Computational Methods for Process Simulation

Particular emphasis should be placed on keeping force and mass units consistent. Before the conservation principle for total energy is applied to the general control volume illustrated in Figure 1. The net rate of work done by the system is given by W. V is the specific volume. It should be noted that the shear force is not zero at the surface, S. However, the viscous work is zero since it is the product of the system velocity times the viscous force. For the viscous work to be nonzero, the velocity at the control boundary surface, S, must be nonzero.

However, the no-slip condition states that the velocity at the surface S is zero. A second simplification occurs because surface area term is nonzero only at the entrance and exit of the tapered tube. First let's assume that the density is constant. This approach will help to include all significant energy mechanisms in the final model.

A heating coil having an area of 0. Calculate the outlet temperature after 1 hr. It has four significant terms. The first is the rate of accumulation of internal energy within the control volume, the second is the rate of sensible energy entering the system through the inlet flow to the system, the third is the rate of heat transferred to the system through the heating coil, and the final term is the rate of sensible energy leaving the system with the outlet stream.

Again the information-flow diagram is useful since it illustrates the rela- tionships betweeirthe describing equations. Also the cause-and-effect usage of the equations should be noted. The mass balance is used to solve for the total mass of the system and the energy balance for the temperature of the system. This set of differential equations can be solved analytically by separation of variables. It is left as an exercise for the interested reader.

The transient response of the system obtained is shown in Figure 1. The temperature of the tank after 1 hr is The dynamic response of the system with a 50 percent reduction in the heat transfer area is also shown. The temperature after 1 hr is lowered to I hours Time, r hours Figure 1. A slurry reactor shown in Figure 1.

This reactor is a well-stirred tank with catalyst particles dispersed throughout. The rate of such reactions is usually zero order, that is, they do not depend upon the reactant as product concentrations. Develop an analytical solution to the temperature response of the tank when there is a step change in the steam temperature, T s. We will now develop this isothermal version of the energy balance.

The right-hand side of this equation describes the irreversible processes which contribute to the change of state. For an isothermal process, equation 1. If we substitute equations 1. Using these two relations, we get the general form of the mechanical energy equation which is given in equation 1. To calculate the Gibbs free energy, G, we can use the following identity Denbigh, We want to compute the horsepower needed to pump water in the piping system shown in Figure 1.

Water is to be delivered to the upper tank at a rate of 5. All of the piping is ID smooth circular pipe. The pump efficiency is unity. Piping System for Example 1. The mechanical energy balance equation 1. Force and velocity are both vector quantities in Euclidian three space. Two types of forces make up the force f x.

They are body forces and surface forces. The main body force of consequence is that due to gravity. These are pressure forces, shear or viscous forces, and structural or resultant forces. The force f xd is the integrated x directed drag, friction or shear force. This force is normally defined in terms of the friction factor as Bird et al. The structural force, or resultant of the forces acting on the system is r x. This force appears when the control volume cuts through solid objects such as the containing wall of a pipe. Most applications fall into one of three classes: Calculate viscous frictional losses for a given flow.

Calculate the force for a given flow. Calculate the pressure change for a given flow. The first class of problems, calculating the viscous frictional losses, involves the use of the mechanical energy equation since these terms appear explicitly in that equation. The second class of problems, cal- culating a force, involves the use of the momentum balance since forces enter the momentum balance explicitly. Class three problems, calculat- ing the pressure, can be approached by using either the momentum or mechanical energy equations since pressure terms appear in both equa- tions. The choice of which equation to use depends upon what type of information is specified.

If viscous frictional losses are given, then the mechanical energy balance is used. If force information is specified, then the momentum balance is used. Water is flowing at 9. The upstream ID is 7. Calculate the resultant force on the nozzle when it discharges to the Macroscopic Mass, Energy, and Momentum Balances 35 atmosphere.

Consider that the nozzle is attached at its upstream end and that frictional forces are negligible. The control volume for this system as shown in Figure 1. The momentum balance equation 1. We will use equation 1. To compute the pressure drop, we use the mechanical energy balance equa- tion 1. Since the resultant force 36 Computational Methods for Process Simulation on the nozzle is negative, it is acting in the direction opposite to the flow. The pipe wall is therefore in tension.

A jet ejector or jet pump is a device with no moving parts that is widely used for moving liquids between tanks and lifting corrosive liquids. The ba- sic configuration is shown in Figure 1. The high velocity fluid or jet is introduced into a slow flowing or suction line. The mixing is rather chaotic, but a few diameters downstream, the flow is again uniform.

