Homework #3 GEOL 4880 Humphrey fall 2012

1. a) from v = square root(2gDh): 1.4m/s

b) t_c = rg sin a D, where the D is the thickness of the plug: 2 x 103Pa

c) m_e = rg sin a D2/2v, where D is the thickness of the viscous layer and v is the surface velocity: ~360 Kg/ms

d) sin a ~ t_c / ( rg D), where D is the stopped thickness: ~ 7.5degrees

2. a) 2500

b) 20

c) depending on the estimate of precipitation you found, anywhere from 1 week to 1 month.

d) residence time 1 to 2 years.

e) From the first homework we know the area of the continents and the flux of sediment to the sea. The residence time depends on how deep you consider the continents to extend below sealevel. Answers in the range of 100Myr are typical, however we know the cores of continents are Billions of years in age. This is a typical result: for well mixed reservoirs the residence time we calculate is close to the actual average time material spends in the reservoir (average age) and also close to the average time spent in the reservoir by material leaving (transit time), while for poorly mixed reservoirs the residence time is less than the average age, and greater than the average transit time.

f) The hydraulic gradient is approximately rise over run, or slope or 1000m in 10km, or about 0.1. With a conductivity of 1m/day, the water is flowing at about 0.1m/day, or it takes about 100,000 days to get from the input to output. This is the residence time ( or about 300years ). The total input is given by the area times the input rate (2inches/year). I chose to work in a meter width of the aquifer. It is about 4km wide in the recharge zone. Thus the total input per unit width is 5cm/year * 4000m = 200m³/year per meter width (north-south width). This gives a volume of stored water in a 1meter wide slice of 200m³/year * 300yrs = 60,000m³ per 1 meter slice. To get the storage for the entire aquifer you just need to multiply by the north-south width, which on the map I gave you is about 20km. This gives a storage of on the order of 6x10⁸ m³, which you should compare with something like the Seminoe reservoir.

3. a) The shear stress as a function of depth, FIGURE

b) 0 from the sfc down to half depth, and a straight line increase below that to a max of a little less than 3 per sec. (I did not mark this question)

c) velocity as a function of depth, FIGURE

4a) velocity is easily computed from the time-position data. About 8m/s

b) Surprisingly not. Although 8m/s is fast, the acceleration of a mass on a 31degree slope (ignoring viscousity) is 5 m/s2, so that a scale time to reach terminal/steady state velocity is about 2secs, which means it probably got up to speed well before the first sensor

c) Depth at 20s at 90 meters is 0.12m, Basal normal stress is 2.4KPa. Since the total normal stress is rho g d cos(alpha), rho must be about 2400kg/m3. Note this difficulty in this question: should you use the 0.12 meter depth at 20sec or the 0.15 meter depth at 60 secs? Turns out you get similar answers.

d) at 60 secs Depth 0.15, Normal stress 3.1KPa, Pw 1.9, assume completely saturated. There are many ways to try to answer this question. The 2 methods I tried were:

1) Assume the gravel is made up of rock with a typical density 2700kg/m3, and with a bulk density of 2400kg/m3 from question 4c, then you can solve the bulk density equation ( bulk density = rock density*porosity + (1-porosity)*water density) to get the porosity of a little less than 20% The major error in this approach is that the water probably has a significantly higher density than 1000kg/m3 because of suspended fines that take time to settle.

2) Alternatively you can use the difference between the pore pressure at 20sec and 60sec. During this time the flow has stopped and the flow material changes from fluid supported to matrix supported. While the flow is fluid supported, the normal stress and pore pressure at the same. When it stops and is matrix supported, the difference between the pore pressure and the normal stress is the effective stress. The change in the effective stress (about 1200KPa) is just that the water is not supporting the weight of the rock. Therefore the change in effective stress is the buoyant weight of the rock (ie density of rock minus density of water). In other words, the change in effective stress is the porosity*(buoyant weight of the rock), since the buoyant density is about 1700kg/m3, then the porosity is about 30%.

In reality, the data is probably not good enough to justify either approach.

e) t_c = rho g h sin alpha ~ 500Pa

f ) vis = rho g sin(alpha) d^2/(2*v) ~ 10, which is 4 orders of magnitude more than water

g) While the flow is moving the density includes the weight of the rocks, since the rocks are being suspended by the water! The scale is not the time scale to drain water, but the time scale for rocks to settle in the flow. See also the stuff in D2 above.