Conduction Heat Transfer Reading Problems 10-1 10-6 10-20, 10-48, 10-59, 10-70, 10-75, 10-92 10-117, 10-123, 10-151, 10-156, 10-162 11-1 11-2 11-14, 11-20, 11-36, 11-41, 11-46, 11-53, 11-104 Fourier Law of Heat Conduction We can examine a small representative element of the bar over the range x x + x. Just as in the case of a thermodynamic closed system, we can treat this slice as a closed system where no mass flows through any of the six faces. insulated x+ Dx x=l Q x+ Dx x x x=0 Q x A g From a 1 st law energy balance: E t = Q x Q x+ x + Ẇ Using Fourier s law of heat conduction we can write the general 1-D conduction equation as Q x = ka T x ( k T ) x x }{{} longitudinal conduction + ġ }{{} internal heat generation = ρc T t }{{} thermal inertia 1
Special Cases 1. Multidimensional Systems: The general conduction equation can be extended to three dimensional Cartesian systems as follows: ( k T ) + ( k T ) + ( k T ) + ġ = ρc T x x y y z z t 2. Constant Properties: If we assume that properties are independent of temperature, then the conductivity can be taken outside the derivative 2 T + ġ k = 1 T α t 3. Steady State: If t then all terms t 0 2 T = ġ k 4. No Internal Heat Generation: Poisson s Equation 2 T = 0 Laplace s Equation Thermal Resistance Networks Thermal circuits based on heat flow rate, Q, temperature difference, T and thermal resistance, R, enable analysis of complex systems. This is analogous to current flow through electrical circuits where, I = V /R Thermal Electrical Role Q, heat flow rate I, current flow transfer mechanism T in T out, temperature difference V in V out, voltage difference driving force R, thermal resistance R, electrical resistance impedance to flow Resistances in Series The total heat flow across the system can be written as Q = T 1 T 2 R total where R total = 4 R i i=1 2
Resistances in Parallel For systems of parallel flow paths as shown below L k 1 R 1 Q k 2 R 2 Q k 3 R 3 T 1 T 2 In general, for parallel networks we can use a parallel resistor network as follows: R 1 R total T 1 R2 T 2 T 1 T 2 = R 3 1 = 1 + 1 + 1 + and Q = T 1 T 2 R total R 1 R 2 R 3 R total 3
Thermal Contact Resistance the thermal interface between contacting materials offers resistance to heat flow the total heat flow rate can be written as Q total = h c A T interface The conductance can be written in terms of a resistance as h c A = Q total T = 1 R c Table 10-2 can be used to obtain some representative values for contact conductance. Cylindrical Systems For steady, 1D heat flow from T 1 to T 2 in a cylindrical system as shown below k r 2 r 1 T 1 Q r r A=2prL T 2 L we can write Q r = T ( ) 1 T 2 ln(r2 /r 1 ) ( ) where R = ln(r2 /r 1 ) 2πkL 2πkL 4
Composite Cylinders R total = R 1 + R 2 + R 3 + R 4 = 1 h 1 A 1 + ln(r 2/r 1 ) 2πk 2 L + ln(r 3/r 2 ) 2πk 3 L + 1 h 4 A 4 Example 5-1: Determine the temperature (T 1 ) of an electric wire surrounded by a layer of plastic insulation with a thermal conductivity if 0.15 W/mK when the thickness of the insulation is a) 2 mm and b) 4 mm, subject to the following conditions: Given: Find: I = 10 A T 1 =??? ɛ = ɛ 1 ɛ 2 = 8 V when: D = 3 mm δ = 2 mm L = 5 m δ = 4 mm k = 0.15 W/mK T = 30 C h = 12 W/m 2 K 5
Critical Thickness of Insulation Although the purpose of adding more insulation is to increase resistance and decrease heat transfer, we can see from the resistor network that increasing r o actually results in a decrease in the convective resistance. R total = R 1 + R 2 = ln(r o/r i ) 2πkL + 1 h2πr o L Could there be a situation in which adding insulation increases the overall heat transfer? r cr,cyl = k h [m] 6
Spherical Systems We can write the heat transfer as Q = 4πkr ir o (r 0 r i ) (T i T o ) = (T i T o ) R where R = r o r i 4πkr i r o T i r i r o T o Notes: 1. the critical thickness of insulation for a spherical shell is given as r cr,sphere = 2k h [m] Heat Transfer from Finned Surfaces The conduction equation for a fin with constant cross section is ka c d 2 T x 2 }{{} longitudinal conduction hp (T T ) = 0 }{{} lateral convection The temperature difference between the fin and the surroundings (temperature excess) is usually expressed as θ = T (x) T 7
