AME$60634$$ Int.$HeatTrans.$ 1-D Steady Conduction: Plane...

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AME 60634 Int. Heat Trans. D. B. Go 1 1-D Steady Conduction: Plane Wall Governing Equation: Dirichlet Boundary Conditions: T (0) = T s,1 ; T ( L ) = T s,2 d 2 T dx 2 = 0 q x = kA dT dx = kA L T s ,1 T s ,2 ( ) Solution: Heat Flux: Heat Flow: temperature is not a function of k T ( x ) = T s ,1 + T s ,2 T s ,1 ( ) x L q x = k dT dx = k L T s ,1 T s ,2 ( ) Notes: A is the cross-sectional area of the wall perpendicular to the heat flow both heat flux and heat flow are uniform independent of position (x) • temperature distribution is governed by boundary conditions and length of domain independent of thermal conductivity (k) heat flux/flow are a function of k

Transcript of AME$60634$$ Int.$HeatTrans.$ 1-D Steady Conduction: Plane...

Page 1: AME$60634$$ Int.$HeatTrans.$ 1-D Steady Conduction: Plane Wallsst/teaching/AME60634/lectures/AME... · 2018-05-23 · • both heat flux and heat flow are uniform ! independent of

AME  60634    Int.  Heat  Trans.  

D.  B.  Go    1    

1-D Steady Conduction: Plane Wall Governing Equation:

Dirichlet Boundary Conditions: T (0) = Ts,1 ; T (L) = Ts,2

d 2Tdx2

= 0

qx = −kA dTdx

=kAL

Ts,1 −Ts,2( )

Solution:

Heat Flux:

Heat Flow:

temperature is not a function of k

T(x) = Ts,1 + Ts,2 −Ts,1( ) xL

ʹ′ ʹ′ q x = −k dTdx

=kL

Ts,1 −Ts,2( )

Notes: •  A is the cross-sectional area of the wall perpendicular to the heat flow •  both heat flux and heat flow are uniform è independent of position (x) •  temperature distribution is governed by boundary conditions and length of domain è independent of thermal conductivity (k)

heat flux/flow are a function of k

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AME  60634    Int.  Heat  Trans.  

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1-D Steady Conduction: Cylinder Wall Governing Equation:

Dirichlet Boundary Conditions:

T(r1) =Ts,1 ; T(r2) =Ts,2

Notes: •  heat flux is not uniform è function of position (r) •  both heat flow and heat flow per unit length are uniform è independent of position (r)

Solution:

Heat Flux:

Heat Flow: €

T(r) =Ts,1 −Ts,2ln r1 r2( )

ln rr2

⎝ ⎜

⎠ ⎟ + Ts,2

ʹ′ ʹ′ q r = −k dTdr

=k Ts,1 −Ts,2( )r ln r2 r1( )

qr = −kA dTdr

= 2πrL ʹ′ ʹ′ q r =2πLk Ts,1 −Ts,2( )ln r2 r1( )

ʹ′ q r =qr

L=2πk Ts,1 −Ts,2( )ln r2 r1( ) heat flow per unit length

heat flux is non-uniform

heat flow is uniform

1rddr

kr dTdr

!

"#

$

%&= 0⇒

ddr

r dTdr

!

"#

$

%&= 0

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AME  60634    Int.  Heat  Trans.  

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1-D Steady Conduction: Spherical Shell Governing Equation:

Dirichlet Boundary Conditions:

T(r1) =Ts,1 ; T(r2) =Ts,2

Solution:

Heat Flux:

Heat Flow:

T(r) = Ts,1 − Ts,1 −Ts,2( )1− r1 r( )1− r1 r2( )

ʹ′ ʹ′ q r = −k dTdr

=k Ts,1 −Ts,2( )

r2 1 r1( )− 1 r2( )[ ]

qr = −kA dTdr

= 4πr2 ʹ′ ʹ′ q r =4πk Ts,1 −Ts,2( )1 r1( )− 1 r2( )

Notes: •  heat flux is not uniform è function of position (r) •  heat flow is uniform è independent of position (r)

heat flux is non-uniform

heat flow is uniform

1r2

ddr

kr2 dTdr

!

