PhyHWsolutions4

# PhyHWsolutions4 - Find the rate with which falling water...

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Find the rate with which falling water must supply energy to the turbine? Part 1: A generator at Hoover Dam can supply 379 MW of electrical power. Assume that the turbine and the generator are 86.5 % efficient. The acceleration of gravity is 9.8 m/s2 : Find the rate with which falling water must supply energy to the turbine. Answer in units of MW. Part 2: The energy of the water comes from the change of the potential energy, mgh. What is the needed change in potential energy each second? Answer in units of J/s. Part 3: If the water falls 17.2 m, what is the mass of the water that must pass through the turbine each second to supply this power? Answer in units of kg/s. << A generator at Hoover Dam can supply 379 MW of electrical power. Assume that the turbine and the generator are 86.5 % efficient. The acceleration of gravity is 9.8 m/s2 : Find the rate with which falling water must supply energy to the turbine. Answer in units of MW. >> Efficiency = output/input Input = output/efficiency Input = 379/0.865 Input = 438.15 MW << The energy of the water comes from the change of the potential energy, mgh. What is the needed change in potential energy each second? Answer in units of J/s. >> Conversion factor : 1 watt = 1 joule/sec. Therefore, if the input energy = 438.15 MW, this is equivalent to 438,150,000 watts which is numerically equal to 438,150,000 joules/sec. << If the water falls 17.2 m, what is the mass of the water that must pass through the turbine each second to supply this power? Answer in units of kg/s >> Energy = 438,150,000 = m(9.8)(17.2)

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and solving for "m", m = 438,150,000/(17.2 * 9.8) m = 2,599,371.14 kg./sec. How much charge passes a cross- sectional area of the circuit in this time? The compressor on an air conditioner draws 91 A when it starts up. The start-up time is about 0.5 s. How much charge passes a cross-sectional area of the circuit in this time? Answer in units of C. C = i*t = 91*0.5 = 45.5 Coulombs 91 amps is 91 coulombs per second, by definition. That is 45.5 coulombs in 0.5 seconds. What is the minimum voltage placed across the arms that would produce a current that could be felt by a person?
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