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# notes MAE 3311

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This 32 page Class Notes was uploaded by Bupe Chinukwe on Sunday February 21, 2016. The Class Notes belongs to MAE 3311 at University of Texas at Arlington taught by Dr. Hani in Spring 2016. Since its upload, it has received 15 views. For similar materials see Thermodynamics 2 in Engineering and Tech at University of Texas at Arlington.

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Date Created: 02/21/16

Energy Equation ● Ein - Eout + Egen = Estored – Does this refer to a Control Volume? Jeff Luttrell MAE 3314 Spring 2016 1 Fourier's Law of Conduction ● In conduction, the flow of heat energy is proportional to the space derivative of the temperature – Heat Flow vs Heat Flux dT qx=−k A Eq. 2.1 dx qx dT q''x= =−k Eq. 2.2 A dx Jeff Luttrell MAE 3314 Spring 2016 2 Newton's Law of Cooling ●In convection, the flow of heat energy is proportional to a temperature difference – Heat Flow vs. Heat Flux q q''= =h(T −Ts) ∞ A q=h A(T −T ) Eq. 1.3a s ∞ Jeff Luttrell MAE 3314 Spring 2016 3 Heat Diffusion Equation ●Derived from the Energy Equation ● Per coordinate system – Cartesian (shown below) Eq. 2.19 – CylindricaEq. 2.26 – Spherical Eq. 2.29 ∂T ∂T ∂T ∂T ∂ k + ∂ k + ∂ k +q˙ρC p ∂x ( )∂ x) (∂ y ∂ y ∂z ∂z ∂t Jeff Luttrell MAE 3314 Spring 2016 4 Chapter 3 One Dimensional Steady State Conduction ●Start with the full Heat Diffusion Equation, Cartesian Coordinates and Simplify –1-D –Steady State –No heat generation ∂ ∂T ∂ ∂T ∂ ∂T ∂T ( ) ( ) ( ) k + k +q˙ρC p ∂x ∂ x ∂ y ∂ y ∂z ∂z ∂t d dT dx ( dx =0 Jeff Luttrell MAE 3314 Spring 2016 5 3.1.1 1-D, SS, Heat Diffusion ● If conductivity is also constant, and we integrate twice – Result is a linear equation ● Linear temperature variation d T dT ∫ 2dx= =∫0dx=C1 (d x dx dT ∫ dx dx=T (x) = ∫1dx=C1∗x+C 2 Eq. 3.2 T(x) =C1∗x+C 2 Jeff Luttrell MAE 3314 Spring 2016 6 Applications ● Walls and windows of buildings ● Micro-chips on PC boards Jeff Luttrell MAE 3314 Spring 2016 7 Use BC's to Solve ● T(0) = T s,1 ● T(L) = T Variable s,2 Eq. 3.3 T x s,1 T(x)=(T s,2−T s,1) +T s,1 L T s,2 Known Values x x = L Jeff Luttrell MAE 3314 Spring 2016 8 Temperature -> Heat Flow ● With Fourier's Law, find the heat flow q =−k A dT x dx This should look familiar dT = d (T −T ) +T Note: dx dx [ s,2 s,1L s,1] qxdoes not vary with x (conservation of energy) dT = (T s,2T s,1 What about Heat Flux? dx L What about Thermal So ... qx=− k A (Ts,2−T s,1 Resistance? L Eq. 3.4 Jeff Luttrell MAE 3314 Spring 2016 9 3.1.2 Thermal Resistance ● Use Electrical Analogy – Current --> Heat Flow – Voltage Difference --> Temperature Difference – Resistance --> Thermal Resistance – (V1 - V2) = IR --> (T1-T2) = q * R th Jeff Luttrell MAE 3314 Spring 2016 10 Thermal