Process Analysis and Modeling



Differential equatıons

The most common mathematical structure employed in mathematical models of chemical engineering profession involve differential equations. These equations describe the rate of change of a dependent variable with respect to an independent variable. They can be expressed in any of the three coordinate systems mentioned earlier in Section 2.1.2. The independent variables may also include time as the fourth element in the description of a problem. Let’s first discuss the mathematical origin of differential equations.

1 Mathematical origin of differential equations

An algebraic equation that involves a parameter describes a family of curves. For example, equation 5.1 describes the family of curves shown in Figure 1.

[pic] (5.1)

[pic]

Figure 1 Family of curves for the algebraic equation [pic](Mickley, Sherwood & Reed)

Using differentiation and algebraic manipulation equation 5.1 can be expressed as a differential equaiton free of the parameter c. Let’s take the derivative of both sides with respect to x in this equation.

[pic] (5.2)

Substituting the value of c, [pic],obtained from the original equation, the family of curves is expressed as

[pic] (5.3)

Equation 5.1 is known as the primitive of the differential equation represented by equation 5.3. If there are two parameters in the primitive equation we need to differentiate twice to eliminate the two parameters. Solving a differential equation means finding the primitive either by analytical or numerical means. So, usually during a modeling exercise the modeler comes up with one or more differential equations that describes the system. These equations are then solved to obtain the algebraic representation, the primitive equation, for the model.

Q.5.1 What is the equivalent differential equation describing the primitive equation expressed by [pic]?

2 Definition of terms

There are several important terms used in the context of differential equations. These terms classify the various differential equations into groups according to certain features. Most of the time, this classification enables one to select a suitable solution method.

1 Order

The order of a differential equation is the order of the highest derivative in the equation.

2 Degree

The degree is the power the highest order derivative is raised after the equation is cleared of fractions and rationalized.

3 General solution

The solution of a differential equation that involves constants is the general solution. In the previous section, equation 5.1 is the general solution or the primitive to differential equation 5.3. Applying various boundary and/or initial conditions the particular curve out of the family of curves can be specified. The order of a differential equation determines how many separate initial and boundary values can be specified. For example, a second order differential equation needs either one initial and one boundary value or two boundary values. If there are more than that the problem might be overspecified.

4 Particular solution

When all the initial and boundary values are applied to a general solution, the constants are specified and a particular solution is obtained. For example, [pic] is a particular solution to the differential equation 5.3.

3 Classification of differential equations

The two top most classes for differential equations are Ordinary Differential Equations (ODE) and Partial Differential Equations (PDE). An ODE is a differential equation that has only one independent variable in its definition, whereas a PDE has more than one. Examination of the physical setting and problem description usually gives clues about the system and hint if the system should be modeled by ODEs or PDEs. In this chapter we will concentrate on ODEs. The most general form of an ODE with one independent variable x and one dependent variable y can be expressed as

[pic] (5.4)

Any y value that satisfies equation 5.4 is said to be the solution of the differential equation. The most general algebraic solution to this differential equation that ties y and x is

[pic] (5.5)

It is mathematically proven that each equation of the form 5.4 must have a solution of the form 5.5, whether we are able to find it or not. Let’s examine the solution methods for first order ODEs.

1 First order and first degree ODEs

Any first oder and first degree ODE can be expressed as

[pic] (5.6)

Depending on the functional form of M(x,y) and N(x,y) in equation 5.6, there are various solution methods.

1 Separable equations

If M and N in equation 5.6 are functions of only x and y, respectively, the equation can be restated as

[pic] (5.7)

The solution to equation 5.7 is obtained simply by integrating both sides of the equation.

[pic] (5.8)

2 Equations that become separable after change of variable

There is no universal method that can find the right variable change and transforms the problem to a simpler form. However, let’s illustrate the idea with an example. Consider the equation below

[pic] (5.9)

Equation 5.9 can be easily solved by the use of variable change xy=v.

Q.5.2 What is the solution to equation 5.9 after the variable change xy=v?

3 Homogeneous equations

A function f(x) is said to be homogeneous of the n-th degree if when x multiplied by a value t results in a funtional value [pic]. Simillarly, If a function is homogeneous in x and y the below equation should hold.

