Verification Case 48
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.