Engineering 25 Prob 8-8 Solution Tutorial Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering-25: Computational Methods 1 Bruce Mayer, PE ENGR-25_Prob_6-12_Solution.ppt The Conduction Eqn The General Equation: [FlowRate] = [Conductance]·[PressureChange] Electricity → Ohm’s Law I Amps I G V G Siemens V Volts Heat → Fourier’s Law Qth Watts Qth U th T Engineering-25: Computational Methods 2 U th W C T C Bruce Mayer, PE ENGR-25_Prob_6-12_Solution.ppt The Conduction Eqn The General Equation: [FlowRate] = [Conductance]·[PressureChange] Fluids → Poiseuille's s Law Q f kg s Q f C f P C f kg Pa s P Pa Diffusion → Fick’s Law m kg s m U D C Engineering-25: Computational Methods 3 UD m s C kg m 3 3 Bruce Mayer, PE ENGR-25_Prob_6-12_Solution.ppt U vs R CONDUTANCE and RESISTANCE are simply INVERSES 1 1 R Rth G U th 1 1 Rf RD Cf UD R V I So U I V What are the UNITS of “R19” Insulation? Engineering-25: Computational Methods 4 Bruce Mayer, PE ENGR-25_Prob_6-12_Solution.ppt ANCE vs IVITY ConductANCE from ConductIVITY: σ G dV J dx A I V x V JA A x I G V ConductANCE from ConductIVITY: k Uth dT q k dx kA Q th T x Engineering-25: Computational Methods 5 T qA kA x Q th U thT Bruce Mayer, PE ENGR-25_Prob_6-12_Solution.ppt ANCE vs IVITY ConductANCE from ConductIVITY: D UD dC J D dx C JA DA x DA m C x m U DC Note that “IVITY” is a MATERIAL Property that is INdependent of material GeoMetry and/or Physical Size Engineering-25: Computational Methods 6 Bruce Mayer, PE ENGR-25_Prob_6-12_Solution.ppt ANCE vs IVITY • i.e.; “IVITY” is intrinsic or inherent to the NATURE of the Material “ANCE”, on the other hand, depends on “IVITY” and the Physical SIZE & SHAPE of the Material Object (W/m·K) Engineering-25: Computational Methods 7 Bruce Mayer, PE ENGR-25_Prob_6-12_Solution.ppt P8.8: Series Heat Flow Thermodynamically Heat Flows: HiTemp→LoTemp In this Case the Flow Path Heat FLow The Conduction Model Ti T1 T2 T2 q q • The SAME amount of heat, q, flows thru ALL Resistances – i.e.: q = ΔTk/Rk; for any k Engineering-25: Computational Methods 8 To Bruce Mayer, PE ENGR-25_Prob_6-12_Solution.ppt Put Graphic(s) Below on a Blank, wide Screen Heat FLow Ti T1 T2 T2 q Engineering-25: Computational Methods 9 To q Bruce Mayer, PE ENGR-25_Prob_6-12_Solution.ppt P 8-8 Engineering-25: Computational Methods 10 Bruce Mayer, PE ENGR-25_Prob_6-12_Solution.ppt Engineering-25: Computational Methods 11 Bruce Mayer, PE ENGR-25_Prob_6-12_Solution.ppt Engineering-25: Computational Methods 12 Bruce Mayer, PE ENGR-25_Prob_6-12_Solution.ppt Engineering-25: Computational Methods 13 Bruce Mayer, PE ENGR-25_Prob_6-12_Solution.ppt The MATLAB Code % Bruce Mayer, PE % ENGR25 * 31Oct11 % Prob 8.8: Series Thermal Resistance % file = Prob8_8_1110_Soln.m % % Rearrange the Continuity Eqns listed % Eqns 1&2 => (R1 + R2)*T1 - R1*T2 = R2*Ti % Eqns 2&3 => R3*T1 - (R2 + R3)*T2 + R2*T3 = 0 % Eqns 3&4 => -R4*T2 + (R3 + R4)*T3 = R3*To % % Know Ti & To; thus have 3-Eqns in 3-Unknowns % %The Knowns & Constants R = [0.036, 4.01, 0.408, 0.038]; % in Kelvins/(Watt/sq-m) = °C/(Watt/sq-m) Ti = 20; To = -10; % in °C Area = 10; % in sq-m % % the Calc § A = [R(1)+R(2),-R(1),0;R(3),-(R(2)+R(3)),R(2);0,- R(4),R(3)+R(4)]; b = [R(2)*Ti;0;R(3)*To]; % display('T1, T2, T3 in °C') T = A\b % in °C disp(' ') display('Heat Flux in W/sq-m') q = (T(1) - T(2))/R(2) % in W/sq-m disp(' ') display('Heat Flow in W') Q = Area*q % % for fun make a BAR chart of interface temperatures bar(T), xlabel('Interface Location'),... ylabel('Interface Temperature (°C)'),... title('P8-8: Ti = 20 °C * To = -10 °C'), grid Engineering-25: Computational Methods 14 Bruce Mayer, PE ENGR-25_Prob_6-12_Solution.ppt The Results T1, T2, T3 in C T = 19.7596 -7.0214 -9.7462 Heat Flux in W/sq-m q = 6.6785 Heat Flow in W Q = 66.7854 Engineering-25: Computational Methods 15 Bruce Mayer, PE ENGR-25_Prob_6-12_Solution.ppt Bar Chart Most Temp-Drop Occurs Across the Insulation P8-8: Ti = 20 °C * To = -10 °C 20 Interface Temperature (°C) 15 10 5 0 -5 -10 1 Engineering-25: Computational Methods 16 2 Interface Location 3 Bruce Mayer, PE ENGR-25_Prob_6-12_Solution.ppt