Solution Exercise 8

From the photograph it can be inferred that H = 1 m, so:

\displaystyle Ra=\frac{\Delta \rho gkH}{\mu D_{C}}=\frac{235\times 9.81\times 1.25\times 10^{-10}\times 1}{1\times 10^{-3}\times 5.0\times 10^{-10}}=6\times 10^{5}

Return to Exercise 8

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Variable-Density Groundwater Flow Copyright © 2022 by Vincent E.A. Post and Craig T. Simmons. All Rights Reserved.