分析传热传质学

Review

for

Mid

Exam2Midterm

Exam

22014.12.150

ponitsFor

one

hourLabel

the

following

statements

as

either

true

(T)or

false

(F)Short

Response

Sectiontative

SectionCoverage

of

midexam2Lecture

5

(transient

conduction)-

Lecture

10(solutions)Lecture

11

(external

flow)

and

Lecture

12(internal

flow)

will

not

be

in

exam

2.You

can

turn

in

thei.e.

Dec.

4

(thursday)after

the

examTransient

conductionOverviewTransient

conduction (unsteady

state

conduction)a

heat

transfer

process

in

which

the

temperature

varies

with

both

time

andlocation

in

a

medium.is

initiated

whenever

a

steady-state

system

experiences

a

change

in

operatingconditions

and

proceeds

until

a

new

thermal

steady

state

is

achieved.The

temperature

variation

strongly

depends

on

Bi

For

Bi

<<1

For

moderate

to

large

values

of

Bi

For

Bi

>>

1To

use

the

lumped

capacitance

method

The

very thing

that

you

should s

to

calculate

the

Biot

number!!!

If

Bi

<

0.1,

the

error

with

using

the

lumped

capacitance

method

is

small1/

hAL

/

kA

RcondRconv

hL

kBiLumped

CapacitanceiWhen

is

the

steady

state?The

physical

meaning

of

t?When

t

=

t,

what

is

?

exp

hAs

t

exp

t

i

Ti

T

T

T

cV

t

Define:

thermal

time

constanttshA

cV

cV

1

RC

thermallumped

capacitance

hA

t

t

s

VchAst

Bi

FoL2cDimensionless

timeFo

tFourier

numberexpBi

Foi

T

Ti

T

Tytical

solution

to

the

Heat

Equation

for

aPlane

Wallwith

Symmetrical

Convection

ConditionsIf

the

lumped

capacitance

approximation

can

not

be

made,

consideration

mustbe

given

to

spatial,

as

wellas

temporal,

variations

in

temperature

during

thetransient

process.For

a

plane

wall

with

symmetrical

convectionconditions

and

constant

properties,

without

generation,

the

heatequation

andinitial/boundary

conditions

are:2T

1

Tx2

tT

x,0

Tix0

0Tx

h

T

L,t

T

xLk

Tx

Existence

of

seven

independent

variables:T

T

x,t,Ti

,T

,

k,,

hHow

may

the

functional

dependence

be

simplified?of

HeatEquation

andInitial/Boundary

Conditions:Dimensionless

temperature

difference:

*

T

T

i

Ti

T

xx

*

LDimensionless

space

coordinate:Dimensionless

time:t

*

tL

2

FoFo

the

FourierNumber

*

f

x

*,

Fo,

BiNon-dimensionalization

expBi

Foi

Ti

T

T

TRecall

Fo

in

the

result

oflumped

capacitance

methodT

T

x,t,Ti

,T

,

k,,

h112T

1

Tx2

tT

x,0

Tix0

0TxxLk

Tx

h

T

L,t

T

i

*Lx

*

xt

*

L2t

Fo

*

T

T

i

i

T

T2

(T

T

)

1

(T

T

)x2*(x*,0)

1

t2

1

x2

t2

(*i

)

1

(*i

)x2

t2

(*)

(*)x*2Fox

*

0*

*

0xx

*

1

*

Bi

*

(1,t

*)x

*The

BiotNumber:kBi

hLNumerical

Methods

for

Two-Dimensional,

Steady-State

ConductionProcedure:–Step

1:

Represent

the

physical

system

by

a

nodal

network

(ormesh,

grid).–Step

2:

to

obtain

a

finite-difference

equation

(algebraic)

foreach

node

of

unknown

temperature.Method

1:

Finite-difference

form

of

the

heat

equation

(mathematics)Method

2:

Theenergy

balance

method

(physics)–Step

3:

