p0602

# p0602 -

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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 ...
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## This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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