Unruled Notebook

Entries from July 2008

What is a Porous Medium

July 30, 2008 · 4 Comments

What is a porous medium? A popular definition is “A porous medium is a solid structure with interconnected voids.”

Suppose we take three different substances made of two materials, Material 1 and Material 2. Material 2 is the void in the above definition. Given in Figure 1 is the two-dimensional cut-section view of those three substances. Which one of the three substances is a porous medium?

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Figure 1: Porous Medium Example

According to the above definition, all three substances can be treated as porous medium. However, the configuration in Fig. 1a can easily be considered as two separate homogeneous materials for analyzing transport phenomena in and between the constituents.

A rigorous definition for porous medium suited for analyzing transport phenomena is “A porous medium is a region in space comprising of at least two homogeneous material constituents, presenting identifiable interfaces between them in a resolution level, with at least one of the constituent remaining fixed or slightly deformable.”

The requirement that one material should be stationary is for convenience of analyzing transport phenomena in the porous medium, when the other constituent is moving. The other constituent can also remain immobile. For instance, a composite material like Fig. 1c with both orange and blue representing two fixed solid materials can be treated as a porous medium. A closed region with only internal pores is also a porous medium according to our refined definition. But access to these internal pores, by physical or indirect means, is essential to perform further analysis in transport phenomena.

The Wikipedia definition [2] presumes flow through the porous medium defined. So is the definition ‘A porous medium is a solid structure with interconnected voids’. This requirement of flow is not mandatory for the definition. Observe the inclusion of material constituents in our definition, to suggest the voids need not be empty.

Also, unlike in Fig. 1, there can be more than two material constituent arranged in any disorderly fashion, forming a porous medium according to our definition. One restriction is, in a given resolution level, distinct interfaces should demarcate these constituents. An example is a two phase flow through a fixed metal mesh, an often occurring configuration in heat exchanging engineering appliances.

Dry air as a pure homogeneous mixture of many chemical compounds and elements at the human eye visual resolution level, is not formed by two or more constituents demarcated by identifiable interfaces. According to our definition it need not be defined as a porous medium. If on the other hand, at the same visual resolution level, if a region in space contains air plus moisture (water vapor) with identifiable interfaces, with one of these constituents fixed, then that mixture could be identified as a porous medium.

Also, the concept of homogeneity is dependent on the visual resolution level. To treat dry air as a porous medium according to our definition, we need to go into a finer resolution level. Compared to the human eye level, in a sufficiently finer resolution scale, dry air can bee seen to be made of a collection of separate homogeneous constituents (like atoms of oxygen, nitrogen etc.) with possibly identifiable interfaces between them. At such a resolution, if one of the constituents is assumed stationary, air could be identified as a porous medium. However, if our goal is to model transport phenomena, there is not much use defining air as a porous medium this way. Instead, a continuum model of air would suffice.

Based on the above discussions, it is obvious a metal mesh insert or a layer of sand is a porous medium. What about the example pictures given in Fig. 2; do they satisfy the definition of a porous medium? They all do, under some assumptions.

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Figure 2: Non-traditional Examples of Porous Media

All of them have at least two constituents with an identifiable interface between them even at the level of human eye visual resolution. For the cooked rice, the solid constituent doesn’t remain solid always. Before it is cooked, the rice is dry and solid and is surrounded by air in the interstices. However, when this air is replaced and saturated with water and cooked, the rice swells and some water percolates into the rice. The final cooked product, although can be assumed to be conserving mass, would occupy perhaps a larger volume. The volume change is both because of the phase change in the water and in the once solid rice. However, if we have the cooked rice-water mix in a fixed container and allow the water to drain off, the remaining cooked rice would be surrounded by air in the interstices as before. But the rice would have swelled and so the voids would occupy lesser volume. This can also be stated as the volumetric porosity of the rice-air porous medium has reduced in time. The picture not only shows these voids but also the bigger pores that were formed when the water vapour evaporated into the surrounding.

The bread is a porous medium where the pores are not always connected. The pores on the surface and those that are internal – what we would see when we slice it – do not have a connection between them. If we blow air gently from one side, it need not necessarily come out on the other side of the loaf or slice. This sort of porous medium, if the solid dough of bread is replaced by a metal, can create local hot spots due to lack of cooling thorough flow.

Unlike the cooked rice and bread, the human hair scalp and washing net porous medium have mobile solid constituents. This allows the possibility of several geometrical structures out of the same porous medium configuration. Usually, the hair or the solid of the washing net is assumed stationary, when such configuration are treated as porous media for analyzing flow or heat transport. One such analysis is given in reference [3]. Observe also that in all the examples in Fig. 2, the solid constituent is slightly deformable, a phrase we used in the definition. This deformability can be modeled using perturbation methods and in principle, doesn’t alter much the form of the governing conservation equations for studying transport phenomena.

