Heat Transfer - California State University,lcaretto/me375/finalReview.pdf · Final Review May 16,...

Final Review May 16, 2006

ME 375 Heat Transfer 1

Review for Final ExamReview for Final Exam

Larry CarettoMechanical Engineering 375

Heat TransferHeat Transfer

May 16, 2007

2

Outline Basic equations, thermal resistance Heat sources Conduction, steady and unsteady Computing convection heat transfer

Forced convection, internal and external Natural convection

Radiation properties Radiative Exchange

3

Final Exam Wednesday, May 23, 3 5 pm Open textbook/one-page equation sheet Problems like homework, midterm and

quiz problems Cumulative with emphasis on second

half of course Complete basic approach to all

problems rather than finishing details of algebra or arithmetic

Basic Equations Fourier law for heat conduction (1D)

( ) ( )L

TTkAAqQorL

TTkq 2121 === &&&

Convection heat transfer)( = TThAQ ssconv&

Radiation (from small object, 1, in large enclosure, 2)

( )42411121, TTAQrad = &

5

Heat Generation Various

phenomena in solids can generate heat

Define as the heat generated per unit volume per unit time

gene&

Figure 2-21 from engel, Heat and Mass Transfer

6

Rectangular Energy Balance

genzyx

p ezq

yq

xq

tTc &

&&&+

=

heat generatedStored energy

heat inflow heat outflow

genp ezTk

zyTk

yxTk

xtTc &+

+

+

=

+=

Uses Fourier Law

=

Tkq&



7

Cylindrical Coordinates

gen

p

ezTk

z

Tkrr

Tkrrr

tTc

&+

+

+

=

211

dzrdrTkdAqQd rr

== &&


8

Spherical Coordinates

gen

p

eTkr

Tkr

rTkr

rr

tTc

&+

+

+

=

22

2

22

sin1

sinsin1

1


9

1-D, Rectangular, Heat Generation Temperature profile for generation with T

= T0 at x = 0 and T = TL at x = L

( )

+==

LTT

kLe

kxe

kdxdTkq Lgengen 0

222 &&

&

( ) ( )L

TTkLxeq Lgen +

= 022&

&

( )L

xTTkxLe

kxe

TT Lgengen += 02

0 22&&

10

Plot of (T - T0)/(TL - T0) for Heat Generation in a Slab

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x/L

Tem

pera

ture

Diff

eren

ce R

atio

H = 0H = .01H = .1H = 1H = 2H = 5H = 10

10

( )0

2

TTkeL

HL

gen

=

&

11

Slab With Heat GenerationBoth boundary temperatures = TB

1

1.2

1.4

1.6

1.8

2

2.2

2.4

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Dimensionless Distance, x/L

T / T

B

H = 0H = .01H = .1H = 1H = 2H = 5H = 10

B

gen

kTeL

H&2

=

12

Thermal Resistance Conduction

( )kALR

RTTQ

LTTAkQ cond

cond=

=

= 2121 &&

Convection( )

hAR

RTT

QTThAQ convconv

fsfs

1=

== &&

Radiation

( ) radrad hATTTTTTAR 1221122323112111

=+++

=F



13

Composite Materials II

14LrhAh 1111 2

11

=

LrhAh 4242 211

=

1

2

1ln1

rr

Lk

2

3

2ln1

rr

Lk

3

4

3ln1

rr

Lk

Composite Cylindrical

Shell


15

Fin Results Infinitely long fin

Heat transfer at end (Lc = A/p)

( )mL

xLmTTmL

xLmTT cbc

cb cosh

)(coshcosh

)(cosh =

==

( ) mLTThpkAQ bcx tanh0 = =&

( ) ckAhpxbmxb eTTTTe ==( )== == TThpkAqAQ bcxcx 00 &&

16

Fin Efficiency Compare actual

heat transfer to ideal case where entire fin is at base temperature

( )

==

TThAQQ

Q

bfin

fin

fin

finfin

&

&

&

max,

pLA cfin =

for uniform cross section

Figure 3-39 from engel, Heat Transfer

17

Overall Fin Effectiveness Original area, A = (area

with fins, Afin) + (area without fins, Aunfin)

( )( )( )

