One-Dimensional, Steady-State Conduction without Thermal Energy Generation

1. One-Dimensional, Steady-StateConduction withoutThermal Energy Generation Chapter Three Sections 3.1 through 3.4 Lecture 4

2. Methodology Methodology of a Conduction Analysis • Specify appropriate form of the heat equation. • Solve for the temperature distribution. • Apply Fourier’s law to determine the heat flux. Simplest Case: One-Dimensional, Steady-State Conduction with NoThermal Energy Generation. • Common Geometries: • The Plane Wall: Described in rectangular (x) coordinate. Area perpendicular to direction of heat transfer is constant (independent of x). • The Tube Wall: Radial conduction through tube wall. • The Spherical Shell: Radial conduction through shell wall.

3. The Plane Wall Standard Approach • Simplifying the heat equation; • Solving the heat equation; • Applying the b. c.’s; • Determine heat rate/flux with Fourier’s Law.

4. The Plane Wall 1-D, Steady-state, no heat generation T(x) = C1 x + C2

5. The Plane Wall T(x) = C1 x + C2 Two boundary conditions (b.c.’s) are needed. T(0) = T s,1 and T(L) = T s,2 C2 = T s,1 at x=0 (b.c.-1) T s,2 = C1 L + C2 at x=L (b.c.-2) Or C1 = ( T s,2 -T s,1 )/L T(X) = ( T s,2 -T s,1 )x/L + T s,1

6. The Plane Wall T(X) = ( T s,2 -T s,1 )x/L + T s,1 qx and qx” are constants, independent of x.

7. The Plane Wall • Heat Equation: (3.1) • Boundary Conditions: • Temperature Distribution for Constant : (3.3) Plane Wall • Consider a plane wall between two fluids of different temperature: • Implications:

8. (3.5) (3.4) • Thermal Resistances and Thermal Circuits: Conduction in a plane wall: (3.6) Convection: (3.9) (3.12) (3.11) Plane Wall (cont.) • Heat Flux and Heat Rate: Thermal circuit for plane wall with adjoining fluids:

9. Radiation Resistance: (1.9) • Contact Resistance: Plane Wall (cont.) • Thermal Resistance for Unit Surface Area: Values depend on: Materials A and B, surface finishes, interstitial conditions, and contact pressure (Tables 3.1 and 3.2)

10. (3.14) • Overall Heat Transfer Coefficient (U) : A modified form of Newton’s law of cooling to encompass multiple resistances to heat transfer. (3.17) (3.19) Plane Wall (cont.) • Composite Wall with Negligible Contact Resistance: For the temperature distribution shown, kA > kB < kC.

11. Note departure from one-dimensional conditions for . • Circuits based on assumption of isothermal surfaces normal to x direction or adiabatic surfaces parallel to x direction provide approximations for . Plane Wall (cont.) • Series – Parallel Composite Wall:

12. Porous Media • Saturated media • consist of a solid • phase and a single • fluid phase. • Unsaturatedmedia • consist of solid, liquid, • and gas phases. • Porous Media • The effective thermal conductivity of a saturated medium depends on the solid (s) material, its porositye, its morphology, as well as the interstitial fluid (f) (Fig.a). (3.21) • The value of keff may be bracketed by describing the medium with a series resistance analysis (Fig. b) and a parallel resistance analysis (Fig.c). • The value of keff may be estimated by (3.25)

13. The Tube Wall (3.28) What does the form of the heat equation tell us about the variation of with r in the wall? How does vary with ? • Temperature Distribution for Constant : (3.31) Tube Wall • Heat Equation: Is the foregoing conclusion consistent with the energy conservation requirement?

14. Conduction Resistance: (3.33) Tube Wall (cont.) [W/m2] • Heat Flux and Heat Rate: [W/m] [W] (3.32) Why doesn’t a surface area appear in the expressions for the thermal resistance?

