7/28/2019 Problem Solving Through Similitude
1/12
DATE: 24/4/2013
TITLE: PROBLEM SOLVING THROUGH SIMILITUDE
OBJECTIVE: To measure and calculate in different way
(1) Seepage (water leakage) under dam.
(2) To calculate the uplift force on the dam.
INDRODUCTION:
Permeability, as the name implies (ability to permeate), is a measure of how easily a fluid can
flow through a porous medium. In this case, the porous medium is sand and the fluid is water.
Generally, larger the soil grains have larger voids and the high permeability. Therefore, gravels
are more permeable than silts. Hydraulic conductivity is another term used for permeability.
Flow of water through soils is called seepage. Seepage takes place when there is difference in
water heads (hL) on the two sides of the structure such as a dam as shown in Fig. 1. The head (h1,
h2) at a point is the level to which the free water rises at the point above the datum. In designing a
dam it is important to consider seepage under it and the uplift force acting on it.
Figure 1: Seepage beneath a concrete dam
DATA:Prototype dam (real dam) is as follows:
A long dam of dimension 1:100 model.
Head difference (hL) = 20m
Length of the dam = 50 m
Permeability of soil underneath the dam kp = 1x 10-6 m/sec
Calculate all quantities for total length of the dam (eg. seepage m 3/s).
Figure 2: The prototype dam.
Datum
h2
h1
7/28/2019 Problem Solving Through Similitude
2/12
APPARTUS:
a) Fluid mechanics laboratory.
1: 100 model placed in the sand (Permeability km= 2x 10-3 m/sec).
Length of model dam is 20cm.
b) Geotechnical laboratory.
1:250 model made of conducting paper to give equi-potentials (voltage) in the
model.c) Solution of Laplace equation of the problem through solving governing
equations using computer software.
PROCEDURE:
A) Fluid lab
-Establish steady flow through the dam.
-Measure the seepage flow through the dam.
-Introduce the dye at several places and draw the seepage flow lines
underneath dam.
-Take the reading of pressure tapping under the dam.
B) Geotechnical lab
-Apply a voltage difference of 10V across the conducting paper.
-Obtain constant equi-potential lines at 1V intervals.
-Measure voltage at points simulating the pressure tapping points.
OBSERVATION:
Water level at the upstream (H1) =124.mm
Water level at the downstream (H2) =26...mm
Table 1: The seepage underneath dam model.
Trials 1 2 3
Time taken /sec 25.91 32.08 29.78
Collected volume of water /ml 540 700 660
y
x
7/28/2019 Problem Solving Through Similitude
3/12
Table 2: The pressure head under the model dam.
Pressure tapping points 1 2 3 4 5 6
Pressure head /mm 91 98 111 123 129 140
Table 3: The flow lines underneath dam model.
Flow line 1 Flow line 2 Flow line 3 Flow line 4
x/mm y/mm x/mm y/mm x/mm y/mm x/mm y/mm
380 260 310 265 240 260
430 160 320 200 280 150
530 125 440 110 380 90
670 115 560 85 450 60
790 110 710 75 780 35
825 165 790 80 840 64
860 107
Table 4: The voltage at points simulating the pressure tapping points.
Simulated tapping
points
1 2 3 4 5 6
Voltage /V 3.4 4.0 4.6 5.2 5.8 6.6
CALCULATIONS:
1) Draw the flownets for cases (B).
2) Draw the flowlines for case (A) and transfer them using correct scale in
dotted lines to flownets drawn for case (B).
2) Determine seepage flow through the dam.
3) Calculate the uplift force on the dam.
4) Compare with computer software solution.
7/28/2019 Problem Solving Through Similitude
4/12
Scale of case (A) model is 1: 100.
Scale of case (B) model is 1: 250.
Therefore, Specimen calculation for value of X in case (A) at the point 1
X = 380 x (100/250)
= 152 mm
Table 1: The coordinate of the flowlines in case of (A) and (B)
Point No
Flow line 1 Flow line 2 Flow line 3
Case (A) Case (B) Case (A) Case (B) Case (A) Case (B)
X/mm Y/mm X/mm Y/mm X/mm Y/mm X/mm Y/mm X/mm Y/mm X/mm Y/mm
1 380 260 152 104 310 265 124 106 240 260 96 104
2 430 160 172 64 320 200 128 80 280 150 112 60
3 530 125 212 50 440 110 176 44 380 90 152 36
4 670 115 268 46 560 85 224 34 450 60 180 24
5 790 110 316 44 710 75 284 30 780 35 312 14
6 825 165 330 66 790 80 316 32 840 64 336 26
7 860 107 344 43
7/28/2019 Problem Solving Through Similitude
5/12
Fig 1: The flow lines underneath dam model (A)
7/28/2019 Problem Solving Through Similitude
6/12
7/28/2019 Problem Solving Through Similitude
7/12
CALCULATION
2)
CASE (A)
Assume a datum at the base of the dam.
Head different in the Prototype dam Hp = huphdp
= (11.5 + 11.5) -3
= 20 m
Head different in the model dam Hm = humhdm
= (115 + 124) (30 + 26)
= 183mm
Seepage flow through 0.2m length of the model dam
[q]m for 0.2m length = (540/25.91) + (700/32.08) + (660/29.78)
3
= (20.84 + 21.82 + 22.16) / 3
= 22 ml/sec
Seepage flow through 1m length of the model dam
[q]m for 1m length = 22 x (1/0.2)
= 110 ml/sec/m
[q]m[q]P[k]P [k]m
7/28/2019 Problem Solving Through Similitude
8/12
Therefore, Seepage flow through 1m length of the Prototype dam is given by
q = (Nf/ Nd) kH [L2T
-1]
where: q = rate of flow or seepage per unit width
Nf = number of flow channels
Nd = number of equipotential drops
H = total head loss in flow system
K = hydraulic conductivity
From similarity condition,
[q]p /[q]m = [kH (Nf/ Nd)]p/[kH (Nf/ Nd)]m
Since flownets are identical in prototype and model, therefore
(Nf/ Nd)p = (Nf/ Nd)m
[q]p = [q]m [kH]p / [kH]m= 110 x (1x 10-6 x 20) / (2x 10-3 x 0.183)
= 6.01 x 10-6 m3/sec/m
Length of the Prototype dam is 50 m.
