Week 2 Lecture_Conduction with Heat Generation PDF

ME3407 Fluid Dynamics and Heat Transfer Week 2 2.1 Critical radius and variable thermal conductivity 2.2 Heat Conduction with Internal Heat Sources 2.3 General Heat Conduction Equation - the Heat Diffusion Equation...

ME3407 Fluid Dynamics and Heat Transfer Week 2 2.1 Critical radius and variable thermal conductivity 2.2 Heat Conduction with Internal Heat Sources 2.3 General Heat Conduction Equation - the Heat Diffusion Equation 1 Week 2.1 Critical radius and variable thermal conductivity 2  Pipe covered by a layer of insulation Islington Communal Heating provides heating for more r1, T1 than 600 homes. r2,T2 Bunhill heat network, cheaper greener energy | T∞,1 Islington Council Q h1 T∞,2 h2 r3, T3 𝑄𝑄̇ 𝑇𝑇∞,1 𝑇𝑇1 𝑇𝑇2 𝑇𝑇3 𝑇𝑇∞,2 𝑇𝑇∞,1 − 𝑇𝑇∞,2 𝑄𝑄̇ = 𝑅𝑅𝑡𝑡𝑡𝑡𝑡𝑡 𝑇𝑇1 =? ̇ 1 𝑇𝑇1 = 𝑇𝑇∞,1 − 𝑄𝑄𝑅𝑅 1 ln(𝑟𝑟2 ⁄𝑟𝑟1 ) ln(𝑟𝑟3 ⁄𝑟𝑟2 ) 1 𝑇𝑇2 =? ̇ 1 +𝑅𝑅2 ) 𝑇𝑇2 = 𝑇𝑇∞,1 − 𝑄𝑄(𝑅𝑅 𝑅𝑅𝑡𝑡𝑡𝑡𝑡𝑡 = + + + 2𝜋𝜋𝑟𝑟1 𝐿𝐿ℎ1 2𝜋𝜋𝜋𝜋𝑘𝑘1 2𝜋𝜋𝜋𝜋𝑘𝑘2 2𝜋𝜋𝑟𝑟3 𝐿𝐿ℎ2 𝑇𝑇3 =? 𝑇𝑇3 = 𝑇𝑇∞,2 + 𝑄𝑄𝑅𝑅̇ 4 Will a layer of insulation always reduce heat transfer? 3 Critical thickness (radius) for maximum heat transfer 1 𝑅𝑅𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,1 = 2𝜋𝜋𝑟𝑟1 𝐿𝐿ℎ1 𝑇𝑇∞,1 , ℎ1 Thin-walled 𝑇𝑇∞,2 , ℎ2 ln(𝑟𝑟2 ⁄𝑟𝑟1 ) copper tube 𝑅𝑅𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑟𝑟1 2𝜋𝜋𝜋𝜋𝜋𝜋 covered by a layer of insulation 1 𝑟𝑟2 𝑅𝑅𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,2 = 2𝜋𝜋𝑟𝑟2 𝐿𝐿ℎ2 𝑟𝑟1 is fixed. 𝑅𝑅𝑡𝑡𝑡𝑡𝑡𝑡 is a function of 𝑟𝑟2. As 𝑟𝑟2 increased, 𝑅𝑅𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 increases, but 𝑅𝑅𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,2 decreases. We can use 𝑟𝑟 as 𝑟𝑟2. 𝑅𝑅 𝑟𝑟 = 𝑅𝑅𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,1 + 𝑅𝑅𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑅𝑅𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,2 1 ln(𝑟𝑟 ⁄𝑟𝑟1 ) 1 = + + 2𝜋𝜋𝑟𝑟1 𝐿𝐿ℎ1 2𝜋𝜋𝜋𝜋𝜋𝜋 2𝜋𝜋𝜋𝜋𝜋𝜋ℎ2 4 Critical thickness (radius) for maximum heat transfer 𝑅𝑅 𝑟𝑟 = 𝑅𝑅𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,1 + 𝑅𝑅𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑅𝑅𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐,2 - Copied from the previous page 1 ln(𝑟𝑟 ⁄𝑟𝑟1 ) 1 = + + 2𝜋𝜋𝑟𝑟1 𝐿𝐿ℎ1 2𝜋𝜋𝜋𝜋𝜋𝜋 2𝜋𝜋𝜋𝜋𝜋𝜋ℎ2 𝑑𝑑𝑑𝑑 To find critical points of 𝑅𝑅 𝑟𝑟 , we take: =0 𝑑𝑑𝑑𝑑 (Later we can classify the critical points as local minimum or local maximum.) 