Microsoft Excel for Chemical Engineers Notes (by Moataz and Mohammed)
February 28, 2017 | Author: Madhusudhan Reddy Pallaka | Category: N/A
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CAIRO UNIVERSITY FACULTY OF ENGINEERING CHEMICAL ENGINEERING DEPARTMENT
INTRODUCTION TO MICROSOFT EXCEL WITH APPLICATIONS IN CHEMICAL ENGINEERING Eng. Moataz Bellah Mahmoud Mousa Eng. Mohammed Gamal Abdel Nasser
July 2009
Table of Contents Microsoft Excel 2007 Basics .................................................................................................................... 1 Inserting Charts ...................................................................................................................................................... 1 Data fitting using least square method ......................................................................................................... 5 Array calculations .................................................................................................................................................. 9 Matrices ................................................................................................................................................................... 10
Thermodynamics ..................................................................................................................................... 14 Equations of state ................................................................................................................................................ 14 Solving equations of state using excel ........................................................................................................ 16 Vapor Pressure Data Representation Using Equations ........................................................................ 17
Chemical Reaction Engineering........................................................................................................... 21 Batch reactor yield optimization ................................................................................................................... 21 Chemical reaction equilibrium ....................................................................................................................... 23 Example Using Excel........................................................................................................................................... 24 Regressing rate constants in rate equation from experimental data .............................................. 26 Multiple regression using Excel.................................................................................................................... 26 Non linear regression ....................................................................................................................................... 30 Nonlinear Regression Using Excel .............................................................................................................. 30
Fundamentals of Chemical Engineering ........................................................................................... 33 Material Balance................................................................................................................................................... 33 Material balance for non-reactive system .................................................................................................. 35 Material balance for a reactive system without recycle ......................................................................... 37 Material balance for a reactive system with recycle ............................................................................... 39 Energy Balance ..................................................................................................................................................... 43 Energy balance on a non-reactive system .................................................................................................. 50 Energy balance on a reactive system ........................................................................................................... 53
Fluid Mechanics ........................................................................................................................................ 59 Terminal Velocity of Falling Particles .......................................................................................................... 59 Emptying Tank...................................................................................................................................................... 62 Runge-Kutta method .......................................................................................................................................... 62 Pipeline optimization ......................................................................................................................................... 64
PHASE EQUILIBRIA AND MASS TRANSFER .................................................................................. 68 Introduction ........................................................................................................................................................... 68 Bubble point Calculations................................................................................................................................. 70 Dew point Calculations ...................................................................................................................................... 71 Flash Calculations ................................................................................................................................................ 71 Performing flash calculations using Microsoft Excel......................................................................... 72
Microsoft Excel 2007 Basics Inserting Charts Microsoft Office Excel 2007 supports many types of charts to help you display data in ways that are meaningful to you. There are many types of charts available in Microsoft excel, and here we will try to show the most commonly used ones in the engineering applications: Line charts Data that is arranged in columns or rows on a worksheet can be plotted in a line chart. Line charts can display continuous data over time, set against a common scale, and are therefore ideal for showing trends in data at equal intervals. In a line chart, category data is distributed evenly along the horizontal axis, and all value data is distributed evenly along the vertical axis.
Pie charts Data that is arranged in one column or row only on a worksheet can be plotted in a pie chart. Pie charts show the size of items in one data series, proportional to the sum of the items. The data points in a pie chart are displayed as a percentage of the whole pie.
Consider using a pie chart when:
You only have one data series that you want to plot.
None of the values that you want to plot are negative.
Exercise: Search the Microsoft Excel help to find out more information about bar charts and how to use it to draw a Gantt chart
XY (scatter) charts (most used in chemical engineering applications) Data that is arranged in columns and rows on a worksheet can be plotted in an xy (scatter) chart. Scatter charts show the relationships among the numeric values in several data series, or plots two groups of numbers as one series of xy coordinates. A scatter chart has two value axes, showing one set of numeric data along the horizontal axis (x-axis) and another along the vertical axis (y-axis). Scatter charts are typically used for displaying and comparing numeric values, such as scientific, statistical, and engineering data. Consider using a scatter chart when:
You want to change the scale of the horizontal axis.
You want to make that axis a logarithmic scale.
There are many data points on the horizontal axis.
You want to effectively display worksheet data that includes pairs or grouped sets of values and adjust the independent scales of a scatter chart to reveal more information about the grouped values.
You want to show similarities between large sets of data instead of differences between data points.
You want to compare many data points.
Some of Microsoft Excel charts (unstacked, 2-D, area, bar, column, line, stock, xy (scatter), or bubble chart) provide an easy means of fitting data and getting the equation that fits the data plotted on the graph using a method called Regression. Fitting data is performed by an option called “Add Trendline”
Choosing the right trendline type for your data When you want to add a trendline to a chart in Microsoft Office Excel, you can choose any one of these six different trend or regression types: linear trendlines logarithmic trendlines polynomial trendlines
power trendlines exponential trendlines moving average trendlines
Linear trendlines A linear trendline is a best-fit straight line that is used with simple linear data sets. Your data is linear if the pattern in its data points resembles a line. A linear trendline usually shows that something is increasing or decreasing at a steady rate. Logarithmic trendlines A logarithmic trendline is a best-fit curved line that is used when the rate of change in the data increases or decreases quickly and then levels out. A logarithmic trendline can use both negative and positive values. Polynomial trendlines A polynomial trendline is a curved line that is used when data fluctuates. The order of the polynomial can be determined by the number of fluctuations in the data or by how many bends (hills and valleys) appear in the curve. An Order 2 polynomial trendline generally has only one hill or valley. Order 3 generally has one or two hills or valleys. Order 4 generally has up to three hills or valleys. Power trendlines A power trendline is a curved line that is used with data sets that compare measurements that increase at a specific rate. You cannot create a power trendline if your data contains zero or negative values. Exponential trendlines An exponential trendline is a curved line that is used when data values rise or fall at constantly increasing rates. You cannot create an exponential trendline if your data contains zero or negative values. Moving average trendlines A moving average trendline smoothes out fluctuations in data to show a pattern or trend more clearly. A moving average uses a specific number of data points (set by the Period option), averages them, and uses the average value as a point in the line. For example, if Period is set to 2, the average of the first two data points is used as the first point in the moving average trendline. The average of the second and third data points is used as the second point in the trendline, etc..
NOTE: Choosing the type of trendline in many cases depends on the case whose data is being fitted, i.e. you may know that your data must fit linear or logarithmic or whatever trendline before they are plotted. A trendline is most accurate when its R-squared value is at or near 1. When you fit a trendline to your data, Excel automatically calculates its R-squared value. If you want to, you can display this value on your chart. The table below summarizes the types of trendlines and the final forms of equations: Use this type To create Linear
A linear trendline by using the following equation to calculate the least squares fit for a line: where m is the slope and b is the intercept.
