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CE 473/573 Groundwater Fall 2010 Comments on homeworks 4 and 5 24. As you have seen on problem 8 and exam 1, you can determine parameters such as the radius of inﬂuence and transmissivity by ﬁtting the theoretical equation to the data and calculating the parameters from the coeﬃcients. In this case, the steady drawdown caused by the well in the conﬁned aquifer is Qw ln s= 2πT or s=− R r , (1) Qw Qw ln r + ln R. 2πT 2πT (2) In Excel, ﬁting a logarithmic curve to equation (2) will give you the coeﬃcient of ln r as well as the intercept. Then calculate T from the slope and R from the intercept. 25. By applying conservation of mass and using the ﬂow between two points in an unconﬁned aquifer, you should ﬁnd h2j = Kj−1/2 h2j−1 + Kj+1/2 h2j+1 2ij (Δx)2 + , Kj−1/2 + Kj+1/2 Kj−1/2 + Kj+1/2 where ij is the recharge rate for the jth cell and Kj−1/2 = 2Kj−1 Kj /(Kj + Kj−1 ). Apply this formula in the interior of the domain. You will ﬁnd a groundwater divide, which means that some water will ﬂow to the drain on the left. To compute the travel time, realize that the average linear velocity will vary because the head does not decrease linearly. Therefore, compute the travel time between two cells as Tj = where Kj−1/2 v = n Δx , v hj−1 − hj Δx , and add the times over the speciﬁed range (in this case, 13.5 m < x < 20 m). 27. In the PMWIN simulation I ﬁnd a ﬂow of 102.5 m3 /d in the left boundary and a ﬂow of 72.5 m3 /d out the right boundary. Excel gives 104 m3 /d in the left boundary and 71 m3 /d out the right boundary (an error of about 3%). My plots are below 400 350 300 y (m) 250 200 150 100 50 0 0 100 200 300 x (m) 400 500 600 100 200 300 x (m) 400 500 600 400 350 300 y (m) 250 200 150 100 50 0 0 28. Most groups computed the correct hydraulic conductivity. Some groups used the result I stated in class to get the formula in the problem statement. No group started from Darcy’s law. 29. No major problems. 31. Remember that ymax is half the maximum width. I computed the downgradient area by plotting the curve x = y/ tan(y/x0 ) and integrating with the trapezoid rule. I ﬁnd an area of 233 m2 . 32. I used six wells pumping at 450 m3 /d on each side and checked the point on the centerline halfway up the short side since that point is farthest from the wells. I computed the head with N R Q0 ln h =H − , πK i=1 ri 2 2 where N is the number of wells and ri is the distance from the ith well to the point in question. For conﬁgurations in which the point of minimum drawdown is unclear, the safe approach is to compute the drawdown at all points in the construction area. 33. No major problems. 35. A few isolated problems. Remember that storativity should probably be less than 5 × 10−3 . Much higher values probably indicate an error. Be sure you can compute drawdown in unsteady ﬂow to a well and apply the Jacob approximation. The spreadsheet does it automatically. 36. Most groups computed the curve properly. I get 0.105 m for part a and 1.8 hours for part b. 37. As Γ increases, the vertical ﬂow becomes more important. Because vertical ﬂow can supply more of the water needed for the well, the drawdown will be less. This behavior is illustrated in the plot of the unconﬁned well function since W decreases as Γ increases. For part b, when the vertical ﬂow is zero (i.e., Kv = 0 or Γ = 0), the aquifer will behave as a conﬁned aquifer because the only source of water is the compressibility of the water and the soil matrix. As Γ goes to zero, the curve approaches the solution for a conﬁned aquifer.