Review:Conduction

CH EN 3453 – Heat Transfer

Breaking News…

• No more homework (yay!)

• Final project reports due today by 8:00 PM– Email PDF version to report@chen3453.com– Review grading rubric on “Project” page of web

site

• Final exam Wednesday, Dec. 17 at 8:00 AM– This room– Exams must be completed by 10 AM

Chapter 1:Introduction to Heat Transfer

• Heat vs. heat flux vs. heat per length

• Conduction– Fourier’s Law– Ranges of k (Table 2.4 + appendices)

• Convection– Newton’s Law of cooling– Ranges of h (Table 1.1)

• Radiation– Stefan-Boltzmann Law– Emissivity, absorptivity

Chapter 2:Introduction to Conduction

• Thermal properties of matter

• Heat diffusion equation:

∂∂x

k∂T∂x

⎛⎝⎜

⎞⎠⎟

+∂∂y

k∂T∂y

⎛⎝⎜

⎞⎠⎟

+∂∂z

k∂T∂z

⎛⎝⎜

⎞⎠⎟

+ q = ρcp

∂T∂t

Example – Book Problem 2.5A solid, truncated cone serves as a support for a system that maintains the top (truncated) face of the cone at a temperature T1, while the base of the cone is at a temperature T2 < T1. The thermal conductivity depends on the temperature according to k = ko – aT, where a is a positive constant. Do the following quantities increase, decrease or stay the same with increasing x?

(1) heat transfer rate qx(2) the heat flux qx"(3) thermal conductivity k (4) temperature gradient dT/dx

Range of Thermal Conductivities

Figure 2.4 Range of thermal conductivity for various states of matter at normal temperatures and pressure.

Thermal Conductivity of Gases

Figure 2.8 The temperature dependence of the thermal conductivity of selected gases at normal pressures. The molecular weight of the gases is also shown.

Heat Diffusion Equation

∂∂x

k∂T∂x

⎛⎝⎜

⎞⎠⎟

+∂∂y

k∂T∂y

⎛⎝⎜

⎞⎠⎟

+∂∂z

k∂T∂z

⎛⎝⎜

⎞⎠⎟

+ q = ρcp

∂T∂t

Example – Book Problem 2.23The steady-state temperature distribution in a one-dimensional wall of thermal conductivity 50 W/m·K and thickness 50 mm is observed to be T(°C) = a + bx2, wherea = 200°C, b = –2000°C/m2 and x is in meters.

(a) What is the heat generation rate q in the wall?

(b) Determine the heat fluxes at the two wall faces

·

Chapter 3:Steady-State Conduction (1-D)

• The plane wall

• Radial systems

• Energy generation

• Extended surfaces

Heat Transfer through a Wall

R1 R2 R3

Example – Book Problem 3.3aThe window of a car is defogged by attaching a transparent, film-type heating element to its inner surface. For 4-mm-thick window glass, determine the electrical power required per unit window area to maintain an inner surface temperature of 15°C when the interior air temp is T∞,i = 25°C and the convection coefficient hi = 10 W/m2·K while the outside air temp is T∞,o = –10°C and ho = 65 W/m2·K.

Complex Heat Transfer

Contact Resistance

A Cylinder

Example – Book Problem 3.52Steam flowing through a long pipe maintains the inner pipe wall temperature at 500 K. The pipe is covered with two types of insulation, A and B. The interface between the two insulating layers has infinite contact resistance. The outer surface is exposed to air (T∞ = 300 K) and h = 25 W/m2·K

(a) Sketch and label the thermal circuit (b) What are the outer surface temps for materials A and B?

Example – Book Problem 3.59A spherical, 3 mm cryogenic probe at temperature –30°C is embedded into skin at 37°C. Frozen tissue develops and the interface between the frozen and normal tissue is 0°C. If the thermal conductivity of frozen tissue is 1.5 W/m·K and heat transfer at the phase front is characterized by a convection coefficient of 50 W/m2·K, what is the thickness of the frozen layer?