On the other hand, if we are considering a strand of human hair in Fig. 2c, then it is not a porous medium. In the human eye visual resolution it is made only of one material with no identifiable interfaces. If we refine the resolution by putting the hair strand under a microscope, then it would show a porous structure even before we reach atomic level. To study the heat and fluid transport through a hair strand at this resolution level, by treating the hair strand as a porous medium may not be worthwhile.

An extension of a porous medium where the solid constituent of the parent porous medium itself is a porous medium, is identified as a bi-disperse porous medium [4]. Possibilities of extensions into several such resolution levels yielding finer and finer (micro and nano level) porous structures have suggested treatment of porous media geometry as fractals [5].

More basic concepts and terminology that characterize a porous medium are to be understood, before one formulates the governing conservation equations of mass, momentum and heat for analyzing transport phenomena in such porous media. Some of the terminology are porosity, connectivity, consolidation, percolation, tortuosity, homogeneity, isotropy and anisotropy, permeability, form coefficient, dispersion. These are discussed in the next essay.

References

  1. D. A. Nield and A. Bejan, Convection in Porous Media, 2006, Springer.
  2. The Wikipedia definition http://en.wikipedia.org/wiki/Porous_medium
  3. A. Bejan, Surfaces covered with hair: optimal strand diameter and optimal porosity for minimum heat transfer. Biomimetics 1 (1991), pp. 23-38.
  4. bi-disperse-porous-media
  5. B.M. Yu and J.H. Li, Some fractal characters of porous media, Fractals 9 (2001), pp. 365-372.

Categories: Fluid Sciences · Lecture Notes · Porous Medium · Science Notes
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The Koch Curve

July 29, 2008 · 2 Comments

The Koch Curve or the von Koch snowflake was discovered by Helge von Koch (1870-1924) in 1904. It is a closed fractal curve of infinite length within a finite region of space, enclosing a finite area. The construction of such a curve and some of its unique properties is explained in this note.

Take an equilateral triangle as shown below in Figure 1, with sides of, say, 1 unit each. This is the zeroth iteration of the Koch curve.

For the first iteration we split each side of the equilateral triangle into 3 equal parts, with the middle 1/3rd being replaced essentially by another smaller equilateral triangle made of side of 1/3 unit. The resulting figure is shown in Fig. 1b.

What we have done in the first iteration is to replace each side made of 1 unit stick by four, 1/3 unit sticks, arranged such that their end points remain fixed between the original 1 unit length of iteration zero. This procedure automatically fixes the angles of the protruding equilateral triangle in Figure 1b, and thus increases the original area confined by the equilateral triangle of Figure 1a, to that of Figure 1b.

Figure 1

By following the previous procedure, in the second iteration, we could now generate Figure 1c out of Figure 1b, with each of the sides of Figure 1b now being replaced in their middle thirds with even smaller equilateral triangles of sides made of 1/9 unit sticks.

Results of iteration three, four and five are shown in Figure 1d, e and f respectively.

The procedure could be repeated many times, to reach the von Koch Snowflake, when the iteration tends to infinity. There is an applet available at http://www.efg2.com/lab, for trying out more iterations and other Koch type of curves, starting from other basic geometrical figures instead of the equilateral triangle we used here.

Intuitively we could see that this resulting figure would have infinite length for its sides – if we could possibly draw it – while enclosing a finite area as shown in Fig. 1. How much would be the area when we actually do the infinite iterations?

To answer this let us put this procedure in a mathematical footing involving simple algebra.

The derivation for calculating the enclosed area done here follows the explanation given in [1]. Similar explanations are available in the Wikipedia page [2].

Let after n iterations (n = 0)

  • Nn = number of sides
  • ln = length of each side
  • Ln = length of perimeter = Nn × ln

For instance, if l0 = 1, N0 = 3, then L0 = 3.

Since we increase in every iteration the number of sides by a factor of four (compare Figure 1a and 1b) and since N0 = 3, then it means

Nn = 3 x 4n, where n = 0, 1, 2, ……

Further, for each iteration, the length of a side decreases by a factor of 3. for instance we used a stick of length unit 1 to form the side of Figure 1, while in Figure 2 each side of Figure 1 is replaced by 4 sticks of length unit 1/3 of the original. Since l0 = 1, we could write then,

l_n = 1\cdot\left(\frac{1}{3}\right)^n = \frac{1}{3^n}, n=0,1,2,\cdots

The perimeter after the n’th iteration becomes

L_n = N_nl_n = 3 \cdot 4^n\cdot \frac{1}{3^n} = 3 \cdot (\frac{4}{3})^n

which, in other words, means

lim _{n \rightarrow \infty }[L_n] = lim _{n \rightarrow \infty } \left[ 3 \cdot (\frac{4}{3})^n \right] = \infty

So, the length of this closed curve does actually goes to infinity. Now, for the area it encloses.

