Tutorial Aspen Hysys 1b1k6m

on the keyboard; Press keys <Shift> on the keyboard.

To observe the Temperatures, Pressures, and total Flow rates of all material streams in the flowsheet.

Note that Æ

These hot keys give you a quick overview of the distribution of temperature, pressure, or flow in a flowsheet. The * in front of a value indicates that it is a specified variable, one you supplied a number for.

Press keys <Shift> on the keyboard.

To display the material stream Names again.

2. Place the cursor on stream S2 for a moment; Place the cursor on reactor R1 for a moment.

To observe the fly-by information displayed in a small white box for each object.

Note that Æ

The hot keys and fly-byes provide you with specific information about some operating conditions. How can you get a table of information displayed in the PFD window, like the material stream info that appears in the workbook?

3. Click the middle Maximize Window icon ‡ in the upper-right part of the PFD window.

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To expand the PFD window to fit the full area of the HYSYS desktop.

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Tutorial 2.9 Object inspect an open area of the PFD and select Add Workbook Table from the dropdown menu.

To open the Select Workbook Page window. Stream pages that appear in the Workbook are displayed in a list.

Click on Streams in the list, if necessary, and then click the Select button.

To have a stream table placed in the PFD window.

Note that Æ

This table is the same as what appears in the Streams page of the Workbook. Whenever you make changes in the content or format of the Streams page in the Workbook, they will automatically be reflected in this PFD table.

4. Click the Zoom All button in lower left of the PFD window located between the – and +.

To position the entire flowsheet and the stream table in the PFD window.

Click and drag the stream table so that it is centered under the flowsheet in the PFD.

To give it a professional appearance for documentation purposes.

Click the Zoom All button in lower left of the PFD window located between the – and +.

To position again the entire flowsheet and the stream table in the PFD window.

Note that Æ

You may sometimes what to print the PFD window but only with the flowsheet and not the stream table. Before you do the print, you can hide the stream table by object inspecting it and selecting the Hide option. You can always retrieve hidden objects by object inspecting the PFD and selecting the Reveal Hidden Objects option.

Note that Æ

You could print the PFD now, but it needs a title, the assignment, your name, and the date. In other words, you need to document your work.

G. Add some documentation text. The add text function allows you to place text such as titles and dates in the PFD. This section will show you how to add a title to the PFD and size the title to fit on one line. You will also add a second line for the assignment, your name, and the date. 1. Place the cursor on the arrow Πat the top of the PFD vertical scroll bar and scroll the flowsheet and stream table down about 2 inches. Click the Add Text icon in the PFD button bar.

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To make room for some textual information which will appear above the flowsheet. To enter the Add Text mode. The cursor will now appear with a white box.

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Tutorial 2.9 2. Click the cursor anywhere above the flowsheet. Type Cumene Production from Propylene and Benzene and click the OK button. Do not use the <Enter> key while typing. Note that Æ

To open the Text Props View window. To create a text box on the PFD with a title for the flowsheet. The text is surround by a white rectangle and each word of the title appears to be on a new line. However, the text box needs to be elongated to produce a one-line title. To enter the size mode, which allows you to resize any object in the PFD that is selected.

Click the Size Mode icon in the PFD button bar. Note that Æ

Place the cursor on the right, white square of the selected text; wait for the cursor to change to symbol ↔, then drag the white square to the right until the text appears as one line.

The white rectangle around the text has two small, white squares on the right and left sides. These are the resize indicators. If the text is not selected, just click on it to select it. To elongate the text into a one-line title.

Note that Æ

If you have made a spelling error in the title, then object inspect the title and select View Properties… from the drop-down menu, in order to correct your error.

Note that Æ

To give this title more prominence, you what to increase the text font size and change its color.

3. Object inspect the text box and select Change Font… from the drop-down menu.

To open the Font window. To increase the prominence of the title.

Increase the text size to 20 and click OK. Elongate the text to a one-line title again. Object inspect the text box and select Change Colour… from the drop-down menu.

To open the Color window.

Select a red color and click the OK button.

To change the text color of the title.

Click and drag the text box.

To center the title over the flowsheet. Use the Zoom All button to help you with this task. Note that Æ

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You are to leave enough room between the title and the flowsheet, so that you can add another but smaller line of text.

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Tutorial 2.9 4. Click Add Text icon in the PFD button bar; Click the cursor anywhere below the title.

To enter the Add Text mode and open the Text Props View window.

Type Tutorial 2.9, your name, date and click the OK button.

To create a text box between the title and the flowsheet containing the assignment, your name, and the date.

Click and drag the right side of the text box.

To elongate it into one line of text.

Click and drag the text box.

To center it under the title.

5. Click the Zoom All button in lower left of the PFD window located between the – and +.

To position the title, the flowsheet and the stream table in the PFD window.

Choose File/Printer Setup/Graphic Printer… from the menu bar

To select your printer destination for graphical output.

Object inspect an open area of the PFD and select Print PFD from the drop-down menu.

To send the contents of the PFD to your selected printer.

Note that Æ

6. Click the Save Case icon in the HYSYS button bar.

You have just documented and printed this tutorial assignment. You should document all work that you do in HYSYS. Always supply, at least, a title, assignment number, name, and date. To save all of the changes you have made to the PFD, so far, as a simulation case file in one of your personal folders.

H. Connect and disconnect PFD objects. Process streams and process units can be connected and disconnected, either through object property windows or with interactive tools available on the PFD. While using the interactive tools and also specifying additional information, you may inadvertently cause a consistency error. HYSYS issues these kinds of errors when a process unit simulation is over-specified; that is too much information is known for the process unit calculation, and HYSYS can not resolve the inconsistency. You must do that by specifying only the correct amount of information. In the PFD of file pfdtools.hsc, the flowsheet simulation for the heater E1 and reactor R1 is represented mathematically as follows:

ΨS 2 , ΔPE 1 , Q E1 = heater ΨS1 , TS 2 , PS 2 ΨS 3 , ΔPR1 , Q R1 = reactor ΨS 2 , TS 3 , PS 3

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Tutorial 2.9 where heater or reactor is a function whose variables on the left are calculated by HYSYS once those variables on the right are known. The vector Ψi is a short notation to represent the temperature, pressure, flow rate, and chemical composition of stream i; that is, the process state variables for that stream. Since the process state of stream S1 and the temperature and pressure of stream S2 are known, the heater calculation is done first by HYSYS, in order to determine the remaining conditions of stream S2. The reactor calculation is done second by HYSYS, because the temperature and pressure of stream S3 are known and the conditions of stream S2 were determined by the heater simulation. This calculation order of heater followed by reactor is called forward propagation because it follows the material flow of the streams. Another way of looking at the forward propagation is that the outlet streams of a process unit are calculated once its inlet streams and equipment parameters are known. This section shows you how to disconnect heater E1 from outlet stream S2 and then reconnect them. You will also create and fix a consistency error associated with process unit E1. Proceed as follows: 1. Click the drag zoom icon in the button bar at the top of the PFD window. Click and drag a box that surrounds just the flowsheet, and then release the mouse button.

To choose the heater and reactor and their attached streams. HYSYS will automatically zoom in on the area included in the box you have just outlined. To enter the break connection mode. The cursor will now appear with an X icon.

2. Click the Break Connection button in the PFD button bar.

To break the connection of the outlet stream on heater E1.

Click the blue line between heater E1 and stream S2.

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To enter the zoom mode. The cursor will now appear with a magnifying glass icon.

Note that Æ

Stream S2 remains connected to the reactor; however, it is light blue indicating that it is no longer fully specified. Stream S3 is also light blue because not enough conditions are specified to do the reactor calculations. When stream S2 was originally connected to heater E1, the HYSYS simulation of heater E1 had calculated the unknown conditions of stream S2 and thus made them known information to the reactor unit.

Note that Æ

A conversion reactor in a HYSYS simulation may produce a vapor outlet stream (S3), liquid outlet stream (S4), or both. In this simulation, the reaction is in the vapor phase only; however, HYSYS requires that a liquid product stream be defined. We will ignore stream S4 in our discussion, since its flow rate

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Tutorial 2.9 is zero. Note that Æ

You will now change the propagation order in the flowsheet by specify the process state of stream S2. Starting with stream S2, HYSYS will do a forward calculation on reactor R1 first, and then it will try to do a backward calculation on heater E1. This propagation order is represented mathematically as follows:

ΨS 3 , ΔPR1 , Q R1 = reactor ΨS 2 , TS 3 , PS 3 ΨS 1 , ΔPE 1 , Q E 1 = heater 2 ΨS 2 , TS 1 , PS1 where reactor or heater2 is a function whose variables on the left are calculated by HYSYS once those variables on the right are specified. The vector Ψi is a short notation to represent the temperature, pressure, flow rate, and chemical composition of stream i. The reactor function above depicts a forward calculation order, while the heater2 function represents a backward calculation order (i.e., knowing an outlet stream calculate the inlet stream conditions).

HYSYS does a process unit calculation as soon as enough information is known, and then it propagates the simulation both forward and backwards until either it can not propagate further (because not enough information is known) or a consistency error occurs. 3. Double click the stream S2 icon in the PFD.

To open the property window for stream S2. Only the temperature and pressure of S2 are specified, as indicated by the blue values.

Double-click on the Molar Flow cell of stream S2.

To open the Input Composition for Stream window.

Click Mole Flows in the Composition Basis area.

To select the composition basis as component molar flow rates.

Enter 200 in CompMoleFlow cell for benzene. Enter 150 in CompMoleFlow cell for propene. Enter 5 in CompMoleFlow cell for propane Enter 0 in CompMoleFlow cell for cumene.

To specify the kgmoles/hour for benzene. To specify the kgmoles/hour for proplyene. To specify the kgmoles/hour for propane. To specify the kgmoles/hour for cumene.

Click the OK button in the Input Composition for Stream window. (If it’s not visible, scroll to find it.)

To return you to the Worksheet/Conditions page of the stream property window. The remaining cells of stream S2 have values and

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Tutorial 2.9 its object status is green. To close the stream property window.

Click the Close button. Note that Æ

4. Position the cursor over the green light and red light in the HYSYS button bar.

In the PFD, streams S2 and S3 are dark blue indicating that HYSYS has successfully done the forward calculations on reactor R1. To view the popup of “Solver Active” for the green light and “Solver Holding” for the red light. The background colors of these two buttons indicate the state of the calculation solver for HYSYS.

Note that Æ

Currently, the background color of white for the green light and gray for the red light means the solver is active; that is, the green light is on and the red light is off. Thus, whenever you change any blue value, the calculations are done automatically.

Note that Æ

When the green light has a gray background and the red light has a white background, the solver is in the holding state; that is, the red light is on and the green light is off. Thus, whenever you change any blue values, the calculations are not done at all, until you click the green light to turn it on. Whenever HYSYS is doing the calculations and detects some error, it automatically stops the calculations, turns the red light on and the green light off, and issues an error message. When an error event happens, you must diagnosis it, make the necessary corrections, and turn the green light on to restart the calculations.

5. Hold the Ctrl key down on the keyboard and then proceed with the next action. Note that Æ

Place the cursor on the stream S2 icon. Click and drag the blue box that appears on the left of the stream icon to the blue box that appears at the outlet of heater E1.

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To enter temporarily the PFD attach mode and get an arrow cursor without cross hairs. The attach mode allows you to connect unit operations and streams, interactively. If you want the attach mode to be on for an extended period of time, then click the Attach Mode icon in the button bar at the top of the PFD window, instead of using the Ctrl key. To attach stream S2 to heater E1. Stream S2 now becomes the outlet stream of heater E1.

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Tutorial 2.9 Note that Æ

A Consistency Error window appears and the calculations have been stopped, as noted by the red STOP icon on the button bar. Once you fix the consistency error, you will click the green GO icon to continue the calculations.

Note that Æ

The Consistency Error window states that a new value for a component mole fraction of stream S1 was calculated by heater E1, but a specified, blue value already existed for that mole fraction. You must figure out what caused this inconsistency and correct it.

Note that Æ

The last thing you specified in the PFD was stream S2. HYSYS always starts with the latest change you make and propagates the calculations from that point in the PFD. You observed that HYSYS did the calculations on reactor R1 immediately after you specified stream S2 because enough information was known. HYSYS then stop its forward propagation because a process unit was not connected to stream S3.

Note that Æ

When you connected stream S2 to heater E1, HYSYS assumed you wanted a backward propagation done on heater E1, because you last changed stream S2. Mathematically, HYSYS tried to do the following calculation: ΨS 1 , ΔPE 1 , Q E 1 = heater 2 ΨS 2 , TS 1 , PS 1

After HYSYS completed this backward calculation for the conditions of stream S1, it observed that some conditions of this stream had previously been specified with blue values. This detection by HYSYS caused the consistency error.

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Note that Æ

Before you disconnected stream S2, HYSYS did a forward calculation on heater E1, because the process state of stream S1 was specified. After you disconnected stream S2, specified its process state, and reconnected it, HYSYS tried to do a backward calculation on heater E1. The resulting consistency error indicates that the simulation on heater E1 is over-specified. Too much information is known.

Note that Æ

At this point, you have two choices to correct the consistency error. You can de-specify the

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Tutorial 2.9 total and component flow rates of either stream S2 or S1. You decision is based on whether you want to do a backward or forward calculation on heater E1. For this tutorial, you want to do a backward calculation; that is, calculate the total flow rate and composition of stream S1. 6. Click the Close button in the Consistency Error window.

To close the Consistency Error window so you can fix the error. If you used the Ctrl key to attach stream S2 to heater E1.

Proceed to Step 7. Note that Æ

However, if you used the Attach Mode icon in the button bar at the top of the PFD window to make the connection, then click the Attach Mode icon to exit the PFD attach mode and then get an arrow cursor with cross hairs You must exit the attach mode before you can change any process variables in the PFD. To open the property window of heater E1.

7. Double click on heater E1 icon in the PFD. Note that Æ

The object status of heater E1 is a yellow “Unknown Duty”, which implies that the calculation for this stream could not be done because of the consistency error.

Click the Worksheet/Conditions page, if necessary.

To view the conditions of stream S1.

Click in the Molar Flow cell of stream S1 and then hit the key.

To delete the specified molar flow rate of stream S1 and replace it with a blue <empty>.

Double-click on the Molar Flow cell of stream S1.

To open the Input Composition for Stream window.

Click the Erase button in the Composition Controls area and then click the OK button.

To erase the specified mole fractions of stream S1 and replace them with four blue <empty>’s.

Click the Close button.

To close the property window of heater E1 and return to the PFD window. Note that Æ

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In the PFD, stream S2 is fully specified because its material stream is dark blue. Stream S1 is only partially known because its material stream is light blue. Since the temperature and pressure of stream S1 are known, HYSYS should have done a backward calculation on heater E1. However, HYSYS

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Tutorial 2.9 is currently in the hold state, as implied by the red STOP icon on the button bar. 8. Click the Solver Active icon on the button bar; that is, the green GO icon.

To continue the simulation calculations on heater E1 based on a backward propagation.

Note that Æ

You have resolved the consistency error, and HYSYS has successfully done the simulation calculations, because all material streams are dark blue and all energy streams are dark maroon in the flowsheet of the PFD.

Note that Æ

HYSYS always activates the red STOP icon on the button bar whenever a consistency error occurs. After you have fixed a consistency error, you must to click the green GO icon, in order to have HYSYS continue the simulation calculations.

I. Copy a PFD to a Word document. After you complete your work on a HYSYS assignment, you may want to paste either part of the PFD or the whole PFD into a page of a Word document that you are preparing for your formal solution to an assignment. Proceed as follows to capture the current PFD, paste it as a picture into a Word document, and then manipulate this picture to see only the flowsheet figure: To bring the PFD to the top of the desktop, in order to prepare it for capture to the clipboard.

1. Click the PFD icon on the button bar, if necessary. Click the Zoom All button in lower left of the PFD window located between the – and +.

To place the entire flowsheet in the PFD window.

Click the cursor in an empty area, then Move the mouse out of the PFD window.

To de-activate any thing that is selected, and to remove the cross hairs from the PFD.

Press the Print Screen (or PrtScrn) key in the upper-right portion of the keyboard.

To capture the complete contents of the monitor screen as a picture image in the computer's clipboard.

Note that Æ

2. Choose Start/Word 2003 from the Windows desktop. Close the Getting Started box on the far right,

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These above tasks complete what you have to do in the HYSYS program. You will now copy the captured picture image and manipulate it using the Microsoft Word program. To start the Microsoft Word program, which will open with a new document. To display fully the word document window.

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Tutorial 2.9 if it is visible. Click the Show/Hide ¶ icon on the button bar.

To show the paragraph marks.

Press the Enter key on the keyboard twice.

To add two more lines to the new document.

Place the cursor in the second line.

To identify the insertion point for the picture.

Select Edit/Paste from the Word menu bar.

To insert the picture from the clipboard into the Word document.

Object inspect the pasted picture and then select Show Picture Toolbar from the dropdown menu.

To get the popup Picture Toolbar of buttons.

Select the Crop button in the Picture Toolbar.

To get a cursor that looks like the Crop button icon.

Place the Crop cursor over the little black square in the middle of the left edge of the picture. Click and drag this edge to the left edge of the flowsheet.

To remove the left portion of the picture up to the left edge of the flowsheet.

Place the Crop cursor over the little black square in the middle of the right edge of the picture. Click and drag this edge to the right edge of the flowsheet.

To remove the right portion of the picture up to the right edge of the flowsheet.

Place the Crop cursor over the little black square in the middle of the top edge of the picture. Click and drag this edge to the top edge of the flowsheet.

To remove the top portion of the picture up to the top edge of the flowsheet.

Place the Crop cursor over the little black square in the middle of the bottom edge of the picture. Click and drag this edge to the bottom edge of the flowsheet.

To remove the bottom portion of the picture up to the bottom edge of the flowsheet.

Click the Set Transparent Color button in the Picture Toolbar bar and then click its cursor in the lower left area of the cropped picture.

To turn the background color in the picture to white.

Close the Picture Toolbar.

To remove it from the document window. . To select the flowsheet picture and center it within the line of the Word document.

3. Click the cursor inside of the cropped picture, and click the Center icon on Word button bar. Object inspect the flowsheet picture and then select Borders and Shading… from the dropdown menu.

To get the Borders window.

Select Box and 1½-pt width, and click the OK button.

To place a border around the flowsheet picture.

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Tutorial 2.9 Object inspect the flowsheet picture again and select Format Picture… from the drop-down menu.

To get the Format Picture window.

Select the Size tab, change the Scale Height and Width to 100%, and click the OK button.

To enlarge the flowsheet picture.

Click the cursor somewhere outside the picture.

To deactivated the flowsheet picture.

Note that Æ

You can now place a centered figure title on the next line right under the flowsheet picture. To exit the Microsoft Word program and not save the Word document.

4. Choose File/Exit from the menu bar, then click the No button.

J. Close the simulation case. You will close the file of your simulation case and then possibly exit HYSYS. Proceed as follows: 1. Choose File/Close Case from the menu bar, then click the No button. or Press the keys simultaneously, then click the No button.

To close the current simulation file and not save it.

2. Choose File/Exit from the menu bar or Press keys on the keyboard.

To exit the HYSYS program, if you do not plan to do another process simulation case.

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Overview This chapter provides five problem assignments to help you develop your abilities and confidence to simulate individual process units using the HYSYS process simulation software. Over a five-week period, you will have weekly assignments that focus on finding the material and energy balance requirements for the following individual process units: Problem HY.1 HY.2 HY.3 HY.4 HY.5

Description Process Stream Simulation Pump Simulation Cooler Simulation Mixer/Tee Simulation Reactor Simulation

Each weekly problem will be assigned in a separate memorandum. Once you’ve completed these assignments, you will have a mathematical understanding of how HYSYS does its calculations for each process unit. This chapter assumes you have completed certain tutorials found in Chapter 2. They are Tutorials 2.1 to 2.6 and 2.9. Each assignment in this chapter identifies those tutorials that you should complete before you try to solve the problem. While solving a problem, you will need to consult Appendices B, C, etc. in this handbook for information on certain HYSYS simulation modules. Each appendix or module provides a mathematical explanation of how HYSYS does its calculations for that process unit. A module includes a description, a conceptual model, a mathematical model, variable descriptions, example mathematical algorithms, and several HYSYS simulation algorithms. Each assignment identifies which appendix you must consult. Microsoft Word files of the five assignments in this chapter are available on the departments network file server (R:) in folder chem_engineering/public/HYSYS Manual/Chap 3. Your instructor will give you directions on how to access this folder.

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Problem HY.1 Process Stream Simulation The HYSYS simulation module for a process stream is fully defined in Appendix C of this handbook. This process stream module contains a process description, process diagram, assumptions, mathematical model, variable descriptions, mathematical algorithms, and some HYSYS simulation algorithms in functional form. After you read this information about a process stream, you are to practice the HYSYS session below and then do the simulation exercise. For your solution of Problem HY.1, you are to provide a printed copy of the HYSYS flowsheet and the worksheet datasheet for the process stream in the simulation exercise only. Furthermore, you are to provide answers to the questions in the simulation exercise.

HYSYS Session This session will show you, in general , how to do a HYSYS simulation for a process stream. It assumes you are familiar with the material in Tutorials 2.1 to 2.5 of this handbook. You are to find the vapor fraction and heat flow in BTU/hr of a binary mixture of toluene and hexane at 1 atm and 100°C. This mixture is flowing at 100 kgmol/h. The stream is labeled feed and its conceptual diagram is: TF = 100 ° C

TF = 100 ° C

PF = 1 atm n& F = 100 kgmol / h

PF = 1 atm n& F = 100 kgmol / h

feed

z F ,TL = 0.7

z F ,TL = 0.7

z F , HX = 0.3

z F , HX = 0.3

Using the information from above, you are to practice a HYSYS simulation by doing the following general tasks: • • • • • • •

Create a new HYSYS file. Name it xxx_HY1, where xxx are your initials. Choose a property package. Select the SRK property package, which is an equation of state. Specify the two chemical components. Create a process stream and name it feed. Specify the state of this process stream; that is, its temperature, pressure, flow rate, and composition. Change the HYSYS preferences to display your desired property units. Observe the HYSYS calculated results, which appear in black. Note that values you supplied appear in blue.

