# CE 473/573 Groundwater Fall 2010 Comments on homework 1

```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 &lt; 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 &times; 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%.
```