a) Calculate the
Reynolds number for typical groundwater flow
in a clean sand. Use as characteristic length L a typical pore
size
for sand (Fig) and
determine the velocity U from Darcy's
law. Assume a hydraulic
gradient
of 1/100, a typical hydraulic conductivity (Fig),
and density/viscosity (Fig).
Do you expect to find laminar or turbulent flow?
I assume a pore diamter of 0.5 mm, hydraulic conductivity of K=10-4 m/s and a porosity of n=0.3 (clean sand), a viscosity of m=1.3*10-3 Pa*s (10oC), a density of 1000 kg/m3. Please note that this is just a rough estimate, therfore it is not necessary to chose exactly the same numbers. With Darcy's law the velocity is U = K/n * 1/00 = 3.3*10-6 m/s (= 100 m/year). The Reynolds number then is: R= ULr/m = 3.3*10-6m/s * 0.5*10-3m * 1000 kg/m3 / 1.3*10-3 Pas = 1.3*10-3. This Reynolds number is safely in the laminar domain (Fig).
b) Imagine flow of water through a
limestone cave. Could the
flow be turbulent there? Make a quantitative argument.
Most of you have seen a limestone cave before. Water flowing through the cave is dissolving the limestone and creates wide pathways. I am assuming a flow velocity similar to that in small streams: U = 0.5m/s, a length scale (diameter of the flowpath) of 1 m. Using the definition of the Reynolds number we are getting R = ULr/m = 0.5 m/s * 1m * 1000 kg/m3 / 1.3*10-3 Pas = 3.9*105, indicating that we can easly have turbulent flow in limestone caves.
2) (12
points) Flood estimate
A construction project is planned for a small non-monitored stream
near Oracle Arizona. This stream is typically dry during most of the
year,
but floods during extreme presipitation events. The planning agency is
concerned that their construction site might be flooded during extreme
years and wants to know how high the water might be able to rise in the
stream. One way is to estimate the maximum amount of precipitation to
be
expected in a day, assume that all the water makes it into the stream
(no
evaporation) and then convert the discharge rate into a depth using
Manning's
equation.
Info you'll need:
b)
Calculate the maximum discharge in the stream assuming that all
that
precipitation comes down and enters the stream in 2 hours. Express the
result in metric SI units.
Discharge in the river can then be calculated: Qmax = 42.3 miles2 * (1609.3 m/mile)2 * 2.17 inch * 0.0254 m/inch / 2 hours / 3600 s/hour = 839 m3/s.
c) Assume
that the channel has a V-shape cross section, with a 90o
angle at the bottom (draw a little sketch). Let us call the length of
the
sides of the "V": L and the depth of the channel: h. Then RH
= area/wetted perimeter = L2/(2*2*L) = L/4.
Remember
that Q=A*U, with A being L2/2. Rearange
Mannings equation to
get L. Then convert L into h, using L2 = 2h2,
valid
for this particular triangle. So, how deep can the stream get and how
far from
the center do you need to be in order to avoid flooding?
Mannings equation for this example then reads:
2Q/L2 = kappa/n*(L/4)2/3S1/2
The equation can then be solved for L: L =
(2*Q*n*42/3 / (kappa*S1/2))1/(2+2/3)
=> L = 40.2m => h = L/21/2
= 28.4m