Logarithmic
A logarithmic trendline by using the following equation to calculate the least squares fit through points: where c and b are constants, and ln is the natural logarithm function.
Polynomial
A polynomial or curvilinear trendline by using the following equation to calculate the least squares fit through points: where b and
Power
are constants.
A power trendline by using the following equation to calculate the least squares fit through points: where c and b are constants. This option is not available when your data includes negative or zero values.
Exponential
An exponential trendline by using the following equation to calculate the least squares fit through points: where c and b are constants, and e is the base of the natural logarithm. This option is not available when your data includes negative or zero values.
Moving average
A moving average trendline by using the following equation:
Data fitting using least square method Microsoft Excel proved to be very effective in curve fitting, but it does so for five forms of equations which were mentioned previously. If the equation that you need to fit your data to is not one of these five forms then Microsoft Excel will not be able to fit your data automatically using the Trendline option. In these cases the user must be able to fit his data to the form he needs by himself. There are many methods to do so, the most common and easy to use is what so called “Least square method”. To understand this method let‟s see the next example: Consider the following experimental data relating any two variables (say x and y) X 3.4 Y
7.1
16.1 20.0 23.1 34.4 40.0 44.7 65.9 78.9 96.8 115.4 120.0
9.59 5.29 3.63 3.42 3.46 3.06 3.25 3.31 3.50 3.86 4.24 4.62
4.67
And that from your knowledge about your experiment you know that the relation must take the form: Y=aX+b/X+c And you now need to fit these data to that equation to get the values of a, b and c, also to calculate the R-squared. This method is based on a simple idea, which is assuming values for the variables a, b and c, then calculating values of Y‟s based on the assumed values, then comparing the calculated and the real values of Y‟s, and finally changing the values of assumed a, b and c till the difference between the real and calculated vales of Y‟s is minimum. This is simply done by building a simple table, two columns for the experimental data available (X‟s and Y‟s ), another column for the calculated values of Y‟s ( based on the values of X ).
Remember to add the dollar signs on typing the names of cells containing assumed vales of a, b and c before dragging. After dragging the table will be in the form shown below:
Now it‟s now required to minimize the difference between the calculated and the real values. So a new column is introduced to calculate the difference between the calculated and the real values.
Now it‟s required to minimize all the values in the last column by varying the values of the variables a, b and c. Microsoft Excel has a tool that performs this form of iterations which is “Solver”. It‟s available under the Data toolbar in the analysis tab.
On clicking the solver tab a window will appear having the shape:
This tool can change the value of multiple cells to set the value of ONLY ONE cell to a maximum, minimum or a specific value. And we need now to change the value of the three cells (a, b and c) to set the values of 13 cells to zero, which is not available using this tool. A trick can be done to avoid this problem. If the sum of the differences is calculated and then is set to the value of zero then consequently the values of all the 13 cells will be minimized, so a cell for the sum values in the last column is set and calculated. And is now the cell that will be set to zero. But remains a small problem, you may notice that some of the values in the last column are positive and some are negative, so the summation may be zero but the entire cells in the column may not be zeros. So a new column is added where the SQUARES of the values of the difference column are calculated, so all the values will be positive and this problem is no more present. And that‟s why this method is called Least Square Method. Finally the table will have the form:
Now we are now ready to use the solver.
By clicking solve the final results appear:
It‟s now clear that the values in columns D and E are minimized. Now to ensure the accuracy of this fitting the R-squared value must be calculated. R-squared value is calculated from the following relation:
Then finally the shape of the table will be as follows:
And the R-squared value is found to be 0.9967 which is a very good fit.
Array calculations An array formula is a formula that can perform multiple calculations on one or more of the items in an array. Array formulas can return either multiple results or a single result. For example, you can place an array formula in a range of cells and use the array formula to calculate a column or row of subtotals. You can also place an array formula in a single cell and then calculate a single amount. To understand the concept on arrays well, let‟s consider the least square method example discussed previously. On calculating the value of the sum of square of the differences between the calculated and the real values we had to add a column to calculate the differences then another column to calculate the square of the differences, and finally a cell to calculate the summation of the squares. Array calculations provide a quicker means to do these 3 or 4 steps in only one step. In any cell type the following formula “=SUM(C2:C14-B2:B14)” then press Ctrl+Shift+Enter and compare it with the value in cell D17, you will find that they both have the same value, this means that this formula calculated the difference between the values in two columns and got the sum of these values. On typing the following formula: “=SUM((C2:C14-B2:B14)^2)” you will be able to calculate directly the sum of squares of the differences.
Matrices Matrix calculation is a special type of array calculations, as Microsoft Excel deals with matrices as arrays. Microsoft Excel can perform many matrices operations such as adding, subtracting, multiplying matrices, also getting the inverse of a matrix and the value of its determinant. Before discussing how these operations are performed using Microsoft Excel, it‟s recommended to revise quickly the matrix operations and calculations. Consider a system of linear equations like : a1X+b1Y+c1Z=L a2X+b2Y+c2Z=M a3X+b3Y+c3Z=N where X,Y and Z are variables a1,b1,……., b3,c3 are their coefficients ( constants ) L,M and N are constants And it‟s required to solve these equations to get the values of X,Y and Z. One of the methods of solving this system of equations is using Matrices. This can be simply performed by putting the system of equations in the form of matrix as follows a1
b1
c1
a2
b2
c2
a3
b3
c3
Which takes the general form:
X ×
Y Z
L =
M N
AX=B
Before solving such a system, we must check first if it‟s solvable or not, which can be verified by calculating the value of the determinant of the matrix “A”. If its value ≠ 0 then it‟s solvable, if its value=0 then it‟s not solvable. After performing this check the system has to be solved. This was done by performing the following steps: AX=B -1
A AX=A-1B X=A-1B So getting the final solution requires first getting the inverse of the matrix “A”, then multiplying it by matrix “B” to get the final solution. NOTE:on multiplying two matrices, the number of columns of the first matrix has to be equal to the number of rows of the second one.
Now, consider a set of linear equations that we have to solve using Microsoft Excel, like: X-Y-Z+3 L=1 2X+4 Y+3 Z+L=2 2.25 X+Y+2 Z+2 L=3 X+1.5 Y+Z+2 L=4 This will take the matrix form as follows: 1
-1
-1
3
2
4
3
1
2.25
1
2
2
1
1.5
1
2
X ×
Y Z L
1 =
2 3 4
This is introduced to Microsoft Excel as an array:
Then we have to check if the determinant “A” equals zero or not, Microsoft Excel can calculate the value of the determinant of a matrix using the function “MDETERM” Entering the following formula MDETERM(array region) then –as in any array calculation- pressing Ctrl+Alt+Enter calculates the value of the determinant.