After you specify the state of the process stream, HYSYS immediately calculates all of the other properties of that streams (such as mass flow rate, volumetric flow rate, vapor fraction, etc.) using the SoaveRedlich-Kwong (SRK) equation of state. The answers for two of the stream properties are: Vapor Fraction = 0.783 = -1.230E6 BTU/h Heat Flow

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Problem HY.1 A value for the vapor fraction between zero and one indicates that the process stream exists as a two-phase system (vapor and liquid) at the specified temperature, pressure, and composition.

Simulation Exercise What are the temperature in °F and the heat flow in BTU/hr of a process stream containing a tertiary mixture of benzene, hexane, and toluene? The stream is at 2 atm and has a vapor fraction equal to zero. The chemical component flows are as shown below. Vf = 0.0

Vf = 0.0 TF =

?

TF =

?

PF = n& F =

2 atm

PF = n& F =

2 atm

?

feed

?

n& F , BZ = 40 kgmol / h

n& F , BZ = 40 kgmol / h

n& F , HX = 70 kgmol / h

n& F , HX = 70 kgmol / h

n& F ,TL = 120 kgmol / h

n& F ,TL = 120 kgmol / h

While you are completing your HYSYS simulation on the above process stream problem, please answer the following questions: 1. Which HYSYS simulation algorithm in Appendix C of the Process Stream Module would you use to solve the above problem? What temperature does this algorithm calculate? 2. Which assumption(s) the fact that a process stream has uniform properties throughout its length? 3. What steps from the Mathematical Algorithm A would you use to calculate the mass fractions from the mole fractions? that HYSYS has done this calculation correctly? 4. Calculate the enthalpy of the stream in BTU/lb using the appropriate mathematical model equation(s). 5. What are the definitions of the dew-point temperature and bubble-point temperature? 6. How might you find the dew-point temperature of this stream? What is this value? For the above process stream problem, please print its HYSYS flowsheet and its worksheet datasheet.

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Problem HY.2 Pump Simulation The HYSYS simulation module for a pump operation is fully defined in Appendix E of this handbook. This pump module contains a process description, process diagram, assumptions, mathematical model, variable descriptions, mathematical algorithms, and some HYSYS simulation algorithms in functional form. After you read this information about a pump operation, you are to practice the HYSYS session below and then do the simulation exercise. For your solution of Problem HY.2, you are to provide a printed copy of the HYSYS flowsheet and the design and worksheet datasheets for the pump operation in the simulation exercise only. Furthermore, you are to provide answers to the questions in the simulation exercise.

HYSYS Session This session will show you how to do a HYSYS simulation for the pump operation. It assumes you are familiar with the material in Tutorials 2.1 to 2.5 of this handbook. You are to find the power in watts to compress an equimolar mixture of n-hexane and n-octane at 25°C from 1 atm to 4 atm. This liquid mixture is flowing at 100 lb-moles per hour. The pump is labeled P200, and its adiabatic efficiency is 70 percent. The conceptual diagram is: TE = ? TI = 25 ° C PI = 1 atm n I = 100 lbmol / h z I , HX = 0.5 z I ,OC = 0.5

W A = ? Inlet

Exit P-200

ε = 70%

PE = 4 atm n E = ? z E , HX = ? z E ,OC = ?

Using the information from above, you are to practice a HYSYS simulation by doing the following general tasks: • • • • • • • • • •

Create a new HYSYS file. Name it xxx_HY2, where xxx are your initials. Choose a property package. Select the SRK property package, which is an equation of state. Specify the two chemical components. Create a pump and name it P-200. Create the pump inlet, outlet, and energy streams. Name them Inlet, Exit, and Wa, respectively. Supply the adiabatic efficiency of 70%. Specify the state of the inlet stream; that is, its temperature, pressure, flow rate, and composition. Specify the exit pressure. Change the HYSYS preferences to display your desired property units. Observe the HYSYS calculated results, which appear in black. Note that values you supplied appear in blue.

After you specify the state of the inlet stream, the exit pressure, and the pump efficiency, HYSYS immediately calculates all of the other properties of the two streams (such as mass flow rate, volumetric flow rate, vapor fraction, heat flow, etc.) using the Soave-Redlich-Kwong (SRK) equation of state. Also, HYSYS calculates the pump power to be: Beef, Inc.

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3. Process Unit Assignments

Problem HY.2 Power = 0.81 watts A positive value for the pump power indicates that energy must be added to the process stream to increase its pressure.

Simulation Exercise What adiabatic efficiency and power in kilowatts are required to compress an equimolar mixture of n-hexane and n-octane from 25°C and 1 atm to 40°C and 400 atm? This liquid mixture is flowing at 100 lb-moles per hour. TE = 40 ° C TI = 25 ° C PI = 1 atm n I = 100 lbmol / h z I , HX = 0.5 z I ,OC = 0.5

W A = ? Inlet

Exit pump

ε=?

PE = 400 atm n E = ? z E , HX =

?

z E ,OC =

?

While you are completing your HYSYS simulation on the above pump problem, please answer the following questions: 1. Which HYSYS pump simulation algorithm (pumpa, pumpb, etc. in Appendix E) would you use to solve this problem? What are the given variables and their values? What are the calculated variables and their values? 2. After you examine the process states (i.e., the temperature, pressure, flow rate, and composition) of the inlet and exit streams, please answer the following questions. Why does the exit temperature increase slightly? What is unique about the molar flow rate and composition? What equations in the math model and steps in the algorithm reflect it? What assumptions this uniqueness? What are the operating conditions that would invalidate each assumption? 3. What is the ideal work expressed in units of horsepower? 4. In the math algorithm, what variables are only functions of the material state (i.e., the temperature, pressure, and composition) of the liquid? 5. What does the assumption of “adiabatic process” imply? Is this a valid assumption? 6. What is the energy relative imbalance (%RIB) ? Show your calculations. The energy %RIB equals 100*(energy flow in - energy flow out) / (energy flow in). For the above pump problem, please print its HYSYS flowsheet and its design and worksheet datasheets.

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3. Process Unit Assignments

Problem HY.3 Cooler Simulation The HYSYS simulation module for a cooler operation is fully defined in Appendix G of this handbook. This cooler module contains a process description, process diagram, assumptions, mathematical model, variable descriptions, mathematical algorithms, and HYSYS simulation algorithms in functional form. After you read this information about a cooler operation, you are to practice the HYSYS session below and then do the simulation exercise. For your solution of Problem HY.3, you are to provide a printed copy of the HYSYS flowsheet and the design and worksheet datasheets for the cooler operation, as well as the plot generated by the case study, in the simulation exercise only. Furthermore, you are to provide answers to the questions in the simulation exercise.

HYSYS Session This session will show you how to do a HYSYS simulation for a cooler and perform a case study on the operation. It assumes you are familiar with the material in Tutorials 2.1 to 2.5 of this handbook. You are to find the duty in BTU/h needed to cool a mixture of ethanol and water from 200°C and 583 kPa to a saturated vapor at 570 kPa. This stream is flowing at 900 kgmoles per hour. The cooler is labeled E-200, and its pressure drop is 13 kPa. The conceptual diagram is: TI = 200 ° C PI = 583 kPa n I = 900 kgmol / h z I , ET = 0.6

Vf = 1.0

Q = ?

TE =

Inlet

Exit E-200

z I ,WA = 0.4

?

PE = 570 kPa n E = ? z E , ET =

?

z E ,WA =

?

Using the information from above, you are to practice a HYSYS simulation and case study by doing the following general tasks: • • • • • • • • •

Beef, Inc.

Create a new HYSYS file. Name it xxx_HY3, where xxx are your initials. Choose a property package. Select the PRSV property package, which is an equation of state. Specify the two chemical components. Create a cooler and name it E-200. Create the cooler inlet, outlet, and energy streams. Name them Inlet, Exit, and Q, respectively. Specify the state of the inlet stream; that is, its temperature, pressure, flow rate, and composition. Specify the exit pressure and vapor fraction. Change the HYSYS preferences to display your desired property units. Observe the HYSYS calculated results, which appear in black. Note that values you supplied appear in blue.

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3. Process Unit Assignments

Problem HY.3 After you specify the state of the inlet stream and the exit pressure and vapor fraction, HYSYS immediately calculates all of the other properties of the two streams (such as mass flow rate, volumetric flow rate, vapor fraction, heat flow, etc.) using the PRSV equation of state. Also, HYSYS calculates the cooler heat duty to be:

= 3.8309x10^6 BTU/hr Q should have a negative value to indicate that Based on thermodynamic sign conventions, this Q energy is removed from the process stream. However, a positive duty for a cooler operation in HYSYS indicates energy is being removed from the stream. The case study tool allows you to monitor the steady state response of key process variables to changes in your process. Basically, the case study allows you to do “what if” analyses. In this session, you are observing how the duty of the cooler varies when the exit temperature is changed from 200°C to 10°C. The following general tasks must be completed to perform the case study. Detailed instructions for each of these tasks is given in Tutorial 2.4 of this handbook. • • • • •

Insert the variables into the DataBook for your case study. These two variables are the heat duty of cooler E-200 and the temperature of the exit stream. Begin your “what if” analysis. Activate the IND check box for temperature and the Dep check box for the heat duty. Supply a Low Bound, High Bound, and Step Size for the temperature; that is, 10°C, 200°C, and 5°C, respectively. Begin the calculations and generate the plot of heat duty versus temperature.

Simulation Exercise What heat duty in kJ/h is removed from an equimolar mixture of n-hexane and n-octane to cool it from 300°C at 2 bar to a saturated vapor? a saturated liquid? a subcooled liquid at 100°C? What are the final temperatures of the exit stream in the first two cases? This liquid mixture is flowing at 100 lb-moles per hour. The cooler has a 0.2 bar pressure drop. Perform a case study to observe the effect of cooler duty as the independent variable versus the exit temperature as the dependent variable. Be sure to define a range of exit temperatures that includes both the single-phase regions (vapor and liquid) and the two-phase region (vapor-liquid). TI = 300 ° C PI = 2 bar n I = 100 lbmol / h

TE = Inlet

Cooler

z I , HX = 0.5 z I ,OC = 0.5

Beef, Inc.

Vf =

Q = ?

ΔP = 0.2 bar

3-7

Exit

PE = ? n E = ? z E , HX = ? z E ,OC = ?

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3. Process Unit Assignments

Problem HY.3 While you are completing your HYSYS simulation on the above cooler problem, please answer the following questions: 1. Which HYSYS heater/cooler simulation algorithm in Appendix G would you use to solve the exit phase problems? to solve the exit temperature problem? Which HYSYS heater/cooler simulation algorithm would you use to solve the case study problem? 2. Compare the cooler operation to the pump operation. Which assumptions for the two operations are the same? How do they differ? Why are some assumptions different? 3. The total energy of a process stream is composed of its molar enthalpy, kinetic energy, and potential energy. For the subcooled exit stream, what percentage of its total energy is potential energy? kinetic energy? (hint, a trick question) 4. Which equations in the mathematical model would you use to calculate the molar enthalpy of the exit stream given the conditions of the inlet stream and the duty of the cooler? Use these equations to find the exit stream molar enthalpy for a cooling operation with the inlet stream in this problem, and a duty of 2000 watts. Check that HYSYS calculates the same result. 5. In your printed case-study plot, label the dew-point temperature and bubble-point temperature on the temperature axis. After doing this task, you should notice three distinct areas on the plotted curve. Place a label on the liquid portion, vapor-liquid portion, and vapor portion of this plot. This temperature-heat-duty plot is for a multicomponent mixture. What would the vapor-liquid portion of this plot look like, if the mixture contained only one chemical component (e.g., pure n-hexane)? 6. What is the energy relative imbalance (%RIB) for an exit temperature of 100°C? Show your calculations. The energy %RIB equals 100*(energy flow in - energy flow out) / (energy flow in). For the above cooler problem at an exit temperature of 100°C, please print its HYSYS flowsheet, its design and worksheet datasheets, and the plot generated by the case study.

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3. Process Unit Assignments

Problem HY.4 Mixer/Tee Simulation The HYSYS process simulator can solve the material and energy balances of many unit operations that are interconnected by process streams. For example, two or more process streams can be fed to the HYSYS mixer operation to form one exit stream. This exit stream can then be fed to a HYSYS tee operation to split it into several streams. Problem HY.4 uses this example to illustrate how to do a process simulation that has several unit operations in a process flow diagram (PFD). The HYSYS simulation module for a stream mixer operation is fully defined in Appendix D of this handbook. The stream mixer module contains a process description, process diagram, assumptions, mathematical model, variable descriptions, mathematical algorithm, and HYSYS simulation algorithms in functional form. No simulation module exists for the tee operation in this handbook. After you read the information about the stream mixer, you are to practice the HYSYS session below and then do the simulation exercise. For your solution of Problem HY.4, you are to provide a printed copy of the HYSYS flowsheet and the design and worksheet datasheets for the mixer and tee operations in the simulation exercise only. Furthermore, you are to provide answers to the questions in the simulation exercise.

HYSYS Session This session will show you how to do a HYSYS simulation for a stream mixer and a tee, and how to attach two process units. It assumes you are familiar with the material in Tutorials 2.4, 2.5, and 2.9 of this handbook. First, a pure heptane stream is mixed with a pure octane stream. The resulting binary mixture is then split into two streams with different flow rates. You are to find the molar flow rate in kgmol/h and the mole fractions of the tee’s two exit streams. The pure heptane stream is flowing at 100 kg-moles per hour, and the pure octane stream is flowing at 200 kg-moles per hour. Both pure streams are at ambient conditions (25°C and 1 atm). The tee exit streams are also at ambient conditions, which assumes no pressure drop across the mixer or the tee. One tee exit stream is to contain 40% of molar flow rate of the tee inlet stream. The conceptual diagram is: TE1 = ? PE1 = ? n E1 = ?

TH = 25 ° C PH = 1 atm n H = 100 kgmol / h z H , HP = 1.0

Exit1

Heptane M-200

TO = 25 ° C PO = 1 atm nO = 200 kgmol / h

Mix

Octane

zO,OC = 1.0

z E1,OC = ?

T-200

Exit2

n E1 = 0.4 n M

z E1, HP = ?

TE 2 = ? PE 2 = ? n E 2 = ? z E 2, HP = ? z E 2,OC = ?

Using the information from above, you are to practice a HYSYS simulation by doing the following general tasks: Beef, Inc.

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3. Process Unit Assignments

Problem HY.4 • • • • • • • • • •

• •

Create a new HYSYS file. Name it xxx_HY4, where xxx are your initials. Choose a property package. Select the PRSV property package, which is an equation of state. Specify the two chemical components. Create a mixer and name it M-200. Create two mixer inlet streams and one mixer outlet stream. Name them Heptane, Octane, and Mix, respectively. Specify the state of the inlet streams using the Worksheet/Conditions page; that is, supply the temperature, pressure, flow rate, and composition of the heptane stream and the octane stream. Specify the pressure of the mixer outlet stream using the Design/Parameters page. To do this, select Set Outlet to Lowest Inlet. In this problem, the mixer outlet will have the same pressure as the inlet streams. Create a tee and name it T-200. Attach the mixer outlet stream as the inlet to the tee and create the tee exit streams. Name the exit streams Exit1 and Exit2. Specify the flow ratios using the Splits page. If the tee has N exit streams, then you must specify the flow ratios for N-1 of them. An exit stream’s flow ratio indicates what fraction of the tee’s inlet stream is to appear in that exit stream. A flow ratio has a value between zero and one, and all flow ratios sum to one. Enter 0.4 for the Exit1 flow ratio. HYSYS calculates an Exit2 ratio of 0.6. Change the HYSYS preferences to display your desired property units. Observe the HYSYS calculated results, which appear in black. Note that values you supplied appear in blue.

After you specify the given process variables, HYSYS immediately calculates all of the other properties of the streams using the PRSV equation of state. HYSYS calculates the Exit1 and Exit2 flow rates and compositions to be: Property Molar flow rate Mole fraction heptane Mole fraction octane

Exit1 120 kgmol/h 0.3333 0.6667

Exit2 180 kgmol/h 0.3333 0.6667

Simulation Exercise What are the exit stream molar flow rates (lbmol/hr) and mass compositions in the following process? A binary mixture of methanol and ethanol at 45°C and 300 kPa is mixed with a pure water stream at ambient conditions. The methanol and ethanol component flow rates are 50 kgmol/h and 75 kgmol/h, respectively. The water flow rate is 100 kgmol/h. The resulting 3component stream is then split into two exit streams. The total molar flow rates for these two exit streams are in a 3:1 ratio, as shown in the diagram below. For the mixer, the outlet pressure is equal to the lowest inlet pressure.

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3. Process Unit Assignments

Problem HY.4 TE1 = ? PE1 = ? n E1 = ?

TA = 45 ° C

zE1, ME = ?

PA = 300 kPa n A, ME = 50 kgmol / h

Exit1

n A, ET = 75 kgmol / h Alcohol M-200

TW = 25 ° C PW = 1 atm nW = 100 kgmol / h zW ,WA = 1.0

Mix

Water

zE1,WA = ?

T-200

Exit2

n E1 3 = n E 2 1

z E1, ET = ?

TE 2 = ? PE 2 = ? n E 2 = ? zE 2, ME = ? zE 2, ET = ? zE 2,WA = ?

While you are completing your HYSYS simulation on the above problem, please answer the following questions: 1. Which HYSYS stream mixer simulation algorithm in Appendix D would you use to solve the mixer section of this problem? What other unit parameter, which is not referred to in the algorithm, must you specify in HYSYS? 2. What are the temperature in °F, mass density in kg/m3, molar enthalpy in kcal/kgmole, and molar volume in m3/kgmole of the tee’s inlet and outlet streams? What is unique about these values? What are the temperature, mass density, molar enthalpy, and molar volume, in the same units, of the mixer’s inlet and outlet streams? How do these values for the mixer streams differ from those for the tee? Why? 3. No simulation module exists for the tee operation in the blue HYSYS manual. Write the mathematical model for a tee. (Note that the tee operation is analogous to taking a liquid mixture in a large beaker and pouring it into two smaller beakers. What is true about the state of the material in all three beakers?)

4. What is the material relative imbalance (%RIB) for the process flowsheet? Show your calculations. The material %RIB equals 100*(mat’l flow in - mat’l flow out) / (mat’l flow in). You are to draw an overall system boundary around the flowsheet, which contains the mixer and tee operations. The only material and energy streams you are to consider in your imbalance calculation are those that cut your overall system boundary. Therefore, you would not consider stream Mix in your calculations. 5. What is the energy relative imbalance (%RIB) for the process flowsheet? Show your calculations. The energy %RIB equals 100*(energy flow in - energy flow out) / (energy flow in).

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3. Process Unit Assignments

Problem HY.4 For the above problem, please print its HYSYS flowsheet and the design and worksheet datasheets for the mixer and tee.

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3. Process Unit Assignments

Problem HY.5 Reactor Simulation The HYSYS simulation module for a chemical reactor operation is fully defined in Appendix D of this handbook. The chemical reactor module contains a process description, process diagram, assumptions, mathematical model, variable descriptions, mathematical algorithm, and HYSYS simulation algorithms in functional form. After you read this information about the reactor, you are to practice the HYSYS session below and then do the simulation exercise. For your solution of Problem HY.5, you are to provide a printed copy of the HYSYS flowsheet and the design, reactions, and worksheet datasheets for the chemical reactor in the simulation exercise only. Furthermore, you are to provide answers to the questions in the simulation exercise.

HYSYS Session This session will show you how to do a HYSYS simulation for a chemical reactor. It assumes you are familiar with the material in Tutorials 2.5 and 2.6 of this handbook. Acrylonitrile is produced by the reaction of propylene, ammonia, and oxygen:

2 C3 H6

+

2NH 3

+ 3O2

→

2 C3 H3 N

+ 6 H2 O

where 30 molar percent of the propylene is converted. A 45 mole % propylene and 55 mole % ammonia stream at 25 °C and 1 atm is fed to the reactor. The oxygen is fed to the reactor through an air stream also at 25°C and 1 atm. The feed stream is flowing at 22 kgmol/h, and the air stream is flowing at 78 kgmol/h. Assume an adiabatic reactor with no pressure drop. You are to find the temperature of the reactor product stream in °C. Also, you are to find the dew-point temperature (Vf = 1) of the reactor product stream. The conceptual diagram is: TF = 25 ° C PF = 1 atm n F = 22 kgmol / h

TP = ? PP = 1 atm n P = ?

z F , PY = 0.45 z F , AM = 0.55

Feed R-200 R-200

TA = 25 ° C PA = 1 atm n A = 78 kgmol / h

Product

z P, PY = ? z P, AM = ? z P ,O 2 = ?

Air

ε = 30% of PY

z A,O 2 = 0.21

z P, N 2 = ? z P, AN = ? z P,WA = ?

z A, N 2 = 0.79

Using the information from above, you are to practice a HYSYS simulation by doing the following general tasks: • Beef, Inc.

