On the past two weeks, I have done an experiment on hydrostatics, or is also known as fluid statics (fluid at rest) within the fluid mechanics field of study. This condition explains that in a stable condition, the fluid is at rest. The use of fluid in doing work is known as hydraulics, and the science of fluid in motion is known as fluid dynamics.
The natural nature of fluids are they cannot remain stationary under the application of shear stress. However, fluid can apply force normal to any surface contacting it.
If the fluid is considered as a solid object such as a cylinder, the pressure acting on a surface is the same as the pressure on the opposite side of the object, but in a different direction. This condition can be applied to any surface on the imaginary fluid shape. This thus defines that the pressure on a fluid is isotropic, meaning that the force/pressure in any direction applied on the liquid is the same in all directions. Hydrostatic Pressure Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the gravitational pull. The fluid is known as hydrostatic fluid.
The pressure can be calculated from the control volume analysis of a small cube of fluid. It is known that pressure is force applied per unit area P = F/A, and the onlyforce acting on any such small cube of fluid is the weight of column above it, we can calculate the hydrostatic pressure by: The sumary of the theory is the force on any flat surface is the average pressure acting on the submerged surface multiplied by the area of the submerged surface. F = ? gXA Where: ? = water density g = acceleration due to gravity X = vertical distance from free surface to centroid of A
We know that the magnitude of the distributes force F, which may be considered as a small series of small forces spread over the submerged surface. The sum of the moments of all these small forces about any point must be equivalent to the moment about the same point of the resultant force Fr acting through the point of application, also known as the center of pressure. Taking the moments about O : Force on strip ? F=x? g ? A Moment of force an strip ? M=x2? g ? A But we know that : Sum of [x2 ? A] = 2nd moment of area (I? ) Therefore total moment = ? gI?
Therefore Frz = ? gI? and since Fr = F = ? g A X z= ? gI?? gAX= I? AX= 2nd moment of area about ‘oo’1st moment of area about ‘oo’ z= I? AX from parallel axis theorem I? =Igg+ AX2 Therefore, substituting z= Igg+ AX2AX z= IggAX+ X Xc=z+q For a partially submerged plate, the same equations apply except that the area of the plate varies. (A = br) Since Igg = br312 And substituting A = br and X = r2 in the equation for z: z= 23r It can be clearly seen that the centre of pressure is always two-third down the section of the submerged part of the plate. Xc= 23r+q Procedure: – The quadrant is placed on the two dowel pins and the clamping screw is fastened to the balance arm using the clamping screw. L, a, depth d, and width b, of the quadrant end face are measured. With the Perspex tank on the bench, the balance arm is balanced on the knife edges (pivot). The balance pan is hung from the end of the balance arm. A length of hose is connected from the drain cock to the sump and a length from the bench feed to the triangular aperture on the top of Perspex tank The tank is then levelled using the adjustable feet and spirit level.
The counter balance weight is moved until the balance arm is horizontal. The drain cock is closed and water is admitted until the level reaches the bottom edge of the quadrant. A weight is placed on the balance pan, and water is slowly added into the tank until the balance arm is horizontal. The water level on the quadrant and the weight on the balance pan is recorded. Fine adjustment of the water level can be achieved by overfilling and then slowly draining using the stop cock. The above are repeated for each increment of weight until the water level reached the top of the quadrant end face.
Then each increment of weight is removed, noting the weights and water levels until the weights have been removed.
RESULTS AND DISCUSSION
a= 0. 099m b= 0. 075 m d= 0. 100m l= 0. 274 m ?= 1000 kg/m
Weight of load, m(kg ms-2)
Filling tank height of water(m)
Draining tank height of water (m)
Average height of water,y (m)
Wetted surface area, yb (m )
Hydrostatic pressure, m/yb (Pa)
Graph of my2 against y Where the slope is -? b2L and the intercept should be ? b2L(a+d) Therefore ?b2L = -93. 9097 ?b2L(a+d) = 101. 7810
www. wikipedia. org on hydrostatic pressure
www. scribd. com on hydrostatic pressure lab report
Lab demonstrator’s explanation
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