The Columbia River drains into the Pacific ocean at a rate of7, 500 m^3 of water
ID: 481943 • Letter: T
Question
The Columbia River drains into the Pacific ocean at a rate of7, 500 m^3 of water/sec. The river is fresh water, and the ocean is salt water, which is about 1 M in ions total (not worrying about which kind). You may assume a temperature of 298 K for this problem. If there were a semi-permeable membrane that let water through but not salt, how great would the osmotic pressure be between the ocean and the river? One atmosphere of osmotic pressure w rise water 32 feet or 10 meters. Using the osmotic pressure how high would the ocean rise compared to the fresh water? This membrane then could act like a dam with the height given by part b). How much energy would be captured by a turbine run by the water falling over a barrier such as this? What is the potential energy output for one cubic meter of water to fall 200 meters? A typical power plant generates 600 MW. If the Columbia drains 7, 500 m^3/sec of water, how many power plants could be replaced by a 200 meter high dam at the mouth of the river?Explanation / Answer
a) Osmotic pressure = concentration X R X T
osmotic pressure = 1 M X 0.0821 L-atm / mol K X 298 = 24.47 atm
b) one atomospheric pressure raised the water level = 10 Meters
so 24.47 atm pressure will raise water level to = 10 X 24.47 meters = 244.7 meters
c) Potential energy = mass X g X h
h = height
g = acceleration due to gravity = 9.8 / sec2
Potential energy = 1000 X 9.8 X 244.7 = 2398060 Joules / m^3 = 2.398 MJ / m^3
d) Potential energy = Mass X g X h
Potential energy = 1000 x 9.8 X 200 = 1.96 MJ / m^3
e) Power = ( Potential energy / m^3 ) X (speed of river / second) = 1.96 MJ / m^3 X7500m^3 / seconds
Power = 14.7 X 10^3 MJ / sec
A typical power plant generates 600 MW
So number of power plants needed = 14.7 X 10^3 MJ / sec / power generated by one plant
Number of power plants = 24.5 (approx 25)
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