Now it‟s verified that the system of equations is solvable. Now we need to get the inverse of the matrix “A”, this is calculated using the function “MINVERSE” Entering the following formula MINVERSE(array region) –after highlighting the array of the inverse matrix–then pressing Ctrl+Alt+Enter, calculates and shows the inverse of the matrix.
Finally to get the solution we have to multiply A-1 by B To do this use the function MMULT, Entering the following formula MMULT(array1,array2) – after highlighting the array of the product – then pressing Ctrl+Alt+Enter, calculates and shows the matrix X, which is the solution.
Other matrix operations can be easily performed using excel such as addition subtraction of matrices, multiplying matrix by a constant, ... etc.
Thermodynamics This chapter introduces the ideal gas equation of state, then describes how computer programs such as Excel, use modified equations of state to easily and accurately solve problems involving gaseous mixtures. Also in this chapter the vapor pressure data representation is discussed, then the use of Excel to determine constants of different equations, relating vapor pressure to temperature, is illustrated for optimum representation to the experimental data.
Equations of state Solving equations of state allows us to find the specific volume of a gaseous mixture of chemicals at a specified temperature and pressure. Without using equations of state, it would be virtually impossible to design a chemical plant. By knowing this specific volume, you can determine the size – and thus cost – of the plant, including the diameter of pipes, the horsepower of compressors and pumps, and the diameter of distillation towers and chemical reactors. The ideal gas equation of state, which relates the pressure, temperature, and specific volume, is a familiar equation:
The term p is the absolute pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the absolute temperature. The units of R have to be appropriate for the units chosen for the other variables. This equation is quite adequate when the pressure is low (such as one atmosphere). However, many chemical processes take place at very high pressure. For example, ammonia is made at pressures of 220 atmospheres or more. Under these conditions, the ideal gas equation of state may not be a valid representation of reality. Other equations of states have been developed, usually in conjunction with process simulators, to address chemical processes at high pressure. There are two key features: (1) the equation can represent the real p–V–T behavior; and (2) the parameters must be easily found, including for mixtures. This last criterion is no small requirement. There are more than 25 million chemicals, leading to an infinite number of different mixtures. Obviously, you cannot look up the properties of all those mixtures on the Web. The first generalization of the ideal gas law was the van der Waals equation of state:
INTRODUCTION TO MICROSOFT EXCELermodynamics In this equation, the b accounts for the excluded volume (a second molecule cannot use the same 15
space already used by the first molecule), and the a accounts for the interaction force between two molecules. This extension is just a first step, however, because it will not be a good approximation at extremely high pressures. The Redlich–Kwong equation of state is a modification of van der Waal‟s equation of state:
Where,
In these equations, Tc is the critical temperature (in absolute terms), Pc is the critical pressure. The Redlich–Kwong equation of state was modified further by Soave to give the Redlich–Kwong– Soave equation of state, which is a common one in process simulators:
Now the parameter a is given by a different formula,
The is the „acentric‟ factor, which is a tabulated quantity for many substances. Thus, the value of can be computed for each chemical and reduced temperature. The Peng–Robinson equaton is another variation:
All these equations can be rearranged into a cubic function of specific volume. The form of the Redlich–Kwong and Redlich–Kwong–Soave equation of state is
When given the temperature and pressure of a gaseous mixture, and the parameters a and b, then to find the specific volume you would have to solve the cubic equation of state for specific volume
. This represents one algebraic equation in one unknown, the specific volume.
For a pure component, the parameters a and b are determined from the critical temperature and critical pressure, and possibly the acentric factor. These are all tabulated quantities, and there are even correlations for them in terms of vapor pressure and normal boiling point, for example. For mixtures it is necessary to combine the values of a and b for each component according to the
INTRODUCTION TO MICROSOFT EXCELermodynamics composition of the gaseous mixture. Common mixing rules are shown in next equations, in which 16
the ys are the mole fraction of each chemical in the vapor phase:
where for Redlich-Kwong or for Redlich-Kwong-Soave
Thus, the only difference between the problem for a pure component and that for a mixture is in the evaluation of the parameters a and b. Here is the mathematical problem need to be solved: Given a set of chemicals, temperature and pressure, find the specific volume of the mixture. To do this, you must find the critical temperature and pressure of each chemical. Once you have the parameters, you must solve the cubic equation, which is a nonlinear equation in one variable. Excel allows us to easily solve for the specific volumes. However, one advantage of process simulators is that the physical properties of many components are saved in a database that users can access. The next section illustrates how to use Excel to solve equations of state.
Solving equations of state using excel Problem Calculate the molar volume and compressibility factor for gaseous ammonia at a pressure P = 40 atm and a temperature T = 400 K using the van der Waals and Redleich-Kwong equations of state. Step 1 You must first find the critical temperature and pressure; Perry‟s Chemical Engineers‟ Handbook gives Tc = 405.4 K and Pc = 111.3 atm for ammonia Step 2 Calculate values of a and b for Redlich-Kwong using the following equations:
Van der Waals parameters are a= 0.4233 J.m3/mol2, b= 3.737E-5 m3/mol
INTRODUCTION TO MICROSOFT EXCELermodynamics Step 3 Prepare the spreadsheet shown in next figure. The title, name, and data will be useful when 17
you come back to the problem at a future date Step 4 You enter the parameters in the parameter box. The cells containing the critical parameters and the temperature and pressure can be named Tc, Pc, T, and P, respectively. That way, the equation for f (v) will be easier to understand Step 5 The lower box gives the equations actually used as well as the results. Use the Goal Seek command to make f (v) (cell B15) equal to zero by changing
(cell B16)
Step 6 For reference, the result for an ideal gas is also shown
Exercise
a) Repeat the calculations for the following reduced pressures: Pr = 1, 2, 4, 10, and 20.
Vapor Pressure Data Representation Using Equations Vapor pressure is the pressure of a vapor in equilibrium with its non-vapor phases. All liquids have a tendency to evaporate to a gaseous form, and all gases have a tendency to condense back into their original form. At any given temperature, for a particular substance, there is a pressure at which the gas of that substance is in dynamic equilibrium with its liquid form. This is the vapor pressure of that substance at that temperature. The equilibrium vapor pressure is an indication of a liquid's evaporation rate. It relates to the tendency of molecules and atoms to escape from a liquid or a solid. A substance with a high vapor pressure at normal temperatures is often referred to as volatile. The atmospheric pressure boiling point of a liquid is the temperature where the vapor pressure equals the ambient atmospheric pressure. The higher the vapor pressure of a liquid at a given temperature, the lower the normal boiling point (i.e., the boiling point at atmospheric pressure) of the liquid.