Create a new HYSYS file. Name it xxx_HY5, where xxx are your initials. 3-13

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3. Process Unit Assignments

Problem HY.5 • • • • • • • •

• • •

Choose a property package. Select the PRSV property package, which is an equation of state. Specify the six chemical components. Go to the Reactions tab of the Simulation Basis Manager and add the reaction components. Add the conversion reaction. Specify the stoichiometry and basis. Be sure to enter a propylene conversion of 30%, not 0.30%. Define the Global Rxn Set to include this conversion reaction. Connect the Global Rxn Set to the current Fluid Package. Enter the simulation environment and create a conversion reactor named R200. Create two inlet streams, Feed and Air, and one reactor outlet vapor stream, Product, and one outlet liquid stream, noLiquid. The only product in this case is a vapor. This is an adiabatic reactor, so no energy stream is required. Do not supply an energy stream name. Specify the process state of each of the two reactor feed streams. Set the reactor pressure drop to zero and select the Global Rxn Set. Observe the HYSYS calculated results, which appear in black. Note that values you supplied appear in blue.

After you specify the feed and air streams and the chemical reaction, HYSYS immediately calculates the product stream state using the PRSV equation of state. For the product stream, its temperature is: T = 425.25°C The process state of the reactor product stream is calculated by HYSYS. The stream variables cannot be changed by the , because their values appear in black. To find the dewpoint temperature of the product stream, you must create a new stream with the same flow rate, composition and pressure of the product stream. First, place a new stream on the PFD and name it junk. Second, double click on this new stream to open its property window. Select the Util tab, press the Copy Stream Specs From... button, and select the Product stream. Select the Cond tab and set the vapor fraction (Vf = 1) to find the dew point. HYSYS calculates a dew-point temperature of: T = 43.73°C

Simulation Exercise For the simulation exercise, consider the same reaction and reactor feed streams that were used in the HYSYS session. Now the reactor is no longer adiabatic. Instead, it is an isothermal process, meaning the product stream is at the same temperature as the feed streams, 25°C. What is the reactor duty in kJ/h for this isothermal process?

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3. Process Unit Assignments

Problem HY.5 TF = 25 ° C PF = 1 atm n F = 22 kgmol / h

TP = 25 ° C

Q = ?

PP = 1 atm n P = ?

z F , PY = 0.45 zF , AM = 0.55

Feed R-200 R-200

TA = 25 ° C PA = 1 atm n A = 78 kgmol / h

Product

z P, PY = ? z P, AM = ? z P ,O 2 = ?

Air

ε = 30% of PY

z A,O 2 = 0.21

z P, N 2 = ? z P, AN = ? z P,WA = ?

z A, N 2 = 0.79

While you are completing your HYSYS simulation on the above problem, please answer the following questions: 1. For the adiabatic reactor simulation in the HYSYS session, only a vapor product stream was required. For the isothermal case, both a vapor and a liquid product stream is required, as indicated by the red status bar. Why? Above what reactor exit temperature will only a vapor product stream be required? 2. What is the acrylonitrile composition in the liquid product stream in mole fraction? in mass fraction? in parts per million (ppm)? in kg/m3? in kgmol/m3? in molarity (M)? Use the Workbook/Setup... option to add the actual volume flow to the workbook. 3. What assumptions were used to solve this reactor simulation problem? Compare these assumptions to those of the chemical reactor module in Appendix H of this handbook. Are the assumptions the same? If not, how do they differ? 4. Write the overall mole balance equation containing R for the acrylonitrile reaction problem. What are the reaction constant and its units for acrylonitirle production? 5. What is the energy relative imbalance (%RIB)? Show your calculations. The energy %RIB equals 100*(energy flow in - energy flow out) / (energy flow in). 6. What is the material relative imbalance (%RIB) on a total molar basis? Show your calculations. The total molar %RIB equals 100*(total material flow in - total material flow out) / (total material flow in). 7. What is the material relative imbalance (%RIB) on a total mass basis? Show your calculations. How does this compare with the %RIB on a molar basis? Explain. For the above problem, please print its HYSYS flowsheet and the reactor design, reactions, and worksheet datasheets.

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4. Flowsheet Development Assignments

Overview This chapter provides seven problem assignments to help you develop a process flowsheet to make styrene monomer from toluene and methanol by analyzing individual process units and then connecting these individual units to form the complete flowsheet. Over a sevenweek period, you will have weekly assignments that focus on particular sections of the flowsheet, beginning with the reaction section. In these assignments, you will conduct a process simulation on each of the following flowsheet sections using the HYSYS process simulation software: Problem SM.1 SM.2 SM.3 SM.4 SM.5 SM.6 SM.7

Description Styrene Monomer Reaction Section Reactor Effluent Cooling/Decanting Section Methanol Recycle Purification Section Toluene Recycle Purification Section Toluene/Methanol Feed Preparation Section Recycle Mixing and Preheating Section Styrene Monomer Purification Section

Each weekly problem will be assigned in a separate memorandum. Once you’ve completed

these assignments, you will have synthesized the process flowsheet and determined its processing requirements for material and energy.

This chapter assumes you have completed certain tutorials found in Chapter 2. They are Tutorials 2.1 to 2.6 and 2.9. Also, the problem assignments in Chapter 3 on process units should be completed before you begin your analysis of the process flowsheet to produce styrene monomer from toluene and methanol. Technical materials needed to solve the problems in this chapter are provided in Appendices A, B, C, etc. of this handbook. Appendix A provides complete technical data for the production of styrene monomer from toluene and methanol. Appendices B, C, etc. contain simulation modules for various process unit operations. Each appendix or module provides a mathematical explanation of how HYSYS does its calculations for that process unit. A module includes a description, a conceptual model, a mathematical model, variable descriptions, example mathematical algorithms, and several HYSYS simulation algorithms. Please consult the appropriate appendices to complete an assigned problem. Microsoft Word files of the seven assignments in this chapter are available on the

departments network file server (R:) in folder chem_engineering/public/HYSYS Manual/Chap 4. Your instructor will give you directions on how to access this folder. Beginning with the second problem, you will be provided with the HYSYS solution for the previous problem. You are to begin the next problem by starting with the solution to the previous problem. You can find this previous HYSYS solution in the Blackboard CHEG 200 course under the Assignments section.

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4. Flowsheet Development Assignments

Problem SM.1 Styrene Monomer Reaction Section The heart of a process flowsheet is the reactor. The first step in deg a chemical process flowsheet is to define the reactor operating conditions. These conditions are used to simulate the performance of the reactor. Once the reactor has been simulated, other parts of the flowsheet can be developed. From the reactor performance table found in Appendix A, you are to use one of the inlet temperatures to the adiabatic reactor, in order to complete a reactor simulation that forms styrene monomer from methanol and toluene. The operating conditions in the reactor performance table are the temperature, pressure, conversion and yield for an equimolar feed of methanol and toluene to the reactor. The proposed production rate is 300,000 metric tons per year of styrene monomer. What is the production rate in kgmol/hr for a 95% onstream time (8,320 hours per year)? Study the mathematical algorithm and HYSYS simulation algorithm for a reactor in Appendix H to determine what variables must be specified in order to simulate your reactor. Your study should reveal that the feed flow rate and composition to the reactor must be specified instead of the total and component flow rates out of the reactor. Manually estimate the feed flow rate based on 100 percent conversion, no side reaction for ethylbenzene, and the proposed styrene production rate in kgmol/hr. Please document your calculations. Using your estimated feed flow rate, run HYSYS to simulate the conversion reactor for the production of styrene from toluene and methanol. What chemical components are in the reactor effluent stream and why? The conceptual model for this reactor is as follows:

S10

R1

S11

Use the stream and equipment labels above in your HYSYS process flow diagram (PFD). For the conversion reactor, two chemical reactions will occur—one producing styrene monomer and the other ethylbenzene. Please note that the HYSYS conversion of toluene for the first reaction is equal to [overall conversion*yield] at a specific inlet temperature and that for the second reaction is equal to [overall conversion*(1.00-yield)]. After completing this simulation with the estimated feed flow rate, use the HYSYS Adjust operation to modify the given reactor feed in order to obtain the desired production rate of styrene monomer in kgmol/hr. To learn how to apply the Adjust operation, access the electronic version of the HYSYS Reference Manual from the Start/All Programs menu in Windows. After completing the above assignment, supply the HYSYS flowsheet with a material stream table that includes the actual volumetric flow rate, mass density, and molecular weight. You can use the Workbook/Setup… menu option to add these three items to the table. Also, include a composition table on the flowsheet PFD. The two tables can be added to the PFD by clicking the right mouse button on the PFD. Using the HYSYS text capability, you must add a title, the assignment number, the reactor inlet temperature, your name, and the date to the flowsheet. See Tutorial 2.9 for directions on text documentation. Finally, print the datablock for the reactions only by clicking the right mouse button when the cursor is over the reactor icon.

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4. Flowsheet Development Assignments

Problem SM.2 Reactor Effluent Cooling/Decanting Section In Problem SM.1, you simulated the reactor in the styrene monomer project using data provided by the Research and Development Department of BEEF, Inc. The reactor R1 has only one effluent stream, the vapor stream S11. Using physical properties such as critical temperature, critical pressure, and normal boiling temperature for each reaction component, and document that S11 should indeed be a vapor stream. To the stream phase, it may help to think of a phase (PT) diagram for each component. Physical properties of various chemical components are given in Appendix A. A global flowsheet for a chemical process depicts simply the raw materials entering the flowsheet and the product, byproducts and wastes leaving the flowsheet. Appendix A gives a block flowsheet for the chemical process of converting toluene and methanol to styrene monomer. A global flowsheet for the styrene monomer chemical process is shown below. Recycle Reactants Off Gas Byproduct Raw Materials

Reactor

Effluent

Separation Sequence

Pure Product

Wastewater

This global flowsheet shows the reactor producing an effluent stream, which must be separated to purify the product. The reactor effluent goes through a separation sequence in which the off gas, byproduct, pure product and waste are isolated. Unreacted raw materials are also separated in the sequence. The reactants are then recycled to the reactor. The number of process exit streams— off gas, pure product, waste, etc.—determines the number of separation units required in the sequence. As a rule of thumb, for multiple process exit streams, the number of required separation units is between one and the number of exit streams. The first step in the design of a separation sequence is to decide what the first separation unit is. Some types of separation are phase splitting, distillation and extraction. Phase splits are the cheapest method of separation. Think of your separator funnel in organic chemistry labs to visualize a phase split. Therefore, if possible, a phase splitter is the first separation unit in the separation sequence. If the reactor effluent is all vapor, it must be cooled to allow a phase separation to take place. In the styrene monomer project, the reactor effluent is cooled to allow the formation of three distinct phases, the vapor phase and two immiscible liquid phases. The two liquids are an organic phase and an aqueous phase. To predict which components each phase contains, that “like dissolves like.” The organic phase contains the organics of toluene, ethylbenzene, and styrene monomer. The methanol partitions between both the organic phase and the aqueous phase. More information on the three-phase separator is given in the Design Data section of Appendix A. A three-phase separator is also known as a decanter. Appendix J

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4. Flowsheet Development Assignments

Problem SM.2 provides a mathematical description of how a decanter is modeled. Also, Appendix G describes how a cooling operation functions. You now know that the reactor effluent must be cooled from a vapor stream to a temperature at which three phases exist. You must determine to what temperature the stream is cooled to produce the phase separation. The cooling is typically carried out by a heat exchanger in which the effluent is cooled by a water stream. The cooling water is not directly mixed with the effluent. Rather, the two streams exchange heat so that the hot effluent is cooled while the cold water is heated. It is unfeasible to design a heat exchanger that cools the hot stream the entire way to the temperature of the cold stream. Such a heat exchanger would be infinitely large in area. Typically, it is assumed that the hot stream is cooled to within 5 to 10°C of the initial cold stream temperature. As a rule of thumb, the cooling water is supplied at 31°C, so the reactor effluent is cooled to about 38°C. At 38°C three phases may exist, and they could be separated in a decanter. The next step in the styrene monomer project is to simulate the first separation unit. Use HYSYS to simulate the effluent cooling and three-phase separator. The conceptual model for the reactor, cooler, and decanter is as follows: S13 QE3 S10

R1

S11

S12 E3

F3

S14A

S15

Use the stream and equipment labels above in your HYSYS process flow diagram (PFD). for pressure drops through the reactor (R1), cooler (E3), and decanter (F3) using the data in Appendix A. Begin with the HYSYS solution of Problem SM.1 provided in the Blackboard CHEG 200 course under the Assignments section. Your instructor will give you directions on how to access this folder. At what temperature in °C does two phases (vapor-liquid) start to occur on cooling? Does three phases (vapor-liquid-liquid) start to occur on cooling? On a molar basis, what fraction of Stream S11 after cooling to 38°C goes to the vapor phase of the decanter? To the organic phase? To the aqueous phase? On a mass basis, what fraction of Stream S11 after cooling to 38°C goes to the vapor phase of the decanter? To the organic phase? To the aqueous phase? After completing the above assignment, supply the HYSYS flowsheet with a workbook table for cooling and decanting at 38°C. A workbook table can be added to the PFD by clicking the right mouse button on the PFD and using the drop-down menu. Also, use the HYSYS text capability to add a title, the assignment number, your name, and the date to the PFD. See Tutorial 2.9 for directions on text documentation. Finally, supply the three-phase-separator design and worksheet datasheets.

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4. Flowsheet Development Assignments

Problem SM.3 Methanol Recycle Purification Section In Problem SM.2, you simulated the first separation operation, a three-phase decanter, in the separation sequence of the styrene monomer project. In Problem SM.3, you will simulate the next operation in the separation sequence, a distillation column. A distillation column uses the difference in boiling points to separate the components of a stream. Think of a liquid mixture of two components with each component having different boiling points. If the mixture is heated to a temperature between the boiling points of the two components, the more volatile component (lower boiling point) vaporizes while the less volatile component (higher boiling point) remains in the liquid phase. By collecting the vapor, you have effectively separated the binary mixture into two single phases, the vapor phase containing the more volatile component and the liquid phase containing the less volatile. The same principle can be used to separate streams with more than two components into two mixtures with fewer components. All of the more volatile components will vaporize, and all of the less volatile components will remain in the liquid. Exactly where the separation occurs, i.e. which components vaporize and which do not, depends on the temperature that the original mixture is heated to. A distillation column is a series of stages or trays where each act as a separator according to boiling points as described above. A column separates a liquid feed stream into two liquid streams, the distillate and bottoms. The temperature of each tray gradually decreases as you go up the column. A reboiler at the bottom of the column heats the liquid in the column to a saturated vapor. This vapor then rises through the trays of the column. As the temperature decreases going up the column, the less volatile components begin to condense and fall back down through the column. By the time you reach the top of the column at the lowest temperature, only the most volatile components are vapors and exit the column. This vapor stream is condensed to a saturated liquid and then split into a reflux stream and distillate stream. The reflux stream is sent back to the column. The less volatile components exit from the reboiler in the bottoms stream. To consider distillation as a possible separation operation, you must determine the normal boiling points of the stream components in the feed stream to that column. Find the normal boiling points (nbp) for all chemical components in the styrene monomer problem. List the components and nbp in order of increasing boiling point in the table below. HYSYS contains component properties such as normal boiling point. To find a property value, view your simulation fluid package and go to the Components page. Double click on the desired component in the current component list to open a window listing the component properties. Find the tab that contains the desired property, here the Critical tab for normal boiling point at 1 atm, and read the value and units. To design a distillation column, you pick two components with adjacent boiling points, as depicted in the diagram below. The more volatile component is called the light key (LK) and the less volatile component is called the heavy key (HK). The split in the column is between these two key components. A perfect separation means that all of the LK would be in the distillate, and all of the HK would be in the bottoms. However, like in the case of a three-phase separator, perfect separation is unfeasible. An infinite number of trays in the distillation column would be required to perform a perfect separation. Therefore, a separation level is chosen for the column. Most of the LK exits the column through the distillate, with a little in the bottoms, and most of the HK exits the column through the bottoms, with a little in the distillate. The distillate contains all components more volatile than the LK (non-LK's), and the bottoms contains all components

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4. Flowsheet Development Assignments

Problem SM.3 less volatile than the HK (non-HK's). A simple diagram to show the component flows is shown below. non-LK' s mos t LK little HK non-LK' s LK HK non-HK' s

Column

little LK mos t HK non-HK' s

In the table below, where you entered the components and their boiling points, you are to indicate the feed type of each component, i.e. tell whether each component is a LK, HK, non-LK or non-HK. Indicate the distribution of each component in the table by stating which stream(s) (distillate and/or bottoms) the component appears in, and in what relative amount, such as all, most or little. that the column is being designed to separate methanol from water.

Component

Normal Boiling Point, °C

Feed Type

Distribution

A detailed description of a simple distillation column is given Appendix L of this handbook. What are the three main parts of a distillation column? The overall mathematical model for a column consists of a series of smaller math models, one for each of these column parts. In what order are the overall mathematical model equations solved in the column algorithm? How does the column algorithm differ from the HYSYS simulation algorithm? Why? The diagram for the distillation column used for Problem SM.3 is given below. It contains a partial condenser and several stages above and below the feed stage. However, the simple distillation column in Appendix L has a total condenser, meaning the entire vapor coming off the top of the column is converted to a saturated liquid. To achieve this, all of the components are assumed to be condensable. However, the column in the styrene monomer problem contains hydrogen, which is non-condensable. Note hydrogen's extremely low boiling point in your table above. Because the column contains hydrogen, a partial condenser is used which will allow the hydrogen to exit the column as a vapor in stream V.

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4. Flowsheet Development Assignments

Problem SM.3 Q C

Condenser

Stages

F

V Distillate Vent

RF Reflux

D Distillate

Feed Stage

Feed

Stages

B

Q R

Bottoms Reboiler

The use of a partial condenser instead of a total condenser means that one more variable must be specified for HYSYS to simulate the column. The functional form for the HYSYS simulation algorithm of this column is

⎡⎣ ΨV , Ψ D , Ψ B , Q C , Q R ⎤⎦ = column ⎡⎣ Ψ F , PD , PB , N S , N FS , R, xB , LK , VR ⎤⎦ where Q C is the condenser energy rate or duty, Q R is the reboiler energy rate or duty, Pi is the pressure of stream i, N S is the number of column stages, N FS is the feed stage number, R is the reflux ratio, xB , LK is the mole fraction of the light key (LK) in the bottoms, and VR is the vent ratio. The reflux ratio of R is the reflux flow (RF) rate over the distillate (D) flow rate. The vent ratio of VR is the distillate vent flow (V) over the feed flow (F). The vector Ψi is a short

notation to represent the temperature, pressure, flow rate, and chemical composition of stream i. Shortcut methods exist to estimate column design variables like N S , N FS ,and R. To solve the rigorous HYSYS distillation column, the number of column trays, the tray at which the feed enters the column, and three other variables must be specified. For this problem you will use the reflux ratio, the mole fraction of methanol in the bottoms, and the vent ratio as the three other variables. A shortcut analysis has already been performed on the column, and the following values were found: Number of stages Feed Stage Reflux ratio LK in bottoms Vent Ratio

= = = = =

24 14 3 0.001 1.0e-4

mol. frac.

These estimated variables can then be used as inputs to the rigorous distillation column in HYSYS. The shortcut analysis found the first three items based on a specified mole fraction for the light key in the bottoms stream. Beef, Inc.

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4. Flowsheet Development Assignments

Problem SM.3 Begin your HYSYS simulation for the methanol/water column from the HYSYS solution to Problem SM.2 located in the Blackboard CHEG 200 course under the Assignments section. Your instructor will give you directions on how to access this folder. This SM.2 solution now includes a valve and heater to prepare the aqueous stream for distillation. It also has an unattached rigorous distillation column in the PFD. Use the labels given below to define the rigorous distillation column. S17V

QCC3

S17

S16

C3

S18

QRC3 Double click the rigorous distillation column in the PFD, to open its property window. On the Design/Connections page, enter the column name, the material and energy stream names, and the number of stages and feed stage found from the shortcut method. Make sure the partial condenser is chosen so you can name the vent stream S17V. Enter 123 kPa for the condenser pressure and 134 kPa for the reboiler pressure. The pressures of every stage between the condenser and reboiler will be calculated by HYSYS. Go to the Design/Monitor page and enter values in the “Specified Values” column for the reflux ratio, mole fraction of the LK in the bottoms, and the vent ratio. On the Design/Monitor page, the degrees of freedom box in the lower right corner currently show three. It must be zero in order for the column to be solved. To make this zero, you must tell HYSYS which specifications to use. Click the boxes to the right of the values you specified to make the reflux ratio, mole fraction of the LK in the bottoms, and the vent ratio active specifications. Note that a fourth specification for the mole fraction of the heavy key in the distillate must be inactive. Once enough variables are specified to satisfy the degrees of freedom, the column will iterate to find a solution. The green converged message will show up in your column window to indicate the rigorous distillation simulation is solved. Sometimes the column calculations may not converge. When this happens, deactivate the LK specification and active the HK specification. Then, run the calculations again to see if the distillation simulation will converge. How was the value for the vent ratio estimated? What are the distillate, vent, and bottoms flow rates in kg/h? What are the mole fractions of the HK in the bottoms and the LK in the distillate? Using the Performance/Plots page, print plots of the temperature profile and composition profiles of the LK and HK in the column. Hand in your PFD and the design and worksheet datasheets for C3.

