Heat Transfer/Conduction

Conduction

The Conduction Equation
By using Fourier's Law to perform a heat balance in three dimensions, the following equation can be derived relating the temperature in the system at a given point to the cartesian-coordinates of that point and the time elapsed:
 * $${k}\left({\frac{{\partial ^2}T}{\partial {x^2}}+\frac{{\partial ^2}T}{\partial {y^2}} + \frac{{\partial ^2}T}{\partial {z^2}}}\right) = \rho\cdot C_p \cdot\frac{\partial {T}}{\partial t}$$

The derivation assumes there is no heat generation (which is a function of x, y, z and t and must be added if present) and that the material properties don't change with time or temperature (in this case they must be incorporated into the derivatives).

Steady-state conduction
We have steady-state conduction when none of the terms of the Fourier's law depends on time; so, it is obtained:
 * $${k}\left({\frac{{\partial ^2}T}{\partial {x^2}}+\frac{{\partial ^2}T}{\partial {y^2}} + \frac{{\partial ^2}T}{\partial {z^2}}}\right) = 0$$

The Plane Wall
Heat transfer through a wall is a one dimensional conduction problem where temperature is a function of the distance from one of the wall surfaces. It is assumed that the rest of the surfaces of the walls are at a constant temperature. Heat transfer from the surfaces of the wall takes place through convection by the surrounding air, which causes them to have steady state temperatures of $$ T_{1} $$ and $$ T_{2} $$ on their surfaces. Let us assume that the fluid on the side of the wall with temperature $$ T_{1} $$ is at $$ T_{1, \infin} $$ and has a heat transfer coefficient $$h_{1}$$, and that on the side of the wall with temperature $$T_{2}$$ is at $$ T_{2 , \infin} $$ with heat transfer coefficient $$h_{2}$$, and that $$T_{1 , \infin} \ge T_{2 , \infin}$$. The assumption implies that $$h_{1} \ge h_{2}$$. Since the wall does not store any heat energy, all the heat from the hotter surface is conducted to the cooler surface. Conservation of energy dictates that


 * $${{\nabla} ^ 2}T = 0$$

for a body which generates no heat nor stores any heat, because in such a case temperature only varies with position, not with time. Applying the same to the 1-D case with the direction of the x-axis normal to the wall surface, we get


 * $${-k}{\frac{{d^2}T}{d{x^2}}} = 0$$

On solving and putting the appropriate boundary conditions, (At x=0, $$T=T_1$$ and at x=s, $$T=T_2$$) we get a linear variation for T within the wall thickness $$s$$.


 * $$T(x) = (T_2 - T_1){\frac{x}{s}} + T_1$$

It is evident from the equation that the temperature profile within the wall varies linearly with the distance from the surfaces. Since we have the temperature variation, the conduction rate can be calculated from Fourier's Law.


 * $${q_s} '' = -k \frac{dT}{dx} = \frac{k}{s} (T_1 - T_2)$$

It can be seen from the above equation that the heat flux is independent of x and are constants. This example shows the standard method of solving a conduction problem. First, the temperature profile within the body is found using the equation for conservation of energy and the temperature equation is used to solve for the heat flux by plugging it into the Fourier's Law equation.

In general, we would like to have a material with very low conductivity which is able to withstand great temperatures to build furnaces. In practice, we find that high temperature materials have relatively high thermal conductivity. Thus, furnaces are constructed from several layers, each of a different material. We can use the thermal breakdown temperatures of each material to find the optimum thickness so that the heat loss is minimal. It is easy to see that each material should receive heat at its thermal breakdown temperature and reject heat at the thermal breakdown temperature of the adjacent material.

A hollow cylinder
Let us use the same assumptions as used for the plane wall, except this time analyze what happens when we have heat transfer through an infinitely long hollow cylinder. The cylinder has inner radius R1 and outer radius R2. Under the assumption that the inner temperature remains constant at T1 and outer temperature at T2, the Laplace equation still holds true:


 * $${{\nabla} ^ 2}T = 0$$

This time, however, since we are in a cylindrical geometry, it makes most sense to use cylindrical coordinates to expand the Laplacian into the following:


 * $$ {{\nabla} ^ 2}T = {1 \over r} {\partial \over \partial r}

\left( r {\partial T \over \partial r} \right) + {1 \over r^2} {\partial^2 T \over \partial \theta^2} + {\partial^2 T \over \partial z^2 }. $$

Since the cylinder is infinitely long and symmetric, the partials with respect to z and θ are zero so the equation reduces to:


 * $$ {1 \over r} {\partial \over \partial r}

\left( r {\partial T \over \partial r} \right) =-q/k $$

when there is heat generation the above equation is valid,when there is no heat generation q becomes zero. Solving this equation with the boundary conditions that T(R1) = T1 and T(R2)=T2 shows us that:


 * $$ T(r) = \frac{T_2-T_1}{\ln{\frac{R_2}{R_1}}}(\ln(r)-\ln(R_1))+T_1 $$

Substituting this into Fourier's law for conduction gives us:


 * $$ Q'' = \frac{-k(T_2-T_1)}{r\ln{\frac{R_2}{R_1}}} = \frac{k(T_1-T_2)}{r\ln{\frac{R_2}{R_1}}}=\frac{2k(T_1-T_2)}{\phi\ln{\frac{\phi_2}{\phi_1}}}$$

Notice that, unlike was the case with the flat slab, the heat flux is not independent of radius for a cylinder, which has important implications when designing a cylindrical heat exchanger, for example.

