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ME601 1/18/06Slide 1

Heat Transfer Heat transfer modes:

conduction - heat transfer through a solid

convection - heat transfer from a surface to a fluid

radiation - heat transfer from surface-to-surface via electromagneticwaves

In this class, you will concentrate on solving heat transfer problemsnumerically rather than analytically

However, proper use of numerical software requires that you understandheat transfer theory sufficiently to:

understand the important aspects of your problem (what can andcannot be neglected)

develop appropriate sanity checks for your solution

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ME601 1/18/06Slide 2

Conduction

Conduction is governed by Fouriers law:

x

Tq k A

x

heat transfer in x-direction

thermal conductivity

temperature gradient

x

T

xq

area x-direction

one sanity check is that the temperature distribution makes sense relative to Fouriers law

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ME601 1/18/06Slide 3

Conduction Resistances

TH TC

L

The analysis of conduction withinsimple geometries can be conciselysummarized in the form of thermal

resistances

For example, the analysis of 1-D,

steady state conduction through aplane wall leads to:

A

TH

TCq

H Ck A

q T T

L

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ME601 1/18/06Slide 4

Resistance Concept

H Ck A

q T TL

heat transfer temperature difference

1 21

e

I V VR

Ohms LawHeat Transfer

current voltage difference

electrical resistance(thermal resistance)-1

1

H C

pw

q T TR

length you have to conduct

conductivity area available for conductionpw

LR

k A

whereRpw is the thermal resistance associated with conduction through a plane wall:

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ME601 1/18/06Slide 5

Conduction through Other Shapes

ln

2

out

in

cyl

r

rR

L k

1H C

cyl

q T T

R

where

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ME601 1/18/06Slide 6

Conduction through Other Shapes

1 1 1

4sph

in out

Rk r r

1H C

sph

q T TR

where

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7/15ME601 1/18/06Slide 7

Convection Resistance

surface temperature = Ts

fluid temperature = Tf

convq

Newtons law of cooling:

conv s s f q h A T T

heat transfer temperature difference

(thermal resistance)-1

1

conv s f

conv

q T TR

1conv

s

Rh A

whereRconv is the thermal resistance associated with convection:

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8/15ME601 1/18/06Slide 8

Radiation Resistance

surface emissivity, esurface temperature = Ts

surrounding surfaces at temperature = Tsur

radq

heat transfertemperature difference

(thermal resistance)-1

1

rad s f rad

q T TR

32 21 1

4rad

ss s sur s sur

RA TA T T T T e e

whereRrad is the thermal resistance associated with radiation:

4 4rad s s sur q A T T e

2 2rad s s sur s sur s sur q A T T T T T T e

-82 4W5.67x10 m K

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9/15ME601 1/18/06Slide 9

Why Should I Care?

Understanding you can quickly discern what is controlling the heat

transfer in your problem by estimating thermal resistances

Analytical Verification you can sometimes simplify your numericalproblem to a point where it can be exactly verified against these

analytical, resistance solutions

Sanity Check you can always see if your numerical solution makessense by calculating thermal resistances

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10/15ME601 1/18/06Slide 10

Heater in a Bathroom Floor

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11/15ME601 1/18/06Slide 11

Resistance Network Model

air at Tair,1 = 15C

air at Tair,2 = 5C

plywood

studs

airdrywall

strip heater

lineoleum

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12/15ME601 1/18/06Slide 12

Th

Tair,2= 5C

hq

Rp=

0.002 K/W

Rs =0.15 K/W

Rair=

0.46 K/W

Rd=0.014 K/W

Rconv =

0.011 K/W

RL =

0.016 K/W

Tair,1= 15C

Rconv =

0.011 K/W

TL

1q2q

Resistance Values

What can you learn from this?

most of the heat will go to the bathroom

most of the heat will go through the studs

the characteristics of the drywall, plywood and basement convection do not matter

the characteristics of the studs, linoleum, and bathroom convection do matter

you can estimate the heater temperature and heater power required

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13/15ME601 1/18/06Slide 13

Bracket

k = 14 W/m-K

thickness = 1 cm

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14/15ME601 1/18/06Slide 14

FE Analysis

predicted heat flow is 1.84 W

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ME601 1/18/06Slide 15

Resistance Analysis Sanity Check

2

19cm m K 100 cm K 136

14 W 1 cm m Wbracket

LR

k A

200 C 20 C W1.33W

136 K

H C

bracket

T Tq

R

length you have to conduct

conductivity area available for conductionpw

LR

k A