The Conduction Rate Equation

The conduction rate equation, Fourier’s Law, was previously discussed in this blog entry. We can now consider its origin. Fourier’s Law is phenomenological, i.e., it is developed from observed phenomena rather than being derived from first principles, thus, we consider the rate equation as a generalization based on a lot of experimental evidence.

Let us consider the steady-state conduction experiment below where a cylindrical rod of known material is insulated on its lateral surface, while its end faces are maintained at different temperatures, where T1 > T2.

The temperature difference causes conduction heat transfer in the positive x-direction. We are able to measure the heat transfer rate q_x, and we seek to determine how q_x depends on the following variables:

• T, the temperature difference;
• x, the rod length;
• A, the cross-sectional area.

We can imagine first holding ΔT and Δx constant and varying A. Thus, we find that q_x is directly proportional to A. Similarly, holding ΔT and A constant, we see that q_x varies inversely with Δx. Finally, holding A and Δx constant, we find that q_x is directly proportional to ΔT. The collective effect is then:

Even if we change the material from a metal to a plastic, this proportionality will remain valid. But we would also find that, for equal values of A, Δx, and ΔT, the value of q_x would be smaller for the plastic than for the metal. This suggests that the proportionality may be converted to an equality by introducing a coefficient that is a measure of the material behavior. Thus:

where k, the thermal conductivity (W/m.K) is an important property of the material. Evaluating this expression in the limit as x → 0, we obtain the heat rate:

The minus sign is needed because heat is always transferred in the direction of decreasing temperature.

Fourier’s law, as written in Equation 2, suggests that the heat flux is a directional quantity. In particular, the direction of q_x′′ is normal to the cross-sectional area A. Or, more generally, the direction of heat flow will always be normal to a surface of constant temperature, called an isothermal surface.

The following figure shows the direction of heat flow q_x′′ in a plane wall for which the temperature gradient dT/dx is negative.

From Equation 2, it follows that q_x′′ is positive. Note that the isothermal surfaces are planes normal to the x-direction.

Recognizing that the heat flux is a vector quantity, we can write a more general statement of the conduction rate equation (Fourier’s law) as follows:

It can be understood through Equation 3 that the heat flux vector is in a direction perpendicular to the isothermal surfaces. An alternative form of Fourier’s law is therefore:

where q_n′′ is the heat flux in a direction n, which is normal to an isotherm, and n is the unit normal vector in that direction as shown below:

The heat transfer is sustained by a temperature gradient along n. Note also that the heat flux vector can be resolved into components such that, in Cartesian coordinates, the general expression for q′′ is:

Thus, from equation 3:

Each of the above expressions relates the heat flux across a surface to the temperature gradient in a direction perpendicular to the surface. It is also implied in Equation 3 that the medium in which the conduction occurs is isotropic. For such a medium, the value of the thermal conductivity is independent of the coordinate direction. 

Relationship Between Heat Transfer and the First Law of Thermodynamics

In this topic we are interested in the efficiency of heat engines. We are going to build upon the knowledge of thermodynamics and show how heat transfer plays an integral role in managing and promoting the efficiency of a wide range of energy conversion devices. 

Remember that we have defined a heat engine previously as any device that continuously or cyclically operates and converts heat to work. Power plants and thermoelectric devices are examples of heat engine. 

It is extremely important to improve the efficiency of heat engines. For example, an efficient combustion engine consumes less fuel to produce a given amount of work thus reduces emissions of pollutants. More efficient thermoelectric devices can generate more electricity from waste heat. 

The second law of thermodynamics can be represented in a number of distinct but comparable ways and is frequently employed when efficiency is an issue. The KelvinPlanck statement is very relevant to the operation of a heat engine. It states:

"It is impossible for any system to operate in a thermodynamic cycle and deliver a net amount of work to its surroundings while receiving energy by heat transfer from a single thermal reservoir"

Remember that a thermodynamic cycle is a process for which the initial and final states of the system are identical. Consequently, the energy stored in the system does not change between the initial and final states, and the first law of thermodynamics reduces to W = Q.