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4. Flowsheet Development Assignments

Problem SM.4 Toluene Recycle Purification Section In Problem SM.3, you simulated the distillation column that recovered methanol in the aqueous stream from the three-phase decanter. This recovered methanol will eventually be recycled in a later problem. The next operation in the separation sequence of the styrene monomer project is another distillation column. In Problem SM.4, you will simulate a distillation column to separate reactants from products in the organic stream leaving the three-phase decanter. The organic stream contains both methanol and toluene that are to be recycled to the reactor. The organic stream also contains the styrene monomer product and the byproduct ethylbenzene. As with column C3 in Problem SM.3, the first step in simulating the distillation column is to find the normal boiling points (nbp) for all chemical components in the styrene monomer problem and list the components and nbp in order of increasing boiling points in the table below. To find the normal boiling point of the components in HYSYS, view your simulation fluid package and go to the Components page. Double click on the desired component in the current component list to bring up a window listing the component properties. Find the tab that contains the desired property, here the Critical tab for normal boiling point, and read the value and units. In the table below, where you entered the components and their boiling points, you are to indicate the feed type of each component, i.e. tell whether each component is a LK, HK, non-LK or non-HK. Indicate the distribution of each component in the table by stating which stream(s) (distillate and/or bottoms) the component appears in, and in what relative amount, such as all, most or little. If necessary, refer to the Problem SM.3 handout to refresh yourself with the concept of key components. that the column is being designed to separate the reactants from the products.

Component

Normal Boiling Point, °C

Feed Type

Distribution

As in the case of methanol column C3, the feed to this column, called C1, contains the uncondensable, hydrogen, which means a partial condenser must be used. Column C1 is depicted on the next page. What is the functional form for the HYSYS simulation algorithm for this column? Define each of the variables in this functional form. (HINT: Refer to the Problem SM.3 handout) Shortcut methods exist to estimate column design variables like number of stages, feed stage location, and reflux ratio. To solve the rigorous HYSYS distillation column, the number of column trays, the tray at which the feed enters the column, and three other variables must be specified. For this problem you will use the reflux ratio, the mole fraction of toluene in the Beef, Inc.

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4. Flowsheet Development Assignments

Problem SM.4 bottoms, and the vent ratio as the three other variables. A shortcut analysis has already been performed on the column, and the following values were found: Number of stages Feed Stage Reflux Ratio

= = =

LK in bottoms Vent Ratio

= =

28 13 3

0.001 5.0e-3

mol. frac.

These estimated variables are the inputs to the rigorous distillation column in HYSYS. Begin your HYSYS simulation for the column from the HYSYS solution to Problem SM.3 located in the Blackboard CHEG 200 course under the Assignments section. Your instructor will give you directions on how to access this folder. The SM.3 solution now includes a valve and heater to prepare the organic stream for distillation. It also has an unattached rigorous distillation column in the PFD to process the organic stream (S22). Use the labels given below to define the rigorous distillation column. S23V

QCC1

S23

S22

C1

S24

QRC1 Double click the rigorous distillation column in the PFD, to open its property window. On the Design/Connections page, enter the column name, the material and energy stream names, and the number of stages and feed stage found from the shortcut method. Make sure the partial condenser is chosen so you can name the vent stream S23V. Enter 79 kPa for the condenser pressure and 99 kPa for the reboiler pressure. The pressures of every stage between the condenser and reboiler will be calculated by HYSYS. Go to the Design/Monitor page and enter values in the “Specified Values” column for the reflux ratio, mole fraction of the LK in the bottoms, and the vent ratio. On the Design/Monitor page, the degrees of freedom box in the lower right corner currently show three. It must be zero in order for the column to be solved. To make this zero, you must tell HYSYS which specifications to use. Click the boxes to the right of the values you specified to make the reflux ratio, mole fraction of the LK in the bottoms, and the vent ratio active specifications. Note that a fourth specification for the mole fraction of the heavy key in the distillate must be inactive.

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4. Flowsheet Development Assignments

Problem SM.4 Once enough variables are specified to satisfy the degrees of freedom, the column will iterate to find a solution. The green converged message will show up in your column window to indicate the rigorous distillation simulation is solved. Sometimes the column calculations may not converge. When this happens, deactivate the LK specification and active the HK specification. Then, run the calculations again to see if the distillation simulation will converge.

How was the value for the vent ratio estimated? Recall that the vent ratio is the molar ratio of the vent stream flow rate to the feed stream flow rate. You can use the component flow rates of the non-condensibles in the feed and the total feed rate to estimate the vent ratio. This estimation assume all non-condensibles (methanol and hydrogen) will exit the column in the vent stream. What are the distillate, vent, and bottoms flow rates in kg/h? What are the mole fractions of the HK in the bottoms and the LK in the distillate? Using the Performance/Plots page,

print plots of the temperature profile and composition profiles of the LK and HK in the column. Hand in your PFD and the design and worksheet datasheets for C1.

Colum n C1 is designed to separate the reactants from the products in order to recycle the reactants. The C1 distillate stream contains the recycled reactants. The recycle stream must be prepared, in order to be mixed with fresh raw materials before entering the reactor. The fresh and recycled reactant streams are mixed as saturated vapors at 570 kPa. Add a pump and heater to distillate stream S23 to reach the saturation point at 570 kPa in stream S26 below. You must decide whether to heat the stream first, or to pump it first. Compressing a gas takes much more energy than pumping a liquid. Therefore, if you heat the stream to a saturated vapor first, extra energy will be needed to compress the vapor to the desired pressure. As a heuristic rule, pump the liquid first and then heat to a saturated vapor. Don't forget to for the pressure drop across the heater using the data in Appendix A. The desired pressure of 570 kPa is the final pressure, after the pump and heater. Use the following labels in your simulation: QE5

WP4

S25

S23

S26 E5

P4 C1

Hand in the design and worksheet datasheets for P4 and E5.

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4. Flowsheet Development Assignments

Problem SM.5 Toluene/Methanol Feed Preparation Section In Problem SM.1, you simulated the reactor to produce styrene monomer from toluene and methanol. In SM.1 you assumed a single reactor feed stream that contained the two reactants at the optimum reactor conditions of 512.5°C and 400 kPa. Most likely, the raw materials (i.e., the two reactants) will actually be available individually at some other conditions. In this case, pure methanol and pure toluene are both available at ambient conditions, 25°C and 1 atm. In Problem SM.5, you are to simulate the preparation of the methanol and toluene. The pure methanol and pure toluene streams are to be mixed as saturated vapors at 570 kPa. This means that each raw material must be compressed and heated separately before being mixed. Compression of a gas requires considerably more energy than compression of a liquid. As a general heuristic rule, you should first pump the liquid to the desired pressure and then heat the high-pressure liquid to a saturated vapor. You are to simulate the preparation of the two pure raw material streams, one for methanol and one for toluene, from ambient conditions to saturated vapors at 570 kPa. From Problem SM.1, the amount of reactants needed to produce 300,000 metric tons per year of styrene monomer was determined to be 584 kgmol/h of methanol and 584 kgmol/h of toluene. You are to begin a new case file for the HYSYS simulation in this problem, using the PRSV property package with two chemical components—toluene and methanol. You can assume a 75% adiabatic efficiency for the pumps. After you have pumped and heated the raw material streams to the desired conditions, you are to mix the two streams to create a single reactor feed stream. Use the following labels in your HYSYS simulation: QE1

WP1 S2 S1 toluene

S3

P1 E1 M1

QE2

WP2

S6

S5 S4 methanol

S7

P2 E2

Using

the Performance/Plots page, produce graphs of temperature versus heat duty (T vs. for heater E1 and also heater E2. Be sure to include the liquid, vapor-liquid, and vapor regions in each plot. On each cooling curve, label the two sensible heat regions and the latent heat region. What are the dew-point and bubble-point temperatures of stream S2 and also stream S5? What is unique about the dew point and bubble point for each pure chemical compound? Explain. Q, a cooling curve)

Using the Gibbs phase rule (see F&R, 3rd Ed., p. 247), find the degrees of freedom for when a phase transition occurs. How does a cooling curve depict this transition? (HINT: Look at the slope of the curve during a phase transition.) Hand in your PFD and your two cooling curve graphs. Beef, Inc.

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4. Flowsheet Development Assignments

Problem SM.6 Recycle Mixing and Preheating Section In Problem SM.5, you simulated the compression and vaporization of the two fresh reactant streams, pure toluene S1 and pure methanol S4, for the styrene project. The feed to the reactor also contains recycled material from columns C1 and C3, as depicted in the block flowsheet of Appendix A. The recycle streams must be mixed with the fresh reactant stream before entering the reactor, creating a recycle loop in the process. In Problem SM.6 you are to simulate the recycle loop. Begin with the solution to Problem SM.5 located in the Blackboard CHEG 200 course under the Assignments section. Your instructor will give you directions on how to access this folder. The SM.5 solution now includes the fresh reactant compression, vaporization, and mixing, as well as the rest of the process flowsheet from the reactor to the distillation columns C1 and C3. However, the fresh reactant stream is not yet connected to the reactor. The recycled streams from the distillation columns are pumped and heated to bring them to saturated vapors at 570 kPa before mixing with the fresh reactants. Create another mixer in the flowsheet to combine the fresh reactant stream with the two recycle streams. This stream then must be prepared to enter the reactor. Add a heater to heat the stream to the reactor temperature 512°C. The heater used to reach such a high temperature is a fired heater, basically a furnace that burns natural gas. Use the usual HYSYS heater with a pressure drop of 170 kPa. This pressure drop will set the reactor feed stream S9 to the reactor pressure of 400 kPa. Use the following labels in the PFD:

QFH1 S26 S7

M2

S8

S9 FH1

S21

Now stream S9 represents the reactor feed. However, the process state of S9 is calculated by assuming the conditions of reactor feed S10, which you did in Problem SM.1. To close the recycle loop, you must that the process state of stream S9 exiting the fired heater FH1 is identical to the reactor feed stream S10. Open the property windows of streams S9 and S10 and compare their temperatures, pressures, flow rates, compositions, and heat flows. The iterative method of successive substitution is used to converge the process state of the reactor feed stream. To assume a new state for the reactor feed stream of S10, copy the conditions of S9 into S10 using the Define From Other Stream… button in the property window of stream S10. This action will cause HYSYS to recalculate all of the process units and produce a different calculated S9. The process states of S9 and S10 should now be closer to each other. Again copy the conditions of S9 into S10 to assume a new process state for the reactor feed stream. You could continue this process, manually performing the successive iterations until the process state of S9 is identical to that of S10. After performing three (3) manual iterations, you are to hand in the worksheet datasheets for streams S9 and S10.

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4. Flowsheet Development Assignments

Problem SM.6 HYSYS has an operator that will perform automatically the successive iterations for you. Add a recycle operation between streams S9 and S10 as shown below.

S9

R

S10

RCY-1 HYSYS will now automatically iterate on the process state of these two streams to solve the recycle loop. When the recycle operator has converged, you will get a green converged message at the bottom of the recycle property window. After convergence, what are the component molar and mass flow rates of the reactor feed, stream S10? Hand in your PFD and the design and worksheet datasheets for the recycle operator RY.

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4. Flowsheet Development Assignments

Problem SM.7 Styrene Monomer Purification Section In Problems SM.1 through SM.6, you simulated the production of styrene monomer from methanol and toluene, including the separation and recycle of unused reactants. In the resulting flowsheet produced by solving these problems, the bottoms stream from the separation column C1 contains both the product styrene monomer and the by-product ethylbenzene. A distillation column is needed to purify the desired styrene monomer to produce a product stream that meets the design specification. In Problem SM.7, you are to simulate the purification of the styrene monomer. Begin with the solution to Problem SM.6 located in the Blackboard CHEG 200 course under the Assignments section. Your instructor will give you directions on how to access this folder. The SM.6 solution now includes a valve and cooler to prepare the organic stream from the decanter for separation, as well as column C2 for the separation of ethylbenzene from styrene monomer. You need to provide the specifications to simulate the separation in column C2 that operates with a total condenser. The specifications you will use are: number of trays, feed tray location, reboiler and condenser pressures, reflux ratio, and light-key (LK) composition in the bottoms. The number of trays and feed tray location are already specified in the SM.6 solution file. The following information is given to you: Reboiler Pressure Condenser Pressure Reflux Ratio

= = =

83 kPa 31 kPa 35

You must determine the LK composition in the bottoms. that column C2 is being designed to separate ethylbenzene from styrene. The LK composition in the bottoms is a mass fraction, and the bottoms stream contains the product styrene monomer. Using the technical data provided in Appendix A, determine the design specification of the LK in the bottoms stream. What is the LK mass fraction in the bottoms, and what is the feed ratio? Finally, enter this column specification and the other ones above for C2 and then have HYSYS solve the column operation for you. The distillation column produces distillate and bottoms streams at pressure below ambient. The product and by-product must be stored at ambient conditions until they are needed for sales in the marketplace. Simulate the pumping and cooling of both streams to ambient conditions, ing for a pressure drop in each cooler using information provided in Appendix A. Use the labels given in the diagram below:

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4. Flowsheet Development Assignments

Problem SM.7 QE6

WP5 S28

S27

S30 P5

C2

S31

E6

QE7 WP6 S29

S32 E7

P6

S33

Once you have completed the above simulation of the entire flowsheet, you are to perform an overall energy and material balance check. To help you perform these balances, use the tables that are provided on the last page of this problem assignment. What are the material relative imbalances (%RIB’s) on a total mass and molar basis for the entire flowsheet? What is the energy relative imbalance (%RIB’s) for the entire flowsheet? Show your calculations. Hand in your completed tables, the PFD, and the C2 design and worksheet datasheets. Finally, calculate the net profit for the production of styrene monomer from toluene and methanol. The net profit in $/yr is: sales of product

+

sales of by-product

+

fuel-value credit

-

cost of raw mat'ls

-

cost of utilities

-

annualized capital cost

Use the economic information provided in Appendix A for the necessary sales and cost prices and the fuel-value credit. To get the fuel credit, duplicate the vapor streams from the decanter and the one distillation column in the PFD, and then mix these duplicate streams to form one off-gas stream, which will be given a credit as fuel. To determine this credit, you will need to find the lower heating value of this off-gas stream. HYSYS has already calculated this heating value. You need to add it to the HYSYS Workbook window through the Workbook/Setup menu option.

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4. Flowsheet Development Assignments

Problem SM.7

Unit Operation Energy Flows In Unit Operation

Duty or Work, kJ/h

Unit Operation

P1

CE3

E1

P3

P2

E4

E2

P5

FH1

P6

P4

C1 reboiler

E5

C2 reboiler

CE1

C3 reboiler

Duty or Work, kJ/h

Distillation column reboiler and condenser duties can be found on the Summary tab of the column property window.

Unit Operation Energy Flows Out Unit Operation

Duty, kJ/h

Unit Operation

E3

C1 condenser

CE2

C2 condenser

E6

C3 condenser

Duty, kJ/h

E7

Stream Energy and Material Flows Duty or Work

Energy Flow, kJ/h

Material Flow, kg/h

In or Out?

S1 S4 S13 S23V S18 S31 S33

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Appenidx A.

Styrene Monomer Production Introduction†

A two-step process starting with benzene and ethylene produces 90% of the styrene monomer marketed in the United States. First, benzene is alkylated with ethylene to form ethylbenzene. After purification, the ethylbenzene is catalytically dehydrogenated to produce styrene. The dehydrogenation step is endothermic and requires a large quantity of steam mixed with the ethylbenzene to maintain the desired reaction temperature, to depress coking of the catalyst, and to dilute the reaction concentration to enhance the reaction equilibrium. Chemists in our Research and Development (R&D) Department of BEEF, Inc. have discovered a catalyst, which will produce styrene from toluene and methanol in one step, and steam addition is not required. Some byproduct ethylbenzene is also produced which can be sold to conventional styrene producers. This catalyst discovery might give our clients the opportunity to develop a new, low-cost route to styrene.

Proposed Styrene Process Chemical engineers in our R&D Department of BEEF, Inc. have done some pilot-plant studies on this new one-step process. They have defined the following preliminary block flowsheet for this process: toluene recycle H2 fuel toluene

decanter

column

organic furnace methanol

reactor aqueous

methanol methanol recycle recycle column

ethylbenzene column

waste water

styrene monomer

In this flowsheet, toluene and methanol feeds at 25°C and 1 atm are compressed and heated too saturated vapors at 570 kPa. These two feeds are then mixed with a toluene recycle and a methanol recycle (both of which are saturated vapors at 570 kPa) to form the feed stream to the reactor. This feed stream is superheated in a fired furnace and then fed to the catalytic reactor where the following vapor-phase reactions take place: ____________________ †

The material in this appendix has been extracted from the 1985 Student Contest Problem—Styrene from Toluene and Methanol published by the American Institute of Chemical Engineers (AIChE). The reactor performance table was modified by adding a fifth temperature column of 540ºC.

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Appenidx A. C7H8

+

CH3OH

toluene

C7H8

Styrene Monomer Production ↔

methanol

+

CH3OH

toluene

methanol

C8H8

+

styrene

↔

C8H10

H2O water

+

ethylbenzene

+

H2 hydrogen

H2O water

In a preliminary design analysis, one could assume that other byproduct formation and polymerization of styrene monomer are negligible and that the catalyst does not coke or deactivate with time. The reactor effluent stream is condensed with cooling tower water and cooled to 38°C, forming three phases—vapor, organic, and aqueous—in a decanter. The vapor stream from the decanter contains mostly hydrogen, and it could be used as a fuel. The aqueous stream contains primarily methanol and water, and it is sent to a methanol distillation column. This column’s product stream is the recycled methanol, while its bottoms stream is wastewater, which is eventually discharged at 25°C and 1 atm. The organic stream from the decanter contains mostly toluene, ethylbenzene, and styrene monomer. It is sent to a toluene distillation column. This column’s product stream is the recycled toluene stream containing some methanol, while its bottoms stream contains mostly ethylbenzene and styrene monomer, which are sent to the styrene distillation column. In the styrene column, the product stream is mostly ethylbenzene, and the bottoms stream is mostly styrene monomer. Both of these streams are then cooled to 25°C and 1 atm before each enters a storage tank.

Technical Data A. Reactor Performance Chemical engineers in our R&D Department have taken the following data for determining the adiabatic reactor performance. Linear interpolation can be used between temperatures for intermediate values. 480 495 510 525 Inlet Temperature, °C Inlet pressure, kPa abs. 400 400 400 400 Conversion 0.68 0.71 0.76 0.82 Yield 0.87 0.83 0.78 0.72 Rate 36 73 130 190 Conversion = moles toluene reacted/moles toluene fed. Yield = moles styrene formed/moles toluene reacted. Rate = gmoles toluene reacted/m3 catalyst/min.

540 400 0.88 0.66 250

In collecting these reactor performance data, the pilot-plant experiments used only stoichiometric feed to the reactor. Therefore, any design study should be based only on stoichiometric feed (i.e., equal moles of toluene and methanol).

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Appenidx A.

Styrene Monomer Production

B. Physical Properties Some physical properties are given in the table below. These values have been rounded off for manual calculations. All pressures are absolute. Property Molecular Weight Normal Boiling Point, °C Critical Temperature, °C Critical Pressure, kPa Critical Compress. Factor Ideal Gas at 25°C, kcal/mol: Heat of Formation Free Energy of Formation

Methanol 32 65 239 8094 0.230

Water 18 100 374 22054 0.232

Toluene 92 111 321 4233 0.270

Ethylbenzene 106 136 344 3599 0.262

Styrene 104 145 364 3674 0.252

Hydrogen 2 -253 -240 1296 0.318

-48.1 -38.8

-57.80 -54.64

11.95 29.16

7.12 31.21

35.22 51.10

0.00 0.00

C. Thermodynamic Model For a preliminary design study, all necessary thermodynamic calculations for physical properties (such as density and molar enthalpy) and for phase equilibria (such as vapor-liquid or vapor-liquidliquid) can be done using an equation of state. The Peng-Robinson Stryjek-Vera (PRSV) equation of state is recommended for the analysis of the manufacture of styrene monomer from toluene and methanol. The PRSV equation is an improvement on the Peng-Robinson (PR) equation of state, and it extends the application of the PR method to moderately non-ideal systems.

Design Data

(Including Simplifying Assumptions)

A. Material Balance The proposed plant capacity is 300,000 metric tons per year of crude styrene monomer, which includes 300 ppm of contained ethylbenzene. The onstream time is 95% (8,320 hours per year). Yield losses due to trace byproducts can been ignored. Other assumptions are: 1. Impurities in purchased methanol and toluene are negligible. 2. Water, ethylbenzene and styrene monomer recycled to the reactor feed are at small enough concentrations to through as inerts.

B. Three-Phase Separator The reactor effluent condensed with cooling-tower water forms three phases: vapor, organic, and aqueous phases (Phase phenomena are given below). In a preliminary design study, the vapor phase could be given a fuel-value credit. The organic phase must be processed to recover unreacted toluene and methanol for recycle and to purify the styrene and ethylbenzene streams to meet design specifications. Also, the aqueous phase must be processed to recover unreacted methanol. Beef, Inc.

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Appenidx A.

Styrene Monomer Production

The lowest acceptable process outlet temperature for all water-cooled heat exchangers is 38°C and is limited by the cooling tower water supply temperature.

Vapor Phase • •

Phase equilibrium for all condensibles in the three-phase decanter are approximated using the PRSV equation of state. The off-gas will be given a credit as fuel at its lower heating value. This value is the calories evolved from complete combustion of all of the components in the stream at 25°C, when the final state of all of the water formed and originally present in the fuel is vapor.

Aqueous Phase • •

Except for methanol, negligible organics will partition into the aqueous phase. Methanol is to be recycled as a saturated vapor at 570 kPa.

Organic Phase • • •

Negligible water will partition into the organic phase. Mostly methanol partitions into both the organic and aqueous phases. Toluene/methanol are to be recycled as a saturated vapor at 570 kPa.

C. Distillation Nominal atmospheric distillations will operate at 136 kPa (k = kilo, Pa = Pascals) top tray pressure and 123 kPa condenser outlet pressure. Avoid column-operating pressures above nominal atmospheric. Allow 5 kPa pressure drop between the top of the column and the condenser outlet for vacuum columns. Do not exceed 145°C in any column with more than 50 mass % styrene monomer (SM) in the bottoms, in order to minimize SM polymerization. Use the shortcut methods of Fenske for minimum stages, Underwood for minimum reflux, and Gilliland’s correlation for operating reflux before doing a rigorous column calculation. liquids.

