For 15 years, we have given chemical engineering lectures on transport phenomena at the graduate level and on heat transfer at the undergraduate level with the following approach:

1) Start with transient or steady-state macroscopic balances;

2) Continue with microscopic transient balances and conclude with microscopic steady-state balances; and

3) Start with macroscopic balances; we believed that this facilitated learning because in general, macroscopic problems are easier to solve than microscopic problems, and less information is required to obtain such solutions.

In Table 1, we present a summary of the old syllabus of the heat transfer course.

For a long time, we have been thinking about starting the heat transfer course with transient processes, and we currently teach it with the following approach:

1) We begin with microscopic transient balances.

2) We continue with microscopic steady-state balances.

3) We conclude with macroscopic transient or steady-state balances.

4) We begin with microscopic balances because we believe that with this method, the students can develop a solid knowledge of transport phenomena, in comparison with starting with macroscopic balances. In addition, we can solve macroscopic problems by integrating basic microscopic equations etc.

5) We begin with a simple transient problem to reduce the possible “initial shock” that we could cause the students to experience. After the students become accustomed to the methodology, it is possible that it would “seem natural” to them to begin with an analysis of a transient problem.

6) We easily understand any steady-state process after understanding it as a global process, from time zero until it reaches a steady state.

7) The initial phase of a process is generally a transient problem, as are some accidents, unexpected actions or operating errors in industry.

8) The integration of dynamic or transient simulations into the chemical engineering curriculum has aroused considerable interest due to the increasing use of simulators for designing processes, control, training and operational support in industry (Komulainen, Enemark-Rasmussen, Sin, Fletcher, & Cameron, 2012: p. e153) .

In Table 2, we present a summary of the current heat transfer syllabus.

We highlight some aspects of the problem of this section:

1) We discuss the current problem of giving students a first experience with transient phenomena that uses all the intuitive knowledge that they bring with them.

2) In the development of the problem, we discuss in simple terms the concept of heat conduction in a slab and the basis of conductive heat flux.

3) In the figures presented, we show and analyze both the temperature (T) at each point (x) in time (t) and the flux’s value and direction (q_x), i.e., whether the flux is positive, negative or zero.

Analyze the following heat transfer problem in a slab that is initially at a constant temperature (T₀) and is subjected to another constant temperature (T₁) on both surfaces at the points and when the process begins. The slab, which has length (2 L) and coordinates in the interval −L ≤ x ≤ +L, has constant density ρ (kg/m³), constant thermal conductivity k [W/(m∙˚C)] and constant specific heat Cp [J/(kg∙˚C)]. The initial condition at t = 0 s and the boundary conditions are presented below for T₀ > T₁ (Bird, Stewart, & Lighfoot, 2002, 2014; Carslaw & Jaeger, 1980; Crank, 1976; Luikov, 1968) :

Physically, the process occurs in the following sequence:

1) We present the situation of the slab before the process begins at t < 0⁺ s in Figure 1.

2) At time t = 0⁺ s, the slab is subjected to a temperature T₁ < T₀ at its ends, x = −L and x = +L.

In Figure 2, we present the slab at the beginning of the cooling process, i.e., at t = 0⁺ s. At time t = 0⁺ s, the transient state begins with T = T₁ at x = −L and x = +L; at all other points, the slab is at the initial temperature (T₀). Instantaneously, energy flow from the two edges of the slab, x = −L and x = +L. When t > 0 s, the slab is in a transient state and has a temperature profile, , that is a function of time (t) and position (x). There is a diffusive heat flux,

, of the type. In other words, there is a flux “profile”, , that is a function of time (t) and position (x).

3) Continuation of the transient state with a maximum at the center of the slab, x = 0 m.

Shown in Figure 3 is the maximum at the center of the slab, x = 0 m; each

maximum has a different temperature (T) at each time (t). In other words, at point x = 0 m, which has the maximum temperature, there is no heat flux; the temperature is.

4) For a very long time, i.e., as time tends to infinity, the temperature throughout the slab remains constant (T₁).

