Dynamic Details 1

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Dynamic Details

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WorkshopThis module examines some of the advanced features in Aspen HYSYS Dynamicsthat can be enabled in order to more accurately model existing equipment. Thesedetails include valve and actuator characteristics, nozzle locations and heat lossparameters.

Starting with the dynamic simulation prepared in the previous workshop, additionalinformation will be added to select operations so as to provide a more representativemodel of the process.

Learning ObjectivesAfter you have completed this module, you will be able to:

Understand the effect of inherent and installed valve characteristics onprocess controlIncorporate vessel heat loss models into a dynamic simulationAnalyze the impact of nozzle location on vessel behavior

PrerequisitesBefore beginning this section you need to know how to:

Set up Strip ChartsAnalyze a Pressure-Flow network

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Characteristics of a Valve and Its ActuatorThe following valve behaviors have been observed in the field:

All feed valves take one minute to close from fully open.There is a 5 second delay in the response of the valve connected to Charlie.The valve connected to Bravo does not close completely; there remains a2% leak.The liquid valve on the HPSep is an Equal Percentage Valve.

Using the features of Aspen HYSYS Dynamics, each of the above observations canbe realistically represented. Before modeling the observed behavior; however, wewill make a simple model consisting of two identical valves with identicalcontrollers connected.

1. Open up the file EM2-Dyn1.hsc and save it as EM3-Valve1.hsc.

2. Add two new valves to the simulation. Specify streams 1 and 3 as inlet streamsand 2 and 4 as outlet streams.

3. Enter a dynamic pressure specification of 64.5 psia (445 kPa) for each feedstream and 14.7 psia (101.3 kPa) for each product stream.

4. Size the valves to handle 10,000 lb/hr (4535 kg/h) with a 50 psi (345 kPa)pressure drop when the valve opening is 50%.

5. Connect a flow controller to each valve to control the mass flow of the feedstreams. Provide identical controller tunings and ranges according with thefollowing table. Place the controllers in Auto mode.

Controller Settings

Connections

Controller Name FIC-104 FIC-105

Process Variable Source 1, mass flow 3, mass flow

Output Target Object VLV-104 VLV-105

Parameters

Action Reverse

Range PV Minimum 0 lb/hr (0 kg/h)

Range PV Maximum 19,800 lb/hr (9000 kg/h)

Mode Automatic

SP 10,000 lb/hr (4535 kg/h)

Kc 0.25

Ti 0.10

Td -

Stream dynamic specification canbe introduced on the Worksheet PFSpecs window of each unitoperation.

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Actuator Linear RateThe Dynamics | Actuator page on the Valve property view provides the user withthe ability to rigorously model valve dynamics.

In reality, changes that occur in the actuator are not observed instantaneously in thevalve. Moreover, changes in the output signal of a controller (OP) do notinstantaneously translate to changes in the actuator. Because the actuator and valveare both physical objects, they take a finite amount of time to move to theirrespective desired positions. The resulting interaction between these components canhave a dramatic effect on the observed dynamic behavior of the control valve.

The actuator mode defines the relationship between the desired actuator position andcurrent actuator position. In Aspen HYSYS Dynamics, the desired actuator positioncan be set by a PID Controller or Spreadsheet operation (i.e. the controller outputfixes the desired actuator position). Depending on the actuator mode, the currentactuator position can behave in one of the following three ways:

InstantaneousFirst OrderLinear

Figure 1

An actuator is a devicewhich applies a force inorder to cause movementin the valve.The valve opening has adirect impact on the flowthrough the valve. Thisrelationship is a function ofthe valve type and thepressure of the surroundingpieces of equipment.

In instantaneous mode, theactuator movesinstantaneously to thedesired actuator positiondefined by the controller.If First Order has beenselected as the mode, theactuator responds with afirst order lag according tothe provided Actuator TimeConstant.If Linear mode is activated,the actuator moves at aconstant rate according tothe Actuator Linear Rateprovided (%/second).

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1. Open the property view for VLV-105 and select the Dynamics | Actuator tab.Change the mode to Linear and keep the default rate of 0.01 %/sec for theActuator Linear Rate.