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The feed, distillate, and bottoms streams of a distillation column are to be saturated

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Appenidx A.

Styrene Monomer Production

D. Design Specifications Some design specifications for the aromatic and wastewater streams are: Recycle Methanol Recycle Toluene EB Byproduct Crude SM Product Waste Water

No specified limit on toluene. No specified limit on methanol. 4 wt % EB maximum. 5 wt % maximum for sum of EB and SM. 0.8 wt % toluene maximum. 3 wt % SM maximum. 300 ppm EB maximum. (ppm is parts per million by weight) Governmental standards on all pollutants.

The Environmental Protection Agency (EPA) standards for water pollution are given as the maximum parts per million (ppm on mass basis) for any one day. These standards are: 80 ppm for toluene, 60 ppm for methanol, 108 ppm for ethylbenzene, and 108 ppm for styrene monomer.

E. Equipment Pressure Drop For a preliminary design study, the following pressure drops may be assumed: Fired heater Reactor Heat exchangers* (shell and tube sides) Condensers under vacuum Other major equipment Distillation Trays: 1.0 kPa per theoret. stage for pressure columns. 0.6 kPa per theoret. stage for vacuum columns.

66 kPa 70 kPa 13 kPa 5 kPa 13 kPa

*Includes condensers, vaporizers, interchangers and all other exchangers except condensers operated under vacuum.

Negligible pressure drop through piping may be assumed.

F. Compression or Pump Efficiency The isentropic compression efficiency can be assumed to be 80%. The combined mechanical and electrical efficiency is approximately 90%.

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Appenidx A.

Styrene Monomer Production

Economic Data A. Manufacturing Costs Since plant startup is targeted for 1998, the manufacturing costs given below are based on that year. (Basis of units: K = thousands, M = millions, 1998 dollars) Raw Materials: M ethanol $0.19/kilogram Tol uene $0.42/kilogram Credits: Off-gas from three-phase separator $3.10/M kilojoules Utilities: Natural gas* $4.40/M kilojoules Steam : 2865 kPa, sat’d. $17.30/K kilograms 625 kPa, sat’d. $12.20/K kilograms Cooling water $0.03/K liters Inlet temp., avg. 31°C Outlet temp., avg. 41°C maximum El ectricity $0.065/kWH Condensate and Boiler feed water $2.50/K liters *Assume 90% efficiency for the fired heater fuel usage. Onstream time is 95% or 8320 hours per year.

B. Product Sales The gross profit is an initial economic indicator. It is defined as the product sales plus any credits minus the raw material costs. The product values given below are based on 1998 prices. Product Values: Crude Styrene product Ethylbenzene byproduct

$0.91/kilogram $0.57/kilogram

Forecasting costs over the lifetime of a new project is a difficult task. The cost data given here are tentative and appropriate only for preliminary economic evaluations. All costs are for 1998 and apply to the Houston Gulf Coast area, where the plant will be located.

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Appendix B. HYSYS Simulation Modules The HYSYS software system is an integrated engineering environment for the development and analysis of chemical process flowsheets. It provides you with: • steady-state modeling and optimization for process design, and • dynamic modeling for process controllability and control strategy development. HYSYS is licensed for educational use only in our university by Aspen Technology, Inc. of Cambridge, MA. This BEEF handbook focuses on steady-state modeling, often called chemical process simulation. A chemical process flowsheet is a conceptual representation of the transformation of raw materials into products through a series of process unit operations connected by process streams. Appendices C, D, etc. present simulation modules for the material and energy balances of some standard process unit operations. In general, a process unit transforms the material ing through it, and this transformation is represented by a set of algebraic equations, called a mathematical model. This model presents the material balances, energy balance, and thermodynamic relationships for the process unit. Since the math model has more variables than equations, its degrees-of-freedom (DOF) tells you the number of variables that must be specified to solve the equations. The order in which the model equations are solved—the math algorithm—depends on which variables are specified. In HYSYS more than one combination of specifications can be used to solve the mathematical model of a process unit. For example, in a pump simulation you can specify the inlet stream and the outlet pressure or pressure drop. You could also specify outlet information instead of inlet. This type of specification is known in HYSYS as backward propagation. HYSYS can back-calculate for the inlet conditions of a process unit given the outlet conditions and appropriate information to fulfill the degrees of freedom. This flexibility comes from the underlying mathematical equations that exist for each process unit and the different ways these equations can be solved. HYSYS incorporates the mathematical models for many process unit operations and knows which mathematical algorithm of a process unit to use according to which variables you have specified. Some of the physical operations ed by HYSYS are as follows: • • • • • • • • •

material stream energy stream component splitter compressor / expander cooler / heater heat exchanger LNG exchanger mixer plug flow reactor

• • • • • • • • •

pipe segment pump reactor operations separator / 3-phase separator / tank separation column shortcut column solid separator operations tee valve

The HYSYS Reference Manuals describe these process units in detail but do not provide an analysis of their mathematical models. These HYSYS manuals exist as Adobe Acrobat Reader files. You can access them through the Windows Start/Run… prompt. Type in the “Open:” box of the Run window \\eng-file1\engapps\Hysys-Docs and then click the OK button. Beef, Inc.

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Appendix B. HYSYS Simulation Modules Appendices C, D, etc. of this handbook present the mathematical models and some of their math algorithms for ten standard process units, in order to help you understand how HYSYS (or any simulation package, including hand calculations) solves the material and energy balances of these process units. The process unit modules in these appendices are for the following flowsheet operations: †

Module Process Stream Stream Mixer Pump Valve Heater/Cooler Chemical Reactor Two-Phase Separator Three-Phase Separator Component Splitter Distillation Column

Description Contains chemical components flowing at a certain state Mixes two or more process streams to make one stream Increases the pressure of a liquid process stream Decreases the pressure of a process stream Heats or cools a process stream Reacts the chemical compounds to form desired products Separates a process stream into vapor and liquid streams Separates a process stream into vapor, organic, and aqueous streams Splits a process stream into two streams at different temperatures Separate a process stream through a series of equilibrium stages

The format for each module has been standardized, in order to aid your learning process. This format is as follows: 1. 2. 3. 4. 5. 6. 7.

Description Process Diagram Assumptions Mathematical Model Variable Descriptions Mathematical Algorithms HYSYS Simulation Algorithms

The description explains the purpose of the process unit operation. The process diagram depicts the flow of material and energy. The assumptions list the conditions under which the mathematical model is applicable. The mathematical model presents the material balances, energy balance, and thermodynamic relationships for a process unit; that is, the algebraic equations that model a process unit simulation. The mathematical algorithms are representative examples of how the mathematical model could be solved for a specified set of variables that satisfies the degrees of freedom. The HYSYS simulation algorithms are representative examples of what you can specify to do a process unit simulation. These examples include calculations for both forward and backward propagation of information flow. You may have to consult the HYSYS Reference Manuals to discover all of simulation algorithms ed by HYSYS for a process unit operation. ____________________ †

Microsoft Word (.doc) files and Acrobat Reader (.pdf) files are available on the departments network file server (R:) in folder chem_engineering/public/HYSYS Manual for all materials in this HYSYS manual, including the modules in the appendices. Your instructor will give you directions on how to access this folder. If some equations in a Word document happen not to display correctly, then use the (.pdf) file version of the document instead.

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Appendix C. Process Stream Module Description This simulation module models a process stream containing material that is composed of multiple chemical compounds or components. A process stream moves (i.e., flows) this material from one process unit operation to another in a process flowsheet, connecting the two. In a process simulation, the material in a process stream is assumed to have uniform temperature, pressure, flow rate, and composition. These four quantities are referred to as the process state of a material stream. The flow rate and composition can be expressed in of molar, mass, or volumetric quantities. The mathematical model for a multi-component process stream is given below. In this model, the isothermal, single-phase stream has uniform and ideal mixing, no pressure drop, and no chemical reaction. The independent set of equations contains the relationships between molar, mass, and volumetric quantities, and the functions for pure component densities and molecular weights. To solve the equations in this model, (nc+3) design variables must be specified, as indicated by the degrees-of-freedom analysis below. Many mathematical algorithms can be derived from this model to do the material stream calculations. These algorithms differ in their given variables and their solution procedures. Two such algorithms are shown below. The first math algorithm is based on knowing the total molar flow rate and composition, while the second is based on knowing the component molar flow rates only. As indicated below, the second algorithm is just an extension of the first algorithm. Other possible mathematical algorithms for a single-phase stream ed by the HYSYS simulation system are summarized below. This module closes with a brief overview of how HYSYS simulates a process material stream that is multi-phase; that is, vapor and liquid coexisting in equilibrium within the stream. You may need to consult the HYSYS Reference Manuals for further details on multi-phase equilibrium in a process material stream.

Process Diagram Ti Pi ni Zi

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Assumptions Ti

stream i

Pi ni Zi

1. 2. 3. 4. 5.

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single phase isothermal no pressure drop uniform and ideal mixing no chemical reaction

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Appendix C. Process Stream Module

Mathematical Model (1)

M i = mi ni

( 2)

ρi = mi Vi

( 3)

Mi =

nc

∑ j =1

or 1 /

zi , j ⋅ M j

( 4)

M j = molwt [ pure j ]

(5)

ρi = 1 / ∑ ( wi , j / ρ j )

nc

∑(w

i, j

j =1

)

for j = 1, 2,

nc

nc

or

j =1

(6)

ρ j = density [Ti , Pi , pure j ]

(7)

∑( y j =1

)

or M i i, j ⋅ ρ j

, nc

nc

∑ j =1

cˆi , j or

for j = 1, 2,

, nc

wi , j = zi , j ⋅ M j / M i

for j = 1, 2,

, nc

(8)

yi , j = wi , j / ρ j ⋅ ρi

for j = 1, 2,

, nc

(9)

cˆi , j = ρi ⋅ zi , j M i

for j = 1, 2,

, nc

(10 )

cˆi′, j = ρi ⋅ wi , j

for j = 1, 2,

, nc

(11)

ni , j = ni ⋅ zi , j

or cˆi , j ⋅ V

for j = 1, 2,

, nc

(12 )

mi , j = mi ⋅ wi , j

or cˆi′, j ⋅ V

for j = 1, 2,

, nc

(13 )

Vi , j = Vi ⋅ yi , j

for j = 1, 2,

, nc

(14 )

Hˆ i = hmix ⎡⎣Ti , Pi , Z i ⎤⎦

(15 )

Ei = ni ⋅ Hˆ i

(

)

# vars

=

10 ⋅ nc

+

9

# eqns

=

9 ⋅ nc

+

6

=

⋅ nc

+

3

DOF 1

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/Mj

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nc

∑ cˆ′ j =1

i, j

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Appendix C. Process Stream Module

Variable Descriptions Ti Pi

is is

the temperature of process stream i, K. the pressure of process stream i, kPa.

ni mi Vi

is is is

the bulk molar flow rate of the process stream i, kgmol/h. the bulk mass flow rate of the process stream i, kg/h. the bulk volumetric flow rate of the process stream i, m3/h.

Mi Mj

is is

the bulk molecular weight of process stream i, kg/kgmol. the molecular weight of pure component j, kg/kgmol.

ρi ρj

is is

the bulk mass density of process stream i, kg/m3. the mass density of pure component j at Ti and Pi , kg/m3.

nc

is

the number of chemical components or compounds in the mixture.

zi, j wi, j yi, j

is is is

the bulk mole fraction of component j in process stream i, mol j/mol mix i. the bulk mass fraction of component j in process stream i, mass j/mass mix i. the bulk volume fraction of component j in process stream i, vol j/vol mix i.

ci , j cˆi′, j

is is

the bulk molar concentration of component j in process stream i, kgmol/m3. the bulk mass concentration of component j in process stream i, kg/m3.

ni, j mi, j Vi , j

is is is

the bulk molar flow rate of component j in process stream i, kgmol/h. the bulk mass flow rate of component j in process stream i, kg/h. the bulk volumetric flow rate of component j in process stream i, m3/h.

Hi Ei

is

the bulk molar enthalpy of process stream i, kJ/kgmol.

is

the bulk energy or heat flow of process stream i, kJ/h.

Zi Wi Yi

is

the bulk mole fractions of all nc-components in stream i.

is

the bulk mass fractions of all nc-components in stream i.

is

the bulk volume fractions of all nc-components in stream i.

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Appendix C. Process Stream Module Mathematical Algorithm A ⎡⎣ mi , Vi , Wi , Yi ⎤⎦

= streama ⎡⎣Ti , Pi , ni , Z i ⎤⎦

(11)

1.

ni , j

⇐

( 4)

2.

Mj

⇐

(6)

3.

ρj

( 3)

4.

Mi

ni ⋅ zi , j

for j = 1, 2,

, nc

molwt [ pure j ]

for j = 1, 2,

, nc

⇐

density [Ti , Pi , pure j ]

for j = 1, 2,

, nc

⇐

∑z

nc

i, j

j =1

⋅M j

(1) (7) (12 )

5.

mi

⇐

ni ⋅ M i

6.

wi , j

⇐

zi , j ⋅ M j / M i

for j = 1, 2,

, nc

7.

mi , j

⇐

mi ⋅ wi , j

for j = 1, 2,

, nc

(5)

8.

ρi

⇐

1/

for j = 1, 2,

, nc

nc

∑(w j =1

( 2)

9.

(8)

mi / ρi

i, j

/ ρj

)

Vi

⇐

10.

yi , j

⇐

(w

(13 )

11.

Vi , j

⇐

Vi ⋅ yi , j

for j = 1, 2,

, nc

(9) (10 )

12.

cˆi , j

⇐

ρi ⋅ zi , j / M i

for j = 1, 2,

, nc

13.

cˆi′, j

⇐

ρi ⋅ wi , j

for j = 1, 2,

, nc

(14 )

14.

Hˆ i

⇐

hmix ⎡⎣Ti , Pi , Z i ⎤⎦

(15 )

15.

Ei

⇐

ni ⋅ Hˆ i

i, j

)

/ ρ j ⋅ ρi

Mathematical Algorithm J mi , Vi , Wi , Yi , ni , Zi

(11)

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= streamj Ti , Pi , ni ,1 , ni, 2 , ..., ni ,nc

1.

ni

⇐

ni,1 + ni ,2 + ... + ni,nc

2.

zi , j

⇐

ni, j / ni

3.

mi , Vi , Wi , Yi

for j = 1, 2,

, nc

⇐ streama Ti , Pi , ni , Zi

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Appendix C. Process Stream Module Simulation Algorithms If the process state of a material stream is fully defined by knowing its temperature, pressure, total flow rate, and composition, then all of its unknown properties can be calculated, as depicted in the HYSYS simulation algorithms below:

mi , Vi , Wi , Yi

=

streama Ti , Pi , ni , Zi

mi , Vi , Yi , Zi

=

streamb Ti , Pi , ni , Wi

mi , Vi , Wi , Zi

=

streamc Ti , Pi , ni , Yi

ni , Vi , Yi , Zi

=

streamd Ti , Pi , mi , Wi

ni , Vi , Wi , Yi

=

streame Ti , Pi , mi , Zi

ni , Vi , Wi , Zi

=

streamf Ti , Pi , mi , Yi

mi , ni , Wi , Zi

=

streamg Ti , Pi , Vi , Yi

mi , ni , Wi , Yi

=

streamh Ti , Pi , Vi , Zi

mi , ni , Yi , Zi

=

streami Ti , Pi , Vi , Wi

If the process state of a material stream is fully defined by knowing its temperature, pressure, and component flow rates, then all of its unknown properties can be calculated, as depicted in the HYSYS simulation algorithms below:

ni , mi , Vi , Zi , Wi , Yi

= streamj Ti , Pi , ni ,1 , ni, 2 , ..., ni ,nc

ni , mi , Vi , Zi , Wi , Yi

= streamk Ti , Pi , mi ,1 , mi , 2 , ..., mi,nc

ni , mi , Vi , Zi , Wi , Yi

= streaml Ti , Pi , Vi,1 , Vi , 2 , ..., Vi,nc

If the process state of a material stream is multi-phase—either vapor-liquid or vaporliquid-liquid; that is, two or more distinct phases coexist in equilibrium, then three additional HYSYS simulation algorithms exist to determine the stream’s unknown properties as outlined in the remaining pages of this appendix.

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Appendix C. Process Stream Module Vapor-Liquid Equilibrium The above mathematical model for a process stream assumes its material is single phase—either all liquid or all vapor. At certain temperatures and pressures, the material in a process stream can be multi-phase—either vapor-liquid or vapor-liquid-liquid; that is, two or more distinct phases can coexist in equilibrium. Our discussion here focuses on vapor-liquid equilibrium (vle). HYSYS indicates multi-phases through the vapor fraction of a process stream. Vapor fraction ( Vf ) is the ratio of moles in the vapor phase over the total or bulk moles of the process stream; that is, it is what fraction of the total exists in the vapor state. For example, a Vf = 0.4 implies that 40% of the total moles is vapor, while 60% is liquid. The vapor fraction’s range is:

0 ≤ Vf ≤ 1. A calculated vapor fraction of zero indicates that the bulk material is all liquid. A calculated vapor fraction of one implies that the bulk material is all vapor. A calculated value between zero and one means a vapor and liquid are coexisting in equilibrium. The vapor-liquid equilibrium for a multicomponent mixture is best illustrated by a temperature-versus-composition (TXY) diagram for a binary or 2-component system. A general representation of a TXY diagram is as follows:

Pressure = 130 a

Saturated Vapor Curve

120

Vapor Region

Vf = 1

Tdp

b

L

Ti

Pi

V

T

Tbp

Vf = 0

c

Vapor-Liquid Region

80

70

60

d

Liquid Region

0

0.1

0.2

0.3

0.4

0.5

zi,j

xi,j

Saturated

0.6

0.7

Liquid

0.8

Curve

0.9

1

yi,j

Mole Fraction of Component j

For process stream i at a specified pressure of Pi and bulk composition of Zi , a typical condition of vapor-liquid equilibrium at temperature Ti is represented by line segment LTV in the TXY Beef, Inc.

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Appendix C. Process Stream Module diagram. The end points of this line are called the saturated-liquid and saturated-vapor points, and they represent the compositions of the two phases in equilibrium: xi , j for the liquid phase, yi, j for the vapor phase, and component j is the more volatile component in the binary system. The vapor fraction for this equilibrium is given by line segment LT over LV , or in mathematical , it is: total balance :

ni = nSV + nSL 1.0 =

component j balance :

nSV n + SL ni ni

⇒ 1.0 = V f + L f

ni zi , j = nSV yi , j + nSL xi , j zi , j = V f ⋅ yi , j + (1 − V f ) ⋅ xi , j

vapor fraction ratio :

Vf =

for each j for each j

zi , j − xi , j yi , j − xi , j

The ratio for Vf in of mole fractions is called the reverse-lever rule, because the contribution of the vapor phase at Point V is given by the line segment on the opposite side of Point T. The Vf equation is gotten by algebraically combining the total and component balances. In the TXY diagram, the vertical Path a → b → c → d depicts what would happen to process stream i if it were cooled at the specified pressure and bulk composition. Point a indicates that stream i would be in the vapor region, while Point d indicates the liquid region. Point b corresponds to a vapor fraction of one, and it is called the dew-point temperature Tdp . This point is when the first drop of liquid would form while the vapor was cooled. Point c corresponds to a vapor fraction of zero, and it is called the bubble-point temperature Tbp . This point is when the first bubble of vapor would form while the liquid was heated. When the following is true about the temperature of process stream i:

Tbp ≤ Ti ≤ Tdp , you know that vapor-liquid equilibrium exist in process stream i. In HYSYS, you can specify the vapor fraction instead of the temperature or pressure of a process stream. Three examples of vapor-liquid equilibrium (vle) calculations ed by HYSYS are:

Tdp , Xi , Yi = vle Vf = 1.0, Pi , Zi Tbp , Xi , Yi = vle Vf = 0.0, Pi , Zi Teq , Xi , Yi = vle Vf = 0.6, Pi , Zi where Teq is the equilibrium temperature of process stream i for the given vapor fraction. If you were to specify Ti instead of Pi in the above three functional forms, you would be calculating the dew-point pressure, bubble-point pressure, and equilibrium pressure, respectively, for process stream i.

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Appendix C. Process Stream Module When the material in process stream i is multi-phased, its bulk properties are related to the saturated liquid and vapor properties through the reverse-lever rule, as was the case for the bulk composition above. Some common bulk molar properties for a vapor-liquid equilibrium (vle) system are given below.

Vˆi = V f ⋅ VˆSV + (1 − V f ) ⋅ VˆSL

where

Hˆ i = V f ⋅ Hˆ SV + (1 − V f ) ⋅ Hˆ SL

VˆSV = vmixSV ⎣⎡Ti , Pi , Yi ⎦⎤

Hˆ SV = hmixSV ⎣⎡Ti , Pi , Yi ⎦⎤

VˆSL = vmixSL ⎣⎡Ti , Pi , X i ⎤⎦

Hˆ SL = hmixSL ⎡⎣Ti , Pi , X i ⎤⎦

Ti Pi

is is

the temperature of process stream i, K. the pressure of process stream i, kPa.

Vf Lf

is

the molar vapor fraction of process stream i, nSV / ni .

is

the molar liquid fraction of process stream i, nSL / ni .

ni nSV nSL

is is is

the bulk molar flow rate of process stream i, kgmol/h. the molar flow rate of the saturated vapor in stream i, kgmol/h. the molar flow rate of the saturated liquid in stream i, kgmol/h.

nc

is

the number of chemical components or compounds in the mixture.

