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					                       Chapter 8
         Transportation and Metering of Fluids

     transporting fluids from one place to another
     measuring their rates of flow

PIPE, FITTINGS, AND VALVES
Pipe and tubing
 nominal diameter    the actual outside diameter
 schedule number     the wall thickness of pipe
Selection of pipe sizes
   For turbulent flow of liquids in steel pipes larger than 1 in.
   (25 mm) in diameter, the optimum velocity is
                                    0 .1
                              12 m
                      Vopt    0.36           (8.1)
                               
      where     Vopt = optimum velocity, ft/s
                 
                 m = mass flow rate, lb/s
                  = fluid density, lb/ft3
Joints and fittings  to join pieces of tubing or pipe
     thick-walled tube connected by
             screw fittings
             flanges
             welding
  thin-walled tubing are joined by
         soldering
         compression
         flare fittings

  brittle materials (glass, carbon, cast iron) joined by
          flanges
          bell and spigot joints

Screw fittings  the ends of pipe are threaded
          the thread are tapered
          farthest from the end of the pipe are imperfect
          a tight joint is formed when the pipe is screwed
           into a fitting
  tape of polytetrafluoroethylene (teflon) is wrapped
   around the thread to ensure a good seal.
  screw fitting are more weak than the pipe (use
   higher schedule number)
  standardized for pipe size up to 12 in not usually
   larger than 3 in.

Flanges
      matching disk or rings of metal bolted together
      compressing a gasket between their faces
      attached to the pipe by screwing them or by
       welding or by brazing
   blind flange or a blank flange
       a flange with no opening
       used to close a pipe

Welding
       joining pieces of large steel pipe
       for high pressure service


flanged and screw joints    sources of emission of
                             volatile matters
               welding      leak proof
Allowances for expansion
  pipe is subjected to varying temperatures and pressures
    cause the pipe to expand and contract
 fixed supports are not used (may tear loose, bend, or
  break)
 so the pipe rests loosely on rollers
 or is hung from above by chains or rods
 provision is made for taking up expansion by
 bends or loops in the pipe, by packed expansion joints
 by bellows or packless joints or
 by flexible metal hose
Prevention of leakage around moving parts
      Common devices for minimizing leakage while
permitting relative motion are:
          stuffing boxes



          mechanical seals
Stuffing boxes
- provide a seal around a rotation shaft that moves axially




   FIGURE 8.1
   Stuffing boxes: (a) simple form; (b) with lantern gland.
Mechanical seals
    the sliding contact is between a ring of graphite and
      a polish metal surface, usually of carbon steel
     require less maintenance than stuffing boxes




         FIGURE 8.2
         Mechanical seal.
VALVES  to slow down or stop the flow of a fluid
Gate valves  diameter of the opening = dia. of the pipe
    direction of flow does not change
    small pressure drop
    not recommended for controlling flow
    fully open or fully closed
Globe valves  so called because in the earliest designs
                 the valve body was spherical
    widely used for controlling flow
     the fluid passes through a restricted opening and
      changes direction several times
     pressure drop is large
FIGURE 8.3
Common valves:
(a) gate valve;
(b) globe valve;
(c) control valve with
    pneumatic valve
    activator.
Plug cocks and ball valves

 plug cock
       as in a lab stopcock
       fully open to fully closed
       pressure drop is minimal when it is fully open


 ball valve
        sealing element is spherical
        occasionally applied in flow control
Check valves
    permits flow in one direction only
    opened by the pressure of the fluid
    when the flow stops, the valve automatically closes
     by gravity or by a spring pressing against the disk




 FIGURE 8.4
 Check valves: (a) lift check; (b) ball check.; (c) swing check.
Pumps
   transportation of liquids through pipes and channels
   increase the mechanical energy of the liquid
         velocity       pressure    elevation
   positive-displacement pumps and
   centrifugal pumps
Positive-displacement
   apply pressure directly to the liquid by
         reciprocating piston
         rotating members
Centrifugal pump
   generate high rotational velocities
   convert kinetic energy of the liquid to pressure energy
Developed head




 FIGURE 8.5
 Pump flow system.
Eq. (4.65) can be written


        p              2
                           p
                    b Vb   a         a Va 
                                            2

