Heat transfer Heat = amount of energy that is transferred from one system to another (or between system and surroundings) as a result of temperature difference The origin of energy transfer is the random motion of molecules Heat is transferred by - thermal conduction - convection - radiation
Fourier s Law of Thermal Conduction T(x) J x J x 1 A dq dt x dt/dx x J x = heat flux, dq x /dt = the rate of heat flow A = cross-sectional area HEAT HOT δt δx dq dt COLD Fig. 2.19: Heat flow in a metal rod heated at one end. Consider the rate of heat flow, dq/dt, across a thin section δ x of the rod. The rate of heat flow is proportional to the temperature gradient δ T/δ x and the cross sectional area A. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 2002) http://materials.usask.ca A J x = dq dt dt = κ dx dt/dx = temperature gradient κ = thermal conductivity [κ] = W m -1 K -1 or W m -1 0 C -1 Reminder: dc Γ = D dx Fick s First Law
Thermal Conductivities of various materials
Thermal Conductivities of various materials Strong metallic bonding Strong covalent bonding Weak Van-der-Waals bonding
Wiedemann Franz - Lorenz Law Thermal conductivity, κ (W K -1 m -1 ) 450 400 300 200 100 0 κ σ = T C WFL Be W Mg Mo Ni Brass (Cu-30Zn) Bronze (95Cu-5Sn) Steel (1080) Pd-40Ag Hg 0 10 20 30 40 50 60 70 Electrical conductivity, σ, 10 6 Ω -1 m -1 Al Ag-3Cu Ag-20Cu Au Ag Cu κ σt =C WFL =2.45 10 8 W Ω K 2 κ = thermal conductivity σ = electrical conductivity T = temperature C WFL = Lorenz number Fig. 2.20: Thermal conductivity, κ vs. electrical conductivity σ for various metals (elements and alloys) at 20 C. The solid line represents the WFL law with C WFL 2.44 108 W Ω K-2. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 2002) http://materials.usask.ca
Wiedemann Franz - Lorenz Law Thermal conductivity, κ (W K -1 m -1 ) 450 400 300 200 100 0 κ σ = T C WFL Be W Mg Mo Ni Brass (Cu-30Zn) Bronze (95Cu-5Sn) Steel (1080) Pd-40Ag Hg 0 10 20 30 40 50 60 70 Electrical conductivity, σ, 10 6 Ω -1 m -1 Al Ag-3Cu Ag-20Cu Au Ag Cu Fig. 2.20: Thermal conductivity, κ vs. electrical conductivity σ for various metals (elements and alloys) at 20 C. The solid line represents the WFL law with C WFL 2.44 108 W Ω K-2. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 2002) http://materials.usask.ca Thermal conductivity, κ (W K -1 m -1 ) κ σt =C WFL =2.45 10 8 W Ω K 2 κ = thermal conductivity 50000 10000 σ = electrical Copper conductivity T = temperature 1000 Aluminum C WFL = Lorenz number 100 Brass (70Cu-30Zn) Al-14%Mg 10 1 10 100 1000 Temperature (K) Thermal conductivity vs. temperature for two pure metals (Cu and Al) and two alloys (brass and Al-14%Mg). Data extracted from Thermophysical Properties of Matter, Vol. 1: Thermal Conductivity, Metallic Elements and Alloys, Y.S. Touloukian et. al (Plenum, New York, 1970).
Thermal Conductivities of various materials Strong metallic bonding Strong covalent bonding Weak Van-der-Waals bonding
Thermal conduction in metals and some insulators Metals Ag, Cu, Al... Insulators with very strong covalent bonding C (diamond), BeO (beryllia),... HEAT HOT COLD Equilibrium Hot Cold Energetic atomic vibrations Fig. 2.22: Conduction of heat in insulators involves the generation and propogation of atomic vibrations through the bonds that couple the atoms. (An intuitive figure.) From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 2002) http://materials.usask.ca Electron Gas Vibrating Cu + ions Heat is transferred by conduction electrons Heat is transferred as atomic vibrations due to strong bonding between atoms
Parabolic heat equation D th 2 T 2 x = T t where D th = κ cρ thermal diffusitivity ρ = density c = specific heat capacity Heat flow in Heat flow out = = Rate of heat accumulation in volume δx Rate of heat accumulation in volume δx = Heat flow in Heat flow out = J x (x)- J x (x+δx) = J x x T x ρ c t 2 T x = κ δx x δ δ 2 C x t C x t D 2 (, ) (, Reminder: = ) Fick s Second Law 2 x t
Fourier s Law Q = Aκ ΔT L = ΔT (L/κA) J x = dq dt dt = κ dx J x 1 A dq dt x Q = rate of heat flow or the heat current, A = cross-sectional area, κ = thermal conductivity (material-dependent constant of proportionality), ΔT = temperature difference between ends of component, L = length of component Ohm s Law I = ΔV R = ΔV (L /σa) I = electric current, ΔV = voltage difference across the conductor, R = resistance, L = length, σ = conductivity, A = cross-sectional area