Zi zi, j

is

the bulk mole fractions of all nc-components in stream i.

is

the bulk mole fraction of component j in process stream i; vector Z i means all elements zi,1 , zi, 2 , …, zi ,nc .

Yi yi, j

is

the sat’d vapor mole fractions of all nc-components in stream i.

is

the sat’d vapor mole fraction of component j in process stream i; vector Yi means all elements yi,1 , yi , 2 , …, yi,nc .

Xi xi , j

is

the sat’d liquid mole fractions of all nc-components in stream i.

is

the sat’d liquid mole fraction of component j in process stream i; vector Xi means all elements xi ,1 , xi, 2 , …, xi ,nc .

is

the bulk molar volume of process stream i, m3/kgmol.

is

the molar volume of the saturated vapor in stream i, m3/kgmol.

is

the molar volume of the saturated liquid in stream i, m3/kgmol.

is

the bulk molar enthalpy of process stream i, kJ/kgmol.

is

the molar enthalpy of the saturated vapor in stream i, kJ/kgmol.

is

the molar enthalpy of the saturated liquid in stream i, kJ/kgmol.

Vˆi Vˆ

SL

VˆSV

Hi HSV HSL

The value of any bulk vle property must fall between the values for the saturated liquid and vapor properties. If you expand the property view of a process stream, HYSYS will display its bulk, saturated liquid, and saturated vapor properties for the vapor-liquid equilibrium that exists in that process stream.

Beef, Inc.

C-8

7/29/98

Appendix D. Mixer Module Description A mixer operation is used to combine two process streams to form one process material stream. A pipe tee is used to accomplish the mixing operation. The two inlet streams to the pipe tee are usually of the same phase, either liquid or vapor mixtures. The pressure of the exit stream from the pipe tee is at the lowest pressure of the two inlet streams. If the inlet streams behave as ideal mixtures (i.e., no heat of mixing effects), the exit temperature will lie between the two inlet temperatures. However, this temperature may be quite different than those of the inlet streams when mixing effects are significant. The conceptual diagram for the mixer operation is given below for a steady-state system. The system is the mixtures of chemical compounds (or components) ing into, through, and from the pipe tee. The mathematical model given below for the mixer operation balances the material and energy flows of the system. This adiabatic unit operation occurs at steady state with no chemical reaction, and the kinetic and potential energy changes are negligible. No shaft work exists with this process operation. The independent set of equations contains the total and component material balances, the three sets of composition equations, the energy balance, the molar enthalpies of the three process streams, and the pressure relationship for the exit stream. The exit pressure is set to the lowest pressure of the two inlet streams, in order to eliminate the potential of back flow to the inlet streams. To solve this set of equations, (2·nc+6) variables must be specified, as indicated by the degrees-of-freedom analysis in the math model. From this mathematical model, many mathematical algorithms can be derived for doing process simulation calculations. These algorithms differ in their given (or design) variables and their solution procedures. Two such algorithms are detailed below—knowing the process states of two out of the three streams. The process state of a material stream is its temperature, pressure, total flow rate, and composition. The process state of the third stream is calculated using the solution procedure defined in a math algorithm. Other possible simulation algorithms ed by the HYSYS software are summarized below.

Process Diagram TF PF

Assumptions 1. 2. 3. 4. 5. 6.

inlet F

nF TE

ZF Exit

PE nE

TG PG

continuous process steady state no chemical reaction nelgect KE and PE changes adiabatic no shaft work

ZE inlet G

nG ZG

Beef, Inc.

D-1

7/29/98

Appendix D. Mixer Module Mathematical Model (1)

nF + nG − nE = 0

(2)

nF , j + nG , j − nE , j = 0

for j = 1, 2,

, nc

(3)

nF , j = n F ⋅ z F , j

for j = 1, 2,

, nc

(4)

nG, j = nG ⋅ zG, j

for j = 1, 2,

, nc

(5)

nE , j = n E ⋅ z E , j

for j = 1, 2,

, nc

(6)

nF ⋅ H F + nG ⋅ HG − nE ⋅ H E = 0

(7)

H F = hmix TF , PF , Z F

(8)

HG = hmix TG , PG , ZG

(9)

H E = hmix TE , PE , Z E

( 10 )

PE = min PF , PG # vars = 6 ⋅ nc + 12 # eqns = 4 ⋅ nc +

6

DOF = 2 ⋅ nc +

6

Variable Descriptions Ti Pi ni ni , j

is is is is

the temperature of process stream i, K. the pressure of process stream i, kPa. the bulk molar flow rate of process stream i, kgmol/h. the bulk molar flow rate of component j in process stream i, kgmol/h.

nc

is is

the number of chemical components or compounds in the mixture. the bulk mole fractions of all nc-components in stream i.

is

the bulk mole fraction of component j in process stream i; vector Z i means all elements zi ,1 , zi , 2 , …, zi ,nc .

Hi

is

the bulk molar enthalpy of process stream i, kJ/kgmol.

Ψi

is

a short notation for Ti , Pi , ni , and Zi of process stream i; that is, the process state of stream i.

Zi zi , j

Beef, Inc.

D-2

7/29/98

Appendix D. Mixer Module

Mathematical Algorithm A ΨE

= mixera ΨF , ΨG

(5)

1. 2. 3. 4. 5.

nE nF , j nG, j nE , j zE, j

⇐ ⇐ ⇐ ⇐ ⇐

nF + nG z F , j ⋅ nF zG , j ⋅ nG nF , j + nG , j nE , j / nE

( 7)

6.

HF

⇐

hmix TF , PF , Z F

(8)

7.

HG

⇐

hmix TG , PG , ZG

(6)

8.

HE

⇐

(10 )

9. 10.

(1) ( 3) (4) (2)

dH

F

, nc

for j = 1, 2,

, nc

for j = 1, 2,

, nc

for j = 1, 2,

, nc

i

⋅ nF + HG ⋅ nG / nE

⇐ min PF , PG PE Iterate on TE in

b g ⇐ H until f bT g = 0 f TE

(9)

for j = 1, 2,

E

− hmix TE , PE , Z E

E

Mathematical Algorithm B ΨG

= mixerb ΨE , ΨF

(4)

1. 2. 3. 4. 5.

nG nF , j nE , j nG, j zG , j

⇐ ⇐ ⇐ ⇐ ⇐

nE − nF z F , j ⋅ nF z E , j ⋅ nE nE , j − nF , j nG , j / nG

( 7)

6.

HF

⇐

hmix TF , PF , Z F

(9)

7.

HE

⇐

hmix TE , PE , Z E

(6)

8.

HG

⇐

(10 )

9. 10.

(1) ( 3) ( 5) (2)

(8)

dn

E

for j = 1, 2,

, nc

for j = 1, 2,

, nc

for j = 1, 2,

, nc

for j = 1, 2,

, nc

i

⋅ H E − nF ⋅ H F / nG

PG ⇐ min PE , PF Iterate on TG in

b g ⇐ H until f bT g = 0 f TG

G

− hmix TG , PG , ZG

G

Beef, Inc.

D-3

7/29/98

Appendix D. Mixer Module

HYSYS Simulation Algorithms If the process states of any two streams are fully defined (i.e., the temperature, pressure, flow rate, and composition of each stream are known), then the conditions of the third stream can be

calculated, as depicted in the HYSYS simulation algorithms below:

TE , PE , nE , Z E

= mixera TF , PF , nF , Z F , TG , PG , nG , ZG

TG , PG , nG , ZG

= mixerb TF , PF , nF , Z F , TE , PE , nE , Z E

TF , PF , nF , Z F

= mixerc TG , PG , nG , ZG , TE , PE , nE , Z E

Many more algorithms can be used to solve the above mathematical model for a mixer unit operation. The degrees-of-freedom shows that (2·nc+6) variables must be specified to solve the equations in the math model. Any combination of two temperatures, two pressures, two flow rates, and two compositions between the three process streams will solve the mixer module in HYSYS. Below are two examples.

Beef, Inc.

TE , PE , nG , ZG

= mixerd TF , PF , nF , Z F , TG , PG , nE , Z E

TE , PG , nF , Z F

= mixere TF , PF , nG , ZG , TG , PE , nE , Z E

D-4

7/29/98

Appendix E. Pump Module Description A pump operation is used to increase the pressure of a liquid material stream that is flowing from one process unit to another in a process flowsheet. Power (energy/time) in the form of electric energy drives a motor coupled to a steel drive shaft. The drive shaft connected to impellers imparts energy to the liquid in order to increase its pressure. The temperature of the liquid increases slightly, because of the effects of fluid friction. The conceptual diagram for the pump operation is given below for a steady-state system. The system is a liquid mixture of chemical compounds (or components) ing into, through, and from the pump. The mathematical model given below for the pump operation balances the material and energy flows of the system. This adiabatic unit operation occurs at steady state with no chemical reaction, and the kinetic and potential energy changes are negligible. The liquid is considered incompressible (i.e., at constant density); a good assumption for any liquid well removed from its critical point. The independent set of equations in the math model contains the total and component material balances, the energy balance, the molar enthalpies of the two process streams, the adiabatic efficiency, the ideal work based on the mechanical-energy balance for a frictionless fluid, the pressure change, and the inlet mixture density and molecular weight. The adiabatic efficiency relates the ideal to the actual work and has a typical value of 75% for most liquids. To solve this set of equations, (nc+5) variables must be specified, as indicated by the degrees-of-freedom analysis in the math model. From this mathematical model, many mathematical algorithms can be derived for doing process simulation calculations. These algorithms differ in their given (or design) variables and their solution procedures. Two such math algorithms are detailed below—for knowing the process state of the inlet stream and two additional variables. The unknown variables are calculated using the solution procedure defined in a math algorithm. The process state of a material stream is its temperature, pressure, total flow rate, and composition. Other possible simulation algorithms ed by the HYSYS software are summarized below.

Process Diagram TE

WA TI PI nI

Inlet

Assumptions

Exit pump

PE nE ZE

1. 2. 3. 4. 5. 6.

continuous process steady state no chemical reaction neglect KE and PE changes adiabatic incompressible liquid

ZI

Beef, Inc.

E-1

7/29/98

Appendix E. Pump Module Mathematical Model (1)

nI − nE = 0

(2)

nI ⋅ z I , j − nE ⋅ z E, j = 0

( 3)

n I ⋅ H I − n E ⋅ H E + WA = 0

(4)

H I = hmix TI , PI , Z I

( 5)

H E = hmix TE , PE , Z E

(6)

ε = 100 ⋅ WI / WA

(7)

WI = ΔP ⋅ n I ⋅ M I / ρ I

(8)

ΔP = PE − PI

(9)

ρ I = liqden TI , PI , Z I

( 10 )

M I = molwt Z I

for j = 1, 2,

, nc

# vars = 2 ⋅ nc + 14 # eqns =

nc +

9

DOF = 1 ⋅ nc +

5

Variable Descriptions Ti Pi ni

is is is

the temperature of process stream i, K. the pressure of process stream i, kPa. the bulk molar flow rate of process stream i, kgmol/h.

nc

is is

the number of chemical components or compounds in the mixture. the bulk mole fractions of all nc-components in stream i.

is

the bulk mole fraction of component j in process stream i; vector Z i means all elements zi ,1 , zi , 2 , …, zi ,nc .

Hi WA

is

the bulk molar enthalpy of process stream i, kJ/kgmol.

is is

the actual work or power of the pump, kJ/h. the adiabatic efficiency of the pump (0 to 100), percent.

WI ΔP Mi

is is is is

the ideal work or power of the pump (ε = 100%), kJ/h. the pressure drop between the exit and inlet streams, kPa. the molecular weight of process stream i, kg/kgmol. the liquid density of process stream i, kg/m3.

Zi zi , j

ε

ρi

Beef, Inc.

E-2

7/29/98

Appendix E. Pump Module Mathematical Algorithm A = pumpa TI , PI , n I , Z I , ε , PE

TE , ΔP, n E , Z E , WA (1 )

1.

nE

⇐

nI

(2)

2.

z E, j

⇐

z I, j

(8)

3.

ΔP

⇐

PE − PI

(9)

4.

ρI

⇐

liqden TI , PI , Z I

(10 )

5.

MI

⇐

molwt Z I

( 7)

6.

WI

⇐

ΔP ⋅ n I ⋅ M I / ρ I

(6)

7.

WA

⇐

100 ⋅ WI / ε

(4)

8.

HI

⇐

hmix TI , PI , Z I

( 3)

9.

HE

⇐

dn ⋅ H

10.

I

, nc

i

I

+ WA / n E

E

− hmix TE , PE , Z E

Iterate on TE in

b g ⇐ H until f bT g = 0 f TE

( 5)

for j = 1, 2,

E

Mathematical Algorithm E ΔP, n E , Z E , WA , ε

Beef, Inc.

= pumpe TI , PI , n I , Z I , TE , PE

( 1)

1.

nE

⇐

nI

(2)

2.

z E, j

⇐

z I, j

(9)

3.

ρI

⇐

liqden TI , PI , Z I

(10 )

4.

MI

⇐

molwt Z I

(4)

5.

HI

⇐

hmix TI , PI , Z I

( 5)

6.

HE

⇐

hmix TE , PE , Z E

( 3)

7.

WA

⇐

nE ⋅ H E − nI ⋅ H I

(8)

8.

ΔP

⇐

PE − PI

( 7)

9.

WI

⇐

ΔP ⋅ n I ⋅ M I / ρ I

(6)

10.

ε

⇐

100 ⋅ WI / WA

for j = 1, 2,

E-3

, nc

7/29/98

Appendix E. Pump Module HYSYS Simulation Algorithms If the process state of the inlet stream is fully defined (i.e., TI , PI , n I , Z I are known), only two additional variables are required to calculate all unknowns, as depicited in the HYSYS simulation algorithms below:

TE , ΔP, n E , Z E , WA

=

pumpa TI , PI , n I , Z I , ε , PE

TE , PE , n E , Z E , WA

=

pumpb TI , PI , n I , Z I , ε , ΔP

TE , PE , n E , Z E , ΔP

=

pumpc TI , PI , n I , Z I , ε , WA

PE , n E , Z E , WA , ε

=

pumpd TI , PI , n I , Z I , TE , ΔP

ΔP, n E , Z E , WA , ε

=

pumpe TI , PI , n I , Z I , TE , PE

TE , PE , n E , Z E , ε

=

pumpf TI , PI , n I , Z I , WA , ΔP

TE , ΔP, n E , Z E , ε

=

pumpg TI , PI , n I , Z I , WA , PE

If the process state of the exit stream is fully defined (i.e., TE , PE , n E , Z E are known), only two additional variables are required to calculate all unknowns, as depicited in the HYSYS simulation algorithms below:

Beef, Inc.

TI , PI , n I , Z I , WA

=

pumph TE , PE , n E , Z E , ε , ΔP

TI , ΔP, n I , Z I , WA

=

pumpi TE , PE , n E , Z E , ε , PI

TI , PI , n I , Z I , ΔP

=

pumpj TE , PE , n E , Z E , ε , WA

PI , n I , Z I , WA , ε

=

pumpk TE , PE , n E , Z E , TI , ΔP

ΔP, n I , Z I , WA , ε

=

pumpl TE , PE , n E , Z E , TI , PI

TI , PI , n I , Z I , ε

=

pumpm TE , PE , n E , Z E , WA , ΔP

TI , ΔP, n I , Z I , ε

=

pumpn TE , PE , n E , Z E , WA , PI

E-4

7/29/98

Appendix F. Valve Module Description A valve operation is used to decrease the pressure of a process material stream. The process simulation of a valve can determine the process state of the exit stream given the process state of the inlet stream or determine the process state of the inlet stream given the process state of the exit stream. The process state of a material stream is its temperature, pressure, total flow rate, and composition. The conceptual diagram for the valve operation is given below for a steady-state system. The system is the mixture of chemical compounds (or components) ing into, through, and from the valve. The mathematical model given below for the valve operation balances the material and energy flows of the system. This adiabatic unit operation occurs at steady state with no chemical reaction, and the kinetic and potential energy changes are negligible. The independent set of equations in the math model contains the total and component material balances, the energy balance, the molar enthalpies of the two process streams, and the definition of pressure drop. To solve these equations, (nc+4) variables must be specified, as indicated by the degrees-of-freedom analysis in the math model. From this mathematical model, many mathematical algorithms can be derived for doing process simulation calculations. These algorithms differ in their given (or design) variables and their solution procedures. Two such math algorithms are shown below—one for the process state of the inlet stream given, and the other for the exit stream given. The unknown variables are calculated using the solution procedure defined in a math algorithm. Again, the process state of a material stream is its temperature, pressure, total flow rate, and composition. Other possible simulation algorithms ed by the HYSYS software are summarized below.

Process Diagram TI PI

Inlet

valve

Assumptions TE Exit

PE

nI

nE

ZI

ZE

Beef, Inc.

F-1

1. 2. 3. 4. 5. 6.

continuous process steady state no chemical reaction neglect KE and PE changes adiabatic no shaft work

7/29/98

Appendix F. Valve Module Mathematical Model (1)

n I − nE = 0

(2)

nI ⋅ z I , j − nE ⋅ z E , j = 0

( 3)

n I ⋅ H I − nE ⋅ H E = 0

(4)

H I = hmix TI , PI , Z I

( 5)

H E = hmix TE , PE , Z E

(6)

ΔP = PI − PE

for j = 1, 2,

, nc

# vars = 2 ⋅ nc + 9 # eqns =

nc + 5

DOF = 1 ⋅ nc + 4

Variable Descriptions

Beef, Inc.

Ti Pi ni

is is is

the temperature of process stream i, K. the pressure of process stream i, kPa. the bulk molar flow rate of process stream i, kgmol/h.

nc

Zi zi , j

is is

the number of chemical components or compounds in the mixture. the bulk mole fractions of all nc-components in stream i.

is

the bulk mole fraction of component j in process stream i; vector Z i means all elements zi ,1 , zi , 2 , …, zi ,nc .

Hi ΔP

is is

the bulk molar enthalpy of process stream i, kJ/kgmol. the pressure drop between the exit and inlet streams, kPa.

Ψi

is

a short notation for Ti , Pi , ni , and Zi of process stream i; that is, the process state of stream i.

F-2

7/29/98

Appendix F. Valve Module Mathematical Algorithm A ΨE

= valvea ΨI , ΔP

(1)

1.

nE

⇐

nI

(2)

2.

zE, j

⇐

zI, j

(6)

3.

PE

⇐

PI − ΔP

(4)

4.

HI

⇐

hmix TI , PI , Z I

( 3)

5.

HE

⇐

HI

6.

Iterate on TE in

b g ⇐ H until f bT g = 0 f TE

(5)

for j = 1, 2,

E

, nc

− hmix TE , PE , Z E

E

Mathematical Algorithm E ΨI

= valved ΨE , ΔP

(1)

1.

nI

⇐

nE

(2)

2.

zI, j

⇐

zE, j

(6)

3.

PI

⇐

PE + ΔP

(5)

4.

HE

⇐

hmix TE , PE , Z E

( 3)

5.

HI

⇐

HE

6.

Iterate on TI in

(4)

for j = 1, 2,

b g ⇐ H − hmix T , P , Z until f bT g = 0 f TI

I

I

I

, nc

I

I

Beef, Inc.

F-3

7/29/98

Appendix F. Valve Module HYSYS Simulation Algorithms If the process state of the inlet stream is fully defined (i.e., the temperature, pressure, flow rate and composition are known), only one additional variable is required to calculate all unknowns, as depicted in the HYSYS simulation algorithms below:

TE , PE , nE , Z E

=

valvea TI , PI , nI , Z I , ΔP

TE , ΔP, nE , Z E

=

valveb TI , PI , nI , Z I , PE

PE , nE , Z E , ΔP

=

valvec TI , PI , nI , Z I , TE

If the process state of the exit stream is fully defined (i.e., the temperature, pressure, flow rate and composition are known), only one additional variable is required to calculate all unknowns, as depicted in the HYSYS simulation algorithms below:

Beef, Inc.

TI , PI , nI , Z I

=

valved TE , PE , nE , Z E , ΔP

TI , ΔP, nI , Z I

=

valvee TE , PE , nE , Z E , PI

PI , nI , Z I , ΔP

=

valvef TE , PE , nE , Z E , TI

F-4

7/29/98

Appendix G. Heater/Cooler Module Description A heater or cooler operation is used to increase or decrease the energy content of a process material stream. A heater adds energy to the material in a process stream to increase its temperature. A cooler takes away energy from the material in a process stream to decrease its temperature. Neither unit operation affects the stream bulk and component flow rates or composition. The energy added or subtracted from the stream by a heater or cooler, respectively, is known as the duty of that unit operation. A cooler is essentially a heater with a negative duty value, and vice versa. The conceptual diagram for the heater/cooler operation is given below for a steady-state system. The system is the mixture of chemical compounds (or components) ing into, through, and from the heater or cooler. The mathematical model given below for the heating operation balances the material and energy flows of the system. This model also represents the cooling of a process stream if the duty is a negative number. The unit operation occurs at steady state with no chemical reaction, and the kinetic and potential energy changes are negligible. The independent set of equations in the math model contains the total and component material balances, the energy balance, the molar enthalpies of two process streams, and the definition of pressure drop. To solve these equations, (nc+5) variables must be specified, as indicated by the degrees-of-freedom analysis in the math model. From this mathematical model, many mathematical algorithms can be derived for doing process simulation calculations. These algorithms differ in their given (or design) variables and their solution procedures. Two such math algorithms are shown below—one for the process state of the inlet stream and exit temperature and pressure given, and one for the process state of the inlet stream, duty and exit pressure given. The unknown variables are calculated using the solution procedure defined in a math algorithm. The process state of a material stream is its temperature, pressure, total flow rate, and composition. Other possible simulation algorithms ed by the HYSYS software are summarized below.

