Flux density help please (seawater flowing in a tube)

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The discussion focuses on calculating the flux density of salt in a tube connecting seawater and freshwater, emphasizing that the question lacks sufficient data for a definitive answer. Participants highlight the importance of considering diffusion coefficients and the limitations of using velocity and concentration due to the absence of advection. Dimensional analysis is mentioned as a method to argue the problem's complexity. Clarification is sought regarding whether tabulated values can be utilized in the calculations. Overall, the conversation underscores the need for more information to accurately determine the flux density.
olibee
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Homework Statement
this is a practice problem set
Relevant Equations
I know flux = concentration x velocity but I am not quite understanding if that is applicable here
What is the flux density of salt in a horizontal tube 10 cm in length connecting seawater (salinity = 30 g/l) to a tank of freshwater (salinity ~ 0) assuming no advection occurs?
 
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An answer cannot be given based on only the data you have given. This is easy to argue with dimensional analysis. Is this the full question as stated? Are you allowed to look up tabulated values such as the diffusion coefficient of salt in water?
 
And you are right. Velocity * concentration does not apply because that is based on an advective flux and it is explicitly stated that there is no advection.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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