CE En 544 - Exam #1 - SEEPAGE KEY

Winter Semester 2018

Essay Questions

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18. (4 pts) (seep-welleq) Two wells are placed in an unconfined aquifer under the following conditions:

k = 20 ft/day
R = 2000 ft
Q1 = 30 gpm
Q2 = 20 gpm
Ho = 30 ft

If the drawdown at a point halfway between the wells is 10.0 ft, how far apart are the wells?

k=20/(60*24)=0.0139 ft/min Q1 = 30/7.48 = 4.01 ft^3/min Q2 = 20/7.48 = 2.67 ft^3/min h=30-10 = 20 ft h=(H^2-1/(pi*k)*sum(qi*ln(R/ri))^0.5 20=(30^2-1/(pi*0.0139)*(4.01*ln(2000/ri)+2.67*ln(2000/ri)))^0.5 solving for ri: ri = 76.5 ft distance = 2*76.5 = 153.0 ft.

19. (2 pts) (seep-goveq) Describe the difference between an analytical and numerical solution to a profile modeling problem.

An analytical solution is a direct solution to the governing equation in the form of an analytical function. Typically, analytical solutions are only feasible for simple problems. Simplifying assumptions are often used. A numerical solution is found using a scheme such as finite difference or finite element which produces an approximate solution over the problem domain. It can be used on a much wider variety of problems.

20. (4 pts) (seep-analytical) A flow path computed from a particle tracking analysis is shown below. The
particle travels from the river to point A. The length of the flow path is approximately 985 m. The aquifer has a relatively uniform thickness. The elevation of point A is about 705 m. Estimate the travel time (in days) from the river to point A.

vs = ki/ne x/t = k*(dh/dl)/ne t = (x*ne)/(k*dh/dl) t = (985*.21)/((1e-3/100)*(750-705)/985) t = 452771667 sec t = 452771667/(3600*24) t = 5240 days

21. (4 pts) (seep-welleq) A well is placed in a 5 m thick confined aquifer. The thickness of the aquifer is relatively constant. The well is pumped at 1.3 m^3/min for several days until the well reaches steady state. The radius of influence of the well is estimated to be 400 m. The radius of the well is 20 cm. Before pumping, the elevation of the potentiometric surface is 120.0 m. fter the pumping reaches steady state, the elevation of the water in the well is 112.8 m.

a. What is the average horizontal hydraulic conductivity in the aquifer [m/d]?
b. A monitoring well is placed 100 m from the pumping well, estimate the drawdown in the monitoring well [m].

a. Q=1.3*60*24 = 1872 m^3/d k=(q*LN(R/rw))/(2*pi*D*(H-Hw)) k=(1872*LN(400/0.2))/(2*pi*5*(120-112.8)) k=62.9 m/d b. h = H + q/(2*pi*k*D)*ln(r/R) h = 120+1872/(2*pi*62.9*5)*ln(100/400) = 118.69 s =120-118.69=1.31 m