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## Heat Transfer

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**T1**T2 Q Heat Transfer • In the previous sections, we learned that heat can flow from one object to another (e.g., from the warmer to the cooler object) • In this section, we investigate the details of how this heat is transferred • Heat is transferred by three methods - Convection - Conduction - Radiation**Convection**• Heat is transferred by the bulk movement of a fluid (liquid or gas) • Consider a liquid (water). We know from thermal expansion, that its volume V increases with temperature. • Given the density =m/V, it will decrease with temperature since the mass is constant. • If we heat water in a pot, we apply the heat at the bottom. For some finite mass of the water, its volume increases and density decreases**The water above is cooler and therefore has a larger**density (more mass per unit volume). It sinks while the warmer water rises • As the warmer water rises, it transfers heat to the surrounding cooler water until it is the same temperature as the coldest water T1,1 Q Q T2,2 Q • This water then sinks, because all of the water below it is warmer a Convection Current • Convection currents exist in the ocean (Gulf Stream), the sun and stars, …**This kind of convection is called Natural Convection**• Forced Convection – some external device (fan, water pump) establishes the convection currents • Unfortunately, the mathematical description of convection is beyond the scope of this course Conduction • Heat is transferred directly through the material or from one material to another (solids, liquids, and gases) k A T1 Q T2 L T1 T2**The heat conducted is**where L=length heat flows along [m] A=cross sectional area [m2] t=time over which heat is transferred [s] k=thermal conductivity of the material, see Table 16-3, has units of [W/(m C°)] or [J/(s m C°)] • Large k – more heat flow – thermal conductor (good electrical conductors are good thermal conductors – metals) • Small k – heat flow is restricted – thermal insulators**Example Problem**Three building materials, plasterboard [k1=0.30 J/(s m C°)], brick [k2=0.60 J/(s m C°)], and wood [k3=0.10 J/(s m C°)], are sandwiched together. The temperature at the inside and outside surfaces are 27 °C and 0 °C, respectively. Each material has the same thickness and cross-sectional area. Find the temperature (a) at the plasterboard-brick interface and (b) at the brick-wood interface. Solution: Consider conductive heat flow at each interface Given: A1=A2=A3=A, L1=L2=L3=L**Find T12 and T23**Q1=Q2=Q3 a) T12 T23 Inside,0 Outside,4 1 2 3 T0=T01=27 °C Q1 Q2 T4=T34=0 °C Q3 L L L Two unknowns? Stop here go to part b)**Radiation (Thermal)**• Heat is transferred by electromagnetic waves: microwaves, radio waves, infrared radiation, visible (optical), ultraviolet radiation, x-rays, gamma rays • ``Radiation’’ is more general than nuclear radiation (alpha, beta, gamma particles) • Every object emits radiation and absorbs radiation • The temperature, surface area, and surface properties of an object effects the amount of heat that is emitted or absorbed and the wavelength**An object which absorbs 100% of the radiation is known as a**blackbody – a perfect absorber. • An object which is a perfect absorber is a perfect emitter. The radiation emitted by a blackbody is called blackbody radiation. • The absorption/emission properties of an object are described by its emissivity, 0<e1 and is unitless. e=1 for a black-body. • The heat due to radiation is where = Stefan-Boltzmann =5.669x10-8 J/(s m2 K4) and T must be in [K]**Example Problem**The filament of a light bulb has temperature of 3.0x103°C and radiates 60 W of power. The emissivity of the filament is 0.36. Find the surface area of the filament. Solution: Given: P=Q/t=60 W, e=0.36, T=3000 °C=3273.15 K Remember that 1 W= 1 J/s