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CHAPTER |
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One-dimensional, steady |
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state heat conduction |
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2.1 Introduction
In this chapter, we first derive the general three-dimensional heat conduction equation in Cartesian coordinates with the different kinds of boundary conditions associated with it. The solution to the heat conduction equation gives us the temperature field, T (x, y, x, t), in Cartesian coordinates. Once we have this, we can calculate the heat flux anywhere by using the pertinent rate law, which, in this case, happens to be the Fourier’s law of heat conduction. Following this derivation, we look at engineering problems that one can solve with a one-dimensional heat conduction approach. Furthermore, we will be looking at situations involving steady state conditions, that is, when the temperature does not change with time and end the chapter with an elaborate treatment of fins.
2.2 Three-dimensional conduction equation
In what follows, the three-dimensional conduction equation is derived from first principles. The goal is to relate all fluxes and energy exchanges to temperature or its gradients, so that we finally get an equation in temperature. A schematic representation of a three-dimensional control volume is shown in Fig. 2.1.
The major assumptions are
1.Homogeneous material-i.e., material properties do not change in space
2.No bulk motion-this removes convective effects
3.The material obeys Fourier’s law of heat conduction
In Fig. 2.1 Qi’s represents the rate of heat transfer in a given direction Qx = qx.A.
Qy = qy.A and Qz = qz.A.
At any instant of time, by invoking the first law of thermodynamics with È denoting energy rate, we have
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(2.1) |
Ein + Egenerated = Eout + Estored |
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In Eq. (2.1) the subscripts are self-explanatory and it is instructive to observe that the units of all the terms in Eq. (2.1) is W, since this is a heat rate equation.
The individual terms in Eq. (2.1) can be written
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(2.2) |
Ein = (qx + qy + qz ) A |
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Heat Transfer Engineering. http://dx.doi.org/10.1016/B978-0-12-818503-2.00002-2 |
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Copyright © 2021 Elsevier Inc. All rights reserved.
16 CHAPTER 2 One-dimensional, steady state heat conduction
FIGURE 2.1
Schematic representation of a three-dimensional control volume.
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x y z |
(2.3) |
Egenerated = qv |
Here qv is heat generated per unit volume (W/m3), usually assumed to be uniform.
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(2.4) |
Eout = (qx+dx + qy+dy + qz+dz ) A |
In Eq. (2.4), qx+dx |
can be expanded using the first term of Taylor’s series as follows |
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(by neglecting higher order terms) |
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qx+dx ≈ qx + |
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∂ |
(qx ) |
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(2.5) |
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∂x |
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The rate of change of enthalpy is given by |
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∂T |
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Estored = mcp |
∂t |
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(2.6) |
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Substituting Eqs. (2.2)–(2.6) in Eq. (2.1), we have |
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(q |
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z = (q |
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+ q |
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+ q |
z+dz |
) A + mc |
∂T |
(2.7) |
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p ∂t |
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qv x |
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y zA = (qx + x |
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y + qz+ |
z − qx |
− qy − qz ) A + mcp ∂T / ∂t |
(2.8) |
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According to Fourier’s law, the heat flux in the x-direction is given by |
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qx = −kx |
∂T |
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(2.9) |
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∂x |
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The heat flux in the y-direction is |
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qy = −ky |
∂T |
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(2.10) |
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∂y |
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2.2 Three-dimensional conduction equation |
17 |
and the heat flux in the z-direction is
qz = −kz ∂T
∂z
Eq. (2.8) then becomes,
qv x y z = |
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∂ |
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−k |
∂T |
y z + |
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−k |
∂T |
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x z |
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∂x |
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∂x |
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∂y |
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+ |
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∂ |
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−k |
∂T |
x y + ρ x y zcp |
∂T |
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∂z |
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(2.11)
(2.12)
For a homogenous material, the thermal conductivity is constant in space (i.e., kx = ky = kz = k). In the view of this, Eq. (2.12) can be rewritten as
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∂2 T |
+ k |
∂2 T |
+ k |
∂2 T |
+ q |
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∂T |
(2.13) |
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∂x2 |
∂y2 |
∂z2 |
v |
p ∂t |
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Equation 2.13 may be rewritten as
∂2 T |
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∂2 T |
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∂2 T |
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v |
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1 ∂T |
(2.14) |
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∂x2 |
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k |
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α ∂t |
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Here α is thermal diffusivity of the material and is given by k/ρ cp. It is instructive to obtain the units of α.