𝑑𝑑𝑑𝑑 1 1 1 1 =0+ + − 2 =0 𝑑𝑑𝑑𝑑 2𝜋𝜋𝜋𝜋𝜋𝜋 𝑟𝑟 2𝜋𝜋𝜋𝜋ℎ2 𝑟𝑟 𝑘𝑘 A single critical radius is found: 𝑟𝑟𝑐𝑐 = ℎ2 5 Show that Heat transfer is maximum at the critical radius Note: 1. Below rc, increasing the insulation thickness will increase heat transfer. 2. Above rc, increasing the insulation thickness will reduce heat transfer. 3. Natural convection h = 10 W/m2K, insulation K = 0.5 W/mK, rc = 0.05 m 4. If it is not a thin-walled pipe, the critical radius is the same. 6  Conduction with a variable conductivity k Cartesian: k = a + bT, where a and b are constant Fourier Law dT K Q = − KA dx L x2 T2 Q ∫ dx = − A ∫ (a + bT )dT T1 Linear x1 T1 A  b 2 Q = a (T − T ) + (T − T2 ) 2  x2 − x1  1 2 1 T2 2  Q 𝐴𝐴 𝑏𝑏 𝑄𝑄̇ = 𝑎𝑎 + 𝑇𝑇1 + 𝑇𝑇2 𝑇𝑇1 − 𝑇𝑇2 𝑥𝑥2 − 𝑥𝑥1 2 X1 x2 𝐴𝐴 ̇ 𝑄𝑄 = 𝑘𝑘 𝑇𝑇 − 𝑇𝑇2 𝑥𝑥2 − 𝑥𝑥1 𝑚𝑚 1 Where 𝑘𝑘𝑚𝑚 = 𝑎𝑎 + 𝑏𝑏𝑇𝑇𝑚𝑚 and 𝑇𝑇𝑚𝑚 = (𝑇𝑇1 + 𝑇𝑇2 )/2 𝑇𝑇1 − 𝑇𝑇2 𝐿𝐿 𝑅𝑅𝑚𝑚 = = 𝑄𝑄̇ 𝑘𝑘𝑚𝑚 𝐴𝐴 7 Conduction with a variable conductivity k Cylindrical: Example k = 3 + 0.1 T, where T is in °C, r1 = 0.05m, r2 = 0.08m, T1 = 80°C and T2 = 35°C. Calculate heat transfer rate per 1 m of pipe length. dT dT Q = − KA = −(3 + 0.1T )(2πr ⋅ 1) r1 dr dr r r2 dr 2 T Q ∫ = −2π ∫ (3 + 0.1T )dT  r1 r T1 2π  0.1 2 r2 Q = 3(T − T ) + (T − T2 ) = 5264 (W ) 2  ln(r2 / r1 )  1 2 1 2  Note that we can simplify the equation with km evaluated at 𝑇𝑇𝑚𝑚 = (𝑇𝑇1 + 𝑇𝑇2 )/2 𝑟𝑟2 2𝜋𝜋𝑘𝑘𝑚𝑚 ln 𝑄𝑄̇ = (𝑇𝑇1 − 𝑇𝑇2 ) 𝑟𝑟1 𝑟𝑟2 And the thermal resistance is 𝑅𝑅𝑚𝑚 = ln 2𝜋𝜋𝑘𝑘𝑚𝑚 𝑟𝑟1 8 Week 2.2 Heat Conduction with Internal Heat Sources 9 Heat Conduction without Internal Heat Sources ̇ 1) 𝑄𝑄(𝑥𝑥 ̇ 𝑄𝑄(𝑥𝑥) L Energy balance in the control volume between x1 and x: ̇ 1) 𝑄𝑄̇ 𝑥𝑥 = 𝑄𝑄(𝑥𝑥 Q is constant along x in the Fourier’s equation. 𝑑𝑑𝑑𝑑 Fourier’s Law of Conduction: 𝑄𝑄̇ = −𝑘𝑘𝑘𝑘 𝑑𝑑𝑑𝑑 10 Heat Conduction with Internal Heat Sources Q is not constant along x in the Fourier’s equation. 