We know from our high school trigonometry that the area of a triangle can be calculated using the Heron’s formula

\sqrt {s(s-l_n)(s-l_n)(s-l_n)} = \sqrt { \frac {3}{2}l_n \times \frac{1}{2}l_n \times \frac{1}{2}l_n \times \frac{1}{2}l_n } = \frac {\sqrt {3}}{4}(l_n)^2

where ln is the length of a side, with s = 1/2 x (ln + ln + ln) = 3/2 x ln.

Using the above formula, for Figure 1, the zeroth iteration of the Koch curve, we could write

A_0 = \frac{\sqrt{3}}{4}

By comparing Figure 1 and Figure 2 we could see that we are increasing the area by adding 3 x 4 n-1 equilateral triangles (a triangle for each side) with ln = 1/3n. For example, with n = 1, we increase from A0 to A1, by adding three equilateral triangles, each with l1 = 1/3.The area of each of these triangles is obviously,

\frac{\sqrt{3}}{4}\left(\frac{1}{3^n}\right)^2 = \frac{\sqrt{3}}{4}\cdot\frac{1}{9^n}

and so for the figure after n iterations

A_n = A_{n-1} + 3 \times 4^{n-1} \times \frac {\sqrt {3}}{4} \times \frac {1}{9^n}

= A_{n-1} + A_0 \times \frac{1}{3} \times (\frac {4}{9})^{n-1}

For clarity, if we write explicitly for the first few n values

n = 1: A_1 = A_0 + A_0 \times \frac{1}{3}\left(\frac{4}{9}\right) ^0 = A_0\Big[1+\frac{1}{3}\left(\frac{4}{9}\right) ^0\Big]

n = 2: A_2 = A_1 + A_0 \times \frac{1}{3}\left(\frac{4}{9}\right) ^1 = A_0\Big[1+\frac{1}{3}\left(\frac{4}{9}\right) ^0 + \frac{1}{3}\left(\frac{4}{9}\right)\Big]

n = 3: A_3 = A_2 + A_0 \times \frac{1}{3}\left(\frac{4}{9}\right) ^2 = A_0\Big[1+\frac{1}{3}\left(\frac{4}{9}\right) ^0 + \frac{1}{3}\left(\frac{4}{9}\right) ^1 + \frac{1}{3}\left(\frac{4}{9}\right) ^2 \Big]

Therefore, in general we could write

\lim_{n\to\infty} A_n = A_0 \Big[1+\frac{1}{3}\left(\frac{4}{9}\right) ^0 + \frac{1}{3}\left(\frac{4}{9}\right) ^1 + \frac{1}{3}\left(\frac{4}{9}\right) ^2 + \cdots \Big]

= A_0 \Big[1 + \frac{1}{3} \Big( 1 + \frac{4}{9} + \left(\frac{4}{9}\right) ^2 + \cdots \Big) \Big]

The expression inside the braces of the above equation can be seen to be in a geometric progression, which then could be simplified as

\lim_{n\to\infty} A_n = A_0 \Big[1+\frac{1}{3} \cdot \frac{1}{1 - \frac {4}{9}} \Big] = \frac {8}{5} A_0

So, the area enclosed by the closed curve of infinite length is actually only 60 percent more than that of the original area of the equilateral triangle we started in Figure 1. A remarkable property indeed.

Categories: Lecture Notes · Science Notes
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Convection Carnot Engine

July 28, 2008 · Leave a Comment

ResearchBlogging.orgConvection can be distinguished into two types: forced and natural or free [1]. When thermal gradients cause a density gradient that result in local bulk motion in a fluid due to buoyancy forces, natural convection is realized [2]. The natural convection phenomenon can be thought as a Heat Engine [3]. A heat engine is a device that performs the conversion of heat energy to mechanical work by exploiting the temperature gradient between a hot source and a cold sink.

The analogy between natural convection and heat engine is explained in the figure below.

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Figure 1: Convection Carnot Engine Analogue


Heat engines run with a working fluid that undergo a thermodynamic cycle repeatedly to generate power. A thermodynamic cycle is a closed set of linked processes, when executed once, changes the thermodynamic state of a system (working fluid) through change in properties like pressure and temperature, bringing the system back to its original state. The conversion of heat (from, say, fuel) to work (that we use) happens in this cycle.

Natural convection can be thought of as a continuous thermodynamic cycle undergone by the convecting (working) fluid. Let us see how. In an ideal situation, the processes that constitute the natural convection cycle inside an enclosure are as follows:

  1. Process 1: Here the fluid packet considered to experience the force imbalance is composed of mass say dm. The actual movement of this packet is because it receives heat energy at the bottom from a heat source isothermally because of which it immediately expands. In effect, it receives heat energy by undergoing an isothermal expansion process.
  2. Process 2: With this energy, it increases its volume and as seen earlier experiences a force imbalance resulting in the displacement, which causes further expansion as, the packet raises. This results in an adiabatic expansion process in which the packet moving up can perform some work.
  3. Process 3: As it goes up the packet loses the remaining heat energy to the surrounding at the top to maintain equilibrium. This it does as an isothermal compression process as it cools and contracts at the top.
  4. Process 4: Finally this cooled packet is what we get circulated back to the bottom, if we have to account for the lost mass in the bottom by mass conservation principle. This happens through an adiabatic compression process, as the packet further contracts as it comes down.