α = |
W /mK |
= m2 /s |
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kg/m3 J /kgK |
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We see that α has the units of m2/s. The s−1 clearly shows that it is related to “some rate.” That “some” is nothing but the diffusion of heat. It is now evident that α gives us an idea of the rate of diffusion of heat in a medium.
In other words, the quantity α signifies how quickly heat penetrates a solid body. The α for steel is far higher compared to that of wood. A wooden ladle in a boiling soup would not be as hot to feel as a steel ladle when other conditions are the same (i.e. when felt at the same time).
Eq. (2.14) is called the three dimensional transient heat conduction equation with heat generation. The three terms on the left-hand side of Eq. (2.14) represent the net diffusion of heat in the three directions x, y, and z, respectively. The fourth term represents the uniform volumetric heat generation due to either nuclear fusion, metabolism (say cancer) or an exothermic chemical reaction. Eq. (2.14) has the units K/m2. Mathematically, the equation is a second-order linear partial differential equation (PDE) in space, a first-order PDE in time, and supports six boundary conditions (two each on x, y, and z) and one initial condition (in time).
Mathematicians refer to such equations as IVBP problems, with the abbreviation denoting initial value boundary problems. Even kx, ky, and kz need not be equal; as in the case of orthotropic materials where the governing equation is still linear. However if kx, ky, and kz are functions of space (x, y, z) or temperature, the governing equation is no longer linear, and its solution is nontrivial.
18 CHAPTER 2 One-dimensional, steady state heat conduction
For steady state heat transfer, ∂T / ∂t = 0, and Eq. (2.14) becomes
∂2 T + ∂2 T + ∂2 T + qv
∂x2 ∂y2 ∂z2 k
Eq. (2.15) is called the Poisson’s equation. For steady (∂T / ∂t ) = 0 and no heat generation
= 0 |
(2.15) |
(qv = 0), Eq. (2.14) becomes
∂2 T |
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= 0 |
(2.16) |
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∂z2 |
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Eq. (2.16) is called the Laplace equation and can be written compactly as 2T = 0 where 2 is frequently referred to as the Laplacian operator or simply Laplacian.
For one-dimensional, steady state heat conduction and without heat generation, we have
d2T |
= 0 |
(2.17) |
dx2 |
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By using a similar control volume approach, the heat conduction equation in cylindrical coordinates can be derived in the form shown in Eq. (2.20).
1 ∂
kr r ∂r
∂T +
∂r
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∂T |
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+ q |
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∂T |
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k |
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k |
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p ∂t |
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(2.18) |
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r2 ∂φ |
∂φ |
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∂z |
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In Eq. (2.18), r is the radius, φ is the azimuthal angle and z is the axial direction. (Please see Fig. 2.2A). In an analogous fashion, the heat conduction equation in spherical coordinates can be derived in the form shown in Eq. (2.19).
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∂ |
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∂T |
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kr |
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∂T |
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k sinθ |
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sinθ ∂θ |
∂θ |
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∂T |
+ qv = ρcp |
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(2.19) |
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sin θ ∂φ |
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In Eq. (2.19), r is the radius, θ is the zenith angle and φ is the azimuthal angle. (Please see Fig. 2.2)
In compact, coordinate-free form, the governing equations (Eqns. 2.14, 2.18 and 2.19) can be written as
FIGURE 2.2
Schematic of coordinate systems (A) cylindrical (B) spherical.