  Wp   b  gZb              gZa           (8.2a)
                    2                2 
                                           

or in fps units,


         p gZ  V 2   p gZ  V a  2

  Wp   b  b  b b  a  a  a              (8.2b)
           gc   2g c    gc   2g c 
                                     
The quantities in the parentheses are total heads, H

                                         2
                       p            V
                 H         gZ                (8.3a)
                                    2
       and
                                         2
                     gZ  V
                      p
                H                            (8.3b)
                   gc 2 gc
    Ha = total suction head dimension work per unit mass
    Hb = total discharge head
                 H b  H a H
            Wp                               (8.4)
                           
                                     2
              H p       V
                   Z                        (8.5a)
              g g       2g
                                         2
           Hg c pgc      V
                    Z                       (8.5b)
            g    g       2g

In Eqs (8.5a) and (8.5b) each term has the dimension of length
Power requirement
  The power supplied to the pump, PB
                                         
                                 m H
                      PB  m Wp                  (8.6)
                                   
                  
        where     m is the mass flow rate
                             
        define:       Pf  m H                   (8.7)

  The power delivered to the fluid, Pf
                                      Pf
  From Eqs (8.6) and (8.7)       PB              (8.8)
                                      
  For fans, use average density,  = (a+b)/2 for 
Suction lift and cavitation
      If the suction pressure is only slightly greater than the
vapor pressure, some liquid may flash to vapor inside the
pump, a process called cavitation.
   greatly reduces the pump capacity
   causes severe erosion.
       If the suction pressure is actually less than the vapor
pressure, there will be vaporization in the suction line, and no
liquid can be drawn into the pump.
To avoid cavitation
    the pressure at the pump inlet must exceed the vapor
      pressure by a certain value
   called the net positive suction head (NPSH)
   NPSH = 2-3 m (5 to 10 ft) for small centrifugal pumps
         = up to 15 m (50 ft) for very large pumps.
For a pump taking suction from a reservoir

                 1  p a  p v 
           NPSH     h fs   Za
                                                   (8.9a)
                 g             
or in fps units,
                 g c  pa  p v 
           NPSH       h fs   Za
                                 
                                                    (8.9b)
                 g              

      where    pa' = absolute pressure at surface of reservoir
               pv = vapor pressure
               hfs = friction in suction line

NPSHR = Minimum required NPSH specified by manufacturers
EXAMPLE 8.1.
Benzene at 100F (37.8C) is pumped through the system of Fig. 8.5
at the rate of 40 gal/min (9.09 m3/h). The reservoir is at atmospheric
pressure. The gauge pressure at the end of the discharge line is 50
lbf/in.2 (345 kN/m2). The discharge is 10 ft, and the pump suction is 4
ft above the level in the reservoir. The discharge line is 1½-in.
Schedule 40 pipe. The friction in the suction line is known to be 0.5
lbf/in.2 (3.45 kN/m2), and that in the discharge line is 5.5 lbf/in.2 (37.9
kN/m2). The mechanical efficiency of the pump is 0.60 (60 percent).
The density of benzene is 54 lb/ft3 (865 kg/m3), and its vapor pressure
at 100F (37.8C) is 3.8 lbf/in.2 (26.2 kN/m2).

Calculate (a) the developed head of the pump and
           (b) the total power input.
           (c) If the pump manufacturer specifies a required NPSHR
of 10 ft (305 m), will the pump be suitable for this service?
Solution.
(a) The pump work Wp is found by using Eq. (4.65). The upstream
    station a is at the level of liquid in the reservoir, and the
    downstream station b is at the end of the discharge line, as
    shown in Fig 8.5. When the level in the tank is chosen as the
    datum of heights and it is noted that Va = 0, Eq. (4.65) gives
                                          2
              p b gZb  b V b        p a
       Wp                      hf 
                   gc   2gc              
    The exit velocity Vb is found by using data from App. 3. For
    a 1½-in. Schedule 40 pipe, a velocity of 1 ft/s corresponds to a
    flow rate of 6.34 gal/min, and

