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ENGR 215 ~ Dynamics Sections 14.4. Power. Power is the flow of energy per unit time Units Light Bulb: 100W 1 W = 1 Nm/s Car Engine: 250 Hp 1 Hp = 0.707 Btu/sec = 550 ft lb / sec = 746 W.
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Power • Power is the flow of energy per unit time • Units • Light Bulb: 100W • 1 W = 1 Nm/s • Car Engine: 250 Hp • 1 Hp = 0.707 Btu/sec = 550 ft lb / sec = 746 W
Lecture Example 1: How much power can a 20 amp circuit at 120 volts supply in your home at 66% of full load? How many horsepower is that equivalent to?
Efficiency • Efficiency = Power Output / Power Input
Lecture Example 2: Determine the average power (kW) required to accelerate a 1300-kg car from 0 to 100 km/h in 10 seconds. Assume a constant acceleration and the zero frictional losses. Also, determine the maximum power required.
Lecture Example 3: Determine power (kW) required for a 2500-kg Ford F-250 pickup truck to tow a 2000-kg trailer up a hill with 50 m rise for every 1000 m of road at 60 km/h. Also calculate your answer in horsepower.
Lecture Example 4: Trevor who weighs 150 lb challenges your instructor who weighs 180 lb to a bike race up a 1 km long hill with a rise of 100m for every 1000 m of road (5.74% grade). Both your instructor and Trevor are capable of generating 250W of power. By what distance does you instructor lose the race? Assume the bike weighs 22 lb and that air resistance is negligible.
Lecture Example 4.1: How much difference in time does a make if Trevor carries a full water bottle (500ml = 0.5 kg) up the hill in comparison to if he had dumped his water at the bottom of the hill.
Lecture Example 5: A river flowing steadily at a rate of 240 m3/s is considered for hydroelectric power generation. It is determined that a dam can be built to collect water and release it from an elevation difference of 50 m to generate power. Determine how much power (MW) can be generated from this river water after the dam is built.
Lecture Example 6: The material hoist and load have a total mass of 800 kg and the counterweight has a mass of 150 kg. If the upward speed of the hoist increases uniformly from 0.5 m/s to 1.5 m/s in 1.5 s, determine the average power generated by the motor during this time. The motor operated with and efficiency equal to 80%.Solve using P=Fv.Solve using P =ΔE/ Δ t