Transcript ES 403 Heat Transfer - Embry–Riddle Aeronautical University

Heat Conduction
• In the intro to Heat Transfer, the conductive heat
transfer was presented as:
T
q x  kA
x
• However, this is true only in one dimension. In
general, heat flux is a vector having 3 components.
• A more general equation for heat transfer is in terms
of the gradient of temperature. In Cartesian
coordinates:
q  kAT  i qx  jqy  kqz
T
q x  kA
x
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T
q y  kA
y
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T
q z   kA
z
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Heat Conduction (cont.)
• Since we will be dealing with other coordinates also,
realize that the gradient takes on other forms:
z
Cartesian
T
T
T
T  i
j
k
x
y
z
Cylindrical
T
1 T
T
T  i
j
k
r
r 
z
Spherical
j
x
z
r


T
1 T
1 T
T  i
j
k
r
r 
r sin  
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k
y
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r

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i
k
i
k
j
i
j
Heat Diffusion Equation
• To develop a generalized governing law for heat
conduction, we begin by consider the fluxes in an
element as shown:
qz
qz  dz
z
qy  dy
dz
z
y
x
dx
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y
qx
qx  dx
x
qy
qx
qy
dy
qz
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Heat Diffusion Equation (cont.)
• A statement of energy conservation for this control
mass (similar to the 1st Law) would be:
Rat e of change  Rat e of flux  Rat e of energy
of energyin t he  of energyinto  productionin 

 
 

element
  theelement   theelement 
• Lets consider this statement term by term. First, the
rate of change of energy:
Rate of change 
of energyin the  c T dxdydz
p


t
element

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Heat Diffusion Equation (cont.)
• Note that this equation uses Q  mcp dT and thus
assumes constant pressure processes.
• Next consider the heat generation term:
Rate of energy
productionin   qdxdydz


 theelement 
q  heat generationper unit volume
(W/m3 )
• Sources of heat generation will be: electrical
resistance, chemical reactions, nuclear reactions, or
even radiation absorption like in a microwave.
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Heat Diffusion Equation (cont.)
• Lastly, consider the heat flux through the element:
Rate of flux 
of energyinto  Heat influx  Heat outflux


 theelement 
 qxdydz  qydxdz  qzdydx 




qy 

qx 
qz 

dxdydz   qy 
dy dxdz   qz 
dz dxdy
 qx 
x 
y
z 





 qx qy qz 
dxdydz
 


y
z 
 x
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Heat Diffusion Equation (cont.)
• However, using the definition of qx, qy, and qz this can
be rewritten as:
qx qy qz  
T   
T   
T 
    k


  k
    k

x
y
z x 
x  y 
y  z 
z 
    k  T 
• Finally, putting all these together, we get the heat
diffusion equation:
T
c p
   k  T   q
t
• Note that this is a 2nd order differential equation!
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Heat Diffusion Equation (cont.)
• One common simplification of this equation is to
assume that the material conductivity, k, is
independent of position.This is valid if:
• objects are made of a single material
• k is not a strong function of T or T is nearly constant
• In this case, the diffusion equation can be written as:
1 T
q
2
 T 
 t
k
– where
k

 thermaldiffusivity (m2 /sec)
c p
2      Laplacedifferential operation
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Heat Diffusion Equation (cont.)
• In the three coordinate systems of interest, the
Laplace operator takes the form:
Cartesian
 2T  2T  2T
T 2  2  2
x
y
z
2
Cylindrical
 2T 1 T 1  2T  2T
T 2 
 2
 2
2
r
r r r 
z
2
Spherical
1 2
1
 
