Engineering 25
Prob 8-8
Solution
Tutorial
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering-25: Computational Methods
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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
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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
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UD  m s
C  kg m 3
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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
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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
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 T 
 qA  kA 

 x 

Q th U thT
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 DC
 Note that “IVITY” is a
MATERIAL Property that is
INdependent of material
GeoMetry and/or Physical
Size
Engineering-25: Computational Methods
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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
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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
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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
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To
q
Bruce Mayer, PE
ENGR-25_Prob_6-12_Solution.ppt
P 8-8
Engineering-25: Computational Methods
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Bruce Mayer, PE
ENGR-25_Prob_6-12_Solution.ppt
Engineering-25: Computational Methods
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Bruce Mayer, PE
ENGR-25_Prob_6-12_Solution.ppt
Engineering-25: Computational Methods
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Bruce Mayer, PE
ENGR-25_Prob_6-12_Solution.ppt
Engineering-25: Computational Methods
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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
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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
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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
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2
Interface Location
3
Bruce Mayer, PE
ENGR-25_Prob_6-12_Solution.ppt