2. Set up a strip chart to monitor the Mass Flow of each of the two feed streams aswell as the Actuator Desired Position and Percentage Open for thecorresponding valves. Configure the Data Logger to record 3600 points with a 5second sample interval. Run the integrator until all the variables stabilize.

3. Stop the integrator and change the flow set point for FC-104 and FC-105 to5,000 lb/hr (2250 kg/h).

4. Start the integrator and observe the response.

Figure 2

Figure 3

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According to the observed characteristics noted at the beginning of the module, the threevalves connected to the feed streams can move from fully open to closed in 1 minute. Whatshould be the desired Actuator Linear Rate for these control valves?______________________________________________

5. Enter an Actuator Linear Rate of 1.667%/second for VLV-105 (correspondingto the desired 100% in 1 minute) and restore the original set point for FC-104and FC-105..

6. Start the integrator and observe the response.

7. Enter the desired Actuator Linear Rate values for the Alpha, Bravo and Charlievalves (1.667 %/sec) and switch from Instantaneous to Linear mode.

Stickiness Time ConstantIn reality, a valve does not respond instantaneously to changes in the actuator. A firstorder lag can be modeled in the response of the actual valve position to changes inthe actuator position. The Valve Stickiness Time Constant allows the user to specifythe time constant used to model the delayed valve response caused by a stickyactuator. The Valve Stickiness Time Constant can be specified on the Dynamics |Actuator page of the Valve property view.

1. Set up VLV-104 and VLV-105 so that both actuators are operating in Linearmode with an Actuator Linear Rate of 1.667 %/second.

2. Enter a Valve Stickiness Time Constant of 20 seconds for VLV-105.

Figure 4

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3. Change the set points for FC-104 and FC 105 to 5,000 lb/hr (2250 kg/h) and15,000 lb/hr (6800 kg/h) respectively.

4. Set up a strip chart to include the Actuator Current Position and compare this tothe Percentage Open for each valve.

5. Start the integrator and observe the response.

Figure 5

6. According to the valve characteristics noted at the beginning of the module, thevalve on Charlie is not responding instantly, showing a 5-second delay. Makethe appropriate change to VLV-102 to reproduce this behavior.

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Leaky ValvesLeaky valves can be modeled by specifying a non-zero value for the minimum valveposition.

1. Delete the Valve Stickiness Time Constant in VLV-105 so that the actuatorconfiguration in VLV-104 and VLV-105 are identical.

2. Return the controller set points for FIC-104 and FIC-105 back to 10,000 lb/hr(4535 kg/h), and run the integrator until the flow through both valves stabilizes.

3. Enter 2% for the Min Valve Position for VLV-105.

4. Change the controller set point for FIC-104 and FIC-105 to a small numbersuch as 300 lb/hr (135 kg/h) and observe the response.

Figure 6

5. Make the appropriate change to VLV-101 to introduce a 2% leak.

6. Start the integrator and run the simulation until all the variables stabilize, thensave the case as EM3-Valve2.hsc

Save your case!

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Inherent Flow CharacteristicsThe inherent flow characteristic of a valve is the relation between the valve positionand volumetric flow rate when the pressure drop across the valve is constant. Theinherent flow characteristic is expressed as a fraction since it compares the currentvolumetric flow rate to the maximum flow rate through the valve, as shown in theequations below:

L

PxfQ )(Cv

maxmax

)(A

AQ

Qxf

Q = volumetric flow rate (USGPM)Cv = flow of water (60 F) through a fully open valve when the pressure drop is 1 psif(x) = inherent flow characteristicx = valve position (% open)

A = valve areaL = liquid specific density

What would be the Cv of a valve that allows a maximum flow of 5 USGPM of water with apressure loss of 2 psi? ________________________

What would be the Cv if the valve outlet was stream HpLiq but all other parameters werethe same? __________________________________________________

The most frequently encountered valves in terms of operating characteristics areLinear, Equal Percentage and Quick Opening. All three types are implemented inAspen HYSYS, however, the first two are much more common in control systems.

