problem solving through similitude

7/28/2019 Problem Solving Through Similitude

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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


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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


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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.


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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


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Fig 1: The flow lines underneath dam model (A)


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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


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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


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3)

CASE (A)

Specimen calculation for 2nd tapping point in case (A)


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.


[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


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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).


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.


[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


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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


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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:

problem solving through similitude

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