Chapter (8)
Instructor : Dr. Jehad Hamad
2015-2014
Laplace’s Equation of Continuity
qy
Elemental Cube:
dy
Saturation S=100 %
Void ratio e= constant
Laminar flow
q in = q out
Continuity:
qx +
qx
dx
qy +
∂q y
∂y
dy
∂q y
∂q x

+
+
q x + q y −  q x + ∂x dx q y ∂y dy  = 0


∂q x
∂x
∂q y
dx + ∂y dy
=0
∂q x
∂x
dx
Laplace’s Equation
• Continuity:
• Darcy’s law:
∂q x
∂x
∂q y
dx + ∂y dy
=0
qx = kx ⋅i ⋅ A = kx ⋅
• Replacing:
0 = kx ⋅
0 = kx ⋅
• if kx=ky
(isotropy):
0=
∂ 2h T
∂x 2
+
∂ 2h T
⋅ dy ⋅1
2
dx ⋅ dy ⋅1 + k y ⋅
2
+ ky ⋅
∂x
∂ 2h T
∂x
∂h T
∂x
∂ 2h T
∂y 2
∂ 2h T
∂y 2
∂ 2h T
∂y
2
dy ⋅ dx ⋅1
Laplace’s Equation!
Laplace’s Equation
• Typical cases
– 1 Dimensional:
linear variation!!
– 2-Dimensional:
– 3-Dimensional:
0=
∂ 2h T
∂x 2
cons tan t =
∂hT
∂x
=i
+
∂ 2h T
hT = a + b ⋅ x
0=
∂ 2h T
0=
∂ 2h T
∂x
∂x
2
2
+
∂y 2
∂ 2h T
∂y
2
+
∂ 2h T
∂z 2
Laplace’s Equation Solutions
•
•
•
•
•
•
Exact solutions (for simple B.C.’s)
Physical models (scaling problems)
Approximate solutions: method of fragments
Graphical solutions: flow nets
Analogies: heat flow and electrical flow
Numerical solutions: finite differences
Flow Nets
• The procedure consists on drawing a set of
perpendicular lines: equipotential and flow lines.
• These set of lines are the solution to the Laplace’s
equation.
• Identify boundaries:
– First and last equipotential line
– First and last flow lines
Seepage Calculation from a Flow Net
Seepage Calculation from a Flow Net
Seepage Calculation from a Flow Net
How to draw flow nets
a
b= ∆l
Flow Nets
∆h= equipotential drop
∆q
Flow channel
• gradient:
Equipotential lines
• flow per channel:
• total flow:
a
b= ∆l
∆h ∆h h N e
i=
=
=
b
b
∆l
h
∆h
N
⋅A = k⋅ e ⋅A
∆q = k ⋅
∆l
b
a Nf
q = ∆q ⋅ N f = k ⋅ h ⋅ ⋅
b Ne
Flow Nets
Number of flow Channels Nf
Number of potential drops Nd
Seepage Calculation from a Flow Net
Example
Example
Uplift Pressure
Example – Flow Calculation
Nf = 3
Nd = 10
q = kH Nf/Nd = 10-5 ×8.5 × 3/10 = 2.55 × 10-5 m3/s/m
Example - Uplift Pressure
1 2
3
4
5
6
Potential Drop = H/Nd = (10 – 1.5)/10 = 0.85 m
Point 1: P1 = [(10 + 3) – 0.85]γw = 12.15 γw kN/m2
Point 2: P2 = [(10 + 3) – 2×0.85]γw = 11.3 γw kN/m2
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35
36
38
39
40
Mathematical Solutions
Previous Example
Flow Net Method
q = 3×2.45×10-5 m3/s/m
Nf
Δq
q = 7.35×10-5 m3/s/m
q/kH
Mathematical Method
q = 0.5×4.2×10-5×3.5 m3/s/m
From Figure
k
H
q = 7.35×10-5 m3/s/m
S/T΄
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44
45
46
For the proper selection of the filter
material, two conditions should be kept in mind.
1. The size of the voids in the filter material should be small enough to
hold the larger particles of the protected material in plac
e.
2. The filter material should have a high permeability to prevent buildup
of large seepage forces and hydrostatic pressures in the filters.
Based on the experimental investigation of protective filters, Terzaghi
and Peck (1948) provided the following criteria to satisfy the above
condition
to satisfy condition 1
D15F/D85B ≤ 4–5 to satisfy condition 1
D15F/D15B ≥ 4–5 to satisfy condition 2
D15F = diameter through which 15% of filter material will pass
D15B = diameter through which 15% of soil to be protected will pass
D85B = diameter through which 85% of soil to be protected will pass