INTRODUCTION TO MICROSOFT EXCELermodynamics The vapor pressure of any substance increases non-linearly with temperature and is independent of 18
the applied pressure. Also it differs from one substance to another thus it can be a considered a characteristic physical property for substances. There is a large number of saturation vapor pressure equations used to calculate the pressure of water vapor over a surface of liquid water. The two most commonly used equations are 1) Clausius-Clapeyron
where P is the vapor pressure in mmHg and T is the temperature in °C. Note that the denominator is just the absolute temperature in K. Both A and B are the parameters of the equation which are typically determined from experimental data. 2) Antoine equation
where typically P is the vapor pressure in mmHg and T is the temperature in °C. Note that this equation has parameters A, B, and C which must be determined by experimental data. Also a simple polynomial can be used to relate vapor pressure with temperature. This can be written in general form for this problem as
Where a0... an are the parameters (coefficients) to be determined by regression and n is the degree of the polynomial. Typically the degree of the polynomial is selected which gives the best data representation when using a least-squares objective function. So the main task to be done here is to find a good values for the equation constants (whatever it was Antoine, clausius-clapeyron or a simple polynomial equation) from a set of experimental results. After determining the constants you have to compare the equation results with the experimental data to assess the equation reliability.
INTRODUCTION MICROSOFT EXCELermodynamics Problem 19
TO
Next table presents data of vapor pressure versus temperature for benzene. Some design calculations require these data to be accurately correlated by various algebraic expressions which provide P in mmHg as a function of T in °C. Vapor Pressure of Benzene Temperature, T, (0C) Pressure, P, (mm.Hg) -36.7 1 -19.6 5 -11.5 10 -2.6 20 7.6 40 15.4 60 26.1 100 42.2 200 60.6 400 80.1 760 (a) Determine the constants for best fit of the experimental data using a polynomial equation. Also Determine the degree of polynomial which best represents the data. (b) Determine the constants for best fit of the data using Clausius-Clapeyron equation. (c) Determine the constants for best fit of the data using Antoine equation. Step 1 The spreadsheet was set up with the data for temperature and pressure entered in the indicated columns Step 2 Columns for temperature in K and Log P were developed Step 3 Columns were then added for each of the curve fits that were to be carried out along with the parameters that would be manipulated to generate the calculated curves. These are shown above the calculated columns as parameters. Step 4 A row of sum of squares cells were added along the bottom. The contents were set up as array formulas and entered with Crtl+Shift+Enter. For example for the calculated ClasiusClapeyron column: in the sum of squares cell the following was entered: SUM ((E16:E25-D16:D25) ^2) followed by Crtl+Shift+Enter. The function will be written between brackets { } Step 5 The Solver program was then utilized for each of the curve fits. For example for the Clausius-Clapeyron fit E5 and E6 (A and B) were manipulated until E27 (the sum of squares) was minimized.
20
INTRODUCTION MICROSOFT EXCELermodynamics
TO
Exercise The vapor pressure of a pure liquid as a function of the absolute temperature can be expressed as: log (P0) = ) The total vapor pressure of a three component mixture is given by: P= x1P01+x2P02+x3P03 Where xi is the mole fraction of component i. Set up a worksheet to use with solver to find the normal boiling point (the temperature at which P= 760mm.Hg) of a three component mixture. Use the following data a
b
x
Benzene
7.8413 1750
0.5
Toluene
8.0884 1985
0.3
Ethyl benzene
8.114
0.2
2129
Chemical Reaction Engineering Batch reactor yield optimization A chemical plant uses the following process in its production: A volume “V” m3 of solution of compound “A” having a concentration of a0 gmoles/m3 is allowed to react for tr hours; the reactor is then emptied, cleaned and recharged for another cycle. The reaction occurring is a first order reaction with rate constant of 1. It is required to find the optimum reacting time tr for maximum reactor product at different combinations of solution volume, concentration and cleaning time as shown in the next table. (1)
(2)
(3)
(4)
5
5
10
10
a0 (gmol/m )
0.1
0.2
0.1
0.1
tc (hrs)
0.5
0.5
0.5
1
V (m3) 3
Step1 You must first develop an equation for the reaction yield to be optimized. In general for first order reaction,
But,
and the amount of product formed is the difference between initial and final moles of “A”
For optimization purposes it is better to optimize the product yield per unit process time, thus
22
Chemical Reaction Engineering
Step 2 Enter the values in the table as shown in the next figure
Step 3 In yield row the yield equation need to be optimized must be written. The equation entered in cell B8 is shown in row 9 as appears in the next figure.
The presence of dollar sign on the left of cell numbers is to allow dragging of the function to other cells in the same row without changing the cells numbers in the functions. Step 4 Now, initial guesses for “tr” must be entered in other cells then solver is used to find the optimum values. For example in the first case, Solver will try to find optimum value for cell B8 by changing B10. The final sheet form is shown in the next figure.
23
Chemical Reaction Engineering
Chemical reaction equilibrium aA+bB ↔ cC+dD
This is a reversible reaction. When the forward reaction happens at the same rate as the backward reaction, the reaction progress stops and the reaction reaches equilibrium. If the reaction is elementary, then the concentration of reactants and products are related to each other according to:
The number KC is called the equilibrium constant and is a function of temperature. a ,b,c and d are the stoichiometric coefficients. There is another useful definition of the equilibrium constant based on pressure rather than concentration and is used when dealing with gases since the pressure is usually more convenient to measure than concentration. Under conditions of low pressure (~10atm.) the gas can be considered to behave ideally and follows the ideal gas low PV=nRT Here P is the total pressure. In the cases of several components, each has a partial pressure, all of which sum up to the total pressure. P=PA+PB+PC+PD For each component, we can write the ideal gas law as follows:
If we plug in the partial pressures in the definition of KC above we get:
Using the realtion pi=yiPT
24
Chemical Reaction Engineering
Gibbs free energy is related to KP by the relation:
N.B.: Of course, the reaction rate may not be infinitely high, and you may use a catalyst to speed up the reaction. However, even if the reaction rate is increased, you can never go beyond the composition determined by chemical reaction equilibrium. Here is how to set up that problem mathematically.
Example Using Excel When the carbon monoxide and hydrogen enter the reactor in a 1:2 ratio, use Excel to find the equilibrium conversion when the pressure is 50 atm and KP=0.0016. Consider the methanol formation reaction CO + 2H2 ↔ CH3OH At equilibrium:
Thermodynamic data give the value of KP=0.0016 atm-4 at reaction conditions and 50atm pressure. If you start with a stoichiometric mixture of carbon monoxide and hydrogen, what will the equilibrium composition be?
Step 1: We will start by making a quick material balance by writing 3 component material balance equations Step 2: You can solve for the equilibrium using Excel, by putting the equilibrium equation in one cell, a guess of “r” in another cell, and use the Goal Seek to make the former cell zero by changing “r”. The final results are shown in next figure.
