1. Chapter 3 Part 1 One-Dimensional, Steady-State Conduction

2. Thermal Resistance x T Hot airCold air T1T1 T2T2 T3T3 T4T4 L q1q1 q2q2 q3q3 h1h1 k h2h2 Accumulation of Energy = Energy In - Energy Out + Energy Generated Therefore:q 1 = q 2 =q 3

3. + Since:q 1 = q 2 =q 3 =qand A 1 = A 2 =A 3 =A we have

4. Solving for q This is analogous to electrical resistance placed in series, Hot airCold air T1T1 T2T2 T3T3 T4T4 L q R2R2 R3R3 h1h1 k h2h2 R1R1

5. So,and Therefore Equation 3.3 General form Cartesian Coordinate Heat transfer rate Thermal resistance (conduction) Thermal resistance (convection) Equation 3.1 Equation 3.2

6. Composite Wall (series) Hot air Cold air T1T1 T2T2 T5T5 T6T6 LALA q R2R2 R3R3 h1h1 h2h2 R1R1 R5R5 R4R4 T4T4 T3T3 LBLB LCLC ABC kAkA kBkB kCkC

7. So, Therefore,, and

8. Composite Wall (series-parallel) T2T2 T1T1 LALA q RERE RFRF L F = L G LHLH E F H kEkE kFkF kGkG G kHkH RGRG RHRH A1A1 A2A2 A3A3

9. So Therefore,, and R1R1 R2R2 =Recall R3R3 Whereso Equation 3.4

10. Overall Heat Transfer Coefficient For conveniences we can define the overall heat transfer coefficient as Which yields an expression analogous to Newton’s law

11. Example 3.1 A leading manufacturer of household appliances is proposing a self-cleaning oven design that involves use of a composite window separating the oven cavity from the room air. The composite is to consist of two high temperature plastic (A and B) of thicknesses L A = 2 L B and thermal conductivities k A = 0.15 Wm -1 k -1 and k B = 0.08 Wm -1 k -1. During the self cleaning process the oven wall and air temperatures, T w and T a are 400 o C, while room temperature T ∞ is 25 o C. The inside convection and radiation heat transfer coefficients h i and h r, as well as the outside convection coefficient h o, are each approximately 25 Wm -2 k -1. What is the minimum window thickness L = L A + L B, needed to ensure a temperature that is 50 o C or less at the outer surface of the window? This temperature must not be exceeded for safety reasons.

12. Oven cavity A, k A = 0.15 Wm -1 k -1 B, k B =0.08 Wm -1 k -1 LALA LBLB T s,o <50 o C T s,i Hot air Cold air Composite Window L A = L B T w = 400 o C h r = 25 Wm -2 k -1 T a = 400 o C h i = 25 Wm -2 k -1 T ∞ = 25 o C h o = 25 Wm -2 k -1

13. Solution Assumptions: 1.Steady-state 2.Conduction through the window is one dimensional 3.Contact resistance is negligible 4.Radiation exchange between window outer surface and surrounding is negligible Energy balance Accumulation of Energy = Energy In - Energy Out + Energy Generated

14. Solution Energy In = q B = q A = q r +q i Energy Out = q o so q B = q A = q r +q i = q o

15. Solution

16. Rewrite as

17. Solution Replace with values

18. Solution Thermal Circuit Energy In = Energy Out =

19. Solution Thermal Circuit

22. Direct Application of Fourier’s Law Conditions:Steady-state no heat source or sink one dimensional system Consequences:For any differential element dx, q x = q x+dx. Even if the area is a function of x and the conductivity is a function of T. Hence we may use Fourier’s law in the integral form without knowing the temperature distribution Furthermore if the area and the conductivity are constant

23. Example Calculate the heat rate through a pyroceram cone of circular cross section with a diameter D = ax = 0.25x. The small end is located at x 1 = 0.50 mm and the large end at x 2 = 250 mm. The end temperatures are T 1 = 400 K and T 2 = 600 K. The lateral surfaces are well insulated. x x 2 = 0.25mx 1 = 0.05m T 1 = 400K T 2 =600K

