CE 473/573 Groundwater Fall 2011 Comments on homework 4 24. 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). Some groups did not account for the change in conductivity at the edge of the ﬁeld. Your plot of the water table should show a change in slope at the edge of the ﬁeld because the ﬂow from the ﬁeld should equal the ﬂow to the river, but the conductivities are diﬀerent. 26. 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 (or eastern) boundary. I suspect that a few groups did not specify the boundary conditions (or boundary types) correctly. 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 27. Most groups determined 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 R Qw ln s= , (1) 2πT r or Qw Qw ln r + ln R. (2) 2πT 2πT 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. s=− 28. Most groups got this problem. Just remember that s2 = H 2 − h2 .