Solve

the

resulting

set

of

algebraic

equations

for

theunknown

nodal

temperatures.The

Nodal

Networkidentifies

discrete

points

at

which

the

temperature

is

to

bedetermined;

uses

anm,n

notation

to

designate

their

location.What

is

represented

by

thetemperature

at

a

nodal

point,as

for

example,

Tm,n?m

-

1/2n

-1/2deAverage

temperature

ofelement

volumeincrementl

or

ElementeNodal

network

x

incrementnodal

poionrt

noyCntorolvuomStep

2:

Develop

the

finite-difference

equationsMethod

1:

Finite

–difference

form

of

the

heat

equation02T

2T

qx,

y

m,n2T

T

Tm1/2,n

x

xx2

y2

kApproximatem

-

1/2m

+

1/2what

if

x

0?m1/2,n

x

Tm1,n

Tm

,nxm1/2,nx2

T

Tm

,n

Tm1,nxm1/2,nx

TxEin

Eg

0Consider

application

to

an

interior

nodal

point(one

that

exchanges

heat

by

conduction

with

four,equidistant

nodal

points):4x

y

0i1where,

for

example,Tm1,n

Tm,nqm1,nm,n

k

y

x?Est(assume

that

conduction

transfer

occurs

exclusively

through

lanes)Method

2

: The

Energy

Balance

MethodNo

need

to

derive

based

on

the

heat

i.e. erning

equation.As

aconvenience

that

obviates

the

need

topredetermine

the

direction

of

heatflow,

assume

all

heat

flows

are

into

the

nodal

region

of

interest,

and

express

allheat

rates

accordingly.

Hence,

the

energy

balance

es:Control

volumeSurfacem

-

1/2Nodes

at

surfaceConsider

an

external

corner

with

convection

heat

transfer.

(no

generation)qm1,nm,n

qm,n1m,n

qm,n

0withx

,ym1,nm,n1T

Tkkhx

2

T

2hx1

T

0m,nyk

2

Tm1,n

Tm,nxxk

2Tm,n1

Tm,nym,nhxT

T

2

hm,n

yT

T

2Example某二 态热传递。材料导热系数λ。导热材料的上表面温度恒定为t0。其余三面与温度tf的流体进行对流换热，表面对流换热系数hf。现若节点划分 ，各网格边长均为Δx。试由 展开法(Finite

–difference

method)写出节点1的方程，由热平衡法(EnergyBalance

Method)写出节点14的方程。Introduction

to

ConvectionFlow

and

Thermal

ConsiderationsWhen

a

moving

fluid

comes

into

contact

with

a

surface

at

sometemperature

difference

(e.g.

heated

surface,

cold

fluid)

the

fluid

willtransfer

heat

to/from

the

surface

in

a

process ogous

toconduction

(randommotion

of

the

fluid)advect

heat

away

from

the

point

of

contact

by

its

motion

(bulkmotion

ofthe

fluid)Advection

=

bulk

fluid

motion

onlyThe

combined

effect

of

these

two

phenomena

is

called

convectionRate

equationNewton’s

Law

of

Cooling21heat

flux[W/m2]q"

h

Ts

Theat

transfer

coefficient[W/m2-K]surface

temperature[K]fluid

temperature[K]q

Ah

Ts

Tnotes:the

heat

transfer

coefficientFunction

of

fluid

properties,

flow

conditions,

and

geometryin

order

to

study

convection

heat

transfer,

we

must

also

study

fluiddynamicsConvection

Heat

Transfer

CoefficientsLocal

Heat

Flux

and

Coefficient:qs

h

Ts

T

Average

Heat

Flux

and

Coefficient

for

a

Uniform

Surface

Temperature:q

hAs

Ts

T

q

A

qdAs

TssssAhdA

TsAssAh

1

hdAFor

a

flat

plate

in

parallel

flow:1LLoh

hdxWhatPARAMETERS

affect

h

?Fluid

properties

(temperature

dependent)Flow

conditionsGeometry…h

f

(u,

k,

cp

,

,,

,l....)23CLASSIFICATIONSMECHANISMFree

(Natural)

