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Forms of energy

•   Any fluid in motion possesses 4 kinds of energy
     Form of Energy              Example
o    Potential or head     - water at top of
2.   Kinetic or velocity   - water doing work
                             on water wheel
3.   Static or Pressure    - Pressure exerted
                             inside line
4.   Heat                  - Heat generated in
                             line due to friction
                Examples of conversion of energy

Pumps        - mechanical to velocity;
               mechanical to pressure
Steam siphon  - Pressure to velocity
Water hammer - Velocity to pressure
Filling tank  - Pressure to velocity to potential
                              Total Energy

•   The total energy that a fluid in motion has at any time is equal to the
    sum of the form listed above. Because energy can neither be
    created or destroyed, it is therefore possible to convert one form of
    energy to another, such as velocity into pressure.
                              Fluid Flow

Bernoulli’s Principle
• when a fluid is in motion, its pressure increases whenever its
  velocity decreases.

•   Likewise its pressure decreases whenever its velocity increases

•   A fluid is a substance which undergoes continuous deformation
    when subjected to a shear stress.
               Factors determining Fluid Flow

•   Property of fluid to resists any force that tends to
    produce flow. This resistance to flow [or movement] is
    known as viscosity

•   It affects the performance of pumps and the pressure
    drop when pumping through a given pipe line.
              Factors determining Fluid Flow

Pipe size
•   For a given through put, the pressure drop increases
    rapidly with decreasing pipe diameter

•  The degree of roughness of the internal wall of the pipe
   affects fluid flow. Old rusty or pitted pipe would offer
   more resistance to flow (higher pressure drop)

Pipe restrictions
•   The number of fittings, valves, etc affects friction loss.
                          Type of Flow

Streamline flow – fluid flows in smooth layers.
                  Flow is in straight unbroken lines
                  Re < 2000

Turbulent flow – there is an irregular random
                 motion of fluid particles
                 Re > 4000

Transition – between above.
                    Reynolds Number

Nature of flow in pipe depends on pipe diameter, the density and
viscosity of the flowing fluid, and the velocity of flow.
     Re = dvp

                   Where; Re - Reynolds number
                            d - internal diameter of pipe
                            v - kinematic viscosity
                            p - weight density of fluid
                            u - dynamic viscosity
The term kinematics refers to the quantitative description of fluid
motion or deformation.
Kinematic viscosity v = u/p     u = viscosity
                                p = density

•   In plant operations, we quite often want the flow to be from a low to
    a high pressure area. This requires use of pumps or compressors.

•   In the operation of pumps, pressure drop is of utmost importance.
    To operate a centrifugal pump, for example, the suction line can be
    and often is more important than discharge line. Although the
    discharge line may be designed so that the pump is capable of
    putting the required amount of liquid through it,

• If the suction line is not large enough or is improperly
  installed, pumping difficulties will be experience.

• If pressure drop is such that liquid starts to vaporize when it
  reaches the pump, or if suction conditions are not correct,
  then pump will not fill up properly and will become ‘gas
• The pumping rate and discharge pressure will fall off, and pump
  will begin to cavitate and heat up.
                           Flow of Gases

•   Gases are compressible [ liquids are non-compressible]
•   In moving a gas through a line, the higher the pressure on the
    system, the more pounds of gas that can be transferred for the
    same volume rate of flow.
    The effects on plant operations due to compressibility of a
                             gas are:

•      Increased pressure means reduced equipment size
•      Gas such as air, under pressure can be used to operate
       pneumatic tools
•      Hydrocarbon gases can be compressed, cooled and liquefied
•      Refrigeration
                         Fluid Performing Work

A column of water will exert a pressure of 0.433 psi
for each foot of weight.

• Suppose we build a dam and raised the height of
 the water to 200 feet. That would mean that the
 pressure at the bottom of the dam would be 0.433 x
 200 or 86 psi. This says that if turbines were installed
 at the bottom of the dam we would have a driving
 force of 86 psi to turn the turbines which can be used
 to generate electricity.

•   The siphon is a bent tube with branches of unequal length open at
    both ends and it is used to move a liquid from a higher point to a
    lower point, over an intermediate point higher than either.


