MECH 375, Heat Transfer Handout #5: Unsteady Conduction

Transcription

1 1 MECH 375, Heat Transfer Handout #5: Unsteady Conduction Amir Maleki, Fall 2018

2 2 T H I S PA P E R P R O P O S E D A C A N C E R T R E AT M E N T T H AT U S E S N A N O PA R T I - C L E S W I T H T U N A B L E R A D I AT I O N P R O P E R T I E S. U P O N I N J E C T I O N I N T H E B L O O D S T R E A M, T H E PA R T I C L E S F I N D T H E I R W AY S T O T H E M A L I G N A N T C E L L S A N D A D H E R E T O T H E M. F I N A L LY, T H E T U M O R W I L L B E D E S T R O Y E D W I T H T H E H E AT G E N E R AT E D I N T H E PA R T I C L E S B Y A B S O R B I N G A L A S E R B E A M.

3 Learning Objectives By the end of this lecture, you will be able to Analyze an unsteady problem to see if the spatial variations are important or not. Explain the physical meaning of Biot number. Apply the Lumped Capacitance approach to solve an unsteady heat conduction problem. Use 1-term approximation method to deal with problems where spatial variations are significant. 0 A large margin is intentionally left here for your notes Use similarity solutions to deal with problems that approximately infinite in one or more directions. Construct solutions for 3D transient conduction problems using the product solution approach.

4 Unsteady Heat Transfer Unsteady (or transient or time dependent) heat transfer problems are abundant. Examples are: The vectorial form of unsteady conduction equation with a source term is:.(k T) + S = ρc p dt dt We will follow three approaches to deal with this problem. Lumped Parameter Analysis also known as Lumped Capacitance Analysis One-term approximation Semi-infinite systems The other approaches, which we will not talk about, include Graphical Methods (Bender-Schmidt Plots) Numerical Methods Analytical methods (such as Laplace transform).

5 Lumped Parameter Analysis (LPA) In LPA, we assume the object of study has a uniform tempera-ture (no spatial temperature gradient). We consider a simple case where the temperature of the object is dropping due to convection. We set T T T i T = exp ( ) ha s ρvc t τ = ρvc ha s which is called thermal time constant. Then, T T T i T = exp ( t τ ) The initial temperature difference drops 63% after one time constant. For the initial temperature difference to drop 99%, we need t = 4.6τ to elapse.

6 When is LPA valid? We consider a simple case of a plane wall of thickness L exposed to convection heat transfer h. We can write: q = T 1,s T2,s R cond = T 2,s T R conv T 1,s T2,s T 2,s T = R cond R conv The left hand side the ratio of temperature drop due to conductive resistance over temperature drop due to convective resistance. R cond R conv = L/kA 1/hA = hl k This ratio is dimensionless and is called Biot number. Bi = hl k In LPA, we assume temperature spatial gradient is negligible, therefore we require T 1,s T2,s T 2,s T 1 Bi 1 0 if Bi 1, the temperature drop due to conduction is approximately equal to that of convection. 0 if Bi 1, the temperature drop due to conduction is much greater than that of convection.

7 7 We generalize LPA by defining a Biot number Bi = hl c k The characteristic length L c is defined by L c = V A s Geometry L c Plane wall of thickness L L/2 Cylinder of radius R R/2 Sphere of radius R R/3 The physical intuition of LPA method is that the resistance due to conduction is so much smaller than that of convection that we totally neglect temperature drop due to conduction. The LPA is considerably simpler than any other approach, therefore the first thing one should do is to calculate Bi and if the condition Bi < 0.1 is satisfied, we assume LPA is valid.

8 Example 1 A thermocouple junction in installed in a pipeline to measure the temperature of a gas stream. The thermocouple junction is a sphere with diameter D = 1 mm, and is made of a material with k = 20 W/m.K, c = 400 J/kg.K and ρ = 8500 kg/m 3. The junction is initially at T i = 25 C and is placed in the gas stream with temperature T = 200 C. How long does it take for the thermocouple to reach to T = 199 C?

9 Example 2 The base plate of an iron has a thickness of L = 7 mm and is made from an aluminum alloy with ρ = 2800 kg/m 3, c = 900 J/kg.K, k = 180 W/m.K and ε = An electric resistance heater is attached to the inner surface of the plate, while the outer surface is exposed to ambient air and large surroundings at T = T surr = 25 C. The areas of both the inner and outer surfaces are A s = 0.04 m 2. If an approximately uniform heat flux of q = W/m 2 applied to the inner surface of the base plate and the convection coefficient at the outer surface is h = 10 W/m 2.K, estimate the time required for the plate to reach a temperature of 135 C.

