quiz2_f03_sol - 1 2.25 QUIZ 2, 13 NOVEMBER 2003 ANSWER TO...

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1 2.25 QUIZ 2, 13 NOVEMBER 2003 ANSWER TO PROBLEM 1 Ain Sonin NOTE: All extensive quantities are per unit width , consistent with a 2D problem. The answer given here for part (b) is much more complete than was required for the quiz. Momentum theorem applied to the fixed CV 1 : d ± ² v v dV + ± v v dA = ( F ) CV 1 (1) dt CV 1 x rn CS 1 x rn x First, to obtain the outflow velocity v ( x , t ) between the plates, we apply mass x conservation to CV 1 : d ± dV + ± v rn dA = 0 dt CV 2 CS 2 (2) During the closing stroke stroke , d ± = ² ³ dt (3) and eq. (2) takes the form
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2 d ± x ± x ² ² ³ ³ ± x = 0 (4) ´ + v x dt µ 2 from which, using (3), ± x v ( x , t ) = (5) x 2 ( t ) Substituting next from (5) into (1), we have d L ³ x ´ · ³ L · 2 µ · ¸ ¸ x ¸ wdx + · ¸ ¸ Lw = F + ( p a (6) ³ p exit ) L dt 0 ¹ 2 ¹ ¹ 2 ¹ F is the thrust force exerted in the x-direction on the CV at the “cut” through the supports, which is equal to the thrust force in the –x direction. The second term on the right accounts for the possibility that the pressure at the exit plane differs from the
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quiz2_f03_sol - 1 2.25 QUIZ 2, 13 NOVEMBER 2003 ANSWER TO...

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