Verification Case 48

View Model     Problem Statement

PRODUCT: AFT Fathom

TITLE: FthVerify48.fth

REFERENCE: James John, William Haberman, Introduction to Fluid Mechanics, 2nd Ed., 1980, Prentice-Hall, Page 184-186, Example 6.8

FLUID: Water at 20 deg. C

ASSUMPTIONS: All pipes at same elevation. Common exit pressure is 1 atm.

RESULTS:

Parameter	John & Haberman	AFT Fathom
Flow rate through pipe 1 (m3/sec)	0.095	0.0943
Flow rate through pipe 2 (m3/sec)	0.105	0.1057

DISCUSSION:

The problem statement does not have any pressure information, because it is not required if the only purpose is to calculate flow rates. However, AFT Fathom requires at least one pressure boundary because AFT Fathom offers pressure information in the output for all systems. Therefore, a pressure of 1 atm was assumed for the common junction at the exit of the system.

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