"#

$

%&= 0⇒

ddr

r2 dTdr

!

"#

$

%&= 0

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AME  60634    Int.  Heat  Trans.  

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Thermal Resistance

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AME  60634    Int.  Heat  Trans.  

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Thermal Circuits: Composite Plane Wall

Circuits based on assumption of (a)  isothermal surfaces normal to x direction

or (b) adiabatic surfaces parallel to x direction

Rtot =LEkEA

+kFA2LF

+kGA2LG

⎣ ⎢

⎦ ⎥

−1

+LHkHA

Rtot =

2LEkEA

+2LFkFA

+2LHkHA

⎝ ⎜

⎠ ⎟

−1

+2LEkEA

+2LGkGA

+2LHkHA

⎝ ⎜

⎠ ⎟

−1⎡

⎣ ⎢ ⎢

⎦ ⎥ ⎥

−1

Actual solution for the heat rate q is bracketed by these two approximations

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Thermal Circuits: Contact Resistance

In the real world, two surfaces in contact do not transfer heat perfectly

ʹ′ ʹ′ R t,c =TA −TB

ʹ′ ʹ′ q x⇒ Rt ,c =

ʹ′ ʹ′ R t,cAc

Contact Resistance: values depend on materials (A and B), surface roughness, interstitial conditions, and contact pressure è typically calculated or looked up

Equivalent total thermal resistance:

Rtot =LA

kA Ac

+ʹ′ ʹ′ R t,c

Ac

+LB

kB Ac

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Fins: The Fin Equation •  Solutions

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Fins: Fin Performance Parameters •  Fin Efficiency

–  the ratio of actual amount of heat removed by a fin to the ideal amount of heat removed if the fin was an isothermal body at the base temperature

•  that is, the ratio the actual heat transfer from the fin to ideal heat transfer from the fin if the fin had no conduction resistance

•  Fin Effectiveness –  ratio of the fin heat transfer rate to the heat transfer rate that would exist

without the fin

•  Fin Resistance

–  defined using the temperature difference between the base and fluid as the driving potential

η f ≡qfqf ,max

=qf

hAfθb

ε f ≡qfqf ,max

=qf

hAc,bθb=Rt,b

Rt, f

Rt, f ≡θbqf

=1

hAfη f

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Fins: Efficiency

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Fins: Efficiency

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Fins: Arrays •  Arrays

–  total surface area

–  total heat rate

–  overall surface efficiency

–  overall surface resistance

At = NAf + Ab

qt = Nη f hAfθb + hAbθb =ηohAtθb =θbRt,o

Rt,o =θbqt

=1

hAtηo

N ≡ number of finsAb ≡ exposed base surface (prime surface)

ηo =1−NAf

At

1−η f( )

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Fins: Thermal Circuit •  Equivalent Thermal Circuit

•  Effect of Surface Contact Resistance

Rt,o =1

hAtηo(c )

ηo(c ) =1−NAf

At

1−η f

C1

$

% &

'

( )

C1 =1−η f hAf$ $ R t,c

Ac,b

%

& '

(

) *

qt =ηo(c )hAfθb =θbRt,o(c )

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Fins: Overview •  Fins

–  extended surfaces that enhance fluid heat transfer to/from a surface in large part by increasing the effective surface area of the body

–  combine conduction through the fin and convection to/from the fin

•  the conduction is assumed to be one-dimensional

•  Applications –  fins are often used to enhance convection when h is

small (a gas as the working fluid) –  fins can also be used to increase the surface area

for radiation –  radiators (cars), heat sinks (PCs), heat exchangers

(power plants), nature (stegosaurus)

Straight fins of (a) uniform and (b) non-uniform cross sections; (c) annular fin, and (d) pin fin of non-uniform cross section.

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Fins: The Fin Equation •  Solutions