Chapter 2b

MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould

Chapter 2 – Page 40

HEAT TRANSFER FUNDAMENTALS

q xq x+dx

q conv

Ldx

t

x

Tbbasetemp. cross-sectional

area, A

perimeter, Ph , T

c

II-7. Finned or Extended Surfaces (see text, Sec. 3.6)The heat conducted through a wall must usually be removed by convection.

Recall: , , ΔThq conv surfacewettedA

to increase cooling for fixed T, consider an extended surface.

T ,h

With no fins

conv

wet

surface wetted 1 RTT

hA

TTTThAq bbbconv

w




Increasing convection surface area increases q, but increasing surface area meansincreasing L (length of fin) which increases (i.e., ).

ckALR cond basetip TT

Only if is the addition of fins useful (i.e., internal resistance << external resistance).

convcond RR

The Biot number is a non-dimensional number which is a ratio of the conduction resistance (internal) to the convection resistance.

convth,

condth,

solidofeconductancconvection

RR

kh

khBi

ss

where is a characteristic length, .areawetted

volume

For rectangular fin (will show later)2t

If Bi < 0.1, then fins will enhance the heat transfer (i.e., k ~ large, h ~ small)




1-D Rectangular or Pin Fin (uniform cross-sectional area)

1. Geometry & assumptions

-steady state

-1-D, T(x) only

-no source,

-k is constant

2. Identify control volume Acdx

3. Perform conservation of energy

q 0

dxx

q xq x+dx

q conv


Chapter 2 – 3


Conservation of energy for differential control volume gives

dtdUEqq goutin

0convectionby

lostheat

dxxconducted

heat

xatconducted

heat =

- +

-

dxdTkAq cx

dxdxdTkA

dxd

dxdTkAdx

dxqdqq cc

xxdxx

TxThPdxdqconv




Dividing by dx

or [Eq. 3-67]

Defining (convenience)

02

2 TxThP

dxTdkAc

02

2 TxT

kAhP

dxTd

c

ckAhPm 2

022

2 TxTm

dxxTd

Define the temperature difference as TxTx

Note: dx

xdTdx

xd

and 2

2

2

2

dxTd

dxd

022

2 xm

dxxd

[Eq. 3-69]




The general solution to this linear 2nd order homogeneous ordinary differential equation is

mxmx eCeCx 21 [Eq. 3.71]

To find the temperature distribution in the fin, T(x), and qconv(x) or qbase we need only to apply the two boundary conditions.

Fin Case D (Long fin ) TLT

Governing equation in x > 0

Base B.C. at x = 0

Tip B.C. or as x

022

2 m

dxd

bb TT 0

0x TxT

From tip B.C., C1 must equal 0 so that (x) is bounded as x

mxeCx 2




From base B.C. bC 2 mxbex

or xmx

bb

ckAhP

eeTTTxTx

[Eq. 3.84]

The heat transfer rate can be found by using Newton’s law of cooling

bccbb

b

Lx

xmx

b

L

x

mxb

L

xf

PhkAkAhPPh

mhP

mhP

emhPdxehPdxxhPq

22

000

0

bcf PhkAq [Eq. 3-85]

or using Fourier’s law at the base of the fin, we get the same result.

cbbx

mxbc

xcf PhkAAkmemkA

dxxdkAq

00




Fin Case B (insulated at tip)

022

2 m

dxd 0 x Lin

B.C. at base at x = 0 bb TT 0

B.C. at tip at x = L

General Solution

0dxd

mxmx eCeCx 21

Note: , , 2

coshmxmx eemx

2sinh

mxmx eemx

mxmxmx

coshsinhtanh

Therefore, the general solution can also be written as

mxmx

mxmxmxmx

eCCeCC

eCeCeCeC

mxCmxCx

22

2222

sinhcosh

4343

4433

43




243

2CCC

2

431

CCC

Note: Constants can be related. If we find the solution in one form we could convert it to the other!

The solution to fin case B (insulated tip) is

mL

xLmTTTxTx

bb coshcosh

[Eq. 3-80]

The total heat rate is [Eq. 3-81] mLPhkAq cbf tanh




Fin Case A (Convection at Tip)

0222

2 m

dxd in 0 x L

B.C. at base at x = 0 bb TT 0

B.C. at tip at x = L 0 xhdx

xdk

TThdxdTk Lx

Lx

dxd

dxdT

LxLx TT

0

LxLx

hdxdk at x = L

General solution xLmCxLmCx sinhcosh 65




The solution to fin case A is

mLmkhmL

xLmmkhxLm

TTTxTx

bb sinhcosh

sinhcosh

[Eq. 3.75]

[Eq. 3-77]

The total heat rate

mL

mkhmL

mLmkhmL

PhkAq cbfsinhcosh

coshsinh




When should fins be added to surface?

