Question

A hydrological system consists of five horizontal formations with an equal thickness of 10 m each. The hydraulic conductivities of the formations are 20, 10, 15, 50, and 1 m/day, respectively. Calculate equivalent horizontal and vertical hydraulic s conductivities. If the flow in the uppermost layer is at an angle of 30 away from the normal direction relative to the boundary, calculate flow directions in all the formations.

Answer 1

Answer: Hello The required formula is attached below

answer:

A) KH =  19.2 m/day ,  Kv = 4.045 m/day

B) flow directions : ∝1 = 30°, ∝2 = 16.102°, ∝3 = 23.62, ∝4 = 55.55°,

∝5 = 1.66°

Explanation:

Given data :

Thickness of five horizontal formations ( K )  = 10 m each

Hydraulic conductivities of formations ( b )  =

b1= 20,  b2= 10,  b3= 15,  b4= 50 and b5  = 1        (m/day)

a)Determine the equivalent horizontal and vertical conductivities

kH ( equivalent  horizontal conductivity )

   = ∑ Kb / b  = ( K1b1 + K2b2 + k3b3 + k4b4 + k5 b5 ) / b1 + b2 +b3+b4+b5

input given values into the above equation

kH = 960 / 50  = 19.2 m/day

Kv ( equivalent vertical conductivity )

= ∑( b / (b/k) ) = ( b1 + b2 + b3 + b4 + b5 ) / ( b1/k1 + b2/k2 + b3/k3 + b4/k4 + b5/k5 )

input given values into equation above

Kv = 50 / 12.36 = 4.045 m/day

b) Determine the flow directions in all the formations

given that ∝1 = 30°  and Ki / kj = bi / bj

k1 /k2 = tan ∝1 / tan ∝2

20 / 10 = tan 30 / tan ∝2

∴ ∝2 = 16.102°

K2/k3 = tan 16.102 / tan ∝3

= 10 / 15 = tan 16.102 / tan ∝3

∴ ∝3 = 23.62

K3/k4 = tan ∝3 / tan ∝4

= 15 / 50 = tan 23.62 / tan ∝4

∴∝4 = 55.55°

k4 /k5 = tan ∝4 / tan∝5

= 50 / 1 = tan 55.55 / tan ∝5

∴ ∝5 = 1.66°