In Figure 4, we present the situation in the slab when it reaches a steady state, i.e., after the cooling process. The temperature of the slab is constant as time because the (k) is constant and there is no power generation G_e (W/m³) due to, for example, electrical resistance or chemical or nuclear reactions inside the slab that may influence the temperature profile. The temperature is a linear function of the type, however because at x = −L and x = +L the temperature is (T₁), it will be constant all over the slab, i.e., T = T₁. In other words, the slab will reach a steady state as in which T = T₁.

After the process reaches the steady state, the slab temperature (T) does not change over time (t); it is a constant, T = T₁, at all positions. In this situation, the conductive flux is zero, i.e.,. In other words, the flux is constant and equal to zero, , because (k) is constant and the temperature gradient is zero because the temperature is constant (T = T₁) across the slab. Evidently, the heat flux can be calculated as in both the transient and the steady state but is zero after the steady state is reached.

We highlight some aspects of the problem:

1) The following problem is similar to the previous problem; however we include data and perform calculations.

2) In a previous lecture, we presented the basics of using Excel, and the students performed activities such as obtaining the first root or multiple roots of an equation.

3) In another lecture, we analyzed the microscopic transport balances for a pure fluid, i.e., the continuity, momentum and energy equations.

4) At another convenient time, we presented a summary of the numerical method of lines for solving problems.

Solve the heat transfer problem for a solid slab of width 2 L using the microscopic

continuity equation with constant properties (k, ρ and Cp) and unsteady heat conduction equation in (x) to obtain the following (Bird, Stewart, & Lighfoot, 2002, 2014; Carslaw & Jaeger, 1980; Crank, 1976; Luikov, 1968) :

1) The transient temperature profile, assuming that the slab was at temperature (T₀) for t < 0⁺ s and that at time t > 0 s, it was subjected to temperature (T₁) at x = −L and x = +L.

Assuming that (ρ) is constant and that the slab is stationary, we do not obtain important information from the continuity equation for the slab. The unsteady state equation of heat conduction in (x), assuming that the thermal diffusivity is a constant, i.e., , becomes:

We obtain the transient profile using, for example, separation of variables or Laplace transform, with the initial condition and the two boundary conditions mentioned above [ Bird, Stewart, & Lighfoot, 2002: pp. 376 to 378 , example (12.1-2); Crank , Equation (4.17), 1976: p. 47 ] for as follows:

The terms are called eigenvalues, and when we use them in the infinite series in Equation (5), we retrieve the initial condition and the boundary conditions of the problem. These results exhibit some of the important uses of the eigenvalues that we study in mathematics courses for engineering but that sometimes are not easy to understand.

If we move to the right hand side of Equation (1) and multiply the numerator and denominator by (L²), we obtain, and thus, the Fourier number, , comes naturally from the heat conduction equation. is important to the convergence of the series in Equation (5). In general, if, we can use only the first term in the series in Equation (5), or of similar solutions for common geometries, with a relative percent deviation () in the calculation of (T), i.e., almost always <1%, instead of the infinitely many terms in Equation (5).

To obtain the solution of Equation (4), which is given by Equation (5), we start with Equation (4), which is a valid differential equation for a point, and we integrate it to obtain Equation (5), which is valid for all points in. In other words, we start from a microscopic equation to achieve a macroscopic result.

2) The temperature at any point (x) on the slab as is equal to (T₁).

For infinite time (as), we calculate the exponential term of Equation (5) as, the term finite with for, and from Equation (2), we find that T = T₁ as follows:

This result is important because from it we obtain the temperature at infinite time, i.e., the steady-state temperature, which in this particular situation is constant and equal to (T₁).

3) The temperature at or both for finite time (t) and for.

Knowing that allows us to determine that the temperatures both at any finite time (t) and as using Equation (5) as follows:

With the result presented in Equation (7), T = T₁, we retrieve the boundary conditions at and. In addition, this result is valid for, which is when the slab reaches a steady state.

4) The temperature at time t = 0 s as equal to (T₀).

