Feb 14 200 Engineering Data Transmittal .3 627034

Page 1 Of L FEB 14 200 ENGINEERING DATA TRANSMITTAL .3 627034

2. To: (R~lvlngOrgmk8tbn) 3. From: (Originin.(ing Organlptbn) 4. RdnW ED1 No.: SNF Project FDH - Engineering Labs N/A

~ 7. Purehv 0rd.r No.: 5. pmjlRoglD.p(lMv.: 8. Ddgn AumOMybd@nAgmUCog. Engr.: SNF N/A W-441 F. J. Heard 0. Equlp./ComponMNo.: 8. CirlQMOrRuMrkB: N/A Transmittal of SNF-5331, "MOW (Version 2) - Comnercial Software ~,~~~fi,~~: Validation and Configuration Control" CW 142K 12. MaJorAwn. Dwg. No.: 11. R.alver Ramah: N/A 13. PmWFwnlAppliatlMl No.:

Kl-7400-172-1 - -Jan’-06-00 12:lOP Applied Flow Technology 719 686 1001 P.01

Applied Flow Technology Ship: 400 W. Highway 24. Suite 201, Woodland Park, CO 80863 USA Mail: P.O. Box 8358. Woodland Park, CO 80866-6358 USA (719) 686-1000 I FAX (719) 686-1001 INTERNET: www.aft.com. [email protected] Facsimile Cover Sheet

To: I Fred Heard I From: 1 Tom Glassen I Company: Fluor Hsnford , Fax: SO93723655 Fax: (719) 686-1001 Phone: 109-376-6562 Phone: (719) 686-1000 , Pagos: 3 Data: Jan. 6,2000

Best regards,

Tom Glassen Director of Operations & Business Development 719 686 1001 P.02 *Jar1’-06-00 12:lOP Applied Flow Technology UL,UIIUU 1L:Sl -372 3365 FDE Boo2

1 APPUCATlON FOR PERMISSION TO USE I YOUR COPYRIGHTED MATERIAL TO: Tom GLaww, ADP lied Flow technoloav D.C.: Januarv 1. 2000 P.0. BOX 6358 WuuJlaun Park, CO 00866-6550

Permirsbn is requestad to repmduoc the hllDwin~copyrightad materid from: Problem description and result suIIwary fox verification Problena 1 throuah 21. Subject - doEuuwLts ucsuDDlied in elcctroruc form with fihnamrr: vczifvl.doc throuah ~vurSfv2l.doe. Subiact documents will be DriktEd and attached as a tuDvortino aDDendir. S@l,rWnrhorn texl (spec*fy by date d WE.page. paragraph, or illustratbn: Wdesired. attach a copy of the mrtorirl in quwion):

Tlcb ofwark or pmj& in whlch this material wUI k included: W (Vozolon 2) - Cmrcial Software Validaeion and Contiauration Control

~ Widpublication date: Januarv 2000 Author: I. J. Heard Publieher of applicable): rnc. If the copyrighted meterlol is not lo be used in a pubkhed work. plesse pmvlde a brief desuiptin of how h is to I k usad. I SNF-5331, Rev. 0

ARROW (Version 2) - Commercial Software Validation and Configuration Control

r. J. lhrrrd Fluor DadHanford, & p Richland, W 99352 US. Department of Energy Contract DE-ACOB-WRL13200

EDTECN: 627034 uc: 510 orgcode: 82600 Chalpe Code: 105559 /9Ay 0 /-IM F5Qo'f I B(LR Code: FM04J1110 Total Pages: I 37 P kyWOrds: software validation, configuration control, computer program flow network, piping, compressible, ARROW, Applied Flow Technology

Abstrsd: ARROW (Version 2), a compressible flow piping network modeling and analysis computer program from Applied Flow Technology, was installed for use at the U.S. Department of Energy Hanford Site near Richland, Washington. A series of sample problems were performed to validate the software and confirm proper installation. The sample probluns are compared to published results.

Approved For Public Release SNF-5331, Rev. 0

ARROW (Version 2) - Commercial Software Validation and Configuration Control

Prepared by:

F. J. Heard Fluor Daniel Hanford, Inc.

December 1999 SNF-5331,Rev. 0

ABSTRACT

ARROW (Version 2) a compressiblejlowpiping network modeling and analysis programfrom Applied Flow Technology was installedfor use at the Department of Emrgy (DOE) Hanford site near Richlanri, Washingtoa A 5erie5 of tests were performed to validate the sojbare and confirmproper installatioa The sample problems ate compared with published results.

i SNF-5331,Rev. 0

TABLE OF CONTENTS

1.0 INTRODUCTION AND PURPOSE...... 1

2.0 DESCRIPTION AND APPLICATION ...... 2

......

3.0 CONFIGURATION CONTROL...... 4 3.1 HARDWARE ...... 3.2 LICENSEKEY...... 3.3 PROBLEM 3.4 USERMANUAL ...... 4.0 VALIDATION ...... 6 4.1 INTERNALDATABASES ...... 4.1.1 Fluids ...... 6

5.0 RESULTS...... 9 5.1 ACCEPTANCECRITERIA...... 9 5.2 TESTRESULTS ...... 5.2.1 Sample Problem # 5.2.2 Sample Problem #2...... 5.2.3 Sample Problems ...... 5.2.4 Sample Problem 5.2.5 Sample Problem # ...... 5.2.6 Sample Problem 5.2.7 Sample Prablem ...... 5.2.8 Sample Problem 5.2.9 Sample Problem #9 5.2.10 Sample Proble 5.2.11 Sample Proble 5.2.12 Sample Problem #I2 5.2.13 Sample Problem #13...... 5.2.14 Sample Problem # 14...... 5.2.15 Sample Problem #I5 ...... 5.2.16Sample Proble ...... 5.2.17 Sample Problem #I 7 5.2.18 Sample Proble ...... 5.2.19Sample Problem#19 5.2.2OSample Problem #20...... 5.2.21 Sample Problem #21 ...... 19 6.0 SUMMARY AND CONCLUSIONS...... 20

7.0 REFERENCES...... 21 APPENDIX A - SAMPLE PROBLEM DESCRIPTIONS, INPUT, AND OUTPUT ...... 22

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TABLES

Table 3-1 Hardware and System Requirements...... 4

Table 4-1 Listing of Sample Problems Supplied with ARROW., , .. . . . , . . . , , . , , . , , , , , , , , , , , ., , , , , , , .7 Table 5-1 Comparison of Results for Sample Problem #1 ...... _...... _._...... 10 Table 5-2 Comparison of Results for Sample Problem #2 10 Table 5-3 Comparison of Results for Sample Problem #3 ...... 1 1

Table 5-4 Comparison of Results for Sample Problem #4 . . . . . , ...... 11 Table 5-5 Comparison of Results for Sample Problem #5 ...... _12 Table 5-6 Comparison of Results for Sample Problem #6 ...... 12 Table 5-7 Comparison of Results for Sample Problem #7 ...... 13 Table 5-8 Comparison of Results for Sample Problem #8 ...... 13 Table 5-9 Comparison of Results for Sample Problem #9 14 Table 5-10 Comparison ofResults for Sample Problem #10 14 Table 5-1 1 Comparison of Results for Sample Problem #11 15

Table 5-12 Comparison of Results for Sample Problem #I2 , ,...... IS Table 5-13 Comparison of Results for Sample Problem #13 ...... _._...... 16 Table 5-14 Comparison of Results for Sample Problem #14 16 Table 5-15 Comparison of Results for Sample Problem #15 17 Table 5-16 Comparison ofResults for Sample Problem #16 Table 5-17 Comparison of Results for Sample Problem #17 Table 5-18 Comparison ofResults for Sample Problem #18 Table 5-19 Comparison ofResults for Sample Problem #19

Table 5-20 Comparison of Results for Sample Problem #20 , , ...... , ...... , ...... 19

Table 5-21 Comparison of Results for Sample Problem #21 . ._., ...... _...... _...... _. ._.... 19

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1.0 INTRODUCTION AND PURPOSE

The purpose of this report is to document the validation of AFT ARROW (Version 2) computer code consistent with the requirements of “F-PRO-309 (Revision 0) “Computer Software Quality Assurance Requirements” for commercial-off-the-shelf (COTS) software.

AFT ARROW, hereafter referred to as ARROW, was supplied by Applied Flow Technology Corporation in Woodlawn Park, Colorado for use at the Department of Energy’s Hanford site near Richland, Washington. “Applied Flow Technology”, “AFT”, and “AFT ARROW are registered trademarks of the Applied Flow Technology Corporation.

Software validation is the process of evaluating a system or component during or at the end of the development process to determine whether it satisfies specified requirements. Validation involves running test cases to ensure that the computed output meets specified expectations and requirements, including numerical correctness of the results based on comparisons with alternate calculations (e.g., hand calculations, analytical solutions, or acceptance testing).

1 SNF-533 1, Rev. 0

2.0 DESCRIPTION AND APPLICATION

ARROW is piping network analysis tool for use with compressible.flows. ARROW is intended as an aide for network analyses and not as a replacement for other design and analysis methods, including sound judgement, hand calculations, or independent checks. ARROW is supplied by Applied Flow Technology Corporation in Woodlawn Park, Colorado. “Applied Flow Technology”, “AFT”, and “AFT ARROW are registered trademarks of the Applied Flow Technology Corporation.

2.1 Overview

ARROW uses a graphical interface that is based on drag-and-drop operations inherent to most Windows based applications. (Windows is a registered trademark of the Microsoft Corporation.) This allows the user to build a piping network using selections from a palette of components. Date is entered or edited for each of the components by double-clicking the component of interest. Additional global editing features are available from the menu or tool bars that simplify making large-scale changes tot he model.

ARROW has five integrated subordinate windows. The five windows can be viewed as an integrated environment for entering, processing, analyzing, and documenting the results of a piping network simulation. The user will work from one of the five windows at all times. Of the five windows, two are input windows and three are output windows

2.1.1 Input Windows

The two input windows are the Workspace window and the Model Data windows. These two windows, one graphical and the other text-based, process the input data for the model.

2.1.2Output Windows

The three output windows are the Output, the Graph Results, and the Visual Report. The Output window is text-based. The Graph Results and Visual Report windows are graphical. These three windows are essential for reviewing the results of the analysis, spotting analysis errors, and preparing the results for documentation.

2.2 Modeling Capabilities

ARROW is a compressible steady-state one-dimensional Newtonian fluid network flow analysis program and can be used to model a variety of engineering systems including:

Open and closed (recirculating) systems Network systems that branch or loop, with no limit on the number of loops Pressure driven systems Compressor or fan driven systems, including multiple compressors or fans in parallel andor in series

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Compressors or fans with variable speed, controlled discharge pressure and controlled flow Systems with pressure andor flow control valves Systems with valves closed and compressordfans turned off Systems with non-ideal gases Systems that experience sonic choking, including multiple sonic choke points Systems with non-reacting flow stream mixing and user defined mixtures (with optional Chempak) Systems with elevations changes or rotation such in turbomachinery

ARROW also provides hundreds of standard loss models based on well known references such as; Crane (1988), Idelchik (1994), and Miller (1990), but also allows the user to enter there own loss data. Variable or dynamic loss coefficients are supported. Variable loss coefficients may depend on flow, area ratios, and/or flow splits (such as tee’s and wye’s), and the angle of the connecting piping. ARROW also provides automatic calculation of friction factors using the well-known Colebrook-White correlation. Also provided is an extensive database of thermophysical properties for numerous fluids and gases

ARROW uses a Newton-Raphson method to solve the fhdamental equations of pipe flow that govern conservation of mass, momentum, and energy. Solutions are obtained by iterative matrix methods that have been optimized for speed and convergence.

2.3 Engineering Assumptions

ARROW is based on the following fimdamental fluid mechanics assumptions

Compressible flow All gases are superheated Steady-state conditions One-dimensional flow No chemical reactions Supersonic flow does not exist in the system

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3.0 CONFIGURATION CONTROL

ARROW was purchased from Applied Flow Technology located in Woodlawn, Colorado. ARROW is currently licensed for a single user on a single personal computer within a Windows 9x or NT environment. ARROW is considered commercial-off-the-shelf (COTS) software.

3.1 Hardware

ARROW was installed on an AST Bravo with a 200-MHz Pentium processor and 128 MB of main memory. The property tag is WC65744. The operating system is Windows 95 Version 4.00.950B. ARROW was installed by running the automated SETUP program located on the installation disk.

The ARROW installation program is very simple and standard for most Windows based applications. SETUP asks you where to unzip the program and where (i,e., pathname) to install the program. The executable files supplied with ARROW are complied code. No source code is supplied. SETUP handles all the file manipulations, short cuts, desktop icons, and registry entries required by Windows. ARROW was installed using the following default pathname: c:\ARRow.

To install and run AFT ARROW, the computer must meet the following minimum requirements:

Table 3-1. Hardware and System Requirements

Feature Required Recommended IBM - Compatible Processor 80486 Pentium or higher Hard Disk Space I5 Mb Floppy Drive 3% -Inch Monitor Resolution VGA SVGA or higher Random Access Memory (RAM) 32Mb 48Mb Mouse Yes Math Co-processor No Yes MS-Windows Windows 95 or 98 Windows NT

3.2 License Key

A license key is required to activate the AFT ARROW program for use. The license key is supplied by the vendor on a separate floppy disk. The user must supply the license key when requested during setup. The license key is copied from the floppy and stored within a hidden and encrypted file somewhere on the system. ARROW will check for the existence and content of the license file each time the program is activated.

The license key contains information pertaining to the registered purchaser of the program, authorized number of users, and time remaining for the annual software upgrade maintenance (SUM) and help agreement if purchased with ARROW.

4 SNF-5331,Rev. 0

3.3 Problem Reporting

AFT ARROW provides an extensive context searchable online help manual and access to additional help, frequently asked questions (FAQ’s), and modeling tips via the internet (www. aft. com).

Modeling problems, bug reports, and general pleas for assistance can be supplied to AFT via support ([email protected]). A bug report is first verified and then acknowledged usually within two to three business days. The response may or may not solve the problem, but may provide a workaround until the problem is fixed. If requested, a special version of ARROW containing the bug fix (when available) can be supplied to the customer. If the error(s) persist, a special version of ARROW containing an extensive set of debugging tools can be supplied for use by the customers in an attempt to resolve the problem.

Users are notified via email when an update to ARROW is available. Minor updates are received several times a year. Major updates are received approximately once a year. Most of the minor revisions address bugs that have been found and corrected. The major once-a-year revisions usually deal with major enhancements and new capabilities that have been added. Whether or not to update depends on the time remaining in the annual SUM agreement, current workscope, and the nature of the bug fixes or enhancements.

3.4 User Manual

The user manual for AFT ARROW is a soft cover booklet running about 250 pages. The user manual is also available via online help. A single copy of the user manual was included with the software and is currently located at 2440 Stevens Center Room 1704. Additional copies can be purchased from Applied Flow Technology.

5 SNF-5331,Rev. 0

4.0 VALIDATION

The ARROW user’s manual contains a detailed discussion of the fundamental concepts and conventions used in processing piping systems and how to interpret the database information imbedded within ARROW (see Section 2. I). ARROW assumes that the user has a good general understanding of engineering hydraulics. ARROW is designed for use by persons who possess a level of knowledge consistent with that obtained from an undergraduate engineering course in the analysis of piping systems and fluid networks.

4.1 Internal Databases

Most engineering analyses involve processing very large amounts of information, some of which is critical, but most of which is of lessor importance. The key to analyze accurately and efficiently is the proper identification of the critical information. Two databases embedded within ARROW manage information associated with the fluid properties and pressure loss factors.

4.1.1 Fluids

Density and dynamic viscosity are two fluid properties required for all analyses. Fluid vapor pressure is optional. ARROW provides a fluid database for numerous common fluids. If the working fluid is not in the database, the user can enter fluid properties into a custom database for later recall.

Spot checks of the properties for air, helium, hydrogen, oxygen, nitrogen, and water against the values summarized in HOLMAN (1990) showed very good agreement.

4.1.2 Losses

ARROW provides hundreds of standard loss models based on well known references such as; Crane (1988), Idelchik (1994), and Miller (1990), but also allows the user to enter there own loss data. Variable or dynamic loss coefficients are supported. Variable loss coefficients may depend on flow, area ratios, and/or flow splits (such as tee’s and wye’s), and the angle of the connecting piping. ARROW also provides automatic calculation of friction factors using the well-known Colebrook-White correlation.

Spot checks of representative loss coefficients for valves and fittings against values summarized within Crane (1988) showed very good agreement.

4.2 Test Problems

Twenty-one sample problems (input files) are supplied with ARROW. All twenty-one input files are installed in the VERIFY sub-directory below the C:\ARROW directory. The sample problems model systems that range in size from one pipe to seventy pipes. All of the sample problems are for systems with solutions available in published literature. Sample problem 5 requires the optional Chempak module for modeling mixtures and could not be run.

6 SNF-5331,Rev. 0

Table 4-1 lists the sample problems and provides a brief description of each model. Appendix A summarizes the model, the reference, and the output test results. Appendix A contains copyrighted material. This material is printed with permission from Applied Flow Technology Corporation.

Table 4-1. Listing of Sample Problems Supplied with ARROW

Number Description File Name

7 SNF-5331, Rev. 0

Number Description File Name

8 SNF-5331. Rev. 0

5.0 RESULTS

Twenty-one sample problems were successhlly executed and inspected. The input and output files are attached in Appendix A.

5.1 Acceptance Criteria

The following acceptance criteria were used to evaluate the results ofthe sample runs:

1) Each of the sample problems must execute and converge using the default or input supplied convergence criteria.

2) Each of the sample problems, with the exception of sample #5, was compared to the published solutions.

5.2 Test Results

The pertinent results from each of the sample problems are summarized within the following tables and discussed within the following paragraphs. Most of the differences in flow rates and head losses were due to a combination of items: tighter tolerances (both global and individual pipes/junctions), friction factor models, and the rigorous treatment of pump heads (fitted polynomial versus linear interpolation). These differences were judged to reflect the improvements associated with the general state of knowledge now available for use within ARROW. These differences should not reflect adversely given the relatively simple hand calculations and techniques that were used to derive the published solutions.

5.2.1 Sample Problem #I

Sample problem # 1 consists of one pipe and two junctions. Sample problem #1 models simple compressible adiabatic flow in a pipe from ANDERSON (1982). Air is the working gas.

The pipe is 30 meters in length with a hydraulic diameter of 0.15 meters. The friction factor is supplied as an explicit f (0.02). The problem states that the irdet mach number is 0.3, the inlet pressure PI is1 atm, and the inlet temperature TI is 273 "K. From the Ideal Gas Law: density, sonic speed, and mass flow rate can be calculated. Once the mass flow rate is known, it can be applied as an exit flow demand.

The results for sample problem #I are presented in Table 5-1. The corresponding results from ANDERSON (1982) are shown. Appendix A contains a detailed description of the sample problem and listings of both the input and output results.

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Table 5-1. Comparison of Results for Sample Problem #1

Parameter Anderson AFT Arrow 2.0 I MZ- Mach number at exit I 0.475 I 0.4757 I PZ- Static pressure at exit (atm) 0.624 0.6239

TZ- Static temperature at exit (OK) 265.8 266.0 P,z - Stagnation pressure at exit (atm) 0.728 0.7279

5.2.2 Sample Problem #2

Sample problem # 2 consists of one pipe and two junctions. Sample problem #2 models compressible flow in a pipe from SAAD (1993). Methane is the working gas.

As specified, inlet conditions are known and the outlet conditions are sonic. The pipe length that yields sonic flow is the objective. With the known inlet conditions, an implied mass flow rate exists. To pose the problem in AFT Arrow terms; a few simple calculations are needed to obtain the mass flow rate. Once obtained, it is applied as a flow demand at the exit. The problem states that the inlet velocity VIis 30 ds,the inlet pressure PIis 0.8 MPa, and the inlet temperature TIis 320 OK, From the ideal gas law, density, sonic speed and mass flow rate can be determined. Once determine the mass flow rate can be applied as an exit demand. AFT ARROW does not solve for pipe length. Therefore, several manual iterations are required with different pipe lengths to obtain sonic flow (h4 = 1) at the exit. The results for sample problem #2 are presented in Table 5-2. The corresponding results from SAAD (1993) are shown. Appendix A contains a detailed description of the sample problem and listings of both the input and output results.

Table 5-2. Comparison of Results for Sample Problem #2

Parameter Saad AFT Arrow 2.0 Maximum pipe length (m) 4024 4029 49

f’2 - Static pressure at exit (kPa) 48 2 48 10

VZ- Velocity at exit (ds) 433.17 433.13

TZ- Static temperature at exit (OK) 278.43 277.88

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5.2.3 Sample Problems #3

Sample problem #3 consists of one pipe and two junctions. Sample problem #3 models 600-psig steam at 850 OF flowing through a system with 400 feet of horizontal 6-inch Schedule 80 pipe at a rate of 90,000 pounds per hour.

The results for sample problem #3 are presented in Table 5-3. Comparison is made to the published results from CRANE (1988). Appendix A contains a detailed description of the sample problem and listings of both the input and output results.

Table 5-3. Comparison of Results for Sample Problem #3

5.2.4 Sample Problem #4

Sample problem #4 consists of one pipe and two junctions. Sample Problem #4 models air flowing at 65-psig steam and 110 "F through 75 feet of I-inch Schedule 40 pipe at a rate of 100 standard cubic feet per minute (scfm).

The results for sample problem #4 are presented in Table 5-4. Comparison is made to the published results from CRANE (1988). Appendix A contains a detailed description of the sample problem and listings of both the input and output results.

Table 5-4. Comparison of Results for Sample Problem #4

Volumetric flow rate at inlet (ft3/min) 20.2 20.21 Volumetric flow rate at outlet (ft3/min) 20.9 20.93 Velocity inlet (ftlmin) 3,367 3,368 Velocity inlet (ftlmin) 3,483 3,487

5.2.5 Sample Problem #5

Sample Problem 5 could not be 'mn because it requires the optional Chempak module. Table 5-5 was purposely left blank.

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The results for sample problem #5, as supplied by Applied Flow Technology, are presented in Appendix A. Comparison is made to the published results from CRANE (1988).

Table 5-5. Comparison of Results for Sample Problem #5

5.2.6 Sample Problem #6

Sample problem #6 consists of one pipe and two junctions. Sample Problem #6 models saturated steam at 170-psia feeding a pulp digester through and 30 feet of I-inch Schedule 40.

The results for sample problem #6 are presented in Table 5-6. Comparison is made to the published results from CRANE (1988). Appendix A contains a detailed description of the sample problem and listings of both the input and output results.

Table 5-6. Comparison of Results for Sample Problem #6

Exit static enthalpy (Btullbm) NIA 1,196 1,147.6 Exit static temperature (“F) NIA 317 240.7

5.2.7 Sample Problem #7

Sample problem #7 consists of one pipe and two junctions. Air supplied at a pressure of 19.3- pisg and a temperature of 100 “F is measured at a point 10 feet from the outlet of %-inch Schedule 80 pipe discharging to the atmosphere.

The results for sample problem #7 are presented in Table 5-7. Comparison is made to the published results from CRANE (1988). The Crane prediction indicates that this pipe will have subsonic velocity at the exit and hence will not choke. However, a more proper formulation of this problem shows that sonic choking will occur. This is a good example of the limitations of simplified methods such as Crane. The discrepancy comes from how to handle the exit loss of the air as it discharges to atmosphere. The Crane solution takes the appropriate K factor, equal to 1, and lumps it together with the pipe friction to obtain an overall K factor of 7.04. However, if the K factor is applied at the discharge tank and not averaged along the pipe, the flow chokes. The predicted choked flow rate is 64.3 scfm. This is not drastically different from Crane’s prediction, and well within typical engineering uncertainty. However, the difference, which is small here, could be larger in other applications.

Appendix A contains a detailed description of the sample problem and listings of both the input and output results.

