extended surfaces - cld.pt filenature and rationale of extended surfaces • an extended surface...

Extended SurfacesExtended Surfaces

Nature and Rationale of Extended Surfaces• An extended surface (also know as a combined conduction-convection system

or a fin) is a solid within which heat transfer by conduction is assumed to be one dimensional, while heat is also transferred by convection (and/orradiation) from the surface in a direction transverse to that of conduction.

– Why is heat transfer by conduction in the x-direction not, in fact, one-dimensional?

– If heat is transferred from the surface to the fluid by convection, what surface condition is dictated by the conservation of energy requirement?

– What is the actual functional dependence of the temperature distribution inthe solid?

– If the temperature distribution is assumed to be one-dimensional, that is,T=T(x) , how should the value of T be interpreted for any x location?

– How does vary with x ?,cond xq

– When may the assumption of one-dimensional conduction be viewed as anexcellent approximation? The thin-fin approximation.

• Extended surfaces may exist in many situations but are commonly used asfins to enhance heat transfer by increasing the surface area available forconvection (and/or radiation). They are particularly beneficial when is small,as for a gas and natural convection.

h

• Some typical fin configurations:

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

The Fin Equation

22

2 0− =d

mdx

θ θ (3.64)

How is the fin equation derived?

( )2

2 0∞− − =c

d T hPT T

kAdx(3.62)( )

2

2 0∞− − =c

d T hPT T

kAdx(3.62)

• Assuming one-dimensional, steady-state conduction in an extended surfacesurface of constant conductivity and uniform cross-sectional area , with negligible generation and radiation , the fin equationis of the form:

( )k ( )cA

0q⎛ ⎞=⎜ ⎟⎝ ⎠

• ( )0radq′′ =

• Assuming one-dimensional, steady-state conduction in an extended surfacesurface of constant conductivity and uniform cross-sectional area , with negligible generation and radiation , the fin equationis of the form:

( )k ( )cA

0q⎛ ⎞=⎜ ⎟⎝ ⎠

• ( )0radq′′ =

or, with and the reduced temperature ,( )2 / cm hP kA≡ ∞≡ −T Tθor, with and the reduced temperature ,( )2 / cm hP kA≡ ∞≡ −T Tθ

• Solutions (Table 3.4):

Base (x = 0) condition( )0 b bT Tθ θ∞= − ≡

Tip ( x = L) conditions( )A. : Conv ect /i n |o x Lkd dx h Lθ θ=− =

B. : / |Adiabati 0c x Ld dxθ = =

( )Fixed temperC. : atu re LLθ θ=( )D. (Infinite fin >2.65 ): 0mL Lθ =

• Fin Heat Rate:

( )0|ff c x A s

dq kA h x dAdxθ

θ== − = ∫

Fin Performance Parameters• Fin Efficiency:

,max

f ff

f f b

q qq hA

ηθ

≡ = (3.86)

How is the efficiency affected by the thermal conductivity of the fin?Expressions for are provided in Table 3.5 for common geometries.fη

( )1/ 2222 / 2fA w L t⎡ ⎤= +⎣ ⎦

( )/ 2pA t L=( )( )

1

0

212f

I mLmL I mL

η =

• Fin Effectiveness:

Consider a triangular fin:

,

ff

c b b

qhA

εθ

≡

• Fin Resistance: with , and /f ch k A Pε ↑ ↓ ↑ ↓ (3.85)

,1b

t ff f f

Rq hAθ

η≡ = (3.92)

Fin Arrays• Representative arrays of

(a) rectangular and(b) annular fins.

– Total surface area:t f bA NA A= + (3.99)

Number of fins Area of exposed base (prime surface)

– Total heat rate:

,

bt f f b b b o t b

t oq N hA hA hA

Rθ

η θ θ η θ= + ≡ = (3.101)

– Overall surface efficiency and resistance:

,1b

t ot o t

Rq hAθ

η= =

(3.103)

( )1 1fo f

t

NAA

η η= − − (3.102)

• Equivalent Thermal Circuit :

• Effect of Surface Contact Resistance:

( )( ),

bt t bo c

t o c

q hARθ

η θ= =

( )1

1 1f fo c

t

NAA C

ηη

⎛ ⎞= − −⎜ ⎟

⎝ ⎠(3.105a)

( )1 , ,1 /f f t c c bC hA R Aη ′′= + (3.105b)

( )( )

,1

t o cto c

RhAη

= (3.104)

Problem 3.116: Assessment of cooling scheme for gas turbine blade.Determination of whether blade temperatures are lessthan the maximum allowable value (1050 °C) for prescribed operating conditions and evaluation of bladecooling rate.

