Fluid Mechanics Pp

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  • 1. Hydraulics
    By: Engr. Yuri G. Melliza
  • 2. FLUID MECHANICS
    Properties of Fluids
    1. Density ()
    2.Specific Volume ()
    3.Specific Weight ()
  • 3. 4. Specific Gravity or Relative Density
    For Liquids: Its specific gravity (relative density) is equal to the ratio of its density to that of water at standard temperature and pressure.
    For Gases:Its specific gravity (relative density) is equal to the ratio of its density to that of either air or hydrogen at some specified temperature and pressure.
    where: At standard condition
    W = 1000 kg/m3
    W = 9.81 KN/m3
  • 4. 5. Temperature
    6. Pressure
    If a force dF acts on an infinitesimal area dA, the intensity of pressure is,
    where: F - normal force, KN
    A - area, m2
  • 5. PASCAL’S LAW: At any point in a homogeneous fluid at rest the pressures are
    the same in all directions.
    y
    P3 A3
    A
    P1 A1
    x

    B
    C
    z
    P2 A2
    Eq. 3 to Eq. 1
    P1 = P3
    Eq. 4 to Eq. 2
    P2 = P3
    Therefore:
    P1 = P2 = P3
    Fx = 0 and Fy = 0
    P1A1 – P3A3 sin  = 0  1
    P2A2 – P3A3cos = 0  2
    From Figure:
    A1 = A3sin   3
    A2 = A3cos   4
  • 6. Atmospheric pressure:The pressure exerted by the atmosphere.
    At sea level condition:
    Pa = 101.325 KPa
    = .101325 Mpa
    = 1.01325Bar
    = 760 mm Hg
    = 10.33 m H2O
    = 1.133 kg/cm2
    = 14.7 psi
    = 29.921 in Hg
    = 33.878 ft H2O
    Absolute and Gage Pressure
    Absolute Pressure:is the pressure measured referred to absolute zero and using absolute zero as the base.
    Gage Pressure:is the pressure measured referred to atmospheric pressure, and using atmospheric pressure as the base
  • 7. Pgage
    Atmospheric pressure
    Pvacuum
    Pabs
    Pabs
    Absolute Zero
    Pabs = Pa+ Pgage
    Pabs = Pa - Pvacuum
  • 8. 7. Viscosity: A property that determines the amount of its resistance to
    shearing stress.
    moving plate
    v
    v+dv
    dx
    x
    v
    Fixed plate
    S dv/dx
    S = (dv/dx)
    S = (v/x)
     = S/(v/x)
    where:
     - absolute or dynamic viscosity
    in Pa-sec
    S - shearing stress in Pascal
    v - velocity in m/sec
    x -distance in meters
  • 9. 8. Kinematic Viscosity: It is the ratio of the absolute or dynamic viscosity to
    mass density.
     = / m2/sec
    9. Elasticity: If the pressure is applied to a fluid, it contracts,ifthe pressure is
    released it expands, the elasticity of a fluid is related to the amount of deformat-
    ion (contraction or expansion) for a given pressure change. Quantitatively, the
    degree of elasticity is equal to;
    Ev = - dP/(dV/V)
    Where negative sign is used because dV/V is negative for a positive dP.
    Ev = dP/(d/)
    because -dV/V = d/
    where:
    Ev - bulk modulus of elasticity, KPa
    dV - is the incremental volume change
    V - is the original volume
    dP - is the incremental pressure change
  • 10.



    r
    h
    10. Surface Tension: Capillarity
    Where:
     - surface tension, N/m
     - specific weight of liquid, N/m3
    r – radius, m
    h – capillary rise, m
    Surface Tension of Water
  • 11. FREE SURFACE
    h1
    1•
    h2
    h
    2•
    Variation of Pressure with Elevation
    dP = - dh
    Note:Negative sign is used because pressure decreases as elevation increases and pressure
    increases as elevation decreases.
  • 12. Pressure Head:
    where:
    p - pressure in KPa
     - specific weight of a fluid, KN/m3
    h - pressure head in meters of fluid
    MANOMETERS
    Manometer is an instrument used in measuring gage pressure in length of some liquid
    column.
    • Open Type Manometer : It has an atmospheric surface and is capable in measuring
    gage pressure.
    • Differential Type Manometer : It has no atmospheric surface and is capable in
    measuring differences of pressure.
  • 13. Fluid A
    Fluid A
    Open
    Fluid B
    Manometer Fluid
    Manometer Fluid
    Open Type Manometer
    Differential Type Manometer
  • 14. Open
    Open
    Fluid A
    x
    y
    Fluid B
    Determination of S using a U - Tube
    SAx = SBy
  • 15. S
    S
    S
    Free Surface