+=

TThATTAAh

QQ

bfinno

bunfinfinfin

finno

fin &

&

Figure 3-45 from engel, Heat Transfer

+==

finno

unfin

finno

finfin

finno

fintotal A

AAA

QQ

&&

18

Lumped Parameter Model

cpp Lch

chAb

==

V

Assumes same temperature in solid Use characteristic length Lc = V/A

( ) ( ) ( ) +== TeTTToreTTTT btibti

Must have Bi = hLc/k < 0.1 to use this



19

Transient 1D Convection

Figure 4-11 in engel, Heat and Mass

Transfer

All problems have similar chart solutions20

Slab Center-line (x = 0) Temperature Chart Figure 4-15(a) in engel, Heat and Mass Transfer

21

Chart II Can find T at any

x/L from this chart once T at x = 0 is found from previous chart

See basis for this chart on the next page

=

TTTT

00

Figure 4-15(b) in engel, Heat and Mass Transfer

22

Approximate Solutions Valid for for > 0.2

Values of A1 and 1 depend on Bi and are different for each geometry (as is Bi)

=

=

0101

21

rrJeA

TTTT

i

=

=

01

1

01 sin

21

rr

rreA

TTTT

i

Cylinder

Sphere

Slab

11 cos21

=

= eATTTT

i

23

Semi-Infinite Solids Plane that

extends to infinity in all directions

Practical applications: large area for short times Example: earth

surface locallyFigure 4-24 in engel, Heat and Mass Transfer 24

Multidimensional Solutions Can get multidimensional solutions as

product of one dimensional solutions All one-dimensional solutions have initial

temperature, Ti, with convection coefficient, h, and environmental temperature, T, starting at t = 0

General rule: twoD = onetwo where oneand two are solutions from charts for plane, cylinder or sphere



25

Multidimensional Example

( )

( )

( )

slabinfinitei

cylinderinfinitei

cylinderfinitei

TTTtxT

TTTtrT

TTTtxrT

=

,

,

,,

x = a/2

x = -a/2Figure 4-35 in engel, Heat

and Mass Transfer

26

Flow Classifications Forced versus free Internal (as in pipes) versus external (as

around aircraft) Entry regions in pipes vs. fully-developed

Unsteady (changing with time) versus unsteady (not changing with time)

Laminar versus turbulent Compressible versus incompressible Inviscid flow regions ( not important) One-, two- or three-dimensional

27

Flows Laminar

flow is layered, turbulent flows are not (but have some structure)

Figures 6-9 and 6-16. engel, Heat and Mass Transfer

28

Boundary Layer

Region near wall with sharp gradients Thickness, , usually very thin compared to

overall dimension in y directionFigure 6-12 from engel, Heat and Mass Transfer

29

Thermal Boundary Layer Thin region near

solid surface in which most of temperature change occurs

Thermal boundary layer thickness may be less than, greater than or equal to that of the momentum boundary layer

Figure 6-15. engel, Heat and Mass Transfer 30

Dimensionless Convection Nusselt number, Nu = hLc/kfluid

Different from Bi = hLc/ksolid Reynolds number, Re = VLc/ = VLc/ Prandtl number Pr = cp/k (in tables) Grashof number, Gr = gTLc3/2

g = gravity, = expansion coefficient = (1/)(/T)p, and T = | Twall T |

Peclet, Pe = RePr; Rayleigh, Ra = GrPr



31

Characteristic Length Can use length as a subscript on

dimensionless numbers to show correct length to use in a problem ReD = VD/, Rex = Vx/, ReL = VL/ NuD = hD/k, Nux = hx/k, NuL = hL/k GrD = 2gTD3/2, Grx = 2gTx3/2,

GrL = 2gTL3/2

Use not necessary if meaning is clear

32

How to Compute h Follow this general pattern

Find equations for h for the description of the flow given

Correct flow geometry (local or average h?) Free or forced convection

Determine if flow is laminar or turbulent Different flows have different measures to

determine if the flow is laminar or turbulent based on the Reynolds number, Re, for forced convection and the Grashof number, Gr, for free convection

33

How to Compute h Continue to follow this general pattern

Select correct equation for Nu (laminar or turbulent; range of Re, Pr, Gr, etc.)

