Responses to Homework Set #4

1a. Flow net solution I got that looked best was:

n_f=4 stream tubes

n_d=3 equipotential drops

Equation to use-- K=Q/(H_T * b * [n_f/n_d])

Solution: K~52,900 ft/d

1b. Bozic Lake to Tolley River—use map scale to determine shortest (worst case distance) ~2500 ft

Equation to use— v = (K/n) * (dh/dL) and time = velocity/distance

Solution: v~814 ft/d so that time ~ 3.07 days

1c. The flow is from the lake to river, based on higher head in the lake. No contamination will flow upgradient in this case.

2. The key is to compute average area through which flow is occuring. Scale the largest and the smallest widths, and approximate some average value. You will need to do the same for the variable lengths of the flowlines, which range from the shortest (worst case) to the longest (best case).

Solution: K~61,000 ft/d

Benefits and drawbacks: Darcy’s law requires average values to be determined to accomodate the orthogonality of our equation. It’s easier and faster than flow net solution, but less accurate because of the curvature of the flow boundaries.

3. Plug and chug in Darcy’s Law.

Q = KA dh/dL velocity same equation as 1b.

Solution Q~40 ft³/d v~1.8 ft/d

4. Bonus points (30 pts)

Need to consider the z component of Bernoulli formula + pressure component, and have to convert pressure of saline head to fresh-water head.

Equation for pressure component: P=rgh_P

Solution: for saline pressure head, = 74.36 m, which converts to 78.51 m of fresh water head at 14^o C. Add 78.51 to z, which was given as 45.9, total fresh-water head ~124 m

5. Steady-state aquifer test determination using Thiem equation (given in handout). Can work this either of 2 ways, graphically, or solely computationally.

Equation: T = (Q/2 * p *[ s₁-s₂]) * ln (r₁/r₂) verify all units are the same, and make conversions accordingly ;

Equation if you use semi-log plot of drawdown verses radius:

T=(2.303 * Q/[2 * p * Ds]) when Ds is selected at radial distance of 1 log cycle. verify all units are the same, and make conversions accordingly ;

Solution: T~7700 ft²/d

6. The solution to these types of problems is classified as time-distance-drawdown, and is based on the Theis equation. Given T, S, and radial distance from pumping centers at a specific time, these allow computation of drawdown for a specified stress (discharge).

Equations: s=(Q/[4 * p *T]) * W(u) W(u) is determined by first calculating u.

u=[(r²S)/(4Tt)] relation between u and W(u) is given by the table provided with handouts. Since you know r² , S, T, and t, solve first for u, use table to compute W(u), then plug in values to determine drawdown s for various radial distances.

for r=50 ft, s=0.168 ft; for r=150 ft, s=0.147 ft; for r= 500 ft, s=0.124 ft; for r=1000 ft, s=0.110 ft; for r=3000 ft, s=0.089 ft; for r=10,000 ft, s=0.066 ft.

7. Another question dealing with density differences, this is also referred to as the Ghyben-Herzberg formula. It allows determination of the depth to the sharp interface between saline and fresh water below a fresh water lens that floats on more dense saline water.

Equation: z= (r_w/[r_s -r_{w] )} * h

Solution: for z = 17.9 ft, h ~536.5 ft

8. Jacob straight-line solution is a modification of the Theis equation requiring later aquifer-test data to be plotted on semilog paper. s is plotted on arithmetic axis, and t on the logarithmic axis. Connect data points with a straight line, so that the line for the latest data points is continued up to the 0.0 drawdown. This intersection is the value t₀, used in the calculation of the storage coefficient. Ds is the drawdown taken one log cycle apart on the plotted data.

Equations: T=(2.303 * Q/[4 * p * Ds]) S=([2.25 * T * t_0] / r² ) T=K/b

Solution : T~1.0 ft²/min ; S~1.6 * 10^-5 [unitless] ; K~0.02 ft/min