The Sphere

qr =4πk Ts,1 − Ts,2( )1 / r1( ) − 1 / r2( )

Rt ,cond =14πk

1r1−1r2

⎛⎝⎜

⎞⎠⎟

Review of Conduction…

Page 126

Extended Surfaces (fins)

Figure 3.12 Use of fins to enhance heat transfer from a plane wall.(a) Bare surface. (b) Finned surface.

Fins• Fin effectiveness

– Increase in heat transfer relative to heat transfer that would occur without the fin

– Consider only the base area of the fin

• Fin efficiency– Actual heat transfer relative to theoretical

maximum– Maximum assumes entire fin is at base

temperature

Example – Book Problem 3.116Turbine blade mounted on proposed air-cooled rotating disc (Tb = 300°C) in a gas turbine with gas stream at T∞ = 1200°C.

(a) If max allowable blade temperature is 1050°C and blade tip is assumed to be adiabatic, will the air cooling approach work?

(b) What is the rate of heat transfer from blade to coolant?

h = 250 W/m2·K

= 50 mm

k = 20 W/m·KAc = 6x10–4 m2

P = 110 mm

300°C

1200°C

Fin Efficiencies

Fin Efficiencies

Fin Efficiencies

Modified Bessel function of the first kind(Appendix B.5)

Modified Bessel function of the second kind(Appendix B.5)

Fin Efficiencies, continued

Chapter 4:Steady-State Conduction (2-D)

• Shape factors

• Finite-difference equations

• Graphical methods

Shape Factors

Shape Factors, Cont.

Graphical Method - Plotting Heat Flux

1. Consider lines of symmetry and choose sub-system if possible.

2. Symmetry lines adiabatic and count as heat flow lines.

3. Identify constant temperature lines at boundaries. Sketch isotherms between the boundaries.

4. Sketch heat flow lines perpendicular to isotherms, attempting to make each cell as square as possible.

Graphical Solution…

Chapter 5:Unsteady-State Conduction

• Lumped analysis and the Biot number

• Spatial effects

• Semi-infinite solids

• Constant surface temp. and const. heat flux

Review: The Biot Number

• If Bi < 0.1 then the lumped capacitance approach can be used– Eq. 5.5 to find time to reach a given T– Eq. 5.6 to find T after a given time– Eq. 5.8a to find total heat gain (loss) for given time

• L depends on geometry– General approach is L = V/As

• L/2 for wall with both sides exposed• ro/2 for long cylinder• ro/3 for sphere

– Conservative approach is to use the maximum length• L for wall• ro for cylinder or sphere

Bi = hL

k

Lumped Capacitance

• Time to reach a given temperature

• Temperature after a given time

• Heat gain (loss) after a given time

t = ρVchAs

lnθiθ

where θ ≡ T − T∞

T − T∞Ti − T∞

= exp −hAsρVc

⎛⎝⎜

⎞⎠⎟t

⎡

⎣⎢

⎤

⎦⎥

Q = ρVc( )θi 1− exp −tτ t

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

where τ t =1hAs

⎛⎝⎜

⎞⎠⎟ρVc( )

Spatial Effects(When lumped analysis cannot be used)

Dimensionless Variables

Temperature: θ*  ≡  θθi = T − T∞

Ti − T∞

Position: x* ≡xL

Time: t* ≡ αtL2

Solving with Spatial Effects (Bi > 0.1)

• Approximate solution (when Fo > 0.2)– Nondimensionalize temperature, position, time– Look up C1 and ζ1 from Table 5.1

• Plane wall:

• Cylinder:

• Sphere:

θ* = C1 exp −ζ12Fo( )cos ζ1x*( )

θ* = C1 exp −ζ12Fo( )J ζ1r

*( )

θ* = C1 exp −ζ12Fo( ) 1

ζ1r* sin ζ1r

*( )

Table 5.1 – ζ1 and C1 vs. Bi

5.7$–$The$Semi,Infinite$Solid• An$analy7cal$solu7on$can$be$obtained$using$this$idealiza'on;$not$a$prac7cal$concept

Equation 5.26

Equation 5.54