Fourier s Law Q = Aκ ΔT L = ΔT (L/κA) = θ Q = rate of heat flow or the heat current, A = cross-sectional area, κ = thermal conductivity (material-dependent constant of proportionality), ΔT = temperature difference between ends of component, L = length of component Ohm s Law I = ΔV R = ΔV (L /σa) = R I = electric current, ΔV = voltage difference across the conductor, R = resistance, L = length, σ = conductivity, A = cross-sectional area
Definition of Thermal Resistance Q = ΔT θ Q = rate of heat flow, ΔT = temperature difference, θ = thermal resistance Thermal Resistance θ = L Aκ θ = thermal resistance, L = length, A = cross-sectional area, κ = thermal conductivity
Analogy between thermal and electrical phenomena THERMAL PHENOMENA ELECTRICAL PHENOMENA Q = rate of heat flow I = Current ΔT = temperature difference ΔV = bias (voltage) Q = ΔT θ Θ = thermal resistance Heat reservoir ΔT Absolute Hot zero Heat generator Q A L (a) Cold Q R = resistance Q = ΔT/θ EMF (Electromotive Force) Ground ΔT Current supply θ (b) Q I = ΔV R Fig. 2.23: Conduction of heat through a component in (a) can be modeled as a thermal resistance θ shown in (b) where Q = ΔT/θ. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 2002) http://materials.usask.ca
Analogy between thermal and electrical phenomena THERMAL PHENOMENA Q = rate of heat flow ΔT = temperature difference Θ = thermal resistance Heat reservoir Absolute zero Heat generator ELECTRICAL PHENOMENA I = Current ΔV = bias (voltage) R = resistance EMF (Electromotive Force) Ground Current supply
Analogy between thermal and electrical phenomena THERMAL PHENOMENA Q = rate of heat flow ΔT = temperature difference Θ = thermal resistance Heat reservoir Absolute zero Heat generator ELECTRICAL PHENOMENA I = Current ΔV = bias (voltage) R = resistance EMF (Electromotive Force) Ground Current supply T i T 0 Q =(T i -T 0 )/Θ IR 2
Analogy between thermal and electrical phenomena THERMAL PHENOMENA Q = rate of heat flow ΔT = temperature difference Θ = thermal resistance Heat reservoir Absolute zero Heat generator C = thermal capacitance ELECTRICAL PHENOMENA I = Current ΔV = bias (voltage) R = resistance EMF (Electromotive Force) Ground Current supply C = capacitance δ Q = CδT δ Q = CδV Q = δt C δt I = δv C δt
Analogy between thermal and electrical phenomena. Equivalent circuit of transistor THERMAL PHENOMENA ELECTRICAL PHENOMENA Q = rate of heat flow I = Current ΔT = temperature difference ΔV = bias (voltage) Θ = thermal resistance R = resistance C = thermal capacitance C = capacitance Heat reservoir EMF (Electromotive Force) Absolute zero Ground Heat generator Current supply
Transistor specifications: estimation of required heat sink BJT(2N3716) P d = 15 W T j = 110 0 C θ jc = 1.17 0 C/W θ cs = 0.5 0 C/W θ sa =?? T 0 = 25 0 C θ ja = θ jc +θ cs + θ sa P d Tj T0 Tj T0 = Q' = = θ θ + θ + θ ja jc cs sa 0 0 Tj T0 110 C 25 C 0 θ ja = = = 5.67 C / W P 15W d θ ca = θ ja -θ jc - θ cs = 5.67-1.17 0.5 = 4 0 C/W
Q ' ' Q = κ (2πrL) dt dr ' dr Q = κ (2πL) dt r Q Q b ' ' a dr r = κ (2πL) T 0 T i dt b ln( ) = κ 2πL( Ti T0 ) a κ 2πL( Ti T0 ) = b ln( ) a θ = T i T ' Q 0 = b ln( ) a 2πκL a = 5 mm b = 3 mm ρ = 27 nω m aluminum κ = 0.3 W m -1 K -1 polyethylene I = 500 A L = 1 m ' 2 ρl Q = I = 85. 9W 2 πa b ln( ) θ = a = 0.25 2 πκ L ' 0 0 ΔT = Qθ = 21.5 C T i = 41.5 C 0 C / W
Transistor specifications: derated power T j = 150 0 C θ jc =? P max =? θ cs = 0 θ sa = 5 0 C/W T 0 = 25 0 C
Transistor specifications: non-steady-state regime
Stefan s law P = εσ ( T s S radiated 4 T 4 0 ) σ s = 5.67 10-8 Wm -2 K -4 Stefan s constant, ε = emissivity of the surface, S = surface area emitting the radiation, T = temperature of the surface, T 0 = ambient temperature Effective thermal resistance T>>T 0
Emissivities of different materials
What is the temperature of the filament? 120 V 0.333 A 40 W P = 40 W V = 120 V L = 38.1 cm D = 33 μm ρ (273K) = 5.51 10-8 Ωm ρ (T)~ T 1.2 Power radiated from a light bulb at 2408 C is equal to the electrical power dissipated in the filament.