𝑑𝑑𝑑𝑑 Fourier’s Law of Conduction: 𝑄𝑄̇ = −𝑘𝑘𝑘𝑘 Energy balance in the control volume between x1 and x: 𝑑𝑑𝑑𝑑 ̇ 1 ) + 𝑄𝑄̇ 𝑔𝑔𝑔𝑔𝑔𝑔 = 𝑄𝑄(𝑥𝑥) 𝑄𝑄(𝑥𝑥 ̇ Q is a function of x. 𝑑𝑑 𝑄𝑄̇ 𝑑𝑑 𝑑𝑑𝑑𝑑 = − (𝑘𝑘 ) ̇ 1) 𝑄𝑄(𝑥𝑥 𝐴𝐴𝐴𝐴𝐴𝐴 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 ̇ 𝑄𝑄(𝑥𝑥) 𝑑𝑑 𝑑𝑑𝑑𝑑 𝑞𝑞𝑞 = − (𝑘𝑘 ) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 L ′ 𝑑𝑑 𝑄𝑄̇ 𝑑𝑑𝑄𝑄̇ where 𝑞𝑞 = = 𝐴𝐴𝐴𝐴𝐴𝐴 𝑑𝑑𝑉𝑉 q ' is the internal heat source (W/m3) In electrical conductors: Heat generated/unit volume = ρ*i2 where ρ * = Resistivity (ohm m) and i = current density (Amp/m2) 11 Heat Conduction with Internal Heat Sources K Temperature distribution in a plane wall when k is constant is: 𝑑𝑑 𝑑𝑑𝑑𝑑 x 𝑘𝑘 = −𝑞𝑞 ′ 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 x=0 x=L 𝑘𝑘 = −𝑞𝑞 ′ 𝑥𝑥 + 𝑐𝑐 𝑑𝑑𝑑𝑑 𝑞𝑞 ′ 𝑥𝑥 2 𝑇𝑇 𝑥𝑥 = − + 𝑐𝑐1 𝑥𝑥 + 𝑐𝑐2 𝑘𝑘 2 For boundary conditions of T(0)=T1 and T(L)=T2, 𝑐𝑐2 = 𝑇𝑇1 (𝑇𝑇2 − 𝑇𝑇1 ) 𝑞𝑞𝑞𝑞𝑞 𝑐𝑐1 = + 𝐿𝐿 2𝑘𝑘 The maximum temperature is at dT/dx = 0, i.e. x = (k⋅c1)/q’ When T1 = T2, the maximum temperature is at the midplane q' L2 Tmax = T (L / 2) = + T1 8K 12 Heat Conduction with Internal Heat Sources Conduction in a plane wall with uniform heat generation. (a) Asymmetrical boundary conditions. (b) Symmetrical boundary conditions. The maximum temperature exists at the location where dT/dx =0. A boundary condition can be either surface temperature or surface heat flux. 13 Heat Conduction with Internal Heat Sources Example: A flat plate of thickness 0.1 m has an internal heat source of 16 KW/m3. The conductivity of the plate is 10 W/mK. Heat loss at one side of the plate is 1200 W/m2 and the temperature at the second side of the plate is kept at 60 °C. Find 1) the temperature at the first side; 2) the maximum temperature in the plate; and 3) the heat transfer at the second side of the plate. Solution: Known: q’ = 16 KW/m3, k=10W/mK, q(x=0)=-1200 w/m2, T(x=L)=60, L=0.1 m Find: 1) T(x=0), 2) Tmax, 3) q(x=L) q' x 2 Temperature distribution is: T ( x) = − ⋅ + c1 x + c2 K k 2 𝑑𝑑𝑑𝑑 ′ x Heat transfer inside the plate is: 𝑞𝑞̇ 𝑥𝑥 = −𝑘𝑘 𝑑𝑑𝑑𝑑 = 𝑞𝑞 𝑥𝑥 − 𝑘𝑘𝑐𝑐1 x=0 x=L 14  Heat Conduction with Internal Heat Sources q' x 