                            1 ft

                            2 ft

          A                                       B

•   Since the upward force at A is greater than B, the flow will be
    from A to B. Suppose the tube is filled with water and placed in
    the vessels as shown in the figure above. The water in leg CA
    exerts a head or downward pressure of 1 foot of water or 0.433
    psi. The water in branch CB exerts a head of 3 x 0.433 or 1.299
    psi. Therefore, the net downward pressure causing flow from A
    to elevation B is 0.866 psi.
                     Fluid Rates Measurement

•   When pressure is converted to velocity, pressure is reduced. The
    difference in pressure between 2 points in the line of flow is the
    principle on which common flow meters operate.
•   Orifice plate
•   Pitot tube
•   Venturi tube
                   Pressure in Stationary Fluids

Force - means a push or pull

Weight - means heaviness

Pressure - The push or pull or weight or weight per unit area
           of the surface acted upon.

Pressure = __ Force_____           =   __ Weight____
           Area Acted upon             Area Acted Upon

Water weight 62.4 pounds per cubic ft, 1 foot high water would therefore
exert a pressure of:

   62.4lbs   x 1 ft =     0.433 lbs
    ft3        144           ins2
                   Pressure Produced by Liquids
•   Since all fluids have weight, they create a pressure against the walls
    that hold them.
•   The pressure exerted by a liquid at any given point or location in a
    vessel depends upon the height of liquid above it.
•    This pressure is independent of the shape of the vessel
•   The pressure on the btm of all vessels below will be the same
Pressure Produced by Liquid (con’t)
• Different liquids weigh different amount for the same volume and
   therefore would create different pressures. 1 cubic foot of water
   weighs 62.4 pounds. If this cubic foot of water were put in a box 1 foot
   high, 1 foot wide, and 1 foot long, the total force on the bottom of the
   box would be 62.4 pounds. This total force would be distributed over
   an area one foot square or 144 square inches. The pressure on a
   given square inch would be 62.4 pounds divided by 144 square inches
   [0.433 psi].
•   Pressure Produced by Liquid (con’t)

•   If a box were filled with water that was 2 feet high and held 2 cubic feet
    the pressure would equal 62.4 x 2 /144 or 0.866 psi.

•   Each foot of water height will develop 0.433 psi. For example a 100 foot
    tower, if filled with water, would create 43.3 psi at the bottom of the

•   Weight of one cubic foot of material is called the density of that
                 Pressures Produced By Gases

•   The deep layer of air which blankets the earth exerts a pressure
    much like the water pressure at the bottom of the ocean. This
    pressure is known as atmospheric pressure and is about 14.7 PSI at
    sea level. At higher elevation, the atmospheric pressure falls.

•   The weight of air causes an atmospheric pressure of 14.7 psia. Less
    than 14.7 psia is a vacuum.
                          Gauge Pressure

•   Most of pressure gauges are calibrated so that “0” on the gauge is
    atmospheric pressure.

•   Reading therefore is called PSIG
                        Standard Pressure

•   The volume of a given sample of a gas has almost no meaning
    unless the pressure is specified.
•   For uniformity, 760 mm of mercury, or 1 atm, has been adopted as
    the standard pressure.
•   All specific properties of gases such as density – are always at
    standard pressure.

•   When a pressure is less than atmospheric pressure, it is called a
    vacuum. A vacuum is a lack of air or fluid.
Laws Governing Gas Under Pressure
                              Boyle Law

•   At constant temperature, the volume of a fixed weight of a given gas
    is inversely proportional to the pressure under which it is measured.

    P x V = K (at constant temp)

    P1V1 = P2V2 at constant temperature

If the pressure is doubled, volume is halved
If the pressure is halves, volume is doubled
                             Charles’ Law

•   At constant pressure, the volume of a fixed weight of gas is directly
    proportional to the absolute temperature

    V = kT (at constant pressure)

    V1 = T1 (at constant pressure)
    V2 T2
                  Dalton’s Law of Partial Pressure

•   Each gas in a gaseous mixture exerts a partial pressure equal to the
    pressure which it would exert if it were the only gas present in the
    same volume, and the total pressure of the mixture is the sum of the
    partial pressures of all the component gases.

