Introduction to Heat and Mass Transfer. Week 8

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1 Introduction to Heat and Mass Transfer Week 8

2 Next Topic Transient Conduction» Analytical Method Plane Wall Radial Systems Semi-infinite Solid Multidimensional Effects

3 Analytical Method Lumped system analysis valid only when temperature gradients within the solid ignored In general, we must solve heat diffusion equation with appropriate spatial coordinates along with the transient term For one-dimensional heat conduction with constant properties and no thermal energy generation T 1 T x t Need two boundary conditions and one initial condition

4 Analytical Method (contd.) T(x,0) = T i T(r,0) = T i h, T h, T h, T h, T Sphere L x x * = x/l T(r,0) = T i r o r r * = r/r o Large Plane Wall h, T Long Cylinder r o r * = r/r o

5 Analytical Method (contd.) The solution consists of spatial and temporal variables as well as thermal properties and other parameters Non-dimensional analysis helps in a more simplified solutions via casting relevant variables into groups We will consider the following non-dimensional groups:» Temperature» Spatial Coordinate» Temporal Coordinate T T T T i x i x L t t L r r r o

6 Analytical Method (contd.) With the above non-dimensional groups, we need to solve: x t Boundary conditions can also be non-dimensionalized using the same groups In each case, we obtain solutions that depend on constants which are functions of Biot number These constants are listed in Table 5.1 for the cases of large plane wall, infinite (long) cylinder and sphere

7 Plane Wall Exact solution can be written as: C n plane wall planewall C exp Fo cos x n1 n n n 4sin n sin n n plane wall tan Bi n n For Fo > 0., the above infinite series converges and an approximate solution can be given as: plane wall C exp Fo cos x planewall planewall exp cos x 0 1 C Fo

8 Plane Wall (contd.) Total energy transfer from the plane wall at any time t can be written as: where,» Q 0 = Maximum possible energy transfer Maximum possible energy transfer occurs at infinitely long time Q planewall Q sin Q mc T T 0 p i

9 Radial Systems Exact solutions for infinite (long) cylinder and sphere available just like plane wall For Fo > 0., the infinite series solutions converge and an approximate solutions can be given as: cylinder C exp Fo J r cylinder cylinder exp J r sphere C C sin r 1 exp Fo 1 1 r 1 Fo C exp Fo sphere 0 sin r r 1 1 sphere 0 1 1

10 Radial Systems (contd.) Total energy transfer from the infinite cylinder or sphere at any time t can be written as: where, Q sphere Q Q 1 J cylinder 0 Q sin cos » Q 0 = Maximum possible energy transfer Maximum possible energy transfer occurs at infinitely long time

11

12 補充! Analytical Method (contd.) Use of Heisler-Grober charts (plane wall)

13 Analytical Method (contd.) Use of Heisler-Grober charts (plane wall)

14 Analytical Method (contd.) Use of Heisler-Grober charts (plane wall)

15 Example Steel is annealed to relieve stresses to make it less brittle. Consider a 100-mm thick plate that is initially at an uniform temperature of 300C and is heated (on both sides) in a gas-fired furnace for which the surrounding temperature is 700C and heat transfer coefficient is 500 W/m -K.» How long does it take to attain a minimum temperature of 550C in the plate? Use of one-term approximation equations!

16 Example A spherical hailstone 5 mm in diameter is formed in a cloud at -30C. The stone begins to fall through warmer air at 5C. Convection heat transfer coefficient can be considered to be 50 W/m -K and hailstone can be modeled as ice.» How long does it take before the outer surface of the hailstone begins to melt?» What is the temperature of hailstone s center at this instant?» How much energy (J) is transferred to the hailstone? Use of Heisler-Grober charts!