L

t

x w

wtwtwP 122

For thin fin (t << w):and wP 2 wtAc

Calculation of characteristic length for thin fin

22volume

surfacewetted

tw

wtPLLwt

A




Consider result for convective heat transfer at tip case

mL

mkhmL

mLmkhmL

PhkAq cbfsinhcosh

coshsinh [Eq. 3-77]

Note that for thin rectangular cross-section fin!kth

kAPhm

c

2

To find effect of fin length, L, on the heat transfer find from (3-77).

For all values of L > 0 the results show:

gives adding length produces insulating effect!

gives adding length produces no cooling effect!

gives adding length produces a cooling effect and thus there is an advantage to adding a fin.

12

k

ht

12

k

ht

12

k

ht

0dL

dq f

0dL

dq f

0dL

dq f

dLdq f




Thus for economic reasons, it is best to leave the primary surface (i.e. base area) free of fins unless: 1

2

kht

or introducing the Biot number: 1.0~

khBi

Add fins if: k large or h small

• For forced convection where the fluid is liquid or for condensing vapors don't add fins

1

2CmW000,101000~ 2 k

hth

• For surfaces exchanging heat with a low speed gas, , add fins.Cm

Wh

2205~

For these reasons fins are found on air-cooled engines, boiler air preheaters and economizers, air conditioner evaporators & condensers, auto radiators, etc., but not for water to water heat exchangers, condensers, etc.




Fin Effectiveness, f

bbc

f

ow

ff hA

qqq

,/finwithout transferheatfinthroughtranferheatactual

For long rectangular fin – case D (simple), 212121

Bi12

htk

hAkP

cf

In general, fins should not be used unless εƒ ≥ 2 - 3. This requires that Bi ≤ 0.1 - 0.25.

Fin Efficiency, fmax

bTtempbasefinatwerefinentireiffinthroughtransferheatidealfinthroughtranferheatactual

qq f

f

.

where, bfbf hATThAq max

and wetted fin surface areafA

[Eq. 3-87]

Fin Performance Parameters




For a rectangular cross-section fin with an insulated end

mLPhkAq cbf tanh

PLAf

bPLhq max

mL

mLPhL

mLkAPLh

mLPhkA c

b

cbf

tanhtanhtanh

[Eq. 3.92]

[Eq. 3.81]




Knowing the fin efficiency allows one to calculate qf as bffff hAqq max

This relationship is plotted in Figures 3.19 and 3.20 on p. 166-167.

f

Lc3/2(h/kAp)1/2

Lc and Ap are geometry dependent and are defined in Figures 3.19 and 3.20.

In addition Fig. 3.20 for circumferential fins requires evaluation of r2cProcedure for use of Figures 3.19 and 3.20

1. Calculate Lc3/2(h/kAp)1/2 (and r2c/r1 for circumferential fins)

2. Read fin efficiency, f, from appropriate figure 3. Calculate qf using

bfff hAq




Figure 3.19




Corrected length and the insulated tip approximation

Consider a rectangular profile fin with convection at tip

The heat transfer from this fin can be approximated to good accuracy by “unwrapping the tip area” and using the much simpler insulated tip fin formula. A corrected length accounts for the heat transfer from the tip of the actual fin.

Insulated tip

L t/2Lc




Recall for tip convection:

mL

mkhmL

mLmkhmL

PhkAq cbfsinhcosh

coshsinh [Eq. 3-77]

We can use the simpler insulated tip formula if we correct the length:

ccbf mLPhkAq tanh [Eq. 3-93]




q f

q b

Fin array analysis

ufft qqq For a fin lattice (array)

If there is thermal contact resistance between the base and the fin, draw the thermal circuit and apply the proper rate equations written in terms of their thermal resistances!

quf




Example Problem 2.5Given: An aluminum (k = 240 W/m-K) rectangular cross-section fin of length

1.5 cm, thickness 0.25 cm and .

Find: The fin efficiency, f, for a this fin.

h Wm C

280 2




Example Problem 2.5 (cont’d)




Example Problem 2.6

Given: Circular aluminum fins of constant rectangular profile are attached to a tube of outside diameter D = 5 cm. The fins have thickness t = 2 mm, height L = 15 mm, thermal conductivity k = 200 W/m-C, and spacing 8 mm. The tube surface is maintained at a uniform temperature Tb = 200 °C, and the fins dissipate heat by convection into the ambient air at T = 30 °C with heat transfer coefficient h = 50 W/m2-C.

r o

r i

L

t

s

Tb

Find: Net heat transfer per 1m length of tube.




Figure 3.20

Chapter 2b

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