P1 2MEC-2MED Fluida Statis

7/25/2019 P1 2MEC-2MED Fluida Statis

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Dr., I Gede Rasagama, S.Si., M.Pd.

POLITEKNIK MANUFAKTUR NEGERI BANDUNG

Jl. Kanayakan No. 21, Dago

Bandung 40135

Introduction

Pressure force on a fluid element

Gage pressure and vacuum pressure

Hydrostatic pressure distribution

The mercury barometer

Hydrostatic pressure in gases

Manometry

Hydrostatic forces on plane surface

Hydrostatic forces on curved surfaces

THE MIND MAP:

POLITEKNIK MANUFAKTURNEGERI BANDUNG

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Called the hydrostatic condition:

the fluid velocity is zero

the pressure variation is due only to the weight ofthe fluid.

Consider a small wedge of fluid at rest of size x, z, sand depth b into the paper.

There is no shear stress by definition, and pressure isassumed to be identical on each face (small element).

a. Introduction


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Since the element is at rest, summation of all forcesmust equal zero.


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This means:1) There is no pressure change in the horizontaldirection.2) There is a vertical change in pressure proportional tothe density, gravity and depth change in the fluid (i.e.the weight of the column of the fluid above the point).

Note: in the limit as the fluid wedge shrinks to a point,z goes to zero, we have:

Thus, pressure in a static fluid is a point property.POLITEKNIK MANUFAKTUR

NEGERI BANDUNG

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Assume the pressure vary arbitrarily in a fluid,p=p(x,y,z,t).

Consider a fluid element of size x, y, zas shown in

Fig. 2. The net force in the xdirection is given by:


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b. Pressure force on a fluid element


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Fig. 2: Net force in the xdirection due to pressure variation.In a similar manner, net forces acting in yand zdirections can becalculated. The total net force vector,due to pressure, is:


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Notice that the term in the parentheses is the negative vector gradientof pressure and the term dx dy dz =dV, is the volume of the element.Therefore, one can write:

where fpress is the net force per volume. Notice that the pressuregradient (not pressure) causing a net force that must be balanced by

gravity or acceleration and/or other effects in the fluid.Note: the pressure gradient is a surface force that acts on the sides of theelement. That must be balanced by gravity force, or weight of theelement, in the fluid at rest.In addition to gravity, a fluid in motion will have surface forces due toviscous stresses. Viscous forces, however, for a fluid at rest are zero.The gravity force is a body force, acting on the entire mass of theelement:


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The actual pressure at a given position is called the absolute

pressure, and it is measured relative to absolute vacuum. The measure pressure may be either lower (called vacuum

pressure) or higher (gage pressure) than the local atmosphere.


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c. Gage pressure and vacuum pressure

For a fluid at rest, the summation of forces acting on the element mustbe balanced by the gravity force.

This is a hydrostatic distribution and is correct for all fluids at rest,regardless of viscosity.Recall that the vector operator expresses themagnitude and direction of the maximum spatial rate of increase of

the scalar property (in this case pressure).Note: is perpendicular everywhere to surface of constant pressurep. In other words, in a fluid at rest will align its constantpressuresurfaces every where normal to the localgravity vector. Or, thepressure increase will be in the direction of gravity (downward).However, in our customary coordinate z is upwardand the gravityvector is:

whereg=9.807 m/s2.POLITEKNIK MANUFAKTUR

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d. Hydrostatic pressure distribution


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For this coordinate, the pressure gradient vectorbecomes:

Since pressure is only a function of z (independent of xand y), we can write:

As a result, we can conclude: pressure in a continuouslydistributed uniform static fluid varies only withvertical distance and is independent of the shape of thecontainer. The pressure is the same at all points on agiven horizontal plane in a fluid.


