 2024 by Artwork Flow.pdf
marketingartwork
 
Skeleton Culture Code
Skeleton Technologies
 
PEPSICO Presentation to CAGNY Conference Feb 2024
Neil Kimberley
 
Content Methodology: A Best Practices Report (Webinar)
contently
 
How to Prepare For a Successful Job Search for 2024
Albert Qian
 
Social Media Marketing Trends 2024 // The Global Indie Insights
Kurio // The Social Media Age(ncy)
 
Trends In Paid Search: Navigating The Digital Landscape In 2024
Search Engine Journal
 
5 Public speaking tips from TED - Visualized summary
SpeakerHub
 
ChatGPT and the Future of Work - Clark Boyd
Clark Boyd
 
Getting into the tech field. what next
Tessa Mero
 
Google's Just Not That Into You: Understanding Core Updates & Search Intent
Lily Ray
 
How to have difficult conversations
Rajiv Jayarajah, MAppComm, ACC
 
Ad

TRANSIENT AND STEADY STATE HEAT CONDUCTION WITH NO LATERAL CONVECTION SOLVED USING THE INTEGRAL EQUATION TRANSFORM.pdf

  • 1. TRANSIENT STATE AND STEADY STATE HEAT CONDUCTION WITH NO LATERAL CONVECTION By Wasswa Derrick Wasswaderricktimothy7@gmail.com ONE DIMENSIONAL HEAT CONDUCTION RECTANGULAR CO-ORDINATES. The governing equation is: 𝛼 𝜕2 𝑇 𝜕𝑥2 = 𝜕𝑇 𝜕𝑡 To get the steady state temperature, we simply substitute 𝜕𝑇 𝜕𝑡 = 0 𝑖𝑛 𝑠𝑡𝑒𝑎𝑦 𝑠𝑡𝑎𝑡𝑒 And get: 𝜕2 𝑇 𝜕𝑥2 = 0 We integrate the above equation to get the variation of temperature with distance x i.e., 𝜕𝑇 𝜕𝑥 = 𝑐1 𝑻 = 𝒄𝟏𝒙 + 𝒄𝟐 The condition above that the temperature is linear in distance x will help us to solve for the transient state. At 𝑥 = 𝑥1, 𝑇 = 𝑇1 ∴ 𝑇1 = 𝑐1𝑥1 + 𝑐2 … (𝐴) At 𝑥 = 𝑥2, 𝑇 = 𝑇2 ∴ 𝑇2 = 𝑐1𝑥2 + 𝑐2 … . (𝐵) (𝐴 − 𝐵) 𝑙𝑒𝑎𝑑𝑠 𝑡𝑜 𝑐1 = 𝑇1−𝑇2 𝑥1−𝑥2 From A 𝑐2 = 𝑇1 − 𝑐1𝑥1 = 𝑇1 − 𝑇1−𝑇2 𝑥1−𝑥2 𝑥1 Substituting in the general solution, we get
  • 2. 𝑇 = 𝑐1𝑥 + 𝑐2 𝑇 = 𝑇1 − 𝑇2 𝑥1 − 𝑥2 𝑥 + 𝑇1 − 𝑇1 − 𝑇2 𝑥1 − 𝑥2 𝑥1 𝑇 − 𝑇1 = 𝑇1 − 𝑇2 𝑥1 − 𝑥2 (𝑥 − 𝑥1) Since 𝑥2 > 𝑥1 as shown in the diagram above, we can rearrange and get 𝑇 − 𝑇1 = −( 𝑇1 − 𝑇2 𝑥2 − 𝑥1 )(𝑥 − 𝑥1) From 𝑄 = −𝑘𝐴 𝜕𝑇 𝜕𝑥 𝜕𝑇 𝜕𝑥 = −( 𝑇1 − 𝑇2 𝑥2 − 𝑥1 ) Upon substitution, we get: 𝑄 = 𝑘𝐴( 𝑇1 − 𝑇2 𝑥2 − 𝑥1 ) 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: 𝜶 𝝏𝟐 𝑻 𝝏𝒙𝟐 = 𝝏𝑻 𝝏𝒕 We are going to deal with a variety of boundary conditions and initial condition.
