Chapter 2 - Part 1 Pressure

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    CHAPTER 2

    Part 1:PressurePart 1:Pressure

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    Fluid Statics Fluid Statics deals with problems

    associated with fluids at rest.

    In fluid statics, there is no relativemotion between adjacent fluid layers.

    Therefore, there is no shear stress inthe fluid tr in to deform it.

    The only stress in fluid statics isnormal stress

    Normal stress is due to pressure

    Variation of pressure is due only to the

    weight of the fluid fluid statics is onlyrelevant in presence of gravity fields.

    Applications: Floating or submergedbodies, water dams and gates, liquidstorage tanks, etc. 2

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    Pressure Pressure is defined as anormal

    force exerted by a fluid per unitarea.

    Unit of pressure is N/m2, which iscalled pascal (Pa).

    P = F/A

    Since the unit Pa is too small forpressures encountered in practice,

    kilopascal(1 kPa = 103 Pa) and

    megapascal(1 MPa = 10

    6

    Pa) arecommonly used.

    Other units includebar,atm,

    kgf/cm2

    , lbf/in2

    =psi.3

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    Pressure at a PointPressure ?

    Indicating the normal force per unitarea at a given point acting on agiven plane within the fluid mass ofinterest.

    ow t e pressure at a po nt var eswith the orientation of the planepassing through the point ?

    In fluid at rest, pressure at any pointis the same at all direction.

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    The pressure at a point in a fluid atrest, or in motion, is independent ofthe direction as long as there are noshearing stresses present.

    Pressure at any point in a fluid is thesame in all directions.

    Pressure has a magnitude, but not aspecific direction, and thus it is ascalar quantity.

    The result is known asPascals lawnamed in honor ofBlaise Pascal(1623-1662).

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    Absolute, gauge, andvacuum pressures

    Actual pressure at a give point

    is called the absolute pressure.

    Most pressure-measuring

    devices are calibrated to read

    ,

    therefore indicate gauge

    pressure,

    Pgauge= Pabs - Patm.

    Pressure below atmospheric

    pressure are called vacuum

    pressure, Pvac= Patm - Pabs. 6

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    Absolute, gauge, and

    vacuum pressures

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    Atmospheric Pressure, Patm

    * Pressure due to weight of airabove it.

    * Standard value (1 atm =)

    10.35 mH2O water (34 ftH2O)

    760 mmHg

    14.7 si

    101.3 kN/m2

    can be measured by P = gh

    * Fluid pressure at free surfaceis equal to atmosphericpressure.

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    Gauge Pressure, Pg

    * Measured using pressure gauge.

    * Can be positive (above atm pressure) ornegative (below atm pressure)

    * Gauge negative pressure is referred tosuction pressure ofvacuum pressure.

    ero pressure means e pressure sequal to atmospheric pressure.

    * Gauge pressure units:

    N/m2

    gaugepsig

    kPa gauge

    barg

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    Pressure Head of Fluid

    * Pressure head of fluid is a pressure and it isinterpreted as the height of a column offluid of specific weight required.

    * A basic equation is a relationship amongpressure, density and depth. P = gh

    * Consider an element of fluid as shownbelow.

    dA : cross sectional area of element : fluid density

    P : pressure

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    In equilibrium;

    Upward Force = Downward Force

    (P + dP) dA = PdA + mg

    dPdA = mg

    But;

    m = v (v = element volume = dAdh)

    dPdA = vg = gdAdh

    dP = gdh

    Integrated it;

    P2 - P1 = g (h2 - h1)

    =2

    1

    2

    1

    h

    h

    P

    PdhgdP

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    But;

    h2 - h1 = h

    Then

    P2 - P1 = g h

    If h1 = 0 and P1 = 0 (atmospheric pressure), then

    P2 = gh

    Based on the above equation, pressure isproportional to depth ( P h ) regardless of shapeof container.

    O In pressure head, the above equation become

    and its unit is in e.g.; mmHg or mmH20

    21 pph

    =

    12

    Specific weight

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    Variation of Pressurewith Depth

    Pressure in a fluid at rest is

    independent of the shape of the

    container.

    Pressure is the same at all points on

    a horizontal plane in a given fluid.

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    Absolute pressure (atm)

    on diver at 100 ft?

    ( ),2 3 21

    998 9.81 1003.28

    1

    gage

    kg m mP gz ft

    m s ft

    atm

    = =

    Scuba Diving and HydrostaticPressure

    100 ft

    1

    Danger of emergency

    ascent?

    ,2 ,2

    . .

    101.3252.95 1 3.95

    abs gage atm

    kPaP P P atm atm atm

    = + = + =

    2

    1 1 2 2

    1 2

    2 1

    3 . 9 54

    1

    P V P V

    V P a tm

    V P a tm

    =

    = =

    Boyles lawIf you hold your breath on ascent, your lungvolume would increase by a factor of 4,would result in embolism and/or death.

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    Example;

    Change a pressure of 350 kN/m2 gauge into pressure

    head of water and mercury.

    Water

    OHmsmxmkg

    mNx

    g

    Ph

    water

    223

    23

    68.35/81.9/1000

    /10350===

    ercury

    Hgmsmxmkg

    mNx

    g

    Ph

    mercury

    62.2/81.9/13600

    /1035023

    23

    ===

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    Transmission of Fluid Pressure

    * If a fluid remaining constant in the horizontal direction,the pressure applied to a confined fluid increases thepressure throughout by the same amount. This iscalled Pascals Law.

    * The transmission of fluid pressure throughout astationary fluid is the principle upon which manyhydraulic devices are based.

    P1 = P2

    Note : The pressure force exerted by the fluid is

    always normal to the surface at the

    specified points

    11

    222211 F

    A

    AFpAFpAF ===

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    Pascals Law

    Pressure applied to aconfined fluid

    increases the pressure

    throughout by the

    same amount.

    In picture, pistons areat same height:

    Ratio A2/A1 is called

    ideal mechanical

    advantage

    1 2 2 21 2

    1 2 1 1

    F F F AP P

    A A F A= = =

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    Example;Example;

    Dimension of hydraulic jack is shown in figure below.

    If a force of 100 N applied onto the jet handle,determine a maximum force F2 would be support.

    Free Body Diagram of handle

    100 N x 33 cm - F1 x 3 cm = 0

    F1 = (100 N x 0.33 m/0.03 m) = 1100 N

    100 N

    F1

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    Pressure at small piston (left site);

    Based on transmission principle;

    26

    2

    1

    1

    1m/N10x22.6

    4

    d

    N1100

    A

    FP =

    ==

    Therefore;

    kN22.124

    dx10x22.6APF

    2

    6

    222===

    19

    P1 = P2 = 6.22 x 106

    N/m2

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    ExampleExample;;

    Oil with a specific gravity of 0.9 is used

    in a hydraulic apparatus as shown in

    figure below. If a gauge indicate that a

    pressure of 2.15 bar, determine thevalue of W so that the system is in

    equilibrium condition.

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    At x x

    2

    2325

    '

    /232658

    )2)(/81.9)(/1000(9.0/1015.2

    mN

    msmmkgmNx

    ghPP oilgaugex

    =

    +=

    +=

    AreaxessureForce Pr=

    Based on transmission principle;

    Px = Px' = 232658 N/m2

    N66.730916

    )2(4

    xm/N232658 22

    =

    =

    kg30.74507

    s/m81.9

    N66.730916W

    2==

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