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Fig. 4: Hydrostatic pressure is only a function of the depth of the fluid, = = . However, = = Because point D, although at thesame level, lies beneath a different fluid,mercury. The free surface of the

container is atmospheric and forms a horizontal line.Note: In most engineering applications, the variation in acceleration of gravity(g) due to different heights is less than 0.6% and can be neglected.For liquids, which are incompressible, we have:

The quantity, p is a length called thepressure head of the fluid.


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e. The mercury barometer

Mercury has an

extremely smallvapor pressure atroom temperature(almost vacuum),thusp1 = 0. Onecan write:

or

At the sealevel, theatmospheric pressurereads, 761 mmHg

Gases are compressible with density nearlyproportional to pressure, thus the variation in densitymust be considered in hydrostatic calculations. Usingthe ideal gas equation of state, = :

After integration between points 1 and 2 and also

assuming a constant temperature at both pointsT1=T2=T0(isothermal atmosphere), we find:

Note that the atmospheric pressure is nearly zero(vacuum condition) at z = 30 km.


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f. Hydrostatic pressure in gases


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It is shown that a change in elevation of a liquid is equivalent to achange in pressure, = /. Thus a static column of one or

multiple fluids can be used to measure pressure differencebetween 2 points. Such a device is called manometer.


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f. Manometry

Two roles for manometer analysis:

1) Adding/ subtractingas moving down/up in afluid column.

2) Jumping across Utubes: any two points at the sameelevation in a continuous mass of the same staticfluid will be at the same pressure, thus we can jumpacross Utubes filled with the same fluid.


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Consider a plane panel of arbitrary shape completely submerged in a liquid.

Fig. 6: hydrostatic force and center of pressure on a plane submerged in aliquid at an angle .


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g. Hydrostatic forces on plane surface

If h is the depth to any element area dA, the local pressure is:

The total hydrostatic force on one side of the plane is givenby:

We also have: = sin . After substitution, we get:

Since, = sin,

It means, the force on one side of any plane submerged

surface in a uniform fluid equals the pressure at the platecentroid times the plate area, independent of the shape of theplate or angle .To balance the bendingmoment portion of the stress, theresultant force F acts not through the centroid but below ittoward the high pressure side. Its line of action passesthrough the centre of pressure CP of the plate (xCP, yCP).


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To find the center of pressure, we sum moments of the elementalforcepdA about the centroid and equate to the moment of theresultant force, F:

The term = 0by definition of centroidal axes. Using thedefinition of the area moment of inertia about centroidal x axis, =

2, after some simplifications:

The negative sign shows that yCPis below the centroid at a deeperlevel and depends on angle and the shape of the plate (Ixx).

Following the same procedure, we find:

Note: for symmetrical plates, Ixy = 0 and thus xCP= 0. As a result,the center of pressure lies directly below the centroid on the yaxis.


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Fig. 7: Centroidal moments of inertia for variouscrosssections.


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The easiest way to calculate the pressure forces on a curved surface is to compute thehorizontal and vertical forces separately.

Fig. 8: Calculating horizontal and vertical pressure forces on an immersed curvedsurface.

Using the freebody diagram shown in Fig. 8b, one can find:

The horizontal force, FH equals the force on the plane area formed by the projectionof the curved surface onto a vertical plane normal to the component.

The vertical component equals to the weight of the entire column of fluid, bothliquid and atmospheric above the curved surface.

For the surface shown in Fig. 8:


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h. Hydrostatic forces on curved surfaces

Anda sedang mengukur tekanan pada kedalaman 10cm dalam tiga wadah yang berbeda. Urutkan nilaitekanan dari yang terbesar ke yang terkecil:

a. 1-2-3

b. 2-1-3

c. 3-2-1

d. sama pada ketiganya


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TES KONSEP 1


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1. Andaikan anda punya sepotong baja. Akankah baja

ini mengapung di atas air? Mengapa?

2. Bagaimana sebuah kapal yang terbuat dari bajadapat mengapung?


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Pertanyaan:


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P1 2MEC-2MED Fluida Statis