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Table 5-7. Comparison of Results for Sample Problem #7

Parameter Crane AFT Arrow 2.0 Mass flow rate (scfm) 62 I 63 3

5.2.8 Sample Problem #8

Sample problem #8 consists of two pipes and three junctions. Sample problem #8 represents a branching system where the two pipe lengths necessary to obtain a measured pressure and choked flow are unknown. Air is the working gas.

ARROW does not solve for pipe length. However, the length of Pipe #1 can be manually changed to obtain the measured pressure. The resulting mass flow rate is then used as input for Pipe #2.

The results are presented in Table 5-8. Comparison is made to the published results from FOX (1985). The results agree very closely. Appendix A contains a detailed description of the sample problem and listings of both the input and output results.

Table 5-8. Comparison of Results of Sample Problem #8

* AFT Arrow does not have the ability to solve for pipe length, so the length was manually input. With this known length, the AFT Arrow Mach number at Mz should agree with Fox & McDonald’s, and it does. The resulting mass flow rate is then used as input for AFT Arrow pipe #2.

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5.2.9 Sample Problem #9

Sample problem #9 consists of one pipe and two junctions. As specified, inlet conditions are known and outlet conditions need to be determined. With the known inlet conditions, an implied mass flow rate exists. To pose the problem in AFT Arrow terms a few simple calculations are needed to obtain the mass flow rate. Once obtained, it is applied as a flow demand at the exit. The results are presented in Table 5-9. Comparison is made to the published results from LINDEBURG (1984). Appendix A contains a detailed description of the sample problem and listings of both the input and output results.

Table 5-9. Comparison of Results of Sample Problem #9

Pz - Pressure at exit (psia) 10.26 10.09

5.2.10 SampleProblem #IO

Sample problem #10 consists of one pipe and two junctions. Methane is the working gas. The problem states that the pipe length is 75 miles and 40 inches in diameter with an explicit friction factor of 0.00936. The inlet pressure PIis 650-psia. The outlet pressure PZis 450-psia. Determine the mass flow rate.

The results are presented in Table 5-10. Comparison is made to the published results from LINDEBURG (1984). Appendix A contains a detailed description of the sample problem and listings of both the input and output results.

Table 5-10. Comparison of Results for Sample Problem #lo

Parameter Lindeburg AFT Arrow 2.0 Mass flow rate (Ibds) 456 478

14 SNF-5331, Rev. 0 temperature depends on the inlet temperature and the thermodynamic process in the pipe, which is adiabatic. Therefore, to solve this problem the inlet static temperature at J1 must be guessed until the pipe delivers 293 K at the exit.

The results are presented in Table 5-1 1. Comparison is made to the published results from SAAD (1993). Appendix A contains a detailed description of the sample problem and listings of both the input and output results.

Table 5-11. Comparison of Results for Sample Problem #11

Parameter Saad AFT Arrow 2.0 M2 -Mach number at exit 0.685 0.689 MI -Mach number at inlet 0.347 0.347

PI- Static pressure at inlet (kPa) 306 308.1

TI - Static temperature at inlet (“K) 312.76 312.9

5.2.12 Sample Problem #12

Sample problem #12 consists of one pipe and two junctions. Air is the working gas. Inlet conditions are known. Air flows isentropically through a convergent nozzle which feeds an insulated constant are duct.

The results are presented in Table 5-12. Comparison is made to the published results from SAAD (1993). Appendix A contains a detailed description ofthe sample problem and listings of both the input and output results.

Table 5-12. Comparison of Results for Sample Problem #12

Parameter Saad AFT Arrow 2.0 I Mass flow rate when choked (kg/s) I 2.11 I 2.10 I I MI -Mach number at inlet I 0.603 I 0.604 I PI - Static pressure at inlet (ma) 2.106 2.114

TI- Static temperature at inlet (“K) 427.8 429.6

P2,choke - Static back pressure for choking ma) 1.203 1.217

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5.2.13 Sample Problem #I3

Sample problem #13 consists of two pipes and four junctions. Natural gas is the working gas. Inlet conditions are known. Outlet conditions are to be determined. With the known inlet conditions, a few simple calculations are needed to obtain the implied mass flow rate. Once obtained, it is applied as an outlet flow demand.

The first pipe represents the pipe from point 1 to 2. The second pipe represents the pipe from point 1 to the choking pint. AFT Arrow does not explicitly solve for pipe length. To obtain the maximum pipe length, different lengths must be guessed with lengths that exceed sonic flow discarded. The results are presented in Table 5-13. Comparison is made to the published results from SAAD (1993). All results agree closely. The AFT Arrow static pressure below which choking occurs is the pipe exit static pressure. Appendix A contains a detailed description of the sample problem and listings of both the input and output results.

Table 5-13. Comparison of Results for Sample Problem #13

A42 - Mach number at exit 0.38 0.3818

Ll.2 -Length to reach 500 kPa (m) 544.79 558.5 LT*- Length to reach sonic choke point (m) 710 709.8 PT*- Static pressure choke point (kPa) 114.46 114.9 UT*-Mach number at isothermal choke point 0.874 0.8704

5.2.14 Sample Problem #I4

Sample problem #14 consists of one pipe and two junctions. The objective is to solve for the choked flow rate. Air is the working gas.

The results are presented in Table 5-14. Comparison is made to the published results from PERRY. Appendix A contains a detailed description of the sample problem and listings of both the input and output results.

Table 5-14. Comparison of Results for Sample Problem #14

Parameter Perry AFT Arrow 2.0 Mass flow rate (kg/s) 2.7 2.716 SNF-5331, Rev. 0

5.2.15 Sample Problem #15

Sample problem #I5 consists of one pipe and two junctions. Air is the working gas. Inlet conditions are known. Outlet conditions need to be determined. With the known inlet conditions, a few simple calculations are needed to obtain the implied mass flow rate. Once obtained, it is applied as an outlet flow demand.

The results are presented in Table 5-15. Comparison is made to the published results from JANNA (1983). Appendix A contains a detailed description ofthe sample problem and listings of both the input and output results.

Table 5-15. Comparison of Resuits for Sample problem #15

A42 - Mach number at exit 0.14 0.1355 PZ- Pressure at exit (psia) 9.63 9.836 TZ- Temperature at exit (“R) 528.8 528.9

5.2.16 Sample Problem #16

Sample problem #I6 consists of one pipe and two junctions. Air is discharged from a high- pressure tank through a 4-inch (Schedule 40) vent line into the atmosphere. The mass flow is given as 72,000 Ibdhr. Stagnation conditions within the tank are Po= 600-psig (constant) and To= 120 OF. The equivalent length of the pipe is 90 feet. The results are presented in Table 5-16. Comparison is made to the published results from NAYYAR (1992). Appendix A contains a detailed description of the sample problem and listings of both the input and output results.

Table 5-16. Comparison of Results for Sample Problem #16

I A41 - Mach number at valve I 0.317 I 0.318 I PI- Pressure at valve (psia) 128.46 128.66 I

5.2.17 Sample Problem #17

Sample problem #17 is identical to sample problem #16, except that the mass flow rate is doubled.

17 SNF-5331. Rev. 0

The results are presented in Table 5-17. Comparison is made to the published results from SAAD (1992). Appendix A contains a detailed description of the sample problem and listings of both the input and output results.

Table 5-17. Comparison of Results for Sample Problem #17

MI- Mach number at valve 0.3 I7 0.318 PI-Pressure at valve (psia) 256.93 257.32

5.2.18 Sample Problem #I8

Sample problem #18 is identical to sample problem #16, except that the pipe length is doubled

The results are presented in Table 5-18. Comparison is made to the published results from SAAD (1992). Appendix A contains a detailed description of the sample problem and listings of both the input and output results.

Table 5-18. Comparison of Results for Sample Problem #IS

PI -Pressure at valve (psia) 168.52 168.69

5.2.19 Sample Problem #I9

Sample problem #I9 is identical to sample problem #16, except that the air temperature is 500

OF.

The results are presented in Table 5-19. Comparison is made to the published results from SAAD (1992). Appendix A contains a detailed description of the sample problem and listings of both the input and output results.

Table 5-19. Comparison of results for Sample Problem #I9

18 SNF-5331, Rev. 0

5.2.20 Sample Problem #20

Sample problem #20 is identical to sample problem #16, except that the pipe diameter has been decreased to 2-inches.

The results are presented in Table 5-20. Comparison is made to the published results from SAAD (1992). Appendix A contains a detailed description ofthe sample problem and listings of both the input and output results.

Table 5-20. Comparison of Results for Sample problem #20

PI- Pressure at valve (psia) 661.96 662.56

5.2.21 Sample Problem #21

Sample problem #21 is identical to sample problem #16, except that the pipe diameter has been increased to &inches.

The results are presented in Table 5-21, Comparison is made to the published results from SAAD (1992). Appendix A contains a detailed description ofthe sample problem and listings of both the input and output results.

Table 5-21. Comparison of Results for Sample Problem #21

Parameter Nayyar AFP Arrow 2.0

I MI- Mach number at valve I 0.4096 I 0.4142 I I PI - Pressure at valve (psia) I 25.15 I 24.95 I

19 SNF-5331, Rev. 0

6.0 SUMMARY AND CONCLUSIONS

The ARROW (Version 2) compressible flow network analysis program has been installed and successfully validated for use. Based on a direct comparison of twenty of twenty-one sample problems to published results, the ARROW computer program is judged acceptable for use at the Hanford site. Furthermore, the use of an encrypted license key to assures proper installation and configuration control

The pertinent results from each of the sample problems have been summarized within the preceding tables. It must be noted that in most cases very good agreement was found. However, a few cases of significant differences between ARROW and the published results were found. Most of the differences in flow rates and pressures were due to a combination of items: tighter tolerances (both global and individual pipes/junctions), friction factor models, and the rigorous treatment component losses

These differences were judged to reflect the improvements associated with the general state of knowledge now automated and available for use within ARROW. These differences should not reflect adversely given the relatively simple hand calculations and techniques that were used to derive the published solutions.

20 SNF-533 1, Rev. 0

7.0 REFERENCES

Anderson, J. D., “Modern Compressible Flow”, McGraw-Hill, 1982

“Flow of Fluids Through Valves, Fittings, and Pipe”, Technical Paper Number 410, Crane Company, Chicago, Illinois, 1991.

Fox, R. W. and A. T. McDonald, “Introduction to Fluid Mechanics”, Third Edition, 1985, p. 376.

Holman, J. P., “Heat Transfer”, Seventh Edition, McGraw-Hill, Inc., New York, New York, 1990.

Idelchik, I. E., “Handbook of Hydraulic Resistance”, Third Edition, CRC Press, Ann Arbor, Michigan, 1994.

Janna, W. S., “Introduction to Fluid Mechanics”, PWS Publishers, Belmont, CA.

Lindeburg, M. R., “Mechanical Engineering Review Manual for the P. E. Exam”, 7th Ed., 1984.

Miller, D. S., “Internal Flow Systems”, Second Edition, Published by BHRA (Information Services), Cranfield, UK, 1990.

Marks Standard Handbook for Mechanical Engineers, 8th ed., 1978, T. Baumeister, editor.

Nayyar, M. L., “Piping Handbook”, Sixth Edition, McGraw-Hill, New York, 1992.

Perry, R. H. and D. W. Green, “Perry’s Chemical Engineer’s Handbook”, Seventh Edition

Saad, M. A,, “Compressible Fluid Flow”, Second Edition, Prentice Hall, Englewood Cliffs, New Jersey, 1993.

21 SNF-533 1, Rev. 0

APPENDIX A - SAMPLE PROBLEM DESCRIPTIONS, INPUT, AND OUTPUT

Appendix A contains copyright material. Problem descriptions were printed with permission from Applied Flow Technology Corporation.

22 SNF-5331, Rw. 0

Parameter Anderson AFT Arrow 2.0 Mt- Mach number at exit 0.475 0.4757 P2- Static pressure at exit (atm) 0.624 0.6239 T, - Static temperature at exit (deg. K) 265.8 266.0 Poz- Stagnation pressure at exit (atm) 0.728 0.7279

DISCUSSION As specified, inlet conditions are known and outlet conditions need to be determined. With the known inlet conditions, an implied mass flow rate exists. To pose the problem in AFT Arrow terms, a few simple calculations are needed to obtain the mass flow rate. Once obtained, it is applied as a flow demand at the exit. The problem states that the inlet Mach number is 0.3, P,= 1 atm. T,= 273 K. From the ideal gas law, density, sonic speed and mass flow rate are: 1 atm m=--5- = 1.293 kg/xn3 Rq (.2868&)(273K)

273 K = 33 1.2 m/s (sonic vc..,city) liz=plQA=pl(Mlal)A=( 1.293- 3 (0.3) ( 331.2- 3: -0.15*m2 1=2.27kg/s Note that the friction factor in Anderson is the Fanning friction factor. To obtain the Darcy-Weisbach friction factor used in AFT Arrow, multiply the Fanning friction factor by 4.

J1 J2

printed with permission. 23 SNF-533 1, Rev. 0

AFT Arrow 2.0 Input (1) 127199 1O:lO AM Numatec Hanford Corporation Verify1.aro - Modern Compressible Flow. Example 3.4. pg 76

John D. Anderson Modern Compressible Flow, 1982. McGraw-Hill page 76, example 3.4 Air flow in a pipe - Example assumes adiabatic flow and air is a perfect gas

See Veriyl .doc MS Word file for comparison with published resuns

Title: Verify1 .aro - Modem Compressible Flow, Example 3.4, pg 76

Number Of Pipes= 1 Number Of Jundions= 2

Length March Solution Method with Mach Number Limits Segments Per Pipe= 2 Mach Number Increment= .01 PressureTolerance= .wO1 relative change Mass Flow Rate Tolerance= .IC01 relative change Enthalpy Tolerance= ,2001 relative change ConcentrationTolerance= .wO1 relative change Flow Relaxation= (Automatic) Pressure Relaxation= (Automatic)

Fluid Database: AFT Standard Fluid= Air Max Fluid Temperature Data= loo0 deg. K Min Fluid Temperature Data= 200 deg. K Molecular Weight= 28.97 amu Gas Constant= 0.2868 kJlkg-K Ciiical Pressure. 37.25 atm Critical Temperature= 132.41 deg K Acentric Factor= ,021 Equation of State= Ideal Gas Enthalpy Model= Reference Specific Heat Ratio Accuracy= High Atmospheric Pressure. 1 atm Gravitational Acceleration= 1 g's Standard Pressure= 14.696 psia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Number= 4wO Laminar Flow Below Reynolds Number= 2300 Specific Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g Standard Pressure= 14.696 psia Standard Temperature= W deg. F Turbulent Flow Above Reynolds Number= 4002 Laminar Flow Below Reynolds Number= 2300

Pipe Input Table

Pipe Name Pipe Length Length Hydraulic Hydraulic Roughness Roughness Losses (K) Initial Flow Defined Units Diameter Diam. Units Units

Pipe Initial Flow Junctions Geometry Material Size Type Special Units (Up,Down) Condition 11 1,2 1 Cylindrical Pipe 1 Unspecified I None

Assigned Flow Table

24 SNF-5331,RH. 0

AFT Anow 2.0 Input (2) 12n1991010AM Numatec Hanford Corporation Verify1 .am- Modern Compressible Flow, Example 3.4, pg 76

Assigned Flow Name Ow Database Special Fluid Type Flow Flow Temperature Temperature Defined Source Condition Units Units 2 1 AssignedFlowI Yes1 None1 Air1 OuMowI 2.271 kg/secl 273 1 deg. K

Assigned Pressure Name Object Database Special Fluid Pressure Pressure Temperature Defitmd Source Condition Units 1 I Assianed~rtnrrurel Yes1 N/AI Air1 I1 atm I 273

Assigtmd,~mssure I TO~TI:: I LWS~~ , Factor d .K AFT Arrow 2.0 Output (1) Numtec Hanfcfd Corporation Verifyl .am - Modem Compreuible Flow, Example 3.4, pg 76

John D. Anderson Modem Compressible Flow, 1982. McGraw-Hill page 76, example 3.4 Air flow in a pipe - Example assumes adiabatic fkw and air is a perf& gas Sea Verifyl .doc MS Word file for comparison with publishad reauks Tale: Verifyl .am - Modern Comprsgible Flow, Example 3.4, pg 76 Analysisrunon: 12n19910:11:29AM Input File: C:\AFT Prcducts\AFTAnow\Verh3cation\vsrifyl .am

Execution Time= 0.27 seconds Total Number Of Pressure Iterations- 0 Total Number Of Flow Iterations= 2 Total NumhOf Enthalpy Iterations- 2 Number Of Pipes= 1 Numkr Of Junctions. 2

Length March Solution Method wkh Mach Number Umih Segments Per Pipe= 2 Mach NumhIncrement= .01 Pressure Tolerance- .wO1 ratalive change Mass Flow Rate Tolerance= .wO1 relative change Enthalpy Tolerance= .wO1 relative change Flow Relaration- (Automatic) Pressure Relaxation- (Automate)

Fluid Datatase: AFT Standard Fluid- Air Max Fluid Temperature Data= 1000 deg. K Min Fluid Temperature Data= 200 deg. K Mokular Weight= 28.97 amu Gas Conslant= 0.2868 kJkg-K Critical Pressure= 37.25 atm Critical Temperature= 132.41 deg. K Acentric Factor= ,021 Equation of State- Ideal Gas Enthalpy Model- Reference Spacifm Heat Ratio Accuracy. High Atmospheric Pressure- 1 atm Gravitationel Acceleration- 1 g's Standard Pressure- 14.696 psia Standard Temperature- 60 deg, F Turbulent Flow AboM Reynold Nunbow 4ooo Laminar Flow Bdow Reyndda Nunhr= 2202 Specific Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g Standard Pressure= 14.696 pia Standard Temperature- 60 deg. F Turbulent Flow Above Reynolds Nunber= 4wO Laminar Flow Bdow Reynolds Nunhr= ZJOO Overall Della Pressure = -0.03409 MPa Total Inflow= 2.270 kg/sec Total Outflow 2.270 kglsec Total Energy Inflow 992.7 kW Total Energy Outflow= 992.7 kW Total Heat Transferred Into System= O.Oo0 kW

Maximum Pressure is 0.1013 MPa at Junction 1 Ink Minimum Pressure is 0.06322 MPa at Junction 2 Inlet Maximum Static Temperature is 4.703 deg. C 111 Junc(ion 2 Inlet Minimum Static Temperature is 4.699 deg. C et Junction 1 Ink I Pipe OUtDut Table

26 Pips Mach P Static P Stag. TStag. TStatic P Stag. Mass Rho Vel. Vel. #OU out Out Out Out In Flow Static In In Sonic In (atm) (atm) (deg. K) (deg. K) (atm) (kglsec) (kg/m3) [meterdsec) (meterdsec) 1 0.4757 0.6239 0.7279 277.9 266.0 1.064 2.270 1.293 99.33 330.4

All Junction Table

Jct PStag. PStatic TStatic TStag. 1 1 In I In 1 In I In -1 I (atm) I (atm) 1 (deg. K) I (deg. K) 1 11.06441 l.OMx)[ 273.01 277.8 2 I 0.7279 I 0.6239 I 266.0 I 277.9

n ~

SNF-5331, Rev. 0

Parameter Saad AFT Arrow 2.0 Maximum pipe length (m) 4024 40 2 9.4 9 P2 -Static pressure at exit (@a) 48.2 48.10 V2-Velocity at exit (ds) 433.17 433.13 T2 - Static temperature at exit (deg. K) 278.43 277.88

DISCUSSION: As specified, inlet conditions are known and the outlet conditions are sonic. The pipe length that yields sonic flow is the objective. With the known inlet conditions, an implied mass flow rate exists. To pose the problem in AFT Arrow terms, a few simple calculations are needed to obtain the mass flow rate. Once obtained, it is applied as a flow demand at the exit. The problem states that the inlet velocity VIis 30 ds,PI = 0.8 MPa, TI= 320 K. From the ideal gas law, density, sonic speed and mass flow rate are: m=--9- 0.8 MPa = 4.823 kg/m 3 Rq (.5179&)(32OK) m = &VIA = (4.823- 3 30- :I(: -0.32m2 I=10.2265 kg/s AFT Arrow does not solve for pipe length. To obtain the maximum pipe length, different lengths must be guessed with lengths that exceed sonic flow discarded. Note that the friction factor in Saad is the Fanning friction factor. To obtain the Darcy-Weisbach friction factor used in AFT Arrow, multiply the Fanning friction factor by 4.

Printed with permission. 28 AFT Arrow 2.0 Input (1 ) 12f7199 10:13AM Numatec Hanford Corporation Verify2.m - MnU'mne Exampie from Saad pp. 215216

Michel A. Saad Compressible Fluid Flow, 2nd Edition, Prentice-Hall, Englewucd CIM.. NJ. 1993 page 215, example 5.3 Methane flow - Example assummethane is a pede3 gm and llow is adiabatic See Verify2.doc MS Word file for capri.on with published reaults Tlle: Verify2.am - Methane Example from Saad pp. 215216

Number Of Pipes= 1 Number Of Junctions= 2

Length March Solution Method with Mach NumkUmb Segments Per Pipe= 50 Mach NumkIncrement= .01 Pressure Tolerance- .WO1 r&We changa Mass Flow Rate Tolerance= .OW1 relatiw change Enthalpy Tolerance= .WO1 nlatiw change Concentration Tolerance= .WO1 rdatw chanw Flow Relaxation= (Automatic) Pressure Relaxation= (Automatic)

Fluid Database: AFT Standard Fluid- Methane Max Fluid Temperature Data- 400 dog. K Min Fluid Temperatun Data= 150 deg. K Molecular Weight= 16.043 amu Gas Constant= 0.51790380 kJkg-K Critical Pressure= 4.5988 MPa Critical Temperature= 190.556 deg. K Acentric Fact- ,208 Equation of State= Ideal Gas Enthalpy Model= Reference Specific Heat Ratio Accuracy= nigh Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g's Standard Pressure= 14.696 pia Standard Temperature- 60 deg. F Turbulonl Flow AhReynold. Number= 40W Laminar Flow Below Reynolds Number= 2300 Specific Heat Ratio Accuracy= High Atinospharic Pressure= 1 atm Gravitational Acceleration= 1 g Standard Pressure- 14.696 pia Standard Temperatun= 60 deg. F Turbulent Flow Above Reynok Nmk=40W Laminar Flow Below Reynolds Numbsc= 2300

Pipe Name Pip Length Len@ Hydnulii Hydraulic Rwghness Roughness Losses(K) InliilFlow Defined Unb DmnWer Dhm.Unb Unb 1 Pipe I yes1 4029.491 metem] 0.3 I nmtera O.WOl5I EpdDl 01

Pip InXilFlow Junctions Geometry M.L.rbl Size Typ Special Unls (Up.Down) Condluon 1 1.2 I Cyllndkal Pipe I Unrwctfhd I None

Assianed Flow Table

29 SNF-533 I, Rsv. 0

Assigned Flow Name Ob+acl Databmw Spacial Fluid Type Flow Flow Temperature Defined Soum Condition Units 2 I AssignedFlowl Yea None1 Melhanel OuMowl 10.22651 Wec I 320

r Assigned Pressure Name Objecl Databasa Special Fluid Pressure Pressure Temperature Defmed Swm Condition Units 1 I Assigned Pressure I Yea I NlAI Methane1 0.8 I MPa I 320

30 SNF-533 1, Rw.4

AFT Arrow 2.0 Output (1) 12/7/99 Numatec Hanford Corporation Verfy2.aro - Methane Example from Saad pp. 21 5216

Michel A. Saad Compressible Fluid Flow, 2nd Edition. Prentice-Hall. Englewood Cliis, NJ, 1993 page 215, example 5.3 Methane flow - Example assumes methane is a perfect gas and flow is adiabatic

See Verify2.doc MS Word file for comparison with published results

Title: Verify2.aro - Methane Example from Saad pp. 215216 Analysisrunon: 12/1199101319AM Input File: C:WFT ProductsWFTArrow\Veflcation\verify2.aro

Execution Time= 0.83 seconds Total Number Of Pressure Iterations= 0 Total Number Of Flow Iterations= 2 Total Number Of Enthalpy Iterations. 2 Number Of Pipes= 1 Number Of Junctions= 2

Length March Solution Method whh Mach Number Limits Segments Per Pipe= 50 Mach Number Increment= .01 Pressure Tolerance= ,2001 relative change Mass Flow Rate Tolerance= .wO1 relative change Enthalpy Tolerance= ,2001 relative change Flow Relaxation= (Automatic) Pressure Relaxation= (Automatic)

Fluid Database: AFT Standard Fluid= Methane Max Fluid Temperature Data= 400 deg. K Min Fluid Temperature Data= 150 deg. K Molecular Weight= 16.043 amu Gas Constant= 0.51790380 kJkg-K Critical Pressure= 4.5988 MPa Critical Temperature= 190.555 deg. K Acentric Factor= ,008 Equation of State= Ideal Gas Enthalpy Model= Reference Specific Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g's Standard Pressure= 14.696 psia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Number= 4000 Laminar Flow Below Reynolds Number 2300 Specific Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g Standard Pressure= 14.696 psia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Number= 4w0 Laminar Flow Below Reynolds Number= 2300 Overall Delta Pressure = -0.71410561 MPa Total Inflow 10.226500 kg/sec Total Outflow= 10.226500 kgkec Total Energy Inflow= 10,556.053 kW Total Energy Outflow= 10,536.053 kW Total Heat Transferred Into System= O.M)(xKxxI kW

Maximum Pressure is 0.8MNxXx)l MPa at Junction 1 Inlet Minimum Pressure is 0.048102446 MPa at Junction 2 Inlet Maximum Static Temperature is 48.621952 deg. Cat Junction 2 Inlet Minimum Static Temperature is 47.045990 deg. C at Junction 1 Inlet

PiDe Outout Table

31 SNF-5331, Rw. 0

Pipe Length Mass Flow Vel. In vel. out Mach C In Mach # Out P Static In P Static T Static In Out (meters) (kdwc) (dedaoc) (dedaoc) (kPa) (kPa) (deg. K) 1 40294900 10.226500 29.991524 433.13260 0.064798974 0.99336141 800.00006 48,102448 320.oooO3

Jct P Static In P Static Out T Static In T Static Out

IkPa) (kPa) (dw. K) (den. F) 1 8w.000061 800.000061 320.owoO 116.3JOM)Z 2 48.102448 48.102448 277.87512 40,505234

32 SNF-5331,Rev. 0

Parameter Crane AFT Arrow 2.0 Static pressure drop (psi) 40.1 I 41.59

~~ _____

DISCUSSION: Crane does not make a distinction between static and stagnation pressure, and it appears that static pressure is usually assumed. Therefore, the inlet pressure of 600 psig was assumed to be static pressure. To make a one-to-one comparison, the K factors and friction factor used in Crane was used directly in AFT Arrow. The K factors were modeled as Additional Losses, which evenly spreads the effect of resistance across the entire pipe. In practice, the velocity changes in the pipe can yield different answers for fitting pressure losses depending on where they are actually located. The AFT Arrow model uses the Redlich-Kwong real gas equation of state for steam. Note that the Crane formula underpredicts the pressure drop (by about 4%), which in most applications is not conservative.