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2.2 Three-dimensional conduction equation |
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2T + |
qv |
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∂T / ∂t |
(2.20) |
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α |
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for constant thermal conductivity, k.
2.2.1 Boundary conditions
For any given problem, a complete mathematical description requires imposing the appropriate boundary conditions. Some possibilities are:
1.Dirichlet boundary condition: A known value is imposed on boundary for temperature as shown in Fig. 2.3A. Also known as boundary condition of first kind
The conditions for Fig. 2.3A are as follows
at x = 0;T = T1 |
(2.21) |
a x = L;T = T2 |
(2.22) |
2.Neumann boundary condition: Gives a condition for the first derivative of temperature and is shown in Fig. 2.3B. Also known as boundary condition of second kind
∂T |
(2.23) |
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at x = 0;q = −k ∂x (Neumann) |
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at x = L;T = T2 (Dirichlet) |
(2.24) |
3.Robin boundary condition (or) mixed condition (or) third kind of boundary condition:
This concerns convection or radiation or both at the surfaces in question. In Fig. 2.3C, the key point is that both T1 and T2 are unknown.
FIGURE 2.3
Schematic representation of different boundary conditions in heat conduction (A) Dirichlet
(B) Neumann and (C) Robin or mixed.
20 CHAPTER 2 One-dimensional, steady state heat conduction
at x = 0; q = −k |
∂T = h (T |
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(2.25) |
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∞,1 |
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at x = L; q = −k |
∂T |
= h2 (T2 |
− T∞, 2 ) |
(2.26) |
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From Eqs. 2.25 and 2.26, it is intuitively apparent that a heat conduction equation with Robin conditions will be a lot harder to solve.
2.3 Steady state, one-dimensional conduction in a few commonly encountered systems
Let us now look at simple yet potent solutions to one-dimensional, steady conduction in three representative geometries.
1.Plane wall
2.Cylinder
3.Sphere
They are not only amenable to a “clean” mathematical analysis, but offer considerable insights into the engineering of several heat transfer systems.
2.3.1 Heat transfer in a plane wall
Consider a simple solid plane wall (or slab) that is infinitely long with a thickness L, with Dirichlet boundary conditions, as shown in Fig. 2.4.
FIGURE 2.4
Schematic representation of the plane wall under consideration in section 2.3.1.
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2.3 Steady state, one-dimensional conduction |
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The following assumptions are made:
a.T = f(x) alone.
b.Steady state prevails.
c.No heat generation in the solid.
d.Constant thermophysical properties for the solid.
The governing Eq. (2.14) reduces to the following.
d2T = 0 dx2
Integrating Eq. (2.27) twice, we have
dTdx = c1
T = c1x + c2
In Eq. (2.29), c1 and c2 are constants.
Boundary condition 1: At x = 0, T = T1 From Eq. (2.29), we have
(2.27)
(2.28)
(2.29)
c2 = T1 |
(2.30) |
Boundary condition 2: At x = L; T = T2
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(2.31) |
c1 |
= T2 − T1 |
(2.32) |
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Substituting for c1 and c2 in Eq. (2.29), we obtain the following expression for temperature, T, across the slab.
T = T2 − T1 x + T |
(2.33) |
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Eq. (2.33) tells us that the temperature profile is linear across the slab. This is sometimes referred to as LTP (Linear Temperature Profile). This will not be the case when the thermal conductivity is varying (see Problem 2.2 at the end of the chapter). Differentiating Eq. (2.33) with respect to x gives the following expression.
dT = T2 |
− T1 |
(2.34) |
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The heat transfer through the plane wall is calculated by the Fourier’s law of heat conduction.
