ENGR-25_Prob_6-12_Solution.ppt1

Bruce Mayer, PE Engineering-25: Computational Methods

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

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.ppt10

Bruce Mayer, PE Engineering-25: Computational Methods

P 8-8

ENGR-25_Prob_6-12_Solution.ppt11

Bruce Mayer, PE Engineering-25: Computational Methods

ENGR-25_Prob_6-12_Solution.ppt12

Bruce Mayer, PE Engineering-25: Computational Methods

ENGR-25_Prob_6-12_Solution.ppt13

Bruce Mayer, PE Engineering-25: Computational Methods

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

ENGR-25_Prob_6-12_Solution.ppt16

Bruce Mayer, PE Engineering-25: Computational Methods

Bar Chart

Most Temp-Drop Occurs Across the Insulation

1 2 3-10

-5

0

5

10

15

20

Interface Location

Inte

rfac

e T

empe

ratu

re (

°C)

P8-8: Ti = 20 °C * To = -10 °C