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FLUIDS MECHANICS Prepared by: Nor hayati Bt Muhd Shamshudin
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FLUIDS MECHANICS

FLUIDS MECHANICSPrepared by: Nor hayati Bt Muhd ShamshudinINTRODUCTIONDENSITYDefinition of Density, :The density, of a material is its mass, m per volume, V = m/VSI unit: kg/m3Fluids with different properties have different density

PRESSUREDefinition of PressurePressure, P is force, F per area, A: P = F/APressure is often given in the terms of Pascal (Pa).1 Pa = 1 N/m21 atm = 1.013 x 105 Pa = 760 mm Hg1 mm Hg = 1 torr = 133.32 PaSI unit: N/m2

PRESSURE & DEPTH

The top surface of the fluid is open to the atmosfera, with pressure Pat. If the cross-sectional area of the container is A, the downward force exerted on the top surface by the atmosfera is Ftop = PatA At the bottom of the container, the downward force is Ftop plus the weight of the fluid. Recalling that m =VV = hAHence, W = mg = pVg = p(hA)gFbottom = Ftop + W= PatA + (hA)gPbottom = Fbottom/A= [PatA + (hA)g] / A= Pat + hg

EXAMPLE 1The Titanic was found in 1985 lying on the bottom of the North Atlantic at a depth of 2.5 miles. What is the pressure at this depth? ( = 1025 kg/m3)

EXAMPLE 2A cubic box 20.00 cm on a side is completely immersed in a fluid. At the top of the box the pressure is 105.0 kPa; at the bottom the pressure is 106.8 kPa. What is the density of the fluid?

BAROMETERBarometer is an equipment which is used to measure atmospheric pressure.A fluid that is often used in barometer is mercury. Mercury has a density, = 1.3595 x 104 kg/m3Pat = hg = (1.3595 x 104 kg/m3)(0.760 m)(9.81 m/s2) = 1.013 x 105 Pa

EXAMPLE 3A U-shaped tube is filled mostly with water, but a small amount of vegetable oil has been added to one side. The density of the water is 1000 kg/m3 and the density of the vegetable oil is 920 kg/m3. If the depth of the oil is 5.00 cm, what is the difference in level, h between the top of the oil on one side of the U and the top of the water on the other side?

PASCALS PRINCIPLEAn external pressure applied to an enclosed fluid is transmitted unchanged to every point within the fluid.

A hydraulic lift works according to Pascals principleA force F1 exerted on the small piston causes a much larger force, F2, to act on the large piston.When force F1 exerted on the small piston, this increases the pressure in that cylinder by the amount of

By Pascals Principle, the pressure in cylinder 2 increases by the same amount. Therefore, F2 = (P)A2

EXAMPLE 4To inspect a 14,500 N car, it is raised with a hydraulic lift. If the radius of the small piston is 4.0 cm, and the radius of the large piston is 17 cm, find the force that must be exerted on the small piston to lift the car.

BUOYANT FORCEFluids surrounding an object exert a force in the upward direction. The force is known as buoyant force.Consider a cubical block immersed in a fluid of density, . The surrounding fluid exerts normal forces on all of its faces. The horizontal forces pushing to the right and to the left are equal, hence they cancel and have no effect on the block.

F2 > F1The downward force exerted on the top face is less than the upward force exerted on the lower face. This is because the pressure at the lower face is greater. Thus, the difference in forces gives rise to a net upward force that is known as Buoyant Force.

To calculate the value of buoyant force:Firstly, assume that the cubical block is of length L on the side and the pressure on the top surface is P1.Force that acts on top surface, F1F1 = P1A = P1L2Force that acts on bottom surface, F2F2 = P2A (P2 = P1+gL)F2 = (P1+gL)L2 = P1L2 +gL3 = F1 + gL3Buoyant force, Fb = F2 F1 = gL3Note that gL3 is the weight of fluid that would occupy the same volume as the cube. Therefore, the buoyant force is equal to the weight of fluid that is displaced by the cube.This phenomenon is known as Archimedes Principle.

EXAMPLE 5A piece of wood with a density of 706 kg/m3 is tied with a string to the bottom of a water filled container. The wood is completely immersed, and has a volume of 8.00 x 10-6 m3. What is the tension in the string?

SURFACE TENSIONA fluid tends to pull inward on its surface, resulting in a surface of minimum area.

The surface of the fluid behaves much like an elastic membrane enclosing the fluid. This kind of surface enable the insects to stand on the surface of the water.

