One-Dimensional Steady-State Conduction Chapter 3 One-Dimensional Steady-State Conduction Chapter 3

One-Dimensional Steady-State Conduction Conduction problems may involve multiple directions and time-dependent conditions Inherently complex – Difficult to determine temperature distributions One-dimensional steady-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 Chapter 3

The Plane Wall qx x=L x=0 x 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 qx x x=0 x=L Hot fluid Cold fluid Chapter 3

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 Chapter 3

Thermal Resistance (3.2a) (3.2b) (3.2c) 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: Chapter 3

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: Chapter 3

Thermal Resistance for Plane Wall Cold fluid In terms of overall temperature difference: qx Hot fluid x=L x=0 x Chapter 3

Composite Walls Express the following geometry in terms of a an equivalent thermal circuit. Chapter 3

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

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 Chapter 3

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) Chapter 3

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

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

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 T1 x1 x2 x Chapter 3

Radial Systems-Cylindrical Coordinates Consider a hollow cylinder, whose inner and outer surfaces are exposed to fluids at different temperatures Temperature distribution Chapter 3

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) Chapter 3

Thermal Resistance Based on the previous solution, the conduction hear transfer rate can be calculated: Fourier’s law: In terms of equivalent thermal circuit: Chapter 3

Composite Walls Express the following geometry in terms of a an equivalent thermal circuit. Chapter 3

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: Chapter 3

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. Chapter 3

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 the thermal resistance? Chapter 3

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 Chapter 3

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 = Chapter 3

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%.   Chapter 3

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. Chapter 3

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 Chapter 3