1 CHAPTER 5 POROUS MEDIA. 2 5.1 Examples of Conduction in Porous Media component electronic micro channels coolant (d) coolant porous material (e) Fig.

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Presentation transcript:

1 CHAPTER 5 POROUS MEDIA

2 5.1 Examples of Conduction in Porous Media component electronic micro channels coolant (d) coolant porous material (e) Fig. 5.1

3 5.2 Simplified Heat Transfer Model Assume: Porosity Definition: Porosity (a) (5.1)

4 Model: Pores are straight channels (a) and (b) into eq. (5.1) or (b) (c) (5.2) and solid wall area (5.3) Assume: Porosity is constant

Governing Equation: Cartesian Coordinates One-dimensional transient conduction = surface area = flow rate Fig x q dx = porosity = energy generation Assumptions: (1) Constant porosity (2) Constant flow rate (3) Constant properties (solid and fluid)

6 (4) Solid and fluid at same temperature (5) Negligible changes in kinetic and potential energy Conservation of energy for element dx Use Fourier’s law Subscripts: s = solid, f = fluid (d) (e) Define: Conductivity of the solid-fluid matrix as

7 (5.4) (e) becomes Use (f) to formulate (f) (g) (h) Rate of energy change within the element (i)

8 Define: Heat capacity of the solid-fluid matrixas (5.5) (i) becomes (f), (g), (h) and (j) into (d) (j) (5.6) = thermal diffusivity of solid-fluid matrix, defined as (5.7)

Boundary Conditions (i) Specified temperature or (5.8) At inlet or outlet

10 (ii) Convection at outlet boundary (5.9) (iii) Inlet supply reservoir: Conservation of energy for the control volume shown (5.10)

Cylindrical Systems Conservation of energy and Fourier’s conduction law applied to the element dr: m  r dr Fig. 5.5 (5.10)

Applications Example 5.1: Steady State Conduction in a Porous Plate Plate thickness = L Solution (1) Observations 1-D conduction in porous plate Heated at x = L by convection: h, Coolant reservoir temperature = Design hot side temperature = Determine:

13 depends on (2) Origin and Coordinates See Fig (3) Formulation (i) Assumptions (1) Solid and fluid at same temperature (2) Constant porosity (3) Constant properties (4) Constant flow rate (5) No energy generation (6) Steady state

14 (ii) Governing Equation Eq. (5.6): or (a) is coolant flow parameter defined as (b)

15 (iii) Boundary Conditions (c) Convection at eq. (5.9) (d) = conductivity of the solid material Supply reservoir, eq. (5.10), use definition of

16 (4) Solution Integrate (a) twice Substitute into (e), introduce the Biot number (e) and(f) (g) BC (c) and (d) giveand

17 (h) Determine required for: set x = Land in (g) Solve for, use definitions of and (i)

18 Dimensional check Limiting check: (5) Checking (i) If h = 0 becomes insulated, entire Setting in (g) gives plate is at should be atSetting and in (g) gives (ii) If entire plate is atSetting (iii) If in (g) gives BC (c) shows plate is insulated at entire plate at Settingin (g) gives (iv) If

19 (6) Comments (i) Solution (g) shows that increasinglowers T (ii) Solution depends on two parameters, and (iii) Alternate solution for the required flow rate: Conservation of energy for a control volume from supply reservoir and Use (d) to eliminate and rearrange

20 or Using the definition of

21 Example 5.2:Transient Conduction in a Porous Plate Plate thickness = L L Am/  x 1 T Determine: Transient temperature Reservoir temperature = Initial temperature = (1) Observations One-D transient, porous plate At steady state Sudden change in surface temperature to T 1

22 (2) Origin and Coordinates (3) Formulation (i) Assumptions (1) Solid and fluid atsame temperatures (2) Constant porosity (3) Constant properties (4) Constant flow rate (5) No energy generation (6) Initially flow is established, plate temperature =

23 (ii) Governing Equation (iii) Boundary and Initial Conditions (5.6) L Am/  x 1 T (1) (2) (3)

24 (4) Solution Dimensionless form. Let = dimensionless temperature = distance = time= coolant flow rate parameter

25 Governing equation and boundary and initial conditions become Assume a product solution to (a) (a) (1) (2) (3) (b)

26 (b) into (a), separate variables (c) (d) Solution to (c) depends onand. Only gives characteristic values. Thus solution is given by equation (A-6a) (e) andare constant, is defined as (f)

27 The solution to (d) is (g) is constant. BC (1) gives (h) B.C. (2) gives the characteristic equation (e) and (g) into (b) (i) (j)

28 Initial condition (k) Orthogonality gives the constant. (l) Multiplying both sides of (k) by is (3.7a), shows that the weighting function Comparing (c) with the Sturm-Liouville equation, eq.

29 Integrate from toand apply orthogonality,eq. (3.9), Evaluate the integrals and solve for (m) (5) Checking Dimensional check Limiting check :

30 (6) Comments In applications where coolant weight is an important design factor, weight requirement as determined by a transient solution is less than that given by a steady state solution. The saving in weight depends on the length of the protection period. At, entire plate at. Setting in (j) gives, or.