ForcedGEOMETRYInternal

ExternalSTATESingle-PhaseTwo-Phase24erning

equations

for

convection25MassMOMENTUMEnergy

x

y（u

u

u

v

u

)

dxdyConsult

fluid

dynamics

textbooks

pleaseConvection

term

of

MomentumBody

force

pressureViscousMomentum

equation

(Navier-Stokes

equation)（u

u

u

v

u

)

F

p

(

2u

2u

)xt

x

y

x

x2

y2ρvdxConvectionerning

equationsu

v

0；x

y

p

2u

2u

)x

(

x2y2（v

u

v

v

v

)

F

p

(

2v

2v

)y

x2

y2t

x

yy（u

u

u

v

u

)

Ft

x

yx2T

u

T

v

T

k2

2

T

Tt

x

y

cp

x

y

2

u

2if

with

viscous

dissipation

cp

yBoundary

layer

on

a

plate

in

parallelcoming

flowVelocity

and

thermal

boundary

layerPhysics,

characteristics,

and

definitionsBoundary

Layer

developmentHeat

transferVelocity

:,22Thermal

:2x

y2u

u

xpdpdx2T

2Tx2

y2Boundary

layer

approximations!Boundary

Layer

EquationsNow

it

could

be

solved

!33Consider

concurrent

velocity

and

thermal

boundary

layer

developmentforsteady,pressible

flow

with

constantfluid

properties

,

cp

,

k

andtwo-dimensional,negligible

body

forces.Example2

2对流换热的能量微分方程为（假设：流动是二维的；流体为不可压缩的

型流体；流体物性为常数、无内热源；粘性耗散产生的耗散热忽略不计）t

u

t

v

t

2t

2t

x

y

x

y

cp

分别说明其中各项所代表的物理意义基于上式写出稳态下、边界层内的能量方程。Solutions

for

the

flat

plate

in

parallelflow

(Laminar

flow

region)35RecNon-dimensionalization

and

similarityparameters3637u*

v*

0x*

y**

u*

*

u*

12u*u

x*v

y*

u

Lu

Lx*y*

v*

T

*

y*2y*2ReL1

1

2T

*u*

T

*PrSimilarity

parametersRePrNuSolution

（flat

plate

in

parallel

flow)39xNu

0.332Re1

2Pr1

3x1

1khxx

0.332

u

x

2

3

a

Example管内对流换热，关于

数Pr，以下正确的是：代表了粘性力与惯性力的相对大小代表了流动边界层与热边界层厚度的相对大小

c)其数值与温度相关，但是计算中在变化不大的情况下常取定性温度对应的数值d)在换热准则关系式涉及的三个准则数即Nu,Re,

Pr数中，Pr是唯一一个与传热相关的准则数。More

about

the

solutionswhen

to

usehow

to

usefunctional

form41Structure

of

a

procedure

for

solution

ofconvection

problems42Establish

the

geometry

of

the

situationDetermine

the

fluid,

and

evaluate

the

fluid

propertiesEstablish

the

boundary

conditions

( kind,

…)Establish

the

flow

regime

as

determined

by

the

Reynoldsnumber.Select

the

appropriate

equation,

taking

into

account

theflow

regime

and

fluid

property

restrictions

whi

ayapplyCalculate

the

value(s)

of

the

convection

heat

transfercoefficient

and/or

heat

transfer.Solution1

2

1

3Prx

xNu

0.332Re1

1khxx

0.332

u

x

2

3

a

When

can

you

use

this

correlation?Assumptions:

Steady,

pressible,

laminar

flow

withconstant

fluid

properties

and

negligible

viscous

dissipationand

recognizing

that

dp/dx

=0Classification:

Forced,

externalGeometry:

Flat

plate

in

parrellel

flowPr

>=

0.6,

Re

<

5*105Boundary:

isothermalThe

fluid

properties

to

deter