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Unformatted text preview: 11.58. CHAPTER 11, PROBLEM 58 89 11.58 Chapter 11, Problem 58 Problem: The Flettner-Rotor ship was designed with two rotating cylinders of height H acting as sails. The cylinders have diameter D and rotate with angular velocity Ω. The ship moves with velocity Us = U i, the wind velocity is Uw = −V j and the air density is ρ. (a) Estimate the force on the ship as a function of Ω, D, H , U , V and ρ. (b) Determine the thrust for Ω = 800 rpm, D = 2.85 m, H = 16 m, U = 3 knots, V = 12 knots and ρ = 1.2 kg/m3 . Uw = −V j y Us . . . .......................................................................................................................................................................................................................................................................................................................................................................................................................... .. .. . ... ... .. .... .. . .. . .. .... ................ ................................................................. ................................................................................................................................................................................................................................................................................................... . . . . .. . .. . . .... .. .................................................................................................................................................................................................................................................................................................................................................................................................................................. .. . .. . . . ........ ... .......... .... ..... ..... ....................... ...................... .... ................. . ......... .................. .... ... ... .... .. ....... ............ ....................................................................................................................................................................................................... . . .......... .................................................... .. .. . . . ............................................................................................................................................................................................................................................................................................................................................................................................................................................................... . .. . . ... .. . .. . ........... ....... .... . .. . . ......................................................................................................................................................................................................................................................................................................................................................................................................... .. . . .. ... . ... .. . . . . . ..... . ... ... . ......... ...... . ........ ... ... ... ... ... ... ... . . .. ...................................................................................................................................................................................................................................................................................................................................................................................................................... . . ... ... ..... .. ........ ... . .. .. . . . ... .. . ... ... ...... ........... ....... ................................................. . . . . . .................. . ........................................................................................................................ ...................................................................................................................................................................................................................................................................................................................................................................................................................... . ... . ..... .... . . . . ... . . ... .. . | Ω x | Ω Solution: (a) The Kutta-Joukowski law tells us the force on each cylinder is ρ u × ΓH , where u is the freestream velocity in a coordinate frame where the cylinder is at rest. Using a Galilean transformation, we have u = −Us + Uw = −U i − V j Hence, the total force on the ship is F = −2ρH (U i + V j) × Γ Since the cylinders rotate in the clockwise direction with angular-rotation speed Ω, the circulation vector is D π (π D)(−k) = − ΩD2 k Γ= Ω 2 2 Thus, the force is π F = −2ρH (U i + V j) × − ΩD2 k 2 2 = πρ ΩD H (U i × k + V j × k) = πρ ΩD2 H (V i − U j) 90 CHAPTER 11. POTENTIAL FLOW (b) The thrust is the x component of F. For the given values, Ω = (800 rpm) 2π sec−1 60 rpm = 83.78 sec−1 , V = (12 kt) 0.514 m/sec kt = 6.168 Therefore, the thrust is Fx = π 1.2 kg m3 83.78 sec−1 (2.85 m)2 (16 m) 6.168 m = 253 kN sec m sec ...
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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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