INTRODUCTION Examples of Heat Transfer Problems. Focal Point in Heat Transfer

Transcription

1 INTRODUCTION Examples of Heat Transfer Problems (1) Slide Projector, (2) Ice Storage (3) Re-entry Aerodynamic Heating (4) Pot Handle (5) Under-indo radiator of heating system (6) Refrigerator Focal Point in Heat Transfer Determination of the temperature distribution in a region Based on presentations by professor Latif M. Jiji, The City College of Ne York 1

2 Modes of Heat Transfer (i) Conduction: by molecular or atomic activity (ii) Convection: by mass motion (iii) Radiation: by electromagnetic aves Conduction: Fourier's La A S = all area L L = all thickness Tsi T so T 1 = inside surface temperature T 2 = outside surface temperature Is Q& x proportional to S? Ho? Is Q& proportional to L? Ho? x Is Q& proportional to (T 1 -T 2 )? Ho? x T so 1 T 2 si dx 0 L A S [m 2 ] x Fig Q& q x x

3 Heat Transfer Rate (tepelný tok) λs ( T T ) Q& = 1 2 [W] 1.1. x L λ is called thermal conductivity, a property of material Equation (1.1) is valid for: (i) steady state, (ii) onedimensional conduction and (iii) constant λ. Reformulation of (1.1): Apply (1.1) to an element dx: T 1 T(x), T 2 T(x + dx), δ dx Q & x = λ S T(x) T(x dx + dx) = λs T(x + dx) dx T(x) 3

4 Definition: Heat flux (měrný tepelný tok) (1.2) becomes q = λ x dt dx For 3-D: T T q& = λ, q& = λ, x x y y & Q x = λ S dt dx [W] & (1.4) T & = λ (1.5) z q z q& x (1.2) Q& q& = x x [W/m (1.3) S 2 ] Equation (1.5) is knon as Fourier's la 4

5 Convection: Neton's La of Cooling Convection: Energy is transported by means of mass motion T Classification: (1) Free convection (2) Forced convection T T s q& Heat exchange beteen a surface and a fluid moving over it q& ( T T ) here q& = surface (all) flux [W/m 2 ] (měrný tepelný tok) T = surface temperature T = fluid temperature far aay from the surface 5

6 Rerite: ( ) = α T T This is Neton's la of cooling. q (1.6) NOTE: (1) α is called heat transfer coefficient (součinitel přestupu tepla) (2) It is a defined quantity (3) It depends on geometry, fluid properties and motion (4) To determine α, the temperature distribution in the fluid must be knon (5) Major objective in convection: Determination of α 6

7 Typical Values of α Table 1.1 Typical values of α Important: Use this table as a guide only. Do not use these values to solve problems. Process Free Convection Gases Liquids Forced Convection Gases Liquids Liquid metals α (W/m 2 K) ,000 5,000-50,000 Phase change Boiling liquids Condensation 2, ,000 5, ,000 7

8 Radiation: Stefan-Boltzmann La Transmission by electromagnetic aves No medium is needed. Best in a vacuum Maximum possible radiation: By an ideal surface called blackbody. Stefan-Boltzmann la for blackbody radiation flux: 0 (1.7) E = blackbody radiation flux (zářivost) 0 T = surface temperature, measured in absolute degrees (Kelvin) σ = Stefan-Boltzmann constant 8 4 E = σ T [W]

9 σ = 5.67 x 10-8 W/m 2 -K 4 (1.8) Real surface: E = radiation flux (zářivost šedého povrchu) Emissivity, ε, a surface property defined as E ε = E 0 Combining (1.7) and (1.9) 4 (1.9) E = ε σ T (1.10) 9

10 Q & 12 Energy Exchange Beteen To Bodies: A Simplified Model = Energy exchange beteen to surfaces Absorptivity (absorptance) a : Fraction of radiation incident on a surface hich is absorbed Simplified model: Gray surface: ε = a Special case: A small gray surface enclosed by a much larger surface Q& 12 = ε 1 σs 1 (T 4 1 T 4 2 ( ) 1 = small surface, ( ) 2 = large surface ) (1.11) 10

11 Properties Heat transfer depends on material and surface properties as: Conductivity Density Viscosity Specific heat Emissivity 11

12 Example: Application of Problem Solving Methodology Square transistor is mounted on a circuit board. Transistor surface is cooled by convection. Size, dissipated poer, heat transfer coefficient and ambient temperature are knon. Determine surface temperature. (1) Observations (i) Schematic diagram T Ts transistor board Fig

13 (ii) Dissipated electric energy is removed by convection and radiation (iii) Surface temperature is higher than ambient temperature (iv) Increasing dissipated poer increases surface temperature (2) Problem Definition Find the relationship beteen poer and surface temperature (3) Solution Plan Apply Neton's la of cooling to the surface of the transistor. 13

14 (4) Plan Execution (i) Assumptions (1) Steady state (2) Dissipated electric energy leaves transistor surface (no energy leaves from the back side) (3) Negligible heat loss by radiation (4) Uniform surface temperature (5) Uniform heat transfer coefficient (6) Constant ambient temperature (ii) Analysis Neton's la of cooling: T Ts transistor board Fig

15 α = heat transfer coefficient = 8 [W/m 2 K] q& 2 = surface heat flux, W/m T =? surface temperature, o C T = ambient air temperature = 26 o C q& ( ) = α T T (a) Conservation of energy and assumptions (2) and (3): Dissipated poer = Heat removedfromsurface P = Sq& (b) S = surface area = 1 (cm) 1 (cm) = 1 = m 2 15

16 P = poer dissipated in transistor = 0.04 W (b) into (a) Solving for (d) gives T P S = α (T T ) (c) T T + (iii) Computations T = 26 + P Sα = (d) 0,04 0, = 76 o C 16

17 (iv) Checking Dimensional check: Equation (d) is dimensionally correct. Each term has units of temperature. Qualitative check: Equation (d) behaves correctly: Increasing P increases T T = T + Decreasing the surface area increases Decreasing α increases T P Sα T (d) Limiting checks: Poer off (P = 0): surface temperature = ambient temperature If α = 0 ( no heat is removed) then T 17

18 (5) Learning and Generalizing (i) T < 85 o C (ii) Examine assumptions (2) and (3): (2) no energy leaves from the back side (3) negligible heat loss by radiation Heat removed from the back side loers calculated T Radiation heat loss loers calculated T 18

19 Requirements for Energy Conservation Energy Conservation for a Control Volume a) Balance for Energy flux in every moment, [W]. de du E& ak in + E& g E& out = E& ak = ( = = dt dt m du dt ) [W] b) Balance for amount of heat, [J]. E + E E = ΔE [J] in g out ak 19

20 Requirements for Energy Conservation Energy Conservation for a Control Surface E & in E & out = 0 20