p0602

# p0602 -

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 6.2. CHAPTER 6, PROBLEM 2 675 6.2 Chapter 6, Problem 2 Problem: The pipe cross-sectional area and flow speed at the inlet and the outlet are the same and equal to A and U , respectively for incompressible flow into a 180o bend. Assume the velocity is constant on all cross sections. At the inlet and outlet, determine n, (u · n), and ρu(u · n)dA. U, A . .. . .. . ................................ . .... ......... .......... . .............................................................. ................................................................................................................ . .... ................................................................................................................................. ... ........................................................................................................................................................ .... ........................ .................................................................................................................................................. .................. . . ......... ......... ...... ........... ........................................................................................................................................................................ . . .. ... . ................................................................................................................................................. . ............................................................................................................................................................ ... . ..... ......... ... .. . ... . . ....... ... . . . ................. .. ........... . .. ................................................................................ . ......................................................................................................................................................... .......................................................................................................................................... . ... .. .. ........ .. ............ ............................ .. . .. .. .. ... .......................... . .......................... ......................... . . .. . . . . .. .. .................. .. ................... ................... .. .......................... ............................ .. . ...... ........... .......................................................................................... .... . .. ................................................................................................. ..................................................................................................................................................... .. . .......................................................................................................................................................... .............................................................................................................................................................. . . . . . . . . . . . . . ... .. .. . .. ................................................................................................................................................................................ . .... .......... . . . . . .............................................................................................................................................. .. ... ... ... ..... . ......... . ... . ... ................................................................................................................................................ . .... .......... ............................................................................................................ . ... ..................................................................................... .................. ...... . .. ..................................................... ................................................... y ........ . . . . . . . . . . .............. . . . . ............. x U, A .............................................................................................................................. Solution: At the inlet, the unit normal is n=i The velocity vector at the inlet is u = −U i so that u · n = (−U i) · i = −U ρu (u · n) dA = (−ρ U )(−U )(A) = ρ U 2 A At the outlet, the unit normal is n=i The velocity vector is u = U i so that u · n = (U i ) · i = U ρu (u · n) dA = (ρ U )(U )(A) = ρ U 2 A ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online