CHE218_solutions4

# CHE218_solutions4 - Set 4 2008 4.2(a(b Relative to the gun...

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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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CHE218_solutions4 - Set 4 2008 4.2(a(b Relative to the gun...

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