The fluid packet executes a cycle comprising heating – expansion – cooling – compression the processes that constitute a Carnot Heat Engine Cycle [4]. This heat engine cycle was an idea proposed by Nicolas Leonard Sadi Carnot [5] in 1824. We use this ideal cycle today as a reference theoretical maximum efficiency cycle for evaluating any real heat engine thermodynamic cycle like the Otto cycle or Diesel cycle.

Existence of a work producing potential in the Natural Convection flow inside enclosures can be made evident by inserting a propeller made of a light material across the flow path of the fluid packet. The circulating convection wheel in water can be visualized in the laboratory with strewn aluminum powder. In reality, the work output of this cycle, unfortunately, is sufficient only to accelerate the fluid packet against the viscous drag – the fluid brake opposing motion – it experiences on the way. This is explained in the Free Convection and the Rayleigh Number essay.

We could quantify this idea using simple scaling equations.

The buoyancy force generated in the above enclosure convection system is the chief cause for the work producing potential of the thermodynamic cycle. The cause for and effect of this buoyancy on the convection is explained in the Free Convection For Dummies essay.

As an order of magnitude, this work done by the convecting system as a result of the buoyancy force is given by

W ~ (ρU ΔT ⋅g ⋅α) (1)

where ρ is the density and latexα is the thermal diffusivity of the convection fluid, U is the velocity of the convection wheel, ΔT = (T1 - T2) is the temperature difference across the enclosure as defined in Fig. 1 and g is the acceleration due to gravity.

The heat energy added to the convection system, again as an order of magnitude, is given by

Q1 ~ (ρ ⋅cP ⋅ U ⋅ΔT )∕L (2)

where cP is the specific heat of the convecting fluid at constant pressure.

The efficiency of the Convection Carnot cycle can be defined as the ratio of how much useful work one derives from the rotating convection wheel (given in Eq. 1) and how much heat energy one spends to achieve this (given in Eq. 2). This ratio is given as

W-- ~ gL-α Q1 cP (3)

The notation on the LHS of Eq. (3) corresponds to conventional thermodynamic usage (see figure 1).

Typical to the convection system described in Fig. 1, the W ≪ Q and Q1 ~ Q2, which results in the low efficiency. The fluid packet received some energy in the enclosure bottom, which is heated. Subsequently, this energy will be dispersed from it into the surrounding fluid, depending on the strength of cP and α. Before all the energy from the packet diffuses, the fluid packet raises because of buoyancy. This displacement can lift a weight, i.e. do work. However, the displacement has to overcome the viscous resistance and the thermal diffusion along the way for which some of the energy is used. The remaining energy in the packet can be realized as work output.

The convection situation described obeys the First Law of Thermodynamics. However, after overcoming the fluid viscous dissipative brake and thermal diffusion, the remaining kinetic energy in the packet is almost a small perturbation in the First Law of Thermodynamics. When Convection is modeled as a Carnot engine, it should be completely reversible.

From the Second Law of Thermodynamics one could infer the change in entropy for such an ideal convection cycle is zero. In reality, the viscous dissipation and diffusion/conduction irreversibility generates enough entropy to
restrict the work potential of the convection heat engine. This is embedded in the construction of the dimensionless group called the Rayleigh Number.

A recent research paper by Prof. V. A. F. Costa [6], explores nicely, the role of viscous dissipation in natural convection engine [7] discussed here. Additional references that explain the convection process are provided.

References

  1. Convection that cannot be identified distinctly as either forced or free is mixed convection. Free and Forced convection are explained in this note [Link]
  2. Read 1) Free Convection for Dummies [Link] and 2) Free Convection and Rayleigh Number [Link] for an explanation of natural convection phenomenon.
  3. Heat Engine Link [http://en.wikipedia.org/wiki/Heat_engine]
  4. Carnot Cycle Link [http://en.wikipedia.org/wiki/Carnot_cycle]
  5. Sadi Carnot [Wikipedia Link]
  6. On natural convection in enclosures filled with fluid-saturated porous media including viscous dissipation, V. A. F. Costa, International Journal of Heat and Mass Transfer, Volume 49, Issues 13-14, , July 2006, Pages 2215-2226 [Link to abstract]
  7. Narasimhan, A. (2000). Convective Carnot Engine Physics Education, 35 (3), 178-181 DOI: 10.1088/0031-9120/35/3/307
  8. Physical Fluid Dynamics, D. J. Tritton, (1988), Oxford Science Pub. Amazon Link
  9. Convection Heat Transfer, 3rd ed., A. Bejan, (2004), John Wiley and Sons Amazon Link
  10. Convection – M. G. Velarde and C. Normand, (1980), Scientific American
  11. Rayleigh Benard Convection: Structures and Dynamics – A. V. Getling, (1998), World Scientific Amazon Link

