One-Dimensional Steady-State Conduction: Temperature Profiles & Thermal Resistance

1. Chapter 3 One-Dimensional Steady-State Conduction

2. One-Dimensional Steady-State Conduction • Conduction problems may involve multiple directions and time-dependent conditions • Inherently complex – Difficult to determine temperature distributions • One-dimensionalsteady-state models can represent accurately numerous engineering systems • In this chapter we will • Learn how to obtain temperature profiles for common geometries with and without heat generation. • Introduce the concept of thermal resistance and thermal circuits

3. Cold fluid qx Hot fluid x=L x=0 x The Plane Wall Consider a simple case of one-dimensional conduction in a plane wall, separating two fluids of different temperature, without energy generation • Temperature is a function of x • Heat is transferred in the x-direction Must consider • Convection from hot fluid to wall • Conduction through wall • Convection from wall to cold fluid • Begin by determining temperature distribution within the wall

4. Temperature Distribution • Heat diffusion equation (eq. 2.4) in the x-direction for steady-state conditions, with no energy generation: • qx is constant • Boundary Conditions: • Temperature profile, assuming constant k: (3.1) • Temperature varies linearly with x

5. Thermal Resistance Based on the previous solution, the conduction hear transfer rate can be calculated: (3.2a) Similarly for heat convection, Newton’s law of cooling applies: (3.2b) And for radiation heat transfer: (3.2c) • Recall electric circuit theory - Ohm’s law for electrical resistance:

6. Thermal Resistance • We can use this electrical analogy to represent heat transfer problems using the concept of a thermal circuit (equivalent to an electrical circuit). • Compare with equations 3.2a-3.2c • The temperature difference is the “potential” or driving force for the heat flow and the combinations of thermal conductivity, convection coefficient, thickness and area of material act as a resistance to this flow:

7. Thermal Resistance for Plane Wall Cold fluid In terms of overall temperature difference: qx Hot fluid x=L x=0 x

8. Composite Walls • Express the following geometry in terms of a an equivalent thermal circuit.

9. Composite Walls • What is the heat transfer rate for this system? Alternatively where U is the overall heat transfer coefficient and DT the overall temperature difference.

10. Composite Walls (a) Surfaces normal to the x-direction are isothermal • For resistances in series: Rtot=R1+R2+…+Rn • For resistances in parallel: • Rtot=1/R1+1/R2+…+1/Rn (b) Surfaces parallel to x-direction are adiabatic

11. Example (Problem 3.15 textbook) Consider a composite wall that includes an 8-mm thick hardwood siding (A), 40-mm by 130-mm hardwood studs (B) on 0.65-m centers with glass fiber insulation (D) (paper faced, 28 kg/m3) and a 12-mm layer of gypsum (vermiculite) wall board (C). • What is the thermal resistance associated with a wall that is 2.5 m high by 6.5 m wide (having 10 studs, each 2.5 m high?) (Note: Consider the direction of heat transfer to be downwards, along the x-direction)

12. Contact Resistance The temperature drop across the interface between materials may be appreciable, due to surface roughness effects, leading to air pockets. We can define thermal contact resistance: See tables 3.1, 3.2 for typical values of Rt,c

13. Alternative Conduction Analysis When area varies in the x direction and k is a function of temperature, Fourier’s law can be written in its most general form: • For steady-state conditions, no heat generation, one-dimensional heat transfer, qx is constant.

14. T1 x1 x2 x Example 3.3 Consider a conical section fabricated from pyroceram. It is of circular cross section, with the diameter D=ax, where a=0.25. The small end is at x1=50 mm and the large end at x2=250 mm. The end temperatures are T1=400 K and T2=600 K, while the lateral surface is well insulated. • Derive an expression for the temperature distribution T(x) in symbolic form, assuming one-dimensional conditions. Sketch the temperature distribution • Calculate the heat rate, qx, through the cone. T2

15. Radial Systems-Cylindrical Coordinates Consider a hollow cylinder, whose inner and outer surfaces are exposed to fluids at different temperatures Temperature distribution

16. Temperature Distribution • Heat diffusion equation (eq. 2.5) in the r-direction for steady-state conditions, with no energy generation: • Fourier’s law: • Boundary Conditions: • Temperature profile, assuming constant k: • Logarithmic temperature distribution (see previous slide)

17. Thermal Resistance Based on the previous solution, the conduction hear transfer rate can be calculated: • In terms of equivalent thermal circuit: • Fourier’s law:

18. Composite Walls • Express the following geometry in terms of a an equivalent thermal circuit.

19. Composite Walls • What is the heat transfer rate? where U is the overall heat transfer coefficient. If A=A1=2pr1L: alternatively we can use A2=2pr2L, A3=2pr3L etc. In all cases:

20. Example (Problem 3.37 textbook) A thin electrical heater is wrapped around the outer surface of a long cylindrical tube whose inner surface is maintained at a temperature of 5°C. The tube wall has inner and outer radii of 25 and 75 mm respectively, and a thermal conductivity of 10 W/m.K. The thermal contact resistance between the heater and the outer surface of the tube (per unit length of the tube) is R’t,c=0.01 m.K/W. The outer surface of the heater is exposed to a fluid of temperature –10°C and a convection coefficient of h=100 W/m2 .K. • Determine the heater power per unit length of tube required to maintain the heater at To=25°C.

21. Spherical Coordinates • Starting from Fourier’s law, acknowledging that qr is constant, independent of r, and assuming that k is constant, derive the equation describing the conduction heat transfer rate. What is the thermal resistance? • Fourier’s law:

22. For steady-state, one dimensional conditions with no heat generation; The appropriate form of Fourier’s equation is Q = -k A dT/dr = -k(4πr2) dT/dr Note that the cross sectional area normal to the heat flow is A= 4πr2 (instead of dx) where r is the radius of the sphere

23. Equation 2.3-1 may be expressed in the integral form   = - For constant thermal conductivity, k   Q = = Generally, this equation can be written in terms of Q = where   R =

24. Example: Consider a hollow steel sphere of inside radius r1 = 10 cm and outside radius, r2 = 20 cm. The thermal conductivity of the steel is k = 10 W/moC. The inside surface is maintained at a uniform temperature of T1 = 230 oC and the outside surface dissipates heat by convection with a heat transfer coefficient h = 20 W/m2oC into an ambient at T = 30oC. Determine the thickness of asbestos insulation (k=0.5 W/mK) required to reduce the heat loss by 50%.

25. Example (Problem 3.69 textbook) One modality for destroying malignant tissue involves imbedding a small spherical heat source of radius ro within the tissue and maintaining local temperatures above a critical value Tc for an extended period. Tissue that is well removed from the source may be assumed to remain at normal body temperature (Tb=37°C). • Obtain a general expression for the radial temperature distribution in the tissue under steady-state conditions as a function of the heat rate q. • If ro=0.5 mm, what heat rate must be supplied to maintain a tissue temperature of T>Tc=42°C in the domain 0.5<r<5 mm? The tissue thermal conductivity is approximately 0.5 W/m.K.

26. Summary • We obtained temperature distributions and thermal resistances for problems involving steady-state, one-dimensional conduction in orthogonal, cylindrical and spherical coordinates, without energy generation • Useful summary in Table 3.3