FLUID FLOWObjectives :a) identify the simplifications used in describing ideal fluid flow.b) use the continuity equation and Bernoullis equation to explain common effects of ideal fluid flowThis photograph was taken in a water tunnel using hydrogen bubbles to visualize the flow pattern around a cylinder.The flow was started from rest, and at this instant the pattern shows the development of a complex wake structure on the downstream side of the cylinder. Four characteristics of an ideal fluid :

Condition 1 : Steady flow means that all the particles of a fluid have the same velocity as they pass a given point.

Condition 2 :Irrotational flow means that a fluid element (a small volume of the fluid) has no net angular velocity, which eliminates the possibility of whirlpools and eddy currents. (The is non turbulent.)

Condition 3 : Nonviscous flow means that viscosity is negligible.

Condition 4 : Incompressible flow means that the fluids density is constant.

EQUATION OF CONTINUITYFig. a) mass enters tube,m1 = 1V1= 1(A1x1)= 1(A1v1t)

Fig. b) mass exits tube, m2 = 2V2= 2(A2x2)= 2(A2v2t) since the mass is conserved,m1= m2 1A1v1= 2A2v2or Av= constant equation of continuity

for an incompressible fluid, the density is constant, so

A1v1= A2v2

or Av= constant flow rate equation

By the flow rate equation, the speed of a fluid is greater when the cross-sectional area of the tube through which the fluid is flowing is smaller.Think of a hose that is equipped with a nozzle such that the cross-sectional area of the hose is made smaller.

BERNOULLIS EQUATIONa statement of the conservation of energy for a fluid.

Work-Energy Theorem --- has great generality for fluid flow.

F1 does positive work : in the same direction as the fluids motion

F2 does negative work : opposite to the fluid motion

BERNOULLIS EQUATION CAN BE APPLIED TO MANY SITUATIONS.Venturi meterIn a region of smaller cross-sectional area, the flow speed is greater; the pressure in that region is lower than in other regions.

The pressure P1 is greater than the pressure P2, because v1 < v2.This device can be used to measure the speed of fluid flow.

Airplane liftBecause of the shape and orientation of an airfoil or airplane wing, the air streamlines are closer together, and the air speed is greater above the wing than below it. The resulting pressure difference supplies an upward force, or lift. Streamline flow around an airplane wing.The pressure above is less than the pressure below, and there is a dynamic lift force upward.

3. Fluid flow from a tankIf we assume that the cross-sectional area of the tank is large relative to that of the hole (A2 A1), then the water level drops very slowly and we can assume v2 0. Let us apply Bernoullis equation to points 1 and 2.If we note that P1 = P2 at the hole, we get v1 =

VISCOSITYobjectives : to discuss fluid viscosity. Viscosity is a fluids internal resistance to flow. (All real fluids have a nonzero viscosity.)

Fig. (a) :Streamlines never cross and are closer together in regions of greater fluid velocity. The stationary paddle wheel indicates that the flow is irrotational, or without whirlpools and eddy currentsFig a) internal friction causes the layers of the fluid to move relative to each other in response to a shear stress; called laminar flow.

Fig b) at higher velocities, the flow becomes rotational (the speed of the fluid is less near the walls of the pipe than near the centre because of frictional drag between the walls and the fluid); called turbulent.

The viscosities of some fluids are listed in Table below.

The greater the viscosity of the liquid, which is easier to visualize than that of the gas,the greater is the shear stress required to get the layers of the liquid to slide along each other.

Example : the large viscosity of glycerin compared to that of water.

POISEUILLES LAWWhen a fluid flows through a pipe, there is frictional drag between the liquid and the walls, and the fluid velocity is greater toward the center of the pipe.The flow rate depends on the properties of the fluid and the dimensions of the pipe, as well as on the pressure difference (P) between the ends of the pipe.

refer Fig. b) : this effect makes a difference in a fluids average flow ratewhere,P is pressure difference,r is radius,L is length and is viscosity.TERMINAL VELOCITY

When things move through a fluid, drag increases with speed.

This is because they encounter more fluid per second and hit it harder.

For low velocities, drag is proportional to speed, and for higher velocities it increases with velocity squared. In either case the motion will differ from free fall. The resultant downward force will now be the difference between weight W (= mg) and drag D (which increases with velocity) so the equation of motion is: resultant F = W D= mg D= maso the acceleration is

Or

When v = 0, D = 0 so a = g. (This is the initial acceleration.)As v increases, D increases, so a falls.Eventually a velocity is reached when D = mg, and then a = 0. The object continues to fall at a constant terminal velocity

STOKES LAW In1845 a scientist named George Stokes found that the magnitude of the resistive force, Fr on a very small spherical object of radius r falling slowly through a fluid of viscosity with terminal velocity, vT is given by

Fr = 6rvT

This equation, called Stokes Law, has many important applications.

For example, it describes the sedimentation of particulate matter in blood samples.