Q = −kA dT |
(2.35) |
dx |
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22 CHAPTER 2 One-dimensional, steady state heat conduction
Q = −kA T2 − T1 |
(2.36) |
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Eq. (2.36) can also be written, on purpose (we will see why in the ensuing section), as
Q = |
T1 − T2 |
(2.37) |
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2.4 Electrical analogy and thermal resistance
A class of heat transfer problems can be eminently analyzed using an electrical analogy. In this approach, we consider the flow of heat to be analogous to the flow of electrical current, and the following equivalences apply: Q I, Rth R, ∆T ∆V. Rth here refers to thermal resistance.
A typical resistance circuit for the problem considered above is shown in Fig. 2.5. From Fourier’s law and also from the solution to conduction in a plane wall that
we just saw Eq. (2.37), we know that
Q = |
T |
(2.38) |
L/kA |
From Ohm’s law, we know that because of the existence of potential difference (∆V) between the two ends of an electrical conductor, a current I passes through the conductor against the resistance R, and the relation between the three is
I = |
V |
(2.39) |
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A comparison of Eqs. (2.38) and (2.39) clearly shows that the thermal resistance offered by the plane wall is given by
Rthermal = L/kA |
(2.40) |
From Eq. (2.40), it is clear that thicker the wall more is the resistance and higher the thermal conductivity of a material, lower is the resistance. The equation also confirms that Rthermal varies inversely with the cross sectional area A.
FIGURE 2.5
Schematic representation of an electrical resistance network for solving a heat conduction problem.
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2.5 Heat transfer in cylindrical coordinates |
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2.5 Heat transfer in cylindrical coordinates
Conduction problems in cylindrical geometries are commonly encountered in pipes, tubes, current-carrying conductors, and nuclear fuel rods, to name a few.
The governing given by Eq. (2.18) is reproduced here for the sake of completeness.
1 ∂
kr r ∂r
∂T +
∂r
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∂T |
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∂ |
∂T |
+ q |
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= ρc |
∂T |
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p ∂t |
(2.41) |
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r2 ∂φ |
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For one-dimensional, steady state heat transfer without heat generation, where temperature varies only with radius, r Eq. (2.41) reduces to
1 d |
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For constant thermal conductivity, Eq. (2.42) can be written as
1 d |
dT |
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Consider a cylindrical annulus as shown in Fig. 2.6. Let us assume that boundary conditions for the problem shown in the figure are as follows
At r = r1;T = T1
At r = r2;T = T2
Integrating Eq. (2.44) once, we get
(2.43)
T1 > T2. The
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r dT = c |
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FIGURE 2.6
Schematic representation of a cross-sectional view of a cylindrical annulus undergoing one dimensional steady state conduction.
24CHAPTER 2 One-dimensional, steady state heat conduction
Integrating again, we obtain an expression for T or T(r) as
T = c1 ln r + c2
Applying the two boundary conditions to Eq. (2.47), we have
T1 = c1 ln r1 + c2
T2 = c1 ln r2 + c2
On subtracting Eq. (2.49) from Eq. (2.48), we obtain the expression for c1 as
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T1 − T2 |
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Substituting for c1 in Eq. (2.48); we obtain the expression for c1 as
c2 = T1 − T1 −rT2 ln(r1 ) ln 1
r2
(2.47)
(2.48)
(2.49)
(2.50)
(2.51)
Substituting for both c1 and c2 in Eq. (2.48), we now arrive at the final “usable” form of Eq. (2.47).
T = T1 |
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Therefore from Eq. (2.52), one can clearly see that the temperature distribution in a cylinder or rather a cylindrical annulus is logarithmic.
The heat transfer through the cylinder can be calculated as
Q = −kAn |
dT |
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Q = −k2πr1L T1 − T2 |
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(2.55) |
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ln (r2 |
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2π kL
Please note that Q(at r = r1) = Q(at r = r2). In the above equation, L is the length in the direction perpendicular to the plane of the paper.
Invoking the concept of electrical analogy by looking at Eq. (2.55), we obtain the expression for Rcond,cyl as
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Rcondn,cyl = |
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