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Transient Conduction: Spatial Effects and the Role of Analytical Solutions Chapter 5 Sections 5.4 to 5.7 Lecture 10.

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Presentation on theme: "Transient Conduction: Spatial Effects and the Role of Analytical Solutions Chapter 5 Sections 5.4 to 5.7 Lecture 10."— Presentation transcript:

1 Transient Conduction: Spatial Effects and the Role of Analytical Solutions
Chapter 5 Sections 5.4 to 5.7 Lecture 10

2 Solution to the Heat Equation for a Plane Wall with
Symmetrical Convection Conditions If the lumped capacitance approximation cannot be made, consideration must be given to spatial, as well as temporal, variations in temperature during the transient process. For a plane wall with symmetrical convection conditions and constant properties, the heat equation and initial/boundary conditions are: (5.29) (5.30) (5.31) (5.32) Existence of eight independent variables: (5.33) How may the functional dependence be simplified?

3 Dimensionless temperature difference:
Plane Wall (cont.) Non-dimensionalization of Heat Equation and Initial/Boundary Conditions: Dimensionless temperature difference: Dimensionless space coordinate: Dimensionless time: The Biot Number: Exact Solution: (5.42a) (5.42b,c) See Appendix B.3 for first four roots (eigenvalues ) of Eq. (5.42c).

4 The One-Term Approximation :
Plane Wall (cont.) The One-Term Approximation : Variation of midplane temperature (x*= 0) with time : (5.44) Variation of temperature with location (x*) and time : (5.43b) Change in thermal energy storage with time: (5.46a) (5.49) (5.47) Can the foregoing results be used for a plane wall that is well insulated on one side and convectively heated or cooled on the other? Can the foregoing results be used if an isothermal condition is instantaneously imposed on both surfaces of a plane wall or on one surface of a wall whose other surface is well insulated?

5 Graphical Representation of the One-Term Approximation
Heisler Charts Graphical Representation of the One-Term Approximation The Heisler Charts, Section 5 S.1 Midplane Temperature:

6 Temperature Distribution:
Heisler Charts (cont.) Temperature Distribution: Change in Thermal Energy Storage:

7 Radial Systems Long Rods or Spheres Heated or Cooled by Convection.
One-Term Approximations: Long Rod: Eqs. (5.52) and (5.54) Sphere: Eqs. (5.53) and (5.55) Graphical Representations: Long Rod: Figs. 5 S.4 – 5 S.6 Sphere: Figs. 5 S.7 – 5 S.9

8 The Semi-Infinite Solid
A solid that is initially of uniform temperature Ti and is assumed to extend to infinity from a surface at which thermal conditions are altered. Special Cases: Case 1: Change in Surface Temperature (Ts) (5.60) (5.61)

9 Semi-Infinite Solid (cont.)
Case 2: Constant Heat Flux (5.62) Case 3: Surface Convection (5.63)

10 Example 1 (Problem 5.34) A 100-mm-thick steel plate (=7830 kg/m3, c=550 J/kgK, k=48 W/mK) that is initially at uniform temperature of Ti=200C is to be heated to a minimum temperature of 550C. Heating is affected in a gas-fired furnace, where the products of combustion at T = 800 C maintain a convection coefficient of h=250 W/m2K on both surfaces of the plate, how long should the plate be left in the furnace ? What is the surface temperature?

11 Example 1 Known: Dimension and properties of steel plate,
convection condition. Find: Time needed to reach Tmin of 550C, and T surface Schematic:

12 Example 1 Assumptions: 1-D conduction in the plate, radiation negligible Constant properties. Analysis: Lumped Capacitance Method (Need Bi<0.1)

13 Example 1 Analysis: 2. Approximate Solution (Need Fo>0.2)
The minimum T is at the center of the plate From Table 5.1, Bi=0.26, ζ1=0.488 rad, C1=1.0396 Solve Fo=3.839 >0.2

14 Example 1

15 Example 1 Fo=3.839 =(αt/L2) t =Fo*L2/α
Analysis: Fo=3.839 =(αt/L2) t =Fo*L2/α α = k/(c) = 48/(7830*550)=1.1146x10-5 m2/s t= Fo*L2/α = 3.839*(0.05)2/1.11x10-5 = 861 s

16 Example 1 Analysis: Solve TL =579.2 C

17 Example 1 Analysis: 3. Heisler Chart (Figure D.1, page 948)
The minimum T is at the center of the plate Get Fo value from Figure D.1, page 948 Fo ≈ 3.8, t = 852 s

18 Example 1

19 Example 1 Bi-1 Fo

20 Example 1 Analysis: 4. The surface T can be determined by using Heisler Chart Get θ1/ θ0 value from Figure D.2, page 949 Solve TL =582.5 C

21 Example 1

22 Methods for Solving Transient Problems
Check Bi <0.1? Or Assume Bi<0.1, analytical solution, then check Bi<0.1? Bi>0.1, Fo>0.2?, Approximate solution (P257), Table 5.1 (P274) or Bi>0.1, Fo>0.2, Approximate solution , Heisler Chart (Fig. D1-D9, ) (Pay attention to the B.C. and I.C. in Figure 5.6) 3. Bi>0.1, Fo<0.2, Exact solution (Eqn. 5.39)

23 Problem: Thermal Energy Storage
Problem 5.80: Charging a thermal energy storage system consisting of a packed bed of Pyrex spheres.

24 Problem: Thermal Energy Storage (cont.)

25 Problem: Thermal Energy Storage (cont.)
< < <

26 Problem: Thermal Response Firewall
Problem: 5.93: Use of radiation heat transfer from high intensity lamps for a prescribed duration (t=30 min) to assess ability of firewall to meet safety standards corresponding to maximum allowable temperatures at the heated (front) and unheated (back) surfaces.

27 Problem: Thermal Response of Firewall (cont.)

28 Problem: Thermal Response of Firewall (cont.)

29 Problem: Microwave Heating
Problem: : Microwave heating of a spherical piece of frozen ground beef using microwave-absorbing packaging material.

30 Problem: Microwave Heating (cont.)
Thus The beef can be seen as the interior of a sphere with a constant heat flux at its surface, thus the relationship in Table 5.2b, Interior Cases, sphere, can be used. We begin by calculating q* for Ts=0°C.

31 Problem: Microwave Heating (cont.)
Since this is less than 0.2, our assumption was correct. Finally we can solve for the time: < COMMENTS: At the minimum surface temperature of -20°C, with T∞ = 30°C and h = 15 W/m2∙K from Problem 5.33, the convection heat flux is 750 W/m2, which is less than 8% of the microwave heat flux. The radiation heat flux would likely be less, depending on the temperature of the oven walls.

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