which allows the 1-D fin equation to be written as d 2 θ dx 2 m2 θ = 0 where the fin parameter m is m = ( hp ka c ) 1/2 The solution to the differential equation for θ is θ(x) = C 1 sinh(mx) + C 2 cosh(mx) [ θ(x) = C 1 e mx + C 2 e mx ] Potential boundary conditions include: Base: @x = 0 θ = θ b Tip: @x = L θ = θ L [T -specified tip] θ = dθ = 0 dx x=l [adiabatic (insulated) tip] θ 0 [infinitely long fin] 8
Fin Efficiency and Effectiveness The dimensionless parameter that compares the actual heat transfer from the fin to the ideal heat transfer from the fin is the fin efficiency η = actual heat transfer rate = maximum heat transfer rate when the entire fin is at T b Q b hp Lθ b An alternative figure of merit is the fin effectiveness given as ɛ fin = total fin heat transfer the heat transfer that would have occurred through the base area in the absence of the fin = Q b ha c θ b Transient Heat Conduction The Biot number can be obtained as follows: T H T s T s T = L/(k A) 1/(h A) = internal resistance to H.T. external resistance to H.T. = hl k = Bi R int << R ext : the Biot number is small and we can conclude T H T s << T s T and in the limit T H T s R ext << R int : the Biot number is large and we can conclude T s T << T H T s and in the limit T s T 9
Transient Conduction Analysis if the internal temperature of a body remains relatively constant with respect to time can be treated as a lumped system analysis heat transfer is a function of time only, T = T (t) typical criteria for lumped system analysis Bi 0.1 Lumped System Analysis For the 3-D body of volume V and surface area A, we can use a lumped system analysis if Bi = hv ka < 0.1 results in an error of less that 5% For an incompressible substance specific heat is constant and we can write dt mc }{{} dt = Ah }{{} (T T ) C th 1/R th where C th = lumped capacitance and R th = thermal resistance We can integrate and apply the initial condition, T = T i @t = 0 to obtain T (t) T T i T = e t/(r th C th) = e t/τ = e bt 10
where the time constant can be written as 1 b = τ = R th C th = thermal time constant = mc Ah = ρv C Ah Example 5-2: Determine the time it takes a fuse to melt if a current of 3 A suddenly flows through the fuse subject to the following conditions: Given: D = 0.1 mm T melt = 900 C k = 20 W/mK L = 10 mm T = 30 C α = 5 10 5 m 2 /s k/ρc p Assume: constant resistance R = 0.2 ohms the overall heat transfer coefficient is h = h conv + h rad = 10 W/m 2 K neglect any conduction losses to the fuse support 11
Approximate Analytical and Graphical Solutions (Heisler Charts) The lumped system analysis can be used if Bi = hl/k 0.1 but what if Bi > 0.1 we must find a solution to the PDE where temperature is T (x, t) = f(x, L, t, k, α, h, T i, T ) 2 T x 2 = 1 α T t The analytical solution to this equation takes the form of a series solution T (x, t) T T i T = n=1,3,5... A n e ( nλ L ) 2 αt cos ( ) nλx L By using dimensionless groups, we can reduce the temperature dependence to 3 dimensionless parameters Dimensionless Group Formulation temperature position θ(x, t) = T (x, t) T T i T x = x/l heat transfer Bi = hl/k Biot number time F o = αt/l 2 Fourier number note: Cengel uses τ instead of F o. With this, two approaches are possible 1. use the first term of the infinite series solution. This method is only valid for F o > 0.2 2. use the Heisler charts for each geometry as shown in Figs. 11-15, 11-16 and 11-17 12
First term solution: F o > 0.2 error about 2% max. Plane Wall: θ wall (x, t) = T (x, t) T T i T = A 1 e λ2 1 F o cos(λ 1 x/l) Cylinder: Sphere: θ cyl (r, t) = T (r, t) T T i T = A 1 e λ2 1 F o J 0 (λ 1 r/r o ) θ sph (r, t) = T (r, t) T T i T = A 1 e λ2 1 F o sin(λ 1r/r o ) λ 1 r/r o λ 1, A 1 can be determined from Table 11-2 based on the calculated value of the Biot number (will likely require some interpolation). The Bessel function, J 0 can be calculated using Table 11-3. Heisler Charts find T 0 at the center for a given time (Table 11-15 a, Table 11-16 a or Table 11-17 a) find T at other locations at the same time (Table 11-15 b, Table 11-16 b or Table 11-17 b) find Q tot up to time t (Table 11-15 c, Table 11-16 c or Table 11-17 c) Example 5-3: An aluminum plate made of Al 2024-T6 with a thickness of 0.15 m is initially at a temperature of 300 K. It is then placed in a furnace at 800 K with a convection coefficient of 500 W/m 2 K. Find: i) the time (s) for the plate midplane to reach 700 K ii) the surface temperature at this condition. Use both the Heisler charts and the approximate analytical, first term solution. 13