17 HW # 5 prob. 4 Steel is annealed to relieve stresses to make it less brittle. Consider a 100-mm thick plate that is initially at a uniform temperature of 300C and is heated (on both sides) in a gas-fired furnace for which the surrounding temperature is 700C and heat transfer coefficient is 500 W/m -K.» How long does it take to attain a minimum temperature of 550C in the plate? Use of Heisler-Grober charts!

18 HW # 5 prob. 5 A spherical hailstone 5 mm in diameter is formed in a cloud at -30C. The stone begins to fall through warmer air at 5C. Convection heat transfer coefficient can be considered to be 50 W/m -K and hailstone can be modeled as ice.» How long does it take before the outer surface of the hailstone begins to melt?» What is the temperature of hailstone s center at this instant?» How much energy (J) is transferred to the hailstone? Use of one-term approximation equations!

19 Semi-infinite Solid A solid which extends infinitely in all directions except one particular direction Thermal changes at the surface do not transmit quickly through the remaining body Using a similarity variable (x/t 1/ ), the PDE governing heat transfer through the semi-infinite solid reduced to an ODE expressed in terms of the similarity variable Note that the similarity variable dictates the following peculiar behavior for semi-infinite solids:» As x, we can consider t 0» As t, we can consider x 0

20 Semi-infinite Solid (contd.)

21 Semi-infinite Solid (contd.) Solutions usually obtained in terms of error function and complementary error function x exp erf x u du 0 erfc x Error functions tabulated in Table B- Three common surface conditions of practical interest:» Constant surface temperature» Constant surface heat flux» Constant surface convection 1erf x

22 Semi-infinite Solid (contd.)

23 Semi-infinite Solid (contd.) Constant Surface Temperature Constant Surface Heat Flux T x, t T x s erf T T t Constant Surface Convection i s 1/ q " t " 0 x qx x 0 T x, t T exp erfc i k 4 t k t, T x t Ti x hx h t x h t erfc exp erfc T T t k k t k i

24 Semi-infinite Solid (contd.)

25 HW # 5 prob. 5 A thick oak wall initially at 5C is suddenly exposed to surrounding of combustion products where the temperature is 800C and heat transfer coefficient is 0 W/m -K.» Determine the time of exposure required for the surface to reach the ignition temperature of 400C Use of Fig. 5.8!

26 補充! Multidimensional Effects Transient problems with heat transfer in two or three dimensions can be considered using the solutions obtained for one dimensional transient cases Usually multiple dimensions can be considered as product of the following one-dimensional solutions: S x, t T x, t T T T i semi-infinite solid P x, t T x, t T i T T plane wall C r, t T r, t T i T T infinite cylinder

27 Multidimensional Effects (contd.)

28

29 Multidimensional Effects (contd.)

30 HW # 5 prob. 6 A stainless steel cylinder (k=16.3 W/mC, α=0.44x10-5 m /s, ρ =7817kg/m 3, c= 460 J/kgC) is heated to a uniform temperature of 00C and then allowed to cool in an environment where the air temperature is maintained constant at 30C. The convection heat transfer coefficient may be taken as 00 W/m C. The cylinder has a diameter of 10 cm and a length of 15 cm. Calculate the temperature of the geometric center of the cylinder after a time of 10 min using product solution method and Heisler-Grober charts. Also calculate the heat loss.

31 HW # 5 will due on 11/3 (Thursday), right before the class! Late HW will not be accepted!!

32 Transient Analysis Strategy For a general transient conduction problem, we can follow the following procedure:» Compute Biot number and check whether Bi < 0.1» If Bi < 0.1, then we can usually implement lumped system analysis with reasonable accuracy» Otherwise we compute required constants as function of Biot number and implement approximate 1D analytical solutions for large plane walls (small thickness) or infinitely long cylinders (l/r o >10) or spheres (We can implement Heisler-Grober charts for the these standard shapes)» If multidimensional effects are present, then we consider the given solid as product of relevant 1D analytical solutions for standard shapes

33 Questions What is the physical significance of Biot number? What is the physical significance of Fourier number? What is the product solution method? How can it be used for multidimensional systems?

