2A4_Lecture02_2008 - CHEM ENG 2A04 Heat Transfer Heat Modes...

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Unformatted text preview: CHEM ENG 2A04 Heat Transfer Heat Modes and Importance Modes Wednesday January 9, 2007 Wednesday Heat Transfer Heat Transfer The three principal laws upon which The Engineering studies are derived: Engineering Conservation of MASS (Continuity, Mass MASS transfer) transfer) Conservation of MOMENTUM (Fluid MOMENTUM mechanics, Mass transfer) mechanics, Conservation of ENERGY (Thermodynamics, ENERGY HEAT TRANSFER) HEAT Heat Transfer Heat Transfer Occurs when there is an energy difference in Occurs a medium or between mediums medium All of nature “seeks a lower energy state.” .” In the case of heat transfer: In HOT Flow of Flow Energy Energy COLD Thermodynamics vs. Heat Transfer Thermodynamics: Study of transfer of work and energy between a Study system and its surrounding in equilibrium in How we achieve equilibrium is not a concern of How thermodynamics, but to engineers it is… HEAT TRANSFER: Determination of the rate of energy (mostly Determination rate internal energy) transferred from one system to internal another resulting in a temperature change. temperature Thermo vs. Heat Transfer – Example Consider cooling a steel bar by placing it in water Thermodynamics final temperature of the steel - water system final Heat transfer temperature of the bar and the water as a temperature function of time function how long it will take to reach equilibrium temperature profile within the bar Both are useful Heat transfer concerns energy in transit due to Heat energy a temperature difference temperature Modes of Heat Transfer Conduction Temperature gradient in a stationary medium (solid or fluid, rarely gas) Convection Heat transfer between a surface and a moving fluid at different temperatures Radiation Surfaces of finite temperature emit energy in the form of electromagnetic waves Nomenclature T - Temperature [=] oC, K, oF, R Temperature C, Q - Heat (from 1st law of thermo) [=] Joules, Btu Heat q - Heat transfer rate [=] Joules/s = Watts Heat q′′ - Heat flux [=] W/m2 [=] q′′′ - Rate of heat generation [=] W/m3 [=] Subscripts - x, y, z, r, θ, φ qx - heat is flowing in the x direction heat q′′ - radial heat flux r radial Conduction • Conduction – Place one end of a steel bar in a fire and the other Place in your hand. Even though your hand is not in the fire, it will quickly become hot as a result of conduction through the bar conduction T1 T2 q T1>T2 Fourier’s Law q′′ = −k∇T Convection Convection • Convection – A hot metal plate is exposed to ambient air. Cooling will hot be by convection. be – What happens to the rate of cooling if we move the air in What some way i.e. using a fan? some – Why is this the case? We are not changing the Why temperature of the air in any way. temperature T∞ q Ts Ts > T∞ Newton’s Law of Cooling q′′ = h( Ts − T∞ ) Radiation Radiation • Radiation – Even though there is no medium through which Even the heat can flow, the sun heats the earth. The mechanism for this heating is radiation mechanism – Every solid emits radiant energy. Amount and rate depend on solid temperature Tabs rate Stefan – Boltzmann Law q′′ = σ Tabs 4 Advice It is extremely important that you understand the physical mechanisms that are occurring in physical heat transfer problems in order to be able to identify and use the proper equations to mathematically describe what is happening mathematically Example Example Think about a coffee cup Think Heat is transferred by: Heat – Conduction and possibly convection from the coffee to the cup wall (depending on whether the water inside is in motion.) motion.) – Conduction through the wall of the cup Conduction through – Convection from the outer wall surface to the ambient air (conduction from the wall to the air —a gas — is minimal.) minimal.) How would things change if How you were holding the cup? you Thought Experiments Thought Experiments 1. Air in a room is at Ti=20°C and air outside is Air °C To=-10°C. That means: everywhere I go in the room °C. the temperature is 20°C, whereas everywhere outside 20°C, it is -10°C. -10 What is the wall temperature? Touch the wall…touch What the window. the Draw the temperature profile Thought Experiments Thought Experiments T Ti To inside wall outside x Draw the temperature profile Draw the direction of heat flux q” through the wall Though Experiments Though Experiments The temperature profile should look like T Ti q” To inside wall outside x Note: the temperature varies continuously Thought Experiments Thought Experiments 2. Fill two cups with hot water; one cup made of Styrofoam and one made of wax paper. Water is at 100°C and room temperature is at 20°C. Hold both 100°C cups and draw the temperature profiles cups T Ti To cup wall room r Thought Experiments Thought Experiments The temperature profile should look like Styrofoam cup Ti T Ti T Wax paper cup Close to 100 To cup wall outside To cup wall outside r r Importance of Heat Transfer Importance of Heat Transfer Used whenever we are concerned with the flow Used of heat — impacts on every engineering every discipline discipline Heat Exchanger Fuel Cell PCs Problem 1.31: Power dissipation from chips operating at a surface temperature of 85°C and in an enclosure whose walls and air are at 25°C for (a) free convection and (b) forced convection. Schematic: T A ir = 25 C h = 4 . 2 ( T s - )1 / 4 or W /m 2 -K o sur = 25oC q ra d S u b s tra te C h ip , P e le c o Ts = 85 C , ε = 0 .6 0 q conv L = 15 m m Assumptions: (1) Steady-state conditions, (2) Radiation exchange between a small surface and a large enclosure, (3) Negligible heat transfer from sides of chip or from back of chip by conduction through the substrate. Analysis: A = L2 = ( 0.015m ) =2.25×10-4 m 2 2 4 Pelec = qconv + qrad = hA ( Ts − T∞ ) + ε Aσ ( Ts4 − Tsur ) (a) If heat transfer is by natural convection, 5/ 4 qconv = CA ( Ts − T∞ ) =4.2W/m 2 ⋅ K 5/4 2.25×10-4 m 2 qrad = 0.60 ( 2.25×10-4 m 2 ) 5.67×10-8 W/m 2 ⋅ K 4 Pelec = 0.158W+0.065W=0.223W (b) If heat transfer is by forced convection, ( ) ( 60K ) ( 358 -298 ) K 4 4 5/4 4 =0.158W =0.065W qconv = hA ( Ts − T∞ ) =250W/m 2 ⋅ K 4 ( 2.25×10-4 m 2 ) ( 60K ) =3.375W Pelec = 3.375W+0.065W=3.44W Announcements Announcements Assignment 1: Due January 16 at the end of class. at In-class Quiz: Monday January 21 In-class Monday Office hours & Mid-term dates: to be determined Tomorrow’s tutorial: Please, bring your textbook (if you have got it) you ...
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This note was uploaded on 01/24/2010 for the course CHEM ENG 2A04 taught by Professor Toddhoare during the Winter '10 term at McMaster University.

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