Now let us express this in terms of a heat flux per unit length, so that we know how to use this as an approximation for cylinders that are not infinitely long. We have the cross-sectional area of the cylinder is $$ A = 2{\pi}rL$$, so:


 * $$ Q' = \frac{Q}{L} =\frac{2{\pi}k(T_1-T_2)}{\ln{\frac{R_2}{R_1}}}=\frac{2{\pi}k(T_1-T_2)}{\ln{\frac{\phi_2}{\phi_1}}}$$

Thus the heat transferred per unit length is closely related to the logarithms of the radii.

Multi-layer conduction and electro-thermal analogy
Often, conduction takes place not through a single plane wall or a single cylinder, but through multiple plane layers or through coaxial cylinders. For example, the walls of modern houses are composed by several layers for increasing their insulation: so, we can find the following stratification, from internal towards external face:

(1) internal painting

(2) plaster

(3) bricks, e.g. porotherm type

(4) a thermal insulator, such as polysthirene

(5) a hollow space (air is a good insulator)

(6) again bricks, plaster and painting

So, we have 8 layers, and the solution of the equation


 * $${-k}{\frac{{d^2}T}{d{x^2}}} = 0$$

becomes more complex, because there are more boundary conditions to be respected. A simplified way for solving the problem is the electro-thermal analogy: if we observe both the expression of $$Q$$ for plane wall and for hollow cylinder, we note that they can be written as:


 * $$Q=G\Delta T=\frac{\Delta T}{R}$$

where $$G$$ (or $$R$$, as one prefers), depend on the geometry and contain the conductivity $$k$$. This formula is analogue to Ohm's law:


 * $$I=\frac{\Delta V}{R}$$

So, the thermal problem of conduction can be modeled with thermal resistances (or thermal conductance, if one prefers $$G=1/R$$): each thermal resistance is an object which is kept at different temperatures on its nodes, so heat passes through it; in the same way, if we apply a difference of voltage to an electrical resistance, current passes through it. This analogy is useful because it allows to model a multi-layer problem as a scheme of resistances connected in series. In fact, if electrical resistance A is connected in series with B, the end node of A is at the same voltage of the start node of B; if thermal resistance A (a layer of the wall) is connected in series (is adjacent) to the thermal resistance B (the other layer), on the common node (the interface) it cannot be a thermal gradient. For the i-th plane wall:


 * $$R_i=\frac{s_i}{k_iA_i}$$

where $$A$$ is the area of the wall. So we have:


 * $$Q=\frac{T_{ext}-T_{int}}{\displaystyle\sum_{i=1}^{n}R_i}$$

For coaxial hollow cylinders (e.g. insulated pipes), we can use the same formula, but resistance is given by:


 * $$R=\frac{\ln\frac{\phi_2}{\phi_1}}{2\pi kL}$$

Finite Difference Method
The finite difference method attempts to solve a differential equation by estimating the differential terms with algebraic expressions. The method works best for simple geometries which can be broken into rectangles (in cartesian coordinates), cylinders (in cylindrical coordinates), or spheres (in spherical coordinates). Otherwise, the finite element method should be used. If the finite difference method can be used, it is considerably easier to implement than the finite element method, at the cost of some accuracy.

The simplest estimate for first-order differentials,


 * $$ \frac{\partial x}{\partial y} = \frac{\Delta x}{\Delta y}$$

It can be shown that the error caused by using this approximation is roughly proportional to $$ {\Delta y}^2 $$.

To estimate a second derivative, it is treated as the derivative of the first derivative, and the approximation above is applied twice in succession:


 * $$ \frac{\partial^2 x}{{\partial y}^2} =

\frac{{(\frac{\partial x}{\partial y}})_2-(\frac{\partial x}{\partial y})_1}{\Delta y} = \frac{(\frac{\Delta x}{\Delta y})_2 - (\frac{\Delta x}{\Delta y})_1}{\Delta y} =\frac{x_3-2x_2+x_1}{{\Delta y}^2} $$

It is evident that three points are necessary to estimate a second derivative, whereas only two are necessary to estimate a first-order derivative. This particular method can also be shown to be second-order with $$ {\Delta y}$$.

Transient Conduction
In problems of transient conduction, the complete Fourier's equation must be solved


 * $${k}\nabla^2T = \rho\cdot C_p \cdot\frac{\partial {T}}{\partial t}$$

The solution of this equation is more difficult than steady-state equation, but it's possible in simple cases. Otherwise, numerical methods should be used.

The plane wall
The sun heating the earth is an example of radiant heat transfer. The sun warms the earth without warming the space between the sun and the earth.

Finite Difference Method
finite difference method is used to find for odd shaped bodies