25
Chemical Reaction Engineering
Second column is the initial moles of each species. The third column is computed according to the equations displayed in the column beside it. The equilibrium equation is calculated in C10, according to the formula displayed in cell D10. Goal seek is used to make cell C10 zero by varying cell C9. Now once you have prepared the spreadsheet, it is easy to change the conditions, either the equilibrium constant or the starting moles of various species. Exercise Find the molar flow rates of all species out of an equilibrium reactor used in ammonia production when the inlet moles ratio of nitrogen, hydrogen and ammonia are 1.1, 3, and 0.2 respectively. The equilibrium constant is 0.05atm-4 at 589 K. The reactor is operated at 220 atm.
26
Chemical Reaction Engineering
Regressing rate constants in rate equation from experimental data Engineers often need to express experimental data in terms of an equation. They must decide on the equation and then determine the parameters that provide the best fit to the data. The problem is simplest if the equation is linear. The mission will a little bit harder if the best fit equation is polynomial one. Although this later case can be considered easy, as long as the equation relates 2 variables only. In this part we will discuss how to fit any set of functions of more than 2 parameters either appearing linearly (multiple linear regression), or nonlinearly (multiple non-linear regression).
Multiple regression using Excel Early in the introduction of the course it was shown how to fit a polynomial to data. The same thing can be done when the functions are not simple powers, but are more complicated functions. However, to keep the problem linear, the unknown coefficients must be coefficients of those functions; that is, the functions are completely specified. Multiple regression simply determines how much of each one is needed. Thus, the form of the equation is
The goal is to find the best M values of ai, given the M functions fi (x) and data
yi = y
(xi), i =1,. . ., N. In Excel, you put the x values in a column and create additional columns, with each column being a function, evaluated for the x value in that row. The example used here is to find the constants in a reaction rate formula. The expected expression is
and the goal is to find the values of k, n, and m that give the best fit of the rate for various partial pressures of substances A and B. This form is not linear, which is a requirement of multiple regression, but a transformation can make it linear. Take the logarithm of both sides of the equation.
This equation has the following form: where the dependence upon two or more variables is clear. The data is entered into the spreadsheet, and the various terms are transformed as shown in next table.
Columns A and B are the
partial pressures of the two chemicals for which the rate is measured, as indicated in column C.
27
Chemical Reaction Engineering
Step 1 Obtain columns D, E, and F by taking the logarithm of columns A, B, and C, respectively. Do this in the first cell (D2), copy it across the E and F rows, and copy down the three rows (D2, E2, and F2).
Step 2 Proceed with parameter estimation by choosing „Data/Data Analysis,‟ and then choosing „Regression.‟ If „Data Analysis‟ does not appear in your menu under „Data‟ you will have to add it from Add-Ins. Enter F2:F13 for the y values and D2:E13 for the x values. This tells the computer that you want the best line representing ln(rate) depending linearly upon ln pa and ln pb.
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Chemical Reaction Engineering
Step 3 There are several other options; choose residuals, residual plots, and line fit plots. These are all useful for evaluating the results. You can place the results on another sheet in the same Workbook or on the same sheet by specifying a location. Next table shows part of the output. The best fit is for
The curve fit is then
You might think at this point the correlation is complete. It is not, though, because the data were transformed to make the parameter estimation problem linear. Thus, the statistics are in terms of the transformed problem. It is always a good idea to calculate the curve fit using the original variables.
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Chemical Reaction Engineering
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Chemical Reaction Engineering
Non linear regression Nonlinear regression is a curve fit in which the unknown parameters enter into the problem in a nonlinear way. Nonlinear regression is much more difficult (for the computer), so it is best to always try to manipulate your model into a form that is linear. Sometimes that is not possible, and then nonlinear regression must be used. You need to be aware, though, that the methods described here do not always work. Nonlinear regression uses techniques borrowed from the field of optimization, and it is difficult to construct a method that works every single time for every problem. For nonlinear curve fits it is necessary to use functions such as Solver in Excel. This is demonstrated using the same example given above for multiple regression.
Nonlinear Regression Using Excel Step 1 Place the data on a new sheet as shown in next figure
Step 2 Enter arbitrary values for the parameters k, n, and m. Step 3 In column D, calculate the value of rate using the parameters in C16:C18, the data in columns A and B, and the formula, Eq.
Step 4 Make column E the difference between columns C and D, and then square the result and put it in column F. Step 5 Sum Column F, divide by the number entries [COUNT (F2:F13)] to obtain the (sum of squares)/N, Eq. (E.3).
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Chemical Reaction Engineering
Step 6 The goal is to minimize F17 by choosing values C16:C18. To do that, choose „Tools/Solver.‟ You might have to add it to the Excel program if that was not done when the program was installed. A screen appears in which you insert F17 as the quantity to be affected, and choose „Min‟ as the option. Then insert C16:C18 as the cells to be changed, and click „Solve.‟ The results are shown in next figure.
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Chemical Reaction Engineering
The best correlation is
These numbers are slightly different from those obtained using multiple regression. Multiple regression and nonlinear regression obtained the solution by minimizing two different objective functions.
Fundamentals of Chemical Engineering Material Balance Generally the material balance calculation was simply governed by the equation: In=out + r Where In and out are the flow rates of the component in the inlet and outlet streams to the unit around which material balance in established, is the stoichiometric coefficients
(-ve for reactants and
+ve for products ) if there is a chemical reaction involved in the unit . Generally this equation can be applied to a single unit or group of units. To be able to know if the available information about the unit are sufficient to solve this unit or there are some information needed we calculate the degrees of freedom of this unit as follows: DOF= number of variables- number of equations- number of given variables-number of additional relations The system will be solvable if the DOF=Zero, if the DOF is negative then the system is said to be over specified i.e. more information are given, if the DOF is positive then the system is said to be under specified i.e. more information are needed to solve this unit. For a flow sheet where a number of units are operating depending on each other, a DOF table should be constructed to ensure that the system can be solved using the available data. Unit I
Unit II
----------
Unit N
Process
Overall
No. of variables No. of equations No. given variables
of
No. of additional relations DOF The system will be solvable if the DOF of process equals zero (gray cell) even if the DOF of all units are not Zeroes. Note: The degrees of freedom table not only indicates if the system is solvable or not, but also determines what so called “Strategy of solution” or the way by which the system will be solved and the sequence of units solved.
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Fundamentals of Chemical Engineering
The solution is not usually straight forward, so there are some techniques that are usually used to be able to solve such a system. The main two of them are tearing and carrying over. Tearing: a method where a stream is assumed ( components flow rates ) and recalculated again, then comparing the assumed with the calculated values, and so on till both are equal. Carrying over: a method where all equations of all units are written and the uncalculated values remain unknown till the calculations lead to getting their values, sometimes the equations are put in a matrix form. The most common way used in simulation programs is the tearing method. In practical cases material balance and energy balance are both performed simultaneously, and this – of course- will result in more accurate results, however, more calculations and harder iterations will be needed. But in some cases when the effect of pressures and temperatures on material balance is not significant, or when some equipment lack such information material balance calculations can be performed alone. Microsoft Excel can be used successfully in solving material balance problems, its main advantage is that the user builds the program by himself – not like the other simulation programs where the user only is inputing the data – and this gives a good chance to understand the principals well and know the source of error in calculations if exists.