  • 3. THE INTEGRAL 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 𝜕2 𝑇 𝜕𝑥2 = 𝜕𝑇 𝜕𝑡 Let us change this equation into an integral equation as below: 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 Postulate 1: Any temperature profile of the form 𝑇 = 𝐵𝑒 𝑎𝑥 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 𝑎 = 1 and 𝐵 = 𝑇0 𝑇 = 𝑇0𝑒 𝑥 𝛿 𝜕2 𝑇 𝜕𝑥2 = 𝑇0 𝛿2 𝑒 𝑥 𝛿 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = 𝑇0 𝛿 [𝑒 𝐿 𝛿 − 1] What we have learned from the above integral is that ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = [ 𝜕𝑇 𝜕𝑥 ] 𝐿 0 Which we shall use in later text. Anyway, to continue, let us evaluate ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝛿[𝑒 𝐿 𝛿 − 1]
  • 4. Upon substitution in the heat equation, we get: 𝛼 𝑇0 𝛿 [𝑒 𝐿 𝛿 − 1] = 𝜕 𝜕𝑡 [𝑇0𝛿[𝑒 𝐿 𝛿 − 1]] There’s a common term 𝑇0[𝑒 𝐿 𝛿 − 1] which we drop out and get: 𝛼 𝛿 = 𝑑𝛿 𝑑𝑡 We finally get 𝛿 = √2𝛼𝑡 + 𝑐 We finally substitute in the temperature profile to get: 𝑇 = 𝑇0𝑒 𝑥 √2𝛼𝑡+𝑐 Where c and 𝑇0are evaluated using the boundary and initial conditions. It can be shown that hyperbolic temperature functions like ( 𝑠𝑖𝑛ℎ𝑎𝑥 𝑜𝑟 𝑐𝑜𝑠ℎ𝑎𝑥) 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 𝑥 𝛿 + 𝐵𝑒 −𝑥 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 𝑚 = 1 𝛿 Where 𝛿(𝑡) will be got by solving the heat equation. The table below will give the required temperature profiles for given boundary conditions [1]
  • 6. CASE 1: FIXED END TEMPERATURE The initial and boundary conditions are: 𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 𝑻 = 𝑻𝒃𝟏 − 𝑻∞ 𝒂𝒕 𝒙 = 𝟎 𝑻 = 𝑻𝒃𝟐 − 𝑻∞ 𝒂𝒕 𝒙 = 𝑳 We start with a temperature profile below: 𝑇 − 𝑇∞ = 𝐴𝑒 𝑥 𝛿 + 𝐵𝑒 −𝑥 𝛿 We evaluate the constants A and B using the boundary conditions and get the temperature profile below: 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) 𝑠𝑖𝑛ℎ 𝑥 𝛿 + 𝑠𝑖𝑛ℎ 𝐿 − 𝑥 𝛿 𝑠𝑖𝑛ℎ 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 𝑎𝑡 𝑡 = 0, 𝛿 = 0 and get: 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) 𝑠𝑖𝑛ℎ 𝑥 𝛿 + 𝑠𝑖𝑛ℎ 𝐿 − 𝑥 𝛿 𝑠𝑖𝑛ℎ 𝐿 𝛿 Let us show that the above postulate implies the initial condition 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) 𝑠𝑖𝑛ℎ 𝑥 𝛿 + 𝑠𝑖𝑛ℎ 𝐿 − 𝑥 𝛿 𝑠𝑖𝑛ℎ 𝐿 𝛿 = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ )(𝑒 𝑥 𝛿 + 𝑒 −𝑥 𝛿 ) + (𝑒 𝐿−𝑥 𝛿 + 𝑒 −(𝐿−𝑥) 𝛿 ) 𝑒 𝐿 𝛿 − 𝑒 −𝐿 𝛿 𝑒 −(𝐿−𝑥) 𝛿 = 𝑒− (𝐿−𝑥) 0 = 𝑒−∞(𝐿−𝑥) = 0 Similarly 𝑒 −𝐿 𝛿 = 𝑒 −𝐿 0 = 𝑒−∞𝐿 = 0 So, we are left with