Because of the Kelvin–Planck statement, a heat engine must exchange heat with two or more reservoirs, gaining thermal energy from the higher-temperature reservoir and rejecting thermal energy to the lower-temperature reservoir. Therefore, converting all of the input heat to work is impossible and:

W = Q_in – Q_out,

where Q_in and Q_out are both defined to be positive. i.e.,

Q_in = heat transferred from the high temperature source to the heat engine

Q_out = the heat transferred from the heat engine to the low temperature sink.

The efficiency of a heat engine is the fraction of heat transferred into the system that is converted to work:

For a reversible process the ratio Q_out/Q_in is equal to the ratio of the absolute temperatures of the respective reservoirs (From the 2^nd Law of Thermodynamics. Thus, the efficiency of a heat engine undergoing a reversible process, i.e., Carnot efficiency, η_C, (as previously discussed) is given by:

where T_c and T_h are the absolute temperatures of the low and high temperature reservoirs, respectively. 

The Carnot efficiency is the maximum possible efficiency that any heat engine can achieve operating between those two temperatures. Any real heat engine, which will necessarily undergo an irreversible process, will have a lower efficiency.

From our knowledge of thermodynamics, we know that, for heat transfer to take place reversibly, it must occur through an infinitesimal temperature difference between the reservoir and heat engine. In heat transfer mechanisms, in order for heat transfer to occur, there must be a nonzero temperature difference between the reservoir and the heat engine. This introduces irreversibility and reduces the efficiency. 

Let us now consider a more realistic heat engine model where heat is transferred into the engine through a thermal resistance R_t,h, while heat is extracted from the engine through a second thermal resistance R_t,c where subscripts h and c refer to the hot and cold sides of the heat engine respectively. 

Let us now consider a more realistic heat engine model where heat is transferred into the engine through a thermal resistance R_t,h, while heat is extracted from the engine through a second thermal resistance R_t,c where subscripts h and c refer to the hot and cold sides of the heat engine respectively. This is shown in the following figure:

 

These thermal resistances are associated with heat transfer between the heat engine and the reservoirs across a nonzero temperature difference through mechanisms of conduction, convection and/or radiation. E.g., the resistances could represent conduction through walls separating the heat engine from the two reservoirs

Note that the reservoir temperatures are still T_h and T_c but that the temperatures seen by the heat engine are T_h,i < T_h and T_c,i > T_c, as shown in the diagram above. The heat engine is still assumed to be internally reversible, and its efficiency is still the Carnot efficiency.

However, the Carnot efficiency is now based on the internal temperatures T_h,i and T_c,i . Therefore in order to account for the realistic irreversible process, the efficiency, ηm, is as follows: 

where the ratio Q_out/Q_in, has been replaced by the corresponding ratio of heat rates, q_out/q_in. This replacement is based on applying energy conservation at an instant in time. Utilizing the definition of a thermal resistance, the heat transfer rates into and out of the heat engine are given by:

The above equations can be solved for the internal temperatures to result in:

For the T_c,i equation, q_out has already been related to q_in and η_m, The more realistic, modified efficiency can then be expressed as:

Solving for η_m, results in:

where R_tot = R_t,h + R_t,c.

It is evident that η_m = η_C only if the thermal resistances R_t,h and R_t,c could somehow be made infinitesimally small (or if qin  0). For realistic (nonzero) values of R_tot, η_m < η_C, and η_m further deteriorates as either R_tot or q_in increases. As an extreme case, note that η_m = 0 when T_h = T_c + q_inR_tot , meaning that no power could be produced even though the Carnot efficiency is nonzero.

In addition to the efficiency, another important parameter to consider is the power output of the heat engine, given by:

Maxwell Equation Solved Example

Example

Show that: 

Solution

This example uses the Maxwell Relations Table from the previous blog entry. You can read it here

Lets combine the first and second laws: dU = TdS - pdV

Divide both sides by dV and keep T as a constant:

Note that:

This results in:

Using the Maxwell Relation for A:

This results in:

Since:

This confirms that