1. Redefine stream 1 as a pure water stream at 60 F (15.6 C) and 15.7 psia (108.2kPa).

2. Provide a Cv of 100.0 for VLV-104 and turn off the FIC-104 controller.

3. Add a Transfer Function and select the Percentage Open of VLV-104 as theOP Target.

Transfer Function icon

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Figure 7

4. On the Parameters | Configuration tab, define 0 and 100 as the minimum andmaximum PV range and 0% and 100% as the minimum and maximum OPrange.

5. Enter a value of 0 as the Transfer Function PV.

6. Open the Parameters | Ramp tab and enable Ramp as the Active TransferFunction.

7. Enter 100% as the Ramp Magnitude and 10 minutes as the Ramp Duration.

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Figure 8

8. Add a strip chart entitled Valve Characteristics to track the Percentage Open forVLV-104 as well as the Std Ideal Liq Vol Flow through VLV-104.

9. Run the integrator and press the Start Ramp button.

10. Once the flow has stabilized, press the Reset Ramp button and repeat steps 7and 9 for Equal Percentage and Quick Opening valves. Ensure that the Ramp isreset for each valve type.

After running each of the three scenarios, you will get a plot similar to Figure 9corresponding to each type of valve.

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Figure 9

Installed Flow CharacteristicsThe installed flow characteristic of a valve depends on the pressure drop across theentire flow circuit where the valve is installed.

max

)(Q

QxF

F(x) = installed flow characteristicx = valve position (% open)Q = volumetric flow rate through flow circuitQmax = volumetric flow rate through circuit when the valve is fully open

It is convenient for calculation purposes to define a parameter r that is the ratio ofthe minimum pressure drop across the valve (i.e. when the valve is 100%open)divided by the total pressure drop across entire the flow circuit:

system

valve

PP

r

1. Change the Valve Operating Characteristic of VLV-104 back to Linear, andthen create a copy of the valve.

2. Connect stream 2 as the feed to this new valve and enter a Valve Opening of25%.

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3. Add a product stream to the new valve and move the atmospheric pressurespecification from stream 2 to the new boundary stream.

4. Set the Pressure Specification of stream 1 to 43.5 psia (300 kPa).

5. Click the Reset Ramp button and repeat steps 7 and 9 from the previous sectionusing several different valve positions for the new valve (i.e. 45%, 65%, 85%).

6. Using historical data from the Valve Characteristics strip chart createdpreviously, calculate Q/Qmax and the r parameter for each valve positioninvestigated.

Since the total pressure drop across the system is fixed, opening the new valve willultimately increase the value of r, since the original valve will constitute anincreasingly large proportion of the total pressure drop.

After exporting the strip chart data for each value of r to Excel and normalizing theflows to calculate Q/Qmax, the Linear valve should result in a plot similar to Figure10 shown below.

Figure 10

Repeating this exercise for a Quick Opening and Equal Percentage valve will yieldplots similar to Figures 11 and 12 shown on the following page.

Characteristic Curves Linear Valve

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Figure 11

Diagram

Figure 12

Diagram

Based on the figures above, it is clear that appropriate valve selection will producean installed characteristic that complements the chosen control algorithm. As aresult, the overall gain of the loop will be relatively constant over the controlleroperating range.

What Type of Valve do you think is the most common for control purposes?_______________________________________

7. Define VLV-103 for the HP Liq stream as an equal percentage valve and saveyour case.

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Equipment Location

Static HeadBy default, static head is not included in dynamic calculations. For example, thepressures of the product streams leaving a separator will all be equal. For any unitoperations with hold-up, Aspen HYSYS calculates the static head by considering theequipment hold-up, geometry, and elevation of any attached nozzles. In order toinclude the effect of static head for any unit, the calculation option must be enabledvia the Options page of the Integrator property view.

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1. Open the Integrator property view via the Simulation menu and activate theEnable static head contributions checkbox on the Options tab.

2. Run the integrator and observe any changes resulting from the inclusion of thestatic head calculations; particularly the pressures surrounding the separator.

Whats the Pressure of Stream HP Liq? _________________________

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NozzlesBy default, all unit operations are placed on the ground. The Rating | Nozzles pageof all vessels contains information regarding the base elevation of the unit operationas well as the elevation and diameter of the associated nozzles.

