Objective:

1-To determine the position of the center of pressure on the rectangular face of the toroid.

2-To compare the measured value with that the predicted from the theoretical analysis. Introduction: Design of dams, gates, liquid storage tanks, ships, submarines, buildings, etc, requires the calculation of the magnitude, direction and line of action of pressure forces on plane surfaces. When the pressure is uniform over a surface, such as static pressure on a horizontal surface, the resultant force is equal to area times the pressure and acts through the centroid of the area. However, in many cases, such as liquids acting on a non-horizontal surface, the situation is more complex. A general approach has been developed to estimate resultant force and its line of action. Theory For a static fluid, the shear stress is zero and the only stress is the normal stress, i.e., pressure p. Recall that p is a scalar, which when in contact with a solid surface exerts a normal force towards the surface.

Figure: Hydrostatic pressures and pressure force on a plane surface

The hydrostatic force on any surface is due to the fluid pressure acting on that surface, as shown in the figure. Pressure is a normal stress which is positive when in compression. Since the pressure is everywhere normal to the surface, the resultant pressure force (Fp) is also normal to the surface. The magnitude of Fp is

The resultant pressure force acting at this center of pressure must give the same moment as the distributed pressures

Notations

p =

y = distance measured from level of zero pressure and measured in the plane of the surface.

the horizontal.

y coordinate of the centroid of the surface. pressure at the centroid of the surface. Io = moment of inertia about the level of zero pressure

for the surface on which the pressure force is acting. moment of inertia about the horizontal centroid axis.

Apparatus The apparatus is constructed to permit the measurement of the moment due to the total fluid thrust on a wholly or partially submerged plane surface and comparison to a theoretical value based on fluid depth. A toroid is immersed in a tank containing water and a Hook gauge measures the depth of immersion, the toroid is mounted on a balance and pivoted about the center of curvature of the toroid. Thus, only the vertical face along GC of the toroid has any moment about the balancing point. A rider weight balances the weight of the toroid in the dry, so that the moment of the hydrostatic force on GC is measured by the weight at the pan. The cylindrical sides of the quadrant have their axes coincident with the center of rotation of the tank assembly, which allows the total fluid pressure acting on these surfaces to exert no moment around the center. Therefore the only moment that exists is due to the fluid pressure acting on the plane surface.

Figure: apparatus similar to the one used in the experiment.

Figure : Hydrostatic pressure distribution on the toroid (balanced).

Procedure 1.Measure the dimension a,b,c, and d, see figure3. 2. Level the tank under the toroid and adjust the weight W1to level the balance arm. Carefully admit water to the tank until it just touches the bottom of the guardant .Take the vernier reading> 3. Raise the water level in increments of about 10mm and add weight to the balance pan to level the balance arm for each depth of water. Note the mass M and the depth of immersion h for each reading. O a Mass d y

b a = 11 cm b = 7.5 cm d = 10 cm c = 30 cm

Figure3: Dimensions measured.

Data collected h ( cm

) M ( g

) Immersed Area

(m^2)*.01 Theoretical Ycp(cm)

Experimental Ycp(cm)

M\h^2

1.5 4 0.11 1 1.5 1.8 3 20 0.23 2 3 2.2

4.5 42 0.34 5.25 4.5 2.1 6 82 0.45 7 6 2.3

7.5 122 0.56 8.75 7.5 2.2 9 172 0.68 10.5 9 2.1

h (cm) M(g) Theoretical Ycp(cm)

Experimental Ycp(cm) (h-d/2)

10 214 6.67 10 9.95 11 254 7.52 11 10.95 12 284 8.40 12 11.95 13 324 9.32 13 12.95 14 356 10.27 14 13.95