Resistance ● From V = IR to (T2 - T1) = q R th ● So R =thT2 - T1) / q – What are the UNITS? Degrees/Watts ● C/watt or K/watt ● Conduction thermal resistance – Note: switch T /1T 2 (T 1T ) 2 L R th = – Drop 'sign' qx k A Eq. 3.6 ● Resistance values ALWAYS Positive Jeff Luttrell MAE 3314 Spring 2016 11 Conduction R vs. Electrical R th ● ● Electrical Resistance Conduction Resistance – R proportionaL to – R proportionaL to – R proportiona(1/A) – R proportiona(1/A) L R th k A Jeff Luttrell MAE 3314 Spring 2016 12 Convection Thermal Resistance ●Newton's Law of Cooling q=h A(T −T ) s ∞ (T sT ) ∞ 1 R th = q x hA 1 R th hA Eq. 3.9 Jeff Luttrell MAE 3314 Spring 2016 13 Radiation Thermal Resistance ● 4 Non-linear T terms so... –Linearize per Equation 1.9 for a radiation heat transfer coefficient q =h A(T −T ) rad r s ∞ (T −T ) R = s ∞ = 1 A th ,r q h rad r 1 R th ,r Eq. 3.13 hrA Jeff Luttrell MAE 3314 Spring 2016 14 Radiation Heat Transfer Coefficient Area Emissivity Stefan-Boltzman constant 4 4 Factor q radσ(T −T ) 1 2 2 2 2 2 = Aϵσ(T +T )1T −T2) 1 2 2 2 = Aϵσ(T +T )1T +T2)(T 1T ) 2 1 2 = Ah (T −T ) r 1 2 Jeff Luttrell MAE 3314 Spring 2016 15 Solve like Electrical Resistance Network ?? ● Given – h and h 1 2 – T∞,1 T ∞,1and T ∞,2 – L, A, and k Ts,1 ● Find the heat flow T ● Find T and T s,2 s,1 s,2 T∞,2 x x = L Jeff Luttrell MAE 3314 Spring 2016 16 Electrical Analogy Thermal Resistance ● Rth,1 1/(h1A) ● Rth,2 L/(k A) Rth ,1 Rth ,2 Rth ,3 ● Rth,3 1/(h2A) ● Resistors in Series R th,total th,1R th,2R th,3 1 L 1 R th,total + + h 1 k A h A 2 Eq. 3.12 Jeff Luttrell MAE 3314 Spring 2016 17 Solve for Heat Flow ● R th- (T2 - T1) / q then... (T∞,2−T ∞,1 (T ∞,1T ∞,2) q=− = R th,total R th,total Eq. 3.14 ● What about the surface temperatures? Jeff Luttrell MAE 3314 Spring 2016 18 Surface Temperatures? ?? ● Analogy-wise: – T is like V at s,1 s,1 – T∞,1 T s,2is like V at s,2 ● Can we find these V's? T s,1 ● Then we can find T – s,2 T s,1and T s,2 T∞,2 x x = L Jeff Luttrell MAE 3314 Spring 2016 19 Surface Temperatures ?? Eq. 3.16 q xq =qh1q L h2 similar (T −T ) (T −T ) (T −T ) T∞,1 qx= ∞,1 s,1= s,1 s,2 = s,2 ∞,2 1/(h1A) L/k A 1/(h2A) Ts,1 T s,1= T ∞,1 - q/R th,1 T s,2 T s,2= T s,1- q/R th,2and/or... T = T + q/R T∞,2 s,2 ∞,2 th,3 x x = L Jeff Luttrell MAE 3314 Spring 2016 20 Composite Walls ● A wall with multiple layers – 5 resistors in series T∞,1 L L L 1 2 3 T s,1 R th,total th,1R th,2R th,3R th,4R th,5 Ts,4 1 L 1 L 2 L 3 1 T T R th,total + + + + s,2 s,3 h 1 k A 1 k 2 k 3 h A 2 ● q = (T∞,1- T∞,2/ R th,total T∞,2 x x = L See Figure 3.2 Jeff Luttrell MAE 3314 Spring 2016 21 Parallel Paths Option 1 3 options, different results T∞,1 L1 L2 L3 – 1 : Parallel L2 only Ts,1 Ts,2 1 R L2 (1/R )+(1/R ) L2,A L2,B T∞,2 1 R L2 x x = L k L2, A A k L2,BAB ( L )( L ) Note: Account for AREAs 2 2 Jeff Luttrell MAE 