# Heat Transfer Lecture Slides II ( Dr. Gregory A. Kallio )

# ME 259

Heat Transfer

Lecture Slides II

Dr. Gregory A. Kallio

Dept. of Mechanical Engineering, Mechatronic Engineering & Manufacturing Technology

California State University, Chico

# Steady-State Conduction Heat Transfer

# Incropera & DeWitt coverage:

# –Chapter 2: General Concepts of Heat Conduction

# –Chapter 3: One-Dimensional, Steady-State Conduction

# –Chapter 4: Two-Dimensional, Steady-State Conduction

# General Concepts of Heat Conduction

# Reading: Incropera & DeWitt

# Chapter 2

# Generalized Heat Conduction

Fourier’s law, 1-D form:

Fourier’s law, general form:

-q” is the heat flux vector, which has three components; in Cartesian coordinates:

(magnitude)

The Temperature Gradient

ÑT is the temperature gradient, which is:

–a vector quantity that points in direction of maximum temperature increase

–always perpendicular to constant temperature surfaces, or isotherms

(Cartesian)

(Cylindrical)

(Spherical)

Thermal Conductivity

k is the thermal conductivity of the material undergoing conduction, which is a tensor quantity in the most general case:

–most materials are homogeneous, isotropic, and their structure is time-independent; hence:

which is a scalar and usually assumed to be a constant if evaluated at the average temperature of the material

Total Heat Rate

Total heat rate (q) is found by integrating the heat flux over the appropriate area:

k and Ñ T must be known in order to calculate q” from Fourier’s law

–k is usually obtained from material property tables

–to find ÑT, another equation is required; this additional equation is derived by applying the conservation of energy principle to a differential control volume undergoing conduction heat transfer; this yields the general Heat Diffusion (Conduction) Equation

Heat Diffusion (Conduction) Equation

For a homogeneous, isotropic solid material undergoing heat conduction:

Cylindrical and spherical coordinate system forms given in text (p. 64-65)

This is a second-order, partial differential equation (PDE); its solution yields the temperature field, T(x,y,z,t), within a given solid material

Heat Diffusion (Conduction) Equation

For constant thermal conductivity (k):

For k = constant, steady-state conditions, and no internal heat generation

–this is known as Laplace’s equation, which appears in other branches of engineering science (e.g., fluids, electrostatics, and solid mechanics)

Boundary Conditions and Initial Condition

Boundary Conditions: known conditions at solution domain boundaries

Initial Condition: known condition at t = 0

Number of boundary conditions required to solve the heat diffusion equation is equal to the number of spatial dimensions multiplied by two

There is only one initial condition, which takes the form

–where Ti may be a constant or a function of x,y, and z

Types of Boundary Conditions for Conduction Problems

Specified surface temperature, e.g.,

Specified surface heat flux, e.g.,

Specified convection (h, T¥ given), e.g.,

Specified radiation (e, Tsur given), e.g.,

Solving the Heat Diffusion Equation

Choose a coordinate system that best fits the problem geometry.

Identify the independent variables (x,y,z,t), e,g, is it a S-S problem? Is conduction 1-D, 2-D, or 3-D? Justify assumptions.

Determine if k can be treated as constant and if

Write the general heat conduction equation using the chosen coordinates.

Reduce equation to simplest form based upon assumptions.

Write boundary conditions and initial condition (if applicable).

Obtain a general solution for T(x,y,z,t) by some method; if impossible, resort to numerical methods.

Solving the Heat Diffusion Equation, cont.

Solve for the constants in the general solution by applying the boundary conditions and initial condition to obtain a particular solution.

Check solution for correctness (e.g., at boundaries or limits such as x = 0, t = 0, t ® ¥ , etc.)

Calculate heat flux or total heat rate using Fourier’s law, if required.

Optional: rearrange solution into a nondimensional form

Example:

GIVEN: Rectangular copper bar of dimensions L x W x H is insulated on the bottom and initially at Ti throughout . Suddenly, the ends are subjected and maintained at temperatures T1 and T2 , respectively, and the other three sides are exposed to forced convection with known h, T¥.

FIND: Governing heat equation, BCs, and initial condition

One-Dimensional, Steady-State Heat Conduction

Reading: Incropera & DeWitt,

# Chapter 3

# 1-D, S-S Conduction in Simple Geometries w/o Heat Generation

Plane Wall

–if k = constant, general heat diffusion equation reduces to

–separating variables and integrating yields

–where T(x) is the general solution; C1 and C2 are integration constants that are determined from boundary conditions

1-D, S-S Conduction in Simple Geometries w/o Heat Generation

Plane Wall, cont.

–suppose the boundary conditions are

–integration constants are then found to be

–the particular solution for the temperature distribution in the plane wall is now

1-D, S-S Conduction in Simple Geometries w/o Heat Generation

Plane wall, cont.