J1 J2

* Printed with permission. 33 SNF-533 1, Rcv. 0

AFT Arrow 2.0 Input (1) 1 mi99 10:08AM Numatec Hanford Corporation

Verify3.arO ~ Steam flow in a pipe, Example 4-10from Crane page 4-6

Crane Co. Flow of Fluids Through Valves, Fittings. and Pipe, Technical Paper No. 410,Crane Co., Joliet, IL, 1988 Page 4-6. example 4-10 Steam flow - Example dws not specify the heat transfer conditions, so it was assumed adiabatic in this model

See Verify3.doc MS Word file for comparison wnh published resuits

Title: Veriiy3.aro - Steam flow in a pipe. Example 4-10 from Crane page 4-6 Number Of Pipes= 1 Number Of Junctions= 2

Length March Solution Method with Mach Number Limits Segments Per Pipe= 10 Mach Number Increment= .01 Pressure Tolerance= ,0001 relative change Mass Flow Rate Tolerance= .OW1relative change Enthalpy Tolerance. .OW1 relative change Flow Relaxation= .5 Pressure Relaxation= .5

Fluid Database: AFT Standard Fluid= Steam (Simplified) Max Fluid Temperature Data= 1500 deg. F Min Fluid Temperature Data= Ma deg. F Molecular Weight= 18,016amu Gas Constant= 0,1102Btulbm-R Critical Pressure= 3208.22 psia Critical Temperature= 1165.09 deg. R ¢ric Factor= ,344 Equation of State= Redlich-Kwong Enthalpy Model= Generalized Specific Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm GravitationalAcceleration= 1 g's Standard Pressure= 14.696psia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Number= 4000 Laminar Flow Below Reynolds Number= 2300 Specific Heat Ratio Accuracy= High Mmospheric Pressure= 1 atm GravitationalAcceleration= 1 g Standard Pressure. 14.696psia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Number= 4000 Laminar Flow Below Reynolds Number= 2300

PiDe InDut Table

Pipe Name Pipe Length Length Hydraulic Hydraulic Roughness Roughness Losses (K) Initial Flow Defined Units Diameter Diam. Units Units 1 Pipe I Yes1 4001 feet1 5.761 1 inches 1 0.015 Explicit f I 3.5 1

Pipe Initial Flow Junctions Geometry Material Size Type Special Units (Up.Dom) Condition 1 1,2 1 Cylindrical Pipe 1 Steel I 6 inch 1 schedule 80 None ksianed Flow Table

34 SNF-533 1, Rw. 0

AFT Arrow 2.0 Input (2) IZ7/9910:08AM Numatec Hanford Corporation Verify3.arO - Steam flow in a pipe, Example 4-1 0 from Crane page 4-6

Assigned Flow Name Object Database Special Fluid Type Flow Flow Temperature Defined Source Condtion Unks

1 Assign: Flow 1 Unkr Model 1 Losso I

Assigned Pressure Table

Assigned Pressure Name Object Database Special Fluid Pressure Pressure Temperature Defined Source Condition Units 1 I Assigned Pressure 1 Yes I N/A I Steam (Simplied) I 6W I psig I 850

~ Assigned,Pressure I Temper??YF I Losso 1 Units Factor

35 SNF4331, Icev. 0

AFT Armw 2.0 Output (1) 12f7199 Numatec Hanford Corporation Verify3.aro - Stesrn flow in e pipe, Example 4-10 from Crane page 4-6

Cnne Co. Flow of Fluids Through Valves, Fmings, and Pipe, Technical Papof No. 410, Cnne Co., Jdi,IL. 1988 Page 4-6,example 410 Steam flow - Example does not specify the heat transfer conditionr, sa it ma awumed adiabatic In this malei

Sea Verify3.doc MS Word file for mparlson with puMihd rmuka

Tile: Verify3.aro - Steam flow in a pipe, Example 4-10 from Cram pago 4-6 Analysis run on: 1WB910:04:11 AM Input File: C:WFT ProductsWFT ~Verif~ation\wrify3..aro

Execution Time= 0.39 seconds Total Number Of Pressure Iterations- 0 Total Number Of Flow Iterations= 2 Total Number Of Enthalpy Iterations;. 2 Numbar Of Pip=1 Number Of Junctions- 2

Length March Solution Method with Mach Number Limb Segments Per Pipe= 10 Mach Number Increment= .01 Pressure Tolerance. .wO1 relafive change Mass Flow Rete Toleranc-a= .ooO1 relative change Enthalpy Tolennw- .W1relative change Flow Relaxation= .5 Pressure Relaxation= .5

Fluid Database: AFT Standard Fluid- Steam (Simplified) Max Fluid Temperature Data= 1500 deg. F Min Fluid Temperature Data= MO cbg. F Molecular Weight.- 18.016 amu Gar, Constant= 0.1 102 BNlbm-R Critical Pressure= 3208.22 pia Critical Temperature= 1165.09 deg. R Acentric Factor= ,344 Equation of State= Redlich-Kwong EnV\alpy M+ Generalized Specific Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g's Standard Pressure- 14.696 psia Standard Temperature= 60 deg. F Turbulent Flow Above Reyndds Number= 4WO Laminar Flow Below Reynokls Number= 23W Specific Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm Gravitational Acceleration- 1 g Standard Pressure= 14.696 psia SWnrdTemperature= 60 deg. F Turbulent Flow Above Reynolds Number- 4W0 Laminar Flow Below Reynolds Number= 23W Ownll Deita Pressure = 41.41 psia Total Inflow= 25.00 Ibm/lrec Total OUmw 25.00 lbmlaec Total Energy Inflow;; 35,663BtuEs Total Energy OUMOWP 35,663 6th Total Heat Transferred Into Syatem- 0.000 BhJs Maximum Pressure is 614.7 psia at Junction 1 Inlet Minimum Pressure is 573.1 pia at Jundin 2 Inlet Maximum Static Temprature is 851 .O deg. F at Junchn 1 InM Minimum Static Temperature is 848.0 deg. F at Junction 2 InY

Piw Out~utTable SNF-5331,Rcv. 0

AFT AROW 2.0 Output (2) 1mi99 Numatoo Hanford Cwporation Verify3.aro - Steam ibw in a pip, Example 4-10 from Crane page 4-6

Pipe dP Static Mass Vel. Vel. PStatic PStatic TStatic TStatic f K RID fU Total Flow In Out In Out In Out D+K bid) (Ibmhr) (feet/sec) (fd8d @la) (pain) (deg. F) (deg. F) 1 41.59 90,OM) 167.0 179.2 600.0 558.4 850.0 846.8 0.01500 3.500 12.50 16.00

All Junction Table

37 SNF-533 1, Rev. 0

TITLE: VERIFY4.ARO REFERENCE: Crane Co., Flow of Fluids Through Valves, Fittings, and Pipe, Technical Paper No. 410, Crane Co., Joliet, IL, 1988, Page 4-9, example 4-16 GAS: Air ASSUMPTIONS: Example does not specify the heat transfer conditions, so it was assumed isothermal in this model. RESULTS:

.. , Parameter Crane ; AFT Arrow 2.0 Static pressure drop @si) 2.61 I 2.712 I I Volumetric flow rate at inlet (ft3/min) 1 20.2 1 20.21 Volumetric flow rate at outlet (ft3/min) 20.9 I 20.93 Velocity inlet (Wmin) 3,367 3,368 Velocity inlet (Wmin) 3,483 3,487

DISCUSSION: Crane does not make a distinction between static and stagnation pressure, and it appears that static pressure is usually assumed. Therefore, the inlet pressure of 65 psig was assumed to be static pressure. Note that the Crane formula underpredicts the pressure drop (by about 4%), which in most applications is not conservative.

J1 J2

Printed permission. with 38 SNF-5331,Rev. 0

AFT Arrow 2.0 Input (1) Numatec Hanford Corporation Verify4.aro -Air flow in a pipe, Example 4-1 6 from Crane page 4-9

Crane Co. Flow of Fluids Through Valves, Fmings. and Pipe, Technical Paper No. 410, Crane Co., Joliet, IL. 1988 Page 49. example 4-16 Air flow - Example does not specify the heat transfer condlons, but it appears to use isothermal so this model uses isothermal

See Verify4.doc MS Word file for comparison with published results

Title: Veriiy4.aro -Air flow in a pipe, Example 4-1 6 from Crane page 4-9

Number Of Pipes= 1 Number Of Junctions= 2

Length March Solution Method with Mach Number Limits Segments Per Pipe= 2 Mach Number Increment= .01 Pressure Tolerance= .ooO1 relative change Mass Flow Rate Tolerance= .ooOl relative change Enthalpy Tolerance= .Wa1 relative change Concentration Tolerance= .Wa1 rdative change Flow Relaxation= (Automatic) Pressure Relaxation= (Automatic)

Fluid Database: AFT Standard Fluid= Air Max Fluid Temperature Data= loo0 deg. K Min Fluid Temperature Data= 200 deg. K Molecular Weight= 28.97 amu Gas Constant= 0.06855 Btullbm-R Critical Pressure= 37.25 atm Critical Temperature= 132.41 deg. K Acentric Factor= ,021 Equation of State= Ideal Gas Enthalpy Model= Reference Specfic Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g's Standard Pressure= 14.696 psia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Number= 4Wa Laminar Flow Below Reynolds Number= 2300 Specific Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g Standard Pressure= 14.696 psia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Number= 4OOO Laminar Flow Below Reynolds Number= 2309

Pipe Name Pipe Length Length Hydraulic Hydraulic Roughness Roughness Losses (K) Initial Flow Defined Units Diameter Diam. Units Units 1 Pipe I yes I 751 feet/ 1.0491 inches I O.OW15 feet I 01

Pipe Initial Flow Junctions Geometry Material Size Type Special Units (Up,Down) Condition 1 1,2 I Cylindrical Pipe I Steel 1 1 inch I schedule 40 None

Assioned Flow Table

39 SNF-5331,Rev. 0

AFT Arrow 2.0 Input (2) 12nI99 10:24 AM Numatec Hanford Corporation Veriiy4.aro -Air flow in a pipe, Example 4-16 from Crane page 4-9

Assigned Flow Name Object Database Special Fluid Type Flow Flow Temperature Temperature Defined Source Condition Units Units

Assigned Pressure Name Object Database Special Fluid Pressure Pressure Temperature Defined Source Condnion Units 1 I Assigned Pressure I Yes I N/AI Air] 65 1 psig 1 110

Units Factor SNF-5334,Rav. 0

AFT Anow 2.0 Output (1) 12/7/99 Numatec Hanford Corporation Varify4.aro -Air flow in a pipe, Example 4-16 from Cram? page 4-9

Crane Co. Flow of Fluids Through Vah.Fniigs, and Pip, Technical Paper No. 41 0.Crane Co.. Joliet, IL, 1988 Page 49, example 416 Air flow - Example does nd specify the heat tranafw condhons. but R appears to use idermalso thb deluscs isothermal Sw Verify4.d~~MS Word file for mparison with publiihed results

Tale: Verify4.aro -Air flow in a pip, Example 416 from Crane page 4-9 Analysis run on: 12/7/99 102233AM Input File: C:\AFT ProductsiAFT Arrow\Verf1cation\verify4.aro

Execution Time= 0.1 7 seconds Total Number Of Pressure Iterations= 0 Total Number Of Flow Iterations= 2 Total Number Of Enthalpy Iterations= 2 Number Of Pipes= 1 Number Of Junctions= 2

Length March Solution Method with Mach Number Limits Segments Per Pip= 2 Mach Number Increment= .01 Praswre Tolerance= .ooO1 relativa change Mass Flow Rata Tolerance= .JW1 relative change Enthalpy Tolerance= .OW1 relative change Flow Relaxation= (Automatic) Pressure Relaxation= (Automatic)

Fluid Database: AFT Standard Fluid= Air Max Fluid Temperature Data= loo0 deg. K Min Fluid Temperature Data= 200 rbg. K Molecular Weight= 28.97 amu Gas Constant= 0.06855 Btu/lbm-R Critical Pressure= 37.25 atm Critical Temperature= 132.41 deg. K Acentric Factor= ,021 Equation of State= Ideal Gas Enthalpy Model= Reference Specifi Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g's Standard Pressure= 14.6% psia Standard Ternparature= 60 dog. F Turbulent Flow Abovr, Reynold$ Number= 4OOO Laminar Flow Below Reynolds Number= 2300 Specific Heat Ratio Accuracy= High Atmphwc Pressure= 1 atm Gravitational Acceleration;: 1 g Standard Pressure= 14.696 psia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Nunber= 4OW Laminar Flow Below Reynolds Number- 2333 Overall Deita Pressure = -2.707 psia Total Inflow 0.1272 Ibdsec Total OuMw 0.1272 lbrn/sec Total Energy Inflow- 26.10 8Ws Total Energy OuMow 26.10 6th Total Heat Transferred Into System= 5.740E-04 Btds

Maximum Pressure is 79.70 psia at Jundion 1 InM Minimum Pressure is 76.98 piaat Junction 2 Inlet Maximum Static Temparature is 110.3 dog. F et Junction 2 I Minimum Static Temprature in 110.3 deg. F at Junction 1 11

Piw Out~utTable

41 SNF-5331 & AFT Arrow 2.0 Output (2) 12/7/99 Numatec Hanford Corporation Verify4.aro -Air flow in a pipe, Example 4-1 6 from Crane page 4-9

Pipe dP Static Vol. VOl. Vel. Vel. Mass P Static P Static T Static T Static Total Flow In Flow Out In Out Flow In Out In Out (psid) (fWmin) (ft3lmin) (feethin) (fedmin) (scfm) (psig) (psis) (den. F) (deg. F) 1 2.712 20.21 20.93 3368. 3487. 100.0 65.00 62.29 110.0 110.0

Jct P Static P Static T Static T Static In out In Out (prig) (psig) (deg. F) (deg F) 1 65.00 65.00 110.0 110.0 2 62.29 62.29 110.0 110.0

42 SNF-5331,Rev:O

TITLE: VERIFY5.ARO REFERENCE: Crane Co., Flow of Fluids Through Valves, Fittings, and Pipe, Technical Paper No. 410, Crane Co., Joliet, IL, 1988, Page 4-1 I, example 4-18 GAS: Natural Gas (mole fractions: 75% methane, 21% ethane, and 4% propane) ASSUMPTIONS: Isothermal flow at 40 degrees F RESULTS: Parameter Crane AFT Arrow 2.0 Mass flow rate using standard friction (MMscfd) t 107.8 123.1 Mass flow rate using Weymouth (MMscfd) 105.1 120.0* Mass flow rate using Panhandle (MMscfd) 133.7 118.5*

t Crane’s calculation wes the “Simplified Compressible Flow Formula” * Crane uses the actual Weymouth and Panhandle equations. AFTArrow does have these equations, but insteadsolves the governing mass and momentum equations over pipe sections. The AFTArrow Weymouth and Panhandle solutions above were obtained using the Weymouth and Panhandle friction factor correlation options in AFTArrow rather than the standard Darq- Weisbachfriction factor (as used in the first case above). DISCUSSION: The mixture properties for this example offer an opportunity to use the Chempak mixture capabilities. The problem statement does not say whether the fractions are on a mass or mole basis, but it does say in Crane that the mixture molecular weight is 20.1. This is consistent with mole fraction. AFT Arrow’s output indicates the molecular weight of the mixture is 20.1 1. As noted above (*), AFT Arrow has optional friction factor models for Weymouth and Panhandle (see Crane, page 1-8 or AFT Arrow documentation). These were used for the second and third cases above. However, regardless of what friction factor model is used, AFT Arrow differs from any of the three methods above in that it directly solves the governing equations and it does so over pipe segments. In this model, the pipe was broken into 100 segments. Since it can be easily demonstrated that AFT Arrow’s solution satisfies the mass and momentum equations for this pipe, and the solution differs from the Crane solutions, the Crane solutions do not offer

* Printed with permission. 43 SNF-5331,Rev. 0

TITLE: VERIFY6.ARO REFERENCE: Crane Co., Flow of Fluids Through Valves, Fittings, and Pipe, Technical Paper No. 410, Crane Co., Joliet, IL, 1988, Page 4-13, example 4-20 GAS: Steam ASSUMPTIONS: Example does not specify the heat transfer conditions, but it appears from the calculation procedure that adiabatic flow is assumed. The AFT Arrow model assumes adiabatic. RESULTS: Parameter Crane Modified Crane Sonic AFT Arrow 2.0 Darcy Formula Velocity Formula Mass flow rate (Ibmlhr) 11,780 11,180 11,042 - Exit static enthalpy (Butllbm) NIA 1,196 1,147.6

Exit static temperature (deg. F) NIA 317 240.7

DISCUSSION Crane does not make a distinction between static and stagnation pressure, and it appears that static pressure is usually assumed. However, the problem statement is that the source steam comes from a header, where conditions are more likely stagnation. Therefore, stagnation pressure was assumed in the AFT Arrow model. The Crane Sonic Velocity Formula yields a flow rate prediction that agrees well with the AFT Arrow prediction. The Crane Modified Darcy Formula appears to yield flow rates that are too high. If the inlet conditions in the AFT AKOWmodel are changed to static, the AFT Arrow flow rate prediction increases slightly to 11,359 (Ibmlhr). AFT Arrow uses real gas properties for the steam as specified in the System Properties window. The Crane sonic calculation assumes an isenthalpic process, which is why the exit static enthalpy is assumed to be constant at 1,196 Btullbm. With this assumption, the exit temperature is 3 17 deg. F. However, an isenthalpic assumption turns out to be poor when the Energy Equation is applied. From the First Law,

If the process is adiabatic, the heat transfer is zero and therefore, v,2 v22 h2 =h, +--- 22 Thus, the static enthalpy will drop because of the velocity increase. With this is accounted for, the exit static enthalpy decreases to 1148. This yields an exit static temperature of 240.7 deg. F, which is 76 degrees cooler. J2

* printed with permission. 44 SNF-5331,Rev. 0

AFT how2.0 Input (1) 12n199 11:01 AM Numatsc Hanford Corporation Verify6.am - Steam flow in a pip, Example 4-20 from Crane page 4-1 3

Crane Co. Flow of Fluids Through Valves, Fittings, and Pip, Tschnhl PawNo. 410, Crane Co., Joliet, IL, 19B8 Paga 4-13, exampk 4-2U Sonic &ding mmprison Steam fkw - Example aswmea adiabatic fkw Sea Vdy6.d~MS Word flk, for comphwith publirauita Tlle: Verify6.aro - Steam flow in a pip, Example 4-20 fmm Crane wge 4-13 Number Of Pipes= 1 Number Of Junctions= 2

Length March Solution Mahod with Mach Number Limls Segments Par Pip- 10 Mach Number Incront- .01 PmwreTolerance= ,3001 relatichanaa Mass Flow Rate Tolerams= ,3001 nl& change Enthalpy Tolerams- ,3001 nlativs change Flow Relaxation- .5 Pressure Rdaxation= .5

Fluid DntakAFT Standard Fluid- Steam (Simplified) Max Fluid Tempraturn Data= 15M) &g. F Min Fluid Temperature Data- xx) d8g. F Molecular Weight- 18.016amu Gal Conatant- 0.11Mn Btullbm-R Critical Pressure= 3208.22 paia Critical Temperature= 1165.09deg. R Acentric Fador ,344 Equation of Stmte= Redtch-Kwong Enthalpy Model- r3eneraIizd Specific Heal Ratio Accuracy= High Atmu@~dcPressure- 1 atm GravihtionalAcceleration- 1 g's Standard Pressure= 14.696 pia Standard Tempraturn= 60 deg. F Turbulent Flow Abow Rlryndda Numbar; 4WO Laminar Flow Below Reyndds Nu*- 2300 Specif= Heal Ratio Accuracy- High hoephric Prwsure= 1 am Gravitational Accdmlion= 1 g Standard Pnuura= 14.696pia Standard Tempratura- 60 deg. F Turbulent Flow baReynolds Number= 4wo Laminar Flow Below Reynoldo Number= 2300

Pipe Name Pip Length Length Hydraulic Hydraulic Roughness Roughness Losses(K) InitialFlow Defined Units Diamhr Dh.Unita Unb 1 Pipe I Yea I 301 fwtl 2.0671 inches1 O.wO15 feet I 8.53 I

Pip InitialFlow JUnctions Geometry Matefial Sue TYP Special Unb (UDDOWI) Condition 1 1.21 cylindrical PipI Stdl zinch Schedule40 None

45 AFT Anow 2.0 Input (2) 19ni9911l:Ol AM Numatffi Hanfcfd Corpomtlon Verify6.am - Stom flow In a pip, Example 4-20 from Crane page 413

