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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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? The damage caused by electric shock depends on the current flowing through the body; 1 mA can be felt and 5 mA is painful. Above 15 mA, a person loses muscle control, and 70 mA can be fatal. A person with dry skin has a resistance from one arm to the other of about 1.1 × 10^5 Ohms. When skin is wet, the resistance drops to about 5100 Ohms.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/03/2009 for the course PHY PHY 2049 taught by Professor Robertr.criss during the Fall '08 term at University of South Florida.

Page1 / 8

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online