Therefore, total seepage flow through the Prototype dam
Q = 6.01 x 10-6 x 50
= 3x 10
-4
m
3
/sec
CASE (B)
Seepage flow through 1m length of the Prototype dam
[q]p = kP HP (Nf/ Nd)p
= 1x 10-6 x 20 x (Nf/ Nd)p
From flownet graph 2 (Nf/ Nd)p = 3/10
=0.3
Therefore, [q]p = 1x 10-6 x 20 x 0.3
= 6 x 10-6 m3/sec/m
Length of the Prototype dam is 50 m.
Therefore, total seepage flow through the Prototype dam
Q = 6 x 10-6
x 50= 3x 10-4 m3/sec
7/28/2019 Problem Solving Through Similitude
9/12
3)
CASE (A)
Specimen calculation for 2nd tapping point in case (A)
Assume a datum at the base of the dam.
Water head 2nd tapping point in case (B) When total head different is 0.183 (Hm). Here, upstream
side head (hum) is 0.239m and downstream side head (hdm) is 0.056m.
h2m = measured head + thickness of base
= 0.098 + 0.013
= 0.111m
Therefore, total head at 2nd tapping point = water head + datum head
= 0.111+ 0= 0.111m
Therefore,
Total head at 2nd tapping point in the prototype dam when head different is 20 (Hp). Here,
upstream side head (hup) is 23m and downstream side head is 3m.
From similarity condition,
[h2 - hd]p / HP = [h2 - hd]m / Hm
(h2p3) / 20 = (0.111 0.056) / 0.183
h2p = 0.055 x 20/0.183 + 3
= 9.01 m
Water head at 2nd tapping point of Prototype dam = total head datum head
= 9.01 0
= 9.01m
Table 2: Water head at each tapping points.
Tapping points 1 2 3 4 5 6
Water head in case (A) /m 0.104 0.111 0.124 0.136 0.142 0.153
Total head /m 8.25 9.01 10.43 11.74 12.40 13.60
Water head in prototype /m 8.25 9.01 10.43 11.74 12.40 13.60
Average water head under the the Prototype dam hp = (8.25+9.01+10.43+11.74+12.4+13.6)/6
= 11m
Average water pressure under the Prototype dam P = hpg= 11x 1000 x 9.81
= 108 kPa
7/28/2019 Problem Solving Through Similitude
10/12
The area of the Prototype dam foundation (1m length)
A = l x w
= 1 x (0.29 x 100)
= 29 m2
Uplift force on 1m length of Prototype dam F = PxA= 108 x 29
= 3132 kN
CASE (B)
Specimen calculation for 2nd tapping point in case (B).
Assume a datum at the base of the dam.
Voltage at 2nd simulated tapping point in case (B) when applying 10 V different (Vm). Here,
upstream side voltage (Vum) is 10 V and downstream side voltage (Vdm) is 0 V
[V2 - Vd]m = 4.0 V
Therefore,
Total head at 2nd tapping point in the prototype dam when head different is 20 (Hp). Here,
upstream side head (hup) is 23m and downstream side head is 3m.
From similarity condition,
[h2 - hd]p / HP = [V2 - Vd]m / Vm
(h2p3) / 20 = 4 x 10
h2p = 4 x 20/10 + 3
= 11 m
Water head at 2nd tapping point of Prototype dam = total head datum head
= 110
= 11m
Table 3: Water head at each tapping points.
Simulated tapping points 1 2 3 4 5 6
Voltage in case (B) /V 3.4 4.0 4.6 5.2 5.8 6.6
Total head /m 9.8 11 12.2 13.4 14.6 16.2
Water head in prototype /m 9.8 11 12.2 13.4 14.6 16.2
Average water head under the Prototype dam hw = (9.8+11+12.2+13.4+14.6+16.2)/6 mm= 12.87 m
7/28/2019 Problem Solving Through Similitude
11/12
Average water pressure under the Prototype dam P = hpg
= 12.87 x 1000 x 9.81
=126 kPa
The area of the Prototype dam foundation (1m length)A = l x w
= 1 x (0.29 x 100)
= 29 m2
Uplift force on 1m length of Prototype the dam F = PxA
= 126 x 29
= 3654 kN
4)
From computer software modal analyze
Seepage flow through 1m length of the Prototype dam
[q]p = 5.4x 10-6 m3/sec/m
Total seepage flow through the Prototype dam Q = 5.4 x 10-6 x 50
= 2.7 x 10-4 m3/sec
7/28/2019 Problem Solving Through Similitude
12/12
Fig 3: The variation of water pressure under long dam foundation.
Average water pressure under the Prototype dam P = (85+93+108+123+138+156+168)/7
= 125 kPa
The area of the Prototype dam foundation (1m length)
A = l x w
= 1 x (0.29 x 100)
= 29 m2
Uplift force on 1m length of Prototype the dam F = PxA
= 125 x 29
= 3625 kN
RESULTS
CASE (A) CASE (B) Software model
Seepage flow/ (m3/sec) 3x 10-4 3x 10-4 2.7x 10-4
Uplift force /kN 3132 3654 3625
COMMENTS ON THE RESULT:
Top Related