Compute appropriate temperature for finding properties

Evaluate fluid properties (, k, , Pr) at the appropriate temperature

Compute Nusselt number from equation of the form Nu = C Rea Prb or D Rac

Compute h = k Nu / LC34

Property Temperature Find properties at correct temperature Some equations specify particular

temperatures to be used (e.g. /w) External flows and natural convection

use film temperature (Tw + T)/2 Internal flows use mean fluid

temperature (Tin + Tout)/2

35

Key Ideas of External Flows The flow is unconfined Moving objects into still air are modeled

as still objects with air flowing over them There is an approach condition of

velocity, U, and temperature, T Far from the body the velocity and

temperature remain at U and T T is the (constant) fluid temperature

used to compute heat transfer36

Flat Plate Flow Equations Laminar flow (Rex, ReL < 500,000, Pr > .6)

3/12/12/12

3/12/12/12

PrRe664.0Re33.12

PrRe332.0Re664.02

LLLwall

f

xx

xxwall

f

kLhNu

UC

kxhNu

UC

x

===

=

===

=

Turbulent flow (5x105 < Rex, ReL < 107)

3/18.05/12

3/18.05/12

PrRe037.0Re074.02

PrRe0296.0Re059.02

LLLwall

f

xx

xxwall

f

kLhNu

UC

kxhNu

UC

x

===

=

===

=

For turbulent Nu, .6 < Pr < 60



37

Flat Plate Flow Equations II

( ) 3/18.05/12 Pr871Re037.0Re1742

Re074.0

2===

=

LLLL

wallf k

LhNuU

C

Average properties for com-bined laminar and turbulent regions with transition at xc = 500000 /U Valid for 5x105 < ReL < 107 and

0.6 < Pr < 60

Figure 7-10 from engel, Heat and Mass Transfer 38

Heat Transfer Coefficients Cylinder average h (RePr > 0.2; properties

at (T + Ts)/2 5/48/54/13/2

2/12/1

000,282Re1

Pr4.01

PrRe62.03.0

+

+

+==k

hDNu

Sphere average h (3.5 Re 80,000; 0.7 Pr 380; s at Ts; other properties at T)

[ ]4/1

4.03/22/1 PrRe06.0Re4.02

++== sk

hDNu

39

Other Shapes and Equations

Part of Table 7-1 from engel, Heat and Mass Transfer 40

Tube Bank Heat Transfer

Table 7-2 from engel, Heat and Mass Transfer

41

Key Ideas of Internal Flows The flow is confined There is a temperature and velocity

profile in the flow Use average velocity and temperature

Wall fluid heat exchange will change the average fluid temperature There is no longer a constant fluid

temperature like T for computing heat transfer

42

Area Terms


Acs is cross-sectional area for the flow Acs = D2/4 for circular pipe Acs = WH for rectangular

duct Aw is the wall area for heat

transfer Aw = DL for circular pipe Aw = 2(W + H)L for

rectangular duct

D

W

H

L

L



43

Average Temperature Change Let T represent the average fluid

temperature (instead of Tavg, Tm or ) T will change from inlet to outlet of

confined flow This gives a variable driving force (Twall

Tfluid) for heat transfer Can accommodate this by using the first

law of thermodynamics: = cp(Tout Tin) Two cases: fixed wall heat flux and fixed

wall temperature

T

Q& m&

44

Fixed Wall Heat Flux Fixed wall heat flux, , over given wall

area, Aw, gives total heat input which is related to Tout Tin by thermodynamics

wallq&

( )p

wwallinoutinoutpwwall cm

AqTTTTcmAqQ&

&&&& +===

Outlet can be any point along flow path where area from inlet is Aw

We can compute Tw at this point as Tw = Tout + /hwallq&

45

Constant Wall Temperature

( ) ( ) pw

cmhA

sinsout eTTTT&

=


hAw / cp = NTU, the number of transfer units

This is general equation for computing Tout in internal flows

m&

46

Log-mean Temperature Diff This is usually written as a set of

temperature differences( ) ( ) ( )

=

=

sin

sout

sinsout

sin

sout

inout

TTTT

TTTT

TTTT

TTTLMlnln

( ) )(ln

TLMhA

TTTTTThAQ w

sin

sout

inoutw =

=& engel uses Tlm for LMT

47

Developing Flows

Momentum boundary layer development

Thermal boundary layer development

48

Fully Developed Flow Temperature profile does not change

with x if flow is fully developed thermally This means that T/r does not change

with downstream distance, x, so heat flux (and Nu) do not depend on x

PrRe05.0Re05.0 DL

DL th Laminar entry

lengths Turbulent

entry lengths10Re359.1 41 =

DL

DL ht



49

Eggs from Figure 8-9 in engel, Heat and Mass Transfer

Entry Region Nusselt Numbers

50

Internal Flow Pressure Drop General formula: p = f (L/D) V2/2 Friction factor, f, depends on Re =