  • 7. 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) (𝑒 𝑥 𝛿) + (𝑒 𝐿−𝑥 𝛿 ) 𝑒 𝐿 𝛿 = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ )𝑒 −(𝐿−𝑥) 𝛿 + 𝑒 −𝑥 𝛿 = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ )𝑒 −(𝐿−𝑥) 0 + 𝑒 −𝑥 0 = 𝑒−∞𝑥 We finally get 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ )𝑒−∞(𝐿−𝑥) + 𝑒−∞𝑥 = 0 Since 𝐿 − 𝑥 > 0 Hence at 𝑡 = 0, 𝑇 = 𝑇∞ and hence the initial condition. Let us solve the heat equation analytically using the temperature profile above: 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = [ 𝜕𝑇 𝜕𝑥 ] 𝐿 0 [ 𝜕𝑇 𝜕𝑥 ] 𝐿 0 = (𝑇𝑏1 − 𝑇∞) 𝛿 [( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) coth ( 𝐿 𝛿 ) − ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1)𝑐𝑜𝑠𝑒𝑐ℎ ( 𝐿 𝛿 )] To evaluate ∫ (𝑇)𝑑𝑥 𝑙 0 we need to know T 𝑇 = (𝑇𝑏1 − 𝑇∞) [ ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) 𝑠𝑖𝑛ℎ 𝑥 𝛿 + 𝑠𝑖𝑛ℎ 𝐿 − 𝑥 𝛿 𝑠𝑖𝑛ℎ 𝐿 𝛿 ] + 𝑇∞ ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝛿(𝑇𝑏1 − 𝑇∞) [( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1)coth ( 𝐿 𝛿 ) − ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) 𝑐𝑜𝑠𝑒𝑐ℎ ( 𝐿 𝛿 )] + 𝑇∞𝐿 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝑑 𝑑𝑡 (𝛿(𝑇𝑏1 − 𝑇∞) [( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) coth ( 𝐿 𝛿 ) − ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) 𝑐𝑜𝑠𝑒𝑐ℎ ( 𝐿 𝛿 )]) + 𝑑(𝑇∞𝐿) 𝑑𝑡 But 𝑑(𝑇∞𝐿) 𝑑𝑡 = 0 𝑠𝑖𝑛𝑐𝑒 𝐿 𝑎𝑛𝑑 𝑇∞ 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠 So, we have
  • 8. 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝑑 𝑑𝑡 (𝛿(𝑇𝑏1 − 𝑇∞) [( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) coth ( 𝐿 𝛿 ) − ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) 𝑐𝑜𝑠𝑒𝑐ℎ ( 𝐿 𝛿 )]) Upon substitution in the heat equation, we get: 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 𝛼 𝛿 (𝑇𝑏1 − 𝑇∞) [( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) coth ( 𝐿 𝛿 ) − ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) 𝑐𝑜𝑠𝑒𝑐ℎ ( 𝐿 𝛿 )] = 𝑑 𝑑𝑡 (𝛿(𝑇𝑏1 − 𝑇∞) [( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) coth ( 𝐿 𝛿 ) − ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ + 1) 𝑐𝑜𝑠𝑒𝑐ℎ ( 𝐿 𝛿 )]) The factor (𝑇𝑏1 − 𝑇∞) [( 𝑇𝑏2 −𝑇∞ 𝑇𝑏1 −𝑇∞ + 1) coth ( 𝐿 𝛿 ) − ( 𝑇𝑏2 −𝑇∞ 𝑇𝑏1 −𝑇∞ + 1) 𝑐𝑜𝑠𝑒𝑐ℎ ( 𝐿 𝛿 )] is common on both sides of the equation so we cross it out and get 𝛼 𝛿 = 𝑑𝛿 𝑑𝑡 Where the boundary conditions are: 𝛿 = 0 𝑎𝑡 𝑡 = 0 On solving the above we get: 𝛿 = √2𝛼𝑡 Upon substitution in the temperature profile, we get 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) 𝑠𝑖𝑛ℎ 𝑥 𝛿 + 𝑠𝑖𝑛ℎ 𝐿 − 𝑥 𝛿 𝑠𝑖𝑛ℎ 𝐿 𝛿 OR 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) 𝑠𝑖𝑛ℎ 𝑥 √2𝛼𝑡 + 𝑠𝑖𝑛ℎ 𝐿 − 𝑥 √2𝛼𝑡 𝑠𝑖𝑛ℎ 𝐿 √2𝛼𝑡 In steady state, 𝑡 → ∞ 𝑠𝑜 𝛿 → ∞ Upon substitution we get
  • 9. 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ )𝑠𝑖𝑛ℎ 𝑥 ∞ + 𝑠𝑖𝑛ℎ 𝐿 − 𝑥 ∞ 𝑠𝑖𝑛ℎ 𝐿 ∞ 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) 𝑠𝑖𝑛ℎ0 + 𝑠𝑖𝑛ℎ0 𝑠𝑖𝑛ℎ0 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = 0 0 L’hopital’s rule is then invoked i.e., 𝑙𝑖𝑚 𝑥 → 𝑐 𝑓(𝑥) 𝑔(𝑥) = 𝑙𝑖𝑚 𝑥 → 𝑐 𝑓′ (𝑥) 𝑔′(𝑥) We differentiate the numerate and denominator with respect to 1 𝛿 since it is the one varying and we get 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = 𝑥 𝛿 ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ )𝑐𝑜𝑠ℎ 𝑥 𝛿 + ( 𝐿 − 𝑥 𝛿 )𝑐𝑜𝑠ℎ 𝐿 − 𝑥 𝛿 𝐿 𝛿 𝑐𝑜𝑠ℎℎ 𝐿 𝛿 Upon simplification, we get 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = 𝑥 ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) 𝑐𝑜𝑠ℎ 𝑥 𝛿 + (𝐿 − 𝑥)𝑐𝑜𝑠ℎ 𝐿 − 𝑥 𝛿 𝐿𝑐𝑜𝑠ℎℎ 𝐿 𝛿 We then substitute 𝛿 = √2𝛼𝑡 = ∞ And get 𝑇 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ = 𝑥 ( 𝑇𝑏2 − 𝑇∞ 𝑇𝑏1 − 𝑇∞ ) + (𝐿 − 𝑥) 𝐿 Upon substitution, we get: 𝑻 − 𝑻∞ 𝑻𝒃𝟏 − 𝑻∞ = 𝒙 𝑳 ( 𝑻𝒃𝟐 − 𝑻∞ 𝑻𝒃𝟏 − 𝑻∞ ) + (𝟏 − 𝒙 𝑳 ) Hence, we get a temperature profile linear in x in steady state.
  • 10. CASE 2: CONVECTION AT THE END OF A FINITE METAL ROD The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 −𝒌 𝒅𝑻 𝒅𝒙 = 𝒉𝑳(𝑻 − 𝑻∞) 𝒂𝒕 𝒙 = 𝒍 𝑻 = 𝑻∞ 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 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = cosh[𝑚(𝐿 − 𝑥)] + ( ℎ𝐿 𝑚𝑘 ) sinh[𝑚(𝐿 − 𝑥)] cosh 𝑚𝐿 + ( ℎ𝐿 𝑚𝑘 ) 𝑠𝑖𝑛ℎ𝑚𝐿 Or 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = cosh [ (𝐿 − 𝑥) 𝛿 ] + ( ℎ𝐿𝛿 𝑘 ) sinh[( 𝐿 − 𝑥 𝛿 )] cosh 𝐿 𝛿 + ( ℎ𝐿𝛿 𝑘 ) 𝑠𝑖𝑛ℎ 𝐿 𝛿 To show that the initial condition is satisfied we assume from the above that 𝑎𝑡 𝑡 = 0, 𝛿 = 0. 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = cosh [ (𝐿 − 𝑥) 𝛿 ] + ( ℎ𝐿𝛿 𝑘 ) sinh[( 𝐿 − 𝑥 𝛿 )] cosh 𝐿 𝛿 + ( ℎ𝛿 𝑘 )𝑠𝑖𝑛ℎ 𝐿 𝛿 Becomes: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = cosh [ (𝐿 − 𝑥) 𝛿 ] cosh 𝐿 𝛿 = 𝑒 (𝐿−𝑥) 𝛿 + 𝑒 −(𝐿−𝑥) 𝛿 𝑒 𝐿 𝛿 + 𝑒 −𝐿 𝛿 𝑒 −(𝐿−𝑥) 𝛿 = 𝑒− (𝐿−𝑥) 0 = 𝑒−∞(𝐿−𝑥) = 0 Similarly 𝑒 −𝐿 𝛿 = 𝑒 −𝐿 0 = 𝑒−∞𝐿 = 0 So, we are left with