Tank Name Obm D&baw Special Fluid Pnsaure Pmum Temperature Temperature Balance Defined Swna Condition Unita Unb Energv 11Tank I Yes I N/A I St-m (Simpifid) I 170 I psia I 368.48 I deg. F I No

Tank Balanm (Pip#l) (Pipx2) (Plpa#3) (PipM) (Pipe) (Pipeffi) (Plpe#7j (Pipem) Concenbakn K In. K Out K In. K Out K In. K Out K In, K Out K In. K Out K In. K Out K In. K Out K In. K Out 1 No (1)O.O 2 No (1)O.O

Kin KOut Kin KOut

46 M-5331, Rev.0 AFT AROW 2.0 Output (1 ) 12/1199 Numtec HanfdCorporation VerlW.aro - Steam Row In a pip.. Exampk 4-20 from Crane paga 4-1 3

Crane Co. Flow of Fluids Through Valve% Fmingl, and Pip, Technical Paw No. 410, Crane Co., Jdi. IL, 1988 Page 4-13, example 4-20 Sonic choking mpriwn Steam flow - Example assumadmW Row See Verify6.doc MS Word flle for cunprlwnwith publbhd mulb Title: Verify6.aro - Staam flow in a pip, Example 4-20 from Crane 4-13 Analyak run on: 12/7/99 11:00:22AM input File: C:\AFT PmducbWFT [email protected]

Execution Time= 8.41 and. Total Number Of Profaurn It-- 0 Total Number Of Flow ksrati0ll.- 3 Total NumkOf Enthalpy Itamtiow 3 Number Of Pip=1 Numlmr Of Junction.= 2

Length March Solution Mahod with Mach Number Limits Segment. Per Pip- 10 Mach Number Incnment= .01 Pmaaun Tobranw= .W1rdatii change Maru Flow Rate Tderam= .MM1 dah chwp EhlpyTderam= .W1dative change Flow Relaxation= .5 Pmwre Relaxation= .5

Fluid Database: AFT Standard Fluid- Stem (Slmplii) Max Fluid Temparahrre Data= 1500 dag. F Min Fluid Tmpnturn Data- 2M)deg. F Mdewlar Weight- 18.016 amu Gaa Constant= 0.110227 BluAtfn-R Ctitiil Pnuure- 3208.22 pia Critical Temperature- 1165.09dq1.R hntric Fador. ,344 Equation of State- Redlich-Kmg Enthalpy Model= Gnnaralired spec if^^ Heat Ratio Accuracy= High Atmoapbrk Pressure= 1 atm GRvitaiionalAmaleration- 1 g's Standard Profaura- 14.696 pia Standard Temperature= 60 dog. F Turbulent Flow Above Reynold. Nmb4OOO Laminar Flow Below Rey~ld.Number 2300 Speck Heal Ratio Accuracy- High Atmmpbrk Praswra- 1 atm Gravitational Amaleration- 1 g Standard Profaun- 14.696 @a Standard Tempmature= W deg. F Turbulent Flow Above Reyndd. Nwnber= 4OOO Laminar Flow Below ReynolA Number 2300 OMRll Delta Pressure = -155.300 pah Total lnflowl3.06717 Ibmkc Total Outflow= 3.06717 !bm/aec Total Emrgy lnflowl3666.42 BW Total E~gyOUMOW? 5666.42 Btuh Total Heal Tnnafefred Into System- O.wo00 Btuh

Maximum Pnuunb 170.000 p*. at Junction 1 In!& Minimum Pnuure is 14.7W poi. at Jundon 2 OutY MIuimum Static Tmpnhra is 368.480deg. F at Jundon 1 Ink( Minimum Static Tempnture ia 368.480dag. F at Juhn1 Inkt Sonic Choking Exists at Junclion 2 (Tank)

Piw 0- 47 AFT Armw 2.0 Output (2) NumaIec Hanfcfd CorpMltion VetifyS.am - Steam fiow in a pip, Example 4-20 from Crane page 413

Pip Mass Vel. Out Mach P Stag. P Stag. PSt.tc P Static TStaic TStatic H Stag. H Stag. H Static Flow # Out In Out In Out In Out In Out In (Ilnwhr) (fwthec) (ma) @.I.) bin) (psia) (den. F) (den. F) (Btwlbm) (Btwlbm) (Btwlbm) 1 11,041.8 1547.3 0.994833 170.000 63.0569 164.638 34.9468 362.238 240.738 1195.37 1195.37 1192.60

I (Btwlbm) I 1 I 1147.561 11.85971 8.5Jow)I 3329741 00191182

All Junction Tab*

P Stag. P Stag. P Stalii P Stalii T Stag. TStag. T Static T Static IJdl In I out I In I out I In I out 1 In I out I (Wk) I @o*) I (psis) I (p.ia) I (den. F) I (den. F) I (den. F) I (den. F) 1 I 170.oooO1 170.oooO1 170.waOI 170.woOI 368.4801 368.4801 368.4801 368.480 2 I 63.05691 14.70001 63.05691 14.70001 368.4801 368.4801 368.4801 368.480 SNF-533 1, Rev. 0

Parameter Crane AFT Arrow 2.0 Mass flow rate (scfm) 62.7 63.3 r

DISCUSSION: Crane does not make a distinction between static and stagnation pressure, and it appears that static pressure is usually assumed. From the problem description, the static pressure is clearly appropriate. - ’ The predicted flow rates agree very closely. The Crane prediction indicates that this pipe will have subsonic velocity at the exit and hence will not choke. However, a more proper formulation of this problem shows that sonic choking will occur. This is a good example of the limitations of simplified methods such as Crane. The discrepancy comes from how to handle the exit loss of the air as it discharges to atmosphere. The Crane solution takes the appropriate K factor, equal to 1, and lumps it together with the pipe friction to obtain an overall K factor of 7.04. However, ifthe K factor is applied at the discharge tank and not averaged along the pipe, the flow chokes. The predicted choked flow rate is 64.3 scfm. This is not drastically different from Crane’s prediction, and well within typical engineering uncertainty. But the difference, which is small here, could be larger in other applications.

J1 J2

Printed with permission. :49 SNF-533 1, Rev. 0

AFT Arrow 2.0 Input (1) 12/7/9911:08AM Numatec Hanford Corporation Verify7.aro - Air flow in a pipe, Example 4-22 from Crane page 4-14

Crane Co. Flow of Fluids Through Valves. Fittings, and Pipe, Technical Paper No. 410, Crane Co., Joliet. IL, 1988 Page 4-14. example 4-22 Air flow - Example does not specify the heat transfer conditions. so it was assumed adiabatic in this model See Verify7.doc MS Word file for comparison with published results

Title: Verify7.aro -Air flow in a pipe, Example 4-22 from Crane page 4-1 4

Number Of Pipes- 1 Number Of Junctions= 2

Length March Solution Method with Mach Number Limits Segments Per Pipe= 10 Mach Number Increment= .01 Pressure Tolerance= .WO1 relatie change Mass Flow Rate Tolerance= ,0001 relatie chanae Enthalpy Tolerance= ,0001 relative change Flow Relaxation= .5 Pressure Relaxation= .5

Fluid Database: AFT Standard Fluid= Air Max Fluid Temperature Data= 1OW deg. K Min Fluid Temperature Data= 2W deg. K Molecular Weight= 28.97 amu Gas Constant= 0.0685486 BtuilbmR Critical Pressure= 37.25 atm Critical Temperature= 132.41 deg. K Acentric Factor= ,021 Equation of State= Redlich-Kwong Enthalpy Model= Generalized Specific Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g's Standard Pressure= 14.696 psia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Number= 4OW Laminar Flow Below Reynolds Number= 23W Specific Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g Standard Pressure= 14.696 psia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Number 4000 Laminar Flow Below Reynolds Number= 2300

Pbe InDut Table

Pipe Name Pipe Length Length Hydraulic Hydraulic Roughness Roughness Losses (K) Initial Flow Deflned Units Diameter Diam. Units Units 1 Pipe 1 Yes I 101 feet 0.5461 inches 1 0.00015 I feet I1

Pipe Initial Flow Junctions Geometry Material Size Type Special Units IUp,Down) Condition 1 1,2 1 Cylindrical Pipe I Steel I 1R inch I schedule 80 I None

Assianed Pressure Table

50 SNF-5331,Rev. 0

AFT Arrow 2.0 Input (2) 12r7/99 11:08 AM Numatec Hanford Corporation Verify7.aro - Aii flow in a pipe, Example 4-22 from Crane page 4-14

Name Object Database Special Fluid Pressure Pressure Temperature Defined Source Condnion Units Assigned Pressure Yes N/A Air 19.3 psig 100 Assigned Pressure YeS N/A Air 14.7 psia 100

AssigneF~ Tempe;;; ~ LOSS:^ I Units Factor

d .F

51 SNF-5331,Rev. 0 I AFT Anow 2.0 Output (1) 12/7/95 Numatec HanfdCwpontion Vdy7.aro - Air fkw in a pip, Example 4-22 fmm Crane page 4-14

Crane co. Flow of Fluids Through Valves, F&thgs, and Pip, Technigl Papr No. 410, Crane Co.. Joliat, IL. 1988 Page 4-14, example 4-22 Air fkw - Example does not apecifythe haat bander Mnditlons, M It was ewumd adiabatic in this madel Sea Vdy7.d~~MS Word file for comparkm with publiMrcaub

Title: VeMy7.am - Air flow in a pip, Example 4-22 from Crana paga 4-1 4 Analysis run on: 12n/99 1 1 :0718 AM Input File: C:WFT ProdumWFT Armw\Verificath\verify7.am

Exclcution Time= 0.88 seconds Total Number Of Pressure Iterations=0 Total Number Of Flow Iterations= 5 Total Number Of Enthalpy Iterations- 5 Numbsr Of Pipes- 1 Number Of Junctions= 2

Length March Solution Method wah Mach Number Limb Segment. Per Pip- 10 Mach Number Inaemmt- .01 Prosaura Tderanw- .MM1 dath change Mass Flow Rate Tokmcr .JW1 relath change Enthalpy Tolaranw- .XU1 relatii chaw Flow Relamtion- .5 Pressure Reluxation= .5

Fluid Oatabaae: AFT Standard Fluid= Air MFLX Fluid Tmprature Data= 1MM deg. K Min Fluid Tsmprahlra Data= 200 dog. K Molecular Weight= 28.97 amu 08. Conatant= 0.0685486 Btuflbm-R Critical Pressure- 37.25 atm Crlticai Temperature= 132.41 deg. K Amntric Factor= .M1 Equation of State= RdlichKwong Enthalpy Model- Genemlbd Sp.dflc Heat Ratio Accuracy- High Atnmphuic Pmawn= 1 atm Gravitational Acceluaiitm- 1 g's Standard Pressure= 14.696 pia Standard Tempnture- 60 dag. F Turbulent Flow Above Reyndds Number- d x) Laminar Flow Below Reymlds Number- 2300 Specilk Heat Ratio Accuracy- Huh Atmorpherk Pressure= 1 atm Gravitational Acwleration- 1 g Standard Pressure- 14.696 pia Stadrd Temp.nture= 60 deg. F Turbuhnl Flow Abare Reynokk Number 4000 Laminar Flow Bdcw Reynold. Nu- 23LW Overall Ddtu Pnuura = -17.0789pie Tdal Inflow= 0.08Mi723 Ibinkc Tdal Ouhlow- 0.0805723 Ibm/wc Total Energy Inflows 16.4566BWs Tdal Energy OuMowl16.4566 Btuh, Total Haat Tradermd Into System- 0.mBttu's Maximum Pmsunb 33.9960 pia at Junction 1 Inlet Minimum Pressure is 14.6994 piaat Junction 2 Ink Maximum Static Temperature is 137.944 dag. Fat Junction 2 InY Minimum Static Tempraturn b 107.466hg. F at Junction 1 Inlet

Piw Out~utTabla

52 SNF-533 1, Rev. 0 I AFT Anow 2.0 Output (2) 12/7/95 Numotac Hanford Corporatkn Verify7.am - Air Row in a pip, Example 4-22 from Crane page 4-1 4

Pip Masa Vd.In Vd.0ut Mach Mach P Static P Static dP Static TStatic TStaIic K RID FbW IIn x out In Out Total In Out (scfm) (feafwc) (fdhc) (W) (paw (psid) (dag. F) (dag. F) 1 63.2892 301.94’2 662.431 0.261373 0.588539 33,9960 14.6995 19.2965 1oO.ooO 71.0659 1.wooO 6.09426

I 1 I 7.094261 0.0277289

All Jundin Tabla SNF-533 1, Rev. 0

TITLE: VERIFY8.ARO REFERENCE: Robert W. Fox and Alan T. McDonald, Introduction to Fluid Mechanics, Third Edition, John Wiley & Sons, 1985, Pages 632-633, example 12.8 GAS: Air ASSUMPTIONS: 1) Adiabatic flow, 2) Perfect gas. RESULTS

* AFTArrow does not have the ability to solve for pipe length, so the length was i? With this known length, the AFTArrow Mach number at M2 should agree with Fox & McDonald Lmd it does. The resulling massflow rate is then used as input for AFTArrow pipe #2. DISCUSSION: The predictions agree very closely. The hvo pipes in the AFT Arrow model represent the solutions to stations 2 and 3. Note that the friction factor in Fox &McDonald is the Fanning friction factor. To obtain the Darcy- Weisbach friction factor used in AFT Arrow, multiply the Fanning friction factor by 4.

J1 .I7__ Supply Pi> Location 2

\Pipe L 1-3 J3" Location b3 - Choked

Rintcd with permission. 54 SNF-5331,Rw. 0

AFT Arrow 2.0 Input (1) 12/7199 11:I3 AM Numatec Hanford Corporation Veriiy8.aro - Air flow in a pipe, Example 12.8 from Fox & McDonald page 632-3

Robert W. Fox and Alan T. McDonald Introductionto Fluid Mechanics, Third Edtion, John Wiley & Sons, 1985 Pages 632-633. example 12.8 Air flow. The two pipes represent the solutions to stations 2 and 3

See Veiiy8.doc MS Word file for canparison with published resulis

Title: Veify8.aro ~ Air flow in a pipe, Example 12.8 from Fox & McDonald page 632-3

Number Of Pipes= 2 Number Of Junctions= 3

Length March Solution Method with Mach Number Limits Segments Per Pipe= 10 Mach Number Increment= .01 Pressure Tolerance= .ooO1 relative change Mass Flow Rate Tolerance= .wO1 relative change Enthalpy Tolerance= .ooOI relative change Concentration Tolerance= .wOl relative change Flow Relaxation= .5 Pressure Relaxation= .5

Fluid Database: AFT Standard Fluid= Air Max Fluid Temperature Data= 1OOo deg. K Min Fluid Temperature Data= Ma deg. K Molecular Weight= 28.97 amu Gas Constant= 0.06855 BtuAbm-R Critical Pressure= 37.25 atm Critical Temperature= 132.41 deg. K Acentric Factor= ,021 Equation of State= Ideal Gas Enthalpy Model= Reference Specific Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm GravitationalAcceleration= 1 g's Standard Pressure= 14.696 psia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Number= 4000 Laminar Flow Below Reynolds Number= 2300 Speciflc Heat Ratio Accuracy= High Mmospheric Pressure= 1 atm GravitationalAcceleration= 1 g Standard Pressure= 14.696 psia Standard Temperature= 60 deg F Turbulent Flow Above Reynolds Number= 4003 Laminar Flow Below Reynolds Number 2300

PiDe Input Table

Pipe Name Pipe Length Length Hydraulic Hydraulic Roughness Roughness Losses (K) Initial Flow Defined Units . Diameter Diam. Units Units 1 Pipe L 1-2 Yes 4.29 meters 7.16 mm 0.0235 Explicit f 0 2 Pipe L 1-3 Yes 4.994 meters 7.16 rnrn 0.0235 Explicit f 0

Pipe Initial Flow Junctions Gwrnetly Material Sue Type Special Units (Up.Down) Condition 1 1, 2 Cylindrical Pipe Unspecified None 2 1,3 Cylindrical Pipe Unspecified None

55 SNF-5331,Rev. 0

AFT Armw 2.0 InDut Iz1 1WR911 :I3 AM Numalac Hanfofd Corporation yerifv8.aro - Air flow in a Pipe. Eamde 12.8 from Fox & McDonald Daqe 632-3

Asaigd Flow Name Object Database Special Fluid Type Flow Flow Temperature Defined Source Condition Units 3 I Location 3 -Choked I Yes I None I Air I Outtlow I 3.07637E-03 I kdsec 1 286

Assigned Pressure Name Object Database Special Fluid Pressure Pressure Temperature Temperature Dsfind Source Condition Units Units

Tank Tabh

~ Pressure Temperature Balance Defined Sourca Condaion Units ~ Units 760 deg. K 4 Tank Balance (Pipe #I) (Pipe K?) (Pipe #3) (Pipe #4) (Pipe 15) (Pipe #6) (Pipe #7) (Pipe rite) Concantration K In. K Out K In. K Out K In, K Out K In. K Out K In. K Out K In. K Out K In. K Out K In. K Out

56 ~ SNF-5331.Rw. 0

AFT Arrow 2.0 Output (1) 12/7/99 Numatec Hanford Corporation Verify8.aro - Air flow in a pipe, Example 12.8 from Fox & McDonald page 632-3

Robert W. Fox and Alan T. McDonald Introductionto Fluid Mechanics, Third Edition, John Wiley & Sons, 1985 Pages 632-633. example 12.8 Air flow - The two pipes represent the solutions to stations 2 and 3.

See Verify8.doc MS Word file for comparison with published results

Title: Veriiy8.aro -Air flow in a pipe, Example 12.8 from Fox & McDonald page 6323 Analysisrunon: 12/719911:13:38AM Input File: C:WFT ProductsWFTArrow\Verification\veiiy8,aro

Execution Time= 1.10 seconds Total Number Of Pressure Iterations= 0 Total Number Of Flow Iterations=5 Total Number Of Enthalpy Iterations=5 Number Of Pipes= 2 Number Of Junctions= 3

Length March Solution Method with Mach Number Limits Segments Per Pipe= 10 Mach Number Increment= .01 Pressure Tolerance. .wO1 relative change Mass Flow Rate Tolerance= .ooO1 relative change Enthalpy Tolerance= .ooO1 relative change Flow Relaxation= .5 Pressure Relaxation= .5

Fluid Database: AFT Standard Fluid= Air Max Fluid Temperature Data= 1wO deg. K Min Fluid Temperature Data= 200 deg. K Molecular Weight= 28.97 amu Gas Constant= 0.06855 Btuflbm-R Critical Pressure= 37.25 atm Critical Temperature= 132.41 deg. K kcentric Factor= ,021 Equation of State= Ideal Gas Enthalpy Model= Reference Specific Heat Ratio Accuracy= High 4tmospheric Pressure= 1 atm GravitationalAcceleration= 1 g's Standard Pressure= 14.696 psia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Number= 4w0 Laminar Flow Below Reynolds Number= 2300 Specific Heat Ratio Accuracy= High ktmospheric Pressure= 1 atm Sravitational Acceleration= 1 g Standard Pressure= 14.696 psia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Number= 4000 Laminar Flow Below Reynolds Number= 2300 Total Inflow= 0.01356 lbmlsec rotal Outtlow= 0.01356 Ibm/sec Total Energy Inflow= 2.660 Btu/s rotal Energy outflow= 2.660 Btds Total Heat Transferred Into System= 0.WO BtUs

Maximum Pressure is 14.70 psia at Junction 1 Inlet Minimum Pressure is 2.551 psia at Junction 3 Inlet Maximum Static Temperature is 89.90 deg. F at Junction 2 Inlet Minimum Static Temperature is 73.13 deg. Fat Junction 1 Inlet

%e Output Table SNF-533 1. Rev. 0

AFT how2.0 Output (2) 12l7199 Numatec Hanford Corporation Verify8.am - Air flow in a pip. Exar@a 12.8 from Fox & McDonald page 632-3

Pip TStag. TStaVic TSWc f Rho Out In Out Ststic In (dw.0 (dw. Kl (dw.K) wnl3) 1 296.0 293.9 287.0 0.02350 1.171 2 296.1 293.9 248.7 0.02350 1.171

All Junction TaMs

58 SNF-5331, Rev. 0

TITLE: VERIFY9.ARO REFERENCE:Michael R. Lmdeburg, P.E., Mechanical Engineering Review Manual, Seventh Edition, Professional Publications, Belmont, CA, 1984, Pages 8-1 1, 8-12, example 8.12 GAS: Unspecified except that the kvalue (ix., y) is 1.4 ASSUMPTIONS: 1) Adiabatic flow, 2) Perfect gas, 3) The gas is air, but for purposes ofthe example it only matters that the gas has k = 1.4,4) No temperature was specified, so assume 70 deg. F RESULTS:

Mz-Mach number at exit 0.35 0.356 P, - Pressure at exit @sia) 10.26 10.09

DISCUSSION: As specified, inlet conditions are known and outlet conditions need to be determined. With the known inlet conditions, an implied mass flow rate exists. To pose the problem in AFT Arrow terms, a few simple calculations are needed to obtain the mass flow rate. Once obtained, it is applied as a flow demand at the exit. The problem states that the inlet Mach number is 0.3, PI= 12 psia, TI= 70 F (assumed). From the ideal gas law, density, sonic speed and mass flow rate are:

al=m=J 1.4 [ 53.34 =)29.67 R = 1128. I f~s(sonic velocity) h=plVp4=fi(Mpl)A=I 0.6116- 3 (0.3) ( 1128.1- :I(: -0.32ft2 1=1.4631bm/s With this flow rate at the exit, the predictions agree vely closely. Note that the friction factor in Lindeburg is the Fanning friction factor. To obtain the Darcy-Weisbach friction factor used in AFT Arrow, multiply the Fanning friction factor by 4.

~ ~ Printed with permission. 59 AFT Armw 2.0 Input (1 ) 1217199 11:16 AM Numatec Hanford Corporatiin V+.m -Air f!m in a pip, bmpia8.1 2 from Lindeburg page E11

Mind R. Lindaburg, P.E. Mechanical Enpinearing Review Manual, Sevmth Edition. Profnsional Publiins, Bslmont. CA, 1984 PagwE11, M2.ruQrnple8.12 Gas flow - A~umcldto be air fcf thb mcdd, but fcf puvof the exampk il only matters that the gas has k = 1.4

Sw V+.doc MS Word flle for compariw~with prblkhod mub

TIVerify9.aro -Air flow in a pip, bmpb8.12 fmm Undnburg pmge 8-1 1

Number Of Pipea= 1 Number Of Junctions- 2

Length March Solution Mahod with Mach Number Limit. Sagmanta Per Pipa= 10 Mach Nurnter Incrarnent- .01 PnumTohnm= .OW1 mhtivo change Mass Flow Rate Thnw- .WO1 relative change Enthalpy Tdaranm= .WO1 rdntiw change Concantration Tobmnca= .WO1 relative change Flow Relaxatin- .5 Pressure Relawetion= .5

Fluid Datnbess: AFT Standard Fluid= Air Max Fluid Tempemturn Dab= 1MM dag. K Min Fluid Tampaturn Dah- 200 dw~.K Mokcukr Weight= 28.97 amu Gas Conatant= 53.3424 ft-lbfflbm-R Critical Pnuuw37.25 am Critical Temperature- 132.41 dag. K Aantrlc Fadof= ,021 Equaiin of State- Ideal Gas Enthalpy Modd= Rdwenca S~ICHeat Ratio Aecunoy- HQh Atnmphark Proasurn= 1 am GRvit.tioluI Acdomtii- 1 g's Standard Pnuuw14.696 pda Standard Ternpenturn- 60 dag. F Turbulent Flow Aban Reyndd. Nunbow 4ooo Laminar Flow BdmRaynoh Nu- 2390 Specific Heat Ratio Accuracy= High Atmosphdc Pressurn= 1 ah Grama1&x8leration= 1 g Standard Pnuura= 14.696 pais Standard Temperature= 60 dag. F Turbubnt Flow Above Reynold. Nmb4OOO Laminar Fkw Below Reyndd. Nuh~Po0

Pip Nanw Pip L-h Lsngth Hydraulk Hydraulic Rwghnoas Roughnoas Losses(K) InitiaiFlow Defined Unlt. Dkmau Dh.Unib Unit.