10 10

11 Spatial Effects We will run into examples that LPA does not apply any more. In other words, the spatial gradient inside the object due to conduction is significant and cannot be ignored. For the sake of simplicity, we only consider such problems with 1D spatial gradient. Assuming constant properties: 2 T x 2 = 1 α T t. α = κ ρc p is called thermal diffusivity. The above partial differential equation requires two boundary conditions and one initial condition: For example, for the plane wall shown here, the boundary conditions are α is a measure of how quickly heat can penetrate inside a fluid. and initial condition is

12 Plane Wall The analytical solution is T T T T i = n=1 C nexp ( ζ 2 nfo ) cos (ζ n x/l) Fo = αt L 2 C n = 4 sin ζ n 2ζ n +sin 2ζ n and the descrete eigenvalues ζ n are positive roots of transcendental equation: ζ n tan ζ n = Bi Here Bi is the Biot number defined by Bi = hl k 1-term approximation 0 Fourier number is often called dimensionless time and denoted with t. If Fo > 0.2, we can approximate the above solution with only the first term of the infinite series: This formulation can be applied to a wall of thickness L, where at x = 0 the wall is insulated and is exposed to heat convection at x = L.

13 13

14 Infinite Cylinder The one-term approximation writes T T T T i = C 1 exp ( ζ 2 1 Fo) J 0 (ζ 1 r/r 0 ) Q Q 0 = 1 2C 1exp( ζ1 2Fo ) ζ 1 J 1 (ζ 1 ) Sphere T T T T i = C 1 exp ( ζ 2 1 Fo) 1 ζ 1 r/r 0 sin (ζ 1 r/r 0 ) Q Q 0 = 1 3C 1exp( ζ1 2Fo ) ζ1 2 (sin ζ 1 ζ 1 cos ζ 1 ) C 1 and ζ 1 are computed as a function Bi, which in radial systems is defined here by Bi = hr 0 k

15 Example A solid sphere made of fused-quarts has a thermal diffusivity of α = m 2 /s, a diameter of D = 2.5 cm, and a thermal conductivity of k = 1.52 W/m.K. The sphere is initially at a uniform temperature of T i = 25 C and is suddenly subjected to a convection heating environment characterized by T = 200 C and h = 110 W/m 2.K. Assume radiation heat transfer is negligible. After 4 minute into this heating process, calculate the temperatures at a) the center of the sphere; b) at a depth of 6.1 mm from the surface of the sphere.

16 Semi-infinite solids Semi-infinite geometries also have rather simple analytical solution. Similarity solution starts by defining similarity variable The semi-infinite unsteady heat conduction is solved using the so called similarity solutions Depending on the choice boundary condition, the final solution may vary. We will consider these cases. η = x/ 4αt, which subsequently turns the PDE into a ODE.

17 17 Case 1: Constant surface temperature T(0, t) = T s Case 2: Constant surface heat flux q s = q 0 Case 3: Surface convection k T x = h [T T(0, t)] x=0 T(x, t) T s T i T s ( x = erfc 2 αt ) [ ( )] [ ( hx exp k + h2 αt erfc k 2 x 2 αt + h )] αt k

18 18

19 Example Irreversible thermal injury is associated to death of cells, which occurs if the cells are contact for a temperature T larger than 48 C for a period of t 10s. How long a contact with a device at T = 100 C has to last so that the thermal injury destroys the entire skin layer?

20 Product solution approach We can construct 2D geometries by composing 1D geometries. We consider three base solutions Plane Wall Infinite Cylinder P(x, t) = T(x,t) T T i T C(r, t) = T(r,t) T T i T Semi-infinite Solid S(x, t) = T(x,t) T T i T We can construct various 3D geometries: T T T i T 3D solid = Π j T T T i T intersection solid j

21 Example: A short brass cylinder of diameter D = 10 cm and height H = 12 cm is initially at a uniform temperature T i = 120 C. The cylinder is now placed in atmospheric air at T = 25 C, where heat transfer takes place by convection, with a heat transfer coefficient of h = 60 W/m 2.K. Calculate the temperature at (a) the center of the cylinder and (b) the center of the top surface of the cylinder 15 min after the start of the cooling.

22 22