CE En 547 - Exam #1 KEY

SAMPLE

Essay Questions

Click here to access the Equation Sheet.

22. (head-darcy) An artesian aquifer is overlain by three soil layers and a small pond as shown below. How much water flows into the pond (in ft^3) from the aquifer per square foot of area in one month's (30 days) time?

k1=1e-2/(2.54*12)*3600*24=28.3 ft/d k2=3e-3/(2.54*12)*3600*24=8.50 ft/d k3=1e-2/(2.54*12)*3600*24=1.39 ft/d Keq = sum(H)/sum(Hi/Ki) = (4+5.2+8.1)/(4/28.3+5.2/8.5+8.1/1.39) = 2.63 ft/d i = dh/L = 4/(4+5.2+8.1) = 0.231 A = 1 ft^2 q = kiA = 2.63*0.231*1 = 0.608 ft^3/d t = 30 days Vol = q*t = 18.2 ft^3

23. (modflow) What is a drain? How are drains simulated in MODFLOW?

A drain is a dry stream channel or a man-made object that may influence the grounwater table by draining off water when the water table rises above a certain elevation. Drains are simulated in MODFLOW by assigning an elevation and a conductance to a set of selected cells. If the head is below the elevation, the drain has no effect. If the head rises above the elevation, the water flow out of the aquifer at a rate equal to the conductance times the head difference.

24. (head-darcy) For a soil deposit with horizontal layering of heterogeneous deposits, which is
greater, Keq for the horizontal direction or Keq for the vertical direction? Why?

Keq for the horizontal direction should be greater since the K in this direction will be dominated by the presence of high K zones. Since the flow is parallel to the layering, the water seeks out the path of least resistance and flows mainly through the high K layers. In the vertical case, the water must flow through each layer and therefore the Keq will be dominated by the low K zones.

25. (modflow) Enter a short description for each of the following terms associated with the Evapotranspiration package:

a. Max ET Rate
b. ET Surface
c. Extinction Depth

If the water table is at the ET Surface (typically the ground surface), ET occurs at the Max ET Rate. As the water table drops, the ET rate drops linearly to a value of zero at the Extinction Depth.

26. (gw-concepts) Describe the three basic types of ground water models.

Predictive: used to make a prediction as to how the aquifer will behave in response to some future event. Interpretive: Used to figure out how an aquifer system works or to analyze what happened to an aquifer at some point in the past. Generic: hypothetical model used for academic purposes

27. (head-darcy) The water level in an observation well A is measured at 3785.4 ft. At a point 1238 ft downgradient from that point, the water level in a second well (B) is measured at 3629.3 ft.

a. What is the hydraulic gradient between the well A and well B?
b. Assuming that the hydraulic conductivity is 3.0 ft/day and the effective porosity is 0.3, what is the seepage velocity between the two points? [ft/day]
c. How long would it take for a particle to travel from point A to point B? [days]

Show your work and label your answers.

a. i = (Dh/L) = (3785.4-3629.3)/1238 = 0.126 b. vs = ki/n = (3*.126)/0.3 = 1.26 c. = L/vs = 1238/1.26 = 981.8

28. (modflow) Suppose you wish to represent the soil profile shown below in a MODFLOW model using a single layer. What values would you enter for the equivalent horizontal and vertical hydraulic conductivity? Put your answers in [m/d].

Kh=(0.3*5+0.001*2+0.03*5)/(5+2+5) =0.138 Kv=(5+2+5)/(5/0.1+2/0.0002+5/0.015)=0.00116

29. (head-darcy) For the previous problem, assume you got 0.01 m/d for Kv. Also assume that the effective porosity is 0.2 and that the head at the top of the profile is 1024 m and the head at the bottom of the profile is 1020.5 m. How long would it take [in days] for water to travel from the top of the profile to the bottom?

v = k*i/n = k*dh/dl/n v = (0.01)*(1024-1020.5)/(5+2+5)/0.2 = 0.0146 v = x/t --> t = x/v t = (5+2+5)/0.0146 = 822.9d