#3 - p 2 V 2 2 p 2 V 2 2 = 84.536 V 2 = p 1 V 1 V 1 = 188.9...

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Homework # 2 Solutions     ME 300 Prof. Vanka  Fall 2008 2.55   Mass = 10 kg.  Heat transfer from the mass = 5 kJ/kg. Hence Q = - (10) x 5 = -50 kJ Elevation decrease = decrease in PE = mg  z  = 10(9.7) (50) = 4850 ;    PE  = -4.85  kJ KE = 0.5 x m (30 2 – 15 2) = 0.5 x 10 x (900-225) = 3375 J = 3.375 kJ U = -(10) x 5kJ = -50 kJ  E = U + ∆ KE + ∆ PE = Q – W; W = - 50 – (-50) – 3.375 – (-4.85) = 1.475kJ 2.66 Mass = 1.2 lbm; p 1 V 1 2 = p 2 V 2 2 ; n = 2 Work = (p 2 V 2 – p 1 V 1 )/(1-n) = (p 2 V 2 – p 1 V 1 )/(-1) u 2 – u 1 = 990.58 – 1363.3 = -372.72 Btu/lbm U = 1.2 x 372.72 = - 447.264 Btu Q = -342.9 Btu U = Q – W; W = Q – U = -342.9 + 447.264 = 104.364 Btu p 2 V 2 = p 1 V 1 – 104.364 Btu p 1 V 1 = (500)lbf/in 2 x 144 x 1.701 ft 3 /lbm x 1.2 = 146966.4 lbf. ft p 2 V 2 = 188.9Btu -104.364 = 84.536 Btu p 1 V 1 2 =
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Unformatted text preview: p 2 V 2 2 p 2 V 2 2 = 84.536 V 2 = p 1 V 1 V 1 = 188.9 Btu x 1.701 Hence V 2 = 188.9 x 1.701 / 84.536 = 3.8 ft 3 2.83 = 0.40; Energy generated = 100 MW; a) Value of electricity generated per year = 0.08 x 100 x 1000 x 8000 = $64 million b) Qin = 100/0.4 = 250 MW; Per year = 250 x 8000 x 3600 MJ = 7.2 x 10 9 MJ Fuel cost = 4.50 x 7.2 x 10 9 /10 6 = 32.4 million dollars. c) It is profit, but there are other costs, such as capital costs, operational costs, inefficiencies in transmission, etc. 2.89 Refrigerator, COP = 2.4. Qin = 600 Btu/h. COP = Qin/Work; Work = 600/2.4 = 250 Btu/h = 0.293 x 250 = 73.25 Watts; Energy consumption = 73.25 x 360 / 1000 = 26.37 kW.h Cost = $ 2.10 / month 2.91 COP Heat Pump = 2.5 = Qout / Work = 20 kW / Power; a) Power = 20/2.5 = 8 kW b) Cost = $0.08 x 8 x 200 = $128 per month...
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#3 - p 2 V 2 2 p 2 V 2 2 = 84.536 V 2 = p 1 V 1 V 1 = 188.9...

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