[pic] (5.10)

For the differential equation 5.6 where M(x,y) and N(x,y) are both homogeneous functions in x and y of the same degree the change of variable y=ux leads to a solution as shown.

[pic] (5.11)

Q.5.3 Show the derivation of equation 5.11 with the substitution y=ux.

4 First order first degree equations with linear coefficients

If M(x,y) and N(x,y) are linear functions of x and y so that equation 5.6 can expressed as

[pic] (5.12)

the following substitutions usually transforms equation 5.12 to a homogeneous one.

[pic] (5.13)

The constants m and n are found solving the following equations

[pic] (5.14)

Q.5.4 This method would fail if [pic]. If this is the case one can eliminate y from equation 5.12 by subtituting [pic]. The resulting equaion would be separable in x and w. Proove this method and suggest a general solution.

5 Exact equations

If the partial deriatives of M(x,y) and N(x,y) with respect to y and x, respectively, are the same, the left hand side of equation 5.6 is said to be an exact differential. The solution can be expressed by

[pic] (5.15)

or

[pic] (5.16)

Q.5.4 Consider the following differential equation

[pic]

a) Proove that the left hand side forms an exact differential.

b) What is the general solution?

6 Equations solvable by integrating factors

An integrating factor transforms a differential equation into an equation where the left hand side is an exact differential. Consider the below differential equation

[pic] (5.17)

The left hand side of Equation 5.17 forms an exact differential. However, the same equation can also be simplified by dividind both sides by 2y to give

[pic] (5.18)

Equation 5.18 does not have an exact differential on its left. To solve it we need to realize that 2y is an integrating factor that transforms the left hand side to an exact differential.

7 First order linear equations

The form of these type of differential equations can be rewritten as

[pic] (5.19)

where P and Q are either constants or functions of x only. [pic] is an integrating factor for this differential equationand the solution is

[pic] (5.20)

8 Bernoulli’s equation

The form of these type of differential equations can be expressed as

[pic] (5.21)

where P and Q are either constants or functions of x only. When equation 5.21 is divided by [pic] and the substitution [pic] is used it becomes a linear first order equation and can be solved as described in section 5.3.1.7.

9 Other integrating factors

Finding an integrating factor is usually a difficult task. However, there several standard forms which have known integrating factors. Three of these forms are below.

If [pic] is true, then [pic] is an integrating factor.

If [pic] is true, then [pic] is an integrating factor.

If [pic] and [pic] hold, then [pic] is an integrating factor.

10 Equations of the first order and higher degree

Equations of the first order and higher degree have the following form

[pic] (5.22)

Equation 5.22 can be solvable for dy/dx, y or x. Depending on this criterion there 3 cases to be considered for solution .

Case I. Eqations solvable for dy/dx.

Let’s investigate this case with an example. A typical equation is as follows

[pic] (5.23)

Equation 5.23 can be rewritten as

[pic] (5.24)

Equation 5.24 has the following two solutions

[pic] (5.25)

These two solutions can be combined by multiplication to give the final solution as

[pic] (5.26)

In differential equations of first order and first degree for every value of the independent variable there is only one corresponding slope at that point. However, for equations of n-th degree there are n slopes specified at every point.

Case II. Eqations solvable for y.

Differential equations solvable for y has the general form of

[pic] (5.27)

For solution first use the substitution dy/dx=p and then differentiate with respect to x. The result would have the general form

[pic] (5.28)

If equation 5.28 can be integrated to give an algebraic relation in x, p and c as

[pic] (5.29)

the final solution can be expressed after eliminating p between equations 5.27 and 5.29. If equation 5.28 is difficult to integrate, it is also possible to express the final solution as a parametric one by equations 5.27 and 5.29 where p is the parameter that specifies x and y.

Case III. Eqations solvable for x.

Differential equations solvable for x has the general form of

[pic] (5.30)

A more convenient version of equation 5.30 is obtained after differentiating with respect to y and substituting dx/dy=1/p.

[pic] (5.31)

After the manipulation the solution method outlined in the second case should be followed to obtain either a parametric or algebraic solution.