34 Closure Coverage thus far..» talked about non-dimensionalization of heat diffusion equation to solve transient conduction problems» discussed analytical solutions for large plane wall, infinite cylinder and sphere as well as semi-infinite solid» considered multidimensional heat transfer treatment

35 Closure (contd.) Analytical solutions for one dimensional transient heat conduction through large plane wall, infinite cylinder and sphere using non-dimensional variables Use of Heisler-Grober charts for solving the above heat conduction problems Similarity solutions for semi-infinite solids Multidimensional heat transfer analysis using the method of product solutions f x, t, Bi

36 Next Topic Transient Conduction» Numerical Method Finite Difference Formulation Explicit Method Stability Criterion Implicit Method

37 Numerical Method For transient heat transfer problems:» Lumped capacitance analysis restricted to few cases where temperature gradients absent» Analytical solutions applicable to few simple geometries under benign boundary conditions Numerical methods useful for solving more complicated unsteady problems For two dimensional, transient conditions with constant properties and no thermal energy generation, we solve: 1 T T T t x y

38 Finite Difference Formulation i, j+1 - If Dx = Dy Uniform Mesh i-1, j i, j i+1, j Spatial temperature derivatives just like D steady conduction Dy Dx i, j-1 How do we approximate time derivative of temperature? Explicit Method Implicit Method Two Dimensional Mesh

39 Explicit Method Using forward difference approximation for the temporal temperature derivative, we can write: T t i, j 1 T For explicit method, we use temperature values at previous time step when using central difference approximations for the spatial temperature derivatives T P1 T P i, j i, j Dt 1,, 1, x Dx y Dy i, j explicit T T T P P P i j i j i j T i, j explicit T T T P P P i, j1 i, j i, j1

40 Explicit Method (contd.) Substitute into heat diffusion equation to obtain discretized equation for any interior node: explicit P1 P P P P P T Fo T Fo T T T T 1 4 i, j i, j i1, j i1, j i, j1 i, j1 In the above equation, RHS depends only on temperature at previous time step Separate discretized equations at boundary nodes No solver required because current temperature depends only on previous temperatures Explicit method very easy to implement numerically

41 Stability Criterion Although the explicit method is rather simple it is almost never used in practice since it is conditionally stable Using general stability analysis, we can show that the time step is restricted to obtain bounded solutions von Neumann stability limit for D: Fo explicit 14Fo explicit Dt As Dx refined to reduce error, Dt becomes increasingly explicit restrictive and computationally cumbersome Dx 4

42 Implicit Method Using forward difference approximation for the temporal temperature derivative, we can write: For implicit method, we use temperature values at current time step when using central difference approximations for the spatial temperature derivatives T P1 P1 P1 1,, 1, x Dx y Dy i, j implicit T t T T T i j i j i j i, j 1 T P1 i, j i, j T Dt T i, j P implicit T T T P1 P1 P1 i, j1 i, j i, j1

43 Implicit Method (contd.) Substitute into heat diffusion equation to obtain discretized equation for any interior node: implicit P P 1 P 1 P 1 P 1 P 1 T Fo T Fo T T T T 1 4 i, j i, j i1, j i1, j i, j1 i, j1 In the above equation, RHS depends on temperatures at current time step Separate discretized equations at boundary nodes The resulting linear system of simultaneous equations solved using either direct (e.g. Matrix Inversion) or iterative (e.g. Gauss Seidel) techniques Implicit method rather difficult to implement numerically

44 Implicit Method (contd.) Although implicit method is difficult to code, it is unconditionally stable i.e. large time steps are allowed However, Dt should be chosen such that the underlying physics of the given problem is resolved correctly Numerical methods (explicit or implicit) solve discretized algebraic equations that are approximations of the original PDE the accuracy increases using mesh refinement (smaller Dx) and finer time steps (smaller Dt)

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