EXPERIMENT NO. 2

HYDROSTATIC FORCE ON A SUBMERGED AND SEMI-SUBMERGED FACE

Year and Section: 4CE-A Date Started: September 7, 2015

Group Number: 4 Date Finished: September 7, 2015

Group Members: Date Submitted: September 14, 2015

DURAN, RAPHAEL V. ILOCTO, JOHN JOSEPH A.

ESCUETA, JOHN EZEKIEL Y. INOCENCIO, JOHN ANGELO B.

GATCHALIAN, ROD ANTHONY E. KABIGTING, ANGELO R.

GULA, JOHN GIL R.

1. OBJECTIVE/SThis activity aims to determine the hydrostatic force on a submerged and semi-submerged

rectangular area.

2. APPARATUS AND SUPPLIESApparatus (see figures 6.b):

Center of Pressure ApparatusHydraulics Bench

Supplies (see figures 6.c):Tap Water

3. PROCEDURE1. Set up the apparatus.2. Place a mass of approximately 50 grams on the balance weight holder.3. Open pump delivery valve and allow water into the tank until balance arm is

horizontal, then close the pump delivery valve.4. Read height of water level on the scale of torroid.5. Repeat steps (3) and (4) for various values of weight in the balance pan in about eight

steps up to 400 grams6. Stop hydraulic bench pump7. Disconnect supply hose and allow apparatus to drain

CLEANING PROCEDURE1. Stop hydraulic bench pump2. Disconnect supply hose and allow apparatus to drain3. Carefully wipe the apparatus dry with a soft cloth.

CE 412L: Fluid Mechanics Laboratory | Experiment No. 2: Hydrostatic Force on a Submerged and Semi-Submerged Face

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4. DATA AND RESULTSTable 2.1.1 – Hydrostatic Force on Submerged and Semi-submerged Face

Weight on Balance Arm,

W(kg)

Depth of water above

torroid, h(mm)

Depth of water on torroid,

h(mm)

Reading on the torroid scale,

h+d (mm)

Height of balance arm above water surface,

hs=200(h+d)

0.0615 0 50 50 1500.1115 0 68 68 1320.1615 0 83 83 1170.2115 0 96 96 1040.2615 10 100 110 900.3115 22 100 122 780.3615 35 100 135 650.4115 46 100 146 54

Depth of center of

pressure, ŷ (mm)

Force on end face, F (N)

Moment arm of force in end face,

ŷ+hs (mm)

Observed moment,

F(ŷ+hs) in (N-mm)

Calculated moment,

9.81*W*r in (N-mm)

%Error = (C-O)/O*100

33.33 9.07 x10-4 183.33 0.166 0.166 0%45.33 1.68 x10-3 177.33 0.298 0.301 1.01%55.33 2.50 x10-3 172.33 0.431 0.436 1.16%64.00 3.35 x10-3 168.00 0.569 0.571 0.35%73.89 4.36 x10-3 163.89 0.710 0.710 0.70%83.57 5.23 x10-3 161.57 0.845 0.840 0.59%94.80 6.17 x10-3 159.80 0.983 0.975 0.81%104.68 6.97 x10-3 158.68 1.106 1.110 0.36%

5. FORMULAS AND COMPUTATIONS

FORMULAS:

i) hs=200-(h+d)

ii) yp=(2/3)[(3h2+3hd+d2)/(2h+d)]

iii) M = F(yp+hs)

iv) F= [(1/2)Ɣh][(h+d)2-h2]

v) THEO = 9.81(W)(r)

vi) %error = [(C-O)/O]x100

COMPUTATION

a.) hs=200-(h+d)

1.) 200 – (0+50) = 150mm 5.) 200 – (10+100) = 90mm

2.) 200 – (0+68) = 132mm 6.) 200 – (22+100) = 78mm

3.) 200 – (0+83) = 117mm 7.) 200 – (35+100) = 65mm

CE 412L: Fluid Mechanics Laboratory | Experiment No. 2: Hydrostatic Force on a Submerged and Semi-Submerged Face

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4.) 200 – (0+96) = 104mm 8.) 200 – (46+100) = 54mm

b.) yp=(2/3)[(3h2+3hd+d2)/(2h+d)]