We obtain this temperature using the following procedure. At t = 0 s and x = 0 m, , and from a table, such as the one in Spiegel, Lipschutz, & Liu (2009: p. 135) , we find that the sum of the series

. By substituting this value into the infinite series in Equation (5), we obtain T = T₀ as follows:

The result obtained in Equation (8) justifies the infinite number of eigenvalues (μ_n) needed to retrieve the value (T₀) which is the initial condition. If we had used any value in other than x = 0 m, the result would be the same, but we would have to consider the terms with.

5) The steady-state temperature profile.

We obtain the steady-state temperature profile, , from Equation (5) by assuming, finite, and for; the result, T = T₁ is obtained as follows:

We can obtain the steady-state profile another way. In other words, by solving Equation (1) with the accumulation term equal to zero, i.e., , and then applying the boundary conditions T = T₁ at and, integrating twice and calculating the integration constants (C₁ and C₂) yields, at infinite time, a constant temperature profile, T = T₁:

6) The heat flux, , in transient and steady state.

By deriving the temperature in unsteady state or the steady state using Equation (5) and Equation (6) or Equation (10), respectively, the derivative

and the parameter, we obtain, respectively

Clearly, in Equation (12), we have no flux as because the temperature in the slab is a constant (T₁) and the flux is zero, i.e., q_x = 0 W/m². However, according to Equation (11), the flux is of the type, which changes with position (x) and time (t), however can reach zero as.

7) The temperature at any point on the slab, , using the first term of Equation (5).

We calculate the temperature in the slab of width 2 L = 0.2 m using the first term of Equation (5) at the points, , , and for, T₀ = 400˚C, T₁ = 50˚C, k = 37.7 W/(m∙˚C), ρ = 7.822 kg/m³, Cp = 444 J/(kg∙˚C) and

. We obtain the results below using

, and; we note that the temperature (T) is symmetric around x = 0 m, which is the reference point.

8) The temperature at any point on the slab, , using the first two terms of Equation (2).

We obtain the results below using the first two terms of Equation (5). We detail the calculation for x = −0.1 m, t = 600 s and Fo = 0.651 and present only the results for the other points.

These results are equal to those calculated using the first term of Equation (5) to two decimal places. This is because, which facilitates the convergence of Equation (5). For Fo = 0.651, it is sufficient to use only the first term of Equation (5).

9) The flux at any point on the slab, , using the first term of Equation (11).

We obtain the results presented below for x = −0.1 m, x = −0.05 m, x = 0 m, x = +0.05 m and x = 0.1 m, respectively, for Fo = 0.651, and. The flux q_x = 0 W/m² at x = 0 m, and heat is lost from both surfaces of the slab.

The flux in the slab is negative for x = −0.1 m and positive for x = +0.1 m. It is considered positive on the positive side of the x-axis, i.e., from x = 0 m to x = +0.1 m. Using this sign convention, heat is lost from both surfaces of the slab until we reach the steady state, when the flux becomes zero,

.

10) The flux at any point on the slab, , using the first two terms of Equation (11).

We obtain the flux using the first two terms of the series in Equation (11). The results are listed below; we highlight the calculations for x = −0.1 m, t = 600 s and Fo = 0.651.

We calculate that the deviations of the fluxs of this item in relation to the previous item as satisfying < 1%. Generally, < 5% is sufficient for engineering calculations. In these calculations, we do not consider the experimental error in the properties (k, ρ and Cp), in dimension (x) etc., which must be considered in detailed calculations.

11) Develop a computer program to calculate the transient temperature profile, , and the flux, , at any point (x) and time (t). (You may use a commercial program or an appropriate program obtained from another source).

In Table 3 and Table 4, we present a computer program and a Fortran subroutine, respectively, to calculate the temperature, , and the flux,.

In previous lectures, we discussed the use of the method of lines. There are several references in the literature that introduce the method of lines and dig deeper into the subject (Cutlip & Shacham, 1999; Sadiku & Obiozor, 2000; Schiesser & Silebi, 2009; Wouwer, Saucez, & Vilas, 2014) .