T 
1
 2T
rT   2
T
 sin 
 2 2
2
r r
r sin   
  r sin   2
2
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Heat Transfer Boundary Conditions
• Since heat conduction problems involve differential
equations, we need to consider the different types of
boundary conditions.
• In general, BC’s will be one of 4 types:
– Fixed temperature:
Ts
T(x,t)
• T(xw,t) = Ts
xw
– Fixed heat flux
q”s
• -k(dT/dx)xw=q”s
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T(x,t)
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Heat Transfer BC’s (cont)
– Adiabatic Wall
q”s=0
• -k(dT/dx)xw=0
T(x,t)
– Convectively cooled
xw
Ts
• -k(dT/dx)xw=h(T -Ts)
T
T(x,t)
• Those of you who remember your Diff. Eq. Will
recognize the first BC as a Dirichlet type, the next
two as Neumann type, and the last as being mixed!
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1-D, Steady State Heat Transfer
• In this chapter we will consider steady state heat
transfer in one dimension.
• As a result we can set the left hand side of our
governing equation to zero:
0
T
c p
   k  T   q  0
t
• The direction of heat flux will either be axial as in
either the X or Z directions, or will be radial in the R
direction.
• We will also consider situations which are really
multidimensional, but which can be well modeled
with 1-D approximations.
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The Planar Wall
• The first case of interest is the planar wall made of a
single material as sketched.
• We will not consider heat generation at first, so q  0
and the governing equation is just:
  kT  
d  dT 
k
0
dx  dx 
• Another way to state this equation is:
dT
qx  k
 constant
dx
• Or
TS1
k
A
TS2
q
kA
TS1  TS 2 
qx 
L
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T
L
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x
The Planar Wall (cont)
• A more complicated version of this problem is to
combine multiple walls of different materials.
• For each wall section we have:
kB A
kA A
TS1  T2  qB 
T2  T3 
qA 
LA
LB
kC A
T3  TS 4 
qC 
LC
• But, it must also be true that
qA  qB  qC  qx
kA
kB
LA
LB
TS1
30
LC
T
T2
since energy is not stored inside
the wall.
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kC
T3
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TS4
x
The Planar Wall (cont)
• This give three equations, and 3 unknowns (qx, T2,
T3) if the two wall temperatures are given (TS1,TS2).
• Solving for the qx yields:

TS1  TS 4 
qx 
 LA   LB   LC 


  
  
 k A A   k B A   kC A 
• This result can be generalized to any multilayer
composite wall by introducing the concept of
thermal resistance, Rt.
• Thermal resistance plays the role in heat conduction
as electrical resistance does in electrical conduction.
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The Planar Wall (cont)
• Using an analogy to flow of electrical current, if the
temperature change is the potential difference and q
the flux rate, then:
V
T
L
R
 Rt 

 thermalresistance
i
qx kA
• The total resistance offered by this composite wall is
similar to resistors in series:
T
T T
qx 


Rt , A  Rt , B  Rt ,C Rt Rtot
TS1
Rt , A
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TS 4
T3
T2
Rt , B
Rt ,C
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The Planar Wall (cont)
• This electrical analogy can be extended to other
situations such as the wall shown.
• This case resembles having resistors in series and
parallel:
kB
Rt , B
TS1
Rt , A
TS 2
kA
Rt , D
LA
Rt ,C
• The resistance in this case is give by:
Rt , B Rt ,C
Rtot  Rt , A 
 Rt , D
Rt , B  Rt ,C
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LB
kc
Lc
kD
LD
The Planar Wall (cont)
• And the net heat flux through the wall is just
qx 
T
Rtot
T
qx 
ARtot
or
• A common way of expressing the effect of a
composite wall is to use the Overall Heat Transfer
Coefficient, U, defined by:
qx  UAT
qx  UT
or
• By comparison with the previous equation, it is
obvious that :
1
U
ARtot
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The Planar Wall (cont)
• Lastly, don’t confuse the thermal resistance, Rt, with
the “R” factor quoted for many common construction
materials.
• The “R” factor used in the construction industry is
defined by:
L
" R" 
k
where L is the insulation thickness in inches and k
the material conductivity in BTU inches/(hr ft2 oF).
• Thus the “R” factor is the same as 1/U, but with
British units.
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Contact Resistance
• One assumption that has been made in the previous
development is that heat flow easily between the
different layers of material.
• In fact, when two solids are in
contact, their surface roughness
results in voids separating the two
materials.
TA
• How easily heat flows between the
two materials then depends upon the
amount of direct contact between
them and the fluid filling the gaps.
Voids
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TB
Contact Resistance (cont)
• To account for this effect, the Contact Resistance is
defined by:
1 TA  TB
Rt,c  
qx  hc TA  TB 
or
hc
qx
• Unfortunately, it is very difficult to accurately predict
the contact resistance and experimentation is usually
necessary.
• Factors which effect the how much contact resistance
there is are surface preparation, contact pressure and
the void fluid.
• Two polished surfaces will make better contact, thus
reducing resistance from the voids.
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Contact Resistance (cont)
• Joining two materials by force increases the contact
area through the surface by the deformation of the
faces under load - thus reducing resistance.
• Filling the voids with more conductive fluids such as
oil or thermal greases reduces the resistance of the
voids themselves.
• Finally, placing a thin foil of a highly ductile and
conductive material such as lead or indium between
two surfaces before joining them with force acts to
effectively fill the voids.
• See the book for typical thermal contact resistance
values.
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Radial Flow - Cylinders
• The next case of interest is the radial flow of heat in
a cylindrical system. The best instance of this case is
heat loss from a pipe.
• Without heat generation and with k=constant, the
governing eqn. is:
d 2T k dT
  k T   k 2 
0
dr
r dr
• Another way to state this equation is
to multiply by r and combine terms:
d 2T
dT d  dT 
rk 2  k
  rk
0
dr
dr dr  dr 
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Radial Flow - Cylinders (cont)
• This result is consistent with applying Fourier’s Law
to the case but letting the area be a function of r:
dT
dT
qr  kA
 k 2 rL
 constant
dr
dr
• By separating the variables and integrating:
Ts 2
r2
dT
 qr
Ts1 dr dr  r1 2 rLk dr
r2
 r2 
 qr
TS 2  TS1 
ln 
r1
2 Lk  r1 
• Or:
TS2

