SPOILERS Sample questions for Geology 627 midterm 1, Fall 2003

Note that these are intended to give you an idea of the style and level of complexity of the questions that will appear on the exam. They are not intended to provide exhaustive coverage of the factual, mathematical, and conceptual content of the exam. The exam will include material covered in lectures through October 8, assigned readings (through Chapter 4), Problem Sets 1 and 2 through question 2, and Labs 1-5. Multiple Choice - Mark all answers that are correct. There may be more than one correct answer for some questions. 1. Darcy's law can be expressed as a) Q = -Tw∇h, where Q has dimensions of L3/T and w is the width of the aquifer b) qz = -Kz m dh/dz, where m is the aquifer thickness and qz is the vertical specific discharge c) q = -k/µ grad P, if flow is strictly horizontal and q is the horizontal specific discharge vector d) q = -K grad h, where q is the specific discharge vector 2. If the potentiometric surface is above the land surface, the pressure at midpoint of the screened interval is a) equal to atmospheric pressure b) greater than atmospheric pressure c) less than atmospheric pressure d) equal to the elevation head 3. Major aquifer systems of the U.S. that are subject to saltwater intrusion include a) the Piedmont and Blue Ridge b) the Dakota Sandstone c) the High Plains Aquifer System d) the Atlantic Coastal Plain Aquifers 4. The wet bulk density of a porous medium can be computed using the equation(s) a) ρwb = (ms + mw)/VT where ms and mw represent the masses of solids and water, respectively, in

a water saturated medium and VT is the total volume (solids plus water). b) ρwb = n(ρs + ρw) where n is the porosity and the ρs and ρw represent the densities of the solids

and water respectively c) ρwb = (1-n)ρs + nρw where n is the porosity and the ρs and ρw represent the densities of the

solids and water respectively d) ρwb = ρb + nρw where n is the porosity ρb is the dry bulk density and ρw is the density of water 5. Joints in columnar basalt that formed when the lava cooled are considered a) primary porosity b) secondary porosity c) the specific retention d) the effective porosity 6. At a piezometer nest the depth to water below the land surface in a water table well is less than the depth to water below the land surface in a deeper piezometer. This indicates that this location is in a a) recharge area

b) discharge area c) confined aquifer d) stagnation point Short answer problems 1. A flow rate of 45 ml/minute is measured during a constant head permeameter experiment. In this experiment, the water level in the funnel is 15 cm above the datum and the outlet port is 5 cm above the datum. Without moving the permeameter or the datum, the funnel is then raised so that its water level is 30 cm above the datum. What is the steady flow rate for these new conditions? Head change across sample increases from 10 cm to 25 cm. Hence gradient increase by a factor of 2.5 and the new flow rate is 112.5 ml/minute. 2. Consider the unconfined aquifer system shown below. The aquifer is 2000 m wide and 4000 m long (direction parallel to the river). All inflow to this groundwater system is by areal recharge at a vertical flux, R, of 3.0 x 10-2 m/month. Outflow from the groundwater system in the area shown in the diagram is by discharge to the river at a volumetric flow rate, Q, of 6.0 x 105 m3/month.

a) Is the water table rising or falling? Inflow = R A, where A is the surface area 8 million square meters. Thus, Inflow = m

5104.2 ×3/mo., which is less than Q. Hence W.T. is falling.

b) What is the volume of water per unit time that is being added to or removed from storage in the aquifer? Q – Inflow = m5106.3 × 3/mo. is removed from storage. c) If the aquifer has a specific yield of 0.20, what is the rate (in m/month) at which the water table is rising or falling? Water table is falling at rate of (Q – Inflow)/A = 0.045 m/mo. d) If recharge continues at the specified rate but the discharge to the river adjusts to a steady state condition in which the water table is neither rising nor falling with time, what will be the steady rate of groundwater discharge to the river? Steady state when Q = Inflow = m5104.2 × 3/mo. 3. The horizontal specific discharge, q, in a confined aquifer has a magnitude of 1.5 x 10-7 m/s and a direction of N450W (450 west of north). The aquifer is homogeneous but anisotropic, with the principal axes of anisotropy oriented N-S and E-W. Ky (in the N-S direction) is 1 x 10-4 m/s and Kx (in the E-W direction) is equal to 0.5 times Ky. You may find it helpful to sketch this situation before attempting to answer the questions below. a) Determine the magnitudes of the specific discharge vectors qx and qy. qx = - 1.5 x 10-7 sin 45o = -1.06 x 10-7 m/s qy = 1.5 x 10-7 cos 45o = 1.06 x 10-7 m/s b) Determine the magnitude and direction of the hydraulic gradient dh/dl. Let i = dh/dl. ix = - qx/Kx = 2.12 x 10-3 iy = - qy/Ky = - 1.06 x 10-3 and the magnitude of i = 2.37 x 10-3

Tan θ = 1.06/2.12 and the direction θ = 26.6o S of E. c) If the head at a point A is 100.00 m above sea level, what is the head at a point B located 100 m due south of A? The distance ∆l between head contours that pass through A and B is given by the leg of a right triangle such that ∆l = 100 sin 26.6o = 44.8 m Hence ∆h = (dh/dl) ∆l = 0.106 m and hB = 100.11 m 4. The poroelastic constitutitive equations are:

paapaa

v

vv

2221

1211

+=+=

σζσε

A 1-m3 block of rock has the fluid pressure uniformly reduced by 10 MPa while the block is

constrained horizontally and free to expand vertically. The coefficient Pa10

22 105.1 −×=a

22a

-1. Compressive stress and strain are considered positive. Assume 11a = , , and

. 2112 aa =

1112 aa −=a) What is the increment of fluid content ζ and what is the volume of water released? Because vertical stress is constant, ∆ζ = a22 ∆p = 1.5 x 10-10 Pa-1 (- 107 Pa) = - 1.5 x 10-3 (negative sign means water is released from storage)

And ∆Vw = - 1.5 x 10-3 m3 because V = 1 m3

b) What is the vertical contraction of the rock? Because vertical stress is constant, ∆εϖ = a12 ∆p = - 1.5 x 10-10 Pa-1 (- 107 Pa) = 1.5 x 10-3

(positive sign means aquifer contracts) And ∆b = 1.5 x 10-3 m because b = 1 m