es341 lab 2

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ES-F341 F04 Lab 2 Title: Hydrostatic Forces on Vertical and Inclines Flat surfaces Abdul Sayed, [email protected] Ryan Johnson, [email protected] Mi Chin Yi, [email protected] Grant Cummings [email protected] Cecilia Hull [email protected] Date performed: 09/25/2014 Date due 10/2/2014

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Page 1: ES341 lab 2

ES-F341

F04

Lab 2

Title: Hydrostatic Forces on Vertical and Inclines Flat surfaces

Abdul Sayed, [email protected]

Ryan Johnson, [email protected]

Mi Chin Yi, [email protected]

Grant Cummings [email protected]

Cecilia Hull [email protected]

Date performed: 09/25/2014

Date due 10/2/2014

Page 2: ES341 lab 2

Introduction:

An engineer must take into account hydrostatic forces into structural design

considerations as they impose important forces on common structures such as dams, irrigation

canals, fluid storage tanks, and retaining walls.

Hydrostatic forces can be measured by calculating the pressure distribution on the objects

which is submerged. The main objective is to measure the hydrostatic forces exerted on vertical

and inclined flat surface. Pressure is defined as the force per unit area. Fluid density and fluid

depth are used to determine fluid pressure by Pressure = density x gravity x height.

Limitations in this experiment arise due to the water level not being stable after water

insertion. Angle measurement also created opportunity for error.

Page 3: ES341 lab 2

Equipment & Setup:

Equipment used in this experiment:

Center of Pressure apparatus (Model/Serial Number = 185269, type#1)

Angular Calibrator

Staff Guage

Metal weights

The illustration below details the equipment used in this experiment.

Above schematic of equipment used taken from:

Claydon, J. (2014). Centre of pressure. [online] Jfccivilengineer.com. Available at:

http://www.jfccivilengineer.com/centre_of_pressure.htm [Accessed 2 Oct. 2014].

Page 4: ES341 lab 2

Above is a picture of the equipment used, taken from:

Barve, N. (2014). Scitech Hydrostatic Force & Center of Pressure apparatus. [online]

Sci-tech.biz. Available at: http://www.sci-tech.biz/trainersDetail.php?Hydrostatic-Force-Center-

of-Pressure-apparatus-170 [Accessed 2 Oct. 2014].

Page 5: ES341 lab 2

Procedure:

1. Add water and using weights balance tank.

2. Add increments of water gradually and heften or lessen weights ever so slightly so

that observed angle may approach zero.

3. Make recordings of vertical height of water corresponding to a given load that zeroed

the angle.

4. Continue trial until a total of four data sets are obtained corresponding to heights

of water specified in Table 1.

5. Repeat the above steps but for an angle of 10 degrees.

Results:

Table 1 summarizes the collected measurements. Listed are the mass of the load arm [m],

and the vertical distance between the pivot and the water surface [h].

Table 2 and 3 contain the following data for angles of 0, and 10 degrees, respectively: the

mass of the load [m], the force of water against gate [FL], submerged height of gate [YG],

vertical distance to centroid of gate [hc], area of gate under water [AG], hydrostatic force on gate

[Fw], distance from water surface to center of pressure [Ycp] both theoretically calculated and

experimentally measured.

Data in lines 1-4 of tables are of “partial 1”; water level less than full, “partial 2”; water

level closer to full, “full” ; water level just reaches top of gate, and “above”; water level above

top of gate.

*Formatting for Table 2 and 3 taken from “Lab Handout 2” from Blackboard.

Page 6: ES341 lab 2

Table 1

Part A Part B

m (kg) h (m) m (kg) h (m)

Partial 1 0.050 0.154 0.050 0.152

Partial 2 0.100 0.132 0.100 0.130

Full 0.220 0.0990 0.220 0.0965

Above 0.400 0.0508 0.400 0.0495

Table 2 (θ = 0o)

m (kg)

FL (N)

YG

(m)hC

(m)AG

(m2)FW (N)

YCP,theor.

(m)YCP,exp (m)

0.050 .491 .154 .0770 .00978 7.39 .103 -.137

0.100 .981 .132 .0660 .00838 5.43 .0881 -.129

0.220 2.16 .0990 .0495 .00629 3.05 .0660 .0809

0.400 3.92 .0508 .0254 .00322 .804 .0339 1.18

Table 2 (θ = 10o)

m(kg)

FL (N)

YG

(m)hC

(m)AG

(m2)FW (N)

YCP,theor.