3314 Spring 2016 22 Parallel Paths Option 2 T∞,1 L1 L2 L3 ● 2nd : Parallel walls Ts,1 Ts,2 T ∞,2 x x = L Jeff Luttrell MAE 3314 Spring 2016 23 Parallel Paths Option 3 T∞,1 L1 L2 L3 ● 3rd : Parallel paths Ts,1 Ts,2 T ∞,2 x x = L Jeff Luttrell MAE 3314 Spring 2016 24 Differences of Parallel Options ● The thermal resistance networks are different – Option 1: T L1-L2 interfaceT L2-L3 interfacethe same for both paths --> k L2,A≈ k L2,B ● If k >> k or k << k L2,A L2,B L2,A L2,B 1-D assumption may not be valid – Option 2: Valid if T s,1and T s,lastre the same for both paths – Option 3: Two paths do not affect each other Jeff Luttrell MAE 3314 Spring 2016 25 New Concept Overall Heat Transfer Coefficient ● Users prefer the notation q = U A (T 1 T )2 Eq. 3.17 Constant x Area x Temperature Difference ● R th1/UA Eq. 3.19 – But R ths area embedded ● U = 1/(AR ) th Eq. 3.18 Jeff Luttrell MAE 3314 Spring 2016 26 Overall Heat Transfer Coefficient 1 L1 L 2 L 3 1 Rth + + + + Sum of h1A k1A k2A k A 3 A 2 Resistances A L1 L 2 L3 A Multiply through A Rth +A +A +A + by Area h1A k1A k 2 k3A h2A 1 L1 L 2 L3 1 A Rth + + + + Simplify h1 k1 k 2 k3 h2 1 1 Overall Heat U= A R = L L L Transfer Coefficient th 1 + 1+ 2 + 3+ 1 h1 k1 k2 k 3 h2 Easier Option: Find R Then multiply by A, totalMAE 3314 Spring 2016 27 Using “U” ● Consider a room with 4 walls ● U total U1 + U2 + U3 + U4 U1 U2 ● A = A1 + A2 + A3 + A4 U3 U4 q = U total(T inT )out Jeff Luttrell MAE 3314 Spring 2016 28 3.1.4 Contact Resistance The Causes ● Previously assumed interfaces are perfect ● Many interfaces imperfect – Roughness – Mis-match contours T∞,1 L1 L2 – Debris T s,1 T s,2 Ts,3T ∞,2 JeTf LutTrell MAE 3314 Spring 2016 x x = L9 A B Contact “Resistance” ● Create a “resistance” to account for imperfect contact at the interface R = 1 + L1 +R + L 2 + 1 th h A k A th,ck A h A 1 1 2 2 '' T AT B TA−T B Rth,c Rth,c '' R th,c = Note A andBT – q q A from previous – chart R''th,c--> contact resistance (heat flux) – R --> contact resistance (heat flow) th,c Jeff Luttrell MAE 3314 Spring 2016 30 Contact Resistance Values ● Metallic surfaces R'' th,cn Table 3.1 (p.118) – 1. Contact Pressure/Vacuum conditions (?) – 2. Aluminum w/ interfacial fluids ● Solid/Solid Interfaces in Table 3.2 (p. 119) – Some with “thermal interface materials” (TIM) that fill gaps with high conductivity material Text: Thermal Management of Microelectronic Equipment by Yeh & Chu Jeff Luttrell MAE 3314 Spring 2016 31 Summary ● 1-D Cartesian coord, SS, no generation, constant 'k' – T=f(x) is linear – qxcan be found from T=f(x) – R th electrical analogy ● Series and parallel resistor networks ● Find intermediate temperatures ● U - overall heat transfer coefficient ● Contact resistance Jeff Luttrell MAE 3314 Spring 2016 32

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