Categories: Lecture Notes · Science Notes · Thermal Sciences
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Scale Analysis

July 14, 2008 · 1 Comment

Scale analysis or order of magnitude analysis is a back of the envelope [1] method for deriving the most information with a reasonable understanding of the physics of a phenomenon. In other words, it is a method to yield first cut estimates about the relationships between several parameters involved in a particular problem. Let us take an example [2] to see how this works.

Suppose we have a thin plate of transverse width D (see figure) is immersed in a hotter fluid at time t = 0. The question we have is at what time the center of the plate will “feel the heat” so that the temperature of that location starts rising from then on, from its initial temperature. Quenching [4], a process of engineering interest, is the reverse of this process, where the plate is “cooled” by the fluid.

Figure 1 Schematic of heating a thin plate by hot fluid

To answer this question let us simplify things a bit and assume that the height and thickness of the plate are such that one can consider an unsteady (changing in time) conduction heat transfer process to govern the temperature increase in the plate. The one dimensional heat equation governing the transient conduction heat transfer process in this thin plate is then given by

\rho c_P\frac{\partial T}{\partial t}=k\frac{\partial ^2T}{\partial x^2} \cdots (1)

where T is the temperature, t the time, x the spatial position, c_P is the specific heat capacity at constant pressure, k the thermal conductivity and symbol \rho , the density of the material that undergoes the transient conduction heat transfer.

Owing to the symmetry of the situation (see figure) we can concentrate on one half of the plate (D/2) for making an order of magnitude estimate of each term in Eq. (1) above. The LHS of Eq. (1) for instance can be \textit{scaled} as

\rho c_P\frac{\partial T}{\partial t} \sim \rho c_P\frac{\Delta T}{t} \cdots (2)

Observe the Tilda instead of the equality symbol in Eq. (2). Also, the delta T in Eq. (2) is the temperature difference possible for the system (Fig. 1) in the time t (unknown) and is unknown as such and we can make a guess about it once we write the scale for the RHS of Eq. (1) as well. Proceeding to do so, the RHS of Eq. (1) can be \textit{scaled} as

k\frac{\partial ^2T}{\partial x^2} = k\frac{\partial}{\partial x}\Big( \frac{\partial T}{\partial x}\Big) \sim k\frac{1}{(D/2)}\frac{\Delta T}{(D/2)} = k\frac{\Delta T}{(D/2)^2} \cdots (3)

Observe in Eq. (3) as well there appears a \Delta T whose exact value is unknown. Using Eqs. (2) and (3) in (1), we can find a time scale as

t \sim \frac{(D/2)^2}{\alpha} \cdots (4)

where \alpha is the thermal diffusivity of the material of the plate.

If Eq. (4) is the answer to our problem of when the center of the plate will start feeling the heat or will begin to rise its temperature, then the two unknown \Delta T values in Eq. (2) and (3) must be equal. In fact, it is the only instance when the \Delta T across the spatial distance of D/2 (see figure) will match exactly with the \Delta T for the time duration t (measured from the initial time of t = 0). Hence our assumption about \Delta T while finding the answer in Eq. (4) is correct. The result in Eq. (4) for determining the heat penetration time compares well with the results from the exact analysis to the problem.

And the solution method is that simple.

Let us proceed with some more observations on the basic rules [2] for using this technique on other situations/equations.

It is necessary to define limits for the spatial region in which we perform the analysis. For instance, in the above example the x-axis is extended between 0 and D/2 (and not beyond that).

The philosophy behind this technique is every equation, in a physical sense, is a balance between two dominant scales resulting in the effect or phenomenon that equation is trying to describe. If this is not so, that is, if there are more than a few terms in an equation, one must find out the dominant terms for a particular situation so that the equation could be interpreted (reduced) as a balance between two terms.

For identifying the dominant scales, in general, these rules could be used

For an equation of the form c = a + b, if the order of magnitude of one term is greater than the other, i.e. O(a) > O(b) , then the order of magnitude of the sum c = a + b is determined by the dominant term, i. e. c = a + b can be interpreted in a scaling sense as O(c) = O(a). Similar conclusions hold for c = a – b.

If on the other hand, in an equation c = a + b, O(a) = O(b), then one term cannot be thrown out in comparison with the other. Both of them have to be treated with respect and the only scaling argument one can derive is O(c) \sim O(a) \sim O(b) .

In a product c = ab, the small is as weighty as the big and so the only scaling rule is of the form O(c) = O(a)O(b). Similar arguments hold for fractions (c = a/b) as well.

