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°