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Fundamentals of Chemical Engineering
Material balance for non-reactive system Separation Train
D1
Xylene, styrene, toluene and benzene are to be separated with the array of distillation
7% Xylene 4% Styrene 54% Toluene 35% Benzene
D #2
columns that is shown below where F, D, B, D1, B1, D2 and B2 are the molar flow rates in mol/min. F is the feed flow rate of 70 mol/min a- It‟s required to calculate the flow rates of the top and bottom products of the second and third distillation columns. b- Also it‟s required to calculate the molar flow rates of the four components in both streams B and D.
B1
18% Xylene 24% Styrene 42% Toluene 16% Benzene
F #1 15% Xylene 25% Styrene 40% Toluene 20% Benzene
D2
15% Xylene 10% Styrene 54% Toluene 21% Benzene
B #3
B2
24% Xylene 65% Styrene 10% Toluene 1% Benzene
Part a : Solution of such a system can be performed by solving a set of algebraic equations simultaneously, these equations are material balance equations for each component. Material balance equations on the overall of the three columns will yield: 0.15×70=0.07×D1+0.18×B1+0.15×D2+0.24×B2 0.25×70=0.04×D1+0.24×B1+0.10×D2+0.65×B2 0.40×70=0.54×D1+0.42×B1+0.54×D2+0.10×B2 0.20×70=0.35×D1+0.16×B1+0.21×D2+0.01×B2 This can be shown in the form of a matrix: AX=B
Its solution is as follows X=A-1B
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Fundamentals of Chemical Engineering
So the inverse of the matrix A must be calculated, by using Microsoft excel, a function called “MINVERSE” is used to calculate the inverse of a matrix. Then the inverse is multiplied by matrix “b” using the function “MMULT”. The calculated matrix is then the solution to the above system of equations and required flow rates are calculated. Part b : The two streams B and D can be got from material balance equations around both towers number 2 and 3, For tower 2: D=D1+B1
For tower 3: B=D2+B2
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Fundamentals of Chemical Engineering
Material balance for a reactive system without recycle Get started The following flow diagram represents a part of methanol production process, it‟s produced by the reaction of synthesis gas ( CO and H2 ) according to the following reaction:
CO H2 CH4 CH3OH
3
Catalytic Reactor 1
2
Flash Separator
CO H2 CH4 CH3OH
CO H2 CH4
4
CO H2 CH4 CH3OH
Fresh feed of 1000 Kgmol/hr containing 33% CO, 66.5% H2 and 0.5% CH4 (all in mole %) is introduced to the reactor, this reaction is catalytic and only 40% conversion of CO is achieved. The product is then fed to a separator where methanol is separated from the unreactants, it was reported that the bottom product contains 3% of CO2, 2% of H2, 4% of CH4 and 96% of methanol reactor effluent. You are now required to solve this system to get the flow rates of all components in all streams in addition to the rate of the reaction, once by solving a system of equations (using matrix) and once by unit-to-unit calculations. For the unit-to-unit calculations method: It‟s common to arrange the variables in the form of table where the columns represent the streams and the rows represent the components in each stream as shown: 1
2
3
4
CO H2 CH4 CH3OH Total So each cell represents the flow rate of a component in a stream. The solution requires filling the cells with the material balance equations of the units involved and the additional relations (note that the size of matrix equals the number of variables in the process column). Note that there must be an external cell for the rate of reaction.
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Fundamentals of Chemical Engineering
For example consider the table will be in the next form:
Then the inserted formulas will be as follows:
Material balance equation CO)1=0.33×Stream (1) H)1=0.665×Stream (1) CH4)1=0.005×Stream (1) CO)2=CO)1-r H)2=H)1-2r CH4)2= CH4)1 CH3OH)2= CH3OH)1+r r=0.4× CO)1
In cell Feed B2 B3 B4 Reactor C2 C3 C4 C5 B8
Formula =0.33*E8 =0.62*E8 =0.05*E8 =B2-B8 =B3-2*B8 =B4 =B5+B8 =0.4*B2
And so on for the separation unit. Finally the results appear in all streams. This method has the advantage that you needn‟t build a matrix of a big size especially for systems containing may units and many components. Also it makes the user able to know the location of error ( if exists ) and so easiness in correction and modification. The final results will be:
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Fundamentals of Chemical Engineering
Material balance for a reactive system with recycle For the above example, it‟s clear that nearly half the reactants only reacted and the other half was introduced to the separator, separated from the product. Practically in such processes the unreacted reactants are recycled and mixed with the fresh feed to make use of them, also in such systems where there are some inerts in the feed (like Methane) some of the recycle stream is purged before mixing with the fresh feed to prevent accumulation of inerts in the system. On applying such a concept in the above case we will realise that the flow diagram will be as follows: 7
CO H2 CH4 CH3OH
6
CO H2 CH4 CH3OH
4
Catalytic Reactor 1
2
3
CO H2 CH4
CO H2 CH4 CH3OH
CO H2 CH4 CH3OH
Flash Separator
CO H2 CH4 CH3OH
5
First, the degrees of freedom table must be constructed, it‟ll be as follows: Mixer
Reactor
Flash
Splitter Process
No of var
11
9
12
12
28
No of Eq
4
4
4
4
16
No of rel
0
1
4
4
9
No of G.V
3
0
0
0
3
No of DOF
4
4
4
4
0
It‟s clear that the system is solvable as the degrees of freedom of process is zero, so either a matrix is constructed and a system of equations is solved (28 equations, i.e. square matrix 28×28) and it‟ll be so time consuming, or the easier method unit-to-unit calculations. But note that the solution will no be straight forward as there is no unit that has zero degrees of freedom, so tearing technique can be applied. It should be clear first that a number of variables will be assumed and recalculated again, the assumed variables will be determined based on the degrees of freedom table, as the assumed variables must solve a unit and introduce new information to the next unit and so on till the recycle stream is calculated and the assumed variables are recalculated again.