Sample of calculation

cby

cdab

yM

ydaybcM

ydayybgcMg

yybgFYAF

ydaFcMg

yydaFcW

M

p

cp

p

p

62)(

)31(

2**

)31(*

2***

2/****

)31(**

)32)((**

0

2

2

0

(b) Derivation of the equation;

cbdda

cbddyM

12))(2/1()2/(

3

yAyIy

AyFFLMgc

cp

)2/()2/(12

2

dybddy

bd

addy

ddygbd

addy

ddygbd

2)2/(12)2/(

2612)2/(

2

2

adyddy

dyddygbd )2/(

2)2/(

)2/(12)2/( 2

]2/)[2/(

12

2

dadydgbd

cgbdad

cbddyM

adcbddy

cgbd

1222

22123

3

Conclusions and discussion:

1- The measured values of the position of center of pressure are closed to those calculated from physical dimensions. The difference could be caused because of personal errors in reading and approximations in calculations, and because of the device accuracy.

2- For (M/y2) vs. (y) Theoretically:

Slope = - -ρb/6c = 41.67 kg/m3 Intercept = ρb *(a+d)/2c = 25 kg/m2

Experimentally: Slope = -68.473 kg/m3 Intercept = 27.174 kg/m2 For (M) vs. (y-d/2) Theoretically:

Slope = (ρbd/c) (a+0.5d) = 3.75 kg/m Intercept = ρbd3 /12c = 0.0208 kg

Experimentally: Slope= 3.73 kg/m Intercept = 0.024 kg

3- The pressure forces on the four faces are not zero nor

negligible. They were ignored since (for each two opposite faces) they are equal in magnitude but opposite in direction. So they cancel each other, and this is not a mistake.

4- The line of action of the buoyancy force passes through the

point (0) about which we took the moment, so the moment is equal to zero. And that is why we ignore it.

5- The center of pressure does not depend on the fluid used, but on the dimensions of the toroid.

yAyIy

AyIyy cpcp

where I, y , A are independent parameters of The location of the center of pressure will not be affecte

Hydrostatic Pressure in Liquids

In this Java simulation the hydrostatic pressure of a liquid is measured with an U-tube manometer. On the upper side of the red coloured chamber there is a membrane which is deformed more or less, depending on the pressure. Consequently the pressure of the air in the connected tube (pink) increases so that the liquid sinks in the left part of the U-shaped tube and rises in its right part. The displacement of the liquid's level measures the hydrostatic pressure.

Remark 1: It is supposed that the liquids in the U-shaped tube and in the container are the same. Remark 2: Note that only the hydrostatic pressure of the liquid is registered, not the pressure of the air!

The problem of this simulated experiment is the relation between hydrostatic pressure and depth. You can raise or lower the manometer with pressed mouse button. On the right side it is possible to select one of several liquids. In addition, you can write the values of density and depth directly into the text fields. On the lower right the applet will indicate the measured hydrostatic pressure (in hPa). (1 hPa = 1 Hektopascal = 100 Pa = 100 N/m2)

You can see that the hydrostatic pressure of water increases by about 1 hPa for each cm of depth.

The following formula is more general:

p = g p ... hydrostatic pressure of the liquid g ... gravitational acceleration

h ... depth

Hydrostatic Pressure in a Liquid The pressure at a given depth in a static liquid is a result the

weight of the liquid acting on a unit area at that depth plus any pressure acting on the surface of the liquid.

The pressure due to the liquid alone (i.e. the gauge pressure) at a given depth depends only upon the density of the liquid distance below the surface of the liquid h.

Pressure is not really a vector even though it looks like it in the sketches. The arrows indicate the direction of the force that the pressure would exert on a surface it is contact with.

Pressure can not exert any force parallel to the surface in

which it is contact.

* The pressure at a given depth is independent of direction -- it is the same in all directions. This is another statement of the fact that pressure is not a vector and thus has no direction associated with it when it is not in contact with some surface.

* The pressure on a submerged object is always perpendicular to the surface at each point on the surface.

Pascal's Principle:

Any external pressure applied to a fluid is transmitted undiminished throughout the liquid and onto the walls of the containing vessel.

Example 1: The absolute pressure at a depth h in a liquid open to the atmosphere in increased by the pressure of the atmosphere pushing down on the surface of the liquid.

