Here's an outline of what you have to do:
A watt is one joule per second, W = J/s
A joule is a kg m2 /s2
Work done to lift a mass in a gravitation field = mass X gravitational constant at Earth's surface X change in height
gravitational constant at Earth's surface = 9.8 m/s2
If one kW of electricity is required, which is 1000 J of energy per second, what mass must change height by 10 meters to do this much work? We are assuming perfect efficiency for the moment.
1000 kg m2 / s2 = mass X 9.8 m/s2 X 10 m
Solve for the mass, which will be in kg units. This is the mass of water that will produce 1000 J of work by falling a distance of 10 meters. This is the mass that must fall 10 meters every second to produce 1000 J of electrical energy by the power plant every second, which is 1000 watts or 1 kW.
i) If the power plant is 70% efficient, it is only converting 70% of the energy of the falling water to electrical work. The power plant needs to produce 1000 J of energy per second. Seventy percent of what number equals 1000 J ? That is, 0.70x = 1000 J, x = 1429 J. So, what mass of water must fall 10 m to produce 1429 J of work?
ii) A kWh (kilowatt hour) of electricity is 1000 J/s X 3600 s = 3,600,000 J. The coal-fueled power plant producing this amount of energy produces 2.2 kg of CO2.
If 2.2 kg of CO2 are produced for every 3,600,000 J of energy produced by the coal plant, how many kg of CO2 are produced for every 1000 J of energy? This would be the kg of CO2 produced every second, and 1000 J is the energy demand per second for this particular power plant. Multiply this kg of CO2 by the number of seconds in a year to get the kg of CO2 produced by the coal-fueled power plant per year. This is the amount of CO2 that would not be produced if the power plant was hydroelectric.
Steve
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