(m)YCP,exp (m)

0.050 .491 0.152 .0760 .00965 7.20 .101 -.137

0.100 .981 0.130 .0650 .00826 5.26 .0866 -.0854

0.220 2.16 0.0965 .0483 .00613 2.90 .0644 .0883

0.400 3.92 0.0495 .0246 .00314 .763 .0330 1.23

Page 7: ES341 lab 2

Discussion:

Figure 1: Plot of results from table 2 for Ycp theoretical vs mass and Ycp experimental vs mass

Page 8: ES341 lab 2

Figure 2: Plot of results from table 3 for Ycp theoretical vs mass and Ycp experimental vs mass

Page 9: ES341 lab 2

Questions:

1. Could this apparatus be used for other fluids to determine their center of pressure?

Yes, this apparatus can be used for any incompressible fluid.

2. What are the important forces to consider in your analysis?

The load from the metal weight, as well as the force from the hydrostatic pressure of water

were important forces to consider.

3. List sources of error.

Systematic error due to limitations of equipment, error arising from improperly interpreting

balance as being level, friction between moving parts were all sources of error.

4. Discuss your experimental and theoretical results

There is a considerable error in both experiments therefore, the results are not reliable.

In both cases, the curves do not match so theoretical and experimental data do not agree.

5. Would this experiment be suitable for two immiscible fluids? Explain.

Two fluids, despite if they do not mix, would be impractical to work with in this experiment

since that would add to the complexity of the procedure.

Conclusion:

Results from this experiment showed inverse correlation between theory and data. The

reason for this may be faulty data acquisitioning or erroneous processing.

According to the figures, we can see the results for both parts are not reliable based on the

difference between the theoretical value and the experimental value.

Page 10: ES341 lab 2

References:

Toniolo, H. (2014). [online] Classes.uaf.edu. Available at:

https://classes.uaf.edu/webapps/portal/frameset.jsp?tab_tab_group_id=_2_1&url=%2Fwebapps

%2Fblackboard%2Fexecute%2Flauncher%3Ftype%3DCourse%26id%3D_146492_1%26url

%3D [Accessed 2 Oct. 2014].

Claydon, J. (2014). Centre of pressure. [online] Jfccivilengineer.com. Available at:

http://www.jfccivilengineer.com/centre_of_pressure.htm [Accessed 2 Oct. 2014].

Barve, N. (2014). Scitech Hydrostatic Force & Center of Pressure apparatus. [online]

Sci-tech.biz. Available at: http://www.sci-tech.biz/trainersDetail.php?Hydrostatic-Force-Center-

of-Pressure-apparatus-170 [Accessed 2 Oct. 2014].

Page 11: ES341 lab 2

Appendix:

Observations:

G (height of gate) = 0.1016 meter

B (width of gate) = 0.0635 meter

R1 (distance from pivot to top of gate) = 0.1016 meter

R2 (distance from pivot to bottom of gate) = 0.2032 meter

RL (distance from pivot to weight) = 0.254 meter

*The following definitions and formulae taken from “Lab Handout 2” from Blackboard :

Definitions:

θ = angle of apparatus

RL = length of load arm (m)

R1 = radius to the upper edge (m)

R2 = radius to the lower edge (m)

B = width of gate (m)

G = height of gate (m)

W = specific weight of water (kN/m3)

m = mass on load arm (kg)

FL = force on load arm (N)

FW = resultant force of water against gate

(N)

h = vertical distance between the pivot

and the water surface (m)

hC = vertical distance from the water

surface to the centroid of the submerged

area of the gate (m)

AG = submerged area of the gate (m2)

Ycp = distance, along the angle of

incline, from the water surface to the

center of pressure (m)

Ψ = distance, along the angle of incline,

from the water surface to the centroid of

the submerged area of the gate (m)

YG= submerged height of the gate (m)

IO = second moment area of the

submerged area of the gate (m4)

Page 12: ES341 lab 2

Formulas:

Hc = YG/2

AG = B*YG

Here is a sample calculation for a mass of .05 kg for table 2

FL = mg = 0.05kg * 9.81m/s^2 = 0.491N

AG = B * YG = .0635m * .154m = 0.00978m^2

FW = γW*hC*AG = 9810N/m^3 * .077m * .00978m^2 = 7.39 kN

( [FLRL*cosθ] / Fw ) - h/cosθ = YCP, exp

( [.491*.254*cos(0)] / 7.39 ) - .154/cos(0) = -0.137

YCP, theor=Ψ+I0/(AG*Ψ)= Ψ+ B*YG3/12/(AG*Ψ)

=0.1+0.075*0.13/12/(0.00314*0.1)=0.029m