  • 11. 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 (𝐿−𝑥) 𝛿 𝑒 𝐿 𝛿 = 𝑒 −𝑥 𝛿 = 𝑒 −𝑥 0 = 𝑒−∞𝑥 = 0 Hence at 𝑡 = 0, 𝑇 = 𝑇∞ 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 −𝒌 𝒅𝑻 𝒅𝒙 = 𝒉𝑳(𝑻 − 𝑻∞) 𝒂𝒕 𝒙 = 𝒍 𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 The governing temperature profile is: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = cosh [ (𝐿 − 𝑥) 𝛿 ] + ( ℎ𝐿𝛿 𝑘 ) sinh[( 𝐿 − 𝑥 𝛿 )] cosh 𝐿 𝛿 + ( ℎ𝐿𝛿 𝑘 ) 𝑠𝑖𝑛ℎ 𝐿 𝛿 The governing equation is 𝛼 𝜕2 𝑇 𝜕𝑥2 = 𝜕𝑇 𝜕𝑡 Let us change this equation into an integral equation as below: 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 … … . . 𝑏) 𝛼 ∫ ( 𝜕2 𝑇 𝜕𝑥2 ) 𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 ∫ ( 𝜕2 𝑇 𝜕𝑥2 )𝑑𝑥 𝑙 0 = [ 𝜕𝑇 𝜕𝑥 ] 𝑙 0 = (𝑇𝑠 − 𝑇∞) 𝛿 ( − ℎ𝐿𝛿 𝑘 + (𝑠𝑖𝑛ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑐𝑜𝑠ℎ 𝐿 𝛿 ) 𝑐𝑜𝑠ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑠𝑖𝑛ℎ 𝐿 𝛿 ) 𝑇 = cosh [ (𝐿 − 𝑥) 𝛿 ] + ( ℎ𝐿𝛿 𝑘 )sinh[( 𝐿 − 𝑥 𝛿 )] cosh 𝐿 𝛿 + ( ℎ𝐿𝛿 𝑘 )𝑠𝑖𝑛ℎ 𝐿 𝛿 (𝑇𝑠 − 𝑇∞) + 𝑇∞
  • 12. 𝜕 𝜕𝑡 ∫ (𝑇)𝑑𝑥 𝑙 0 = 𝜕 𝜕𝑡 [𝛿(𝑇𝑠 − 𝑇∞) ( − ℎ𝐿𝛿 𝑘 + (𝑠𝑖𝑛ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑐𝑜𝑠ℎ 𝐿 𝛿 ) 𝑐𝑜𝑠ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑠𝑖𝑛ℎ 𝐿 𝛿 )] + 𝜕(𝑙𝑇∞) 𝜕𝑡 𝜕(𝑙𝑇∞) 𝜕𝑡 = 0 Upon substitution of all the above in the heat equation, we get: 𝛼 (𝑇𝑠 − 𝑇∞) 𝛿 ( − ℎ𝐿𝛿 𝑘 + (𝑠𝑖𝑛ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑐𝑜𝑠ℎ 𝐿 𝛿 ) 𝑐𝑜𝑠ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑠𝑖𝑛ℎ 𝐿 𝛿 ) = 𝜕 𝜕𝑡 [𝛿(𝑇𝑠 − 𝑇∞) ( − ℎ𝐿𝛿 𝑘 + (𝑠𝑖𝑛ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑐𝑜𝑠ℎ 𝐿 𝛿 ) 𝑐𝑜𝑠ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑠𝑖𝑛ℎ 𝐿 𝛿 )] We notice that the term(𝑇𝑠 − 𝑇∞) ( − ℎ𝐿𝛿 𝑘 +(𝑠𝑖𝑛ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑐𝑜𝑠ℎ 𝐿 𝛿 ) 𝑐𝑜𝑠ℎ 𝐿 𝛿 + ℎ𝐿𝛿 𝑘 𝑠𝑖𝑛ℎ 𝐿 𝛿 ) is common and can be eliminated: 𝛼 𝛿 = 𝑑𝛿 𝑑𝑡 We go ahead and solve for 𝛿 provided 𝛿 = 0𝑎𝑡 𝑡 = 0 and get the expression 𝛿 = √2𝛼𝑡 So, the final solution for the finite metal rod with convective flux at the end of the metal rod is: 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = ( 