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Fundamentals of Chemical Engineering
So let‟s assume the flow rates of the components in stream 2 (feed to the reactor), then the sequence of solution will be as follows: 7
1
3 Mixer
Reactor 2 calculated
2 assumed
4 Separator
Splitter
5
6
Now, stream 2 can be totally assumed (flow rates of all four components) then four information are added to the reactor and then can be solved, then stream 3 is specified which will allow the solution of the separator, which in turn will add four information (stream 7) to the mixer and as a result the mixer can be solved and stream 2 can be recalculated. Then the calculated values are compared with the assumed ones, if the difference is in the range of allowable error then it‟s Ok, else the calculated values are used as the assumed ones till the values of the assumed and the calculated flow rates are equal (or the difference between them is nearly zero). To achieve this, a table must be constructed (like that constructed in the previous example) but the only difference is that there will be two columns for the same stream “2” (assumed one) one will be for the assumed value and the other for the calculated one. And the aim is to equalize the terms in both columns. Note that the cells of the assumed flow rates will not contain formulas; it‟ll contain the assumed values, while the cells of the calculated flow rates will contain the formulas. The formulas will be inserted in the same way as the previous example; the table will be as follows:
Fundamentals of Chemical Engineering
41
These formulas will result in the following flow rates:
CO H2
1 330 620
2 assum 600 800
3 360 320
4 349.2 313.6
5 10.8 6.4
6 34.92 31.36
7 314.28 282.24
2 Calc 644.28 902.24
CH4
50
100
100
96
4
9.6
86.4
136.4
CH3OH
0
10
250
10
240
1
9
9
Total
1000
1510
1030
768.8
261.2
76.88
691.92
1691.92
r
240
Feed
1000
Kgmol/hr
It‟s now clear that the flow rates of column “2 assumed” are not equal to those of column “2 calc” so if we can change the flow rates of column “2 assumed” (as they contain values not formulas) so that they are equal to column “2 calc” then these flow rates are the final ones and all the other cells will be automatically calculated. To achieve this we can insert another column in which the square the difference between the assumed and the calculated values is calculated (Why Square Of The Difference???), and the sum of the squares will be calculated below, then the table will be in the form shown:
This will yield the following values:
CO H2
1 330 620
2 assum 600 800
3 360 320
4 349.2 313.6
5 10.8 6.4
6 34.92 31.36
7 314.28 282.24
2 Calc 644.28 902.24
square Diff 1960.7184 10453.0176
CH4
50
100
100
96
4
9.6
86.4
136.4
1324.96
CH3OH
0
10
250
10
240
1
9
9
1
Total
1000
1510
1030
768.8
261.2
76.88
691.92
1691.92
188779246
r
240
Feed
1000
Kgmol/hr
Now we can say that setting the cell “J6” to the value of zero means that the difference between the assumed and the calculated values is zero, or they are equal. The final step is to use the solver to change the assumed values in order to set the cell “J6” to the value of zero.
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Fundamentals of Chemical Engineering
Finally, the solver solution will yield the following results:
CO H2
1 330 620
2 assum 3 4 5 6 7 2 Calc square Diff 692.985693 415.791416 403.317673 12.4737425 40.3317673 362.985906 692.985906 0.01460138 1110.25734 555.868781 544.751406 11.1173756 54.4751406 490.276265 1110.27627 0.13758542
CH4
50
367.668119 367.668119 352.961394 14.7067248 35.2961394 317.665255 367.665255 0.05351814
CH3OH
0
10.3516531 287.54593 11.5018372 276.044093 1.15018372 10.3516535 10.3516535 0.00063926
Total
1000
2181.2628 1626.87425 1312.53231 314.341936 131.253231 1181.27908 2181.27908 0.20634421
r
277.194277
Feed
1000
Kgmol/hr
Now compare the values of the cells in column “2 assum” and column “2 calc”. It‟s clear that assumed and calculated values are the same. Now all flow rates are available in the table.
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Fundamentals of Chemical Engineering
Energy Balance The basic Closed-system energy balance equation (for non reactive systems)
Where Hi, Hj are the specific enthalpies of the stream i (outlet stream) and stream j (inlet stream) [KJ/Kg] Zi, Zj are the elevations of streams i and j i, j are the velocities of streams i and j Q is the heat input to the system W is the work done by the system •
Magnitude of potential and kinetic energies are relatively small in chemical process applications
•
Except pumps, turbines and compressors most chemical process units do not involve work
•
If this is the case, the energy balance will reduce to:
Now, it‟s important to know how the enthalpy term is calculated. Enthalpy: Specific Enthalpy term represents the energy of the stream, and it‟s calculated from:
It‟s known that calculating enthalpy requires setting a reference temperature, i.e. the value of enthalpy varies according to the reference temperature. So, the above equation must be in the following form
For a mixture the specific enthalpy takes the form:
For non-ideal mixtures a term of heat of mixing must be added. Heat capacities are always function of temperature, the general form of the heat capacity is: Cp=C1+C2T+C3T2+C4T3+C5T4
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Fundamentals of Chemical Engineering
So after integration H will have the form (For a single component stream):
The values of constants C1, C2, C3, C4 and C5 vary according to the component, these values are available in many references, the most popular are: -
Coulson and Richardson, “Chemical Engineering”, Volume 6, Appendix D.
-
Robert H. Perry, “Chemical Engineers‟ Handbook”, Chapter 2.
So for a single component stream, say liquid Benzene, to calculate its enthalpy at any temperature we have to get the values of the constants, which will be : C1=1.2944E+05 C2=-1.6950E+02 C3=6.4781E-01 C4=0 C5=0 Where T in Kelvin, Cp in J/Kmol.K (these values are got from chapter 2 in Perry). Then, taking reference temperature 25oC (298oK), the value of the enthalpy of this stream will be calculated as follows:
Substituting the temperature by 70oC (343oK) the solution will be H=6.3795E6 J/Kmol Now, let‟s consider such a stream but with two components, say toluene and o-xylene (500 Kmol/hr each), and we need to calculate the mixture enthalpy at temperature of 80oC taking reference temperature of 298oK, so first the constants of both must be got and they were found to be: Toluene
Xylene
C1
1.4014E+05
3.36500E+04
C2
-1.5230E+02
1.0175E+03
C3
6.9500E-01
-2.6300E+00
C4
0
3.0200E-03
C5
0
0
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Fundamentals of Chemical Engineering
Then the enthalpy of the stream will be the sum of the enthalpies of both components, so each one will be separately calculated then added to each other. The specific enthalpy of Toluene will be 7327606.7 J/Kmol The specific enthalpy of Xylene will be 8508358.9 J/Kmol And the total enthalpy of the stream =500×7327606.7+500×8508358.9=7.918E+09 J/hr =7.918E+06 KJ/hr The same can be applied for three or more components Note: Try to check the values of the constants from Perry to make sure that you are able to get the right values. Calculating Enthalpy of streams using Microsoft Excel: Consider the first case of pure Benzene stream: First: the values of the constants must be entered in the worksheet:
Second: Enter the values of the reference temperature and the operating temperature, enter the values in Celsius and calculate their values in Kelvin.
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Fundamentals of Chemical Engineering
Third: Start writing the equation of enthalpy calculation as a function of temperature.
Fourth: If the stream flow rate is 100 Kmol/hr, then the total enthalpy of the stream can be calculated.