                         40
                 V b        6.31 ft s
                        6.34
 With b = 1.0, Eq. (4.65) gives

W 
     14.7  50144     g
                              10    6.312
                                               
                                                 5.5  0.5144  14.7 144
                                     2  32.17
  p
              54           gc                           54              54
      = 159.9 ftlbf/lb
 By Eq. (8.4) Wp is also the developed head, and
                   H = Hb – Ha = 159.9 ftlbf/lb (477.9 J/kg)
 (b) The mass flow rate is
                    
                       40  54
                   m            4.81 lb s 2.18 kg s
                      7.48  60
      The power input is, from Eq. (8.6),
                       4.81  159.9
                   P 
                    B                2.33 hp 1.74 kW
                        550  0.60
(c) Use Eq. (8.9), pa/ = 14.7144/54 = 39.2 ftlbf/lb. The vapor
    pressure corresponds to a head of
           3.8  144
                      10.1 ft  lb f / lb 30.2 J kg
              54
    The friction in the suction line is
                0.5  144
           hf             1.33 ft  lb f lb 3.98 J kg 
                   54
    The value of the available NPSH from Eq. (8.9),
    assuming g/gc = 1, is

           NPSH = 39.2 - 10.1 - 1.33 – 4 = 23.77 ft (7.25 m)
    The available NPSH is considerably larger than the minimum
    required value of 10 ft, so the pump should be suitable for the
    proposed service.
*Positive-Displacement Pumps*
 Reciprocating pumps - the chamber is a stationary
                         cylinder that contains a piston
                         or plunger
     e.g. piston pumps, plunger pumps, and
        diaphragm pump
 Rotary pumps          - the chamber moves from inlet to
                           discharge and back to the inlet.

1. Piston   - max. discharge pressure is about 50 atm
     liquid is drawn through an inlet check valve into
      the cylinder by the withdrawal of a piston
     then is forced out through a discharge check
      valve on the return stroke
2. Plunger - containing a close-fitting reciprocating
              plunger in a heavy-walled cylinder of
              small diameter 1
   the plunger fills nearly all the space in the cylinder
   discharge pressure 1,500 atm or higher

3. Diaphragm - the reciprocating member is a flexible
              diaphragm of metal, plastic, or rubber.
  handling toxic or corrosive liquids amount up to 100
   gal/min.
  pressure up to 100 atm
FIGURE 8.6
Diaphragm pump.
Volumetric efficiency
     - ratio of the volume of fluid discharge to the volume swept
         by the piston.
Plunger and Diaphragm
    - used as “metering pumps” because the volume flow is
         constant, controllable, and adjustable.
Rotary pumps
    e.g. gear pump, lobe pumps, screw pumps, cam
      pumps, and vane pumps
    contain no check valves
    close tolerances between the moving and stationary
      parts
    operate best on clean, moderately viscous fluid
    discharge pressures up to 200 atm or more
FIGURE 8.6
Gear pumps: (a) spur-gear pump; (b) internal-gear pump.
*Centrifugal pumps*
 the liquid enters through a suction connection concentric
  with the axis of a high-speed rotary element called the
  impeller, which carries radial vanes integrally cast in it.
 the liquid leaving the outer periphery of the impeller is
  collected in a spiral casing called the volute and leaves
  the pump through a tangential discharge connection.




FIGURE 8.8
Single-suction centrifugal pump.
Centrifugal pump theory
 the liquid enters axially at the suction, station a
 in the rotating eye of the impeller, the liquid spreads out
  radially and enters the channels between the vanes at
  station 1.
 it flows through the impeller, leaves the periphery of the
  impeller at station 2.
 collected in the volute, and leaves the discharge at b


    FIGURE 8.8
    Centrifugal pump
FIGURE 8.10
Characteristic curves of a centrifugal pump operating at various speeds.
Comparison of Devices for Moving Fluids
       considering the flow capacity, power requirements, mechanical
       efficiency, reliability, and ease of maintenance
Positive-displacement machines
       - handle smaller quantities of fluids at higher discharge
       pressures than do centrifugal machines.
       - no air binding and usually self-priming.
       - used for controlling and metering flow
       - require considerable maintenance
       - produce the highest pressure
       - cannot be used with slurries.
       - discharge line cannot be closed without stalling or breaking
       the pump, so that a bypass line with a pressure relief valve is
       required
Centrifugal machines
       - deliver fluid at a uniform pressure without shocks or
       pulsations.
       - run at higher speeds than positive-displacement machines
       - the discharge line can be completely closed without
       damage.
       - can handle corrosive liquids and slurries
       - require less maintenance


