engr-25_prob_6-12_solution.ppt 1 bruce mayer, pe engineering-25: computational methods bruce mayer,...
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ENGR-25_Prob_6-12_Solution.ppt1
Bruce Mayer, PE Engineering-25: Computational Methods
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Engineering 25
Prob 8-8SolutionTutorial
ENGR-25_Prob_6-12_Solution.ppt2
Bruce Mayer, PE Engineering-25: Computational Methods
The Conduction Eqn
The General Equation:
Electricity → Ohm’s Law
Volts
Siemens
Amps
V
GVGI
I
Heat → Fourier’s Law
C
CW
Watts
T
UTUQ
Q
ththth
th
[FlowRate] = [Conductance]·[PressureChange]
ENGR-25_Prob_6-12_Solution.ppt3
Bruce Mayer, PE Engineering-25: Computational Methods
The Conduction Eqn
The General Equation:
Fluids → Poiseuille's s Law
Diffusion → Fick’s Law
3
3
mkg
sm
skg
C
UCUm
m
DD
Pa
sPakg
skg
P
CPCQ
Q
fff
f
[FlowRate] = [Conductance]·[PressureChange]
ENGR-25_Prob_6-12_Solution.ppt4
Bruce Mayer, PE Engineering-25: Computational Methods
U vs R
CONDUTANCE and RESISTANCE are simply INVERSES
DD
ff
thth
UR
CR
UR
GR
11
11
SoVIU
IVR
What are the UNITS of “R19” Insulation?
ENGR-25_Prob_6-12_Solution.ppt5
Bruce Mayer, PE Engineering-25: Computational Methods
ANCE vs IVITY
ConductANCE from ConductIVITY: σ G
ConductANCE from ConductIVITY: k Uth
VGIVx
AI
x
VAJA
dx
dVJ
TUQTx
kAQ
x
TkAqA
dx
dTkq
ththth
ENGR-25_Prob_6-12_Solution.ppt6
Bruce Mayer, PE Engineering-25: Computational Methods
ANCE vs IVITY
ConductANCE from ConductIVITY: D UD
CUmCx
DAm
x
CDAJA
dx
dCDJ
D
Note that “IVITY” is a MATERIAL Property that is INdependent of material GeoMetry and/or Physical Size
ENGR-25_Prob_6-12_Solution.ppt7
Bruce Mayer, PE Engineering-25: Computational Methods
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)
ENGR-25_Prob_6-12_Solution.ppt8
Bruce Mayer, PE Engineering-25: Computational Methods
P8.8: Series Heat Flow
Thermodynamically Heat Flows: HiTemp→LoTemp
In this Case the Flow Path
The Conduction Model
• The SAME amount of heat, q, flows thru ALL Resistances– i.e.: q = ΔTk/Rk; for any k
To
q
T2T2Ti T1
q
Heat FLow
ENGR-25_Prob_6-12_Solution.ppt9
Bruce Mayer, PE Engineering-25: Computational Methods
Put Graphic(s) Below on a Blank, wide Screen
Heat FLow
To
q
T2T2Ti T1
q
ENGR-25_Prob_6-12_Solution.ppt14
Bruce Mayer, PE Engineering-25: Computational Methods
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 & ConstantsR = [0.036, 4.01, 0.408, 0.038]; % in Kelvins/(Watt/sq-m) = °C/(Watt/sq-m)Ti = 20; To = -10; % in °CArea = 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 °Cdisp(' ')display('Heat Flux in W/sq-m')q = (T(1) - T(2))/R(2) % in W/sq-mdisp(' ')display('Heat Flow in W')Q = Area*q%% for fun make a BAR chart of interface temperaturesbar(T), xlabel('Interface Location'),...
ylabel('Interface Temperature (°C)'),...title('P8-8: Ti = 20 °C * To = -10 °C'), grid
ENGR-25_Prob_6-12_Solution.ppt15
Bruce Mayer, PE Engineering-25: Computational Methods
The ResultsT1, 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