#2 - Homework#2Solutions:ME300VankaFall2008 Problem2.14

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Homework # 2 Solutions:  ME 300--- Vanka   Fall 2008 Problem 2.14 KE + PE = constant;   z 1   =  0,  V 1   =  200 ft/s.  m = 50 lbm,  g = 31.5 ft/s 2 ½ m V 2  + mgz = ½ m V 1 2  + mgz 1 Solve for V = (V 1 2  – 2 g z) 0.5   When V 2  = 0, z 2  = V 1 2  / (2 g)  =  (200) 2 / (2 x 31.5)  =  634.92 ft. Make a plot, which will be a parabola.  Problem 2.25 Known:  Mass =0.25 kg;  initial specific volume = 0.2 m 3 /kg Work input = -15 kJ. constant pressure = 5 bars; Work =  pdV  =  m  p dv  =  m p (v 2 - v 1 )  =  -15 kJ v 2   =  (-15 kJ) / (0.25 x 5 x 100 kg. kPa) + v 1           =  (-15)  (1000) (N. m) /(0.25 x 5 x 100 x 1000  kg N/ m 2 ) + v 1 v 2   =  -0.12 + 0.2  =  0.08 m 3 /kg;   V 2   =  m  v 2   =  0.25 x  0.08 = 0.02 m 3
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Problem 2.28 Given: N 2  gas, compression, pV 1.35
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#2 - Homework#2Solutions:ME300VankaFall2008 Problem2.14

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