While interpreting the result from a scale analysis, one should be careful about certain things. For instance, the result of such an analysis will be accurate only to an order of magnitude, i.e. if the result predicts a scale of order 1, the actual answer could be anywhere within the next order on both sides (0.1 to 10). This may sometimes lead to erroneous and confusing interpretations. For instance if the result from a scale analysis suggests certain parameter group is of order one, say, C \sim 1 and if the actual answer is found to be C^2 = 0.1 , it doesn’t mean the scaling analysis is necessarily wrong. The correct interpretation lies in the fact that since the scale analysis yields a result of order one for C, it allows C to be even less than 1, which when squared (i.e. C^2 ) could lead to an answer (the correct one that matches with the exact answer, in this case) that is even smaller.

For instance, such an occasion appears when one compares the results from scaling arguments with the more exact results for the prediction of the thermal entrance length in a duct flow exhibiting hydrodynamic and thermal boundary layer growth. The scaling could still be correct; only the interpretation isn’t.

On the other hand, scale analysis as a solution technique finds more use and relevance with differential equations rather with integral equations, a reason for its lack of use in such fields as radiation heat transfer, wherein the phenomenon is modeled more with integral and/or integro-differential equations.

Scale analysis or order of magnitude analysis is practiced both as a research and pedagogical tool for more than a century. The earliest example in the field of fluid flow could be the order of magnitude analysis performed by Ludwig Prandtl [5] to reduce the formidable Navier-Stokes equations [6] into a slightly less formidable boundary layer equations, valid inside a boundary layer [6] – the region of fluid flow near a solid wall, where viscous effects dominate in the fluid.

Many researchers have used this scale analysis with remarkable success in delineating good physical insights about complex problems of scientific relevance. Some applications of this method to a wide spectrum of problems are discussed in [3].

References

  1. http://en.wikipedia.org/wiki/Back_of_the_envelope
  2. The example and the basic rules are discussed in Convection Heat Transfer, by A. Bejan [ Amazon Link]. The same example is also discussed in [3] below. Some observations are mine.
  3. Qualitative Methods in Physical Kinetics and Hydrodynamics, by Vladimir Krainov [Amazon Link].
  4. http://en.wikipedia.org/wiki/Quenching
  5. http://en.wikipedia.org/wiki/Ludwig_Prandtl
  6. http://en.wikipedia.org/wiki/Navier-Stokes_equation
  7. http://en.wikipedia.org/wiki/Boundary_layer

Categories: Lecture Notes · Science Notes · Thermal Sciences
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First Law and Fourier Law

July 12, 2008 · 1 Comment

First Law of thermodynamics relates heat transfer, work transfer and internal energy of a body. Fourier Law of heat conduction proposes how heat transfers in a solid body. We shall heat a block and see how hot the understanding gets.

Observe the figure below. But for the faces at x = 0 and x = L, the rest of the faces (four of them) are “insulated” such that energy cannot cross into or out of this block – our system of interest – as heat. Add to this we are not doing energy transfer as work on this block (like pushing or pulling it; squeezing or shearing it; electrocuting or “magneto-cuting” it; you get the idea). Also, the block isn’t doing any work interaction on us – the surrounding.

fourier_block.png

Figure 1

Applying the First Law of Thermodynamics on this block, since it is a closed system that allows energy interaction with the surrounding (we are heating it, for instance) but not mass flow in or out (the block remains as a solid while we heat and investigate), we can put succinctly what happens to the block as

\delta Q - \delta W = dE (1)

The above equation says the difference in the sum of heat transfer interactions and the sum of work transfer interactions for the block should be equal to the change in the total energy of the block. Or, if there are no energy interactions from or into the block, then the sum total of the energy of the block remains constant. All the terms in Eq. (1) are hence expressed in Joules, the unit of measure for Energy in SI units.

Now we have an equation that characteristically doesn’t contain information about the time involved for the energy interaction processes [as explained here, in an earlier essay]. To bring in this information, let us express Eq. (1) as a rate equation, (i.e. variables expressed as divided per time) resulting in

q - w = \frac{dE}{dt} (2)

In Eq. (2) above each of the term is expressed in Watts ( = Joules / second). Let us take the small elemental slab from the block (the darkened volume in Fig. 1 lying between x and x + dx) and assume two heat transfer interaction take place across the surface area A of the block, one at location x and the other one at (x + dx). As we assume until now no work interactions exist for the block, we an nullify the second term on the LHS of Eq. (2). This would result in

q_{x} - q_{x+dx} = \frac{dE}{dt} (3)

At this stage, if one were to know how to measure the RHS, i.e. the time rate of change of the total energy of the block, then one has a way to measure of what is the heat transfer interaction happening to the block (heated or cooled or neither). The vice versa situation where if one were to measure the LHS, i.e. the heat transfer interactions in time, it would allow one to know of the energy inventory of the block in time.

