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2.5: Untitled Page 17

  • Page ID
    18150
  • Chapter 2

    0.45

    1.5

    1920 

      

     1000 Q

    X

    vs

    597

    L

    (2‐13)

    

    a

    V

     

    a

    Q

    In order that this expression produce correct results, it is absolutely essential that Table 2‐4a. Basic Conversion Factors

    Mass

    Length

    1 pound mass (lbm) = 453.6 gram (g)

    1 inch = 2.54 centimeter (cm)

    1 g = 10‐3 kilogram (kg)

    1 angstrom (Å) = 10‐8 centimeter

    1 ton (short) = 2000 lbm

    1 foot (ft) = 0.3048 meter (m)

    1 ton (long) = 2240 lbm

    1 mile (mi) = 1.609 kilometer (km)

    1 yard (yd) = 0.914 meter (m)

    Force

    1 nanometer (nm) = 10‐9 meter (m)

    pound‐force (lbf) = 32.17 ft lbm s‐2

    1 minute (min) = 60 second (s)

    dyne (dyn) = g cm s‐2

    1 hour (h) = 60 minute (min)

    1 newton (N) = 10 5 dyne (dyn)

    1 day = 24 hours (h)

    1 dyne = 2.248 x 10‐6 pound force (lbf)

    1 poundal = 3.108 x 10‐2 lbf

    Time

    pound‐force (lbf) = 4.448 newton (N)

    1 minute (min) = 60 second (s)

    1 hour (h) = 60 minute (min)

    Pressure

    1 day = 24 hours (h)

    1 atmosphere (atm) = 14.7 lbf /in2

    1 lbf/ft3 = 16.018 kg/m3

    Volume

    1 atm = 1.013 x 10 6 dyne/cm2

    1 in3 = 16.39 cm3

    1 pascal = 1 N/m2

    1 ft3 = 2.83 x 10‐2 m3

    1 atmosphere (atm) = 1.01325 bar

    1 gallon (US) (gal) = 231 in3

    the quantities in Eq. 2‐13 be specified as follows:

    X vs = average drop diameter, m

     (a drop with the same volume

    to surface ratio as the total sum of all drops formed)

     = surface tension, dyne/cm

     = liquid viscosity, P

    a

    V = relative velocity between air and liquid, ft/s

     = liquid density, g/cm3

    Units

    21

    QL = liquid volumetric flow rate, any units

    a

    Q = air volumetric flow rate, same units as Q

    L

    When the numbers associated with these quantities are inserted into Eq. 2‐13, one obtains the average drop diameter, X vs , in micrometers. Such an equation Table 2‐4b. Additional Conversion Factors

    Energy

    Area

    1 calorie (cal) = 4.186 joule (J)

    1 in2 = 6.452 cm2

    1 British Thermal Unit (Btu) = 252 cal

    1 in2 = 9.29 x 10‐2 m2

    1 erg = 10‐7 joule (J)

    1 acre = 4.35 x 10 4 ft2

    1 Btu = 1055 watt sec

    1 ft lbf = 1.356 joule (J)

    Flow Rate

    1 gal/min = gpm 6.309 x 10‐5 m3/s

    Power

    1 ft3/min = 4.719 x 10‐5 m3/s

    1 horsepower (hp) = 745.7 watt

    1 hp = 42.6 Btu/min

    Viscosity

    1 watt = 9.51 x 10‐4 Btu/s

    1 poise (P) = 1 g/cm s

    1 ft lbf /sec = 1.356 watt (W)

    1 poise (P) = 0.10 N s/m2

    1 poise (P) = 1 g/cm s

    Temperature

    1 centipoise (cP) = 10‐3 kg/m s

    1 C = 1.8 F

    K = C + 273.16

    Kinematic Viscosity

    R = F + 459.60

    1 m2/s = 3.875 x 104 ft2/hr

    F = 1.8C + 32

    1 cm2/s = 10‐4 m2/s

    1 stokes (St) = 1 cm2/s

    must always be used with great care for any mistake in assigning the values to the terms on the right hand side will obviously lead to an undetectable error in X vs . In addition to being dimensionally incorrect, Eq. 2‐13 is an empirical representation of the process of atomization. This means that the range of validity is limited by the range of values for the parameters used in the experimental study. For example, if the liquid density in Eq. 2‐13 tends toward infinity,   

    , we surely do not expect that X vs will tend to zero. This indicates that Eq. 2‐13 is only valid for some finite range of densities. In addition, it is well known that when the relative velocity between air and liquid

    22