Due to the symmetry of the temperature profile, by discretizing Equation (4) using centered finite differences and numbering the nodes of half of the slab’s width (L) from i = 1 to i = n, we obtain, for i = 2 to (n − 1)

We write the initial condition for all points satisfying, which is already set in the nomenclature of the computer program, as

We obtain the boundary condition for x = +L, i.e., i = n or the right edge, as

We write the boundary condition at x = 0 m, i.e., i = 1, which, due to the symmetry of the temperature profile, has no flux at that point

. Using the definition of the flux at x = 0 m, ,

i.e., , we obtain

We solve the system of equations, including Equation (33) with the term and Equations (34), (35) and (36) for due to the symmetry with the computer program.

In the program, we write the temperature (T) as the variable (Y) and the derivative as Yprime = dY/dt. We use the method of lines to solve the system of equations and obtain the temperature at all points in and, due to the symmetry, in. We execute the computer program for t = t_end = 600 s, n = 400 and obtain

and. The temperature at x = 0 m, calculated using the analytical solution, is T(1) = 139.34˚C, which, when compared with the numerical solution, T(1) = 138.98˚C, produces RPD = −0.3%. For n = 400 and, we calculate the flux at x = +L, using T(n − 1) = 50.35˚C, T(n) = 50.00˚C and k = 37.7 W/(m∙˚C), as. Comparing

this flux with that of the analytical solution, q_x = 52,905.90 W/m², we obtain RPD = −0.5%. Generally, for engineering calculations, a precision on the order of RPD = 5% is sufficient.

In Table 5 and Table 6, we present the questionnaires that the students responded to in semesters 2015.1 and 2016.1, respectively. In 2015.1, 18 of the 27 students enrolled in Heat Transfer were present for the questionnaire, and 17 responded. In 2016.1, 25 of the 27 students enrolled in the course were present for the questionnaire, and 24 responded. In the following, we present some of the results in the Tables cited above for semesters 2015.1 and 2016.1:

1) In 2015.1 and 2016.1, 65% and 54%, respectively, of the students had little difficulty understanding the theory covered in Heat Transfer.

2) In 2015.1 and 2016.1, 59% and 58%, respectively, had no difficulty calculating microscopic transient balances.

3) In 2015.1, 65% had no difficulty with steady-state calculations, and in 2016.1, 63% had little difficulty.

4) In 2015.1, 76% preferred to perform all the classroom activities each month, but only 46% preferred this option in 2016.1.

5) In 2015.1, 53% had no difficulty understanding the basics because we began with transient problems and then, moved to steady-state problems. In 2016.1, 75% had little difficulty.

6) In 2015.1, 59% preferred that the course began with transient problems, but in 2016.1, 63% would have preferred that it begin with steady-state problems.

7) In 2015.1, 47% understood the basics well, 47% understood the physical sense of the problems well, and 47% connected the basics by solving problems. In 2016.1, only 21% understood the basics well, 25% understood the physical sense of the problems well, and 42% connected the basics by solving problems, which is quite different from 2015.1.

8) In 2015.1, 29% did not understand the basics well, 18% did not understand the physical sense of the problems well, and 35% could not connect the basics by solving a problem. In 2016.1, only 25% did not understand the basics well, 13% did not understand the physical sense of the problems well, and 21% could not connect the basics by solving a problem.

Next, we present the averages of the students’ responses to the questionnaires distributed in semesters 2015.1 and 2016.1 (Table 7). In the calculation, we considered all 41 students that answered the questionnaire. Of the students:

1) 59% had little difficulty understanding the theory covered in Heat Transfer.

2) 59% did not have difficulty calculating microscopic transient balances.

3) 46% did not have difficulty performing steady-state calculations.

4) 59% preferred the classroom activities to traditional monthly exams.

5) 59% had little difficulty understanding the basics because we began with transient problems and then moved to steady-state problems.

6) 41% preferred that the course begin with transient problems, and 51% preferred to begin with steady-state problems.

7) 32% understood the basics well, 34% understood the physical sense of the problems well, and 44% connected the basics by solving problems.

8) 27% did not understand the basics well, 15% did not understand the physical sense of the problems well, and 27% could not connect the basics by solving a problem.