Vacuum pumps
   - a compressor that takes suction at a pressure below
        atmospheric and discharges at atmospheric pressure
Measurement of Flowing Fluids

Full-Bore Meters
   - venturi and orifice meters, rotameters


  Selection  installed cost and costs of operation
             the range of flow rates it can accommodate
             its inherent accuracy
Venturi meter




   FIGURE 8.17
   Venturi meter.

 most commonly used with liquids
 requires less power than other types
 angle of discharge cone is 5o-15o
to minimize boundary layer separation
For incompressible fluid with no friction
Eq. (4.65) becomes

                2         2p a  p b 
                           2
           b V   a V 
                b          a                    (8.22)
                               
                               2
                     Db 
                      Vb   2 Vb
               Va                            (8.23)

                     Da 
 where    Da    = diameter of pipe
          Db    = diameter of throat of meter
               = diameter of ratio Db/Da
                           1            2p a  pb 
           Vb                                              (8.24)
                     b    a4             

Venturi coefficient (Cv )
     - for the small friction loss between locations a and b

                         Cv         2p a  pb 
             Vb                                            (8.25)
                       1     4         
    Cv = the venturi coefficient (determined experimentally)
                  = 0.98 for pipe diameters of 2 to 8 in.
                  = 0.99 for larger sizes
Volumetric and mass flow rates


                       Cv Sb      2p a  pb 
    q  Vb Sb                                    (8.26)
                       1  4          

      where        q = volumetric flow rate
                   Sb = area of throat

      
                      Cv Sb
     m  q                      2p a  pb    (8.27)
                              4
                      1
               
      where   m       = mass flow rate
 Orifice meter

Venturi meter                   Orifice meter

 expensive                      can change to meet the
                                  requirement
 occupies considerable
  space                          larger power consumption
 ratio of throat diameter to
  pipe diameter cannot be
  changed
 the principle is identical
                      C0       2p a  pb 
          u0                                   (8.28)
                   1     4        
where u0 = velocity through orifice (determined experimentally)
       = ratio of orifice diameter to pipe diameter
      pa, pb = pressures at stations a and b
      C0 = the orifice coefficient
          = 0.61 for flange taps and vena contracta taps and
when Re > 30,000                         
                  D0 u 0    4m
            Re0                               (8.29)
                            D.

        where    D0    = orifice diameter
EXAMPLE 8.4. An orifice meter with flange taps is to be installed in a
100-mm line to measure the flow of water. The maximum flow rate is
expected to be 50 m3/h at 15C. The manometer used to measure the
differential pressure is to be filled with mercury, and water is to fill the
leads above the surfaces of the mercury. The water temperature will be
15C throughout. (a) If the maximum manometer reading is to be 1.25
m, what diameter, to the nearest millimeter, should be specified for the
orifice? (b) What will be the power to operate the meter at full load?
Solution (a) Equation (8.29) is used to calculate the orifice diameter.
             The quantities to be substituted are
                   50
              q        0.0139 m3 s
                 3,600
                 62.37  16.018  999 kg m3 (App. 6)
             C0  0.61                   g  9.80665 m s2
From Eq. (2.10),
       p a  pb  9.80665  1.25  13.6  1.0 999 
                               2
                 154 ,300 N m
Substituting these values in Eq. (8.38) gives

                      0.61S0       2  154,300
       0.0139 
                       1  4           999
From which
                                                 2
          S0                       3           D
                    1.296  10                 0

