CHE218_solutions4

CHE218_solutions4 - Set 4 2008 4.2(a(b Relative to the gun barrel 2 ft 2000 mV 2 lbf s 2 s 3 K.E = = 0.02 lbm = 1.242 10 ft lbf = 1.684 kJ 2 2

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Set 4 2008 4.2 (a) Relative to the gun barrel, K . E . = mV 2 2 = 0.02 lbm 2000 ft s 2 2 lbfs 2 32.2lbmft = 1.242 10 3 ft lbf =1.684 kJ (b) V = V 0 - gt ; V = 0 when t = V 0 g = 2000 ft s 32.2 ft s 2 = 62.11s z = Vdt = V 0 t - 1 2 gt 2 = 2000 ft s 62.1s - 1 2 32.2 ft s 2 62.1s ( ) 2 = 6.21 10 4 ft (c) kJ 684 . 1 lbf ft 1242 ft lbm 32.2 s lbf 62130 2 . 32 02 . 0 2 2 = = = = ft s ft lbm mgz PE The fact that the initial kinetic energy and the potential energy at the top of the trajectory are the same is not an accident. Air resistance complicates this (see problem 6.94). 4.3* W = Fdx = mg Δ z = 2.0lbm 6 ft s 2 10ft lbfs 2 32.2 lbmft = 3.73 ft lbf = 5.05 J 4.7* System the pump, steady flow. Start with the enthalpy form of the energy balance (Eqn 4.13) and note dm in = dm out = dm (steady flow). Develop the “steady flow, open system” form by dividing through by dm [Example 4.5, Eqn 4.15]. Eliminate the terms that are 0 you are left with:
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This homework help was uploaded on 04/18/2008 for the course CHE 218 taught by Professor Smith during the Spring '08 term at SDSMT.

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CHE218_solutions4 - Set 4 2008 4.2(a(b Relative to the gun barrel 2 ft 2000 mV 2 lbf s 2 s 3 K.E = = 0.02 lbm = 1.242 10 ft lbf = 1.684 kJ 2 2

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