Introduction to Heat Transfer with Examples - Lecture Slides | ME 3345, Study notes of Heat and Mass Transfer

Download Introduction to Heat Transfer with Examples - Lecture Slides | ME 3345 and more Study notes Heat and Mass Transfer in PDF only on Docsity! K ME S34S Notes Why Is Heat Transfer Important? HVAC (personal comfort) ri i i i Aerospace Material Processing Three basic modes of heat transfer: conduction, convection, and radiation. Heat conduction refers to thermal energy that is transferred as a result of the collision of energetic particles within a solid, stationary fluid, or between them. Dominant in solids and occurs in liquids. Thermal radiation transfers energy between objects by the emission and absorption of electromagnetic waves (it can transfer energy without any intervening medium!). Convective heat transfer or simply convection involves the transfer of energy associated with conduction into a fluid accompanied by fluid motion (advection). Convection: Newton's Law of Cooling In 1701, Isaac Newton performed experiments by placing heated objects in fluids and found that the rate of temperature change of the object is proportional to the temperature difference between the object and the fluid, i.e., aT * «-(T-T. n° oo) The cooling mechanism between fluids in solids in this case was due to convection which can be expressed as: q’ = A(T, -Ty) This is called Newton’s law of cooling, and 4 [W/m?-K] is called convection heat transfer coefficient or film coefficient. Types of Convection: We will concentrate on forced and natural convection Buoyancy-driven flow &) Forced }-— Hot components: on printed circuit boards (a) 2] Vapor bubbles Water Figure 1.5 Convection heat transfer processes. {a) Forced convection. (b) Natural convection. (¢} Boiling. (d) Condensation.  000°001-00SZ uoTesUspUuod Jo SuTpIog asueyo oseyd WIM DorRsAu07 000°0Z-00I spinbry] OSC-S$7 SOSBD UOT}99AU0S pe010,] 000T-0S spibry Sc-C sasen UONIIAUOD 301] OH - ,A/AA) 4 SSI901g NEE ENE RT JHA YIIOO TOJSUBL] 7V9Y WOTIIATIOD ayy Jo sonjea JeordAy, “| wav], In 1879, Joseph Stefan (1835-1893) published an empirical relation that the radiant heat flux from a surface is proportional to 7*. Five years later, Ludwig Boltzmann (1844-1906) theoretically derived the equation, known as the Stefan-Boltzmann law. It expresses the maximum emissive power (radiant heat flux) of a surface as, n _ 4 Grad = Ey 7 of, where o= 5.67 x 10°° W/m?-K‘ is called the Stefan-Boltzmann constant. This equation is only for a perfect of ideal emitter known as a blackbody. E,, is the blackbody emissive power. Temperature is always in K. Radiation and Real Surfaces Most surfaces are not ideal, thus their emissive power is less than a blackbody and do not strictly follow the Stefan- Boltzmann Law. The ratio of the emissive power of a real body to a blackbody is known as emissivity. E= - = EO which is between 0 and 1. - E, oT The emissive power (emitted radiative heat flux) is Grade =E= soT* Irradiation and Absorption Emissive Power speaks of thermal radiation which is leaving a surface. Radiation can also be incident upon a surface and is called Irradiation (G). Has units of W/m?. For an ideal body (black body) all energy is absorbed. For real surfaces, the fraction that is absorbed is described by a material property known as absorptivity: a (between 0 and 1). " _ it _ Prada ~ °Arad,i = aG Methodology for Applications of Conservation Laws 1. The appropriate control volume must be defined and identified. 2. The appropriate time basis must be identified for transient problem. 3. The relevant energy transfer processes must be identified and shown on the control volume. 4. The conservation equations must be written using rate form. Only then can the numerical values be substituted into the equations and to solve for the unknowns. Example: The coating on a plate is cured by exposure to an infrared lamp providing a uniform irradiation of 2000 W/r2. It absorbs 80% of the irradiation and has an emissivity of 0.5. It is also exposed to the surroundings at 30°C and to an air flow at 20°C with a convection coefficient of 15 W/m2-K. Determine the cure temperature of the plate. Known: «=0.8, ¢=0.5 Schematic: | Te G = 2000 Wim? Dur @ t/ Typ =273 +30 =303 K T,,, h irr OF a\ Qconv — And so on. ~<— SAR Note: We must use K for radiation. Assumptions: Schematic: sun - Steady state e@ - Neglect heat transfer to the back “/ Th Girr OF \ conv “? Analysis: ee ee ee — Energy balance (per unit area): Ey = Evy ~ Bout +Ey OG = Grad + Tony = EO(Ts: ~ Tory) + HT, ~ Te.) 0.8 2000 = 0.5x5.67 x10 *(Z* —303*) +15(T, —293) Solve by trial-and-error: 7, =377 K =104°C Discussion: What can we do to decrease T,?