Process Diagram Q

TI PI nI ZI

Beef, Inc.

Assumptions

Inlet

TE Exit

heater

PE nE ZE

G-1

1. 2. 3. 4. 5.

continuous process steady state no chemical reaction neglect KE and PE changes no shaft work

7/29/98

Appendix G. Heater/Cooler Module

Mathematical Model (1)

n I − nE = 0

(2)

nI ⋅ z I , j − nE ⋅ z E , j = 0

( 3)

n I ⋅ H I − nE ⋅ H E + Q = 0

(4)

H I = hmix TI , PI , Z I

( 5)

H E = hmix TE , PE , Z E

(6)

ΔP = PE − PI

for j = 1, 2,

, nc

# vars = 2 ⋅ nc + 10 # eqns =

nc +

5

DOF = 1 ⋅ nc +

5

Variable Descriptions Ti Pi ni

is is is

the temperature of process stream i, K. the pressure of process stream i, kPa. the bulk molar flow rate of process stream i, kgmol/h.

nc

is is

the number of chemical components or compounds in the mixture. the bulk mole fractions of all nc-components in stream i.

is

the bulk mole fraction of component j in process stream i; vector Z i means all elements zi ,1 , zi , 2 , …, zi ,nc .

Zi zi , j

Beef, Inc.

Q Hi ΔP

is

the energy duty of the heater or cooler, kJ/h.

is is

the bulk molar enthalpy of process stream i, kJ/kgmol. the pressure drop between the exit and inlet streams, kPa.

Ψi

is

a short notation for Ti , Pi , ni , and Zi of process stream i; that is, the process state of stream i.

G-2

7/29/98

Appendix G. Heater/Cooler Module

Mathematical Algorithm A ΔP, nE , Z E , Q

= heatera ΨI , TE , PE

(1)

1.

nE

⇐

nI

(2)

2.

zE, j

⇐

zI, j

(6)

3.

ΔP

⇐

PE − PI

(4)

4.

HI

⇐

hmix TI , PI , Z I

(5)

5.

HE

⇐

hmix TE , PE , Z E

(3)

6.

Q

⇐

nE ⋅ H E − n I ⋅ H I

for j = 1, 2,

, nc

for j = 1, 2,

, nc

Mathematical Algorithm E TE , ΔP, nE , Z E

= heatere ΨI , PE , Q

(1)

1.

nE

⇐

nI

(2)

2.

zE, j

⇐

zI, j

(6)

3.

ΔP

⇐

PE − PI

(4)

4.

HI

⇐

hmix TI , PI , Z I

( 3)

5.

HE

⇐

d n ⋅ H + Qi / n

6.

Iterate on TE in

(5)

I

I

b g ⇐ H until f bT g = 0 f TE

E

E

− hmix TE , PE , Z E

E

Beef, Inc.

G-3

7/29/98

Appendix G. Heater/Cooler Module HYSYS Simulation Algorithms If the process state of the inlet stream is fully defined (i.e., the temperature, pressure, flow rate and composition are known), only two additional variable are required to calculate all unknowns, as depicted in the HYSYS simulation algorithms below:

ΔP, nE , Z E , Q

=

heatera TI , PI , nI , Z I , TE , PE

TE , PE , nE , Z E

=

heaterb TI , PI , nI , Z I , ΔP, Q

PE , nE , Z E , Q

=

heaterc TI , PI , nI , Z I , ΔP, TE

PE , nE , Z E , ΔP

=

heaterd TI , PI , nI , Z I , Q, TE

TE , nE , Z E , ΔP

=

heatere TI , PI , nI , Z I , Q, PE

If the process state of the exit stream is fully defined (i.e., the temperature, pressure, flow rate and composition are known), only two additional variable are required to calculate all unknowns, as depicted in the HYSYS simulation algorithms below:

Beef, Inc.

ΔP, nI , Z I , Q

=

heaterf TE , PE , nE , Z E , TI , PI

TI , PI , nI , Z I

=

heaterg TE , PE , nE , Z E , ΔP, Q

PI , nI , Z I , Q

=

heaterh TE , PE , nE , Z E , ΔP, TI

PI , nI , Z I , ΔP

=

heateri TE , PE , nE , Z E , Q, TI

TI , nI , Z I , ΔP

=

heaterj TE , PE , nE , Z E , Q, PI

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Appendix H. Chemical Reactor Module Description HYSYS s the process simulation of many different types of chemical reactors. Our focus here is on the conversion-based model for a reactor. A chemical reactor operation takes the reactants of a feed stream and converts them, usually in the presence of a catalyst, to the desired product, which appears in the effluent stream. Because most reactions do not go to completion, the effluent stream will usually contain some reactants and inert materials from the feed stream as well as by-products and waste materials. The conceptual diagram for the reactor operation is given below for a steady-state system. The system is the mixture of chemical compounds (or components) ing into, through, and from the reactor. The mathematical model given below for the reactor operation balances the material and energy flows of the system. It is illustrated for a specific reaction stoichoimetry —the production of styrene monomer from toluene and methanol with the by-product formation of ethylbenzene. This adiabatic unit operation occurs at steady state with no shaft work, and the kinetic and potential energy changes are negligible. The independent set of equations in the math model contains the total and component material balances, the two sets of composition equations, the reaction conversion and yield equations, the energy balance, the molar enthalpies of the two process streams, and the definition of pressure drop. To solve these equations, (nc+6) variables must be specified, as indicated by the degrees-of-freedom analysis in the math model. From this mathematical model, many mathematical algorithms can be derived for doing process simulation calculations. These algorithms differ in their given (or design) variables and their solution procedures. One such algorithm is shown below with the process state of the feed material stream, reaction conversion and yield, and pressure drop as the specified variables. The process state of a material stream is its temperature, pressure, total flow rate, and composition. Other possible simulation algorithms ed by the HYSYS software are summarized below.

Process Diagram

Assumptions

PF

1. 2. 3. 4. 5.

TE

TF Feed

nF

Reactor

Effluent

PE nE ZE

ZF

continuous process steady state neglect KE and PE changes adiabatic no shaft work

Chemical Reaction Stoichoimetry Rxn 1:

Rxn 2:

Beef, Inc.

C7H8

+

CH3OH

C8H8

↔

+

H 2O

+

H2

toluene

methanol

styrene

water

hydrogen

(TL)

(ME)

(SM)

(WA)

(H2)

C7H8

+

CH3OH

C8H10

↔

+

H 2O

toluene

methanol

ethylbenzene

water

(TL)

(ME)

(EB)

(WA)

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Appendix H. Chemical Reactor Module

Mathematical Model (1)

nF − nE + R1 = 0

( 2)

nF ,TL − nE ,TL − R1 − R2 = 0

( 3)

nF , ME − nE , ME − R1 − R2 = 0

( 4)

nF , SM − nE , SM + R1

(5)

nF , EB − nE , EB +

(6)

nF ,WA − nE ,WA + R1 + R2 = 0

(7)

nF , H 2 − nE , H 2 + R1

=0 R2 = 0 =0

(8)

nF , j = nF ⋅ Z F , j

for j = 1, 2,

, nc

(9)

nE , j = nE ⋅ Z E , j

for j = 1, 2,

, nc

(10 )

ε TL = ( nF ,TL − nE ,TL ) / nF ,TL

(11)

YSM = nE , SM / ( nF ,TL − nE ,TL )

(12 )

nF ⋅ Hˆ F − nE ⋅ Hˆ E = 0

(13 )

Hˆ F = hmix ⎡⎣TF , PF , Z F ⎤⎦

(14 )

Hˆ E = hmix ⎡⎣TE , PE , Z E ⎤⎦

(15 )

ΔP = PF − PE # vars

=

4 ⋅ nc

+

13

# eqns

=

3 ⋅ nc

+

7

=

⋅ nc

+

6

DOF 1

Variable Descriptions Ti Pi ni ni , j

Beef, Inc.

is is is is

the temperature of process stream i, K. the pressure of process stream i, kPa. the bulk molar flow rate of process stream i, kgmol/h. the bulk molar flow rate of component j in process stream i, kgmol/h.

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Appendix H. Chemical Reactor Module is is

the number of chemical components or compounds in the mixture. the bulk mole fractions of all nc-components in stream i.

is

the bulk mole fraction of component j in process stream i; vector Z i means all elements zi ,1 , zi , 2 , …, zi ,nc .

YSM

is is is is

the extent of the styrene monomer reaction (i.e., Rxn 1), 1/h. the extent of the ethylbenzene reaction (i.e., Rxn 2), 1/h. the molar conversion of toluene (moles of toluene reacted per moles of toluene fed). the molar yield of styrene (moles of styrene formed per moles of toluene reacted).

Hi ΔP

is is

the bulk molar enthalpy of process stream i, kJ/kgmol. the pressure drop between the exit and inlet streams, kPa.

Ψi

is

a short notation for Ti , Pi , ni , and Zi of process stream i; that is, the process state of stream i.

nc

Zi zi , j

R1 R2

ε TL

Mathematical Algorithm A ΨE

= reactora ΨF , ε TL , YSM , ΔP

(8)

1.

nF , j

(10 )

2.

nE ,TL ⇐

nF ,TL − nE ,TL ⋅ ε TL

(11)

3.

nE ,SM ⇐

YSM nF ,TL − nE ,TL

(4)

4.

R1

⇐

nE ,SM − nF ,SM

( 5)

5.

R2

⇐

nE , EB − nF , EB

(1)

6.

nE

⇐

R1 + nF

(2)

7.

nE ,TL ⇐

nF ,TL − R1 − R2

( 3)

8.

nE , ME ⇐

nF , ME − R1 − R2

( 7)

9.

nE , H 2 ⇐

nF , H 2 + R1

(6)

10.

nE ,WA ⇐

nF ,WA + R1 + R2

⇐

nF ⋅ x F , j

c

(9)

11.

zE, j

⇐

n E , j / nE

12.

PE

⇐

PF − ΔP

(13 )

13.

HF

⇐

hmix TF , PF , Z F

(12 )

14.

HE

⇐

n F ⋅ H F / nE

15.

Iterate on TE in

b g ⇐ H until f bT g = 0 f TE

E

, nc

for j = 1, 2,

, nc

h

(15 )

(14 )

for j = 1, 2,

− hmix TE , PE , Z E

E

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Appendix H. Chemical Reactor Module

HYSYS Simulation Algorithms If the process state of the feed stream is fully defined (i.e., the temperature, pressure, flow rate and composition are known), only three additional variables plus the reaction stoichoimetries

are required to calculate all unknowns, as depicted in the HYSYS simulation algorithms below:

TE , PE , nE , Z E

= reactora TF , PF , nF , Z F , ε TL , YSM , ΔP

TE , ΔP, nE , Z E

= reactorb TF , PF , nF , Z F , ε TL , YSM , PE

HYSYS can not do back calculations for the conversion reactor. Therefore, the only variation on the simulation algorithms that works in HYSYS is specifying either the effluent pressure or the pressure drop. In HYSYS, the stoichoimetry of each reaction and its associated molar conversion are specified in a reaction set of the Fluid Package. You must define a reaction set for each chemical reaction that occurs in the reactor.

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Appendix I. Two-Phase Separator Module Description A two-phase separator operation is used to separate a feed process stream into two material streams according to phase. A vapor-liquid equilibrium is reached in the vessel at a certain temperature and pressure. When the feed pressure is higher than the vessel pressure, a flash operation occurs in the vessel producing the two phases; otherwise, withdrawing heat from the feed stream produces the two phases. The two phases are then separated into a saturated vapor stream and a saturated liquid stream. The conceptual diagram for the two-phase separation operation is given below for a steady-state system. The system is the mixture of chemical compounds (or components) ing into, through, and from the two-phase separator. The mathematical model given below for the two-phase separation balances the material and energy flows of the system. This unit operation occurs at steady state with no chemical reaction and shaft work, and the kinetic and potential energy changes are negligible. The independent set of equations contains the total material and energy balances, the vapor-liquid equilibrium function (vle), the relationships between vapor fraction and flow rates, the relationships between the outlet temperatures and pressures, the molar enthalpies of the three process streams, and the definition of pressure change. To solve these equations, (nc+5) variables must be specified, as indicated by the degrees-of-freedom analysis in the math model. From this mathematical model, many mathematical algorithms can be derived for doing process simulation calculations. These algorithms differ in their given (or design) variables and their solution procedures. One such algorithm is shown below for the process state of the feed stream given. The unknown variables are calculated using the solution procedure defined in a math algorithm. The process state of a material stream is its temperature, pressure, total flow rate, and composition. Other possible simulation algorithms ed by the HYSYS software are summarized below.

Process Diagram

Assumptions 1. 2. 3. 4. 5.

TV PV

Vapor

nV TF PF

Feed

Q

nF ZF

ZV

sat’d vapor

continuous process steady state no chemical reaction neglect KE and PE changes no shaft work

sat’d liquid

Liquid

TL PL nL ZL

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7/29/98

Appendix I. Two-Phase Separator Module Mathematical Model ( 1)

nF − nV − nL = 0

(2)

nF ⋅ H F − nV ⋅ HV − nL ⋅ H L + Q = 0

( 3)

Vf , YV , X L = vle TV , PV , Z F

(4)

Vf = nV / nF

( 5)

TL = TV

(6)

PL = PV

(7)

H F = hmix TF , PF , Z F

(8)

HV = hmix TV , PV , YV

(9)

H L = hmix TL , PL , X L

( 10 )

ΔP = PF − PV # vars = 3 ⋅ nc + 17 # eqns = 2 ⋅ nc + 12 DOF = 1 ⋅ nc +

5

Variable Descriptions Ti Pi ni

is is is

the temperature of process stream i, K. the pressure of process stream i, kPa. the molar flow rate of process stream i, kgmol/h.

nc

is is

the number of chemical components or compounds in the mixture. the bulk mole fractions of all nc-components in stream i.

is

the bulk mole fraction of component j in process stream i; vector Z i means all elements zi ,1 , zi , 2 , …, zi ,nc .

Yi Xi

is

the vapor mole fractions of all nc-components in stream i.

is

the liquid mole fractions of all nc-components in stream i.

Vf

is is

the molar vapor fraction of the vapor-liquid equilibrium. the energy duty of the two-phase separator, kJ/h.

is is

the molar enthalpy of process stream i, kJ/kgmol. the pressure drop between the exit and inlet streams, kPa.

is

a short notation for Ti , Pi , ni , and composition of process stream i; that is, the process state of stream i.

Zi zi , j

Q Hi ΔP

Ψi

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I-2

7/29/98

Appendix I. Two-Phase Separator Module Mathematical Algorithm A PV , nV , YV , ΨL , Q

= sepa ΨF , ΔP, TV

(10 )

1.

PV

( 3)

2.

Vf , YV , X L

(4)

3.

nV

⇐

V f ⋅ nF

(1)

4.

nL

⇐

nF − nV

( 5)

5.

TL

⇐

TV

(6)

6.

PL

⇐

PV

( 7)

7.

HF

⇐

hmix TF , PF , Z F

(8)

8.

HV

⇐

hmix TV , PV , YV

(9)

9.

HL

⇐

hmix TL , PL , X L

(2)

10.

Q

⇐

nV ⋅ HV + n L ⋅ H L − nF ⋅ H F

⇐

PF − ΔP ⇐ vle TV , PV , Z F

HYSYS Simulation Algorithms If the process state of the feed stream is fully defined (i.e., the temperature, pressure, flow rate and composition are known), only two additional variables are required to calculate all

unknowns, as depicted in the HYSYS simulation algorithms below:

PV , nV , YV , ΨL , Q

=

sepa ΨF , ΔP, TV

ΔP, nV , YV , ΨL , Q

=

sepb ΨF , PV , TV

TV , PV , nV , YV , ΨL

=

sepc ΨF , ΔP, Q

TV , nV , YV , ΨL , ΔP

=

sepd ΨF , PV , Q

HYSYS can also back calculate for the feed stream conditions given certain information about the exit streams and the unit operation. Because of the nature of the vapor-liquid equilibrium, backcalculation requires one more given variable than the forward calculation. So to back-calculate, (nc+6) variables must be specified. These variables are two of the three total flow rates, one exit composition, one exit temperature or pressure, the pressure drop, and the duty of the two-phase separator.

Beef, Inc.

I-3

7/29/98

Appendix J. Three-Phase Separator Module Description A three-phase separator operation is used to separate a feed process stream into three material streams according to phase. A vapor-liquid-liquid equilibrium is reached in the vessel at a certain temperature and pressure. When the feed pressure is higher than the vessel pressure, a flash operation occurs in the vessel producing the three phases; otherwise, withdrawing heat from the feed stream produces the three phases. The three phases are then separated into a vapor stream, a light-liquid (organic) stream and a heavy-liquid (aqueous) stream. The conceptual diagram for the three-phase separation operation is given below for a steady-state system. The system is the mixture of chemical compounds (or components) ing into, through, and from the three-phase separator. The mathematical model given below for the three-phase separation balances the material and energy flows of the system. This unit operation occurs at steady state with no chemical reaction and shaft work, and the kinetic and potential energy changes are negligible. The independent set of equations contains the total material and energy balances, the vapor-liquidliquid equilibrium function (vlle), the relationships between phase fractions and flow rates, the relationships between the outlet temperatures and pressures, the molar enthalpies of the four process streams, and the definition of pressure change. To solve these equations, (nc+5) variables must be specified, as indicated by the degrees-of-freedom analysis in the math model. From this mathematical model, many mathematical algorithms can be derived for doing process simulation calculations. These algorithms differ in their given (or design) variables and their solution procedures. One such algorithm is shown below for the process state of the feed stream given. The unknown variables are calculated using the solution procedure defined in a math algorithm. The process state of a material stream is its temperature, pressure, total flow rate, and composition. Other possible simulation algorithms ed by the HYSYS software are summarized below.

Process Diagram

Assumptions 1. 2. 3. 4. 5.

TV Vapor

PV nV

TF PF nF

ZV

vapor Feed

ZF

organic aqueous

Q

Heavy

Light

TH

TL PL nL

continuous process steady state no chemical reaction neglect KE and PE changes no shaft work

ZL

PH nH ZH

Beef, Inc.

J-1

7/29/98

Appendix J. Three-Phase Separator Module Mathematical Model (1)

nF − nV − nL − nH = 0

(2)

nF ⋅ H F − nV ⋅ HV − nL ⋅ H L − nH ⋅ H H + Q = 0

(3)

Vf , L fL , YV , X L , X H = vlle TV , PV , Z F

(4)

Vf = nV / nF

(5)

L fL = nL / nF

(6)

TL = TH = TV

(7)

PL = PH = PV

(8)

H F = hmix TF , PF , Z F

(9)

HV = hmix TV , PV , YV

( 10 )

H L = hmix TL , PL , X L

( 11 )

H H = hmix TH , PH , X H

( 12 )

ΔP = PF − PV # vars = 4 ⋅ nc + 22 # eqns = 3 ⋅ nc + 17 DOF = 1 ⋅ nc +

5

Variable Descriptions Ti Pi ni

is is is

the temperature of process stream i, K. the pressure of process stream i, kPa. the molar flow rate of process stream i, kgmol/h.

nc

is is

the number of chemical components or compounds in the mixture. the bulk mole fractions of all nc-components in stream i.

is

the bulk mole fraction of component j in process stream i; vector Z i means all elements zi ,1 , zi , 2 , …, zi ,nc .

Yi Xi

is

the vapor mole fractions of all nc-components in stream i.

is

the liquid mole fractions of all nc-components in stream i.

Vf L fL Q Hi ΔP

is is is

the molar vapor fraction of the vapor-liquid-liquid (vll) equilibrium. the molar light-liquid fraction of the vll equilibrium. the energy duty of the two-phase separator, kJ/h.

is is

the molar enthalpy of process stream i, kJ/kgmol. the pressure drop between the exit and inlet streams, kPa.

Ψi

is

a short notation for Ti , Pi , ni , and composition of process stream i; that is, the process state of stream i.

Zi zi , j

Beef, Inc.

J-2

7/29/98

Appendix J. Three-Phase Separator Module Mathematical Algorithm A PV , nV , YV , ΨL , ΨH , Q

= sepa ΨF , ΔP, TV

(12 )

1.

PV

( 3)

2.

Vf , L fL , YV , X L , X H

(4)

3.

nV

⇐

nF ⋅ V f

( 5)

4.

nL

⇐

nF ⋅ L fL

(1)

5.

nH

⇐

nF − nV − nL

(6)

6.

TL , TH ⇐

TV

( 7)

7.

PL , PH ⇐

PV

(8)

8.

HF

⇐

hmix TF , PF , Z F

(9)

9.

HV

⇐

hmix TV , PV , YV

(10 )

10.

HL

⇐

hmix TL , PL , X L

(11)

11.

HH

⇐

hmix TH , PH , X H

(2)

12.

Q

⇐

nV ⋅ HV + nL ⋅ H L + nH ⋅ H H − nF ⋅ H F

⇐

PF − ΔP ⇐ vlle TV , PV , Z F

HYSYS Simulation Algorithms If the process state of the feed stream is fully defined (i.e., the temperature, pressure, flow rate and composition are known), only two additional variables are required to calculate all

unknowns, as depicted in the HYSYS simulation algorithms below:

PV , nV , YV , ΨL , ΨH , Q

=

sepa ΨF , ΔP, TV

ΔP, nV , YV , ΨL , ΨH , Q

=

sepb ΨF , PV , TV

TV , PV , nV , YV , ΨL , ΨH

=

sepc ΨF , ΔP, Q

TV , nV , YV , ΨL , ΨH , ΔP

=

sepd ΨF , PV , Q

HYSYS can also back calculate for the feed stream conditions given certain information about the exit streams and the unit operation. Because of the nature of the vapor-liquid-liquid equilibrium, back-calculation requires one more given variable than the forward calculation. So to backcalculate, (nc+6) variables must be specified. These variables are three of the four total flow rates, one exit composition, one exit temperature or pressure, the pressure drop, and the duty of the three-phase separator.

