Review: Conduction. Breaking News


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1 CH EN 3453 Heat Transfer Review: Conduction Breaking News No more homework (yay!) Final project reports due today by 8:00 PM PDF version to 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
2 Chapter 1: Introduction to Heat Transfer Heat vs. heat flux vs. heat per length Conduction Fourier s Law Ranges of k (Table appendices) Convection Newton s Law of cooling Ranges of h (Table 1.1) Radiation StefanBoltzmann 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 = ρc p T t
3 Example Book Problem 2.5 A solid, truncated cone serves as a support for a system that maintains the top (truncated) face of the cone at a temperature T 1, while the base of the cone is at a temperature T 2 < T 1. 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 q x (2) the heat flux q x " (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.
4 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 = ρc p T t
5 Example Book Problem 2.23 The steadystate temperature distribution in a onedimensional wall of thermal conductivity 50 W/m K and thickness 50 mm is observed to be T( C) = a + bx 2, where a = 200 C, b = 2000 C/m 2 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: SteadyState Conduction (1D) The plane wall Radial systems Energy generation Extended surfaces
6 Heat Transfer through a Wall R 1 R 2 R 3 Example Book Problem 3.3a The window of a car is defogged by attaching a transparent, filmtype heating element to its inner surface. For 4mmthick 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 h i = 10 W/m 2 K while the outside air temp is T,o = 10 C and h o = 65 W/m 2 K.
7 Complex Heat Transfer Contact Resistance
8 A Cylinder Example Book Problem 3.52 Steam 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/m 2 K (a) Sketch and label the thermal circuit (b) What are the outer surface temps for materials A and B?
9 Example Book Problem 3.59 A 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/m 2 K, what is the thickness of the frozen layer? The Sphere ( ) q r = 4πk T s,1 T s,2 ( 1 / r 1 ) 1 / r 2 ( ) R t,cond = 1 4πk 1 1 r 1 r 2
10 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.
11 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 Turbine blade mounted on proposed aircooled rotating disc (T b = 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/m 2 K 1200 C k = 20 W/m K A c = 6x10 4 m 2 P = 110 mm 300 C = 50 mm
12 Fin Efficiencies Fin Efficiencies
13 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
14 Chapter 4: SteadyState Conduction (2D) Shape factors Finitedifference equations Graphical methods Shape Factors
15 Shape Factors, Cont. Graphical Method  Plotting Heat Flux 1. Consider lines of symmetry and choose subsystem 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.
16 Graphical Solution Chapter 5: UnsteadyState Conduction Lumped analysis and the Biot number Spatial effects Semiinfinite solids Constant surface temp. and const. heat flux
17 Review: The Biot Number Bi = hl k 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 Lumped Capacitance Time to reach a given temperature t = ρvc ln θ i where ha s θ Temperature after a given time T T = exp ha s T i T ρvc t θ T T Heat gain (loss) after a given time Q = ( ρvc)θ i 1 exp t τ t 1 where τ t = ha s ρvc ( )
18 Spatial Effects (When lumped analysis cannot be used) Dimensionless Variables Temperature: θ * θ θ i = T T T i T Position: x * x L Time: t * αt L 2 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 * = C 1 exp( ζ 2 1 Fo)cos( ζ 1 x * ) θ * = C 1 exp( ζ 2 1 Fo) J ( ζ 1 r * ) ( ) θ * = C 1 exp ζ 1 2 Fo ( ) 1 ζ 1 r * sin ζ 1r *
19 Table 5.1 ζ 1 and C 1 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
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