The downstream pressure is increased from the upstream suction pressure. We wish to deter- mine the downstream pressure. Either the mechanical energy or momentum balance can be used in principle. However, in this application, there is no easy way to compute the viscous frictional losses. This means that the mechanical energy equation should not be used. Instead, we can assume that because the distance from planes 1 and 2 is short, the drag force term in the momentum balance will not be important. We will therefore use the momentum balance for this problem. At steady state with the gravity, drag, and resultant forces set to zero, equation 1.

What are the errors if the flow is laminar? The bed is 10 cm in diameter and 20 cm high and has a porosity of 0. A research team is designing a flow system for a nuclear reactor to study corrosion problems. The equipment as constructed consists of 70 m to- tal equivalent length of 2. If the bismuth velocity is 0. A conical tank of height 30 cm and radius 10 cm is initially filled with fluid.

After 6 hr the height of the fluid is 27 cm. If the fluid evaporates at a rate proportional to the surface area exposed to the air, find a formula for the volume of fluid in the tank as a function of time. Analysis of the flow through an ejection molder see Figure 1. In your analysis assume: Calculate the average velocity in the nozzle.

The nozzle has a diameter of 0. The pressure at the nozzle entrance is kPa. What is the volumetric flow rate through the nozzle? What is the ram speed? What is the rate of work required to move the ram? Do not neglect frictional terms through the nozzle. A centrifugal pump is being used to supply 6.

The pump suction is 14 kPa and the discharge pressure kPa. The pump operates adiabati- cally. The temperature of the water in and out of the pump is carefully measured and is found to rise from 16 to Calculate the fraction of the energy supplied to the pump that is dissipated through friction. The suction and discharge lines to the pump are the same size.

The strained juice is concentrated in a vacuum evaporator to give an evaporated juice of 58 percent solids. This final concentrated juice contains 42 wt percent solids. Calculate the concentration of solids in the strained juice, the flow rate of the final concentrated juice, and the concentration of the solids in the pulpy juice bypassed. An artificial kidney is a device that removes water and waste metabo- lites from blood. In one such device, the hollow fiber hemodialyzer, blood flows from an artery through the insides of a bundle of hollow cellulose acetate fibers, and dialyzing fluid, which consists of water and various dissolved salts, flows on the outside of the fibers.

Water and waste metabolites — principally urea, creatinine, uric acid, and phos- phate ions — pass through the fiber walls into the dialyzing fluid, and the purified blood is returned to a vein. Suppose that at some time during a dialyzation, the arterial and venous blood conditions are as follows: Calculate the rates at which urea and water are being removed from the blood.

Neglect the urea volume. Suppose we want to reduce the patient's urea level from an initial value of 2. Neglect the loss in total blood volume due to the removal of water in the dialyzer. Agitators keep the contents of each tank uniform in concentration. Calculate the concentration of the second tank at that time. Find analytic solutions to: Macroscopic Mass, Energy, and Momentum Balances 41 a.

The following data have been generated in your laboratory for the de- composition of the very pungent sulfuryl chloride. You are requested to develop a rate expression for this reaction. Try both first- and second-order rate laws to determine which one best fits your data. The sulfuryl chloride can be considered as an ideal gas. Since two moles of gas are generated per mole consumed, the total pressure of the reaction vessel can be related to concentration. A cylindrical tank 1.

The tank is opened to the atmosphere. A discharge hole of The surface of the liquid is located 6. Calculate the time in seconds to do this. Be very careful in defining your control volume and making any simplifying assumptions. One you can make is the follow- ing: Develop analytic solutions to: Startup of an Equilibrium Still. Consider the case of starting the equilibrium still shown in Figure 1. The still is purifying benzene and toluene from a small amount of essen- tially nonvolatile impurity.

No liquid stock is removed from the still during this period. Using one component material balance and an overall material balance will ease the solution to this problem. Solve for the dynamic response of species A for the following set of reversible reactions taking place in a constant volume batch reactor.

Write two component material balances and an overall balance. Through substitution and differentiation of the A component material balance, get one differential equation which is only a function of Na and its derivatives. Solve the differential equation. Be sure to evaluate the integration constants using known boundary condition information. Analysis of a Turbojet Engine Figure 1. As the air enters the diffuser, it slows down and the pressure rises. The compressor further increases the pressure as the air is forced into the combustion chamber, where fuel is added and combus- tion takes place.