Unfortunately, neither of this could be done. We neither have an energy meter, so to say, that directly measures the RHS nor a heat meter, so to say, that directly measures the LHS. So we model and simplify and relate the non-measurable to the measurable. Since the block is a thermodynamic closed which precludes bulk mass exchange with its surroundings (minor few molecules escaping out here and there from the block is excluded in this continuum approach), one can express the RHS of Eq. (3) in a derived form. For doing this we should know we can express the total energy E as

E = \rho (A dx) u (4)

where u is the specific (per unti mass) internal energy of the block and the product of the rest of the symbols in RHS should result in a mass of the block (density times and elemental volume – see Fig. 1). This is possible because we can safely assume the block in Fig. 1 to have negligible Kinetic and Potential energies, at least during the heating or cooling process is going on, just to focus our attention only on that. This results in the total energy to depend only on the change in the internal energy of the block. Further, the change in this internal energy, being something that is not accounted for in KE and PE, can be related to the change in the temperature of the block as

du = c \cdot dT (5)

where c is the specific heat of the block material. Using the above two simplifications, the RHS of Eq. (3) can be rewritten as

\frac{\partial E}{\partial t} = \rho c (A dx) \frac{\partial T}{\partial t} (6)

Now for identifying the measurable quantities. Equation (6) involves in the RHS the density, specific heat, the cross section area of the block (see Fig. 1), the length along which the heat current travels and the change in the temperature in time. All of these are measurable with fair accuracy using our existing technology, at least for the block in consideration. Since the block is a solid, an incompressible substance, one doesn’t have to wonder about whether this specific heat is measured at constant pressure or constant volume. Both of those experiments would result in identical values for ’c’ for solids (for gases these two specific heats will be different). Temperature can be measured using a thermometer or a thermocouple.

Now if one were to use Eq. (6) to measure indirectly (i.e. by measuring the RHS of Eq. (6)) the change in the total energy of the block (the LHS of Eq. (6)) and substitute this in Eq. (3), one could obtain the heat interaction in time for the block. Since Eq. (6) is still in its differential form, we shall keep it that way for the moment and proceed to look at what we can do with the LHS of Eq. (3), the heat interaction.

We are struck with measuring q directly. Unless we know how to do this, there ends, perhaps prematurely, what First Law can tell us about the heating of the block.

Enter Jean Baptiste Joseph Fourier (some argue it should be Enter Jean Baptiste Biot). He showed a way out with his creativity, by assuming the local heat current q to be proportional to the local spatial temperature difference, as in for location x,

fourier

q_{x} \propto - \frac{\partial T}{\partial x} (7)

This proportionality in (7) was later resolved using experiments so that Eq. (7) is written nowadays as

q_{x} = - k \cdot A \cdot \frac{\partial T}{\partial x} (8)

where ‘k’ is a material property (of the block, in our case) called the thermal conductivity. Equation (8) is called the constitutive equation for k because that is the one which defines k and hence using only which one could measure k (as done in experiments to measure k).

However, what Eq. (8) essentially does for us is that it removes the impediment of having to measure q directly and, as suggested by Fourier, allows us to find q by measuring only the local temperatures – using thermometers or thermocouples.

By the way, the negative sign in front of the RHS term indicates the heat current flows in the direction of the negative gradient of the temperature, i.e. from higher temperature to lower temperature. In the Fig. 1, temperature is assumed to decrease along the positive x, the direction of the heat current vector.

Equation (8), the Fourier Law, is distinct from Eq. (2) or (3), the First Law of thermodynamics written for the block.

Proceeding further in rewriting the LHS of Eq. (3), we can see the q at location (x + dx) can be written using a Taylor series expansion truncated to the first two terms as

q_{x + dx} = q_{x} + \frac{\partial q_{x}}{\partial x}dx (9)

What we say in Eq. (9) is that we can find the q at location (x + dx), very small ‘dx’ distance away from x as the q at location x and its spatial change between x and (x + dx) times the distance dx itself. Simple, isn’t it? Of course, the second order variation as in the change in the change of the q at x can be found and added to this, but we assume these variations to be minimal when compared to the first derivative change and neglect them. This is what we meant by saying we truncate the Taylor’s series with the first two terms itself.

