Ex.3 Final Report

Sustainable Renewable Energy Engineering Summer 2016 Fluid Mechanics Lab Experiment 3 Hydrostatic Forces and Center of Pressure Section: 11 - U, T 1:00 p.m. Group: C Student Name ID Mona Ibrahim Albawab U00038472 Mounia Boukhenoufa U14123818 Kholood Abdollah U14121356 Ayat Mohammed Ahmed U00032083 Dr. Mohammed Ali Abdul Kareem Lab Instructor Mohammed Lab Engineer Eng. Monadhel Alchadirchy Experiment Date 21-June-2016 Report Submission 26-June- 2016 Date

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Table of Contents Abstract ……………………………………………………………………………………………………………………………….......... .............. Page 4 Objectives ……………………………………………………………………………………………………………………………….......... .............. Page 4 Theoretical Background ……………………………………………………………………………………………………………………………………… ………..……. Page 4 Apparatus ……………………………………………………………………………………………………………………………………… ………..……. Page 6 Procedure ………………………………………………………………………………. ………………………………………………………………….…. Page 6 Experimental results and completed tables: ………………………………………………………………………………. ………………………………………………………………….…. Page 7 Calculations …………………………………………………………………………….. ………………………………………………………………………. Page 8 Discussion ………….....……………………………………………….. …………......................................................................... Page 11 Conclusion ………………………………………………………………………………………………………………………………. …………………….. Page 11 References ………………………………………………………………………………………………………………………………. …………………….. Page 12

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List of Figures and/or Tables:

Page Number

Figure’s Title

Figure 1: Resultant force on a horizontal plane

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Figure 2: Hydrostatic force on an inclined surface

4

Figure 3: partly submerged surface

5

Figure 4: fully submerged surface

5

Figure 5: apparatus

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Table 1: Results for partial immersion

Table 2: Results for complete immersion

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Abstract: This lab is divided into two parts in the 1st part we will determine the experimental hydrostatic force along with the theoretical one and the center of pressure for a partly 3

submerged body and in the 2nd part we will do the same but this time for a fully submerged body. In both parts we will find the error by comparing the experimental value of the hydrostatic force to the theoretical one. The Study of hydrostatic forces on submerged surfaces is very important for the design of engineering structures such as dams, storage tanks and hydraulic systems. Objectives:  To determine the hydrostatic force acting on a partly or fully submerged body.  To determine the position of the center of pressure.  To compare the experimental values to the exact theoretical values. Theoretical Background: Hydrostatics is the branch of fluid mechanics that studies incompressible fluids at rest.1 Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. Any surface or body that is submerged in water or any other liquid, experiences force acting on it because of the hydrostatic pressure of the fluid.2 Horizontal submerged plane: For a horizontal plane submerged in a liquid (or a plane experiencing uniform pressure over its surface), the pressure, p, will be equal at all points of the surface. Thus the resultant force will be given by Hydrostatic Force = hydrostatic pressure ⨯ Area = ρghA And the center of pressure coincide with the centroid.3 Inclined Submerged Surface:

Figure -1- Resultant force on a horizontal plane

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A) Partial Immersion:

A  by H

y 2

I GG 

by 3 12

Figure -2- Hydrostatic force on an inclined surface

4

HP 

2y 3 MgL

Fexp 

ad  y

2y 3

MgL ad 

y 3

Fth  g  y 2  by

Fth  g H A

Fth 



Figure -3- partly submerged surface

1 gby 2 2 B) Complete Immersion

A  bd H  y

d 2

I GG 

HP 

bd 3 12

d2 d   y  2 12 H  Fexp  

a  d  y   

Fth  g H A

MgL MgL  2 d d2 d d   a    y    2 12 H 2  12 H 

Fth  g  y  d 2  by

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Figure -4- fully submerged surface

Apparatus and Procedure: Experimental Apparatus:  Hydraulic bench  Perspex tank equipped with:  A bridge piece at one end with integral pivot supports whilst at the other end of the tank there is balance arm restraint and datum level indicator  Spirit level attached to the base of the tank. Drain cock at the base of the tank extended to a drain tube.

a

Figure -5- Apparatus

A balance arm pivoted on a knife pivot with a groove to hold the weight hanger (balance pan) on one side and screw bolt for the adjustable counter balance on the other side.  A fabricated quadrant, which has dimensions of 100 mm internal radius, 300 mm outer radius and 75 mm width, acting as the immersed body.  A set of weights that can be attached to the balance ban.  A scale fixed to the quadrant. Experimental Procedure: 1. Ensure the balance arm is resting in the knife pivot. 2. Adjust the adjustable counter balance to balance the pivoted arm assembly to the horizontal position as shown by the beam level indicator. 3. Add very gently, all the provided weights through the balance ban. 4. Wet the surface of the scale to avoid capillary rise by the surface tension on the scale graduation. 5. Add water to the Perspex tank with slow rate until the arm is on the same horizontal position indicated by the beam level indicator. 6. Reduce weight or increase water to bring the arm to the position specified above. 7. Use the drain knob to drain water if necessary. 8. When satisfying the above items read the weights and the water level on the scale. 9. Tabulate your results according to table 1 and 2. 

