iintroduction to heat transfer

By Irene Sanders,2015-04-01 18:38
15 views 0
iintroduction to heat transferto,Heat,heat




    Heat Transfer Modes

    Figure 1: Conduction heat transfer

    Heat transfer processes are classified into three types. The first is conduction, which is defined as transfer of heat occurring through intervening matter without bulk motion of the matter. Figure 1 shows the process pictorially. A solid (a block of metal, say) has one surface

    at a high temperature and one at a lower temperature. This type of heat conduction can occur, for example, through a turbine blade in a jet engine. The outside surface, which is exposed to gases from the combustor, is at a higher temperature than the inside surface, which has

    cooling air next to it. The level of the wall temperature is critical for a turbine blade.

    The second heat transfer process is convection, or heat transfer due to a flowing fluid. The fluid can be a gas or a liquid; both have applications in aerospace technology. In convection heat transfer, the heat is moved through bulk transfer of a non-uniform temperature fluid.

    The third process is radiation or transmission of energy through space without the necessary presence of matter. Radiation is the only method for heat transfer in space. Radiation can be important even in situations in which there is an intervening medium; a familiar example is the heat transfer from a glowing piece of metal or from a fire.

Introduction to Conduction

    We will start by examining conduction heat transfer. We must first determine how to relate the heat transfer to other properties (either mechanical, thermal, or geometrical). The answer to this is rooted in experiment, but it can be motivated by considering heat flow along a ``bar'' between two heat reservoirs at , as shown in Figure 2 It is plausible that the heat

    transfer rate, , is a function of the temperature of the two reservoirs, the bar geometry and the bar properties. (Are there other factors that should be considered? If so, what?). This can be expressed as


    It also seems reasonable to postulate that should depend on the temperature difference

    . If is zero, then the heat transfer should also be zero. The temperature dependence can therefore be expressed as


    Figure 2: Heat transfer along a bar

    An argument for the general form of can be made from physical considerations. One requirement, as said, is if . Using a MacLaurin series expansion, as



If we define and , we find that (for small ),


We know that . The derivative evaluated at (thermal equilibrium) is a

    measurable property of the bar. In addition, we know that if or

    . It also seems reasonable that if we had two bars of the same area, we would have twice the heat transfer, so that we can postulate that is proportional to the area. Finally, although the argument is by no means rigorous, experience leads us to believe that as increases should get smaller. All of these lead to the generalization (made by Fourier in 1807) that, for the bar, the derivative in Equation (4) has the form


    In Equation (5), is a proportionality factor that is a function of the material and the temperature, is the cross-sectional area and is the length of the bar. In the limit for any temperature difference across a length as both , , we can say


    A more useful quantity to work with is the heat transfer per unit area, defined as


    2The quantity is called the heat flux and its units are Watts/m. The expression in (16.6) can

    be written in terms of heat flux as


    Equation (16.8) is the one-dimensional form of Fourier's law of heat conduction. The proportionality constant is called the thermal conductivity. Its units are . Thermal conductivity is a well-tabulated property for a large number of materials. Some values for familiar materials are given in Table 16.1; others can be found in the references. The thermal

    conductivity is a function of temperature and the values shown in Table 16.1 are for room temperature.

    Table 1: Thermal conductivity at room temperature for some metals and non-metals

    Metals Ag Cu Al Fe Steel

     420 390 200 70 50

    Non-metals Air Engine oil Brick Wood Cork

     0.6 0.026 0.15 0.18 0.4-0.5 0.2 0.04

    Steady-State One-Dimensional Conduction

    Figure 3: One-dimensional heat conduction

    For one-dimensional heat conduction (temperature depending on one variable only), we can devise a basic description of the process. The first law in control volume form (steady flow energy equation) with no shaft work and no mass flow reduces to the statement that for all surfaces (no heat transfer on top or bottom of Figure 3). From Equation (6), the heat transfer rate in at the left (at ) is


The heat transfer rate on the right is


Using the conditions on the overall heat flow and the expressions in (16.9) and (16.10)


Taking the limit as approaches zero we obtain




If is constant (i.e. if the properties of the bar are independent of temperature), this reduces



or (using the chain rule)


Equation (16.14) or (16.15) describes the temperature field for quasi-one-dimensional steady

    state (no time dependence) heat transfer. We now apply this to an example.

Example 1: Heat transfer through a plane slab

    Figure 4: Temperature boundary conditions for a slab

    For this configuration (Figure 4), the area is not a function of , i.e. . Equation (5) thus becomes


Equation (16.16) can be integrated immediately to yield




    Equation (16.18) is an expression for the temperature field where and are constants of integration. For a second order equation, such as (16.16), we need two boundary conditions to determine and . One such set of boundary conditions can be the specification of the temperatures at both sides of the slab as shown in Figure 16.4, say ; . The condition implies that . The condition implies that

    , or . With these expressions for and the temperature

    distribution can be written as


    This linear variation in temperature is shown in Figure 5 for a situation in which .

    Figure 5: Temperature distribution through a slab

    The heat flux is also of interest. This is given by


Thermal Resistance Circuits

    There is an electrical analogy with conduction heat transfer that can be exploited in problem solving. The analog of is current, and the analog of the temperature difference, , is voltage difference. From this perspective the slab is a pure resistance to heat transfer and we can define


where , the thermal resistance. The thermal resistance increases as

    increases, as decreases, and as decreases.

    Figure 6: Heat transfer across a composite slab (series thermal resistance)

    The concept of a thermal resistance circuit allows ready analysis of problems such as a composite slab (composite planar heat transfer surface). In the composite slab shown in Figure 6, the heat flux is constant with . The resistances are in series and sum to

    . If is the temperature at the left, and is the temperature at the right, the heat transfer rate is given by


    Figure 7: Heat transfer for a wall with dissimilar materials (parallel thermal resistance)

    Another example is a wall with a dissimilar material such as a bolt in an insulating layer. In this case, the heat transfer resistances are in parallel. Figure 7 shows the physical configuration, the heat transfer paths and the thermal resistance circuit. For this situation, the total heat flow is made up of the heat flow in the two parallel paths,

    , with the total resistance given by


    More complex configurations can also be examined; for example, a brick wall with insulation on both sides (Figure 8).

    Figure 8: Heat transfer through an insulated wall

    The overall thermal resistance is given by


    Some representative values for the brick and insulation thermal conductivity are:

Using these values, and noting that , we obtain

This is a series circuit so

    Figure 9: Temperature distribution through an insulated wall

    The temperature is continuous in the wall and the intermediate temperatures can be found from applying the resistance equation across each slab, since is constant across the slab. For example, to find :

This yields or .

    The same procedure gives . As sketched in Figure 9, the larger drop is across the insulating layer even though the brick layer is much thicker.

    Steady Quasi-One-Dimensional Heat Flow in Non-Planar Geometry

Report this document

For any questions or suggestions please email