p1022

# p1022 - dU/dt = 0 when the terminal velocity is achieved,...

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10.22. CHAPTER 10, PROBLEM 22 1097 10.22 Chapter 10, Problem 22 In general, Bond’s motion downward is governed by the following equation. m dU dt = mg ±,²1 Gravity 1 2 ρ U 2 C D A ± 1 Body drag 1 2 ρ U 2 π 4 d 2 C Dp ± 1 Parachute drag where m is his mass, g is gravitational acceleration, ρ is the density of air, U is velocity, C D is the drag coefficient of his body, A is his frontal area, C Dp is the parachute drag coefficient and d is the parachute diameter. 10.22(a): When the terminal velocity is achieved, dU/dt =0 . Hence, multiplying through by 2 / ( ρ U 2 ) and rearranging terms yields 2 mg ρ U 2 C D A = π 4 d 2 C D p Solving for the parachute drag coefficient, we find C Dp = 4 π d 2 ^ 2 mg ρ U 2 C D A ³ Now, we are given d =4 ft, mg = 180 lb, ρ =0 . 00234 slug/ft 3 and C D A =10 ft 2 .So , the parachute drag coefficient is C Dp = 4 π (4 ft) 2 2 p 180 slug · ft / sec 2 Q p 0 . 00234 slug / ft 3 Q (50 ft / sec) 2 10 ft 2 =4 . 10 10.22(b):

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Unformatted text preview: dU/dt = 0 when the terminal velocity is achieved, Bonds weight balances the drag so that mg = 1 2 } C D A + 4 d 2 C D p ] U 2 Hence, the terminal velocity is U = &gt; : 2 mg C D A + 4 d 2 C Dp = So, for the given values, the terminal velocity is U = &gt; : 2 p 180 slug ft / sec 2 Q p . 00234 slug / ft 3 Q 10 ft 2 + 4 (4 ft) 2 C Dp = = 124 1 + 1 . 257 C Dp 1098 CHAPTER 10. VORTICITY AND VISCOSITY Now, for a conventional parachute, the drag coefficient lies in the range 1 . 20 C D p 1 . 40 Therefore, for the limiting values of C D p , the terminal velocity would be U = l 78 . 3 ft / sec , C D p = 1 . 20 74 . 6 ft / sec , C Dp = 1 . 40...
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## p1022 - dU/dt = 0 when the terminal velocity is achieved,...

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