Q.5.5 Consider the following differential equation of first order and second degree

[pic]

a) Which of the three cases are applicable for this ODE?

b) Which case results into an easier solution path?

11 Clairaut’s equation

These ODEs have the general form of

[pic] (5.32)

Equation 5.32 resembles to Case II studied in Section 5.3.1.10. Following a similar approach let’s differentiate with respect to x and make the substitution dy/dx=p.

[pic] (5.33)

Equation 5.33 rearranged to

[pic] (5.34)

For equation 5.34 to hold either x+f’(p) or dp/dx should equal 0. The case dp/dx=0 leads to the fact p=constant. In that case the general case becomes

[pic] (5.35)

The second solution can be obtained by eliminating p from x+f’(p)=0 and equation 5.32. This solution will satisfy equation 5.32. So, it is a solution to the original ODE. However, as it does not contain any arbitrary constants, it cannot be the general solution. This type of solution is called the singular solution.

2 Second order and first degree ODEs

The general form of the this type of equations is

[pic] (5.36)

It is impossible to discuss all the possible equations that conform to equation 5.36 and suggest a solution method. However, there are several standard forms of equation 5.36 that are important in chemical engineering. These standard forms have well established solution methods. We will first consider equations with missing terms and then second order linear ODEs with constant coefficients.

Case I. Equations not containing y.

1) If dy/dx is not present, the form has to be

[pic] (5.37)

The solution can be found by integrating f(x) twice with respect to x.

2) If dy/dx is present, the form has to be

[pic] (5.38)

The substittion dy/dx=p will create a new first order ODE in p and x that can be solved by previously discussed methods. Please note that d2y/dx2=dp/dx.

Q.5.6 Consider the following ODE of second order and first degree

[pic]

What is the solution for this ODE?

Case II. Equations not containing x.

These ODEs would have the following general form

[pic] (5.39)

Using the substitution p=dy/dx ODE should be first reduced to one of the first order. However, the ODE will be easier to solve if d2y/dx2 is expressed as follows

[pic] (5.40)

Q.5.6 Consider the following ODE of second order and first degree

[pic]

What is the solution for this ODE?

Case III. Linear ODEs of second order and first degree with constant coefficients.

Actually, in the next section we will discuss the generalized linear ODEs. However, as ODEs of second order and first degree with constant coefficients play an important role in the chemical engineering control theory, we found it appropriae to discuss this class here in greater detail. Following the notation we used until now the ODE’s general form can be expressed as

[pic] (5.41)

However, it is a better idea to adopt time, t, as the independent variable and to follow the notation used in the control theory. The general form of the ODE is now expressed by

[pic] (5.42)

The parameters ζ and [pic]are called the damping coefficient and time constant, respectively. The function m(t) is said to be the forcing function. It is better to express the coefficients this way as they both relate to some specific physical facts in the control theory. In this section the details of solution methods are not given. The details will be pieced together when the general linear differential equations are discussed in the next section. For now let’s just concentrate on the tabulated solutions for the five critical regions of ζ around zero and unity. The full solution to the ODE in equation 5.42 requires the complementary solution first. Complementary solution is the solution for the case where the forcing function m(t) is assumed to be zero as shown below.

[pic] (5.43)

This equation, hence the complementary solution, describes the system’s response when there are no external disturbances. When the differential operator D is substituted for the derivative of y with respect to t, the characteristic equation for the ODE is obtained as

[pic] (5.44)

The roots of this characteristic equation determines what the complementary solution will be. Table Table 1 tabulates the five possible cases. For a system to be stable the damping coefficient has to be greater than zero. If it is equal to zero then the system is said to be undamped. It will sustain continuous oscillations. If the damping coefficient is less than one but greater than zero, then the system is underdamped. An underdamped system will go to its stable point through some oscillations. If the damping coefficient is greater than unity the system is called overdamped. An overdamped system will go to its stable point with no oscillations.

Table 1 Solutions for second order linear ODEs with constant coefficients

|Name |ζ |Roots, r1 and r2 |Solution, y(t) |

|Overdamped system |ζ >1 |[pic] |[pic] |

|Critically damped system|ζ =1 |[pic] |[pic] |

|Underdamped system |ζ ................
................

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