    M
    M
    hp
    F
    •C.G.
    •C.G.
    •C.P.
    •C.P.
    yp
    e
    N
    N
    Forces Acting on Plane Surfaces
    F - total hydrostatic force exerted by the fluid on any plane surface MN
    C.G. - center of gravity
    C.P. - center of pressure
  • 16. where:
    Ig- moment of inertia of any plane surface MN with respect to the axis at its centroids
    Ss - statical moment of inertia of any plane surface MN with respect to the axis SS not
    lying on its plane
    e - perpendicular distance between CG and CP
  • 17. Forces Acting on Curved Surfaces
    FV
    Free Surface
    D
    E
    Vertical Projection of AB
    F
    C’
    L
    C
    C
    A
    C.G.
    Fh
    C.P.
    B
    B
    B’
  • 18. A = BC x L
    A - area of the vertical projection of AB, m2
    L - length of AB perpendicular to the screen, m
    V = AABCDEA x L, m3
  • 19. D
    h
    1 m
    D
    P = h
    T
    T
    F
    F
    1 m
    T
    t
    T
    Hoop Tension
    F = 0
    2T = F
    T = F/2  1
    S = T/A
    A = 1t  2
  • 20. S = F/2(1t)  3
    From figure, on the vertical projection the pressure P;
    P = F/A
    A = 1D
    F = P(1D)  4
    substituting eq, 4 to eq. 3
    S = P(1D)/2(1t)
    where:
    S - Bursting Stress KPa
    P - pressure, KPa
    D -inside diameter, m
    t - thickness, m
  • 21. Laws of BuoyancyAny body partly or wholly submerged in a liquid is subjected to a buoyant or upward force which is equal to the weight of the liquid displaced.
    1.
    W
    where:
    W - weight of body, kg, KN
    BF - buoyant force, kg, KN
     - specific weight, KN/m3
     - density, kg/m3
    V - volume, m3
    Subscript:
    B - refers to the body
    L - refers to the liquid
    s - submerged portion
    Vs
    BF
    W = BF
    W = BVB
    BF = LVs
    W = BF
    W = BVB KN
    BF = LVs KN
  • 22. 2.
    BF
    T
    Vs
    W
    where:
    W - weight of body, kg, KN
    BF - buoyant force, kg, KN
    T - external force T, kg, KN
     - specific weight, KN/m3
     - density, kg/m3
    V - volume, m3
    Subscript:
    B - refers to the body
    L - refers to the liquid
    s - submerged portion
    W = BF - T
    W = BVB KN
    BF = LVs KN
    W = BF - T
    W = BVB
    BF = LVs
  • 23. T
    W
    BF
    Vs
    3.
    where:
    W - weight of body, kg, KN
    BF - buoyant force, kg, KN
    T - external force T, kg, KN
     - specific weight, KN/m3
     - density, kg/m3
    V - volume, m3
    Subscript:
    B - refers to the body
    L - refers to the liquid
    s - submerged portion
    W = BF + T
    W = BVB g
    BF = LVs g
    W = BF + T
    W = BVB g
    BF = LVs g
  • 24. W
    T
    Vs
    BF
    4.
    VB = Vs
    W = BF + T
    W = BVB g
    BF = LVs g
    W = BF + T
    W = BVB
    BF = LVs
  • 25. W
    Vs
    BF
    T
    5.
    VB = Vs
    W = BF - T
    W = BVB g
    BF = LVs g
    W = BF - T
    W = BVB
    BF = LVs
  • 26. Energy and Head
    Bernoullis Energy equation:
    2
    HL = U - Q
    Z2
    1
    z1
    Reference Datum (Datum Line)
  • 27. 1. Without Energy head added or given up by the fluid (No work done by
    the system or on the system:
    2. With Energy head added to the Fluid: (Work done on the system)
    3. With Energy head added given up by the Fluid: (Work done by the system)
    Where:
    P – pressure, KPa - specific weight, KN/m3
    v – velocity in m/sec g – gravitational acceleration
    Z – elevation, meters m/sec2
    + if above datum H – head loss, meters
    - if below datum
  • 28. APPLICATION OF THE BERNOULLI'S ENERGY THEOREM
    Nozzle
    Base
    Tip
    Q
    Jet
    where: Cv - velocity coefficient
  • 29. Venturi Meter
    B. Considering Head loss
    inlet
    1
    throat
    exit
    2
    Meter Coefficient
    Manometer
    A. Without considering Head loss
  • 30. Orifice: An orifice is an any opening with a closed perimeter
    Without considering Head Loss
    and from figure: Z1 - Z2 = h, therefore
    1
    a
    h
    Vena Contracta
    By applying Bernoulli's Energy theorem:
    Let v2 = vt
    2
    a
    where:
    vt - theoretical velocity, m/sec
    h - head producing the flow, meters
    g - gravitational acceleration, m/sec2
    But P1 = P2 = Pa and v1is negligible, then
  • 31. COEFFICIENT OF DISCHARGE(Cd)
    COEFFICIENT OF VELOCITY (Cv)
    COEFFICIENT OF CONTRACTION (Cc)
    where:
    v' - actual velocity
    vt - theoretical velocity
    a - area of jet at vena contracta
    A - area of orifice
    Q' - actual flow
    Q - theoretical flow
    Cv - coefficient of velocity
    Cc - coefficient of contraction
    Cd - coefficient of discharge
  • 32. Jet Trajctory
    2
    d
    v sin
    v