𝐜𝐨𝐬𝐡 [ (𝑳 − 𝒙) 𝜹 ] + ( 𝒉𝑳𝜹 𝒌 ) 𝐬𝐢𝐧𝐡 [( 𝑳 − 𝒙 𝜹 )] 𝐜𝐨𝐬𝐡 𝑳 𝜹 + ( 𝒉𝑳𝜹 𝒌 ) 𝒔𝒊𝒏𝒉 𝑳 𝜹 ) OR 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = ( 𝐜𝐨𝐬𝐡 [ (𝑳 − 𝒙) √𝟐𝜶𝒕 ] + ( 𝒉𝑳√𝟐𝜶𝒕 𝒌 )𝐬𝐢𝐧𝐡 [( 𝑳 − 𝒙 √𝟐𝜶𝒕 )] 𝐜𝐨𝐬𝐡 𝑳 √𝟐𝜶𝒕 + ( 𝒉𝑳√𝟐𝜶𝒕 𝒌 )𝒔𝒊𝒏𝒉 𝑳 √𝟐𝜶𝒕 ) For the steady state solution, we set 𝑡 → ∞, but first let us first divide through by 𝛿 and get: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = ( 1 𝛿 cosh [ (𝐿 − 𝑥) 𝛿 ] + ( ℎ𝐿 𝑘 ) sinh [( 𝐿 − 𝑥 𝛿 )] 1 𝛿 cosh 𝐿 𝛿 + ( ℎ𝐿 𝑘 ) 𝑠𝑖𝑛ℎ 𝐿 𝛿 )
  • 13. Now set 𝑡 → ∞ and get: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = ( 0 0 ) L’hopital’s rule is then invoked i.e., 𝑙𝑖𝑚 𝑥 → 𝑐 𝑓(𝑥) 𝑔(𝑥) = 𝑙𝑖𝑚 𝑥 → 𝑐 𝑓′ (𝑥) 𝑔′(𝑥) (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 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = ( (𝐿 − 𝑥) 𝛿 sinh [ (𝐿 − 𝑥) 𝛿 ] + cosh [ (𝐿 − 𝑥) 𝛿 ] + ( ℎ𝐿 𝑘 ) (𝐿 − 𝑥)cosh [( 𝐿 − 𝑥 𝛿 )] 𝐿 𝛿 sinh 𝐿 𝛿 + 𝑐𝑜𝑠ℎ 𝐿 𝛿 + ( ℎ𝐿 𝑘 )𝐿𝑐𝑜𝑠ℎ 𝐿 𝛿 ) We now substitute 𝑡 = ∞ in the formula above and get: 𝑻 − 𝑻∞ 𝑻𝒔 − 𝑻∞ = ( 𝟏 + ( 𝒉𝑳 𝒌 )(𝑳 − 𝒙) 𝟏 + ( 𝒉𝑳 𝒌 ) 𝑳 ) The above is the steady state solution and it shows temperature is linearly negative in x.
  • 14. 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 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 Recall the compact temperature profile that satisfies the boundary and initial conditions is [2]: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = cosh[ (𝐿 − 𝑥) 𝛿 ] cosh 𝐿 𝛿 Again, solving the heat equation using the temperature profile above yields 𝛿 = √2𝛼𝑡 For steady state conditions we set 𝑡 → ∞ And get: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 1 Hence temperature will be uniformly 𝑇𝑠 throughout the metal rod.
  • 15. CASE 4: SEMI-INFINITE ROD CASE The boundary and initial conditions are: 𝑻 = 𝑻𝒔 𝒂𝒕 𝒙 = 𝟎 𝑻 = 𝑻∞ 𝒂𝒕 𝒙 = ∞ 𝑻 = 𝑻∞ 𝒂𝒕 𝒕 = 𝟎 The temperature profile that satisfies the boundary and initial conditions is: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 𝑒 −𝑥 𝛿 Again, solving the heat equation using the temperature profile above yields 𝛿 = √2𝛼𝑡 For steady state conditions we set 𝑡 → ∞ And get: 𝑇 − 𝑇∞ 𝑇𝑠 − 𝑇∞ = 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 𝑚 = √ ℎ𝑃 𝑘𝐴 in steady state.
  • 16. 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. .