VD/ and relative roughness, /D For laminar flows, f = 64/Re

No dependence on relative roughness For turbulent flows

+

=

fD

f Re51.2

7.3log0.21 10

+

11.1

10 7.3Re9.6log8.11 D

f

Colebrook

Haaland

51

Moody Diagram

Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi. Copyright 2005 by John Wiley & Sons, Inc. All rights reserved.

52

Laminar Nusselt Number Laminar flow if Re = VD/ < 2,300 Fully-developed, constant heat flux, Nu

= 4.36 Fully-developed, constant wall

temperature: Nu = 3.66 Entry region, constant wall temperature:

( )( )[ ] 32PrRe04.01

PrRe065.066.3LDLDNu

++=

53

Noncircular Ducts Define hydraulic diameter, Dh = 4A/P

A is cross-sectional area for flow P is wetted perimeter For a circular pipe where A = pD2/4 and P

= D, Dh = 4(D2/4) / (D) = D For turbulent flows use Moody diagram

with D replaced by Dh in Re, f, and /D For laminar flows, f = A/Re and Nu = B

(all based on Dh) A and B next slide54

From engel, Heat and Mass Transfer



55

Turbulent Flow Smooth tubes (Gnielinski)

( )( )( ) ( )



61

Vertical Plate Free Convection Constant wall heat flux

Use = hA(Tw T) to compute an unknown temperature (Tw or T) from known wall heat flux and computed h

Tw varies along wall, but the average heat transfer uses midpoint temperature, TL/2

Use trial and error solution with TL/2 T as T in Ra used to compute h = kNu/L

q&

wall

wallLLwallwall hA

qTTTThAq&

& == 2/2/ )(

62

Vertical Cylinder Apply equations for vertical

plate from previous charts if D/L 35/Gr1/4

For this D/L effects of curvature are not important

Thin cylinder results of Cebeciand Minkowcyz and Sparrow available in ASME Transactions

Cylinder figure from Table 9-1 in engel, Heat and Mass Transfer

63

Horizontal Plate

Hot surface facing up or cold surface facing down

Lc = area / perimeter (As/p) For a rectangle of length, L, and width, W,

Lc = (LW) / (2L + 2W) = 1 / ( 2 / W + 2 / L) For a circle, Lc = R2 / 2R = R/2 = D/4

Cold surface

1173/1

744/1

101015.0101054.0



67

Horizontal Enclosures II

753/1

544/1

10104068.010410195.0



73

Shell and Tube PassesTube flow has three complete changes of direction giving four tube passes

Shell flow changes direction to give two shell passes

Figure 11-5(b) from engel, Heat and Mass Transfer 74

Overall U U is overall heat

transfer coefficient Analyzed here for

double-pipe heat exchanger

UAAUAU

AhR

AhR

iioo

oowall

ii111

11

===

++=


75

Heat Exchange Analysis Heat transfer from

hot to cold fluid TUAQ =&

First law energy balances

( )( )outhinhph

incoutcpc

TTcmQTTcmQ

h

c

,,

,,

==

&&

&&

Assumes no heat loss to surroundings Subscripts c and h denote cold and hot

fluids, respectively Alternative analysis for phase change

76

Parallel Flow Parallel flow

heat exchangerlmTUAQ =&

=

=

2

1

21

1

2

12

lnlnTT

TT

TT

TTTlm


( ) ( )

=

incinh

outcouth

incinhoutcouthlm

TTTT

TTTTT

,,

,,

,,,,

ln

77

Counter Flow Same basic equations

Difference in T1 and T2 definitions

T1 = Th,in Tc,out

T2 = Th,out Tc,in

( ) ( )

=

outcinh

incouth

outcinhincouthlm

TTTT

TTTTT

,,

,,

,,,,

lnFigure 11-16 from engel, Heat and Mass Transfer

==

1

2

12

lnTT

TTUATUAQ lm&

78

Heat Exchanger Problems With Tlm method we want to find U or

A when all temperatures are known If we know three temperatures, we can

find the fourth by an energy balance with known mass flow rates (and cps)