Pip Initial Flow Junctions Gm&y Matofid Size Typo Spackl Units 1Up.Dm) Condbll

Asaimed Flow Tabk SNF-5331,Rcv. 0

AFT Arrow 2.0 Input (2) 1217/9911:16AM Numatec Hanford Corporation Verify9.aro -Air flow in a pipe, Example 8.12 from Lindeburg page 8-1 1

Assigned Flow Name Object Database Special Fluid Type Flow Flow Temperature Defined Source Condition Units 2 I Assigned FIW~ Yes I None1 Air/ OuMowI 1.46261 Ibmlsecl 70 1 Assign:d Flow 1 Tempe:duyF I LossO I Units Model

Assigned Pressure Name Object Database Special Fluid Pressure Pressure Temperature Defined Source Condition Units

61 SI'lF-533 I. Rev. 0 AFT AROW 2.0 Output (1) 1 mt39 Numakc Hanford Corpontion VerifyS.aro - Air Row in a pip, hw8.12 from Lindeburg page 811

Michael R Lidoburg, P.E. Mechanical Enginearing Review Manual, Seventh Edition. Professiwl Publiions, Belmont, CA, 1W ~llgga11,812,examp1a8.12 Gas flow - Aasumed to be air for thh model, but fm pu- of tb example it only matters that the gas has k = 1.4 See Verify9.doc MS Word file for amparison Ahpubliihd mub Tlle: Verify9.ar.a - Air flow in a pip, Example 8.12 from Lindeburg page 8-1 1 hlyaisrunon: 121719911:21:15AM Input Fib: C:\AFT ProductsWFT ~VMcation\wrify9.ero

Execution Time- 0.22 second. Total Number Of Pressure Itere4ioru- 0 Total Number Of Flow Iteratioru= 2 Total Number Of Enthalpy lterationw 2 Number Df Pip- 1 Number Of Junctions- 2

Length March Sdution Mahod with Mach Number Umb Segments PN Pipe- 10 Mach Number Inoramant- .01 Pmasure Tderaw .Mx)1 rdatlve chnngm Mess Flow Rate Tohnw- .ooO1 nlahn chaw Enthlpy TdmnCel .wO1 nlahChMgO Flow Relaxation= .5 Preasura Ralaxation= .5

Fluid Databa8a: AFT Standard Fluid- Air Max Fluid Temperature Data= loo0 dag. K Min Fluid Temperature Data= 200 &J. K Mobouinr Weight- 28.97 umu 011. Constant- 53.3424 R-IbfAbm-R Critical Pnuure= 37.25 atm Crith1Temperature- 132.41 dag. K Awnttic Factor= ,021 Equation of State= Ideal Gas Enthalpy Model- Rderanw Specific Heat Ratio Accuracy- Huh Attxapbrk Pnurure- 1 abn Graviwional Acc&raIin= 1 g'8 Standard Pressure- 14.696 pain Standard Tempraturn= 60 dag. F Turbulent Flow AboM Rayndds Number= 4WO Laminar Flow Bekw Reynolds Nuhr2300 Specik Heat Ratio Accuracy- High Abnomphetic Prslllure- 1 atm Gravitational Accebration= 1 g Standnrd Pnuun14.696 psia Standard Tamparature= 60 deg. F TurbuM Flow Above Reynold. Nmber= 4W0 Laminar Flow Below Reynolds Nu- 2300 OMlall Dalta Pnuure = -1.76051pia Total Inflow= 1.46260 Ibmlm Total OUMw 1.46260 IbnVwc Total Energy lnflowl289.019 Ehrh Tot111 Energy OuMow= 289.019 Etuh Total Heat Trandemd Into Syatam 0.00000 Btuh Maximum Prauure m 12.oMx)pin at Junction 1 In!& Minimum Pressure is 10.0914 psia at Junction 2 lnkt Maximum Static Temparature is 79.3964 dq. F at Jundion 2 lnbt Minimum Static Temperatun h 79.3949 &J. Fat Junction 1 Inlet

Pim Outout Table

62 SNF-5331.Rhr:

Pipe Mach Mach Mass P Static P Static Vel. Rho X In #Out Flow In Out &&In StdiIn

-1 -1 UI Junction TaMe

bia) (B3iFd 1 I 12.m 12.m 2 I 10.0914 10.0914

63 SNF-5331, Rev. 0

TITLE: VERIFYlO.AR0 REFERENCE: Michael R. Lindeburg, P.E., Mechanical Engineering Review Manual, Seventh Edition, Professional Publications, Belmont, CA, 1984, Pages 8-12, 8-13, example 8.14. GAS: Methane ASSUMPTIONS: 1) Isothermal flow, 2) Perfect gas RESULTS:

:.. ,/: . ~ cb::, Parameter’. ’ ’ Lindeburg., AFT Arrow 2.0 :.$&$,. ’ . Mass flow rate (Ibds) 456 I 418 I I I I

DISCUSSION: The predictions agree closely. Part “a”makes a comparison to the Bernoulli equation, which Arrow does not solve. So part “a” was skipped. Note that the friction factor in Lindeburg is the Fanning friction factor. To obtain the Darcy-Weisbach friction factor used in AFT Arrow, multiply the Fanning friction factor by 4.

J1 J2

* Printed with permission. 64 SNF-5331. Lbv. 0

AFT how2.0 Inpi (1) 12/7/9911:31 AM Numatec Hanford Copration

VerifylO.aro ~ Mathane flow in a pip. Example 8.1 3b from Lindeburg page E12

Michael R. Lindeburg, P.E. Mechanical Enginwring Rmivisw Murml, Seventh Edhn, Prof&nal Publications, Belmont, CA, 1984 ~ageaai2,813.examp* 8.13b Mahtane Row - Part "a" vmccinpming to Bamwili, which AROW doe8 not soh, (10 it vias skippsd. This model applies to part "b" SW VdylO.dm MS Word file for mmpriwnwith puMirswb THIa: VerifylO.aro - Methane Row in a pipe, Exsmple 8.1% from Lindeburg page 812 Number Of Pipes- 1 Number Of Junctions= 7.

Length March Solution Method Ah Mach Number Limits Sagmenta Per Pip10 Mach Number Increment= .01 Pnuure Tolerance= .wO1 nlative change Macu Flow Rata Tolerance= .ooO1 raldiva change Enthalpy Tolerance= .wO1 relative change Flow Relaration- .5 Pnurura Relaxation= .5

Fluid Dhba~:AFT Standard Fluid= Methane Max Fluid Temperature Data- 4M)deg. K Min Fluid Ternparatun Data= 150 dag. K Mokcuhr Weight- 16.043 emu Gas Conatant- 96.3243 ft4bffibm-R Critical Pnurun= 4.- MPa Critical Temperature- 190.555 dag. K kmtric Factor- ,008 Equation of State= Rdlii-Kwong Enthalpy Modal- Gmomlizd SpdkHeat Ratio Accuracy- Huh Atmospheric Pressure- 1 atm Gravitatimal Acwlsration- 1 g's Standard Pressure- 14.696 psis Sbndard Temperature= 80 dag. F Turbulent Flow AboM Reynolds Nunbaf- 4OOO Laminar Flow Bdow Reynolds NumbstP 2300 Specific HaR& Accuracy= Huh Abnolph&c Pressure- 1 atm Gravitnlional kcelerntion= 1 g Standard Pnuure- 14.696 pia Standard Tempratura- 60 deg. F Turbulent Flow Above Reynolda Nunb4OOO Laminar Flow Below Reynolds Nu&er= 2300

Pice lnout Tabk

' Pipe Name Pipe Length Lenglh Hydraulk Hydraulic Roughness Roughness Losses(K) InitialFlow I Defined Unit. Dianwter Diam.Units Units 1 Pipe I Yes I 751 miles] 401 inches 0.w9361 Explicitfl 01

Pipe lnilial Flow Juncths Geometry Matehl Si0 Typ Spcial Unls (Up.D~rm) Condtion 1 I 1.2 I Cgindrical PipI unemcmai I I I None

Aaaiand Preaaure Table

65 SNF-S33 1, Rev. 0

AFT Arrow 2.0 Input (2) 12/719911:31 AM Numatec Hanford Corporation VeiifylO.aro - Methane flow in a pipe Example 8.13b from Lindeburg page 8-12

Assigned Pressure Name Object Database Special Fluid Pressure Pressure Temperature Defined Source Condition Units 1 I Assigned PressureI Yes I N/A 1 Methane I 650 I psia I 40

1 AssigneTessure ~ Temperaa;eF ~ Loss: Units Factor 1

de .F

66 SNF-5331,h.B AFT AROW 2.0 Output (1) 1urn9 Nun~atecHanford Corpmtion VerifylO.am - Methane Row m a p+, Exampla 8.13b from Lindeburg page 812

Mkhael R. Lindeburg, P.E Mechanical Enginwring Review MmuaI. Seventh Editkn, Pmfedaionsl Publications, Belrnont, CA, 1984 Pagaa12,813.~mplo8.13b Mehtane flow - Part "a" was compdng to Bmlli,which AROW dcua not &e, 80 it was skipped. This model applies to part "b"

Soo VarifylO.doc MS Word Rla for compahwl(h publiimuka THIO: VarifylO.am - Methene Row in a pip, Ex8mple 8.13b from Lindeburg page a12 Analysis run on: 1WB9 11 :29:53 AM Input File: C:WPrnd~~tslAFT MVerMcetim\wrifylO,aro

Execution Tim= 0.94 second. Total Number Of Pressure Iterations- 0 Total Number Of Flow Iterations- 5 Total Number Of Enthalpy Iterations= 5 Number Of Pipes- 1 Number Of Junctions= 2

Length March Sol* Method with Mach Number Limb Segments Per Pip- 10 Mach Number Incnment= .01 Pr-ure Tolerance- .WO1 rektlva chanw M~MFlow Rata Tokrsncw .ooO1 Michange Enthalpy Tobranca- .ooO1 relative cham Flow Relamtion= .5 PnuunRelaxation= .5

Fluid Mabiue: AFT Standard Fluid- Methane Max Fluid Temperalure Ma= 400 dag. K Min Fluid Tempemtun Data= 150 deg. K Molecular Weight= 16.043 amu Gar Constant- 96.3243 fi-IbfflbmR Crith1 Pnuure- 4.5988 MPa Critical Temperature- 190.555 dag. K Acentric Factor- .OM) Equalim of State= Redlich-Kwong Enthalpy Model= Generaliued SpccKic Heat Ratio Accuracy- Huh Atnm8phheric Pnuure= 1 atm GnviWionalhlamtion= 1 g'e Standard Pnuun- 14.696 pd. Standard Tempantun= 60 dag. F Turbulent Flow Abow Reynolds NmW4WO Laminar Flow Below Reynokk Nuhr- Po0 SpcMc Heat Ratio Accuracy= High Mnmphheric Prwsure= 1 atm Gravitational Accebntion= 1 g Standard Pressure- 14.696 pda Standard Temperature- 60 dag. F rurbukm( FIW R~YW NWW- UMO Laminar Flow Bdow Reymk Nu- 2300 DmllD& Pr-ure = -199.926 paia Tdal inflowl478.029 1bmh.o Total OMW= 47.3.029 ibm~wc Tdal Energy Inflow 182,255. BtuE. rdai EWY OMCW is,ms. EWS rhi eat Trendwed into S~~OIIP3161.81 ~tuh

Maximum Pressure is 65O.W piaat Junction 1 Ink4 Minimum Preasun m 45O.ooO pd. e$ Juncfion 2 lnbt Maximum Statk Tmpnture is 40.0463dag. F at Jut% n 2 it Minimum Static Tempanturo is 40.0187 dag. F at Jundon 1 In

Pim Ot#mi Tab1e

67 SNF-S33 1. Rev. 0

Pipe Maar Vel. in Vel. Out PSI& P Static Flow in Ou(

All Junction Table

68 SNF-5331,Rev. 0

Parameter Saad AFT Arrow 2.0 M2- Mach number at exit 0.685 0.689 MI -Mach number at inlet 0.347 0.347 PI-Static pressure at inlet (Wa) 306 308.1 TI -Static temperature at inlet (deg. K) 312.76 312.9

DISCUSSION: As specified, exit conditions are known and inlet conditions need to be determined for specified volume flow at exit. With the known exit conditions, an implied mass flow rate exists. To pose the problem in AFT Arrow terms, a few simple calculations are needed to obtain the mass flow rate. Once obtained, it is applied as a flow demand at the inlet. The problem states that the exit volume flow rate, Q2,is 1000 m’/min, P2= 150 Wa, T2 = 293 K. From the ideal gas law, density, and mass flow rate are: p2 - 150kPa p2=-- = 1.784 ks/m3 RT2 (.28682)(293 K)

In AFT Arrow, the discharge pressure can be specified. The temperature can be specified at the exit junction, but the actual discharge is what is displayed for the pipe exit. The pipe exit temperature depends on the inlet temperature and the thermodynamic process in the pipe, which is adiabatic. Therefore, to solve this problem the inlet static temperature at J1 must be guessed until the pipe delivers 293 K at the exit. This results in the 3 12.9 K displayed in the above table. All results agree closely. Note that the friction factor in Saad is the Fanning friction factor. To obtain the Darcy-Weisbach friction factor used in AFT Arrow, multiply the Fanning friction factor by 4.

J1 v J2

* printed with permission 69 AFT Anow 2.0 Input (1) 12/7/99 11:46 AM Nurm(bc Hanford Corporation Vdyll.aro-Airnowmmpk5.2fmm seadpp.21%~15

MichlA. Saad Compmaible Fluid Flow, 2nd Edh,Pnntiw-H.ll, Engkwood Clifh. NJ, 1993 Pa@ 213215, mmpla 5.2 Air flow - Example wunmsdimbdic fiow and air b a ptfact gas sea verify1 1.doc MS Word fib for mpatiaon wim pub~hhadrmub Ttle:Verifyll.am-Air~owmmpla5.2fmmSudpp.213215

Number Of Pipes= 1 Number Of Junctions- 2

Length March Solution Mahod wivl Mach Number Limih segnda Par Pip10 Milch Numkr Increment= .01 Pnuurn Tobnnw- ,0001 relath change Mass Flow Rata Tdarsnocr ,0001 nbtb change Enthalpy Tolerance= ,0001 mMichanae Flow Relaxation- .5 Prapure Relaxatinn- .5

Fluid Data!xaa: AFT Standard Fluid- Air Max Fluid Temperature Data- 1000 dag. K Min Fluid Tampunturn Data= 200 deg. K Mctacuku Weight- 28.97 amu Gas Constant- 0.2868 kJbK Critical Pressurn= 37.25 atm Critical Temperature= 132.41 6ag. K AwnMc Factor= ,021 Equation of State- Ideal Gas Enthalpy Model= Rd~enca Specific Heat Ratio Accuncy- High Alnl~~NiCPnuure- 1 atm Grnvilational Acceleration- 1 g’s Standard Pnuure- 14.696 pi. Standmrd Tempaturn= 60 dag. F Turbubnt Flow AboM RaymbNun* 4WO Lamindr Flow Bdow Reynolds Nu- PW Speck Heat Retio Accuracy= Hioh Atmapbrio Pnuuw1 atm Gravitational Accslmtion= 1 g Standard Prp.ursll4.696 pala Standard Tempaturn= 60 deg. F Turbulent Flow AboM Reynolds Nun* 4W0 Laminar Flow Bdow Reynolds Nunhr= 2300

Pip Name Pip Length Lor& Hydraulic Hydnulii Roughness Roughness Losses(K) InitialFlw DdMd Unib Diameter Dh.Unih Unit.

Pip InitilFlow Junctions Geomtty Mat~bl Sb. Typ Spacial Unh (U~.Dorml COndfOn SNF-5331,Rev. 0

AFT Arrow 2.0 Input (2) 12/719911:46AM Numatec Hanford Corporation Verify11 .aro - Air flow example 5.2 from Saad pp. 213-215

Assigned Flow Name Object Database Special Fluid Type Flow Flow Temperature Defined Source Condition Units 1 I Assigned Flow I Yes I None I Air1 Inflow/ 1783.961 I kg/min 1 312.9

! Assign;d Flow ~ Temperra; 1 Loss0 Units Model

Assigned Pressure Table

~~ Assigned Pressure Name Object Database Special Fluid Pressure Pressure Temperature Defined Source Condition Units 2 I Assigned Pressure 1 Yes I N/A/ Air1 150 I kPa I 293 1 Assigne7;re 1 Ternpe;rat;K I LossO 1 Units Factor

71 SNF-9331.h. 0 AFTArmw2.00utpvt (1) 1211199 Nwnatec Hanford Corporation V~ll.am-Airfinummple5.2fmmS.adpp.213-215

Michd A Sed Campasibla Fluid Flow, 2nd Edition. PrentiwHail, Englswood Cllfh, NJ, 199J Page 213-215, example 5.2 Air flow - Example easumas adiaktic finu and air ia a @oct gaa Sw Verify1 1 .doc MS Word file for cornpatisan with put4iiM nwb

TiVerify1 1 .am -Air flow wimple 5.2 fmm Saad pp. 21 2-21 5 Analpi. fun on: 17~7/991 1:36:43 AM Input File: C:WProductsWFT ARau\Verification\vuifyll.am butionTim- 2.85 sewn& Total Number Of Prewurn lterslions- 0 Total Number Of Flow Iterations= 2 Total Number Of Enthalpy Iterations- 2 Number Of Pip=1 Number Of Junctions- 2

!~I&IUI March Solution Method with Mach Number Llmits Sog- Per Pip= 10 Mach Numbar Inemment- .01 Pnuure Tolerance- .ooO1 miahchange Mau Flow Rate Tolerance= .wO1 mlatii changa Enthalpy Tolerance= .ooO1 reldicham Flow Relaxation= .5 Prwaun Relawation= .5

Fluid Database: AFT Standard Fluid- Air Max Fluid Tempenhlre Oata- loo0 deg. K Min Fluid Tempmture Deta- 2W deg. K Moleculw Waight- 28.97 amu Gam Co- 0.2868 kJ&K Critical Prwauw 37.25 atm Crhal Temperature- 132.41 deg. K hntric Factor= ,021 Equation of State- Ideal Gas Enthalpy Modd- Rafem Sp.cmC Heat Ratio Accuracy= High Atmwpheric Prewun- 1 B(m Qmvitatirnal Acceleration= 1 g's Standard Pnuura. 14.696 pia Standard Temperature- 60 dog. F Turbulent Flow Above Reynokk Nwb4OOO Laminar Flow Bdow Reynold. Number- pw Specific Heat Ratio Accuracy- High Atlmqitldc Prewure- 1 atm Gmvitational Acceluatian= 1 g Standard Pnuura= 14.696 psia Standard Temperature- 60 deg. F TUrbUlen( Flow heRaynokk Number= 4OOO Laminar Flow Below Reynolds Number= 2300 OmUDelta Pnuura = -128.9 kPa Total lnflowl29.73 kglwc Total Outflow= 29.73 kghc Total Energy Inllw 14,287 kW Total Energy OUMOWS 14,287 kW Total Heat Tmmformi Into System- 0.ooO kW

Maximum Pnuuni. 33.0 kPa at Junction 1 inlet Minimum Prewure is 150.0 kPa at Junction 2 Inlet Maximum Statk Temperature ia 320.3 dog. Kat Junction Minimum Static Temperatun b 293.0 dog. Kat Ju~m

Piw Out~utTak

72 AFT Armw 2.0 Output (2) Numates Hanford Corpontion V~ll.am-Airflowexample5.2fromSaadpp. 213-215

Pip Mass Vel. Vd. Maoh Mach PStatic P Static TStatic T Static Rhostatic Flow In Out #In #Out In Out In Out out ~ksbsc) (meterdsoc) (malsrs/8ec) fkPa) (kPa) (deg. K) (deg. K) (kgh3) 1 29.73 122.6 235.8 0.3469 0.6891 308.1 150.0 312.9 293.0 1.764

Jd P Static P Static T Static TStatic In Out In Out (kPa) fkPa) (&a .K) fdw.K) 1 308.0 308.0 312.9 312.9 2 150.0 150.0 m.0 268.0 ~

SNF-5331,Rev. 0

Parameter Saad AFT Arrow 2.0 Mass flow rate when choked (kg/s) 2.1 1 2.10 MI-Mach number at inlet 0.603 0.604 PI - Static pressure at inlet (ma) 2.106 2.114 TI-Static temperature at inlet (deg. K) 427.8 429.6 P2.s~kc- Static back pressure for choking (MPa) 1.203 1.217

DISCUSSION: All results agree closely. The AFT Arrow static pressure below which choking occurs is the pipe exit static pressure. Note that the friction factor in Saad is the Fanning friction factor. To obtain the Darcy-Weisbach friction factor used in AFT Arrow, multiply the Fanning friction factor by 4.

