Experiment #3

Center of Pressure on a Submerged Plane Surface

Stephen Mirdo

Performed on October 7, 2010

Report due October 14, 2010

Table of Contents Object ………………………………………..………………………….………….…. p. 1 Theory ………………………………………………………………..………....…pp. 1 - 4 Procedure …………………………………………………………………….………...p. 5 Results ………………………………………………………..……...........……………p. 6 Discussion and Conclusion …………………………………………………….………p. 7 Appendix ……………………………………………………..…….…....………pp. 8 - 10

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Object The object of this experiment was to calculate the hydrostatic force a fluid exerts on a submerged plane surface and then compare the experimental hydrostatic force to the theoretical hydrostatic force.

Theory

A fluid at rest is said to be in a static condition. Consider a small three dimensional “chunk” of a static fluid, called a fluid particle as seen in Figure 1. The fluid particle experiences a constant pressure on all sides acting inward towards a single point at its center. The horizontal x and y components of the pressure are equal and opposite and have a net sum of zero. It can then be said that the pressure does not change in the horizontal direction.

Figure 1: Diagram of a fluid particle with the acting pressures and forces.

The pressure components acting in the vertical z direction are also equal and opposite in magnitude and negate one another. The only component of Figure one that is not negated is the weight of the fluid particle that acts vertically along the z axis. By Newton’s Second Law, the force of weight is expressed as the product of mass, m, and acceleration, a.

F = ma (Equation 1) The mass of the fluid particle can be expressed as the product of its density, ρ, and its volume, V.

m = ρV (Equation 2)

Substituting Equation 2 into Equation 1 for the mass term and letting the acceleration term equal gravitational acceleration yields:

F = ρVg (Equation 3)

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Because the x and y components of the pressure acting on the fluid particle are negated, the volume term of Equation 3 is reduced from a volume to a unit length, dz. As the depth of the fluid particle increases, the pressure acting on the particle also increases due to the weight of the fluid. This is known as the pressure gradient. The change in pressure, dP, can be expressed as a first order, non-homogenous, ordinary differential equation. The change in pressure with respect to the change in height will equal the specific gravity of the fluid particle, ρg.

dP/dz = ρg (Equation 4) Separating and integrating Equation 4 will yield the hydrostatic equation.

Po∫PdP = zo∫zρg dz ΔP = ρgΔz (Equation 5)

Figure 2: Diagram of the center or pressure apparatus, reference heights and acting force. (Adapted from A Manual for the Mechanics of Fluids Laboratory, W.S. Janna,

University of Memphis Department of M.E., 2008) If the hydrostatic equation is applied to an infinite number of points on a submerged surface, such as that of the submerged plane of the torus in Figure 2, it will yield an increasing pressure for an increasing depth. This increase with depth is known as the pressure gradient, which can be seen in Figure 3 below. Inside of the pressure gradient, there is a location where an equivalent resultant force, Rf, will act.

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Figure 3: Diagram of the pressure gradient and the equivalent resultant force

To calculate the theoretical resultant force, Rf, that acts on the submerged plane of

the torus, the hydrostatic equation is applied. Substituting the vertical depth of sinΘzc into the hydrostatic equation (Equation 5), where zc is the distance from the centroid of the plane to the free surface, and integrating with respect to the area of the pressure gradient will yield the equation of the resultant force of the pressure gradient.

∫P dA = ∫ρgsinΘzc dA Rf = ρgsinΘzcA (Equation 6) The components of Equation 6 are as follows:

ρ = the density of the fluid g = gravitational acceleration zc = the distance from the free surface to the centroid of the plane A = the area of the submerged plane

To calculate the distance from the surface of the fluid to the centroid of the

submerged plane of the torus, zc, employ the following equation where Ri is the inner radius of the torus, Ro is the outer radius of the torus and z is vertical height from the pivot to the fluid level.

zc = (Ri – z) + (Ro – Ri)/2 (Equation 7)

After determining the distance from the surface of the fluid to the centroid of the submerged plane in Equation 7, Equation 6 may be used to calculate the theoretical resultant force Rf.

To calculate the experimental resultant force of the fluid in the torus of the center of pressure apparatus, a sum of the moments about the pivot point must calculated. To obtain the distance of the resultant force to the pivot point, zR, employ Equation 8 below where Ixx,c is the second moment of inertia of a square specimen, zc (Equation 7) is the distance from the centroid to the free surface, and A is the area of the submerged plane.

zR = Ixx,c / (zcA) + zc (Equation 8)

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The second moment of inertia formula is as follows, where b is the length of the base of the submerged plane and h is the height of the submerged plane:

Ixx,c = (1/12) bh3 (Equation 9)

To calculate the sum of the moments acting around the pivot of the lever arm of the apparatus, employ Equation 10 below where Rf is the experimental resultant force to be determined, z is the vertical distance from the fluid level to the pivot, zR is the distance from the resultant force to the pivot, W is the weight applied to the weight hanger and L is the moment arm of the apparatus. The equation is set equal to zero and rearranged algebraically in Equation 11 so as to solve for the experimental resultant force the fluid exerts on the submerged plane of the torus.