Schematic:

Assumptions: (1) One-dimensional, steady-state conduction in blade, (2) Constant k, (3)Adiabatic blade tip, (4) Negligible radiation.

Analysis: Conditions in the blade are determined by Case B of Table 3.4.

(a) With the maximum temperature existing at x=L, Eq. 3.75 yields

( )b

T L T 1T T cosh mL

∞

∞

−=

−

( ) ( )1/ 21/ 2 2 4 2cm hP/kA 250W/m K 0.11m/20W/m K 6 10 m−= = ⋅ × ⋅ × × = 47.87 m-1 ( ) ( )1/ 21/ 2 2 4 2cm hP/kA 250W/m K 0.11m/20W/m K 6 10 m−= = ⋅ × ⋅ × × = 47.87 m-1

mL = 47.87 m-1 × 0.05 m = 2.39

From Table B.1, . Hence,coshmL=5.51From Table B.1, . Hence,coshmL=5.51

( ) 1200 300 1200 5 51 1037= + − =o o oT L C ( ) C/ . C

and, subject to the assumption of an adiabatic tip, the operating conditions are acceptable.

(b) With ( ) ( ) ( )1 / 22 4 21 / 2c bM hPkA 250W/m K 0.11m 20W/m K 6 10 m 900 C 517W−= Θ = ⋅ × × ⋅ × × − = −o ,

Eq. 3.76 and Table B.1 yield

( )fq M tanh mL 517W 0.983 508W= = − = −

Hence, b fq q 508W= − =

Comments: Radiation losses from the blade surface contribute to reducing the blade temperatures, but what is the effect of assuming an adiabatic tip condition? Calculatethe tip temperature allowing for convection from the gas.

Problem 3.132: Determination of maximum allowable power for a 20mm x 20mm electronic chip whose temperature is not to exceed

when the chip is attached to an air-cooled heat sink with N=11 fins of prescribed dimensions.

cq

85 C,cT = o

Problem 3.132: Determination of maximum allowable power for a 20mm x 20mm electronic chip whose temperature is not to exceed

when the chip is attached to an air-cooled heat sink with N=11 fins of prescribed dimensions.

cq

85 C,cT = o

Schematic:

Assumptions: (1) Steady-state, (2) One-dimensional heat transfer, (3) Isothermal chip, (4)Negligible heat transfer from top surface of chip, (5) Negligible temperature rise for air flow,(6) Uniform convection coefficient associated with air flow through channels and over outersurface of heat sink, (7) Negligible radiation.

Analysis: (a) From the thermal circuit,

c cc

tot t,c t,b t,o

T T T Tq

R R R R∞ ∞− −

= =+ +

( )2 6 2 2t,c t,cR R / W 2 10 m K / W / 0.02m 0.005 K / W−′′= = × ⋅ =

( )2t,b bR L / k W= ( )W / m K

20.003m /180 0.02m 0.042 K / W⋅= =

From Eqs. (3.103), (3.102), and (3.99) ( )ft,o o f t f b

o t t

N A1R , 1 1 , A N A Ah A A

= = − − = +η ηη

Af = 2WLf = 2 × 0.02m × 0.015m = 6 × 10-4 m2 Ab = W2 – N(tW) = (0.02m)2 – 11(0.182 × 10-3 m × 0.02m) = 3.6 × 10-4 m2 At = 6.96 × 10-3 m2

With mLf = (2h/kt)1/2 Lf = (200 W/m2⋅K/180 W/m⋅K × 0.182 × 10-3m)1/2 (0.015m) = 1.17, tanh mLf = 0.824 and Eq. (3.87) yields

ff

f

tanh mL 0.824 0.704mL 1.17

= = =η

ηo = 0.719,

Rt,o = 2.00 K/W, and

( )( )c

85 20 Cq 31.8 W

0.005 0.042 2.00 K / W− °

= =+ +

Comments: The heat sink significantly increases the allowable heat dissipation. If it were not used and heat was simply transferred by convection from the surface of the chip with

from Part (a) would be replaced by 2100 W/m , 2.05 K/Wtoth K R= =21/ hW 25 K/W, yielding 2.60 W.cnv cR q= = =