Now applying Fourier’s assumption at location (x + dx), we could easily rewrite Eq. (9) as

q_{x + dx} = - k \cdot A \cdot \frac{\partial T}{\partial x} + \frac{\partial }{\partial x} \left(- k \cdot A \cdot \frac{\partial T}{\partial x}\right) dx (10)

Combining Eq. (6), Eq. (8), Eq. (10) and Eq. (3) and rearranging and canceling out some terms appearing on both LHS and RHS, we can write the First Law as applied for the block in Fig. 1, in terms of measurable quantities as

\frac{\partial}{\partial x} \left(k \cdot \frac{\partial T}{\partial x} \right) = \rho c \frac{\partial T}{\partial t} (11)

To make this equation even more general, now we can add one more missing variable, the work term that we kept until now as zero. Work interaction can happen in a block such as that in Fig. 1 by means of stretching or shrinking of the block or by passing electricity through it. Let us say we pass electricity through the block. By doing this we are doing work ‘on’ the block system. The result of this work transfer interaction is to add energy to the block. However, the block could either store it or release it back to the surrounding. This means, it has to be accounted for in the First Law energy balance, as written in Eq. (2).

The work interaction of passing an electric current through the block is modeled usually as a heat generation quantity measured in Watts per unit volume of the block. This heat generation is understood to have its origin in the ohmic losses that the electric current exhibits, resulting in that losses to dissipate out of the block as heat transferred to the surroundings. In other words, the block should get hotter because of electric current passage, if it doesn’t throw out this heat ‘generated’.

Another example where direct heat generating is involved is that of a fissile nuclear material, the fuel inside a nuclear reactor, which generates heat internally because of nuclear fission – which, in turn, we take out by means of an heat exchanger device to generate steam and run turbines to generate electricity.

We are now in a position to write First Law of Thermodynamics for the closed system of the block shown in Fig. 1 as

\frac{\partial}{\partial x} \left(k \cdot \frac{\partial T}{\partial x} \right) + \dot{q} = \rho c \frac{\partial T}{\partial t} (12)

The first term in the LHS of the above Eq. (12) is the longitudinal heat conduction that is the resultant difference between the q at x and (x + dx) location in Fig. 1. The second term is the heat generation, in principle, the work interaction term. Both of these terms bring in or take out energy from the block. The total of these two should be balanced by the energy inventory of the block, represented by the RHS term, also known as the thermal inertia of the block. That is, this term determines how hot the block should get to , for a given energy interaction (on the LHS). Lesser the thermal capacity (density times specific heat), quicker the temperature raise of the block.

Obviously this means, in steady state, where we mean the temperature anywhere inside the block is not changing with time, the RHS should go to zero. Also, if there is no heat generation involved for the block, the second term of Eq. (10) also goes to zero. In other words, the steady state First Law energy balance for the block in the elemental volume between x and (x + dx) simplifies to

\frac{d^{2} T}{dx^{2}} = 0 (13)

This means, in Fig. 1, the energy that enters as heat current q at x (in Watts) promptly leaves at the location (x + dx), leaving no trace on the energy inventory of the elemental block volume, so that it doesn’t register a temperature increase in time.

If we write this for all the three directions x, y and z in Cartesian coordinates, then it would look like,

\frac{\partial ^{2} T}{\partial x^{2}} + \frac{\partial ^ {2} T}{\partial y^{2}} + \frac{\partial ^ {2} T}{\partial z^{2}} = 0 (14)

and if you believe in brevity as the wit of the mathematics soul, then we can express Eq. (14) also as

\nabla ^{2}T = 0 (15).

Remember, we have only formulated the First Law statement in differential form and are yet to solve this equation.

Using the conditions of all the four faces being insulated but for the left and right for the block in Fig. 1, we can safely assume the heat transfer interaction in the block to be one dimensional so that we can use Eq. (11) for analyzing its steady state behavior. To solve Eq. (13), we require at least two conditions, i.e. two correct answers about how temperature behaves with respect to space coordinate x. These two conditions are given for us in Fig. 1, if we take the temperature of the block at x = 0, and x = L to be known. In other words, we are measuring the temperatures at these two locations. This allows us to solve Eq. (13), the solution of which looks like

T = T_0 + (T_L - T_0) \cdot (k / L) (16)

Using this information, we can find out how exactly the Fourier assumption in Eq. (6) looks like. Finding the first derivative of T from Eq. (15) and using it in Eq. (8) we obtain

q^{,,} = \frac{k}{L} \left(T_0 - T_L \right) (17).

Now the heat flux q” (Watts per square meter – q across area A in Fig. 1) can be independently measured by other means. Since the temperatures on the RHS of Eq. (17) are measured at locations x = 0 and x = L, the only unknown ‘k’ the thermal conductivity of the material of the block can be determined this way.

And for doing it, we now have cleverly coupled in Eq. (17), the distinct results of First Law, Eq. (3) and Fourier Law, Eq. (8).

Major References

  1. Heat Transfer by A. Bejan, 1993, John Wiley and Sons [Amazon link for book]
  2. A Heat Transfer Textbook, 3rd edition by J. H. Lienhard V and J. H. Lienhard IV [webpage of the book] Link: http://web.mit.edu/lienhard/www/ahtt.html
  3. Some of my lecture slides and figures
  4. Objectives of Thermodynamics and Heat Transfer

Categories: Lecture Notes · Science Notes · Thermal Sciences
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