HEAT TRANSFER
UNIT – 2
FINS & TRANSIENT CONDUCTION
FINS: EXTENDED SURFACES
Heat transfer in extended surfaces of uniform cross-section without heat generation:
Convection: Heat transfer between a solid surface and a moving fluid is governed by the Newton’s cooling
law: q = hA(Ts-T∞), where Tsis the surface temperature and T∞ is the fluid temperature. Therefore, to
increase the convective heat transfer, one can
• Increase the temperature difference (Ts-T∞) between the surface and the fluid.
• Increase the convection coefficient h. This can be accomplished by increasing the fluid flow over
the surface since h is a function of the flow velocity and the higher the velocity, the higher the h.
Example: a cooling fan.
• Increase the contact surface area A. Example: a heat sink with fins.
Many times, when the first option is not in our control and the second option (i.e. increasing h) is already
stretched to its limit, we are left with the only alternative of increasing the effective surface area by using fins
or extended surfaces. Fins are protrusions from the base surface into the cooling fluid, so that the extra surface
of the protrusions is also in contact with the fluid. Most of you have encountered cooling fins on air- cooled
engines (motorcycles, portable generators, etc.), electronic equipment (CPUs), automobile radiators, air
conditioning equipment (condensers) and elsewhere.
The fin is situated on the surface of a hot surface at Ts and surrounded by a coolant at temperature T∞,
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which cools with convective coefficient, h. The fin has a cross sectional area, Ac, (This is the area through with
heat is conducted.) and an overall length, L.
Note that as energy is conducted down the length of the fin, some portion is lost, by convection, from the sides.
Thus the heat flow varies along the length of the fin.
We further note that the arrows indicating the direction of heat flow point in both the x and y directions. This is
an indication that this is truly a two- or three-dimensional heat flow, depending on the geometry of the fin.
However, quite often, it is convenient to analyse a fin by examining an equivalent one–dimensional system. The
equivalent system will involve the introduction of heat sinks (negative heat sources), which remove an amount of
energy
Equivalent to what would be lost through the sides by convection.
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Across this segment the heat loss will be h⋅ (P⋅ Δx)⋅ (T-T∞), where P is the perimeter around thefin. The equivalent
heat sink would be &q&&(A x)
Equating the heat source to the convective loss:
Substitute this value into the General Conduction Equation as simplified for One-Dimension, Steady State
Conduction with Sources:
which is the equation for a fin with a constant cross sectional area. This is the Second Order Differential Equation
that we will solve for each fin analysis. Prior to solving, a couple of simplifications should be noted. First, we
seethat h, P, k and Ac are all independent of x in the defined system (They may not be constant if a more general
analysis is desired.). We replace this ratio with a constant. Let
Next we notice that the equation is non-homogeneous (due to the T∞ term). Recall that non homogeneous
differential equations require both a general and a particular solution. We can make this equation homogeneous
by introducing the temperature relative to the surroundings:
Differentiating this equation we find:
Differentiate a second time:
Substitute into the Fin Equation:
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This equation is a Second Order, Homogeneous Differential Equation.
Solution of the Fin Equation:
We apply a standard technique for solving a second order homogeneous linear differential equation.
We now have two solutions to the equation. The general solution to the above differential equation will be a
linear combination of each of the independent solutions
Then:
θ = A⋅ em⋅ x + B⋅ e-m⋅ x.
where A and B are arbitrary constants which need to be determined from the boundary conditions. Note that it is
a 2nd order differential equation, and hence we need two boundary conditions to determine the two constants of
integration.
An alternative solution can be obtained as follows: Note that the hyperbolic sin, sinh, the hyperbolic cosine, cosh,
are defined as:
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Generally the exponential solution is used for very long fins, the hyperbolic solutions for other cases.
Boundary Conditions:
Since the solution results in 2 constants of integration we require 2 boundary conditions. The first one is obvious,
as one end of the fin will be attached to a hot surface and will come into thermal equilibrium with that surface.
Hence, at the fin base,
θ(0) = T0 - T∞≡θ0
The second boundary condition depends on the condition imposed at the other end of the fin.
There are various possibilities, as described below.
Very long fins:
For very long fins, the end located a long distance from the heat source will approach the temperature of the
surroundings. Hence,
θ(∞) = 0
Substitute the second condition into the exponential solution of the fin equation:
The first exponential term is infinite and the second is equal to zero. The only way that this equation can be valid is if
A = 0. Now apply the second boundary condition.
θ(0) = θ0 = B⋅e-m⋅0 ⇒B = θ0
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Often we wish to know the total heat flow through the fin, i.e. the heat flow entering at the base (x=0).
The insulated tip fin:
Assume that the tip is insulated and hence there is no heat transfer:
The solution to the fin equation is known to be:
Differentiate this expression.
Apply the first boundary condition at the base:
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So that D = θ0. Now apply the second boundary condition at the tip to find the value of C:
This requires that
We may find the heat flow at any value of x by differentiating the temperature profile and substituting it into the
Fourier Law:
So that the energy flowing through the base of the fin is:
If we compare this result with that for the very long fin, we see that the primary difference in form is in the
hyperbolic tangent term. That term, which always results in a number equal to or less than one, represents the
reduced heat loss due to the shortening of the fin.