–The conduction heat rate is found from Fourier’s law:

–If k were not constant, e.g., k = k(T), the analysis would yield

»note that the temperature distribution would be nonlinear, in general

1-D, S-S Conduction in Simple Geometries w/o Heat Generation

Electric Circuit Analogy

–heat rate in plane wall can be written as

–in electrical circuits we have Ohm’s law:

–analogy:

Thermal Circuits for Plane Walls

Series Systems

Parallel Systems

Thermal Circuits for Plane Walls, cont.

Complex Systems

Thermal Resistances for Other Geometries Due to Conduction

Cylindrical Wall

Spherical Wall

Convective & Radiative Thermal Resistance

Convection

Radiation

Critical Radius Concept

Since the surface areas of cylinders and spheres increase with r, there exist competing heat transfer effects with the addition of insulation under convective boundary conditions (see Example 3.4)

A critical radius (rcr) exists for radial systems, where:

–adding insulation up to this radius will increase heat transfer

– adding insulation beyond this radius will decrease heat transfer

For cylindrical systems, rcr = kins/h

For spherical systems, rcr = 2kins/h

Thermal Contact Resistance

Thermal contact resistance exists at solid-solid interfaces due to surface roughness, creating gaps of air or other material:

Thermal Contact Resistance

R”t,c is usually experimentally measured and depends upon

–thermal conductivity of solids A and B

–surface finish & cleanliness

–contact pressure

–gap material

–temperature at contact plane

See Tables 3.1, 3.2 for typical values

EXAMPLE

Given: two, 1cm thick plates of milled, cold-rolled steel, 3.18mm roughness, clean, in air under 1 MPa contact pressure

Find: Thermal circuit and compare thermal resistances

1-D, S-S Conduction in Simple Geometries with Heat Generation

Thermal energy can be generated within a material due to conversion from some other energy form:

–Electrical

–Nuclear

–Chemical

Governing heat diffusion equation if k = constant:

S-S Heat Transfer from Extended Surfaces (i.e., fins)

Consider plane wall exposed to convection where Ts>T¥:

How could you enhance q ?

–increase h

–decrease T¥

–increase As (attach fins)

Fin Nomenclature

x = longitudinal direction of fin

L = fin length (base to tip)

Lc = fin length corrected for tip area

W = fin width (parallel to base)

t = fin thickness at base

Af = fin surface area exposed to fluid

Ac = fin cross-sectional area, normal to heat flow

Ap = fin (side) profile area

P = fin perimeter that encompasses Ac

D = pin fin diameter

Tb = temperature at base of fin

1-D Conduction Model for Thin Fins

If L >> t and k/L >> h, then the temperature gradient in the longitudinal direction (x) is much greater than that in the transverse direction (y); therefore

Another way of viewing fin heat transfer is to imagine 1-D conduction with a negative heat generation rate along its length due to convection

Fin Performance

Fin Effectiveness

Fin Efficiency

–for a straight fin of uniform cross-section:

–where Lc = L + t / 2 (corrected fin length)

Calculating Single Fin Heat Rate from Fin Efficiency

Calculate corrected fin length, Lc

Calculate profile area, Ap

Evaluate parameter

# Determine fin efficiency hf from Figure 3.18, 3.19, or Table 3.5

Calculate maximum heat transfer rate from fin:

Calculate actual heat rate:

Maximum Heat Rate for Fins of Given Volume

Analysis:

“Optimal” design results:

Fin Thermal Resistance

Fin heat rate:

Define fin thermal resistance:

Single fin thermal circuit:

Analysis of Fin Arrays

Total heat transfer =

heat transfer from N fins +

heat transfer from exposed base

Thermal circuit:

–where

Analysis of Fin Arrays, cont.

Overall thermal resistance:

Example

Given: Annular array of 10 aluminum fins, spaced 4mm apart C-C, with inner and outer radii of 1.35 and 2.6 cm, and thickness of 1 mm. Temperature difference between base and ambient air is 180°C with a convection coefficient of 125 W/m2-K. Contact resistance of 2.75×10-4 m2-K/W exists at base.

Find: a) Total heat rate w/o and with fins

b) Effect of R”t,c on heat rate

Two-Dimensional, Steady-State Heat Conduction

Reading: Incropera & DeWitt

# Chapter 4

# Governing Equation

Heat Diffusion Equation reduces to:

Solving the HDE for 2-D, S-S heat conduction by exact analysis is impossible for all but the most simple geometries with simple boundary conditions.

Solution Methods

Analytical Methods

–Separation of variables (see section 4.2)

–Laplace transform

–Similarity technique

–Conformal mapping

Graphical Methods

–Plot isotherms & heat flux lines

Numerical Methods

–Finite-difference method (FDM)

–Finite-element method (FEM)

Conduction Shape Factor

The heat rate in some 2-D geometries that contain two isothermal boundaries (T1, T2) with k = constant can be expressed as

–where S = conduction shape factor

(see Table 4.1)

Define 2-D thermal resistance:

Conduction Shape Factor, cont.

Practical applications:

–Heat loss from underground spherical tanks: Case 1

–Heat loss from underground pipes and cables: Case 2, Case 4

–Heat loss from an edge or corner of an object: Case 8, Case 9

–Heat loss from electronic components mounted on a thick substrate: Case 10

…