7/29/2019 Heat Transfet Basics
1/15
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
7/29/2019 Heat Transfet Basics
2/15
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
7/29/2019 Heat Transfet Basics
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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:
7/29/2019 Heat Transfet Basics
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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
7/29/2019 Heat Transfet Basics
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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
7/29/2019 Heat Transfet Basics
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:
7/29/2019 Heat Transfet Basics
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
7/29/2019 Heat Transfet Basics
10/15ME601 1/18/06Slide 10
Heater in a Bathroom Floor
7/29/2019 Heat Transfet Basics
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
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