( )( )outhinhph

incoutcpc

TTcmQTTcmQ

h

c

,,

,,

==

&&

&&Can find from two temperatures for one stream and then find unknown temperature

Q&



79

Correction Factors Correction factor parameters, R and P

Shell and tube definitions below

Correction factor charts show diagrams that illustrate the equations for P and R

( )( )

shellp

tubep

intubeouttube

intubeinshell

intubeinshell

intubeouttube

cm

cm

ttTT

TTTT

R

tTtt

TTTT

P

&

&=

=

=

=

=

12

21

,,

,,

11

12

,,

,,

80

Correction Factor Chart I


P

81

Effectiveness-NTU Method Used when not all temperatures are

known Based on ratio of actual heat transfer to

maximum possible heat transfer Maximum possible temperature

difference, Tmax is Th,in Tc,in Only one fluid, the one with the smaller value

of cp, can have Tmax Define Cc = ( cp)c and Ch = ( cp)h

m&m&m&

82

Effectiveness,

In effectiveness-NTU method we find , then find = max Use CminTmax to find max because C1T1

= C2T2 or T2 = C1T1/C2 If T2 = Tmax and C1/C2 > 1, T2 > Tmax CminTmax is maximum heat transfer that

can occur without impossible T < Tc,in

( ) ( )chincinhCCC

TTCQ

QQ ,minmin

,.minmax=

==

&

&

&

Q& Q&

Q&

83

Find Example chart for finding effectiveness from NTU = UA/Cmin and Cmin/Cmax ratio


For NTU = 1.5 and Cmin/Cmax= 0.25, = ?

For NTU = 1.5 and Cmin/Cmax= 0.25, = .7

84

Effectiveness Equations Double pipe parallel flow

( )

ce cNTU

+

=+

11 1

( )

( )cNTU

cNTU

cee

= 11

11

Double pipe counter flowmax

minCCc =

minCUANTU =

Figures from Figure 11-26 from engel, Heat and Mass Transfer



85

Black-Body Radiation Basic black body equation: Eb = T4

Eb is total black-body radiation energy flux W/m2 or Btu/hrft2

is the Stefan-Boltzmann constant = 5.670x10-8 W/m2K4

= 0.1714x10-8 Btu/hrft2R4

Must use absolute temperature Radiation flux varies with wavelength

Eb is flux at given wavelength,

86

Spectral Eb Energy (W/m2)

emitted varies with wavelength and temperature

Maximum point occurs where T= 2897.8 mK

T increase shifts peak shift to lower


InfraredUltraviolet

87

Partial Black-body Power

=1

10

0, dEE bb

=0

4 '1 dET

f b

Black body radiation between = 0 and = 1 is Eb,0-1

Fraction of total radiation (T4) between = 0 and any given is f

Figure 12-13 from engel, Heat and Mass Transfer88

Radiation Tables Can show that f is function of T

( ) ( ) ( ) ( )

=

=

=T

TCTCb TdeTCd

eC

TdE

Tf

05

1

05

14

04 1

11

1122

Radiation tables give f versus T See table 12-2,

page 672 in text Extract from this

table shown at right

89


Radiation in finite band,

( ) ( )TfTfdE

TdE

T

dET

f

bb

b

12

04

04

4

12

2

1

21

11

1

=

==

1)(0)0( == ff90

Emissivity Ratio of actual emissive power to blakc

body emissive power Diffuse surface emissivity does not

depend on direction Gray surface emissivity does not depend

on wavelength Gray, diffuse surface emissivity is the

does not depend on direction or wavelength

Simplest surface to handle and often used in radiation calculations



91 92

Average Emissivity Average over all wavelengths

( ) ( ) ( ) ( )

=

==

05

1

05

14

04 1

11

1122

TdeT

Cde

CT

dET TCTCb

For emissivity with constant values in a series of wavelength ranges

( )( ) ( )

( )( ) ( )

( )( ) ( )

+

+

=

TTC

T

TTC

T

TC eTTdC

eTTdC

eTTdC

2

2

2

1

2

1

2 11

11

11

513

512

05

11

( )[ ] ( ) ( )[ ] ( )[ ]TfTfTfTf 2312211 10 ++=

Applies to other properties as well

93 94

Properties Incoming

radiation properties Reflectivity, Absorptivity, Transmissivity,

Energy balance: + + = 1Figure 12-31 from

engel, Heat and Mass Transfer

95

Data Solar

radiation has effective source temperature of about 5800 K

Figure 12-33 from engel, Heat and Mass Transfer 96

Kirchoffs Law Absorptivity equals emissivity (at the

same temperature) True only for values in a given direction

and wavelength Assuming total hemispherical values of

and are the same simplifies radiation heat transfer calculations, but is not always a good assumption