ΣMo = Rf(z + zR) – WL = 0 (Equation 10)

Rf = WL / (z + zR) (Equation 11)

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Procedure Equipment:

TQ H11 Center of Pressure Apparatus (SN: A0390/10) Weights in any increment Water Experiment:

1) Place the center of pressure apparatus on a level surface. This is important for

obtaining an accurate force balance scenario. 2) Add weight to the weight hanger of the center of pressure apparatus. For

example, add 20 grams. The hanger can be found on the far left of the apparatus as seen in Figure 1.

3) Pour water into the torus until the added weight of the water balances the lever

arm about the pivot to a horizontal position. 4) Record the vertical height of the water in the torus relative to the pivot of the

lever arm. This value will be needed to calculate the moment the hydrostatic force the fluid exerts on the plane at the bottom of the torus.

5) Repeat Steps 2 through 4 for four more weights. It is not important to use an

identical increment of weight for each trial. It is, however, important to reference the vertical height from the free surface of the fluid in the torus to the pivot of the lever arm.

6) Use Equation 11 to calculate the experimental hydrostatic force the water

exerts on the submerged plane of the torus. Use Equation 6 to calculate the theoretical hydrostatic force the water exerts on the submerged plane of the torus.

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Results

Table 1: Dimension of the torus used to calculate theoretical and experimental force of the pressure gradient exerted on the torus.

Base of Submerged Plane (m) 0.075 Height of Submerged Plane (m) 0.1 Moment Arm L (m) 0.2 Inner Radius of Torus (m) 0.1 Outer Radius of Torus (m) 0.2

Table 2: Weights, heights and calculated experimental and theoretical hydrostatic forces

Mass (g) Weight (N) z (m) zc (m) zR (m)

Rf Experimental

(N)

Rf Theoretical

(N) 450 4.41 0.092 0.058 0.0724 5.37 4.27 470 4.61 0.088 0.062 0.0754 5.64 4.56 490 4.81 0.084 0.066 0.0786 5.91 4.86 510 5.00 0.08 0.07 0.0819 6.18 5.15 530 5.20 0.076 0.074 0.0853 6.45 5.44 550 5.40 0.072 0.078 0.0887 6.72 5.74

Note: Theoretical hydrostatic force calculations in Table 2 used values of ρ = 1000 kg/m3 and g = 9.81m/s2. Table 3: Percent error calculation for theoretical and experimental hydrostatic force

Rf Experimental

Rf Theoretical % Error

5.37 4.27 25.9% 5.64 4.56 23.7% 5.91 4.86 21.7% 6.18 5.15 20.0% 6.45 5.44 18.4% 6.72 5.74 17.0%

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Discussion & Conclusion

In summing the moments about the pivot of the apparatus, the buoyant force is neglected. As seen in the apparatus setup in Figure 2, the fluid resides inside the torus. The presence of buoyancy comes from the air outside of the torus. Because the density of air is a mere fraction of that of the material of the torus and the fluid it contains, it can be neglected in the hydrostatic force calculations.

The weight of the torus can also be neglected. Because the center of the curvature of the torus is at the location of the pivot, it is negated. The weight of the torus was not included in the calculations because the device was calibrated with ballast water so as to begin the experiment with a net moment of zero about the pivot.

It was noted that a large discrepancy between the theoretical and experimental values occurred. This is most likely due to errors in measurement of the height of the fluid inside of the torus. Another possible cause could be that the apparatus was not sitting level on the counter where the experiment was performed. If the apparatus is not sitting level, the moment calculations will yield inaccurate results. A leveling device on or near the testing apparatus would aid in ensuring the moment balance is accurate.

Another source of error would be the use of the accepted density of water, 1000 kg/m3, for the theoretical calculation of the hydrostatic force. This accepted value is the density of sea water at 4oC. The water used in this experiment was tap water at approximately 23oC. However, if the actual density of the tap water was used, the theoretical calculations would not differ greatly enough to compensate for the magnitude of the error.

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Appendix Data Usage Sample calculation of the weight of a 450 gram mass block:

W = 450 g * (kg/1000g) * 9.81 m/s2 = 4.41 N Sample calculation of the distance from the surface to the centroid, zc, at a fluid depth of 92mm

zc = [(100mm – 92mm) + (200mm – 100mm)/2] * (m/1000mm) = 0.058 m

Sample calculation of the theoretical resultant force acting on the torus at a weight of 4.81 N and fluid height of 84 mm

Rf, theoretical = 1000 kg/m3 * 9.81 m/s2 * (150mm – 84mm) * 75mm * 100mm = 4.86 N

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Bibliography Introduction to Fluid Mechanics, 3rd Edition

William S. Janna (1993)

A Manual for the Mechanics of Fluids Laboratory William S. Janna (2008)