The combustion mixture passes through the turbine driving the compressor and expands in the exhaust nozzle. It leaves the nozzle close to atmospheric pressure and at a higher velocity than the entering air. Compute the velocity of exit gas. Compute the thrust force that the engine develops. Air flowing at a constant mass rate of 0. The heating is accomplished in a vertical heat exchanger of constant cross-sectional area. The exchanger is 4 m long.

When considering small subsystems of chemical plants, the number of describing relations are small and the develop- ment of a computational strategy is not difficult. Usually the relations can be solved directly by partitioning the equations, that is, solving each equation of the equation set for a single unknown variable in a sequen- tial manner. As equation sets become coupled, namely, as each relation involves more of the unknown variables, the probability that an equa- tion set can be partitioned decreases.

When an equation set cannot be partitioned, the equations must be solved simultaneously or an iterative scheme devised. In this chapter, numerical methods for simultaneously solving sets of linear and nonlinear algebraic equations are first discussed. Matrix techniques are used for linear equations. The mathematical notion of a matrix as a linear transformation is presented as well as the computer techniques such as Gaussian elimination for solving matrix algebra.

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A number of methods are presented for the solution of a single nonlinear equation and sets of nonlinear equations. Algorithms for structuring sets of algebraic equations into simple com- putational strategies are presented. In general, no solution, a number of solutions, or an infinite number of solutions may exist. We will be concerned with the case where m — n and we seek a single, real, physically meaningful solution. Systems of linear equations arise with great regularity in material balance problems.

An example of a two-dimensional linear system is 2a:! For example, consider the following three equations in three unknowns: The equations have been partitioned into a sequential solution strategy. Steady-State Lumped Systems 47 2. We can solve the four relations simultaneously. However, an alter- native approach to the problem is available. If not, a new value of rr 3 would have to be assumed. This procedure is called tearing the equations. The tearing the pick- ing of the iterative or recycle variable is not unique, and various criteria for what we mean by the best tear are possible.

Generally the best tear is dictated not only by the structure of the equations but by the way in which the variables enter into the equations. It is not always clear whether it is more efficient to solve all equations at once or to use tearing techniques. How- ever, for small sets of equations less than 50 equations to a set , si- multaneous solution to the describing equations becomes practical and advantageous.

This is especially true when the equations are linear and matrix techniques can be applied. If the rank is less than the number of unknowns, there are an infinite number of solutions Carnahan et al. Although a large number of techniques exist for solving a well-defined set of linear equations one having a unique solution , the most efficient methods are those based on the method of Gaussian elimination. In order to understand Gaussian elimination, consider the following set of three linear equations: Similarly, we replace the third equation, 2.

Namely, we solve for x 3 from equation 2. The Gaussian elimination procedure can therefore be considered as two operations, a forward pass and a backward pass. The objective of the forward pass is to transform the original matrix into an upper-triangular matrix. The backward pass calculates the unknown variables using the back substitution procedure.

An alternative method of Gauss-Jordan elimination is possible. It is a one pass method that eliminates the need for the back substitution phase of the Gaussian elimination procedure. Essentially, the Gauss- Jordan method transforms the original A matrix into an identity matrix through row operations.

This allows for the direct solution of the un- known x vector. While the Gauss-Jordan method will involve more computational effort for solving a single problem, it is more efficient when we have several right-hand side vectors b that need solution with the same A matrix. The Gauss- Jordan procedure is illustrated below with Ri standing for the original zth row and i? The steps are called factorization of the A matrix or LU Lower triangular, Upper triangular decomposition.

The solution is obtained by using a forward pass and then a backward pass.

Computational Techniques for Process Simulation and Analysis Using MATLAB®

To avoid this, partial pivoting strategies are employed. Partial pivoting means at the k th step we interchange the rows of the matrix so that the largest remaining element in the k th column is used as the pivot element Carnahan et ai, ; Ayvub and McCuen, Some computer programs for the solution of sets of algebraic equa- tions also allow for iterative improvement. The inverse of A, A -1 , can easily be computed from the LU factor- ization.

Its capabil- ities have expanded greatly in recent years and is today a leading tool for engineering computation. The MATLAB software consists of a basic package of mathematical routines as well as optional tool- boxes that cover specific engineering application areas such as process control, neural networks, optimization, signal processing and symbolic mathematics. These routines are some of the best implementations of numerical methods that are currently available.

MATLAB now offers a convenient computing environment for the solution of process simulation problems and will be used extensively in this book. An important command is the abort command AC or control C. Often a mistake has been made and extraneous output is being displayed or you find yourself in a seemingly endless loop.

