SolHW5ME2320 - = 0.776 gen = P elec/P tur = 0.9375 tot = turbine gen = 0.7275 tot = 0.7275 2 – 76 Given D = 100 m V = 8 m/s overall =

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2 – 64 All the electrical energy consumed by the motor, including the power delivered to the fan will eventually dissipate as heat. Thus, all the energy supplied will become heat. Q gen = W in = W shaft / η = (0.25 hp)/0.54 = 0.463 hp = 345 W 2 – 70
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2 – 71 Given: h = 70 m; m = 1500 kg/s; P tur = 800 kW; P elec = 750 kW. Find: η total Solution: P available = mgh = (1500kg/s)(9.81m/s 2 )(70m) P available = 1030.05 kW η turbine = P tur /P available
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Unformatted text preview: = 0.776 η gen = P elec /P tur = 0.9375 η tot = η turbine η gen = 0.7275 η tot = 0.7275 2 – 76 Given: D = 100 m; V = 8 m/s; η overall = 32%; ρ air = 1.25 kg/m 3 ; Find: P gen , E gen in 24 hrs., Revenues if $0.06/kWhr Solution: P gen = 0.5mV 2 η =0.5 ρ VAV 2 = (0.5)(1.25)(.25 π )(100) 2 (8) 3 (0.32) P gen = 804.25 kW E gen = P gen X time = 19301.95 kWhr Revenues/day = $ 1158 2 - 80...
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This note was uploaded on 04/17/2008 for the course ME 232 taught by Professor Monefort during the Spring '08 term at Western Michigan.

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SolHW5ME2320 - = 0.776 gen = P elec/P tur = 0.9375 tot = turbine gen = 0.7275 tot = 0.7275 2 – 76 Given D = 100 m V = 8 m/s overall =

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