• If the world production of neodymium is 7000 tonnes/yr. (7 x 106 kg), what is
ID: 1842076 • Letter: #
Question
• If the world production of neodymium is 7000 tonnes/yr. (7 x 106 kg), what is the maximum annual production of Prius vehicles possible without regeneration?
• If each vehicle has an operating life of 10 years and all the neodymium in each vehicle is recycled, what is the maximum possible annual vehicle production in steady state in 10 years? 20 years?
• What is the depletion time for the neodymium resource if the total reserve is 8 million tonnes (8 x 109 kg) and the only application is for identical Prius motors? Calculate for both with and without 100% regeneration
. • If each vehicle uses 1000 kg of iron, and the world reserves of iron are 300x1012 kg, what is the system [vehicle] resource depletion time without regeneration assuming no other uses for the iron?
• If other applications initially use 10% of the neodymium resource, and all users increase their consumption 1%/year (c = 7 x 106 kg/yr x 0.01/year) what is the resulting depletion time with no regeneration?
• If all users increase their consumption 2%/year (c = 1.4 x 105 kg/yr2) and regeneration begins at zero and increases at 90% of the rate of the increase in consumption, what is the resulting depletion time?
Explanation / Answer
Each Prius vehicle uses neodymium = 2.2 pound or 1 Kg
1.Maximum annual production of Prius vehicle = 7000000/year = 7000000/ year answer
2.Maximum possible annual vehicle production in steady state in 10 years= 70000000
in 20 years= 140000000
3. depletion time without regeneration = 1142.857143 years
depletion time with regeneration uses = y=a(1-E(-nt))
, here a = 7000000 n=100% t=time of depletion
4. total number of vehicle in a year = 7000000 / year
usage of iron per vehicle = 1000 kg
total iron usage in an year= 7000000000 kg
depletion time = 4285.714286 years (without regeneration)
If other applications initially use 10% of the neodymium resource, and all users increase their consumption 1%/year (c = 7 x 106 kg/yr x 0.01/year) what is the resulting depletion time with no regeneration?
y= a(1+n)^t
700000= 700000(1.01)^t
depletion time =231.407 years
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