TS1  TS 2 
qr  2 Lk
lnr2 / r1 
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L
TS1
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Radial Flow - Cylinders (cont)
• Or using the concept of thermal resistance:
lnr2 / r1 
T  T 
Rt 
qr  S 1 S 2
2 Lk
Rt

• Now think of the consequences of these
results.
TS1
– First, the temperature does not vary linearly,
rather it varies with the natural logarithm of
r.
TS2
r1
– Second, while the heat rate is constant, the
heat flux is not, but varies as 1/r.
qr  constant
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qr 
qr constant

A
2 rL
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r2
Radial Flow - Cylinders (cont)
• For composite radial layers, the situation is analogous
to the composite planar wall.
• Thus, for a pipe made of steel coated with a layer of
fiberglass and then plaster:

TS1  TS 2 
qr 
r
Rtot
r2 3
r4
r
1
lnr3 / r2 
lnr4 / r3 
lnr2 / r1 
Rt 


2 Lksteel 2 Lk fiberglass 2 Lk plaster
TS1
• Also, contact resistance will occur
in radial systems as in planar walls.
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Radial Flow - Spheres
• Radial heat flow in spheres is qualitatively similar to
that in cylinders, but the flux area varies as r2.
• Thus, Fourier’s Law is:
dT
dT
qr  kA
 k 4 r 2
 constant
dr
dr
• And after integrating you get:

TS1  TS 2 
qr  4 k
1 / r1   1 / r2 
Rt 
1 / r1   1 / r2 
4 k
• Composite layers are handled as in the previous two
cases.
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Convection Boundaries
• All the cases considered thus far have assumed the
surface temperatures are known.
• In general, however, the temperature of the
surrounding fluid is knows, and hopefully the
convective heat coefficient, h.
• Using the concept of thermal resistance to Newton’s
Law of convective cooling shows that:
h
T  T 
T
qx,conv  hATS  T   S 
TS
Rt ,conv
Rt ,conv 
1
hA
TS
T
Rt ,conv
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Convection Boundaries (cont)
• Thus, for cooling on a planar wall:
h1
T
 T , 2 
qx 
1
L
1


h1 A kA h2 A
 ,1
h2
k
T,1
T,2
L
• While for a pipe:
r2
T
 T , 2 
qr 
1
lnr2 / r1 
1


h1 2r1 L
2Lk
h2 2r2 L
 ,1
h1
r1
T,1
k
h2
T,2
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Critical Radius of Insulation
• An unusual situation arises when applying insulation
to a pipe of small radius.
• If only the insulation layer and the convective
process on the outer surface are considered, the
thermal resistance is:
r2
lnr2 / r1 
1
Rtot  Rt ,cond  Rt ,conv 

2Lk
hL2r2
r1
• The conductive term above increases
with r2, while the convective term
decreases with r2.
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h2
k
Critical Radius of Insulation (cont)
• Thus, a plot of Rtot versus r2 might look like:
Rtot
Rt,cond
Rt,conv
r1
rcr
r2
• The radius at which Rtot reaches a minimum is called
the critical radius of insulation, rcr.
• Below this radius, the addition of insulation is counter
productive since the increase in the outer surface
area (where convection occurs) is more significant
than the insulating properties of the material.
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Critical Radius of Insulation (cont)
• To find the rcr, we find the minimum of this curve:
dRtot
d  lnr2 / r1 
1 
1
1