J1 J2

* Printed with permission. 74 .'sNFJ331.Rev. 0

AFT AROW 2.0 Input (1) 17~719911 :54 AM Numatec Hanford Cocpontlon Verify12.am - Air flow mmpta 5.5 from Sadpp. ZX-227

Mlchel A. Sad Compfassible Fluid Flow, 2nd Edh.PmnIke-Hall, Engkwwd Cliff., NJ, 1993 Page 226-227, mmpb5.5 Air flow - Example assumed adlaWk flow and alr la e pried (pa Sed Varlfyl2.d~~MS Word fib for wmprim with publiiraub Tiib: Verjfyl2.a~- Air flm mmple 5.5 fmm Sad pp. 22S227 Number of Pip- 1 Number of Junctions- 2

Lmgth March Solution Mahod wim Mach Number Limb SqmenEl Per Pipe= 10 Mach Number Increment- .01 Prassura Tobrame= .ooO1 ml&n chga Ma- Flav Rate Tderanoe- .ooO1 rdative change Enthalpy Tole~nnoe=.ooOl ml&n change Flow Rsbxation- .5 Proawn Relaretion= .5

Fluid Database: AFT Standard Fluld- Air Max Flub Tmwntun Data= 1ooO ho.K Min Flud TamhraData- 2M) dag.-K Molecular Wught- 2E 97 mu Gas Conatane 0 2868 kJAq-K Cnbal Pressure= 37 25 8tm Cr*al Tempmhln- 132 41 dag K AoanMc Factor= 021 Equ*bn of State- 16.1 Gu Enthalpy Modd= Rdem Spoufic He# R8tb Accuracy= High Atmapbric Pnarura- 1 atm Gmvlt.tDnsl Acwbration- 1 g's Standard Pmuure= 14 696 pia Standard Tampamturn= 60 dag F Turbulent Flow AboM Reynolds Nunber= 4ooo Laminar Flow B.bw Rey~kbNumber 2300 SP.afr Heat Ratio Accuracy- Hlgh Atmmpherk Preuum= 1 .tm Gmvhtional Accakmtion= 1 g Standard Pnuuw 14 696 pia Standard Tampemturn- W dag F Turbulnt Flow Above Reyndd. Nunkc 4ooo Laminar Flow Below ROYM~Nu- 2390

Pipe Name Pipe Length Le& Hydraulic Hydraulic Roughma Roughnoas Losws(K) lnitbl Flow Defined Unib Dlam~ter Dhm.Uniti Units 1 Pip I y-1 0.61 -I 0.04 metem I 0.02 I Explicit f I 01

Pipe InMalFlow Junctbru GmnWy MatWl Sm Typ Spacial Unb (UD.DM) Condltkn -1 I I, 2 I cvlindrial Pipe I unspeci~I I I Nom SNF-5331,Rev. 0

Assigned Pressure Name Object Database Special Fluid Pressure Pressure Temperature Defined Source Condtion Units 1 Assigned Preasure Yes N/A Air 2.7 MPa 460 2 Assigned Pressure Yes N/A Air 0.1 MPa 293

76 SNFJ331. Rev. 0

AFT Armw 2.0 Output (1) 1mt39 Numatec Hanfd Capwaion Vefifyl2.am - Air flow mmp5.5 fmm Saad pp. 226-227

Michd A Saad Compressible FiuM Flow, 2nd Edition, Pnntica-Hall, Englwocd Cli, NJ, 1993 Page 226227. axample 5.5 Air h- Example aasumes admWflow and air k a petfad gas SWV8rifylZ.docMS WWdfllefMmmprl.onwithpubllahadMulh

Title: Varifyl2.aro ~ Air flow axampla 5.5 fmm Soad pp. p6227 hlyaia run on: 1WB9 11:50:14 AM Input File: C:WFT ProdudaWFT ~V~cation\wrifyl2,aro

ExeAian Time= 1.43 ssconds Total Number Of Pmsaure Itantiom= 0 ToIal Number of Flow Itentbns= 2 Total Number Of Enthalpy Iterationa- 2 Number of Pipea= 1 Number Of Junctions- 2

Length March Solution Mahad with Mach Number Limb Sqwmta Par Pipe= 10 Mach Number lncment- .01 Pmsaura Tc4emw= .ooO1 mblh change Mau Flow Rata Tolerance- .MM1 mblh change Enthalpy Tolerance= .MM1 mblh change Flow Relaxation- .5 Pnuun Relaxation= .5

Fluid Databaas AFT Standard Fluid- Air Max Fluid Temperatun DaIa- loo0 dag. K Min Fluid Tamprature Data= 200 dag. K Molacuhr WaigM= 28.97 amu Ow Constant- 0.2868 kJ/ke-K Critical Pwure= 37.25 atm Critical Temperatun= 132.41 dsg. K Acsntrlc Fador= .M1 Equation of State- Idoal Gas Enthalpy Model- Rafermw Specific Heat Ratio Aoouracy- High Atmoopharic Pressure= 1 atm Gravitational Acc&ration= 1 g's Standard Pmssuw 14.696 pia Standard Temperature- 60 dq. F Turbulent Flow Ahow Reynolds Nunb4WO Laminar Flow blow Reyncida Nurrbw 23M) SpaciTw Heat Ratio Accuracy= High Atmapherk Pmure= 1 atm Gravitational Accskation- 1 g Standard Pmun14.696 pia Standard Temperatun= 60 dq. F TurbuM Flow Ahow Rayndda Numb- 4wo Laminar Flow Bsbw Reyncida Number- 2300 Overall Deltp Pmsaun = -2600. kPa Total Inflow- 2.1W kghc Total OuMw 2.103 kg/sea Total Energy Inflw 1311. kW Total E~rgyOulilcw 1311. kW Total Heat Trandmed Into Syatanr -2.662E-06 kW

Maximum Prwsura h 2280. kPa aI Junction 2 InM Minimum Pnuure is 100.0 kPa at Junotibn 2 Outla Maximum Static Temperatun k 460.0 dag. Kat Junctim 1 Ink Minimum Static Temperature is 293.0 dag. K at Junctkn 2 Ink

"'CAUTION"' Junction 2 h dcwmtmm of a .hock, atatic pmpertia cannot be alculated at junction

Sonic Choking Edat. at Junction 2 (Assigned Pmun)

Piw OutDut Tak& n SNF-5331,Rev. 0

AFT AROW 2.0 Output (2) 1m/99 Nurrmt.c Hanfcfd Cqotutbn Verifyl2.am - Air flow example 5.5 from Saad pp. 226-227

Pip Maw Vd. Vd. Mach Mfsh P Statii P Static TStatic T Static Flow In Out Xh #Out In Out In Out

Jct PStatic PSW TSWi TSWIc In Out In Out (kPa) (kPa1 (dw.K) (dw . K) 1 2114. 2114.3 429.6 429.6 2 2280. 100.0 293.0 293.0 ~

SNF-5331,RCV. 0

TITLE: VERIFY 13.ARO REFERENCE: Michel A. Saad, Compressible Fluid Flow, 2nd Edition, Prentice-Hall, Englewood Cliffs, NJ, 1993, Page 270, example 6.5 GAS: Natural Gas ASSUMPTIONS: 1) Isothermal, 2) Perfect gas, 3) Natural gas can be represented with methane RESULTS:

The problem states that the inlet Mach number is 0.1, PI= I MPa, TI= 293 K. From the ideal gas law, density, sonic speed and mass flow rate are: 4- 1MPa a=-- = 6.585 kgh3 RTI (.5179&)(293K)

al = JYRT~= \/- = 445.9 m/s (sonic velocity) li2 = p1QA = pI(M1al)A = (6.585- 3 (0.1) ( 445.9- 3: -0.0S2m2 1= 1.4764 kg/s The first pipe represents the pipe from point 1 to 2. The second pipe represents the pipe from point 1 to the choking pint. AFT Arrow does not solve for pipe length. To obtain the maximum pipe length, different lengths must be guessed with lengths that exceed sonic flow discarded. All results agree closely. The AFT Arrow static pressure below which choking occurs is the pipe exit static pressure. Note that the friction factor in Saad is the Fanning friction factor. To obtain the Darcy-Weisbach friction factor used in AFT Arrow, multiply the Fanning friction factor by 4.

'Riadnithpcrmissim. 79 SNF-5331,Rw. 0

J1 PI> J2

J3 P2> J4 ~ L1-Choke

80 SNF-5331,Rw. 0

AFT Armw 2.0 Input (1) 12/7/9911:59Ab NudacHanfd Corporation VSrifYl3.am Natural Gas flow example 6.5 from Saad pp. 270

Michel A. Saad Compressible Fluid Flow, 2nd Edition, Prentiw-Hall. Englewood Cliis, NJ. 1993 Page 270,examp 6.5 Natural OM flow - laothsmwl flow, psrf& gas, assumed to be methane See Verifyl3.doc MS Word file for comparison with published resuhs Title: Verifyl3.arO Natural Gas flow example 6.5 from Saad pp. 270

Number Of Pip=2 Number Of Junctions- 4

Length March Solution Mahod with Mach Number Limits Segments Per Pip- 50 Mach Number Incnmcnt= .01 Proasurn Tdsram- .ooO1 relative change Maw Flow Rate Tolerance- .ooO1 relative change Enthalpy Tolennw= .W1rdativs change Flow Relamtbn- .5 PmureRelaxation- .5

Fluid Data-: AFT Standard Fluid= Melhane Max Fluid Temperahlre Data= 4M)deg. K Min Fluid Tempraturn Data- 150 dag. K Mokuler Weight- 16.043 amu Gas Constant= 0.5179 kJ&K Critical Proawn- 4.5988 MPa Critical Tmpratuw 190.555 dog. K hntric Fedor= .WE Equation of State= ldsal OM Enthalpy Model- Refmw Specific Heat Ratio Accuracy= Hah Atmmpheric Pressure- 1 atm Gravitational Acw!antion= 1 g's Standard Pressure- 0.1 MPa Standard Temperature- 293 deg. K Turbulent Flow Above Reyynoldr Nun* 4OOO Laminar Flow Below Reynolds Numbsr= 2300 Specific Heat Ratii Accuracy= High Atmmpiwric Pnuure= 1 atm Gravitational Aocdmtiin- 1 g Standard Proasurn= 0.1 MPa Standard Temperahlra= 293 dog. K Turbulent Flow Abvo Reynolds Number 4ooO Laminar Flow Below Reynolds Number 2300

Pip Name Pip Length Length Hydraulic Hydraulic Roughness Roughness Losses (K) Initial Flow Defined Units Diameter Diam. Units Units 1 L 1-2 YM W meters 0.08 meters 0.008 Explicit f 0 2 L1-Choke Yea 709.814 meters 0.08 meters 0.008 Explicit f 0

Pip InitialFlow Junctions Gwmdry Material Size Type Special Unb IUp,Dorm) Condtion .1 1.2 Cvlindrical Pipe Unspecified None 2 3,4 Cylindrical Pipe Unspecified None

81 SNF-5331,Rw. 0

AFT Anow 2.0 Input (2) 1217El9 1159AM Numatac Hanford Corporation Verify1 3.aro Natural Gas flow example 6.5 from Saad pp. 270

AMikned Flow Name Object Datahse Special Fluid Type Flow Flow Temperature Defined Source Condtion Units 2 Assigned Flow Yes None Methane Outflow 1.4764 kgIsec 293 4 Assigned Flow Yes None Methane Outflow 1.4764 kgIsec 293

Assigned Pressure Name Object Database Special Fluid Pressure Pressure Temperature Defined Source Condition Units 1 Assigned Pressure Yes NIA I Methane I I1 MPa I 293 3 Assigned Pressure Yes NIA I Methane I I1 MPa I 293 I

Assigned Pressure Temperature Loss Units Factor 1 deg. K 0 3 deg. K 0

a2 SNF-533 1, Rev. 0

Execution Time- 1.W wcond. Total Number Of Pressurn Ikeralha=0 Total Number Of Flow Itemtiom= 2 Total Number Of Enthalpy Itera(ii2 Number Of Pips= 2 Number Of Junctions= 4

Length March Solution Mahod with Mach Number Limb I Sqmenta Par Pipe- 50 Mach Number Incmnt= .01 Pruawre Tolerance= .oM)1 Michaw Maw Fkw Rate Tdemnce- .KO1 reMive change Enthalpy Tokranca= .oOol relative chaw Flow Relaxation- .5 Pressure Relaxation= .5

I 63 SNF-533 1. Rev. 0

Pip Rhostatic Vd. Vel. Out SonicIn ScnicOut

Internal Piw Resulh, Pipe 1, Name: L 1-2

84 SNF-5331,Rev. 0 AFT Armw 2.0 Output (3) 1 m199 Numatac Hanford Capontiin VoMyl3.am N.turalQas fbw exampla 6.5 from Saad pp. 270

Internal Pim RMU~Pipe 2, Name: L1-Choke

85 SNF-533 1. Rev. 0

~ ~~ AFT Anow 2.0 Output (4) 12/7/99 Nurnatm Hanfwd Corpantii Verifyl3.aro Nlltunl Gas fiow mmpk 6.5 from Saad pp. 270 SNF-5331,Rev. 0

All Junction Table

I-87 SNF-533 I, Rev. 0

Parameter Tilton AFT Arrow 2.0 Mass flow rate (kgk) 2.7 2.716

J1 J2 SNF-5311, Rov. 0

AFT Armw 2.0 Input (1) 12/7/9912:12PM Numatac Hanford Corporation Verifyl4.am - Air flow Example 8 fmn Perry's pp. 625

RhrtH. Perry and Don W. Green Edtora, Author Jam. N. Tilton Perry's Chemical Engineer's Handbook, Seventh Edition Page 625,example 8 Air flow - Adiabatic Sw Verifyl4.doc MS Word file for comparison with publiiskd muits

Tale: Verifyl4.aro - Air flow Exampb 8 from Perry's pp. 625

Number 01 Pipes= 1 Number M Junctions= 2

Length March Solution Method with Mach Number Limit. Segments Per Pipe= 10 Mach Number Increment= .01 Preesure Tolerance= .ooO1 relmchaw Mass Flow Rate Tolerance= .ooO1 relath change Enthalpy Tolerance= .ooO1 relative change Flow Relaxation= .5 Pressure Rataxation= .5

Fluid Database: AFT Standard Fluid= Air Max Fluid Temperature Data= loo0 deg. K Min Fluid Temperature Data= 2M) deg. K Molecular Weight= 28.97 amu Gas Constant= 0.2868 kJkg-K Crilical Pressure= 37.25 atm Critical Temperature= 132.41 deg. K Acentric Factor= ,021 Equation of State= Ideal Gas Enthalpy Model= Reference Sped0 Heat Ratio Accuracy= Standard Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g's Standard Pmure= 0.1 MPa Standard Temperature= 293 deg. K Turbulent Flow Above Reynolds Number= 4ooo Laminar Flow Below Reynolds Numb- 2300 Specifw Heat Ratio Accuracy- Standard Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g Standard Pressure= 0.1 MPa Standard Temperature= 293 deg. K Turbulent Flow Above Reynolds Nmkr- 4WO Laminar Flow Below Reynolds Number= 2300

Pipe Name Pipe Length Length Hydraulic Hydraulic Rwghness Rpughnesr Losses(K) InitialFlow - Defined Unit. Dmder Dhm.Unita Unit. 1 Pipe I Yes I 101 meters 2.0671 inchI 0.0188 f I 21

Pip Initial Flow Junctions Goodry MatMia1 Ske Type Special Units (UD.DM) Condhn 1 1,~I cylindrical Pipa I SMI 2inchI schedule40 None

Assigned Pressure Table

89 SNF-533 1, RCV.0

AFT Arrow 2.0 Input (2) 12/7/9912:12PM Numatec Hanford Corporation

Verify14.aro ~ Air flow Example 8 from Perry's pp. 625

Assigned Pressure Name Object Database Special Fluid Pressure Pressure Temperature Defined Source Condiiion UnRs 1 Assigned Pressure Yes NIA Air 1 MPa (9) 20 2 Assigned Pressure Yes N/A Air 1 atm 20

Units Factor

de .C SNP-5331. &. AFT AROW 2.0 Output (1) 12n199 Numatec Hanford Ccfporation Verifyl4.am - Air ikw Emmple 8 fmn Parry's pp. 6-25

Rc4aft H. Parry and Don W. Grwn Editon, Author Jaw. N. Tilton Parry's Chemical EnpiweHandbwk, Sewnth Edition Page 6-25,example 8 Air ikw -Adiabatic

SW Venfyl4.doc MS Word file for eompfiaon wlth publiirsrub

Tine: Verify14.am -Air flow Erampb 8 from Peapp. 6-25 Analylhrunon: 12/7BS12:13:43PM Input Fib: C:\AFT Pmducta\AFT Anow\Vaifiation\vsril4.am

Execdin Timi 1.65 MCO~~. Total Number Of Pnuunlteratiw*- 0 Total Number Of Flow Iterations= 2 Total Number Of Enthalpy Iteratiom- 2 Number Of Pipes- 1 Number Of Junctions= 2

Lemh March Solution Mathod with Mach Number Limb Sqmnta Par Pipe= 10 Mach Number Incmnt= .01 PmureTderanrmp .ooO1 dative change Maw Flow Rate Tolerance- .Wo1 &ha chap. Enthalpy Tderanca- ,3301 relative changa Flow Relaxation- .5 PnuunRdaxation- .5

Fluid Databa~:AFT Standard Fluid- Air Maw Fluid Tempsrature Data= 1000 dq. K Min Fluid Temperature Data- 200 @. K Molecular Weight= 28.97 amu Gas Conatant= 0.2868 kJkg-K Critical Preswre= 37.25 atm Critical Temperature- 132.41 dq. K bmtric Factor= ,021 EquaUon of State- Ideal Gas Enthalpy Mdd= Refennw SpeciRc Heat Ratio Accuracy= Standard Atmospheric Pmure- 1 atm Gravitatimal Acderakn= 1 g's Standard Prauure- 0.1 MPa Stardard Temperature= 2% dog. K Turbulonl Flow AboM Reynold. Nunber- 4ooo Laminar Flow Below Reynold. Nunbr= 2300 Spscilic Heal Ratio Accuracy= Standard Atmcapherk Preswre- 1 atm Gravitational Acwleration= 1 g Standard Pmum- 0.1 MPa Standard Temperature- 293 dq. K Turbulent Flow Above Reynolds Nun- 4OOO Laminar Flow Bdow Reyndda Nunhr- 23M) Overall Deb Pmesure - -1.330 MPa Total lnflowl2.716 Total OuMowl2.716 kgba: Total Energy lnflowp 1230. kW TdEnergy 0- 1230. kW Total Heal Transferred Into System= 0.ooO kW Maximum Pnuunia 1.037 MPa at Ju& 1 InY Minimum Pressure ia 0.1013 MPa at Junotion 2 Outld Madmum Static Temperature is 2O.M) dq. C at Jundim Minimum Static Temperaure b 20.00 dog. C at Judi "'CAUTION"' Jundion 2 ia dmwmtmam of. W,dak papi+! cannot be calculated et jUndbn.

Sonic Choking Edst. at Junction 2 (Assigned Pnuum) AFTARDw2.oou(pu( (2) Nunmk Hanfd Corpontian Verlfyl4.am - Air flmv Exampla 8 frun Pony's pp. 625

Pipe Mass Mach Mach PStatic PShIk TStatic TStatic K fL/D fL/ Flow #In #Out In Out In Out D+K (we4 (Wa) (kPa) (daa .K) (dw.K) 1 2.716 0.2948 0.9946 1037. 285.5 288.2 245.4 2.000 3.581 5.581

All Jundion Table SNF-533 1, Rev. 0

TITLE: VERIFYl5.ARO REFERENCE: William S. Janna, Introduction to Fluid Mechanics, PWS Publishers, Belmont, CA 1983, Pages 3 17-3 19, example 8.8 GAS: Air ASSUMPTIONS: 1) Adiabatic flow, 2) Perfect gas

Parameter Janna AFT Arrow 2.0 M2-Mach number at exit 0.14 0.1355 P2 -Pressure at exit (psia) 9.63 9.836 Tz-Temperature at exit (deg. R) 528.8 528.9

DISCUSSION: As specified, inlet conditions are known and outlet conditions need to be determined. With the known inlet conditions, an implied mass flow rate exists. To pose the problem in AFT Arrow terms, a few simple calculations are needed to obtain the mass flow rate. Once obtained, it is applied as a flow demand at the exit. The problem states that the inlet velocity is 100 Ws,PI = 15 psia, TI= 530 R. From the ideal gas law, density, and mass flow rate are: 9- 15 psia = 0.0764 lbdft3 PI =-- ft.lbf RT1 (53.34-)(53OR)1bm.R

With this flow rate at the exit, the predictions agree very closely.

J1 J2

Printed with permission. 93 SNF-533I, Rev. 0

AFT Arrow 2.0 input (1) 127199 12:23 PM Numatec Hanford Corporation Verifyl 5aro -Air flow in a pipe, Example 8.8 from Janna page 317-319

William S. Janna Introductionto Fluid Mechanics, PWS Publishers, Belmont. CA 1983 Pages 317319. example 8.8 Air flow - Adiabatic flow, perfect gas assumption See Verifyl5.doc MS Word file for comparison with published results

Title: Verifyl 5.aro -Air flow in a pipe, Example 8.8 from Janna page 317-319

Number Of Pipes= 1 Number Of Junctions. 2

Length March Solution Method with Mach Number Limits Segments Per Pipe= 2 Mach Number Increment= .01 Pressure Tolerance= .oM)1 relative change Mass Flow Rate Tolerance= ,0001 relative change Enthalpy Tolerance= .WO1 relative change Flow Relaxation= (Automatic) Pressure Relaxation= (Automatic)

Fluid Database: AFT Standard Fluid= Air Max Fluid Temperature Data= 1WO deg. K Min Fluid Temperature Data= 200 deg. K Molecular Weight= 28.97 amu Gas Constant= 0.068549 Btwlbm-R Critical Pressure= 37.25 atm Critical Temperature= 132.41 deg. K Acentric Factor.: ,021 Equation of State= Ideal Gas Enthalpy Madel= Reference Specific Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g's Standard Pressure= 14.696 psia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Number= 40M) Laminar Flow Below Reynolds Number= 2300 Specific Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g Standard Pressure= 14.696 psia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Number= 4OOO Laminar Flow Below Reynolds Number= 2300

Pipe Name Pipe Length Length Hydraulic Hydraulic Roughness Roughness Losses (K) Initial Flow Defined Units Diameter Diam. Units Units 1 Pipe I Yes/ 5001 feet1 2.0671 inches I 0.0175 Explicit f I 01

Pipe Initial Flow Junctions Geometry Material Size Type Special Units (U&Dom) Condnion 1 1,2 1 Cylindrical Pipe 1 PVC 1 2 inch I schedule 40 None

Assianed Flow Table

94 SNF-533 1, Rev. 0

AFT Arrow 2.0 Input (2) 127199 1223 PM Numatec Hanford Corporation Verifyl5aro - Air flow in a pipe, Example 8.8 from Janna page 317-319

Assigned Flow Name Object Database Special Fluid Type Flow Flow Temperature Defined Source Condtion Units 2 I Assigned Flow 1 Yes I None1 Air1 OuttlowI 0.178141 Ibdsec 530

Assigned Pressure Name Object Database Special Fluid Pressure Pressure Temperature Defined Source Condtion Units

Assigned Pressure I Temoerature I Loss I Units Factor 1

95 SNF333 1. RCV.0 AFT AROW 2.0 Output (1) 12f7E9 Numabc Hanford Corponton Verifyl5.am-Airfiowinapip. Emp8.8fmmJannapeOe317-319

William S. Janna introduction to Fluid Mechanb, PWS Publishem, Bdmont, CA 1983 Pages 317319, example 8.8

Air flow ~ Adiabatic flow, perfect gns assumption

SW Verify15.k MS Word file fw comparison wlth publidrd mub We: Verifyl5.aro - Air flow in a pipe, Example 8.8 from Janna page 31731 9 hlysis run on: 1WB9 12:22:35 PM Input File: C:WFT Pmduch\AFT ~Ve~\varfyl5.m

Execution Time= 0.16 wands Total Number Of Pressure Iteratiom= 0 Total Number Of Flaw Iterations- 2 Total Number Of Enthalpy Iteratiow 2 Number Of Pip=1 Number Of Junctions= 2

Length March Sdution Mahod wah Mach Number Limb Sqmenlnt.PM Pip= 2 Mach Number incramant= .01 Pmaeura Tolerance- .ooO1 dative changa Mau Flow Rate Tohmo- ,0001 mlakchange Enthalpy T&ranw= ,0001 relativechange Flow Relaxation- (Automatic) Pressure Relaxation= (Automatic)

Fluid Databass: AFT Standard Fluid- Air Max Fluid Temperature Dah- 1000 deg. K Min Fluid Temperature Data- 200 deg. K Molecular Weight- 28.97 amu G8a Constent- 0.068649 BtUbm-R CriIjcai Preuure= 37.25 akn Critical Temperature- 132.41 deg. K Acenbic Fadot= ,321 Equntion d State- Ideal Gas Enthalpy M&C Rdm~ Speck Heal Ratio Accuracy- High Atmaphark Pressure- 1 atm Gravitational Acceleration= 1 g's Standard Praasuw 14.696 pia Standard Temperature- 60 dag. F Turbulent Flow Above Reynolds Numb4MxI Laminar Flow Below Reynolds Numbar= 2302 Specific Heat Ratio Accuracy- High Atnmpimric Pnssure- 1 atm Gnvhtionai Acceleration- 1 g Standard Pressure= 14.696 pain Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Nunbar= 4ooo Lqminar Flaw Balow Reyynok Number 2302 Ownli Deita Praasure = 5.1205 plia Total inflow 0.17814ibmhec TotalOuMowl0.17814lbmhac Total Enargy lnROw0 34.844 BN. Total Energy OUMOWC 34.844 6th Total Heal Transferrd Into System= 0.WBN.