    1
    3
    v cos
    R = v cos (2t)
    If the jet is flowing from a vertical orifice and the jet is initially horizontal where
    vx = v.
    v = vx
    y
    x
  • 33. Upper
    Reservoir
    Suction Gauge
    Discharge Gauge
    Lower
    Reservoir
    Gate Valve
    Gate
    Valve
    PUMPS: It is a steady-state, steady-flow machine in which mechanical work is added to the fluid in order
    to transport the liquid from one point to another point of higher pressure.
  • 34. 1. TOTAL DYNAMIC HEAD
    4. BRAKE or SHAFT POWER
    FUNDAMENTAL EQUATIONS
    2. DISCHARGE or CAPACITY
    Q = Asvs = Advdm3/sec
    3. WATER POWER or FLUID POWER
    WP = QHtKW
  • 35. 5. PUMP EFFICIENCY
    6. MOTOR EFFICIENCY
    7. COMBINED PUMP-MOTOR EFFICIENCY
  • 36. 8. MOTOR POWER
    • For Single Phase Motor
    • 37. For 3 Phase Motor
    where: P - pressure in KPa T - brake torque, N-m
    v - velocity, m/sec N - no. of RPM
     - specific weight of liquid, KN/m3 WP - fluid power, KW
    Z - elevation, meters BP - brake power, KW
    g - gravitational acceleration, m/sec2 MP - power input to
    HL - total head loss, meters motor, KW
    E - energy, Volts
    I - current, amperes
    (cos) - power factor
  • 38. Penstock
    Headrace
    turbine
    Y – Gross Head
    Tailrace
    HYDRO ELECTRIC POWER PLANT
    A. Impulse Type turbine (Pelton Type)
    1
    2
  • 39. 1
    Headrace
    Generator
    Penstock
    Y – Gross Head
    B
    ZB
    Draft Tube
    2
    B – turbine inlet
    Tailrace
    B. Reaction Type turbine (Francis Type)
  • 40. Fundamental Equations
    Where:
    PB – Pressure at turbine inlet, KPa
    vB – velocity at inlet, m/sec
    ZB – turbine setting, m
     - specific weight of water, KN/m3
    1. Net Effective Head
    Impulse Type
    h = Y – HL
    Y = Z1 – Z2
    Y – Gross Head, meters
    Where:
    Z1 – head water elevation, m
    Z2 – tail water elevation, m
    B. Reaction Type
    h = Y – HL
    Y = Z1 –Z2
  • 41. 2. Water Power (Fluid Power)
    FP = Qh KW
    Where:
    Q – discharge, m3/sec
    3. Brake or Shaft Power
    Where:
    T – Brake torque, N-m
    N – number of RPM
    Where:
    eh – hydraulic efficiency
    ev – volumetric efficiency
    em – mechanical efficiency
    4. Turbine Efficiency
  • 42. 5. Generator Efficency
    6. Generator Speed
    Where:
    N – speed, RPM
    f – frequency in cps or Hertz
    n – no. of generator poles (usually divisible by four)
  • 43. Turbine-Pump
    Pump-Storage Hydroelectric power plant: During power generation the turbine-pump acts as a turbine and
    during off-peak period it acts as a pump, pumping water from the lower pool (tailrace) back to the upper
    pool (headrace).
  • 44. A 300 mm pipe is connected by a reducer to a 100 mm pipe. Points 1 and 2 are at the same elevation. The pressure at point 1 is 200 KPa. Q = 30 L/sec flowing from 1 to 2, and the energy lost between 1 and 2 is equivalent to 20 KPa. Compute the pressure at 2 if the liquid is oil with S = 0.80. (174.2 KPa)
    300 mm
    100 mm
    1
    2
  • 45. A venturi meter having a diameter of 150 mm at the throat is installed in a 300 mm water main. In a differential gage partly filled with mercury (the remainder of the tube being filled with water) and connected with the meter at the inlet and at the throat, what would be the difference in level of the mercury columns if the discharge is 150 L/sec? Neglect loss of head. (h=273 mm)
  • 46. The liquid in the figure has a specific gravity of 1.5. The gas pressure PA is 35 KPa and PB is -15 KPa. The orifice is 100 mm in diameter with Cd = Cv = 0.95. Determine the velocity in the jet and the discharge when h = 1.2. (9.025 m/sec; 0.071 m3/sec)
    PA
    1.2 m
    PB