Figure 14

The elevation of each nozzle attached to the equipment is displayed relative toseveral reference points:

The Ground is a common reference point from which all equipmentelevations can be measured.The Base is defined as the bottom of the piece of equipment.

1. Create a Strip Chart that includes the following variables:

Aqueous and Liquid mass flow rates for stream HPLiqLiquid Percent Level for HPsepHvyLiquid Percent Level for HPsep

2. Run the integrator briefly and observe the results.

3. Change the nozzle location for stream Hp Liq from 0% to 25% and run theintegrator again.

4. Try several different set points for the level controller LC-HP Sep and note anychanges to the Strip Chart after the integrator is run.

Can you explain the Results? ___________________________________________________

Was the controller able to achieve the desired set point all the time?_________________________

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In conjunction with the levels of the individual phases within a vessel, the nozzlediameter and elevation determine the exact nature of the product streams.

Note that the specified nozzle elevation refers to the center of the nozzle opening, notto the bottom or top of the nozzle. If, for example, the product nozzle is locatedabove the liquid level, the exit stream will draw material from the vapor holdup.Additionally, if the liquid level is even with the center of the nozzle, the molefraction of liquid in the product stream will vary linearly with fraction of nozzle areataken up by the liquid.

Move the HPSep liquid product nozzle back to the original location (0%) and returnthe set point for LC-HP Sep back to 50%.

Heat Transfer ModelsIn Aspen HYSYS Dynamics, there are several options that allow user to account forany heat lost by a vessel to the environment. For all unit operations with hold-up, theheat transfer options are located on the Rating | Heat Loss tab. By selecting None asthe heat loss option, it is possible to ignore the heat loss calculation in the energybalance; otherwise the Simple or Detailed model may be used.

Simple ModelThe Simple model uses the following equation to determine heat loss:

)( ambfloss TTUAq

The user can either specify the heat loss value directly or have the heat losscalculated based on the following variables:

The overall heat transfer coefficient (U) and ambient temperature (Tamb) arespecified by the user.Heat transfer area (A), and fluid temperature (Tf) are calculated by AspenHYSYS.

Save your case!

There are severalunderlying assumptionsthat are considered duringa heat loss calculation:

The wall and insulationsurrounding the fluidhave a fixed thermalconductivity and heatcapacityThe temperature acrossthe wall and insulation isassumed to be constant(lumped parameteranalysis).The calculation usesconvective heat transferon the inside andoutside of the vessel.

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Detailed ModelIn the Detailed model, the user supplies both conductive and convective information.

The program calculates the heat loss and supplies a temperature profile from thefluid to the ambient.

1. Enter the following conductive and convective data for the HPSep vessel.

Heat Loss Parameters

Conductive Properties

Material Metal Insulation

Thickness 0.167 ft (0.051 m) 0.098 ft (0.030 m)

Cp 0.113 Btu/lb-F (0.473 kJ/kg-C) 0.82 kJ/kg-C (0.196 Btu/lb-F)

Density 487 lb/ft3 (7801 kg/m3) 32.46 lb/ft3 (520 kg/m3)

Conductivity 26 Btu/hr-ft-F (45 W/m-K) 0.087 Btu/hr-ft-F (0.15 W/m-K)

Convective Properties

Inside Vap Phase U 1.761 Btu/hr-ft2-F (36 kJ/h-m2-C)

Inside Liq Phase U 8.806 Btu/hr-ft2-F (180 kJ/h-m2-C)

Outside U 1.761 Btu/hr-ft2-F (36 kJ/h-m2-C)

Vapor to liquid U 0.8806 Btu/hr-ft2-F (18 kJ/h-m2-C)

The ambient temperature can either be set locally using the Rating | Heat Loss tabin the separator property view, or globally via the integrator settings.

2. From the main menu, select Simulation | Integrator and open the Heat Losstab. Keep the default ambient temperature setting of 77 oF (25 oC) and close theintegrator property view.

3. Activate the Temperature Profile radio button on the Rating | Heat Loss tabof the HPSep property view and then click the Initialize Temperature button.

4. Start the integrator.

Observe the temperature profile of the vessel. Did it immediately reach a steady state value?_________________________________________

Do you see any changes? _________________________

Save your case as EM3-Heatloss.hsc.

Save your case!

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