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Fin Effectiveness:
How effective a fin can enhance heat transfer is characterized by the fin effectiveness, f ε,which is as the ratio of
fin heat transfer and the heat transfer without the fin. For an adiabaticfin:
If the fin is long enough, mL>2, tanh(mL)→1, and hence it can be considered as infinite fin( case D in table)Hence,
for long fins,
Fin Efficiency:
The fin efficiency is defined as the ratio of the energy transferred through a real fin to thattransferred through an
ideal fin. An ideal fin is thought to be one made of a perfect or infiniteconductor material. A perfect conductor has
an infinite thermal conductivity so that theentire fin is at the base material temperature.
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TRANSIENT CONDUCTION
Introduction:
To this point, we have considered conductive heat transfer problems in which the temperatures are
independent of time. In many applications, however, the temperatures are varying with time, and we
require the understanding of the complete time history of the temperature variation. For example, in
metallurgy, the heat treating process can be controlled to directly affect the characteristics of the processed
materials. Annealing (slow cool) can soften metals and improve ductility. On the other hand, quenching
(rapid cool) can harden the strain boundary and increase strength. In order to characterize this transient
behavior, the full unsteady equation is needed:
K
Where α = is the thermal diffusivity without any heat generation and considering spatial variation of
ρCP
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temperature only in x-direction, the above equation reduces to:
For the solution of equation (5.2), we need two boundary conditions in x-direction and one initial
condition. Boundary conditions, as the name implies, are frequently specified along the physical boundary
of an object; they can, however, also be internal – e.g. a known temperature gradient at an internal line of
symmetry.
Biot and Fourier numbers:
In some transient problems, the internal temperature gradients in the body may be quite small and
insignificant. Yet the temperature at a given location, or the average temperature of the object, may be
changing quite rapidly with time. From eq. (5.1) we can note that such could be the case for large thermal
diffusivity α.
For very large ri, the heat transfer rate by conduction through the cylinder wall isapproximately
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Whenever the Biot number is small, the internal temperature gradients are also small and a transient
problem can be treated by the “lumped thermal capacity” approach. The lumped capacity assumption
implies that the object for analysis is considered to have a single mass averaged temperature.
In the derivation shown above, the significant object dimension was the conduction path length𝐿 =
𝑟𝑜 − 𝑟𝑖. In general, a characteristic length scale may be obtained by dividing the volume of the solid by
its surface area:
Using this method to determine the characteristic length scale, the corresponding Biot number may be
evaluated for objects of any shape, for example a plate, a cylinder, or a sphere. As a thumb rule, if the
Biot number turns out to be less than 0.1, lumped capacity assumption is applied.
In this context, a dimensionless time, known as the Fourier number, can be obtained by multiplying the
dimensional time by the thermal diffusivity and dividing by the square of the characteristic length:
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Lumped thermal capacity analysis:
The simplest situation in an unsteady heat transfer process is to use the lumped capacityassumption, wherein
we neglect the temperature distribution inside the solid and only dealwith the heat transfer between the solid
and the ambient fluids. In other words, we areassuming that the temperature inside the solid is constant and is
equal to the surfacetemperature.
The solid object shown in figure 5.2 is a metal piece which is being cooled in air after hot forming. Thermal
energy is leaving the object from all elements of the surface, and this is shown for simplicity by a single
arrow. The first law of thermodynamics applied to this problem is
Now, if Biot number is small and temperature of the object can be considered to be uniform,this equation can
be written as
Integrating and applying the initial condition i T(0) = T ,
Taking the exponents of both sides and rearranging,
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Rate of convection heat transfer at any given time t:
Total amount of heat transfer between the body and the surrounding from t=0 to t:
Maximum heat transfer (limit reached when body temperature equals that of the surrounding):
Use of Transient temperature charts (Heisler’s charts):
The Plane Wall:
In Sections 5.5 and 5.6, one-term approximations have been developed for transient,one-dimensional
conduction in a plane wall (with symmetrical convection conditions)and radial systems (long cylinder and
sphere). The results apply for Fo_ 0.2 and canconveniently be represented in graphical forms that illustrate
the functional dependenceof the transient temperature distribution on the Biot and Fourier numbers.
Results for the plane wall (Figure 5.6a) are presented in Figures 5S.1 through5S.3. Figure 5S.1 may be
used to obtain the midplanetemperature of the wall, T(0, t) _To(t), at any time during the transient process.
If To is known for particular values ofFoandBi, Figure 5S.2 may be used to determine the corresponding
temperature atany location off the midplane. Hence Figure 5S.2 must be used in conjunction withFigure
5S.1. For example, if one wishes to determine the surface temperature (x* 1) at some time t, Figure 5S.1
would first be used to determine Toat t. Figure 5S.2would then be used to determine the surface
temperature from knowledge of To. The
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FIGURE 5S.1 Midplane temperature as a function of time for a plane wall of thickness 2L.
FIGURE 5S.2 Temperature distribution in a plane wall of thickness 2L
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FIGURE 5S.3 Internal energy change as a function of time for a plane wall of thickness 2L
The cylinder:
FIGURE 5S.4 Centerline temperature as a function of time for an infinite cylinder of radius ro
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FIGURE 5S.5 Temperature distribution in an infinite cylinder of radius ro
FIGURE 5S.6 Internal energy change as a function of time for an infinite cylinder ofradiusro
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For sphere:
FIGURE 5S.7 Center temperature as a function of time in a sphere of radius ro
FIGURE 5S.8 Temperature distribution in a sphere of radius ro
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FIGURE 5S.9 Internal energy change as a function of time for a
sphere of radius ro
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