1.) (2/3){[3(0)2 + 3(0)(50) + 502]/[(2(0)+50]} = 33.333

2.) (2/3){[3(0)2 + 3(0)(68) + 682]/[(2(0)+68]} = 45.333

3.) (2/3){[3(0)2 + 3(0)(83) + 832]/[(2(0)+83]} = 55.333

4.) (2/3){[3(0)2 + 3(0)(96) + 962]/[(2(0)+96]} = 64.000

5.) (2/3){[3(10)2 + 3(10)(100) + 502]/[(2(10)+100]} = 73.889

6.) (2/3){[3(22)2 + 3(22)(100) + 502]/[(2(22)+100]} = 83.577

7.) (2/3){[3(35)2 + 3(35)(100) + 502]/[(2(35)+100]} = 94.804

8.) (2/3){[3(46)2 + 3(46)(100) + 502]/[(2(46)+100]} = 104.681

c.) M = F(yp+hs)

1) 9.07 x10-4 (183.33) = 0.166 5.) 4.36 x10-3 (163.889) = 0.710

2) 1.68 x10-3 (177.33) = 0.298 6.) 3.23 x10-3 (161.574) = 0.845

3) 2.50 x10-3 (172.33) = 0.431 7.) 6.17 x10-3 (159.804) = 0.983

4) 3.35 x10-3 (168.00) = 0.569 8.) 6.97 x10-3 (158.681) = 1.106

d.) F= [(1/2)Ɣh][(h+d)2-h2]

1.) [(1/2)(9.81 x10-9)(74)][(0+50)2-502] = 9.07 x 10-4

2.) [(1/2)(9.81 x10-9)(74)][(0+68)2-682] = 1.68 x 10-3

3.) [(1/2)(9.81 x10-9)(74)][(0+83)2-832] = 2.50 x 10-3

4.) [(1/2)(9.81 x10-9)(74)][(0+96)2-962] = 3.35 x 10-3

5.) [(1/2)(9.81 x10-9)(74)][(10+100)2-1002] = 4.36 x 10-3

6.) [(1/2)(9.81 x10-9)(74)][(22+100)2-1002] = 5.23 x 10-3

7.) [(1/2)(9.81 x10-9)(74)][(35+100)2-1002] = 6.17 x 10-3

8.) [(1/2)(9.81 x10-9)(74)][(46+100)2-1002] = 6.97 x 10-3

e.) THEO = 9.81(W)(r)

1.) 61.5 (9.81 x10-3) (.275) = 0.166 5.) 265.5 (9.81 x10-3) (.275) = 0.705

2.) 111.5 (9.81 x10-3) (.275) = 0.301 6.) 311.5 (9.81 x10-3) (.275) = 0.840

3.) 165.5 (9.81 x10-3) (.275) = 0.436 7.) 361.5 (9.81 x10-3) (.275) = 0.975

4.) 211.5 (9.81 x10-3) (.275) = 0.571 8.) 411.5 (9.81 x10-3) (.275) = 1.110

f.) %error = [(C-O)/O] x100

1.) [(0.166-0.166)/ (0.166)] x100 = 0.00% 5.) [(0.705-0.710)/ (0.710)] x100 = 0.704%

CE 412L: Fluid Mechanics Laboratory | Experiment No. 2: Hydrostatic Force on a Submerged and Semi-Submerged Face

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2.) [(0.301-0.298)/ (0.298)] x100 = 1.01% 6.) [(0.840-0.845)/ (0.845)] x100 = 0.592%

3.) [(0.436-0.431)/ (0.431)] x100 = 1.16% 7.) [(0.975-0.983)/ (0.983)] x100 = 0.814%

4.) [(0.571-0.569)/ (0.569)] x100 = 0.351% 8.) [(1.110-1.106)/ (1.106)] x100 = 0.362%

6. FIGURES/DIAGRAMS/GRAPHS

6.a. Figures:

(6.a.1 Empty Center of Gravity Apparatus) (6.a.1 Filled Center of Gravity Apparatus)

6.b. Materials used:

(6.b.1 Center of Gravity Apparatus) (6.b.2 Hydraulics Bench)

6.c Supplies used:

(6.c.2 Tap water)

CE 412L: Fluid Mechanics Laboratory | Experiment No. 2: Hydrostatic Force on a Submerged and Semi-Submerged Face

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6.d Location:

6.e Pictures:

CE 412L: Fluid Mechanics Laboratory | Experiment No. 2: Hydrostatic Force on a Submerged and Semi-Submerged Face

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7. SOURCES OF ERRORS

8. CONCLUSIONS AND REMARKS

CE 412L: Fluid Mechanics Laboratory | Experiment No. 2: Hydrostatic Force on a Submerged and Semi-Submerged Face

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