Problem 1)
An inclined permeameter tube is filled with three layers of soil of different permeability as
shown below. Find the head at points (A – B – C – D) with respect to given datum, for the
following cases:
a. Assume k1 = k2 = k3.
b. Assume 3k1 = k2 = 2k3.
Solution
a. k1 = k2 = k3
: vertical flow ‫ وبالتالي كأوىا وتعامل مع‬، ‫والحظ أن طبقات التربة مرتبة بحيث أن الماء سيىتقل مه طبقة لألخرى‬
qeq  q1  q2  q3
keqieq A  k1i1 A1  k2i2 A2  k3i3 A3
 A1  A2  A3  A
 k1  k2  k3  k  keq
 ieq  i1  i2  i3
20  10  5 5

8  4  16
28
h A  20 m
ieq 
No loss of head as A is first point
 5 
hB  h A  i1 L1  hB  20  
  8  18 .57 cm
 28 
 5 
hc  hB  i2 L2  hc  18 .57  
  4  17 .85cm
 28 
 5 
hD  hc  i3 L3  hD  17 .85  
  16  15cm
 28 
b. 3k1 = k2 = 2k3 → k2 = 3k1 , k3 = 1.5 k1.
K1 ‫ بداللة احداها و هذا ما فعلىاي بالتحويل بداللة‬K ‫ووجد قيم‬
q eq  q1  q 2  q3
k eq ieq A  k1i1 A1  k 2 i2 A2  k 3i3 A3
 A1  A2  A3  A
 k 2  3k1 , k 3  1.5k1
H
k eq 
 k eq 
8  4  16
 k eq  1.4k1
 8   4   16 
   
  

 k1   3k1   1.5k1 
 H1   H 2   H 3 


  
  
 k1   k 2   k 3 
 k eq ieq  k1i1  k 2 i2  k 3i3  1.4k1ieq  k1i1  3k1i2  1.5k1i3
1.4ieq  i1  3i2  1.5i3
5
 0.25
28
1.
i1  1.4ieq  1.4 
2.
i2 
1.4
1.4 5
1
ieq 


3
3 28 12
3.
i3 
1.4
1.4 5 1
ieq 


1.5
1.5 28 6
hA  20m
hB  h A  i1 L1  hB  20  0.25  8  18cm
1
hc  hB  i2 L2  hc  18     4  17.67 cm
 12 
1
hD  hc  i3 L3  hD  17.67     16  15cm
6
Problem 2)
An inclined premeameter tube is filled with layers of soil of different permeability as shown
below.
Find the total head, elevation head and pore water pressure at points ( A – B – C – D ) with
respect to the given datum, Assume “ 3K1 = K2 = 2K3 = 1.5K4 ” .
Soln.:
K1 ‫ بداللة احداها و هذا ما فعلىاي بالتحويل بداللة‬K ‫ووجد قيم‬
K1 = K1
K2 = 3 K1
K3 = 1.5 K1
K4 = 2 K1
K* “ equivalent for the layers 1 and 2 “ =
Keq “ for all layers K*, K3, K4” = 𝐻 ∗
𝐾
∗+
𝐾1∗ 𝐻1 + 𝐾2∗𝐻2
𝐻∗
𝐻
𝐻3
𝐾3
+
𝐻4
𝐾4
=
=
𝐾1∗2+3𝐾1∗2
4
6+8+12
6
8
12
+
+
2 𝐾 1 1.5 𝐾1 2 𝐾1
= 2 𝐾1
= 1.814 𝐾1
qeq = q* = q3 = q4
Keq ieq Aeq = K* i* A* = K3 i3 A3 = K4 i4 A4
Because there is no variation in the area of the premeameter tube, then Aeq = A* = A3 = A4
Keq ieq Aeq = K* i* A* = K3 i3 A3 = K4 i4 A4
1.814 𝐾1 * ieq = 2 K1 * i* = 1.5 K1 i3 = 2 K1 i4
1.814 ieq = 2 i* = 1.5i3 = 2 i4
30−17
ieq = 6+8+12 = 𝟎. 𝟓
 i* = 0.454
 i3 = 0.6047
 i4 = 0.454
1) Total Head :
HA = 30 cm.
HB = HA - i* * 6cm = 30 – 0.454 * 6 = 27.28 cm.
HC = HB – i3 * 8cm = 27.28 – 0.6047 * 8 = 22.44 cm.
HD = HC – i4 * 12cm = 22.44 – 0.454 * 12 = 17 cm.
2) Elevation Head : “the elevation from the datum”
HA = 0.0cm .
“ point A is located on the datum ”
HB = 2.308 cm.
“ from trigonometric relations ”
HC = 5.385 cm.
“ from trigonometric relations ”
HD = 10 cm.
“ from trigonometric relations ”
3) Pore water pressure :
As Total head = elevation head + pressure head “ here is pore water pressure” + velocity head “ ignored
because V  0.0 ”.
HA = Total head @ A – Elevation head @ A = 30 – 0 = 30 cm.
HB = Total head @ B – Elevation head @ B = 27.28 – 2.308 = 24.972 cm.
HC = Total head @ C – Elevation head @ C = 22.44 – 5.385 = 17.055 cm.
HD = Total head @ D – Elevation head @ D = 17 – 10 = 7 cm.