Chapter 3 OneDimensional SteadyState Conduction Chapter 3 1
- Slides: 26
Chapter 3 One-Dimensional Steady-State Conduction Chapter 3 1
One-Dimensional Steady-State Conduction • Conduction problems may involve multiple directions and timedependent conditions • Inherently complex – Difficult to determine temperature distributions • One-dimensional steady-state models can represent accurately numerous engineering systems • In this chapter we will Ø Learn how to obtain temperature profiles for common geometries with and without heat generation. Ø Introduce the concept of thermal resistance and thermal circuits Chapter 3 2
The Plane Wall Consider a simple case of onedimensional conduction in a plane wall, separating two fluids of different temperature, without energy generation Cold fluid • Temperature is a function of x • Heat is transferred in the x-direction Must consider qx – Convection from hot fluid to wall – Conduction through wall – Convection from wall to cold fluid Ø Begin by determining temperature distribution within the wall Chapter 3 Hot fluid x=0 x=L x 3
Temperature Distribution • Heat diffusion equation (eq. 2. 4) in the x-direction for steady-state conditions, with no energy generation: v qx is constant • Boundary Conditions: • Temperature profile, assuming constant k: (3. 1) v Temperature varies linearly with x Chapter 3 4
Thermal Resistance Based on the previous solution, the conduction hear transfer rate can be calculated: (3. 2 a) Similarly for heat convection, Newton’s law of cooling applies: (3. 2 b) And for radiation heat transfer: (3. 2 c) Ø Recall electric circuit theory - Ohm’s law for electrical resistance: Chapter 3 5
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). Ø Compare with equations 3. 2 a-3. 2 c Ø 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: Chapter 3 6
Thermal Resistance for Plane Wall Cold fluid In terms of overall temperature difference: qx Hot fluid Chapter 3 x=0 x x=L 7
Composite Walls ? Express the following geometry in terms of a an equivalent thermal circuit. Chapter 3 8
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. Chapter 3 9
Composite Walls (a) Surfaces normal to the xdirection are isothermal Ø For resistances in series: Rtot=R 1+R 2+…+Rn Ø For resistances in parallel: Rtot=1/R 1+1/R 2+…+1/Rn Chapter 3 (b) Surfaces parallel to xdirection are adiabatic 10
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/m 3) and a 12 -mm layer of gypsum (vermiculite) wall board (C). Ø What is 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) Chapter 3 11
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 Chapter 3 12
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. Chapter 3 13
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 x 1=50 mm and the large end at x 2=250 mm. The end temperatures are T 1=400 K and T 2=600 K, while the lateral surface is well insulated. 1. Derive an expression for the temperature distribution T(x) in symbolic form, assuming one-dimensional conditions. Sketch the temperature distribution 2. Calculate the heat rate, qx, through the cone. T 2 T 1 x 2 Chapter 3 x 14
Radial Systems-Cylindrical Coordinates Consider a hollow cylinder, whose inner and outer surfaces are exposed to fluids at different temperatures Temperature distribution Chapter 3 15
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: v Logarithmic temperature distribution (see previous slide) Chapter 3 16
Thermal Resistance Based on the previous solution, the conduction hear transfer rate can be calculated: • Fourier’s law: Ø In terms of equivalent thermal circuit: Chapter 3 17
Composite Walls ? Express the following geometry in terms of a an equivalent thermal circuit. Chapter 3 18
Composite Walls ? What is the heat transfer rate? where U is the overall heat transfer coefficient. If A=A 1=2 pr 1 L: alternatively we can use A 2=2 pr 2 L, A 3=2 pr 3 L etc. In all cases: Chapter 3 19
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/m 2. K. Ø Determine the heater power per unit length of tube required to maintain the heater at To=25°C. Chapter 3 20
Spherical Coordinates • Fourier’s law: • 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 thermal resistance? Chapter 3 21
For steady-state, one dimensional conditions with no heat generation; The appropriate form of Fourier’s equation is Q = -k A d. T/dr = -k(4πr 2) d. T/dr Note that the cross sectional area normal to the heat flow is A= 4πr 2 (instead of dx) where r is the radius of the sphere Chapter 3 22
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 Chapter 3 Q = where R = 23
Example: Consider a hollow steel sphere of inside radius r 1 = 10 cm and outside radius, r 2 = 20 cm. The thermal conductivity of the steel is k = 10 W/mo. C. The inside surface is maintained at a uniform temperature of T 1 = 230 o. C and the outside surface dissipates heat by convection with a heat transfer coefficient h = 20 W/m 2 o. C into an ambient at T = 30 o. C. Determine thickness of asbestos insulation (k=0. 5 W/m. K) required to reduce the heat loss by 50%. Chapter 3 24
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). Ø Obtain a general expression for the radial temperature distribution in the tissue under steady-state conditions as a function of the heat rate q. Ø 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. Chapter 3 25
Summary • We obtained temperature distributions and thermal resistances for problems involving steady-state, onedimensional conduction in orthogonal, cylindrical and spherical coordinates, without energy generation • Useful summary in Table 3. 3 Chapter 3 26
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