2 Temperature distribution is: T ( x) = − ⋅ + c1 x + c2 K k 2 𝑑𝑑𝑑𝑑 x Heat transfer inside the plate is: 𝑞𝑞̇ 𝑥𝑥 = −𝑘𝑘 = 𝑞𝑞 ′ 𝑥𝑥 − 𝑘𝑘𝑐𝑐1 𝑑𝑑𝑑𝑑 x=0 x=L Use the two boundary conditions to determine c1 and c2: q(x=0)= -1200 w/m2, -1200 = -10 c1, so c1=120 (K/m) 16 ×103 0.12 T(x=L)=60 °C, 60 = − 10 + 120 0.1 + 𝑐𝑐2 , so c2=56 (°C). 2 1) T(x=0)=56°C 𝑑𝑑𝑑𝑑 𝑞𝑞′ 𝑥𝑥 𝑘𝑘𝑐𝑐1 10×120 2) =− + 𝑐𝑐1 = 0, 𝑥𝑥 = = = 0.075 𝑚𝑚 , 𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚 = 60.5℃ 𝑑𝑑𝑑𝑑 𝑘𝑘 𝑞𝑞𝑞 16×103 3) q(x=L)=16 × 103 × 0.1 − 10 × 120 = 400 w/m2 15  Heat Conduction with Internal Heat Sources Fourier’s Conduction Equation of 1D cylindrical system: Q = −kA dT = −k (2πrL) dT dr dr 𝑑𝑑𝑄𝑄̇ 𝑑𝑑 𝑑𝑑𝑑𝑑 Take the derivative of the Fourier’s Law: = −2𝜋𝜋𝜋𝜋 𝑘𝑘 ⋅ 𝑟𝑟 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑄𝑄̇ 𝑑𝑑𝑄𝑄̇ 1 𝑑𝑑 𝑑𝑑𝑑𝑑 q ' is the internal heat source (W/m ) 3 𝑞𝑞 ′ = = 𝑑𝑑𝑑𝑑 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋 = − 𝑟𝑟 𝑑𝑑𝑑𝑑 (𝑘𝑘 ⋅ 𝑟𝑟 𝑑𝑑𝑑𝑑 ) 1 𝑑𝑑 𝑑𝑑𝑑𝑑 Volume of a cylinder: 𝑉𝑉 = 𝜋𝜋𝑟𝑟 2 𝐿𝐿 The heat equation for cylindrical geometries: 𝑘𝑘 ⋅ 𝑟𝑟 = −𝑞𝑞𝑞 Volume element 𝑑𝑑𝑑𝑑 = 2𝜋𝜋𝑟𝑟𝐿𝐿𝑑𝑑𝑑𝑑 𝑟𝑟 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 General solution 𝑑𝑑 𝑑𝑑𝑑𝑑 𝑘𝑘 ⋅ 𝑟𝑟 = −𝑞𝑞 ′ 𝑟𝑟 when r is not zero: 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑟𝑟 2 𝑘𝑘 ⋅ 𝑟𝑟 = −𝑞𝑞 ′ + 𝑐𝑐 𝑑𝑑𝑑𝑑 2 𝑑𝑑𝑑𝑑 𝑞𝑞 ′ 𝑐𝑐1 = − 𝑟𝑟 + 𝑑𝑑𝑑𝑑 2𝑘𝑘 𝑟𝑟 𝑞𝑞′ 2 𝑇𝑇 = − 4𝑘𝑘 𝑟𝑟 + 𝑐𝑐1 ln 𝑟𝑟 + 𝑐𝑐2 16 Example A current of 200 A is passed through a stainless-steel wire ( k = 19 W/m⋅K ) 3 mm in diameter. The resistivity of the steel is 70 µΩ⋅cm, and the length of the wire is 1 m. The wire is submerged in a liquid at 110 oC and experiences a convection heat transfer coefficient of 4 kW/m2⋅K. Calculate the center temperature of the wire. 1 𝑑𝑑 𝑑𝑑𝑑𝑑 Conduction Equation: 𝑘𝑘 ⋅ 𝑟𝑟 = −𝑞𝑞𝑞 𝑟𝑟 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 2 𝐼𝐼 Heat Generation: 𝑞𝑞 ′ = 𝑖𝑖 2 𝜌𝜌∗ = 𝜌𝜌∗ 𝐴𝐴 Boundary Conditions: 𝑑𝑑𝑑𝑑 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑎𝑎𝑎𝑎 𝑡𝑡𝑡𝑡𝑡 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐: 𝑟𝑟 = 0 = 0 𝑑𝑑𝑑𝑑 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡: 𝑇𝑇 𝑟𝑟 = 𝑟𝑟𝑜𝑜 = 𝑇𝑇𝑜𝑜 Surface temperature is determined by convection: 𝑄𝑄̇ 𝑔𝑔𝑔𝑔𝑔𝑔 = 𝑄𝑄̇ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑞𝑞 ′ 𝑉𝑉 = ℎ 𝐴𝐴 𝑇𝑇𝑜𝑜 − 𝑇𝑇∞ 𝑞𝑞 ′ 𝑉𝑉 𝑞𝑞 ′ 𝑑𝑑 𝑇𝑇𝑜𝑜 = 𝑇𝑇∞ + = 𝑇𝑇∞ + ℎ 𝐴𝐴 4ℎ 2 𝑉𝑉 𝜋𝜋𝑟𝑟 𝐿𝐿 𝑟𝑟 𝑑𝑑 ( = = = ) 𝐴𝐴 2𝜋𝜋𝜋𝜋𝜋𝜋 2 4 17 Example A current of 200 A is passed through a stainless-steel wire ( k = 19 W/m⋅K ) 3 mm in diameter. The resistivity of the steel is 70 µΩ⋅cm, and the length of the wire is 1 m. The wire is submerged in a liquid at 110 oC and experiences a convection heat transfer coefficient of 4 kW/m2⋅K. Calculate the center temperature of the wire. Rearrange the Conduction Equation: 𝑑𝑑 𝑘𝑘 ⋅ 𝑟𝑟 𝑑𝑑𝑑𝑑 = −𝑞𝑞 ′ 𝑟𝑟 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑞𝑞 ′ 2 By integration: 𝑘𝑘 ⋅ 𝑟𝑟 = − 𝑟𝑟 + 𝑐𝑐1 𝑑𝑑𝑑𝑑 2 𝑑𝑑𝑑𝑑 Apply the boundary 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑎𝑎𝑎𝑎 𝑡𝑡𝑡𝑡𝑡 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐: 𝑟𝑟 = 0 = 0 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑞𝑞 ′ We have 𝑐𝑐1 = 0. So = − 𝑟𝑟 𝑑𝑑𝑑𝑑 2𝑘𝑘 𝑞𝑞 ′ 2 By integration: 𝑇𝑇 = − 𝑟𝑟 + 𝑐𝑐2 4𝑘𝑘 Apply the boundary at the surface: 𝑇𝑇 𝑟𝑟 = 𝑟𝑟𝑜𝑜 = 𝑇𝑇𝑜𝑜 𝑞𝑞 ′ 2 𝑐𝑐2 = 𝑇𝑇𝑜𝑜 + 𝑟𝑟 4𝑘𝑘 𝑜𝑜 𝑞𝑞 ′ 2 The centre temperature: 𝑇𝑇 𝑟𝑟 = 0 = 𝑐𝑐2 = 𝑇𝑇𝑜𝑜 + 𝑟𝑟 18 4𝑘𝑘 𝑜𝑜 Example A current of 200 A is passed through a stainless-steel wire ( k = 19 W/m⋅K ) 3 mm in diameter. The resistivity of the steel is 70 µΩ⋅cm, and the length of the wire is 1 m. The wire is submerged in a liquid at 110 oC and experiences a convection heat transfer coefficient of 4 kW/m2⋅K. Calculate the center temperature of the wire. 2 𝐼𝐼 Heat Generation: ′ 2 ∗ 𝑞𝑞 = 𝑖𝑖 𝜌𝜌 = 𝜌𝜌∗ 𝐴𝐴 2 200 𝑞𝑞 ′ = 𝜋𝜋 70 × 10−6 × 10−2 = 5.604 × 108 (𝑊𝑊 ⁄𝑚𝑚3 ) 2 4 0.003 𝑞𝑞 ′ 𝑑𝑑 Surface temperature: 𝑇𝑇𝑜𝑜 = 𝑇𝑇∞ + 4ℎ 𝑞𝑞 ′ 𝑑𝑑 5.604 × 108 × 0.003 𝑇𝑇𝑜𝑜 = 𝑇𝑇∞ + = 110 + = 215.1 ℃ 4ℎ 4 × 4000 𝑞𝑞 ′ 2 The centre temperature: 𝑇𝑇 𝑟𝑟 = 0 = 𝑇𝑇𝑜𝑜 + 𝑟𝑟 4𝑘𝑘 𝑜𝑜 2 5.604 × 108 0.003 𝑇𝑇 𝑟𝑟 = 0 = 215.1 + = 231.7℃ 4 × 19 2 19 Heat Conduction with Internal Heat Sources dT dT Fourier Conduction Equation of 1D spherical system: Q = − kA = −k (4πr 2 ) dr dr Derivative of Q : 𝑑𝑑𝑄𝑄̇ 𝑑𝑑 2 𝑑𝑑𝑑𝑑 = −4𝜋𝜋 (𝑘𝑘 ⋅ 𝑟𝑟 ) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 