97

Effect of Temperature Emissivity, , depends on surface

temperature Absorptivity, , depends on source

temperature (e.g. Tsun 5800 K) For surfaces exposed to solar radiation

high and low will keep surface warm low and high will keep surface cool Does not violate Kirchoffs law since

source and surface temperatures differ 98

View Factor, Fij or Fij Fij or Fij is the

fraction of radiation, leaving surface i, that strikes surface j AiFij = AjFji kFik = 1 (enclosure) F12+3 = F12 + F13


Fkk = 0 only if k is a flat surface View factors from equations or charts

99

All Black Surface Enclosure Heat transfer from surface 1 reaching

surface 2 is A1F12T14

Heat transfer from surface 2 reaching surface 1 is A2F21T24 = A1F12T24

Net heat exchange between surface 1 and surface 2: A1F12(T14 T24) Negative value indicates heat into surface 1 For multiple surfaces

( )4411

jiij

N

ji

N

jjii TTFAQQ ==

==

&&

100

Gray Diffuse Opaque Enclosure

Kirchoffs law applies to the average: = at all temperatures

For opaque surfaces = 0 so + = 1 For gray, diffusive, opaque surfaces

then = 1 = 1 Define radiosity, J = Eb + G = emitted

and reflected radiation

( )ii

ii

i

ibiibi

i

iii A

RwhereR

JEJEAQ

=

=

=1

1&

101

Net Radiation Leaving Surface = A(J G) Can show


Q&

( )JEAQ b =

1&

i

iibi R

JEQ

= ,&

i

ii A

R

=1

102

Gray Diffuse Opaque II

011

=

+

=

== i

biiN

j ij

ji

i

ibiN

j ij

ji

REJ

RJJ

RJE

RJJ

Combining two equations for

( )iji

ij

N

j ij

jiN

jjiiji

N

jiji FA

RR

JJJJFAQQ 1

111=

===

===

&&

iQ&

Solve system of N simultaneous linear equations for N values of Ji

Black or reradiating surface ( = 0) has Ji = Ebi = Ti4

iQ&



103

Review Circuit Analogy Look at simple

enclosure with only two surfaces

Apply circuit analogy with total resistance

22

2

12111

1

212112 111

AFAA

EER

EEQ bbTotal

bb

++

=

=&

104

Three-Surface Circuit

Three or more surfaces easirerby system of equations

Exception: = 0

21

212112, RRR

EER

EEQ bbTotal

bbnet ++

=

=

c

&

231312

111

RRR

R

++

=c

3Q&

105

Review Three-Surface Circuit

If 3 = 0, net,12can be found from circuit with two parallel resistances

21

212112, RRR

EER

EEQ bbTotal

bbnet ++

=

=

c

&

Q& Q&

231312

111

RRR

R

++

=c

106

Radiation Exchange Two possible surface conditions: (1)

known temperature, (2) known iQ&

( ) ( ) NiJJFAJEAQN

jjiijiib

i

iii i ,,11 1

K& ==

= =

4

,1,1

111 ibN

ijjjij

i

ii

N

ijjij

i

i TEJFJFi

==

+ ==

i

N

ijjjijii

N

ijjiji QJFAJFA &=

== ,1,1

Solve this set of N simultaneous equations for N values of Ji

(1)

(2)

107

Radiation Exchange II Once all Ji values are known we can

compute unknown values of Ti and For known Ti

For known

iQ&

( ) ( )iii

iiib

i

iii JT

AJEAQi

=

= 411

&

iQ&

4111

iii

iiii

ii

iib QA

JTQA

JEi

&&

+

=

+=

108

Numerical Heat Transfer Finite difference expressions with

truncation error Computers give roundoff error Convert differential equations to

algebraic equations Solve system of algebraic equations to get

temperatures at discrete points Reduce step size for stability

Will not be covered on final

Heat Transfer - California State University,lcaretto/me375/finalReview.pdf · Final Review May 16,...

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