24. Solution Assumptions: Steady State One-dimensional conduction in x direction No energy source or sink Constant properties Based on the assumptions the heat transfer rate is constant along the x direction. Hence from table A.2, pyroceram(500k): k = 3.46 W m -1 K -1 From equation 1.1

25. Where A the cross-section area is

27. Thermal Resistance The Cylinder Knowing the inside surface temperature T i and the outside surface temperature T o of a cylinder and assuming that: conductivity k is constant steady-state no heat source or sink L is much larger than r o heat transfer solely radial than we can use equation 1.1 to determine the heat transfer rate across the cylinder, so: roro riri TiTi ToTo L

28. By separation of variables we find

29. r2r2 r1r1 r3r3 r4r4 For a cylinder with several layer qrqr R2R2 R3R3 R1R1 R5R5 R4R4 T ∞,1 T1T1 T2T2 T3T3 T4T4 T ∞,4 A B C

30. So for A 1 which is define as

31. Similarly for A 2 which is define as And

32. Example The possible existence of an optimum insulation thickness for radial systems is suggested by the presence of competing effects associated with an increase in this thickness. In particular, although the conduction resistance increases with the addition of insulation, the convection resistance decreases due to increasing outer surface area. Hence there may exist an insulation thickness that minimizes heat loss by maximizing the total resistance to heat transfer. a) Is there an optimum thickness associated with the application of insulation to a thin-walled copper tube of radius r i used in the transport of a refrigerant. The temperature of the refrigerant (T i ) is less than the temperature of the ambient air (T ∞ ).

33. Cold air riri r k TiTi T∞T∞ h Assumptions: Steady state One dimensional temperature gradient (radial) Negligible tube wall thermal resistance Constant properties for insulation Negligible radiation q

34. q’ We can define q’ as the heat transfer rate per unit length and R’ t as the total thermal resistance therefore And Optimum at

35. Hence, we have an optimum at at we determine whether it is a minimum or maximum using the second derivative Therefore, we a minimum, which is usually called the critical radius

36. using the typical values and the critical radius is So if, r c is larger than r i, heat transfer will increase with the addition of insulation up to a thickness of. Therefore for r i = 0.005 m R’ t R’ convection R’ conduction R mk/W

37. Thermal Resistance The Sphere Knowing the inside surface temperature T i and the outside surface temperature T o of a sphere and assuming that: conductivity k is constant steady-state no heat source or sink heat transfer solely radial than we can use equation 1.1 to determine the heat transfer rate across the sphere, so: riri roro k TiTi ToTo

39. A spherical, thin walled metallic container is used to store liquid nitrogen at 77 K. The container has a diameter of 0.5 m and is covered with an evacuated, reflective insulation composed of silica powder. The insulation is 25 mm thick, and its outer surface is exposed to ambient air at 300K. The convection coefficient is known to be 20 W m -2 K -1. The latent heat of vaporization and the density of liquid nitrogen are 2x10 5 J kg -1 and 804 kg m -3, respectively. What is the rate of heat transfer to the liquid nitrogen? Example

40. Cold air r 1 =0.25 m r 2 =0.275 m k Liquid nitrogen T ∞,1 = 77K ρ = 804 kg m -3 h fg = 2 x 10 5 J kg -1 T ∞,2 = 300K h=20W m -2 K -1 mh fg Vent

41. Assumptions: Steady state One dimensional temperature gradient (radial) Negligible sphere wall thermal resistance Constant properties for insulation Negligible radiation Thermal circuit q

42. What is mass of liquid nitrogen boiling-off? Performing an energy balance for a control surface about the nitrogen we have

43. Which represents 5.64 kg per day or 7 liters per day

44. One-dimensional, steady-state solutions to the heat equation with no generation Plane WallCylindrical WallSpherical Wall Heat equation Temperature distribution Heat flux Heat rate Thermal resistance