4 𝑑𝑑𝑄𝑄̇ 𝑑𝑑 𝑄𝑄̇ 1 𝑑𝑑 2 𝑑𝑑𝑑𝑑 Volume of a sphere: 𝑉𝑉 = 𝜋𝜋𝑟𝑟 3 𝑞𝑞𝑞 = = = − (𝑘𝑘 ⋅ 𝑟𝑟 ) 3 𝑑𝑑𝑑𝑑 4𝜋𝜋𝑟𝑟 2 𝑑𝑑𝑑𝑑 𝑟𝑟 2 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 Volume element 𝑑𝑑𝑑𝑑 = 4𝜋𝜋𝑟𝑟 2 𝑑𝑑𝑑𝑑 q ' is the internal heat source (W/m3) 1 𝑑𝑑 2 𝑑𝑑𝑑𝑑 The heat equation for spherical geometries: 𝑘𝑘 ⋅ 𝑟𝑟 = −𝑞𝑞𝑞 𝑟𝑟 2 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 20 Heat Conduction with Internal Heat Sources Example: Find the temperature distribution inside a solid ball, which has a constant conductivity, k, and its radius is ro. The surface temperature is To. For constant k, re-arrange the heat equation, we have d 2 dT q' (r ) = − r2 dr dr k dT2 q' r 3 Integrate once, r =− + c1 dr k 3 At the centre of the ball, r=0, dT/dr = 0, so c1 = 0. dT q' r =− dr k 3 q' r 2 Integrate again, T = − + c2 k 6 q ' ro2 At the surface of the ball, T(r=ro)=To, so c2 = To + k 6 q' 2 2 T = − (r − ro ) + To 6k 21 Week 2.3 General Heat Conduction Equation - the Heat Diffusion Equation 22 General Heat Conduction Equation (Cartesian coordinates) Energy balance is 23 Energy quantities are given by: Fourier’s Law X-direction: Y-direction: Z-direction: 24 Net heat transfer in x-direction: 𝜕𝜕 𝜕𝜕𝜕𝜕 𝑞𝑞𝑥𝑥 − 𝑞𝑞𝑥𝑥+𝑑𝑑𝑑𝑑 = 𝑘𝑘 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝜕𝜕𝑥𝑥 𝜕𝜕𝑥𝑥 𝜕𝜕 𝜕𝜕𝜕𝜕 Y-direction: 𝑞𝑞𝑦𝑦 − 𝑞𝑞𝑦𝑦+𝑑𝑑𝑑𝑑 = 𝑘𝑘 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕 𝜕𝜕𝜕𝜕 Z-direction: 𝑞𝑞𝑧𝑧 − 𝑞𝑞𝑧𝑧+𝑑𝑑𝑑𝑑 = 𝑘𝑘 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 25 Heat generation: 𝑞𝑞𝑔𝑔𝑔𝑔𝑔𝑔 = 𝑞𝑞 ′ (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕 Change in the internal energy: = 𝑚𝑚𝑚𝑚 = 𝜌𝜌𝜌𝜌 (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕 so that the general three-dimensional heat-conduction equation is 𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕 𝜕𝜕𝜕𝜕 ′ 𝜕𝜕𝜕𝜕 𝑘𝑘 + 𝑘𝑘 + 𝑘𝑘 + 𝑞𝑞 = 𝜌𝜌𝜌𝜌 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 26 General Heat Conduction Equation (cylindrical coordinates) 1 ∂  ∂T  1 ∂  ∂T  ∂  ∂T  ∂T  kr + 2  k  +  k  + q ' = ρc p r ∂r  ∂r  r ∂φ  ∂φ  ∂z  ∂z  ∂t 27 Tutorial Questions Tutorial 1: Q7, Q8, Q15 – Q21 28