Beef, Inc.

J-3

7/29/98

Appendix K. Component Splitter Module Description A component splitter operation is used to approximate the separation of a material feed stream into a product and bottoms streams; that is, it is the simplest model that one can use for a separation operation. The component splitter solves for the process state of the two exit streams given the process state of the feed stream and the fractional split of each feed component that goes into the product stream. The process state of a material stream is its temperature, pressure, total flow rate, and composition. The conceptual diagram for the component splitter operation is given below for a steady-state system. The system is the mixture of chemical compounds (or components) ing into, through, and from the component splitter. The mathematical model given below for the component splitter operation balances the material and energy flows of the system. This unit operation occurs at steady state with no chemical reaction, and the kinetic and potential energy changes are negligible. The independent set of equations in the math model contains the total and component material balances, the component composition equations, the product mixture equation, the component fractional splits, the energy balance, the molar enthalpies of the three process streams, and the vapor fractions of the two exit streams. To solve these equations, (2nc+7) variables must be specified, as indicated by the degrees-of-freedom analysis in the math model. From this mathematical model, many mathematical algorithms can be derived for doing process simulation calculations. These algorithms differ in their given (or design) variables and their solution procedures. One such math algorithm is shown below—where the process state

of the feed stream, the overhead split fractions, and the temperature and pressure of the two exit streams are given. The unknown variables are calculated using the solution procedure

defined in a math algorithm. Again, the process state of a material stream is its temperature, pressure, total flow rate, and composition. Other possible simulation algorithms ed by the HYSYS software are summarized below.

Process Diagram

Assumptions TP PP

TF PF

Feed

component splitter

Product

XP

nF ZF

Q

nP

Bottoms

1. 2. 3. 4. 5.

continuous process steady state no chemical reaction neglect KE and PE changes no shaft work

TB PB nB XB

Beef, Inc.

K-1

7/29/98

Appendix K. Component Splitter Module Mathematical Model ( 1)

nF − nP − n B = 0

(2)

n F , j − n P , j − n B, j = 0

for j = 1, 2,

, nc

(3)

nF , j = nF ⋅ z F , j

for j = 1, 2,

, nc

(4)

nP , j = nP ⋅ x P , j

for j = 1, 2,

, nc

(5)

n B, j = n B ⋅ x B, j

for j = 1, 2,

, nc

(6)

nP = ∑ nP , j for j = 1, 2,

, nc

nc

j =1

(7)

f j = nP , j / nF , j

(8)

nF ⋅ H F − nP ⋅ H P − n B ⋅ H B + Q = 0

(9)

H F = hmix TF , PF , Z F

( 10 )

H P = hmix TP , PP , X P

( 11 )

H B = hmix TB , PB , X B

( 12 )

Vf , P = vfrac TP , PP , X P

( 13 )

Vf , B = vfrac TB , PB , X B # vars = 7 ⋅ nc + 15 # eqns = 5 ⋅ nc +

8

= 2 ⋅ nc +

7

DOF

Variable Descriptions Ti Pi ni ni , j

is is is is

the temperature of process stream i, K. the pressure of process stream i, kPa. the bulk molar flow rate of process stream i, kgmol/h. the bulk molar flow rate of component j in process stream i, kgmol/h.

nc

zi , j

is is

the number of chemical components or compounds in the mixture. the bulk mole fraction of component j in process stream i; vector Z i means all elements zi ,1 , zi , 2 , …, zi ,nc for stream i.

xi , j

is

the liquid mole fraction of component j in process stream i; vector Xi means all elements xi ,1 , xi , 2 , …, xi ,nc for stream i.

fj

is

the fraction of component j in the feed that goes to the product stream; vector f means all elements f1 , f2 , …, fnc for the nc-components.

Hi Q V f ,i

is is is

the molar enthalpy of process stream i, kJ/kgmol. the energy duty of the component splitter, kJ/h. the molar vapor fraction of the phase equilibrium in process stream i.

Ψi

is

a short notation for Ti , Pi , ni , and composition of process stream i; that is, the process state of stream i.

Beef, Inc.

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Appendix K. Component Splitter Module Mathematical Algorithm A nP , X P , n B , X B , Q

= splittera ΨF , f , TP , PP , TB , PB

( 3)

1.

nF , j

⇐

nF ⋅ z F , j

for j = 1, 2,

, nc

( 7)

2.

nP , j

⇐

f j ⋅ nF , j

for j = 1, 2,

, nc

(2)

3.

n B, j

⇐

n F , j − nP , j

for j = 1, 2,

, nc

(6)

4.

nP

⇐

nc

∑n j =1

P, j

(1)

5.

nB

⇐

n F − nP

(4)

6.

x P, j

⇐

nP, j / nP

for j = 1, 2,

, nc

( 5)

7.

x B, j

⇐

n B, j / n B

for j = 1, 2,

, nc

(9)

8.

HF

⇐

hmix TF , PF , Z F

(10 )

9.

HP

⇐

hmix TP , PP , X P

(11)

10.

HB

⇐

hmix TB , PB , X B

(8)

11.

Q

⇐

n P ⋅ H P + n B ⋅ H B − nF ⋅ H F

(12 )

12.

Vf , P ⇐

vfrac TP , PP , X P

(13 )

13.

Vf , B ⇐

vfrac TB , PB , X B

HYSYS Simulation Algorithms If the process state of the feed stream is fully defined (i.e., the temperature, pressure, flow rate and composition are known) and the component fractional splits are given, only four additional

variables are required to calculate all unknowns, as depicted in the HYSYS simulation algorithms below:

nP , X P , n B , X B , Vf , P , Vf , B , Q

=

splittera ΨF , f , TP , PP , TB , PB

nP , X P , nB , X B , TP , TB , Q

=

splitterb ΨF , f , PP , Vf , P , PB , Vf , B

nP , X P , nB , X B , PP , PB , Q

=

splitterc ΨF , f , TP , Vf , P , TB , Vf , B

nP , X P , nB , X B , TP , Vf , P , Vf , B

=

splitterd ΨF , f , Q, PP , TB , PB

These are but a few of the many simulation algorithms for this unit operation. You can select the four additional variables for among the exit temperatures, exit pressures, exit vapor fractions, and the heat duty. HYSYS can not do back calculations on the component splitter. Thus, you must always specify the process state of the feed stream.

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Appendix L. Simple Distillation Module Description A simple distillation operation is used to separate a feed process stream into two product streams—the distillate and bottoms materials. As illustrated in the conceptual diagram below, a distillation column is composed of several process units—a condenser, a finite number of equilibrium stages (one of which is the feed stage), and a reboiler. A distillation column exploits the relative volatility (i.e., the boiling points) of the chemical compounds in the feed stream. Those compounds that are more volatile (i.e., have a lower boiling point) want to concentrate in the distillate stream, while those compounds that are less volatile (i.e., have a higher boiling point) want to concentrate in the bottoms stream. In the conceptual diagram, the set of process units operates as a steady-state system. The system is the mixture of chemical compounds (or components) ing into, through, and from the distillation column.

Column Diagram QC

Assumptions

Condenser

TD R Stage 1

TF PF nF

F Feed

ZF

Reflux

D Distillate

PD nD

1. 2. 3. 4. 5.

continuous process steady state no chemical reaction neglect KE and PE changes no shaft work

XD

Feed Stage 2 Stage 3

TB B Bottoms

QR Reboiler

PB nB XB

Heat ( QR ) is added to the reboiler to vaporize part of the liquid and produce the vapor flow that travels up the column to the condenser. The other part of the liquid in the reboiler is draw off as the bottoms stream. Heat ( QC ) is extracted in the total condenser to convert the saturated vapor to a saturated liquid. Part of this liquid is then reflux down the column to the reboiler. The other part is drawn off as the distillate stream. A feed stream enters on one of the equilibrium stages. At the temperature and pressure of this stage, a vapor-liquid equilibrium is established, and a vapor stream flows to the stage above while a liquid stream flows to the stage below. At each stage in the column, this same equilibrium process occurs at a different temperature and pressure, causing the components to separate by their difference in boiling points. The more volatile components concentrate in the vapor, and the less volatile components concentrate in the liquid. The temperature and pressure profiles in the column are similar, going from higher values at the reboiler to lower values at the condenser.

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Appendix L. Simple Distillation Module As show in the three-stage diagram above, a distillation column consists of many smaller process units that are tied together. A mathematical model for the whole column is simpler to present by modeling each section of the column—condenser, stages, reboiler—separately and then show how they overlap. The mathematical models for the total condenser, three stages including a feed stage, and the reboiler of a distillation column are given below. The column is at steady state with no chemical reaction or shaft work and negligible kinetic and potential energy changes. In each math model, the independent set of equations contains the material and energy balances, the thermodynamic functions for molar enthalpy, vapor-liquid equilibrium (vle) functions, temperature and pressure relationships and a component mole fraction sum. The degrees-of-freedom (DOF) analyses below show the number of variables that must be specified to solve each model separately. An overall degrees-of-freedom analysis is also shown for all the smaller models combined. The overlapping of the process units causes variables and equations to be duplicated when determining the DOF. Each process unit duplicates the process state variables of two streams (i.e., the temperature, pressure, flow rate, and composition) and the molar enthalpy functions of two streams. To solve the entire distillation column, (nc + 10) variable must be specified. The mathematical algorithm for solving the entire column is given below. Because of the dependence of each process unit on the others, the equations must be solved simultaneously. All unknown variables are iterated in all equations until all the equations equal zero. Other possible simulation algorithms ed by the HYSYS software are summarized below.

Variable Descriptions Ti Pi ni ni , j

is is is is

the temperature of process stream i, K. the pressure of process stream i, kPa. the bulk molar flow rate of process stream i, kgmol/h. the bulk molar flow rate of component j in process stream i, kgmol/h.

nc

zi , j

is is

the number of chemical components or compounds in the mixture. the bulk mole fraction of component j in process stream i; vector Z i means all elements zi ,1 , zi , 2 , …, zi ,nc for stream i.

xi , j

is

the liquid mole fraction of component j in process stream i; vector Xi means all elements xi ,1 , xi , 2 , …, xi ,nc for stream i.

yi , j

is

the vapor mole fraction of component j in process stream i; vector Yi means all elements yi ,1 , yi , 2 , …, yi ,nc for stream i.

R

is is is

the reflux ratio—the reflux flow rate of R over distillate flow rate of D. the molar vapor fraction of the phase equilibrium in process stream i. the equilibrium vaporization ratio or K-value of component j.

Hi Qu ΔPu

is is

the molar enthalpy of process stream i, kJ/kgmol. the energy duty of the condenser or reboiler unit u, kJ/h.

is

the pressure drop of the condenser or reboiler unit u, kPa.

Ψi

is

a short notation for Ti , Pi , ni , and composition of process stream i; that is, the process state of stream i.

V f ,i Kj

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Appendix L. Simple Distillation Module Condenser Diagram QC V1 sat’d vapor

sat’d liquid R

D

Reflux

Distillate

Condenser Mathematical Model (1)

nV 1 − nR − nP = 0

(2)

nV 1 ⋅ yV 1, j − nR ⋅ x R, j − nD ⋅ x D, j = 0

(3)

nV 1 ⋅ HV 1 − nR ⋅ H R − nD ⋅ H D − QC = 0

(4)

HV 1 − hmix TV 1 , PV 1 , YV 1 = 0

(5)

H R − hmix TR , PR , X R = 0

(6)

H D − hmix TD , PD , X D = 0

( 7)

TR − TD = 0

(8)

PR − PD = 0

(9)

R − nR / nD = 0

( 10 )

x R, j − x D, j = 0

( 11 )

TD − vle PD , Vf , D = 0, X D = 0

( 12 )

ΔP − PV 1 + PD = 0

for j = 1, 2,

, nc

for j = 1, 2,

, nc

bubble - point temperature

# vars = 3 ⋅ nc + 15 # eqns = 2 ⋅ nc + 10 DOF = 1 ⋅ nc +

Example knows: TV 1 , PV 1 , nV 1 , YV 1 ,

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PD , R

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Appendix L. Simple Distillation Module Stage 1 Diagram R

V1

Stage 1

V2

L1

Stage 1 Mathematical Model (1)

nV 2 + nR − nV 1 − nL1 = 0

(2)

nV 2 ⋅ yV 2, j + nR ⋅ x R, j − nV 1 ⋅ yV 1, j − nL1 ⋅ x L1, j = 0

(3)

nV 2 ⋅ HV 2 + nR ⋅ H R − nV 1 ⋅ HV 1 − nL1 ⋅ H L1 = 0

(4)

HV 2 − hmix TV 2 , PV 2 , YV 2 = 0

(5)

H R − hmix TR , PR , X R = 0

(6)

HV 1 − hmix TV 1 , PV 1 , YV 1 = 0

(7)

H L1 − hmix TL1 , PL1 , X L1 = 0

(8)

for j = 1, 2,

, nc

yV 1, j − K j ⋅ x L1, j = 0

for j = 1, 2,

, nc

(9)

K j − kvalue T1 , P1 , YV 1 , X L1 = 0

for j = 1, 2,

, nc

( 10 )

TV 1 − TL1 = 0

( 11 )

TV 1 − T1 = 0

( 12 )

PV 1 − PL1 = 0

( 13 )

PV 1 − P1 = 0 nc

( 14 )

∑x j =1

L1, j

− 1.0 = 0 # vars = 5 ⋅ nc + 18 # eqns = 3 ⋅ nc + 11 DOF = 2 ⋅ nc +

Example knows: TV 2 , PV 2 , nV 2 , YV 2 ,

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TR , PR , nR , X R , P1

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Appendix L. Simple Distillation Module Feed Stage 2 Diagram V2

L1

F

Feed Stage 2

Feed

V3

L2

Feed Stage 2 Mathematical Model (1)

nF + nV 3 + nL1 − nV 2 − nL 2 = 0

(2)

nF ⋅ z F , j + nV 3 ⋅ yV 3, j − nL1 ⋅ x L1, j − nV 2 ⋅ yV 2, j − nL 2 ⋅ x L 2, j = 0

( 3)

nF ⋅ H F + nV 3 ⋅ HV 3 + nL1 ⋅ H L1 − nV 2 ⋅ HV 2 − nL 2 ⋅ H L 2 = 0

(4)

H F − hmix TF , PF , Z F = 0

(5)

HV 3 − hmix TV 3 , PV 3 , YV 3 = 0

(6)

H L1 − hmix TL1 , PL1 , X L1 = 0

(7)

HV 2 − hmix TV 2 , PV 2 , YV 2 = 0

(8)

H L 2 − hmix TL 2 , PL 2 , X L 2 = 0

(9)

for j = 1, 2,

, nc

yV 2, j − K j ⋅ x L 2, j = 0

for j = 1, 2,

, nc

( 10 )

K j − kvalue T2 , P2 , YV 2 , X L 2 = 0

for j = 1, 2,

, nc

( 11 )

TV 2 − TL 2 = 0

( 12 )

TV 2 − T2 = 0

( 13 )

PV 2 − PL 2 = 0

( 14 )

PV 2 − P2 = 0 nc

( 15 )

∑x j =1

L 2, j

− 1.0 = 0 # vars = 6 ⋅ nc + 22 # eqns = 3 ⋅ nc + 12 DOF = 3 ⋅ nc + 10

Example knows: TV 3 , PV 3 , nV 3 , YV 3 , TL1 , PL1 , nL1 , X L1 ,

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TF , PF , nF , Z F , P2

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Appendix L. Simple Distillation Module Stage 3 Diagram V3

L2

Stage 3

L3

V4

Stage 3 Mathematical Model (1)

nV 4 + nL 2 − nV 3 − nL 3 = 0

(2)

nV 4 ⋅ yV 4, j + nL 2 ⋅ x L 2, j − nV 3 ⋅ yV 3, j − nL 3 ⋅ x L 3, j = 0

(3)

nV 4 ⋅ HV 4 + nL 2 ⋅ H L 2 − nV 3 ⋅ HV 3 − nL 3 ⋅ H L 3 = 0

(4)

HV 4 − hmix TV 4 , PV 4 , YV 4 = 0

(5)

H L 2 − hmix TL 2 , PL 2 , X L 2 = 0

(6)

HV 3 − hmix TV 3 , PV 3 , YV 3 = 0

(7)

H L 3 − hmix TL 3 , PL 3 , X L 3 = 0

(8)

for j = 1, 2,

, nc

yV 3, j − K j ⋅ x L 3, j = 0

for j = 1, 2,

, nc

(9)

K j − kvalue T3 , P3 , YV 3 , X L 3 = 0

for j = 1, 2,

, nc

( 10 )

TV 3 − TL 3 = 0

( 11 )

TV 3 − T3 = 0

( 12 )

PV 3 − PL 3 = 0

( 13 )

PV 3 − P3 = 0 nc

( 14 )

∑x j =1

L 3, j

− 1.0 = 0 # vars = 5 ⋅ nc + 18 # eqns = 3 ⋅ nc + 11 DOF = 2 ⋅ nc +

Example knows: TV 4 , PV 4 , nV 4 , YV 4 , TL 2 , PL 2 , nL 2 , X L 2 ,

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Appendix L. Simple Distillation Module Reboiler Diagram sat’d vapor V4 L3

sat’d liquid

B Bottoms

sat’d liquid

QR

Reboiler Mathematical Model ( 1)

nL 3 − nV 4 − nB = 0

(2)

nL 3 ⋅ x L 3, j − nV 4 ⋅ yV 4, j − nB ⋅ y B, j = 0

( 3)

nL 3 ⋅ H L 3 − nV 4 ⋅ HV 4 − nB ⋅ H B + QR = 0

(4)

H L 3 − hmix TL 3 , PL 3 , X L 3 = 0

(5)

HV 4 − hmix TV 4 , PV 4 , YV 4 = 0

(6)

H B − hmix TB , PB , X B = 0

(7)

for j = 1, 2,

, nc

yV 4, j − K j ⋅ x B, j = 0

for j = 1, 2,

, nc

(8)

K j − kvalue TB , PB , YV 4 , X B = 0

for j = 1, 2,

, nc

(9)

TV 4 − TB = 0

( 10 )

PV 4 − PB = 0 nc

( 11 )

∑x j =1

( 12 )

L 3, j

− 1.0 = 0

ΔPR − PL 3 + PB = 0 # vars = 4 ⋅ nc + 14

Example knows: TL 3 , PL 3 , nL 3 , X L 3 ,

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# eqns = 3 ⋅ nc +

9

DOF = 1 ⋅ nc +

5

PB , nB

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Appendix L. Simple Distillation Module Column Degrees-of-Freedom Analysis Total Number of Variables Condenser

3⋅nc +

Stage 1

5⋅nc + 6⋅nc +

22

Stage 3

5⋅nc +

18

Duplicates Total Variables

4⋅nc 23⋅nc 8⋅nc 15⋅nc

+ + + +

}

2⋅nc +

8

}

2⋅nc +

8

}

2⋅nc +

8

}

2⋅nc +

8

8⋅nc +

32

18

Feed Stage 2

Reboiler

Duplicates

15

14 87 32 55

Each process unit duplicates the temperature, pressure, flow rate, composition, and enthalpy for two streams (i.e., 2⋅nc + 8).

Total Number of Equations Condenser

2⋅nc +

10

Stage 1

3⋅nc +

11

Feed Stage 2

3⋅nc +

12

Stage 3

3⋅nc +

11

Reboiler Duplicates Total # equations

3⋅nc + 14⋅nc + 14⋅nc +

Duplicates }

2

}

2

}

2

}

2

9 53 8 45

8

Each process unit duplicates the enthalpy functions for two streams.

Overall Degrees of Freedom Number of Variables = Number of Equations = DOF = Example knows: TF , PF , nF , Z F ,

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15⋅nc

+

55

14⋅nc 1⋅nc

+ +

45 10

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Appendix L. Simple Distillation Module

Mathematical Algorithm ΨD , ΨR , ΨB , QC , QR = column ΨF , PD , P1 , P2 , P3 , PB , R, nD Iterate on all unknowns in

f all unknowns

⇐ all equations

until f all unknowns = 0

HYSYS Simulation Algorithms The above mathematical algorithm is for a simple distillation column with three stages and the feed entering on the second stage. HYSYS s the simulation calculations for a rigorous distillation column that may contain up to 200 stages. If the process state of the feed stream to a rigorous column is fully defined (i.e., the temperature, pressure, flow rate and composition are known) and some column characteristics are given, only two additional variables are required to calculate all unknowns, as depicted in the HYSYS simulation algorithms below:

ΨD , ΨR , ΨB , QC , QR

=

columna ΨF , PD , PB , N S , N FS , R, nD

ΨD , ΨR , ΨB , QC , QR

=

columnb ΨF , PD , PB , N S , N FS , R, nB

where N S is the number of column stages, N FS is the feed stage number, and R is the reflux ratio.

The above two column algorithms are but a few of the many HYSYS simulation algorithms for this unit operation. For the rigorous column, HYSYS calculates all of the equilibrium stage pressures from the distillate and bottoms pressures ( PD and PB ). The number of

stages, the feed stage number, and the reflux ratio can be estimated using the shortcut column operation in HYSYS. The last two specified variables in the rigorous column algorithm can be any combination of the reflux ratio, distillate flow rate, reflux flow rate, bottoms flow rate, condenser duty, reboiler duty, and exit component compositions.

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