Hw 2 solutions

Hw 2 solutions - Solution of Homework 2 2-16 Two sites with...

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Solution of Homework 2 2-16 Two sites with specified wind data are being considered for wind power generation. The site better suited for wind power generation is to be determined. Assumptions 1 The wind is blowing steadily at specified velocity during specified times. 2 The wind power generation is negligible during other times. Properties We take the density of air to be ρ = 1.25 kg/m 3 (it does not affect the final answer). Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V 2 /2 per unit mass, and 2 / 2 V m & for a given mass flow rate. Considering a unit flow area ( A = 1 m 2 ), the maximum wind power and power generation becomes kJ/kg 0245 . 0 /s m 1000 kJ/kg 1 2 ) m/s 7 ( 2 2 2 2 2 1 1 1 mech, = = = = V ke e kJ/kg 050 . 0 /s m 1000 kJ/kg 1 2 ) m/s 10 ( 2 2 2 2 2 2 2 2 mech, = = = = V ke e kW 0.2144 kJ/kg) )(0.0245 m m/s)(1 )(7 kg/m 25 . 1 ( 2 3 1 1 1 mech, 1 1 mech, 1 max, = = = = = Ake V e m E W & & & kW 0.625 kJ/kg) )(0.050 m m/s)(1 )(10 kg/m 25 . 1 ( 2 3 2 2 2 mech, 2 2 mech, 2 max, = = = = = Ake V e m E W & & & since 1 kW = 1 kJ/s. Then the maximum electric power generations per year become area) flow m (per h/yr) kW)(3000 2144 . 0 ( 2 1 1 max, 1 max, kWh/yr 643 = = = t W E & area) flow m (per h/yr) kW)(2000 625 . 0 ( 2 2 2 max, 2 max, kWh/yr 1250 = = = t W E & Therefore, second site is a better one for wind generation. 2-39 A ski lift is operating steadily at 10 km/h. The power required to operate and also to accelerate this ski lift from rest to the operating speed are to be determined. Assumptions 1 Air drag and friction are negligible. 2 The average mass of each loaded chair is 250 kg. 3 The mass of chairs is small relative to the mass of people, and thus the contribution of returning empty chairs to the motion is disregarded (this provides a safety factor). Analysis The lift is 1000 m long and the chairs are spaced 20 m apart. Thus at any given time Wind V , m/s Wind turb n
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there are 1000/20 = 50 chairs being lifted. Considering that the mass of each chair is 250 kg, the load of the lift at any given time is Load = (50 chairs)(250 kg/chair) = 12,500 kg Neglecting the work done on the system by the returning empty chairs, the work needed to raise this mass by 200 m is () kJ 24,525 /s m kg 1000 kJ 1 m) )(200 m/s kg)(9.81 (12,500 2 2 2 1 2 = = = z z mg W g At 10 km/h, it will take t == = = distance
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This note was uploaded on 04/16/2008 for the course ME 2322 taught by Professor Oler during the Spring '06 term at Texas Tech.

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Hw 2 solutions - Solution of Homework 2 2-16 Two sites with...

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