0
2


dr2
dr2  2Lk
hL2r2  2Lkr2 hL2r2
• Or simply:
k
r2  rcr 
h
• Two notes:
– First, for the insulation to be effective, the outer radius
might have to be much greater than rcr. As a result,
thin pipes are often left un-insulated!
– Second, for some cases rcr < r1, i.e. any insulation
thickness is effective.
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Heat Generation - Plane Walls
• Let’s revisit the plane wall case, but this time include
heat generation, q .
• The governing equation in this case is:
d 2T q
 0
2
dx
k
• Integrating this equation twice yields a quadratic
equation in x with two unknown constants of
integration:
q 2
T ( x)  
x  C1 x  C2
2k
• The constants must be determined by applying the
appropriate boundary conditions.
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Heat Generation - Plane Walls (cont)
• Let’s consider the simplest case possible, that where
the temperature on both walls are equal at Ts.
• Symmetry dictates that C1=0 and to have
T=Ts at x=L’ (half thickness):
q 2
Ts
C2 
L '  Ts  T0
2k
• Note that C2 is also equal to the
temperature at the center of the wall, T0.
To
T
-L’
• Thus the temperature variation in the wall
is given by:
2
T
(
x
)

T
x
 
q
0
2
2
or

T ( x) 
L '  x   Ts
 

T

T
 L'
2k
s
0
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Ts
L’
x
Heat Generation - Plane Walls (cont)
• In the more general case where the two surface
temperatures are at different values, Ts,1 and Ts,2, the
solution is:
qL '2 
x 2  Ts ,2  Ts ,1 x Ts ,1  Ts ,2
T ( x) 

1  2  
2k  L ' 
2
L'
2
• Which looks like:
Ts2
q  0
q  0 q  0
Ts1
-L’
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T(x)
L’
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Heat Generation - Radial Systems
• Radial systems are very similar. The governing
equation is:
Ts
1 d  dT  q
r
  0
r dr  dr  k
To
• Which after integrating twice yields:
q 2
T (r )   r  C1r  C2
4k
• Once again, symmetry requires C1 = 0 and
C2 must equal centerline temperature and:
q 2
C2 
R  Ts  T0
4k
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R
Heat Generation - Radial Systems (cont)
• Substitution then yields the result that:
qR 2  r 2 
1  2   Ts
T (r ) 
4k  R 
T ( r )  Ts 
r2 
 1  2 
T0  Ts
 R 
or
• In the more general case shown below, the solution
is:
q  2 2
2
2 ln r / r2  
Ts,2
r2
T (r ) 
 r2  r  r2  r1

4k 
lnr1 / r2 

 

lnr / r2 
 T1  T2 
 T2
lnr1 / r2 
r1
Ts,1
Ts,1
• Care to prove it?
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Heat Generation with Convection
• Now let’s add surface convection into these two
problems.
• First, we would like to find the total heat flux out of
the surface. From our previous results:
Plane Wall
dT
qs  kA
dx
Radial System
xL
 qL ' 
qs  kA 

 k 
qs  qAL '  qV 
V   1/2 totalvolume
ES 403 Heat Transfer
qs  kA
dT
dr
r R
  qR 
qs  k 2 RL

 2k 
qs  q R2 L  qV
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Heat Generation with Convection (cont)
• In retrospect, this result makes sense - the heat flux
out must equal the heat generated within!
• However, this heat to the wall must be carried away
by convection (or radiation, possibly).
• Thus:
Plane Wall
qAL  hATs  T 
Radial System
q R2 L  h2 RLTs  T 
q R
Ts 
 T
2h
qL '
Ts 
 T
h
• Notice how these results are independent of the
internal temperature profile!
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Extended Surfaces
• From considering the critical radius of insulation, we
saw the importance of surface area when determine
the total heat transfer in a combined conduction/
convection system.
• In many surfaces the rate of heat transfer can be
enhanced by increasing surface area through the use
of extended surfaces or “fins”
• Consider the case shown below:
T
T
Ts
ES 403 Heat Transfer
h
A
Ts
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h
A
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Extended Surfaces (cont)
• The addition of fins can greatly increase the total
convective surface area.
• However, realize that this added surface area is not
as effective as the original since T decreases from the
base temp, Tb, along the length of the fin!
• Since the heat loss from the fin
is proportional to the difference
(T(x) - T), the base of the fin is
more effective than the root.
• To calculate the net heat flux,
we need to find T as a function
of x.
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Tb
T(x)
T
x
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Extended Surfaces (cont)
• This problem is at least 2 dimensional. However a
very good approximation involves considering heat
flux in only one dimension - base to tip.
• Heat flux normal to this direction can be modeled as
heat loss (negative q ) which is due to convection.
• To see this, consider the following fin geometry:
q (x)
• Let Ac be the fin tip area and
P the length of the perimeter
of this area.
• Then qdV  hdA(T ( x)  T )
or
dx
P
qAc dx  hPdx(T ( x)  T )
ES 403 Heat Transfer
Ac
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Extended Surfaces (cont)
• The governing equation for this quasi 1-dimensional
problem is then:
d 2T q d 2T hP
T ( x)  T   0
  2 
2
dx
k dx
kAc
• This problem can be notationally simplified by letting:
hP
2
and
 ( x)  (T ( x)  T )
m 
kAc
• To get:
d 2
2

