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p0638

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Unformatted text preview: 728 CHAPTER 6. CONTROL-VOLUME METHOD 6.38 Chapter 6, Problem 38 Problem: The velocity profile in a channel changes from u1 (y ) to u2 (y ), where u1 (y ) = 2U y /h − 1 (y/h)2 2 u2 (y ) = λU y /h + (y/h)2 − (y/h)3 and for 0 ≤ y ≤ h, and the profiles are symmetric about the centerline at y = h. U is a constant velocity scale. Assuming the flow is steady and incompressible with density ρ, solve for the constant λ. .................................................................................................................................................................................................................................................................................................................................................................. ... . .... .... . . .. .. . . . . . .. ............... .. ...... .... ........................... .... ........ ........................... .... .. ...... .................... .............................................................................................................................................................................................................................................................................................................................................................................................................................................................. ................................................................................................................................................................................................................................................................................................................................................................. ................................... . . . . . . .. .. . . . . . . . . . . . . . . . .. . ... . . . . . . . . . . . . . . .. . .. .. . . . . . . . . . . ρ u 1 (y ) 2h x u2 (y ) n2 n1 y . .. . .. . . . . . .. ... . ....................................................................................................................................................................................................................................................................................................................................................................... ... .. . .......... ........................................................................................................................................ ........................................................................ .......................... . ...................... .................. . .. .. . .. ... .. ... .. .. . .. .. ...... . .. . .. ..................................................................................................................................................................................................................................................................................................................................................................................................................................... .... .. .................................................................................................. ...... ................................................................................................................ ................................................................................... . ... .... ... . .. .. ........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................................................................................................................................................................ ... . . .... ... ........ .... .... ..... ..... ........................................................................................................... ........................................................................................... . . . ..... ........................................................................................... . . . . . . . . ..... . .... . .. .... . .... . .. Solution: Because the flow is symmetric about the centerline, we select the stationary control volume indicated by the dashed contour in the figure. The control volume includes the lower half of the channel, and the unit normals are parallel to the inlet and outlet velocities. We begin with the mass principle. Since the control volume is stationary, ure = u. Thus, the mass principle tells us that d dt ρ dV + V S ρ u · n dS = 0 Since the flow is steady and incompressible, the instantaneous rate of change of mass in the control volume vanishes. Hence the net flux of mass out of the control volume is zero i e ρ u · n dS = 0 S At the inlet, the velocity vector is parallel to and points in the opposite direction of the outer unit normal so that u1 · n1 = −u1 (y ). The outlet velocity vector is parallel to and points in the same direction as the outer unit normal so that u2 · n2 = u2 (y ) There can be no flow across the centerline and the channel wall, of course, since v = 0 and n = ±j wherefore u · n = 0. Thus, h − 0 h ρ u1 (y ) dy + 0 ρ u2 (y ) dy = 0 Substituting the given velocity profiles and dividing through by ρ yields h − 2U 0 y 1y − h 2h 2 h dy + λU 0 y y + h h 2 − y h 3 dy = 0 6.38. CHAPTER 6, PROBLEM 38 729 This equation can be rearranged to read as follows. h 2U h 0 y 1y − h 2h 2 d y = λUh h y y + h h h 0 2 − y h 3 d y h Next, it is convenient to change integration variables according to η ≡ y/h, wherefore 1 2U h 0 1 η − η 2 dη = λUh 2 1 0 η + η 2 − η 3 dη Dividing through by U h and evaluating the integrals gives 1 1 2 η2 − η3 2 6 η =1 η =0 1 1 1 = λ η2 + η3 − η4 2 3 4 or, 11 111 − =λ +− 26 234 Finally, solving for λ, the solution is 2 λ= 8 7 =⇒ η =1 η =0 2 7 =λ 3 12 ...
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