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CE 473/573 Groundwater Fall 2010 Comments on homework 1 1. The spread was not too large between the favorite and least favorite learning objectives. The results can be seen on the course website. 2. Most groups had the right idea: Because the soil properties do not change, the difference in conductivities is due to diﬀerences in density and dynamic viscosity. Then one can write ρw μoil Koil = Kw ρoil μw Remember that μ is the dynamic viscosity. For SAE 10 oil, my ﬂuids book gives μoil = 0.1 Ns/m2 . Also, be sure to show your work so that I can evaluate it. 4. Although there were some errors in determining an expression for K, the main thing missing from the answers of most groups was a comparison of the likely soil types determined from the value of K and the particle diameter. The former suggests a well-sorted sand, while the latter suggests a silt. One improvement in the model might be to try to relate the pore size and geometry to the particle size. 5. Most groups chose a conservative (i.e., safe) criterion of Re < 1. To be conservative, choose large values of K and particle diameter for sand to maximize the Reynolds number. 8. In class we found that h = h0 exp(−KAt/At L), and we ﬁt this curve to the data and used the ﬁtting coeﬃcient to ﬁnd K. I recommend that method. Using one group’s data, I ﬁnd K = 6 × 10−4 cm/s, which is Verifying Darcy’s law requires showing that the Reynolds number Re = vd/ν is smaller than 1 throughout the experiment. The maximum Reynolds number will occur when the head is greatest, so compute Remax = h0 d vmax d =K . ν L ν As before, if you have no information on the particle diameter, determine the soil type from the value of the hydraulic conductivity and use a large value of the diameter to be conservative. I ﬁnd that the Reynolds number is much smaller than one. Groups had quite a bit of trouble with the uncertainty analysis; be sure to ask me if you have further questions after reading my comments. I ﬁnd that the uncertainty is about 1.5% and that the burette area contributes about 75% of the squared uncertainty and the head contributes 25%.