Madmum Praasura is 15.000 psis al Junction 1 iniat Minimum Pmssure h 9.- p*. a Junction 2 Ink Maximum Static Temperature is 71.152 deg. F at Jundion 2 InY Minimum Static Temperature ia 71.152 dag. F at Ju~vAbn1 inlst

Piw Outout Table SNF-533 1, Rev. 0

AFT how2.0 Output (2) 12m99 Numatec Hanford Corporation Verifyl5.aro -Air flow in a pipe. Example 8.8 from Janna page 317-319

Pipe Mach Mach P Static P Static TStatic TStatic Mass Vel. In Vel. Rho I # In P Out In Out In Out Flow Out Static In I (psia) I (Psia) I (deg. R) I (deg. R) I (Ibm/sec) I (fwtkc) I (feeVsec) I (Ibmnt3) 1 I 0.0889131 0.135451 15.0001 9.83W1 530.001 528.92 0.178141 100.061 152.281 0.076402

Jct P Static P Static T Static T Static In Out In Out (psia) (psia) (deg. F) (deg. F) 1 15.woO 15.0000 70.330 70.330 2 9.8360 9.8360 69.248 69.248

97 Parameter Nayyar AFT Arrow 2.0 MI- Mach number at valve 0.3 17 0.3 18 PI -Pressure at valve (psia) 128.46 128.66

J1 v J2

'Rinted with permission. 98 SNF-533 1, Rev. 0

AFT Arrow 2.0 Input (1) 12f7199 1225 PM Numatec Hanford Corporation Verifyl6.am -Air flow in a pip, Example 88.2 from Nayyar Page B.339-342

Mohlndlw L. Nayyar Pidm Handbook. SihEdition. McGmHill. New York. 1992 ~ig0-~.339.~.542,Exampla 8e.2 Air Row - Exampb assume6 air is I perfect gar and adiabatic flow SaVwifyl6.d~ MS Word fib for mparkon with published raults Tle: Vwifyl6.aro - Air f!w in a pip. Ekample 88.2from Nayyar Page 8.339-342

Number Of Pipas= 1 Number Of Junctions= 2

Length March Solutim Mathod with Mach Number Limb Sqmentu Per Pipe= 10 Mach Number Increment= .01 Pressure Tolsrance- ,3001 relative change Mass Flow Rate Tolerance= ,3001 mlative change Enthalpy Tdmm=,3001 rdaivschange Flow Relaxation- .5 Pressure Relaxation= .5

Fluid Databaw: AFT Standard Fluid= Air Max Fluid Temperature Data= 1wOdeg. K Min Fluid Temperature Data- 200 dcg, K Molecular Weight- 28.97 amu Gas Constant= 0.0685486 BtullbmR Critical Preuure- 37.25 atm Critical Temperature- 132.41 dog. K Acentric Fador= .021 Equation of State- ideal Gas Enthalpy Model- Reference Specific Heat Ratio Accuracy= High Abnwpharic Preuure- 1 atm Gravitatio~lAcoeleration= 1 g'a Standard Pressure= 14.696 p.ia Standard Temperature- 60 deg. F Turbulent Flow Above Reynddr Nunk4OOO Laminar Flow Below Reynold. Nunhr= pw Specific Heat Ratio Accuracy= Huh Atmcapherk Proasurn- 1 atm Gravitational Aaalsration- 1 g Standard Pnuursp 14.696 pia Standard Temperature= 60 dq. F Turbulant Flow Abow Reynolds Nuntar= 4wo Laminar Flow Below Reyno!da Number= 23M)

Pip Name Pipe Length Length Hydraulic Hydraulic Roughness Roughness Losses (K) Initial Flow Ddd Unh Diametlw Dmm.Unh Unh 1 Piw I Yea I 901 fwtl 4.~61 inches I 0.017 Explicit f j 01

Pipe Inkial Flow Junoliom Garnary Material Sie We Special Unh (Up,Dorm) Condition 1 1, z I Cylinddal PIW I SWI 4inchI schedule40 None

Assioned Flow Table

99 SNF-5331,Rev. 0

~~ ~~~ ~ ~~~~~~~ AFT AITOW2.0 Input (2) 12/7/99 12:25 PM NunmtsC Hanfwd Corporation Verifyl6,aro - Air flow in a pipe, Example 68.2 from Nayyar Page 8.339-342

Assigned Flow Name Object Database Special Fluid Type Flow Flow Temperature Temperature Defined Source Condition Units Units 1 I Assigned~~owl yes None/ Air1 lnflowl 720001 Ibmhr1 108.75 I dq. F

I /bsioned~~owI ~wrI

Tank Name Object Database Spacial Fluid Pressure Pressure Temperature Temperature Balance Defined Source Condition Units Units Energy 2 1 Tank1 Yea I NlAl Air1 14.7 I psia 1 1MI deg. F I No

Tank Balance (Pipe #l) (Pipe 12) (Pipe #3) (Pipe #4) (Pipe #5) (Pipe #6) (Pipe #7) (Pipe #8) Concentration K In. K Out K In. K Out K In. K Out K In, K Out K In, K Out K In. K Out K In. K Out K In. K Out SNF-5331, Rev. 0

~~ AFT AROW 2.0 OMpM (1) 1m/99 Nu~natecHenford Corpontin Verifyl6.m - Ah flow In e pip, hpk88.2 fmm Nayyar Page 6.339-342

Mohinder L. Nayyar PiDina Handbwk. SihEdition. Mffiraw-Hill. NmYork. 1992 P;g&B.339. 8.342, Example B6.2 Air now - Example auumer air m e perfact guand dmb.tic (kw

Sw Vwify16.d~~MS Word fib for compriron vui(h publiihd rewb

Tide: Verifyl6.am -Ah flow in a pip, Example 88.2 from Newr Pa90 8.339342 Analysis run on: 12/7/99 122503 PM lnprl Fib: C:\AFT Products\AFT AmwlVeMiwtbn~l6.m

Execution Time- 2.91 seconds Tola1 Number Of PnuunIterations= 0 Total Number Of Fbw Itdons=2 Total Number Of Enthalpy Itetatiins- 2 Number Of Pipes- 1 Number Of Junctions- 2

Len# March Sddion Methodwich Mach Numb Limb Sagmmh POI Pip- 10 Mach Numtu Incremd- .01 PmrunTokrance= .W1rda6w change Mau Flow Rate Toleralrce= .O001 nhtb change Erdhaipy Toleranm= .ooO1 relatii ChmW Flow Rdaxation- .5 PnuunRelaxation- .5

Fluid Databasa: AFT Standard Fluid- Air MuFluid Temperatun Dah= IO00 dag. K Mln Fluid Temperature Data= 200 dog. K Mdecubr Weight= 28.97 amu GuConstant= 0.0685486 EtdbmR Crwal Pmuum- 37.25 rhn Cri(ical Temperature- 132.41 dog. K Acentric Factor ,021 Equhnof State- Ideal Gas Enthalpy Model= Reference Spec& Hmt Ratio Accuncy= High Atmospheric Prnwre- 1 atm Gravitalional Acwlerabn- 1 g's Standard Pressure= 14.696 pa Standard Temperature- 60 &g. F Turbulent Flow Above Reynolds Nunber- 4wo Laminar Flow Eebw Reyndds Nu- 2300 Specific Hwt Ratio Accuracy= High Atmospheric Prwsun= 1 atm Gravitational Acc.*Rlion= 1 g Standard Pmuun= 14.696 pi. Standard Temperature- 60 dag. F TurbulardFk AboM Reynold. Nwbr=4wo Laminar Flow Below Reynold. Nu- 2300 Overall Delta Prnwn= -1 23.230 pia Tot81 I- 2O.oooO lbmhec Total OUmw2O.MxX) 1bmh.C Total Energy InRow= 4149.85 Btuls Total Energy OUMw 4149.85 Btuh Total Heat Transfmed into System= O.MM00 B(uh

Maximum Prwsure is 128.660 pieat Junction 1 InW Minimum Prwsun is 14.7000 psis 8t Junctbn 2 OIRk( Maximum Static Tempraturn is 12O.OOO dog F at JunC(i0n 2 Inlat Minimum Static Tanpae(ure h 119.986 dag. F d JunEUQ) 1 Inlr(

Sonic Choking Exisb at Junction 2 (Tank)

Pim OU~DU~Tab AFT Arrow 2.0 Output (2) 1 m199 Numstac Hanfd Copratim Verifyl6.ara - Air fbw in a pip, Exanpls 88.2 from Nayyar Page 8.339-342

Pip Maas Vel. In Val. Out Mach Mach P Static PStatic TStatic TStatic TStag. TStag. Flow Y In I Out In Out In Out In Out (Ibhr) (feahc) (feat/wc) (pain) (paia) (&a . F) (dog. F) (dog. F) (deg. F) 1 72.OW.O 370.246 1071.85 0.317777 0.534971 128.660 37.9497 108.750 25.6957 119.986 120.141

411 Junction Tabk

102 SNF-5331,Rw. 0

Parameter Nayyar AFT Arrow 2.0 MI- Mach number at valve 0.3 17 0.318 PI-Pressure at valve (psia) 256.93 257.32

DISCUSSION: The problem assumes an unusual inlet boundary condition where the flow rate is known and the stagnation temperature. AFT Arrow uses the static temperature at the inlet because it is typically associated with a flow rate. To match the 120 F stagnation temperature, the inlet static temperature was iterated a few times. The cqnditions result in sonic choking at the discharge. The predictions agree very closely.

J1 v J2 P1>

* Rintcd withpermission. 103 SNF-533 1, Rev. 0

AFT Anan 2.0 IW (1) 1217199 12:27 PM Numatec Hanford Corporation Verifyl7.am - Air flow in a pip. Example 88.3 fmm Nayyar Page 8.342

Mohinder L. Nayyar Pipkg Handbook, Sih EMon, McGw-Hik NwYork, 1982 Paga 8.342, Example 88.3 Air flow - Example asrum- air is a pmfd gaa and sdmbaic flow

Sea Verify17.doc MS Word file for mmpriwn with puMbhed reauiia TI: Verify1 7.am -Air flow in a pip, Example B8.3 horn Nayyar Pago B.342

Number Of Pip=1 Number Of Junctions= 2

Length March Sdutiin Method with Mach Number Lim’b Segments Per Pip10 Mach Number Increment= .01 Pnuura T&ram.IO01 Mhn change Mass Flow Rate Tdmm=.DO01 mlativa chaw Enthalpy Tdanme= .ooO1 &thn chqp Concentrafin Tolerance- .ooO1 dah change Flow Rdaxation- .5 Prowurn Relaxation- .5

Fluid Databaw: AFT Standard Fluid- Air Max Fluid Tlrmpratura Data= 1ooO deg. K Min Fluid Tamparaturn Data= Mo dag. K Molecular Weight= 28.97 amu Gas Constant= 0.0685486 BtdbmR Critical Prsswrn- 37.25 atm Critical Tampratum= 132.41 dq. K Acenth Factor= ,021 Equation of State- Ideal Gaa Enthalpy Modal= Reference SmcHeat Ratio Accuracy= High Atmapheric Pre?aura= 1 atm Gravitational Acceleration= 1 g’s Standard Pressure= 14.696 pia Standard Tmpratura;: €0 dq. F Turbulent Flow AboM Reyndd. Nmh4OOO Laminar Flow Below Reynold8 Nunbr= 2300 Sp#kHedt Ratio huracy=High Atmoophwc Pressure= 1 atm Gravitational Acceleration= 1 g Standard Prowurn= 14.696 pia Stacdard Temperature= 60 dq. F Turbulent Flow Above Reyndds Nmber 4OOO Laminar Flow Below Reynoib Number= 2300

Pip Name Pip Length Lmh Hydnulk Hydnulio Rwghness Roughness Losses(K) Initial Flow Defined Unb Diameter Dhm.Unka Units 1 PipI Yw I 901 fartl 4.~61 lncbl 0.017 Explicit f 1 01

Pip InitiilFh Junctions GeOmary Mataial She TYP Special Units (U~.Down) Condition 1 I 1,2 I Cylindricai PipI SWI 4inchl schedule40 None

104 SNF-533 1. Rsv. 0

AFT ARaw 2.0 Input (2) 12l7199 12:27 PN Numatec Hanford Cwporation Verifyl7.aro - Air Row in a plpe, Example 88.3 from Nayyar Page 8.342

Objed Mksa Speckl Fluid Type Flow Flow Temperature Temperature Mmd Smm cwdillon

Tank Name Objj Database Special Fluid Pnsrurn Pmsaurn Temperature Temperalure Balance Defifined Source Condhn Units Unks Energy 2 Tank1 Yea I I NIA Air1 14.7 I wia I 120 I deg. F I No

Tank Babme (Pipe W1) (Pipe #Z) (Pip #3) (Pip #4) (Pip W) (Pipe #6) (Pipe #7) (Pipe #e) Concentration K In. K Out K In. K Out K In. K Out K in. K Out K in, K Out K In. K Out K In, K Out K In, K Oul -1 -1 2 No 1 (1) 0.0 I Kln KOut KIn KOut

105 SNF-5331,Rev, 0

AFT Anow 2.0 Output (1) 1WB9 Numatec Hanford Cwporatiion Varifyl7.aro -Air flow in a plp, Example B8 3 from Nayyar Page B 342

Mohinder L. Nayyar Piping Handbaok, Sixth Edition, McGmwHill, New York, 1992 P~lgc6.342, Example 68.3 Air llow - Example assumes air is a perfect gas and adiabatic flow See Varifyl7.doc MS Word fila for comparison with publhhd results Tsle: Varifyl7.aro - Air flow in e pip, Exampk 68.3 from Nayyar P~lge6.342 Analysis run on: 1WE9 12:27:42 PM Input File: C:\AFT ProductsMFT A~mwlV~\Wyl7.arn

Execution Time= 2.53 .sconds Total Number Of Pressure Itamtiom- 0 Total Number Of Flow bmtions- 2 Total Number Of Enthalpy Iterations- 2 Number Of Pim= 1 Number Of Junctions= 2

Length March Solution Method with Mach Number Limits Sagments Per Pipe= 10 Mach Number Increment- .01 Pnrsure Tokranca- .ooO1 mlatiw change Mae8 Flow Rate Tolerance- .ooO1 rdath change Enthalpy Tokrance- .ooO1 relative change FkH( Rdaxatkm- .5 Prsclsura Rehxatim- .5

Fluid Databasa: AFT Standard Fluid- Air Man Fluid Tempraturn Data- 1ooO dag. K Min Fluid Tempscaturn Data- 200 Qa. K Mdecular Weights 28.97 amu Gas Constant- 0.0685486 BhJlbmR Critical Pressure- 37.25 atm Critical Temprahrre- 132.41 dag. K Acentric Factor= ,021 Equation of States Ideal Gas Enthalpy Model= Raferenca SpecMc Heat Ratio Accuracy- High Atmapheric Prmsure- 1 atm GNVMOMI Acceleration- 1 g's Standard Prasure= 14.696 psia Standard Tempmture- 60 dag. F Turbulent Flow Above Reynold. Nunbar- 4ooo hmimr Flow Below Rynolds Number- 2300 SpeCitic Heal RatioAccurncy= High Aimu@l&PnUwe=1atm GravitationalAcodeia~ion= 1 g Standard Pnuurcr 14.696 peia Standard Temperature- 60 dag. F Turbulent Flow AboM RaynoIda Nunbar= 4ooo Laminar Flow Bdow Reynolds Nutrbr- 2300 Owrnl Deita Prmsura - -261.161 @a Total Inflow? 4o.oooO IbmE.ac Total OuMowl 4o.Mxx) Ibmlwc Total Energy lmlowl8299.69 Be Total Energy 0- 8299.69 Btuh Total Heat Tranafd into System= 0.MwXx) Ws

Maximum Preasura is 257.321 peia at Junc(ion 1 lnbt Minimum Pressure is 14.7wO pah at Junction 2 Outbt Maximum Static Tempm4ura ie 12o.wO dag. F at Junction 2 lnbt Minimum Static Temperaturn b 11 9.986 dag. F at Junction 1 Inlet Sonk Choking Exists at Junction 2 (lank)

Piw Outwt Tabk 106 SNF-5331,Rev. 0

AFT Anow 2.0 Output (2) 1 2r7B Numakc Hsnford Corpmtion I Verifyl7.ero- Air flow in a pipe, Exmpk 88.3from Nayyar Page 6.342

Pip Mass Vel. In VeI.0ut Mach Mach P Static P Static T Static T Static T Stag. T Stag. Flow #in # Out In Out In Out in (Ibm/hr) (fWsec) (fwhc) (psia) (psia) (dw. F) (deg. F) (deg F) 1 144,OOO. 370.242 1066.13 0.317774 0.988658 257.323 76.4630 108.750 26.6951 119.986 120.137

All Junction Table

I07 SNF-533 1, Rev. 0

TITLE: VERIFY 18.ARO REFERENCE: Mohinder L. Nayyar, Piping Handbook, Sixth Edition, McGraw-Hill, New York, 1992, Page B.342, Example B8.4 GAS: Air ASSUMPTIONS: 1) Adiabatic flow, 2) Perfect gas RESULTS: Parameter Nayyar AFT Arrow 2.0 A41 - Mach number at valve 0.243 0.243 PI-Pressure at valve (psia) 168.52 168.69

DISCUSSION The problem assumes an unusual inlet boundary condition where the flow rate is known and the stagnation temperature. AFT Arrow uses the static temperature at the inlet because it is typically associated with a flow rate. To match the 120 F stagnation temperature, the inlet static temperature was iterated a few times. The conditions result in sonic choking at the discharge. The predictions agree very closely.

J1 v J2

printed with permission. 108 SNF-5331,Rev. 0

AFT Arrow 2.0 Input (1) 12/7/99 1230 PM Numatec Hanford Corporation Verifyl8.aro-AirRowinapipe, Eramp4aB8.4frcfnNayyarPageB.342

Mohinder L. Nayyar Piping Handkmk, SiEM!. MeGRw-Hin, NewYak, 1992 Page 8.342, Example B8.4 Air flow - Example WIU~air m a perfect gas and adiabatic ?low

SeeVarifyl8.docMS Wordfilefacomprbonwith publiihed resub

Tle: Verfyl8.aro -Air in a pip, from Nayyar Page 8.342 flow Exampla 68.4 \ Number Of Pipes= 1 Number Of Junctions= 2

Langth March Solution Methcd with Mach Number Limb Segmerh Per Pipe= 10 Mach Number Increment- .01 Prauure Tolaranw= .W1mlative change Maas Flow Rate Tolera~=.wO1 dative change Enthalpy Tolerance- ,0001 relativechange Concentration Tolerance= .W1&ti change Flow Relaxation= .5 Pressure Relaxation.; .5

Fluid Database: AFT Standard Fluids Air Max Fluid Temperature Data- loo0 deg. K Mln Fluid Tamperaturn Data= 200 deg. K Molecular Waight= 28.97 amu Gas Constam 0.0685486 BlullbmR Critical Pressure= 37.25 atm Critical Temperature= 132.41 dag. K Acentrlc Factors .M1 Equation of Stata= tdaal Oas Enthalpy MUM= Reference SpecRc Heat Ratio Accuracy= Hah Atmospheric Pressure= 1 ah GraMdd Acc&mtii= 1 0'. Standard Pressure= 14.696 pia Standard Tarnpnture- 60 dag, F Turbulent Flow Above Reynolds Nunbar= 4ooo Laminar Flow Below Reyrdds Number.. Po0 Specific Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g Standard Pressure= 14.696 pia Standard Temperature- 60 dag. F Turbuknt Flow Above Reynold. NunW4WO Laminar Flow Below Reyndd. Numbor= 2300

PiDe InDut Table

1 Pipe Name Pip Length Langth Hydraulic Hydraulic Roughmrrs Roughness Losses(K) InlialFiow Defined Unb Dimtar Diam.Unib Unb

Pip InitiilFlow Junctions GaOmary Material Sire Type Special Units (Up.Dorm) Condition

Assinned Flow Tabla ~

SNF-5331. Rev. 0

AFT Arrow 2.0 Input 12/7/99 12:30Ph Numatec Hanford Cowration

Assignad Flow Nama OW Database Special Fluid Type Flow Flow Temperature Temperature D~itwd Sourca Condnicm Units Units

Tank Name Object Database Special Fluid Pressurn Pressure Temperature Temperature Balance Ddnd Soum Condition Units Units Energy '2 Tank1 Yes I N/A Air1 14.7 I psia I 120 I deg. F 1 No

Tank Balanw (Pip#I) (Pip12) (Pip rW) (Pip #4) (Pipe #5) (Pipe #6) (Pipe #7) (Pipe #8) Concsnhatin K In. K Out K In. K Oul K In, K Out K In. K Out K In. K Out K In. K Out K In, K Out K In. K Out 2 No I (l)O,OI

KIn KOut KIn KOut

110 SNF-5331. Rev. 0

AFT Anow 20 output (1) 12n199 Numlltac Hanford Corporation Verifyl8.aro-Airflowlnaplp, ~B8.4fmnNayyarPageB.342

Mohinder L. Nayyar Piping Handbook, Sixth Edition, McGmHIll. New York. 1992 Page 8.342, Example 88.4 Air flow - Example assumea air is a perfect o.. and admMi flow Sw Verifyl8.da: MS Word fila for wmpamOn with publlahad mub Tie: Verifyl8.aro - Air flow In a pip, Example 68.4 fmNayyar Page 8.342 Analyait run on: 12VB12:29:24 PM Input File: C:WT ProductsW ArrowtVorikatbn\wrifyl&am

Execution Tim= 3.13 wcond. Total Number 01 Presaure Iterations= 0 Total Numhar Of Flow Itarafons= 2 Total Number 01 Enthalpy Iterations- 2 Numhar Of Pipes- 1 Numbsr Of Juncliim- 2

Lenoth March Solution Method with Mach Numimr Limb Segmenla Per Pipe= 10 Mach NwnkIrrxenwA= .01 Prewure Tolerance= ,0001 dative change MOM Flow Rate To!annce- ,0001 dative change Enthalpy Toleram- ,0001 rdaliw change Flow Rehxation= .5 Pressure Relaxation- .5

Fluid Databass: AFT Standard Fluid- Air Max Fluid Temperature Data= 1000 deg. K Min Fluid Tempdurn Data= 200 dag. K Molecular Weight= 28.97 amu Gas Conatant- 0.0685486 BNlbmR Crilial Pmaaure= 37.25 atm Crith1 Temperature- 132.41 dag. K Acentric Factor- ,021 Equalion of SWa= Ideal Gas Enthalpy Modal- Reference SWiHeal Ralio Accuracy- Huh Almoapherk Pwaaure= 1 atm Gmvitatio~lAcoelemlim= 1 g's Standard Pnuure- 14.696 pia Standard Tempnture- 60 deg. F Turbulent Flow Abovr, Reynolds Numb4OOO Laminar Flow Bdow Reynold. Numter= 2300 SpcciTic Heal Ratio Accuracy- Hiah Atmospheric Pressure= 1 alm Gravitational Acwlemtion- 1 g Standard Pressure- 14.696 pia Standard Tanpeture= 60 dag. F Turbulent Flow Above Reynolds Nmbm-= 4OOO Laminar Flow Balow Reyynold. Numb.r= 2300 Overall Ddta Pressure - -161.046 pia Total lnflowr 2O.o(300 Ibmlaec Total Ouhlow 2O.Wlbmhec Total Energy lnflawl4150.63 Btuh Total Energy Ouhlw 4150.63 Btuh Total Heat Tranafened Into System- 0.WBtuh

Maximum Pressure it 168.690 psa al Judon 1 lnld Minimum Pnuure ia 14.7000 pala al Junction 2 Outld Maximum Static Temperature ia 120.145 dag. Fat Jundion 1 lnlst Minimum Static Tempaature Is 120.000 dag. F al Junction 2 Ink

Sonic Choking Exists at Junction 2 (Tank)

Piw Omut Table 111 -

I AFT Arrow 2.0 Output 12nB Numalec Hanfwd Corporation

Pip Mass Vel. In Vel.Out Mach Maoh P Static P Static TStatic T Static TStag. TStag. Flow # In # Obi In out In Out In Out (Iknmr) (fwvMc) (fedaoc) (wia) (psia) (deg F) (deg. F) (deg. F) (deg. F) 1 72,000.0 284.747 1070.27 0.243388 0.993082 168.690 38.0391 113.500 26.1325 120.145 120.300

All Junction Table

112 SNF-533 1, Rev. 0

TITLE: VERlFY19.ARO REFERENCE: Mohinder L.Nayyar, Piping Handbook, Sixth Edition, McGraw-Hill, New York, 1992, Page B.342, Example B8.5 GAS: Air ASSUMPTIONS: 1) Adiabatic flow, 2) Perfect gas RESULTS: Nayyar AFT Arrow 2.0 .I I PI- Pressure at vdve @sia) I 165.29 I 165.60 I

DISCUSSION: The problem assumes an unusual inlet boundary condition where the flow rate is known and the stagnation temperature. AFT Arrow uses the static temperature at the inlet because it is typically associated with a flow rate. To match the 500 F stagnation temperature, the inlet static temperature was iterated a few times. The conditions result in sonic choking at the discharge. The predictions agree very closely.