What is the temperature of filament in electric bulb? 120 V 0.333 A 40 W T =?? P = 40 W V = 120 V L = 38.1 cm D = 33 μm σ s = 5.67 10-8 Wm -2 K -4 ε = 0.35-0.39 Power radiated from a light bulb at 2408 C is equal to the electrical power dissipated in the filament. P 4 4 6 2 5 2 = P = εσ S( T T ) S = πdl = 3.14 33 10 0.381m = 3.95 10 m radiated s 0 40W = (0.35)(5.67 10 8 )(3.95 10 5 )( T 4 (293K) 4 ) T = 2673K T W = 3680 K
Emission spectra of heated bodies T=2673 K Sun spectrum -- in outer space -- on equator -- slanting sunlight T=6050 K 250 500 750 1000 1250 1500 Wavelength, nm
How long does it take to light an electric bulb? t f = 0.042 s = 42 ms
Convection SOLID GAS M V v m Gas Atom Fig. 1.23: Solid in equilibrium in air.during collisions between the gas and solid atoms, kinetic energy is exchanged.
Convection : Newton s law of cooling Newton s law of cooling Q ' = hs ( T T s 0 ) Q = heat flow, T s = temperature of the surface, h = coefficient of convective heat transfer, S = surface area, T 0 = ambient temperature
1000 -
P = 750 W S = 1m 0.75m h 6 W m -2 0 C -1 T =?? P Solution ΔT Q = Θ = ' convection = hs ΔT ΔT = P hs = (6Wm 750W = ) 2 (1m 0.75m) 0 83. 3 2 0 1 C => T = 83.3 + 25 = 108.3 0 C
T i = 25 0 C κ = 0.76 Wm -1 0 C -1 T 1 T 2 T o = -40 0 C S =1m 0.75m l = 10 mm h i = 15 W m -2 0 C -1 h o =25 W m -2 0 C -1 T 1 =?? T 2 =?? Q =??
T i = 25 0 C T 1 h i = 15 W m -2 0 C -1 κ = 0.76 Wm -1 0 C -1 h o = 25 W m -2 0 C -1 T 2 T o = -40 0 C S =1m 0.75m l = 10 mm h i = 15 W m -2 0 C -1 h o = 25 W m -2 0 C -1 T 1 =?? T 2 =?? Q =?? T 2 = -18.5 0 C T 1 = -11.2 0 C
ρ =18 nω m I =700 A a = 5 mm b-a = 1.5 mm c-b = 2 mm κ 1 = 0.3 W m -1 0 C -1 κ 2 = 0.25 W m -1 0 C -1 T 0 = 20 0 C h =25 W m -2 K -1 T =?? T h =??
ρ =18 nω m I =700 A a = 5 mm b-a = 1.5 mm c-b = 2 mm κ 1 = 0.3 W m -1 0 C -1 κ 2 = 0.25 W m -1 0 C -1 T 0 = 20 0 C h =25 W m -2 K -1 T =?? T h =?? ASSUMPTIONS
Q ' ' Q = κ (2πrL) dt dr ' dr Q = κ (2πL) dt r Q Q b ' ' a dr r = κ (2πL) T 0 T i dt b ln( ) = κ 2πL( Ti T0 ) a κ 2πL( Ti T0 ) = b ln( ) a Θ = T i T ' Q 0 = b ln( ) a 2πκL a = 5 mm b = 3 mm ρ = 27 nω m aluminum κ = 0.3 W m -1 K -1 polyethylene I = 500 A L = 1 m ' 2 ρl Q = I = 85. 9W 2 πa ΔT ' = Q Θ = 21.5 0 C b ln( ) Θ = a 2πκL = 0.25 0 T i = 41.5 C 0 C / W
Θ = b ln( ) a 2πκL ρ =18 nω m I =700 A a = 5 mm b-a = 1.5 mm c-b = 2 mm κ 1 = 0.3 W m -1 0 C -1 κ 2 = 0.25 W m -1 0 C -1 T 0 = 20 0 C h = T =?? T h =?? Thermal Resistance T i = 58.9 0 C
ρ =18 nω m I =700 A a = 5 mm b-a = 1.5 mm c-b = 2 mm κ 1 = 0.3 W m -1 0 C -1 κ 2 = 0.25 W m -1 0 C -1 T 0 = 20 0 C h =25 W m -2 K -1 T = 58.9 0 C T h =?? h = h =25 W m -2 K -1
T C T0 ρ =18 nω m I =700 A a = 5 mm b-a = 1.5 mm c-b = 2 mm κ 1 = 0.3 W m -1 0 C -1 κ 2 = 0.25 W m -1 0 C -1 T 0 = 20 0 C h =25 W m -2 K -1 T = 58.9 0 C T h =??!!!
T 0 = 25 0 C T j = 100 0 C S = 100 cm 2 =0.01 m 2 ε =0.75 h = 10 W m -2 K -1 Θ jc = 15 0 C/W Θ cs = 1 0 C/W P d =?? Θ radiation = 15.4 0 C/W convective transfer is more important > Θ convection = 10 0 C/W Θ sink 6.0 0 C/W = 5W