Chapter 2b
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Transcript of 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
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 41
HEAT TRANSFER FUNDAMENTALS
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)
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 42
HEAT TRANSFER FUNDAMENTALS
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
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 43
HEAT TRANSFER FUNDAMENTALS
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
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 44
HEAT TRANSFER FUNDAMENTALS
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]
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 45
HEAT TRANSFER FUNDAMENTALS
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
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 46
HEAT TRANSFER FUNDAMENTALS
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
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 47
HEAT TRANSFER FUNDAMENTALS
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
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 48
HEAT TRANSFER FUNDAMENTALS
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
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 49
HEAT TRANSFER FUNDAMENTALS
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
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 50
HEAT TRANSFER FUNDAMENTALS
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
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 51
HEAT TRANSFER FUNDAMENTALS
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 52
HEAT TRANSFER FUNDAMENTALS
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
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 53
HEAT TRANSFER FUNDAMENTALS
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
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 54
HEAT TRANSFER FUNDAMENTALS
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.
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 55
HEAT TRANSFER FUNDAMENTALS
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
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 56
HEAT TRANSFER FUNDAMENTALS
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]
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 57
HEAT TRANSFER FUNDAMENTALS
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
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 58
HEAT TRANSFER FUNDAMENTALS
Figure 3.19
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 59
HEAT TRANSFER FUNDAMENTALS
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
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 60
HEAT TRANSFER FUNDAMENTALS
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]
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 61
HEAT TRANSFER FUNDAMENTALS
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
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 62
HEAT TRANSFER FUNDAMENTALS
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
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 63
HEAT TRANSFER FUNDAMENTALS
Example Problem 2.5 (cont’d)
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 64
HEAT TRANSFER FUNDAMENTALS
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.
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 65
HEAT TRANSFER FUNDAMENTALS
Example Problem 2.6 (cont’d)
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 66
HEAT TRANSFER FUNDAMENTALS
Example Problem 2.6 (cont’d)
MAE 310 course notes – Fall 2011, Copyrighted by R. D. Gould
Chapter 2 – Page 67
HEAT TRANSFER FUNDAMENTALS
Figure 3.20