J1 v J2

* Printed with permission. 113 SNF-5331. Rev. 0

AFT Arrow 2.0 Input (1) 12i20/99 10:41 AM Numatec Hanford Corporation Vwifyl9.aro. Air fiow in a pipe, Example 68.5 from Nayyar Page 6.342

Mohinder L. Nayyar Piping Handhk, Sixth Edition. Mffiraw-Hill, New York, 1992 Page 8.342, 6.343, Example 68.5

Air flow ~ binpleassum68 air ia a perfect gas and adiabatic flow

See Verifyl9.doc MS Word file for comparison with published results Title: Verifyl9.arO - Air flow in a pipe, Example 68.5 from Nayyar Page 6.342 Number 01 Pip=1 Number Of Junctions= 2

Length March Solution Method with Mach Number Limits Segments Per Pipe- 10 Mach Number Incrcmwnt= .01 Pnssure Tolerance= ,0001 relative change Mass Flow Rate Tderance. ,0001 relative change Enthalpy Tolerance- ,0001 relative change Flow Relaxation- .5 Pnssura Relaxation= .5

Fluid Da(sbsse. AFT Standard Fluid= Air Max Fluid Temperature Data= 1000 deg. K Min Fluid Temperatun Data= 200 deg. K Molecular Weight= 28.97 amu Gas Constant- 0.0685486 BtuAbmR Critical Proaswe= 37.25 atm Critical Temperature= 132.41 deg. K Acentric Factor= .M1 Equation of State= Ideal Gas Enthalpy Model= Reference Specific Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm Gravitatinal Acceleration;: 1 g's Standard Pnssure= 14.696 pia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Nwnber= 4Mx) Laminar Flow Below Reynolds Number= 23M) Specific Heat Ratio Accuracy- High Abncqhsric Preawre- 1 atm Gravita(iionnl Acceleration= 1 g Standard Pnssure= 14.696 pia Standard Temperature= 60 dag. F Turbulent Flow AboM Reynolds Number= 4000 Laminar Flow Below Reynolds Number= 2300

Pipe Name Pipe Length Length Hydraulic Hydraulic Roughness Roughness Losses (K) Initial Flow Defined Units Diameter Diam. Units Unik 1 Pipe I Yes I 901 feel 4.0261 inches 1 0.017 Explicit f I 01

Pipe InKial Flow Junctions Geometry Material Size Type Special Units (Up.D~rm) Condition 1 1,2 I Cylindrical Pipe 1 Steel I 4 inch I schedule 40 None

Assiond Flow Table WT ARau 2.0 Input (2) 12tZOB9 10:41 AM Numatffi Hanford Corporation

Veyl9.aro ~ Air flow in a pipe, Example 68.5 from Nayyar Page 8.342

Assigned Flow Nams Obw Database Spacial Fluid Type Flow Flow Temperature Temperature Defned Swrce Condition UnRs Units 1 1 Assiinej~~owj yes None1 Air/ Inflow1 720001 lbmhrl 481 I deg F

Tank Name Object Databaw Spcial Fluid Pressure Pressure Temperature Defined Source Condition Units 2 Tank1 Yell I N/A Air1 14.7 I pia 1 1201 deo. F I No

Tank Balance (Pipe #l) (Pipe Y2) (Pip 13) (Pip#4) (Pipe #5) (Pipe #6) Concentration K In, K Out K In, K Out K In. K Out K In. K Out Kin, K Out K In. K Out 2 No 1 (1)O.OI I Tyk I (Pr#9)1 (Pippllg 1 KIn KOut KIn KOut

I IS AFT Anow 2.0 output (1) 1mEJ9 Numatac Hanford Corporation VNIfyl9.am - Air hvIn a pip, Earn* 88.5 from Nayyar Page 8.342

Mohlnder L. Nawar Piping Handbook. Sixth Edltiion. McG~Hill,Nm York. 1992 Page 8.342.8.343. Example 88.5 Air fbw - Example auumes air is a pelfact gar and adIaMc Row Sw Verifyl 9,doc MS Word file for comparison with publishad result0

Ti&: Verify19 ero - Air flow in a pip. Example 68.5 from Nayyar Page 8 342 Analyab run on: 12RoB9 10:40:24AM Input File: C:\AFT Prodwb\AFT ~Vwlfmtbn\\nrifyl9.aro

Execution Time= 3.35 seconds Tdal Numbec Of Pmsure Iteratlona= 0 Total Number Of Fbw Iterations= 2 Total Number Of Enthalpy Itentionts 2 Number Of Pipes- 1 NumkOf Jundwns- 2

Length March Sdution Method with Mach Numb Limb Sunnanta Per Pim- 10 M&h Number Increment- .01 Pmuum TdUatWe- OOO1 nlabve ChUlgO MauFbw Rate Tolerance- .OOO1 relative cham- Enthalpy Tolerance- .OOO1 relative change Flow Rdaxatwn= .5 Pressure Relaxation= .5

Fluid Database: AFT Standard Fluid- Air MuFluid Temperature Data- 1000 dag. K Min Fluid Temperature Data- 200 dog. K Molecular Weight- 28.97 amu Ga5 Constant- 0.0685486 Et4bmR Critical Pressure= 37.25 atm Critical Temwrsture- 132 41 dm.- K

Aamtric~ ~ Fa&= ~ ~~~ ,021 Equation of Slate- ideal Gas Enthalpy Model= Rdn~enca Speck Heat Ratio Accuracy= Huh Atmoaphenc Pressure- 1 atm GmvitatDnaI Accelaratbn= 1 9's Standard Pressure= 14.696 pia Standard Temperature- 60 dag F Turbulurt Flow AboM Reynolds Number- 4WO Laminar Flow Below Rsynoldr Number 2300 Specfic Heat Ratio Accuracy= High Atmorpheric Prosaura= 1 atm Gravitational Accaleration- 1 g Standard Preasure- 14.696 mia S(sndsrd Tunprahm- 60 dag F Turbulent Flow Above Reynolds Number- 4LlW Laminar Flow Bebw Reynolds Number 2300 Ovwall Daila Preuun -162.824 pia Total I*- 2o.mIbmhc Total OUmw 2o.oooO lbmhec Tdal EnqyInflow= 6011.31 Btds Total Energy Outflow= 601 1.31 8th Total Heat Transferred Into System= 0 00000 EN.

Mmrnum Pmrure I165 604 paat JuncbOn 1 Inlet Minimum Pressure ia 14 7000 m!a at Junction 2 Om Mmmum Static Temperature 18 499.335 dag Fat Junction Inlet Minimum Static Temperature is 120 OOO deg F at Judn Inlet

Sonic Choking Exists at Junction 2 (Tank)

PIW OU(DUt Table

t16 Pip Mass Val. In Vel. Out Mach Mach P Static P Static T Static T Static T Stag. T Stag. Flow # In # Out In Out In out In out (Ibmlhr) (fwt/wc) (fsaEwc) Mia) (psia) (deg. F) (deg. F) (deg F) (deg. F) 1 72.000.0 476.028 1377.07 0.318401 0.994614 165.604 49.0050 481.wO 345.576 499.335 499.740

&IJunclion Tabla

Jct P Static P Static TStatk TStatk In Out In Out fpsia) (psia) (daa. F) (den. F) 1 165.W37 165.W37 481.000 481.000 2 91.8048 14.7000 12O.W 120.000

117 SNF-5331, Rev. 0

Parameter Nayyar AFT Arrow 2.0 MI- Mach number at valve 0.235 0.235 PI-Pressure at valve (psia) 661.96 662.56

~ ~ ~ ~ ~ ~

JI v J2

* printed with permission. 118 SNF-5331.Rw. 0

AFT Armw 2.0 Input (1) 127/99 1239 PM Numatec Hanford Corporation Verify2O.aro -Air flow in a pipe, Example 88.6 from Nayyar Page 8.343

Mohindw L. Nayyar Piping Handbook, Sixth Edition, McGraw-Hill, New York. 1992 Paw 8.343. Example 88.6 Air flow - Example assumes air is a perfect gas and adiabatic flow

Sw Vsrify20.doc MS Word file for comparison with published resuits

Tlle: Verify2O.am -Air flow in a pipe, Example 88.6 from Nayyar Page 8.343

Number Of Pip=1 Number Of Junctions= 2

Length March Solution Method with Mach Number Limits Segments Pw Pipe= 10 Mach Number Increment= .01 PmunTolarance= .ooO1 relative change Mas. Flow Rate Tolerance= .ooO1 relative change Enthalpy Tokrance= .ooO1 relative change Fkw Relaxation= .5 Pmsure Relaration= .5

Fluid Dataabw: AFT Standard Fluid- Air Max Fluid Temperature Data= 1000 deg. K Min Fluid Temperature Data= 200 deg. K Molecular Weight= 28.97 amu Gas Constant= 0.0685486 BtuAbmR Critical Pmsure= 37.25 atm CRical Temwture= 132.41 deg K Aambic Factor= ,021 Equation of State= Ideal Gas Enthalpy Model= Reference Specific Heat Ratio Accuracy= High Atmoqheric Pressure= 1 atm GravitationalAcceleration= 1 g's Standard Pressure= 14.696 pia Standard Temperature= €0 deg. F Turbulent Flow Above Reynolds Number= 4ooo Laminar Flow Below Reynolds Number 2300 Specific Heat Ratio Accuracy= High Atmoclpharic Pressure= 1 atm GravMonal Acceleration= 1 g Standard Pressure= 14.696 psia Standard Temperatures €0 deg. F Turbuhnt Fkw Above Reynolds Number= 4OOO Laminar Flow Below Reynolds Number= 2300

Pipe Name Pipe Length Length Hydraulic Hydraulic Roughness Roughness Losses (K) initial Flow Defined Units Diameter Diam. Units Units 1 Pipe I Yes I 90 feet I 2.0671 inches 1 0.019 Explicit f 1 01

Pip InitiilFlow Junctions Geometry Material Size Type Special Units (Up,Down) Condlion 1 I 1, 2 1 Cylindrical Pipe 1 Steel I 2 inch [ schedule 40 None

Aarianed FIow Table

119 SNF-533 1, Rw. 0

AFT Arrow 2.0 Input (2) 12f7/99 1239 PM Numatec Hanford Corporation

Verify20.aro ~ Air flow in a pipe, Example 68.6 from Nayyar Page 6.343

Assigned Flow Name Object Database Special Fluid Type Flow Flow Temperature Temperature Defined Source Contihion Units Units 1 I Assigned Flow1 Yes 1 None1 Air1 Inflow1 72000] lbmhrl 113.751 deg. F

Model

Tank Name Object Database Special Fluid Pressure Pressure Temperalure Temperalure Balance Defined Source Contihion Units Units Energv 2 Tank1 Yes I NIA Air1 14.7 I psia 1201 deg. F I No

Tank Balance (Pipe XI) (Pipe #2) (Pipe W3) (Pipe #4) (Pipe #5) (Pipe 16) (Pipe X7) (Pipe #e) Concentration K In. K Out K In. K Out K In. K Out K In, K Out K In, K Out K In, K Out K In, K Out K In, K Out I I 2 No I (1) 0, 0 I

Tank (Pipe Y9) (Pipe #lo) K In. K Out K In, K Out

120 SNF-533.1. ROV.0

~ AFTAnav2.ooutput (1) 17J7199 Numatec Hanford Corpontion VerifyZOrrn - Air h h a pip, Exvnpk B8.6 from Nayyar Page 8.343

Mohindar L. Nayyar Piping Handbook. SiEdition. McGraw-Hill, NmYwk. 1$92 P+ B.343, Example 86.6 Air Row - Example assumes air is a prfsct gaa and adiabdc fbw Sw VerifyM.dcc MS Word file for cornprism with puMbW mwlt. Title: VerifyZO.aro - Air flak in a pip, Example 88.6 fmm Nayyar Pago €1.343 Analysis run on: 1WB91237:s PM lnpn File: C:\AFT Prcducts\AFT~VNikation\m.ern

Execution Time= 2.74 -dr Total Number Of Pressure It~~tiom-0 Total Number Of Flow Iterations- 2 Total Number Of Enthalpy Iterations- 2 Number Of Pip=1 Number Of Junctions- 2

Length March Solution Method with Mach Number Limb Segments Par Pip= 10 Mach Number Imd=.01 Prauure Tolerance= .ooO1 rdatii change Ma- Flow Rate Tolerance= .0001 rdative changa Enthalpy Tolerance- .IO01 relatiw change Flow Relaxation- .5 Pressure Reiaxation= .5

Fluid Datebase: AFT Standard Fluid- Air Max Fluid Temperature Data= 1000 dag. K Min Fluid Temperature Date- 200 dag. K Mekular Weight- 26.97 amu Gas Constant- 0.0685486 Btu4bm-R Critical Prmwre= 37.25 atm Critical Temprature= 132.41 dag. K Acentric Factor;; .M1 Equation of State= Ideal Gas Enthalpy Model- Reference Sp.Cinc HdRatia Accuracy= Hoh Atnmaphwc Pressure= 1 atm Gravitational Accdmtion= 1 8's Standard Prauure- 14.636 poia Standard Tempentun- 60 deo. F Turbulent Flow Above Reynolds Nunber- 4ooo Laminar Flow blow Reynolds Numbar= 2500 Spacif~Heat Ratio Accuracy= Huh Atmoopherk Pressure- 1 atm Gravitational Acceleration- 1 g Standard Pressure= 14.636 pia Standard Temperature= 60 dog. F Twbulw\t Flow AbDM Reyndda Nunb4Mx) Laminar Flow Bdow Reynolds Nu- 2500 Overall D& Pnuun= 673.699 pia Total IMw= 20.- Ibmlsec Total OutRowc ZO.Ow0 lbmhrec Total Energy IMw= 41 49.70 Btuh Total Energy OuMw 4149.70 Btuh Total Heat Transferred Into System- O.Ww0 Btuh

Mudmum Pnuunis 662.559 piaat Junction 1 lnla Minimum Pressure ia 14.7000 piaat Junction 2 Outid Maximum Static Temperature b 120.000dog. F id Jumtion 2 Ink4 MmhSt& Tempenhm b 119.955 dag. F d Junction 1 lnht

Sonic Choking Exiats at Junction 2 (Tank)

Piw OWTable 121 SNF-533 I, RCV.

AFT Arrow 2.0 Output (2) 127199 Numaiec Hanford Corporation

Veriiy2O.aro ~ Air flow in a pipe, Example 68.6from Nayyar Page 6.343

Pipe Mass Vel. In Vel.0ut Mach Mach P Static P Static T Static T Static T Slag. T Stag. Flow X In Y Out In Out In Out In Out

(Ibmhr) (feeVsec) (feevSec) (psia) (psia)~ B.F) (deg F) (deg. F) (deg. F) 1 72,000.0 275.155 1062.54 0.235138 0.984739 662.563 145.708 113.750 27.2966 119.955 120.105

All Junction Table

JCI P Static P Static T static T Static ! I In I out I In lout1 (Pam) I (paia) I (deg. F) I (deg F) 1 I 662.5591 €62.55901 113.7501 113,750 2 I 270.3841 14.70001 120.000) 120.000

122 SNF-533 1, Rev. 0

Parameter Nayyar AFT Arrow 2.0 MI- Mach number at valve 0.4096 0.4142 PI-Pressure at valve (psia) 25.15 24.95

DISCUSSION: The problem assumes an unusual inlet boundary condition where the flow rate is known and the stagnation temperature. AFT Arrow uses the static temperature at the inlet because it is typically associated with a flow rate. To match the 120 F stagnation temperature, the inlet static temperature was iterated a few times. The result is sub-sonic conditions at the discharge. The predictions agree very closely.

J1 v J2

* printed with permission. 123

. C....d_L._ SNF-5331.Rev. 0

AFT Arrow 2.0 Input (1) 12/7/99 1241 PM Numatec Hanford Corporation Verify21.aro - Air flow in a pipe, Example 68.7 from Nayyar Page 8.343-344

Mohinder L. Nayyar Piping Handbook, Sixth Edition, McGraw-Hill, New York, 1992 Page 8.343-344, Example 88.7 Air flow - Example assumes air is a perfect gas and adiabatic flow See Verify21.doc MS Word file for comparison with published results

Title: Verify21.aro -Air flow in a pipe, Example 88.7 from Nayyar Page 8.343-344

Number Of Pipes= 1 Number Of Junctions= 2

Length March Solution Method with Mach Number Limits Segments Per Pipe= 10 Mach Number Increment= .01 Pressure Tolerance= ,0001 relative change Mass Flow Rate Tolerance= ,0001 relative change Enthalpy Tolerance= .OW1 relative change Flow Relaxation= .5 Pressure Relaxation= .5

Fluid Database: AFT Standard Fluid= Air Max Fluid Temperature Data= 1WO deg. K Min Fluid Temperature Data= 2W deg. K Molecular Weight= 28.97 amu Gas Constant= 0.0685486 Btuilbm-R Critical Pressure= 37.25 atm Critical Temperature= 132.41 deg. K Acentric Factor= ,021 Equation of State- Ideal Gas Enthalpy Model= Reference Specific Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g's Standard Pressure= 14.696 psia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Number= 40W Laminar Flow Below Reynolds Number= 2300 Specific Heat Ratio Accuracy= High Atmospheric Pressure= 1 atm Gravitational Acceleration= 1 g Standard Pressure= 14.696 psia Standard Temperature= 60 deg. F Turbulent Flow Above Reynolds Number= 4000 Laminar Flow Below Reynolds Number= 23W

Pipe Name Pipe Length Length Hydraulic Hydraulic Roughness Roughness Losses (K) Initial Flow Defined Units Diameter Diam. Units Units

Pipe Initial Flow Junctions Geometry Material Size Type Special Units (Up.Down) Condition

Assianed Flow Table

124 SNF-S~~I,-ROV.0

AFT hmv 2.0 Input (2) 12n/9912'41 PM Numatec Hanford Corporation Verify21.am -Air Row in a pip, Exampla 88.7 from Nayyar Page 8.343344

Atsigned Flow Nama ObW Dnhbaw Spacial Fluid Typ Flow Flow Temperature Temperature Defd Source Condition Units Unhs

Tank Nama Object Database Spacial Fluid PnUure Pressure Temperature Temperature Balance Def~rmd Source Condlimn UnRs Units Energy 2 I Tank1 Yes I N~AI Air1 14.7 I psia 1 1201 deg F I No

Tank Balance (Pip#l) (Pip G!) (Pip #3) (Pip M) (Pip ffi) (Pipe #6) (Pipe #7) (Pipe #B) Coneenhation K In. K Out K In, K Out K In, K Out K In. K Out K In. K Out K In. K Out K In. K Out K In. K Out

Tank (Pip IW) (Pip #lo) K In. K Out K In. K Out 121 I I

125 SNF-533 1, Rev. 0 AFT Anow 2.0 OM fl)., 1 m199 Numates ~unford6&ration Verify21 .am -Air flow in a pip., Exmple 88.7 from Nayyar Page 8.343344

Mohinder L. Nayyar Piping Handbook, Sixth Ed*on, MffiwHill, NmYork, 1992 Page 8.343344, Example 68.7 Air flow - Example assumes air is a prfd gas and adiabatic flow

Sea VeMy21 .dm MS Word file for ampriaon with published mub

TWO VeMy21 am - Ar flow in a pip, Exampb 88.7 from N0yy.r Pap. 6.343-344 Analysisnmon: 12/7/99 12:41.15 PM Input File: C:\AFT ProdudslAFT AmMV&kation\verify21.am

Execution Time= 0.99 wconds Total Number M Pressure ItMations= 0 Total Number Of Flow Itera(iins= 2 Total Numbw Of Enthalpy Iteratins- 2 Number Of Pipas= 1 Number Of Junctions= 2

Length March Solution Mahod Mth Mach NumbLimb Segments Per Pip= 10 Mach Number Increment= .01 Pressure Tolerance;. ,0001 relative change MAUS Flow Rate Tdwanca= ,0001 rektbm chqo Enthalpy Tolerance- ,0001 rdativa change Flow Relaxations .5 Prussure Relaxation= .5

Fluid Database: AFT Standard Fluid= Air Max Fluid Temperature Data= 1000 dag. K Min Fluid Temperature Data= 200 dag. K Mdmlar WdgM= 28.97 amu Gas Constant= 0.0685486 BtullbmR Critical Pnsrure= 37.25 atm Critical Temperature= 132.41 dag. K Acentric Factor= .021 Equation of State= Ideal Gas Enthalpy Model= Reference Specific Heat Ratio Accuracy- High Atmospheric Pnsrure= 1 atm Qmvkationel Accdmtion- 1 9's Standard Pressure- 14.696 pia Standard Tempsrature= 60 deg. F Turbulent Flow Above Reynoldr Nunber= 4Mx) Laminar Flow Belan Reynolds Nun-&r= 23M) Specilic Heat Ratio A~CUNCY=High Atmorrphric Pressure= 1 atm Qmvitationai Acealeration- 1 g Standard Prauuree 14.696 psia Standard Temperature= €0 deg. F Turtxllent Flow Above Reynolds Nunb4Mx) Laminar Flow Below Reynold. Numbar 2300 Overall Deb Pressure = -1 3.3543 pia Totd In(bwl2o.oooO lbmhutc Total Outflow 2O.oooO 1bn-d~~ Total Energy Inflow= 4149.09 Bhrls Total Energy Outflow= 4149.09 Btulr, Total Heat Transferred Into System= 0.00000Btuh

Maximum Pnsrure is 24.9467 piaat Junction 1 Inlet Minimum Pressure ia 14.7000 pia at Jundin 2 Outla Maximum Static Temperature is 12o.WO deg. F et Junction 2 Inlet Minimum Static Temperature is 119.834 deg. F at Junction 1 inla

Pim Out~utTaMe

126 AFT AROW 2.0 Output (2) 12/7/99 Numatsc Hanford Corpontiwn VeIify21 .aro - Air flow in a pipe. Example 88.7 from Nayyar Page 8.343-344

Pip Mass Vel. In Vel. Out Mach Mach P Static P Static TStatic T Static T Stag. T Stag. Fh # In # Out In Out (ihhr) (faethrcrc) (faethrcrc) (psia) (psia) (deg.In F (d$F) ~ (dig?F) 1 (dgF) 1 1 72.000.0 479.282 800.696 0.414173 0.713575 24.9468 14.0338 101.wO)I 67.2485 119.834 119.877

Jct P Static P Static T Static T Static In Out In Out Wi) (psia) (den. F) (dw. F) I I 24.94671 24.94671 IOI.~~IOI.~ 2 I 19.6747) 14.70001 120.000I 120.003

127 DISTRIBUTION SHEET

To FrOm Pape 1 of 1 Distribution ~-FDH - Engineering Labs Pmjed THkMk Order Date December 1999 I SNF-5331 EDTNo. 627034 I

Texl EDTECN Name MSlN VHthAll TexlOnly Appendk only Attach. Onlv B. A. Crea R3-86 X T. Choho R3-86 X Lb-3- X F. J. Heard (Ir) J. J. Irwin R3-86 X C. R. Miska R3-86 X S. L. Mischke R3-86 X M. J. Schliebe L6-13 X J. R. Brehm R3-26 X Central Files B1-07 X I DIMC I H6-15 I X I I I IWE/RL Readina Room I H2-53 1 X 1 I I I SNF Project File W-441 R3-11 X lCVD Library I I I I Startup Librarv I B2-64 I X I I I I I CVD Satellite Librarv I X3-25 I X I I I I