CE 473/573 Groundwater
Fall 2008
Comments on homework 4
23. The best way to approach this problem is to use the refraction relation K1 /K2 =
tan σ1 / tan σ2 and the fact that coarse sand has a much higher conductivity than clay.
26. Only isolated problems.
27. The flow per unit width is Q′ = K(h21 − h22 )/2L − w0 (L/2 − x). For no divide to occur,
the flow must be unidirectional. Therefore, the limiting case is Q′ = 0 at x = 0,
and if w0 < K(h21 − h22 )/L2 , no divide will occur. If the heads are equal, then any
recharge will cause a divide. The limiting recharge increases with K because higher
conductivity soils can conduct the recharged water away faster. For part c, substitute
numbers into the expression for the limiting recharge.
28. Both the water table elevation and the effective conductivity vary little in this case.
The full solution involves the effective conductivity
1
Kef f = (K1 (h − b2 ) + K2 b2 ) .
h
The flow equation is
d
dh
hKef f
= 0.
dx
dx
One integration gives
dh
dh
+ K 2 b2
C1 = K1 (h − b2 )
dx
dx
d h2
dh
= K1
+ (K2 − K1 )b2 .
dx 2
dx
Another integration gives
h2
+ (K2 − K1 )b2 h.
2
Applying the boundary conditions at x = 0 and x = L gives
C1 x + C2 = K1
(1)
h2L − h20
(hL − h0 )
+ (K2 − K1 )b2
2L
L
2
h
C2 = K1 0 + (K2 − K1 )b2 h0 .
2
Recognize that the discharge Q is equal to −C1 . Also, once the constants are computed, then equation (1) can be used to plot the water table elevation and the effective
conductivity vs. x. For part d, the point at which the water table crosses (x = d, say)
into the lower soil is unknown. For x > d, the aquifer is an unconfined aquifer with
conductivity K2 and known boundary heads. For x < d, the effective conductivity
and flow are given above. The key is to match the flows at x = d so that d can be
computed.
C1 = K1
29. In class we found that the equilibration time was Teq = −Sb′ ln α/K ′ , where α is the
fraction of the initial head difference to be considered (0.01 in this case). Answers to
parts a and b were 1.13 d and 0.67 d, respectively. The time decreases for the case
with lower storativity because for the same head change, less volume is required.