    P total = P1 + P2 +P3 …….
                             Raoult’s Law

Ideal solution - exhibits no change in the properties of its constituents
                 beyond that of dilution.
               - For our present purpose an ideal solution is most
                 usefully defined in terms of generalization known as
                Raoult’s law.

Raoult’s Law
- At any given temperature, the vapor pressure of any component of a
  solution (i.e, its partial pressure in the mixed vapor in equilibrium
  with the solution) is equal to the product of the mole fraction of that
  component in the solution and its vapor pressure in thepure liquid
  state at the same temperature.
                            Raoult’s Law

Raoult’s Law (con’t)
  Thus in a solution composed of 2 volatile liquid A and B, the vapor
   pressure of which in the pure state are PAo and PBo respectively, we
   may write:
                P A = X AP Ao
                P B = X BP Bo
Total vapor pressure P of an ideal solution:
                P = X AP Ao + X BP Bo
                         Problem solving

•    A sample of gas occupies a volume of 86.8 ml at a pressure of
     730 mm of mercury and a temperature of 27oC. What will be its
     volume at standard pressure and 27oC?
    Standard pressure = 760 mm hg
    V1 = 86.8 ml             P1 = 730 mm
    V2 = ? Ml                P2 = 760 mm
    Pressure is increased from 730 to 760 mm, the volume is therefore
    decreased, and the original volume must be multiplied by a fraction
    less than 1.

    V2 = 86.8 x 730/760 = 83.4 ml
                           Problem Solving

2.    Calculate the volume that 22.4 liters of a gas measured at the
      standard pressure would occupy at a pressure of 732 mm hg.

     V1 = 22.4 liters              P1 = 760 mm
     V2 = ? Liters                 P2 = 732 mm

     Since the pressure is decreased to 730/760 of its original value,
     the volume is increased by the factor 760/732

     V2 = 22.4 liters x 760/732 = 23.3 liters
                         Problem Solving

3.   A gas measures 83.4 ml at 1 atm pressure and a temperature of
     27 oC. Calculate its volume at standard condition.

     V1 = 83.4 ml             T1 = 27 + 273 = 300 oK
     V2 = ? Ml                T2 = 273 oK

     Since the temperature is decreased, the volume will also be
     decreased. The original volume must therefore br multiplied by a
     fraction less than 1, namely 273/300:

     V2 = 83.4 ml x 273/300 = 75.9 ml
                         Problem Solving

4.   A sample of gas occupies a volume of 86.8 ml at a pressure of 730
     mm and a temperature of 27 oC. What will be its volume at
     standard condition?

     V1 = 86.8 ml            T1 = 300 oK           P1 = 730
     V2 = ? Ml               T2 = 273 oK           P2 = 760

     Pressure is increased from 730 to 760 mm, volume should
     decrease, we therefore multiply the original volume by 730/760.
     Temperature decrease from 300 to 273 oK, likewise volume
     decreases, hence we multiply by the temperature fraction 273/300,

     V2 = 86.8 ml x 730/760 x 273/300 = 75.9 ml.
                         Problem Solving

5. A certain gas measures 546 ml at a pressure of 1 atm and a
   temperature of -80 oC. Calculate the volume it would occupy at a
   pressure of 1.5 atm and a temperature of 30 oC.

   First we change the centigrade temperature to absolute:
   - 80 oC = 80 + 273 = 193 oK
     30 oC = 30 + 273 = 303 oK

    V1 = 546 ml        P1 = 1 atm          T1 = 193 oK
    V2 = ? ml          P2 = 1.5 atm        T2 = 303 oK

    V2 = 546 ml x 1/1.5 x 303/193 = 571 ml
                         Joule-Thomson Effect

•   Gases are cooled upon sudden expansion from high to low pressure,
    even if they do no external work.

•   As gas expands the molecules pull apart from each other, and work
    must be done to overcome the cohesive forces that tend to hold the
    molecules together. Since this work is done at the expense of the
    kinetic energy of the molecules of the gas, the temperature is lowered
    on expansion.
                          Standard Volume

•   Volume measured at standard condition ie temp = 60 oF
•   Pressure = 1 atm (14.7psi)
•   At standard condition one (1) mole of gas occupies 22.4 liters

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