        1  4                              4 1  4
As a first approximation, call 1  4 =1.0. Then
                                       40.6
       D0  40.6 mm                        0.406
                                       100
And
              1  4  1  0.4064  0.986
The effect of this term is negligible in view of the desired precision
of the final result.
     Check the Reynolds number. The viscosity of water at 15C, from
App. 6. is 1.147 cP or 0.001147 kg/ms
                    2
             D       0.041          2
        S0        0
                                0.00132 m2
              4          4
             q     0.0139
        uo                 10.53 m s
             S0 0.00132
The Reynolds number, from Eq. (8.40), is
            0.041 10.53  999
      Re0                      376,000
                0.001147
The Reynolds number is large enough to justify the value of 0.61 for C0.
b) From Fig. 8.18, for  = 0.406, the permanent loss in pressure
   is 81% of the differential. Since the maximum volumetric flow
   rate is 0.0139 m3/s, the power required to operate the orifice
   meter at full flow is


                           0.81 0.0139  154,300
   P  0.81qp a  p b  
                                   1,000
      1.737 kW
Area meters: Rotameters
  - in the orifice, nozzle, or venturi

        the variation of flow rate through a constant area
         generates a variable pressure drop, which is
         related to the flow rate
  - area meters
               pressure drop is constant, the area through
                which the fluid flows varies with the flow
                rate.
               the area is related to the flow rate.
Rotameter

                             consists of a gradually tapered
                              glass tube mounted vertically in
                              a frame with the large end up.
                             fluid flows upward through the
                              tapered tube and suspends
                              freely a float
                             the greater the flow rate, the
                              higher the float rides in the tube
                             can be used for either liquid or
                              gas flow measurement.
FIGURE 8.21
Principle of a rotameter.
Insertion Meters
    sensing element is inserted into the flow stream
    measure the average flow velocity or local velocity at
     one point only.
    the position of the sensing element is important.

Pitot tube
    measure local velocity along a streamline
    consists of two tubes connected to a manometer
    static tube measures the static pressure P0
     (no velocity component perpendicular to its opening)
    the impact tube measures the stagnation pressure of
     the fluid
                                    FIGURE 8.26
                                    Principle of pitot tube.




For incompressible fluids
                            2 ps  po 
                    uo                            (8.35)
                                
EXAMPLE 8.5. Air at 200F (93.3C) is forced through a long,
circular flue 36 in. (914 mm) in diameter. A pitot tube reading is taken
at the center of the flue at a sufficient distance from flow disturbances to
ensure normal velocity distribution. The pitot reading is 0.54 in. (13.7
mm) H2O, and the static pressure at the point of measurement is 15.25
in. (387 mm) H2O. The coefficient of the pitot tube is 0.98.
       Calculate the flow of air, in cubic feet per minute, measured at
60F (15.6C) and a barometric pressure of 29.92 in. (760 mm) Hg.
Solution Assume the Mach number correction is negligible. The
velocity at the center of the flue, which is that measured by the pitot
tube, is calculated by Eq. (8.35) using fps units and a coefficient of
0.98. Equation (8.46) becomes

                            2gc p s  pb 
                  uo  0.98                                    (8.47)
                                  
The necessary quantities are as follows. The absolute pressure at the
instrument is

                 15.25
     p  29.92         31.04 in. Hg
                 13.6
Since 1 lb mole occupies 359 ft3 at 32F and 1 atm, the density of
the air is
          29  492  31.04
                                0.0625 lb ft 3
        359 460  200 29.92 
From the manometer reading

 ps  pb   0.54
                   62.37( g )  2.81(32.174) lb f ft2
               12
By Eq. (8.47), the maximum velocity is


                2 x(2.81 x32 .174 )
   um ax  0.98                      52 .7 ft s
                     0.0625
This is sufficiently low for the Mach number correction to be
negligible. To obtain the average velocity from the maximum
velocity, Fig. 5.8 is used. The Reynolds number, based on the
maximum velocity, is calculated as follows.
From App. 8, the viscosity of air at 200F is 0.022 cP, and


Remax   
          36 12 52.70.0625   670 ,000
            0.022 0.000672 
The ratio V/umax, from Fig. 5.8, is a little greater than 0.86.
Using 0.86 as an estimated value gives


           V  0.86  52.7  45.3 ft s

The Reynolds number Re is 670,0000.86 = 576,000, and
V/umax is exactly 0.86 as estimated. The volumetric flow rate is


                                       520  31.04 
                                2
                    36 
           q  45.3                              60
                    12             4  660  29.92 
              15,704 ft 3
                                         
                                    min 7.41 m s  3
                                                        

				
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