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p1152

# p1152 -

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Unformatted text preview: 11.52. CHAPTER 11, PROBLEM 52 77 11.52 Chapter 11, Problem 52 Problem: In a hurricane, the wind is blowing with velocity U = 50 m/sec, freestream pressure p∞ = 101 kPa and density ρ = 1.20 kg/m3 past a quonset hut, which is a half cylinder of radius R = 4 m and length L = 16 m out of the page. To help the hut survive the hurricane, the internal pressure, pi , is regulated so that there is zero net vertical force on the hut. Using potential-flow theory, determine pi . U n = cos θ + j s nθ p ρp ............................................................................................................................................................................................................................................................................................................................................................... .......................∞............................................................................................... ......... ...... ................... ...... ........................................................................................... .............................................................................................................................................................................................................................................................................................................................................................................................................. ..... ...... ...... ............. ...... ...... ...... ...... .... .... ................................................................................................................................................................................................................................ ............ ... ... .. .. .. . .. ..................................................................................................................................................................................................................................................................................................................................................................................... .. .... .......................................................... .... . ..... .................................................................................. ........................... .......................... ...... ........ ............... ...................................... . . ..... .... . .............. . . . ... . . . . . ............. .............. ...... . . .......................................................................................................................................................................................................................................................................................................................................................................................... ......................................................................................................................................................................................................................................................................................................................................................................... ... . . . .. .. .. .. .. .. . . . . . . . . ............. ................ ...... .............. .......................... .. .... ............... .. . ..... . .. ..... .. . ...... ... .... . .. ... . .. .................................................................................................................................................................................................................................................................................................................................................................................... .......................................................................................................................................................................................................................................................................................................................................... . .................................... ... ........ .. . .. . . . ... . .... ... . . .. ..... .............. ................ ............... ..... ......... .......... ....... .. R Solution: Since the x axis is a symmetry plane for flow past a cylinder in an infinite fluid, we can use the solution obtained by superposing a uniform stream and a doublet. Then, the pressure exerted by the fluid on the outer surface of the hut, pf (θ), will be 1 pf (θ) = p∞ + ρ U 2 1 − 4 sin2 θ 2 The net pressure on the hut is the difference between pf and the internal pressure, p , viz., ∆p = pf (θ) − p The net force on the hut, which has width L out of the page, is π F = −L 0 ∆p n Rdθ = −LR π ∆p [i cos θ + j sin θ] dθ 0 where n is the outer unit normal to the hut. So, the net force on the hut becomes F = −LR i = −LR i −LR j π 0 π 0 π 0 ∆p cos θ dθ − LR j π ∆p sin θ dθ 0 1 p∞ − p + ρ U 2 1 − 4 sin2 θ 2 1 p∞ − p + ρ U 2 1 − 4 sin2 θ 2 cos θ dθ sin θ dθ Now, the horizontal component of the force is exactly zero as the integrals of cos θ and sin2 θ cos θ both vanish. This reflects the fact that the pressure is symmetric about θ = π /2. Also, we know that π sin θ dθ = 2 0 π and 0 sin3 θ dθ = 4 3 78 CHAPTER 11. POTENTIAL FLOW Thus, the vertical force on the hut, Fy , is 1 8 Fy = −2LR p∞ − pi + ρ U 2 + LRρ U 2 2 3 Therefore, after combining like terms, the net force on the quonset hut is 5 F = 2 (pi − p∞ ) + ρ U 2 LR j 3 We are given the fact that the internal pressure is regulated so that there is no net vertical force on the hut. Thus, 5 2 (pi − p∞ ) + ρ U 2 = 0 3 =⇒ 5 pi = p∞ − ρ U 2 6 For the given values, pi = 101 kPa − 5 kg 1.20 3 6 m 50 m sec 2 10−3 kPa kg · m/sec2 = 98.5 kPa Note that this result is completely independent of the hut radius, R, and width L. ...
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