Example 2: A hydraulic pump used to lift a car. When a small force f is applied to a small area a of a movable piston it creates a pressure P = f/a. This pressure is transmitted to and acts on a larger movable piston of area A which is then used to lift a car.

The pressure at a given depth does not depend upon the shape of the vessel containing the liquid or the amount of liquid in the vessel.

Hydrostacic Paradox QT Movie Demonstration Example: If the height of the fluid's surface above the bottom of the five vessels is the same, in which vessel is the pressure of the fluid on the

bottom of the vessel the greatest ? The amount of liquid in each vessel is not necessarily the same.

The pressure P is the same on the bottom of each vessel. Why the pressure does not depend upon the shape of the vessel or the amount of fluid in the vessel rests upon three things: a. Pressure is force per unit area and this is not same as the total weight of the liquid in a vessel. b. A fluid can not support its self without a container. Thus the walls of the container exert a pressure on the fluid equal to the pressure of the fluid at that depth. c. The pressure at given level is transmitted equally throughout the fluid to be the same value at that level.

Explanations:

Vessel A: No matter how wide the vessel, the pressure is just the weight of the fluid above unit area on the bottom. Even if you take the whole weight of the fluid in the container mg and divide by the area of the bottom A, you still get the same results since the vessel is equivalent to a column of water.

Vessel B: Vessel B could be divided into three parts. The fluid in parts 1 and 3 is supported by upward force of the vessel on the fluid. Part 2 could be though of a vertical column of liquid similar to vessel A. One could ask why doesn't the fluid in parts 1 and 3 (which are much bigger) not squeeze column 2 where they meet

(along the dashed line) and there by increase the pressure on the bottom of the column ? The answer is that the fluid in column 2 exerts and equal but opposite pressure outwards on the two other liquids to support itself. From the point of view of column 2, the water outside the dashed lines in sections 1 & 3 could be replace by solid vertical walls (along the dashed lines) and column 2 would still be in equilibrium.

Vessel C: Again we could divide the water into three sections. The middle section is similar to that of vessel A or B. Since the height of the fluid in section 1 or 3 is not high enough to produce the same pressure as the height of the fluid in 2, how does the pressure on the bottoms of section 1 and 3 get to be the same as that of 2 ?

The answer is that top of the container's walls in sections 1 and 3 produce a downward pressure that is equal to the fluid pressure in the middle section at the same level. If you poked a hole in the top of the container in sections 1 or 3, water would fountain upwards from the hole under pressure. From Pascal's principle, this pressure has to be that of the fluid in the middle section at the same level.

Vessel D: Again the center column is similar to vessel A.

The pressure of the vessel's wall creates a pressure that vertically supports the fluid in sections 1 and 3. At the same time the pressure of the walls create a horizontal component

of pressure that sustains the fluid in the center column.

Vessel E: While one can not offer simple arguments like those for the other vessels, the pressure on the bottom is still the same basically because of Pascal's principle.

You go down from the surface to some depth, then move sideways until you can go down again. Repeat the process until you reach the bottom. Since the pressure at the same depth is the same, moving sideways does not change the pressure. Only downwards motion increases the pressure.

hydrostatic pressure 1. n. [Geology]

ID: 277 The normal, predicted pressure for a given depth, or the pressure exerted per unit area by a column of freshwater from sea level to a given depth. Abnormally low pressure might occur in areas where fluids have been drained, such as a depleted hydrocarbon reservoir. Abnormally high pressure might occur in areas where burial of water-filled sediments by an impermeable sediment such as clay was so rapid that fluids could not escape and the pore pressure increased with deeper burial. Synonyms: reservoir pressure See: abnormal pressure, absolute pressure, formation pressure, freshwater, geopressure, geopressure gradient, hydraulic head, hydrostatic head, normal pressure,

(Click to enlarge)