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Experimental results and completed tables: Part a: Partial immersion: Water Temperature = 25oC

Total counter balance mass M (g) Measured depth of water y (mm) Experimental force on area (N)

Fexp 

MgL ad 

Trial 3 120

56

68

75

1.079

1.577

1.917

1.150

1.696

2.063

37.34

45.34

50

6.17%

7.02%

7.08%

Trial 1 250

Trial 2 300

Trial 3 350

111

125

137

4.271

5.206

6.132

4.475

5.502

6.382

1 gby 2 2

Depth of center of pressure (mm)

HP 



Trial 2 100

y 3

Theoretical force on same area (N)

Fth 

Trial 1 70

2y 3

Fth  Fexp Fth

 100%

Table 1: Results for partial immersion

Part b: Complete immersion: Total counter balance mass M (g) Measured depth of water y (mm) Experimental force on area (N)

Fexp 

MgL d d2 a  2 12 H

Theoretical force on same area (N)

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d  Fth  g  y   bd 2  Depth of center of pressure (mm)

HP  y 



74.66

86.12

96.58

4.56 %

5.38 %

3.92 %

2

d d  2 12 H

Fth  Fexp Fth

 100%

Table 2: Results for complete immersion

Calculations: Constants Used: o a= 100 mm = 0.1 m. o b=75 mm = 0.075 m o d= 100 mm = 0.1 m. o L= 285 mm = 0.285 m. o

ρwater @ 25 ° C =997 kg/ m3

Part a: Partial immersion: Fexp 

MgL ad 

Experimental force on area (N)

Fth  Theoretical force on same area (N)

HP 

y 3

1 gby 2 2

2y 3

Depth of center of pressure (mm)



Fth  Fexp Fth

 100%

Error

Trial 1:

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exp=¿

MgL y a+ d− 3 F¿

1 th=¿ ρ g b y 2 2 F¿ p=¿

2y 3

ε

¿ =

=

=

= 1.079 N

1 ⨯997 ⨯9.81⨯ 0.075⨯(0.056)² 2

2 ⨯56 3

=

H¿

0.07⨯ 9.81⨯0.285 0.056 0.1+ 0.1− 3

= 1.150 N

= 37.34 mm

F th −Fexp ∨¿ ⨯100 = Fth

1.150−1.079 ⨯100=¿ 1.150

6.17 %

Trial 2: exp=¿

MgL y a+ d− 3 F¿

1 th=¿ ρ g b y 2 2 F¿ p=¿

2y 3

ε

¿ =

=

=

=1.577 N

1 ⨯997 ⨯9.81⨯ 0.075⨯(0.068)² 2

2 ⨯68 3

=

H¿

0.1⨯9.81 ⨯ 0.285 0.068 0.1+0.1− 3

= 1.696 N

= 45.34mm

F th −Fexp ∨¿ ⨯100 = Fth

1.696−1.577 ⨯100=¿ 1.696

7.02%

Trial 3: MgL

exp=¿

a+ d− F¿

y 3

=

0.12⨯ 9.81 ⨯0.285 0.075 0.1+0.1− 3

= 1.917 N

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1 th=¿ ρ g b y 2 2 F¿ p=¿

2y 3

H¿

ε

¿ =

=

=

1 ⨯997 ⨯9.81⨯ 0.075⨯(0.075)² 2

2 ⨯75 3

= 50 mm

F th −Fexp ∨¿ ⨯100 = Fth

2.063−1.917 ⨯100=¿ 2.063

Part b: Complete immersion

H  y

Fexp 

d 2

MgL d d2 a  2 12 H

d  Fth  g  y   bd 2 

HP  y 



d d2  2 12 H

Fth  Fexp Fth

 100%

Trial 1: H

=

d 0.1 y− =0.111− 2 2

= 2.063 N

= 0.061 m

10

7.08 %

MgL d d² a+ + 2 12

Fexp =

(

th=¿ ρ g y−

¿

0.25 ⨯9.81 ⨯0.285 (0.1)² 0.1 0.1+ + 2 12 ⨯ 0.061

d 0.1 b d=997 ⨯9.81⨯(0.111− ) ⨯0.075⨯0.1 2 2 F¿

)

d d2 100 (100)² H p= y − + =111− + 2 12 2 12 ⨯61

ε

= ¿

= 4.271 N

F th −Fexp ∨¿ ⨯100 = Fth

= 4.475 N

= 74.66 mm

4.475−4.271 ⨯100=¿ 4.475

4.56 %

Trial 2: H

=

d 0.1 y− =0.125− 2 2 MgL d d² a+ + 2 12

Fexp =

(

th=¿ ρ g y−

= 0.075 m

¿

0.3 ⨯9.81 ⨯ 0.285 (0.1) ² 0.1 0.1+ + 2 12 ⨯ 0.075

d 0.1 b d=997 ⨯9.81⨯(0.125− )⨯ 0.075 ⨯0.1 2 2 F¿

)

d d2 100 ( 100)² H p= y − + =125− + 2 12 2 12⨯75

ε

¿ =

F th −Fexp ∨¿ ⨯100 = Fth

=

d 0.1 y− =0.137− 2 2

= 5.502 N

= 86.12 mm

5.502−5.206 ⨯ 100=¿ 5.502

Trial 3: H

= 5.206 N

= 0.087 m

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5.38 %

MgL d d² a+ + 2 12

Fexp =

(

th=¿ ρ g y−

¿

0.35 ⨯9.81 ⨯0.285 (0.1)² 0.1 0.1+ + 2 12 ⨯0.087

d 0.1 b d=997 ⨯9.81⨯(0.137− )⨯ 0.075 ⨯0.1 2 2 F¿

)

d d2 100 (100)² H p= y − + =137− + 2 12 2 12 ⨯87

ε

= ¿

= 6.132 N

F th −Fexp ∨¿ ⨯100 = Fth

= 6.382 N

= 96.58mm

6.382−6.132 ⨯ 100=¿ 6.382

3.92%

References: [1] https://en.wikipedia.org/wiki/Hydrostatics [2] http://www.edinformatics.com/math_science/hydrostatic_pressure.htm http://www.efm.leeds.ac.uk/CIVE/CIVE1400/Section2/Surfaces.htm [3] [4] Lab manual

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