One of the goals of this study is to discuss the possibility of starting the heat transfer course with transient problems. As we previously showed, in 2015.1,

59% of the students preferred that the course began with transient problems, but in 2016.1, 63% of the students would have preferred to begin with steady-state problems. Of the 41 students who responded to the questionnaire, on average, 41% preferred to begin with transient problems and 51% preferred to begin with the steady state. Perhaps the change in the preference for beginning with transient or steady-state problems is related to the practice of most books and professors of beginning with steady-state problems. In addition, it is possible that there could be an initial impact when the course begins with transient problems. In any event, we started with a simple transient problem to try to reduce the “initial shock” that we caused in the students. After the students became accustomed to the methodology, we believed that “it would seem natural” to them to start with transient problems.

Stammitti (2013) updated the teaching and learning experiences of students in the transport phenomena laboratory. His goal was achieved by using computers with Excel spreadsheets and without requiring students to acquire programming skills to use the spreadsheets. In the present investigation, our students created their own Excel spreadsheets, but when it was necessary, we preferred them to write or adapt Fortran programs instead of using commercial simulators. One of our goals is for students to know how to program in Fortran even though commercial simulators are available on the market. In general, we use Imsl subroutines and hope they continue to be available despite their cost; otherwise, we will employ similar subroutines that are free to use. Sometimes, we compare our calculations with those in books or articles, some of which were performed with commercial simulators.

According to Stammitti (2013) , most students felt comfortable with the Excel environment after the laboratory sessions, and only 30% required assistance with the spreadsheets. In the present investigation, on average, 59% of the students did not have difficulty calculating microscopic transient balances and 46% did not have difficulty performing steady-state calculations.

According to his webpage, Lund (2016) came to the conclusion presented next. Based on the students’ performance and his observations, he found that they preferred the method he proposed.

In the present study, on average, 59% of the students preferred to perform four or five classroom activities per month. Cartaxo, Silvino, & Fernandes (2014) asked, in question (5) of their questionnaire, “was the activity more interesting and motivating than the usual methods (exams) used in the course?” They considered the personal motivation of the students to be verified with question (5), to which 82% of the students responded positively to the use of the approach used in the lectures, which was “learning by doing”. We posed a similar question in the present investigation: “Do you prefer four or five activities, or a traditional monthly exam?” On average, 59% of the students preferred to perform all the activities in the classroom.

Question (7) of Regalado-Méndez, Cid-Rodríguez, & Báez-González (2010) is related to the present study. The question is “Do you understand all the concepts?”; the possible answers were “Yes” and “No”. According to Regalado- Méndez, Cid-Rodríguez, & Báez-González (2010), 80% of the students understood the basics. On average, only 32% of our students understood the basics of heat transfer basics well. This is an unsatisfactory and worrying result; we must change the methodology to improve our teaching and ensure that students become “captivated by the learning process” and that they improve their performance.

According to Ganley (2015) , the student feedback regarding the homogeneous kinetics lab-oratory module has been generally very positive. The students’ attitudes, answers in the module and dedication indicate satisfaction with the activities. In Heat Transfer, 59% of our students preferred to perform all the activities in the classroom.

We did not include questions about the reasons for which students preferred to perform all the activities in the classroom on the questionnaire, but we are convinced of the following:

1) Students chose the option to perform all the classroom activities because they were satisfied and almost free of stress and because the activities seemed easier than traditional exams.

2) They seemed to enjoy working in pairs, and perhaps, that will contribute to their professional futures, by decreasing the individualism that sometimes prevails in engineering careers.

3) We promoted “forced study in the classroom” with fewer demands than traditional exams have, but the students were able to comfortably perform almost all the activities.

4) Did the learning become more difficult because the students performed an activity on the same day that a topic was discussed? Apparently, this did not harm the learning. In addition, sometimes, we repeated an activity and subject to review the knowledge and to try to fill any gaps.

5) Students had fun, learned and received feedback on their knowledge in the heat transfer course, as we did.

Items from (1) to (5) mentioned before will be the object of our future research in the subject Heat Conduction.