m
 ( x)  0
2
dx
which has the general solution:
 ( x)  Aemx  Bemx
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Extended Surfaces (cont)
• To solve for the two constants A and B, we need two
boundary conditions.
• At the base, Tx=0 = T0, or
 x0  0  (T0  T )
• At the other end, the tip, the best B.C. would include
the effect of tip heat convection, or:
qconduction xL  qconvection xL   kA dT
 hAc Tx L  T 
c
dx x L
• In practice, it is more reasonable that no heat is lost
through the tip both because Ac is small and the
temperature there is close to T. I.e.
dT
d

0
dx x L dx x L
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Extended Surfaces (cont)
• With this “insulated end” B.C., the solution for A and
B yields:

emx
e mx
coshmL  x 



2 mL
 2 mL
0 1  e
1 e
coshmL
• Of course, the temperature distribution is secondary.
What we really want is the fin heat loss rate:
L
L
0
0
q f   hPT  T dx   hPdx 
or
hP 0
tanh(m L)
m


hP

q f  hPkAc T0  T  tanh L
kAc 

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Fin Effectiveness and Efficiency
• When designing fins, or deciding what fin to use, it is
useful to have some parameters for comparison.
• For fins, there are two parameters commonly used:
effectiveness and efficiency.
– Fin effectiveness, ε f : the ratio of the fin heat transfer rate
to the heat transfer rate that would exist without the fin.
– Fin effiiciency,  f : the ratio of the fin heat transfer rate to
the ideal case where T(x) = T0 all along the fin length.
• Fin effectiveness will tell us how much more effective
having the fin is over not having it.
• Fin efficiency will tell us whether it makes the
optimum use of material.
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Fin Effectiveness and Efficiency (cont)
• For the case just considered, the effectiveness would
be:
qf
kP
f 

tanhm L
hAc (T0  T )
hAc
• Note that for long fins (tanh(mL)1), then three
things improve fin effectiveness:
– A high material conductivity. Pretty obvious!
– A low fluid convectivity. This does not mean that you want h
low, just that fins are more commonly used when h is small
as in natural convection and/or gasses versus liquids.
– A large ratio of tip area to tip perimeter, Ac/P. Thus a wide,
thin fin is preferable. Another way to say this is that you
would rather have two thin fins then one twice as thick
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Fin Effectiveness and Efficiency (cont)
• The fin efficiency for this case is:
qf
tanhmL
f 

hPL(T0  T )
mL
• This result indicates two things.
• The highest efficiency, 1, is achieved for a short fin
since tanh(mL)mL as L is decreased.
• The fin efficiency decreases to a value of zero as the
length increases.
• This is because the region near the tip, where TT,
does not contribute as much as the base sections.
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Other Fin Shapes
• The book gives the results for span efficiency for
other type fins on pages 45 and 46.
• Note that the first case shown is the one we just
solved, but using a corrected length, Lc, not L.
• This slight change improves the comparison to
experiment since the tip is not really insulated!
• Also note that all these solutions just provide the find
efficiency and the fin wetted area, Af.
• To get fin effectiveness, use:
f 
Af
Ab
ES 403 Heat Transfer
f
Ab  fin base area
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Fin Thermal Resistance
• When dealing with complex thermal flow situations, it
is useful to fall back to the electrical analogy.
• For fins, the thermal efficiency is defined by:
Rf 
(T0  T )
qf
• However, also note that the fin effectiveness is
related to the fin thermal efficiency by:
Rb
1
f 
Rb 
Rf
hAb
• Thus use the chart to get efficiency, then from that
find the fin effectiveness and the Rf.
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Overall Surface Effectiveness
• The study of fins so far has concentrated on a single
fin. Now we want to extend this to multiple fins and
the base surface area between them.
• If we define the overall surface
efficiency by:
q
q
o  t  t
qmax hAt 0
qt  totalheat trans
fer At  totalarea
• With just a little bit or work, it
can then be shown that:
NA f
N  # of fins
o  1 
(1   f )
At
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t
T0
w
L
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S
Overall Surface Effectiveness (cont)
• Finally, we can find the overall thermal resistance
by:

1
Rt ,o  0 
qt hAt o
• With the equivalent thermal networks:
T0
NhA  
1
f
T
f
T0
T
hAto 1
h( A  NA )
1
t
f
• One thing to keep in mind - we have assumed h is a
constant. But if you space the fins too closely, h will
actually decrease due to blockage effects!
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