At this point you want to abort the run. Multiple statements can be entered on the same line if separated by semicolons. A line can be continued to the next line by using The help command is an important resource. You can then select the topic on which you require information. Specific topics are then displayed. You can then enter help on the specific topic to learn details about it. In addition to executing commands entered through the keyboard, you can also use commands stored as files. These files all end in the extension.

Short programs are usually created and run directly on-line. In this case, left division is used since the matrix A is to the left of vector b. It is desired to develop the steady-state tray compositions for a six-plate absorption column. It can be assumed that a linear equilibrium relation holds between liquid x m and vapor y m on each plate: The system is shown schematically in Figure 2. To solve the problem, a material balance is written on a representative tray, n, shown in Figure 2.

A typical set of parameters for this problem is a - 0. This file is available from the world wide web un- der: This M- file first defines the constants for the problem. When you run this M-file, you will obtain the results shown as comment lines at the end of this file. Note that to the accuracy given in the default output the results are equivalent using either approach.

Steady-State Lumped Systems 57 Figure 2. A company plans to make commercial alcohol by the distillation process shown in Figure 2. The process contains two distillation columns, both having reflux ratios of three to one. The distillate from the first column is 60 percent alcohol while that from the second column is 95 percent. The bottom stream from the first column contains 80 percent of the organic material feed while the rest leaves the bottom of the second column. No alcohol is present in either of the bottom streams. Solve for the amount of each component in each stream. Component material balances are written for the system.

Alcohol Distillation Process for Example 2. This procedure helps to organize the problem and allows the engineer to catch any redundant dependent equations that may have been written can't put a value on the diagonal. It also aids in assuring that problems will not arise in using partial pivoting algorithms. It is convenient to start with the variables of the first stream and continue in the order of streams. For this example we started with stream 3 and ended with stream The equations are then numbered so as to fill the diagonal elements of the matrix. Again the file is available from the world wide web.

The results that you will obtain by running the file are given as comment lines. The variables have been identified and the units given in these comment lines, although that in formation will not appear when you run the program. Matrix Notation for Example 2. The elements of the diagonal matrix are called the signular values of A. One application of SVD is in determining the rank of a matrix.

In addition, SVD allows us to compute the condition number of a matrix using the function cond. A well-conditioned matrix has a condition number around unity. A very large condition number implies an ill-conditioned matrix. The norm of a matrix uses SVD. The function norm provides several norms see the help norm out- put.

We cannot invert A since it is not a square matrix. If we pre-multiply equation 2. When A is rank deficient, then S T S cannot be inverted because of very small or zero singular values. In this case, we only take the r non-zero singular values so that S becomes an r x r matrix where r is the rank of A.

W. Fred Ramirez (Author of Computational Methods for Process Simulation)

Sparse matrices arise in many engineering problems including those of material and energy balance calculations as illustrated earlier. It is more efficient to define matrices as sparse when appropriate. If you are not sure if a matrix is sparse, then you can use the function issparse A which returns a value of 1 if A is sparse and 0 if it is not sparse.

To define a matrix as sparse, we use the function sparse rowpos, colpos, val, m,n where rowpos are the positions of the nonzero row elements, colpos are the positions of the non-zero column elements, val are the values of the non-zero elements, m is the number of rows, and n the number of columns.

Steady-State Lumped Systems 67 2. The format is suitable for the 7, bisection and tangent methods: This defines the nonlinear relationship between heat capacity 7t and temperature Eq. The format is suitable for the 7. In general, they fall into derivative-free or derivative-based categories. Derivative-free methods are usually more stable, less sensitive to initial guesses, and converge less rapidly than derivative-based methods.

Half-Interval or Bisection Method. Steady-State Lumped Systems 69 For a case such as this, fundamental calculus says that there will be one real root in this interval. Note that we can determine the root graph- ically by plotting the function f x and estimating where it intersects the x-axis. The bisection or half-interval method is a systematic trial-and-error solution consisting of the following steps: Determine the midpoint of the interval, point c.

Repeat the previous step and calculate f x at the midpoint of the new interval at each step of the iteration and discard the half of the interval that does not contain the root. The result is that the interval containing the root will decrease by a factor of two each iteration. Thus by repeated application of this algorithm, the root can be determined to any degree of accuracy. For our example of equation 2. Use 7c "help fzero" for more information. By the principle of similar triangles we have, b 0 - f a f b - 0 Solving for c gives 2.