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Fundamentals of Chemical Engineering
For a case of a multicomponent stream, say the second example (toluene and o-xylene): First: Introducing the constants for both components
Second: Entering the values of operating temperature and reference temperature, in oC and oK.
Third: Entering the formulas of specific enthalpies for both components, you may save time by entering the formula for the first component and put dollar signs at the cells containing temperature, then drag this formula to the next component, this will be so useful and time saving if there is a big number of components.
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Fundamentals of Chemical Engineering
Dragging this formula will yield the following results:
Fourth: Entering the flow rates of both streams to calculate the Enthalpy of each component.
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Fundamentals of Chemical Engineering
Finally: The total enthalpy is calculated as the sum of both enthalpies of toluene and xylene. The final result will be 9.836E9 J/hr
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Fundamentals of Chemical Engineering
Energy balance on a non-reactive system In many cases there are many units in which information are available about the flow rates and the inlet temperatures, and it‟s required to get the outlet temperature, or cases where there is a stream that has to be at specific conditions of temperature and pressure, so other related streams must be adjusted to keep that stream at its required conditions. In such cases the conditions are adjusted by performing energy balance calculations on such a stream. Example: In a process air is required to be heated to a temperature of 100oC. If the available heat source provides 1.8E+06 KJ/sec. The feed flow rate is 1000 gmol/sec (790 gmol N2/sec and 210 gmol H2/sec), what will be the temperature of the feed? T
100oC
N2 O2
N2 O2
On applying the energy balance equation on this system, the equation will be:
Taking Tref=25oC=298oK, and Tout=100+273=337oK
Note that the units have to be homogeneous, i.e. the units of LHS must be KJ/sec as that of Q or the RHS must be J/hr as that of Enthalpies. It‟s now clear that all the parameters of the above equation are known except the inlet temperature, and that this temperature must be adjusted so that the LHS=RHS, so a value will be assumed and then the actual value is got by trial and error (let the assumed value 30oC). Let‟s see how this is performed using Microsoft Excel. First each term has to be calculated alone, i.e. Feed Enthalpy is first calculated, then product Enthalpy. Feed Enthalpy will be easily calculated as what was done in the previous examples based on any assumed value (let it 30oC as mentioned above).
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Fundamentals of Chemical Engineering
The same step is done for the product stream
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Fundamentals of Chemical Engineering
Now only remains to place the Q term in the excel sheet
The calculated value of Q based on the assumed feed temperature of 30oC will be 2.05E+06 J/sec. This value is required to be 1.8E+06 J/sec by changing the value of the inlet feed temperature. Using the solver this can be done in only one step.
By clicking solve, the value of the feed temperature will be adjusted so that Q will be 1.8E+08 J/sec and the feed temperature is now 311.56 oK which is 38.56oC
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Fundamentals of Chemical Engineering
Energy balance on a reactive system For a reactive system the energy balance equation is the same but only one term is added to the equation which is the heat of reaction term. The equation takes the form: Where r is the rate of reaction, Hr is the heat of reaction at the reaction temperature. Note that Hr is negative for exothermic reactions and positive for endothermic reactions. The heat of reaction can be calculated from the heat of formation of products and reactants at the standard temperature (25oC) according to the relation:
Calculate the standard heat of reaction for the oxidation of Pentane: C5H12 + 8O2 5CO2 + 6H2O Given: Hf,C5H10 = - 146.54 KJ/mol o
Hf,O2 o = 0.0 ( because it‟s an element) Hf,CO2 o = - 393.77 KJ/mol Hf, H2O o = - 242 KJ/mol So Hro = (5×-393.77+6×-242) - ( -146.54+8×0 ) = -3274.31 KJ/mol Performing such a step using Microsoft Excel is easy as it‟s only adding and subtracting as shown
To get the heat of reaction at any other temperature, some more calculations must be done. These calculations are based on the next diagram: Reactants (at Reaction temp.)
Reactants (at 298oK)
Trxn
298oK
Example: Calculate the heat of reaction of oxidation of Pentane at 100oC
Products (at Reaction temp.)
Products (at 298oK)
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Fundamentals of Chemical Engineering
To do so we will need the constants for the Cp‟s for all components, and Hr at reference temperature (298oK) Each term of the last equation will be calculated alone and finally they all will be added. First the reactants term :
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Fundamentals of Chemical Engineering
The same step is done for the third term in the equation
Finally the three terms are added to get the heat of reaction at 100oC
The heat of reaction at 100oC is -1223.17 KJ/mol
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Fundamentals of Chemical Engineering
Example: Pentane at 100oC and 1 bar is fed to a heater at a flow rate of 100 mol/sec where complete combustion (reaction with oxygen) occurs according to the reaction: C5H12+8O2 5CO2 + 6H2O Air enters so that the oxygen molar flow rate is eight times that of pentane, and Nitrogen is four times the oxygen. Calculate the temperature of the gases leaving the heater assuming no heat losses from the heater. 100oC C5H12 O2 N2
N2 CO2 H2O
To get the temperature of the exit stream the energy balance equation has to be solved.
As the previous example, each term is calculated separately, taking reference temperature 298 oK. Note that the term Q=0 as there is no heat added or removed from the system. First : Calculate the enthalpy of feed as past examples, note that the flow rate of Pentane is 100Kmol/hr,
flow
rate
of
Oxygen
is
8×100=800
mol/sec
and
the
flow
rate
of
Nitrogen=4×800=3200Kmol/hr.
Second : Heat of reaction term needs the rate of reaction and it equals the number of moles of reactant converted to product per unit time, which –in this case- is the feed flow rate 100 Kmol/hr. And the heat of reaction was calculated in the previous example.
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Fundamentals of Chemical Engineering
Third: Calculate the Enthalpy of product stream based on an assumed temperature (say 120oC) Note that the product flow rates will be calculated based on MB calculations: CO2)out=CO2)in+5×r=500 Kmol/hr H2O)out= H2O)in+6×r=600 Kmol/hr N2)out= N2)in=3200 Kmol/hr Both Pentane and Oxygen are totally consumed.
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Fundamentals of Chemical Engineering
Fourth: the Energy balance equation is now written
Finally use the solver to change the value of the product temperature to make the Energy balance equation =0
Fluid Mechanics Terminal Velocity of Falling Particles A simple force balance on a spherical particle reaching terminal velocity in a fluid is given by
Where Vt is the terminal velocity in m/s, g is the acceleration of gravity given by g = 9.80665 m/s2, is the particles density in kg/m3, r is the fluid density in kg/m3, DP is the diameter of the spherical particle in m and CD is a dimensionless drag coefficient. The drag coefficient on a spherical particle at terminal velocity varies with the Reynolds number (Re) as follows (pp. 5-63, 5-64 in Perry3).
Where
for
Re
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