Pressure versus depth plot

overpressure, reservoir pressure, underpressure 2. n. [Drilling] ID: 1428 The force per unit area caused by a column of fluid. In US oilfield units, this is calculated using the equation: P=MW*Depth*0.052, where MW is the drilling fluid density in pounds per gallon, Depth is the true vertical depth or "head" in feet, and 0.052 is a unit conversion factor chosen such that P results in units of pounds per square in. (psi). See: circulation loss, hydrostatic head, kick, shut-in bottomhole pressure, shut-in pressure, true vertical depth

(Click to enlarge)

Diagram of hydrostatic head and hydrostatic pressure

3. n. [Drilling Fluids] ID: 2002 The pressure at any point in a column of fluid caused by the weight of fluid above that point. Controlling the hydrostatic pressure of a mud column is a critical part of mud engineering. Mud weight must be monitored and adjusted to always stay within the limits imposed by the drilling situation. Sufficient hydrostatic pressure (mud weight) is necessary to prevent an influx of fluids from downhole, but excessive pressure must also be avoided to prevent creation of hydraulic fractures in the formation, which would cause lost circulation. Hydrostatic pressure is calculated from mud weight and true vertical depth as follows: Hydrostatic pressure, psi = 0.052 x Mud weight, lbm/gal x True Vertical Depth, ft. (To convert to SI units, 1.0 psi = 6.9 kPa.) See: equivalent circulating density, kill-weight fluid, lost circulation, mud balance, mud program, mud weight, pressurized mud balance, spotting fluid

Osmosis and Hydrostatic Pressure Simulator

The flow of water across a membrane in response to differing

concentrations of solutes on either side - osmosis - generates a pressure across the membrane called osmotic pressure. Osmotic pressure is defined as the hydrostatic pressure required to stop the flow of water, and thus, osmotic and hydrostatic pressures are, for all intents and purposes, equivalent. The membrane being referred to here can be an artifical lipid bilayer, a plasma membrane or a layer of cells.

The osmotic pressure P of a dilute solution is approximated by the following:

P = RT (C1 + C2 + .. + Cn)

where R is the gas constant (0.82 liter-atmosphere/degree-mole), T is the absolute temperature, and C1 ... Cn are the molar concentrations of all solutes (ions and molecules).

Similarly, the osmotic pressure across of membrane separating two solutions is:

P = RT (delta C)

where delta C is the difference in solute concentration between the two solutions. Thus, if the membrane is permeable to water and not solutes, osmotic pressure is proportional to the difference in solute concentration across the membrane (the proportionality factor is RT).

To get an appreciation for these equations, try the pressure simulator below. You can enter concentrations either as molarity or grams/liter. The choices of solute are:

NaCl: molecular weight 58.44; dissociates in solution into sodium and chloride ions

Sucrose (table sugar): molecular weight = 342.3; does not dissociate Albumin: molecular weight approximately 66,000; a single protein

Note: you can easily enter "impossible" solutions in the simulator, for example, 1 M albumin, which would require dissolving 66 kg of albumin in a liter! The simulator will notify you if the pressure is either less than 1 cm or above 10000 cm. Suggestion: Determine the pressure generated when 8.5 grams/liter NaCl (0.145 M) is on one side and water on the other - this is pretty close to the situation of having blood on one side and water on the other. You should also investigate the question of how much large macromolecules like

albumin contribute to osmotic pressure (blood serum contains roughly 3% albumin).

Assumed temperature is 20C or 293K

A note for the sophisticate: In the discussion and simulator above, it was assumed that each NaCl fully dissociates into two free ions, which is essentially the case for very dilute solutions. However, in more concentrated, physiologic solutions, these two ions attract one another and the "osmotically effective" concentration of solute is less than twice the molarity of NaCl (roughly 0.93X for NaCl). The Van't Hoff equation described above can be modified for additional accuracy by incorporation of an osmotic coefficient I (0.93 for NaCl):

P = I RT (C1 + C2 + .. + Cn) Did you try the blood-water comparision? You should have found that "physiological saline" generates sufficient osmotic pressure to support a column roughly 70 meters high! The calculation performed (without accounting for effective concentration) was: P = 0.082 liter-atm/mole-degree * 997.9 cm/atm * 293 degrees * 0.29 moles

of solute/liter