Sol 3 - .—a M.—_— —_mu—n—n-——I— 8.64...

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Unformatted text preview: _.___. _ _.—a__-. M....—_—__ . _-. —__._mu—n—n_-_-——I— 8.64 Consider the polar—coordinate velocity potential a = Brl'zcostllfl), where s is a constant. [3) Determine whether via: 0. If so, [b] find the associated stream function Mr, 49} and {c} plot the full streamline which includes the x-axis (6': Ill} and interpret. Solution: {a} It is laborious, but the velocity.r potential satisfies Laplace’s equation in polar coordinates: is as is 32¢ . v2 =—— — —— — at: r :3 '3 1.23 A . (I) r3r[r3r]+r2 3r[392] l $ r “fl ) as {a} {h} This example is one of the lialnilglr of “corner flow" solutions in Eq. {3.49}. Thus: yr = armsmaea) Ans. {b} [c] This function represents flow around a 15W“ corner, as shown below. Ans. {c} Flg. PB.E4 3.68 Investigate the complex potential fiusction fie} = Umz + m ln[[z +a).-'[z — all], where m and a are constants, and interpret the flow pattern. Flg. Pass Solution: This represents flow past a Rankine oval, with stream Function identical to flsat given by Eq. {3.29}. ”._.—u - --------r-u-u-—-—-—rq-._.--._.-W—u—u._.—- 1- "i." -—-—--wvwW"—|fi"" 'rr— —- —w§_—Qfi£méfl arm __.. _ — “:1 : _.__'I'-_TE___‘—.+_.,:I.rr{. . _.. _- _ __ it“ #:3513- .itmg_ ....... _ . 5.4552 {H+1}’+‘*2} f _F ' -1}"'”“2}+{H+1 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ n _ _ _ _ _ _ _ _ L. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ L. _ _ _ _ _ _ _ _ _ _ n _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ L. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 5.4551 5.455? -------- ---------------------P--------------------------------P--------------------------------P------------------------------ 5.4555 5.4555 5.4553 2.? .55 2 2.5 .55 2 2.5 3.111} For full}.r developed laminar incompressible flovtr through a straight noncircular duct, as in Sec. 6.3, the Navier—Stokes Equation {4.33} reduce to 32a 32a: trip —+— =——= const {U 3y” 822 six where (y, a} is the plane of the duct cross section and x is along the duct axis. Itifu‘avitgpr is neglected. Using a nonsquane rectangular grid (or, av), develop a finite-clifl’erence model for flfls equation, and indicate hov.r it may.r he applied to solve for flow in a rectangular duct of side lengths n and it. i+l.j+l i-1.§,+1 n+1 an] i+1.j-i Flg. PE.11D Solution: An appropriate square grid is shown above. The finite-difference model is “5+1. j _ 211:4 + “t—t ' “5,141 — 2% + uLj—t 1 I111 2 “+—2w——, or, if nv=nz, to?) trim) Ii dx 1 (as): up “HZ; “Ig+1+u|,J—1+“mg+ul—r,1— I: E Ans. This is “Poisson’s equation,” it looks like the Laplace model plus the constant “source" term involving the mesh size my} and the pressure gradient and viscosity. ...
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