We have developed such an algorithm and it is given in the M-file regfals. The file also gives as an example of the use of the routine the solution to equation 2. Note that just as in the bisection algorithm, the initial two guesses must be such that one gives a positive function evaluation and the other a negative function evaluation.

This allows for the new root to always be an interpolation between the previous values. Note that a bad formulation of the problem 72 Computational Methods for Process Simulation 7, may cause divergence. Steady-State Lumped Systems 73 return end 2. Not all functions g x will converge using direct substitution as shown in Figure 2. The difference between a convergent and divergent system is the slope of g x at the solution condition. For the example of equation 2. Actual implementation is as follows. Iteration T g T 1 2 Divergent 1 Divergent Sometimes it is possible to reformulate a divergent g x function to obtain a new convergent expression.

Iteration T 9 T 1 This method is represented in Figure 2. The Wegstein algorithm can also accelerate the convergence of a normally stable situation. Note in the figure that x 2 is diverging from the solution. Now, we apply the Wegstein formula to the unstable form of the example: The M-file cp2 formulates the explicit unitable form of the problem equation 2. Convergence is only possible for guesses close to the actual root. For example, a initial guess of T — K" will converge as shown in Figure 2.

It is suggested that the reader study the convergence properties of this function by running the problem with different initial guesses for the temperature root. The presence of an optional fourth 7, argument specifies the maximum iteration number. Now, we apply the Wegstein method to the reformulated problem It can be used with the M-file wegstein.

The algorithm converges rapidly and from all initial guesses for the temperature T. The rapidity of convergence of substitution methods depends strongly on the slope of the solution line with respect to the diagonal line. If direct substitution diverges, either the problem must be reformulated or Steady-State Lumped Systems 79 the Wegstein method applied. Equations can have multiple roots. All of these algorithms may find different roots depending on their initial guesses.

Plotting of the function will be helpful in such cases. Wegstein Convergence Figure 2. This defines the nonlinear relationship between heat capacity 7. The format is suitable for the 7i Wegstein method and is a convergent expression using direct 7. Here we want to solve equation 2.

Newton Method for a Single Equation. Again, this method is iterative, but it only requires one initial guess. An important advantage of the Newton method is its rapid convergence. Steady-State Lumped Systems 81 Two disadvantages are that it requires computation of the derivative and that it can be very sensitive to the initial guess tends to blow up with poor initial guesses. It uses a finite difference approximation to the first derivative Figure 2.

The method that will be used for the solution Steady-State Lumped Systems 83 of n simultaneous nonlinear equations in n unknowns is a generalization of the Newton method of the previous section and is called the Newton- Raphson method. This technique tries to improve an initial guess for the solution vector via a linearization procedure.

A set of n equations in n unknowns may be written: If this were the case, 2. Because of the linear approximation of equation 2. A couple of precautions are in order. Systems of nonlinear equa- tions may have many solutions. In that case, it is wise to make the best initial guesses possible and use phys- ical reasoning in interpreting the solution. Also, the Jacobian matrix may become singular as the solution is approached. If this occurs, solu- tion by the Newton-Raphson technique may be impossible, and other, nonderivative methods should be used.

This routine implements an advanced version of the Newton-Raphson method as described above. However, rather than just a linear extrapolation, it implements a mixed quadratic and cubic extrapolation procedure. It is desired to determine the steady-state temperature and concentration in a continuous stirred tank reactor in which an exothermic reaction is taking place. We want to determine conditions for both maximum and minimum cooling. The reactor is shown in Figure 2. Exothermic Continuous Stirred Tank Reactor. It uses the M-file cstr.

Both files are shown in Figure 2. The major difference between a simulation study and a design study is in the type of variables that are specified. All input variables and equipment parameters must be specified in a simulation study and the output variables are calculated. Mathematical Methods and Models in Composites. Practical Seismic Data Analysis. Koenraad George Frans Janssens. Peridynamic Theory and Its Applications. Quantum Computation with Topological Codes.

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Computational Methods for Process Simulation

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Search Results Results 1 -9 of 9. Textbooks may not include supplemental items i. CDs, access codes etc Ergodebooks , Texas, United States Seller rating: Used book in good condition. Has wear to the cover and pages. Contains some markings such as highlighting and writing. This book has hardback covers. In fair condition, suitable as a study copy. Anybook Ltd , United Kingdom Seller rating: First Edition Book condition: Comprehensive reference text contains engineering fundamentals used in developing processing models, and considers both steady state and dynamic systems for spatially lumped and spatially distributed problems.