p1162

p1162 - 96 CHAPTER 11. POTENTIAL FLOW 11.62 Chapter 11,...

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Unformatted text preview: 96 CHAPTER 11. POTENTIAL FLOW 11.62 Chapter 11, Problem 62 Problem: Consider unsteady potential flow for uniform flow of constant velocity U i past a cylinder rotating counterclockwise with an angular rotation rate that varies with time, t, i.e., Ω = Ω(t). (a) Beginning with Γ = C u · ds and letting closed contour C be the cylinder surface, verify that the circulation is Γ = 2π ΩR2 . (b) Solve for the pressure, p, at the front of the cylinder where r = R and θ = π . (c) If p(R, π ) = p∞ + 1 ρΩ2 and Ω(0) = Ωo , solve for Ω(t). 2 y . . ............... ..................... . ................................................................. . ....... .............................................................................. . ....................................................................................... ....... ........ . .... .................................................................................................. ...... ....................................................... . . ............................................................................................................... ............................................................................................................ ... .......... .. ... .. ............................................................................................................ ........................................................................................................ . . ............................................... ... .. ... .. ... .. ............................................................................................................. .................................................................................................................. ... .. .. ........................................ ................................................... ............................................................................................................................ ... .. .... ......................... ........................ .... .... .... . ........................................................................................................... .................................................................................. ... ..... ....... ................................................................................................................................ ..... ..... . .. .... . ............................................................................................................................ .................................................................................................................... ..... ........ ..... .... .... .................................................................................................................... . .................................................. .......................... .... ...................................................................................................................... . .. . ..................................................................................................... ...................................................................................................................... ............................................................................................................... ...... . .. . . ....... . . ...... . ... . . ............................................................................................... ..................................................................................................... .. ................................................................... . ............................................................................................. .................................................................................................. ...... ...................................................................................... ................................................................................ ......................................................................... ................................................................. ......................................................... .. ......................................... ......... U ¡ ¡ ¡R x uθ (R θ) = ΩR − 2U s n θ Ω Solution: (a) Since there is no flow through the cylinder surface, ur (R, θ) = 0 which means the velocity on the cylinder surface is u = uθ (R, θ) eθ . Thus, since ds = Rdθ eθ , we have u · ds = Γ= C 2π = 0 = 2π 0 uθ (R, θ) eθ · Rdθ eθ ΩR2 − 2U R sin θ dθ ΩR2 θ + 2U R cos θ θ =2π θ =0 Therefore, the circulation is Γ = 2π ΩR2 (b) We know from the unsteady equivalent of Bernoulli’s equation that ∂φ 1 p + u · u + = f (t) ∂t 2 ρ 11.62. CHAPTER 11, PROBLEM 62 97 where f (t) is a function to be determined from the boundary conditions. Now, for flow past a rotating cylinder, the velocity potential is φ=U r+ R2 cos θ + R2 Ω(t)θ r =⇒ ∂φ dΩ = R2 θ ∂t dt To determine the function f (t), note that θ = 0 on the x axis. Also, p → p∞ and |u| → U as x → ∞. Hence, we have p∞ 1 = f (t) 0 + U2 + 2 ρ =⇒ 1 p∞ f (t) = U 2 + 2 ρ Thus, we conclude that on the surface of the cylinder, R2 dΩ 1 p 1 p∞ θ + u2 (R, θ) + = U 2 + θ dt 2 ρ 2 ρ Using the fact that uθ (R, π ) = ΩR, we find ρR 2 1 dΩ 1 π + ρ(ΩR)2 + p(R, π ) = ρ U 2 + p∞ dt 2 2 Therefore, the pressure at r = R, θ = π is given by 1 dΩ p(R, π ) = p∞ + ρ U 2 − Ω2 R2 − πρR2 2 dt (c) If p(R, π ) = p∞ + 1 ρΩ2 , then the equation derived in Part (b) for the pressure 2 simplifies to dΩ dΩ Ω2 1 =⇒ =− 0 = − ρΩ2 R2 − πρR2 2 dt dt 2π To solve we rearrange this equation as dΩ dt =− 2 Ω 2π =⇒ − 1 1 t + =− Ω Ωo 2π or 1 1 t 2π + Ωo t 1 + Ωo t/2π = + = = Ω Ωo 2π 2π Ωo Ωo Therefore, the cylinder’s angular velocity is Ω= Ωo 1 + Ωo t/2π ...
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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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