Chapter 13: Heat Transfer  In the previous chapter, we learned that heat can flow from one object to another (e.g., from the warmer to the cooler object)  In this chapter, we investigate the details of how this heat is transferred  Heat is transferred by three methods - Convection - Conduction - Radiation T1T1 T2T2 Q

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

Q  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,1T1,1 T2,2T2,2 Q 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) T2T2 T1T1 L T1T1 T2T2 A Q k

 The heat conducted is where L=length heat flows along [m] A=cross sectional area [m 2 ] t=time over which heat is transferred [s] k=thermal conductivity of the material, see Table 13.1, has units of [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 [k 1 =0.30 J/(s m C°)], brick [k 2 =0.60 J/(s m C°)], and wood [k 3 =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: A 1 =A 2 =A 3 =A, L 1 =L 2 =L 3 =L

Inside, Outside,4 T 0 =T 01 =27 °C T 4 =T 34 =0 °C LLL T 12 T 23 Q1Q1 Q2Q2 Q3Q3 Find T 12 and T 23 Q 1 =Q 2 =Q 3 a) Two unknowns? Stop here go to part b)

Or Since T 34 =0. Now, substitute into part a)

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<e  1 and is unitless. e=1 for a black-body.  The heat due to radiation is where  = Stefan-Boltzmann=5.67x10 -8 J/(s m 3 K 4 ) and T must be in [K]

Example: Problem The filament of a light bulb has temperature of 3.0x10 3 °C and radiates 60 W of power. The emissivity of the filament is Find the surface area of the filament. Solution: Given: P=Q/t=60 W, e=0.36, T=3000 °C= K Remember that 1 W= 1 J/s