15. (3.35) but U itself is tied to specification of an interface. (3.37) Tube Wall (cont.) • Composite Wall with Negligible Contact Resistance For the temperature distribution shown, kA > kB > kC.

16. Spherical Shell What does the form of the heat equation tell us about the variation of with ? Is this result consistent with conservation of energy? How does vary with ? • Temperature Distribution for Constant : Spherical Shell • Heat Equation

17. (3.40) (3.41) Spherical Shell (cont.) • Heat flux, Heat Rate and Thermal Resistance: • Composite Shell:

18. 1-D, Steady-State, No Heat Generation

19. Example 3.1 (pages 104-107) Humans are able to control their heat production rate and heat loss rate to maintain a nearly constant core temperature of Tc= 37°C under a wide range of environmental conditions. This process is called thermaoregulation. From the perspective of calculating heat transfer between a human body and its surroundings, we focus on a layer of skin and fat, with its outer surface exposed to the environment and its inner surface at a temperature slightly less than the core temperature, Ti = 35C=308 K. Consider a person with a

20. Example 3.1 skin/fat layer of thickness L = 3 mm and effective thermal conductivity k = 0.3 W/m K. The person has a surface area A=1.8 m2 and wears a special sporting gear (snow suit and wet suit) made from nanostructured silica aerogel insulation with extremely low thermal conductivity of 0.014 W/mK. The emissivity of the snow and wet suit is 0.95. What thickness of aerogel insulation is needed to reduce the heat loss rate to 100 W (a typical metabolic heat generation rate) in air and water? What are the resulting skin temperatures? (10 C in the surrounding)

21. Example 3.1 Known:Inner surface T of skin/fat layer of know thickness, thermal conductivity, surface area. Thermal conductivity and emissivity of snow and wet suit, ambient conditions. Find:Insulation thickness to reduce the heat loss to 100 W and corresponding skin Ts.

22. Example 3.1 Schematic:

23. Example 3.1 Assumptions: • Steady-state • 1-D conduction through the skin and insulation • Contact resistance negligible • Thermal conductivities are uniform • Radiation exchange between skin and surroundings is between a small surface and a large enclosure at the air temperature • Liquid water is opaque to thermal radiation • Solar radiation is negligible • Body is completely immersed in water in part 2

24. Example 3.1 Analysis: The thermal circuit can be constructed by recognizing that resistance to heat flow is associated with conduction through the skin/fat and insulation layers and convection and radiation at the outer surface.

25. Example 3.1 The total thermal resistance needed to achieve the desired heat loss is found from: The total thermal resistance between the inside of the skin/fat layer and the cold surrounding includes conduction resistance for the skin/fat and insulation layers and an effective resistance associated with convection and radiation, which act in parallel.

26. Example 3.1 Air: The radiation heat transfer coefficient is approximated as having the same value in Example 1.6, hr=5.9 w/m2K. From previous equation we can solve the thickness of insulation Lins

27. Example 3.1 Water: Replace the convective coefficient (h) for water and neglect the radiation, then repeat the above calculation for water:

28. Example 3.1 Skin T: The skin temperature can be easily calculated by considering conduction through the skin/fat layer: Solving Ts from above equation: Is the skin T same for both cases? Why?

29. Problem: Thermal Barrier Coating Problem 3.30: Assessment of thermal barrier coating (TBC) for protection of turbine blades. Determine maximum blade temperature with and without TBC. Schematic:

30. the inner and outer surface temperatures of the Inconel are Problem: Thermal Barrier Coating (cont..) ANALYSIS: For a unit area, the total thermal resistance with the TBC is With a heat flux of <

31. Problem: Thermal Barrier Coating (cont..) Without the TBC, <

32. Problem: Radioactive Waste Decay Problem 3.72: Suitability of a composite spherical shell for storing radioactive wastes in oceanic waters. PROPERTIES: Table A-1, Lead: k = 35.3 W/m∙K, MP = 601 K; St.St.: k = 15.1 W/m∙K.

33. Problem: Radioactive Waste Decay (cont..) <