Page 100 - APPLIED PROCESS DESIGN FOR CHEMICAL AND PETROCHEMICAL PLANTS, Volume 1, 3rd Edition
P. 100

86                       Applied Process Design for Chemical and Petrochemical Plants

             Summation:                                              Pressure drop = AP  = 0.00001799 Kp Q2/d4   (2-52)
                                                                                     = 0.00001799 (52.48) (48.6)
               K = [4.38 + 7.45" + 3.94 + 0.175" + 1.01 = 16.941                       (350)*/ (4.026)4
                                                                                  AP  = 21.2 psi friction pressure loss only
               *Threaded, from Table 2-2.                                              (no elevation change)
               For fittings:
                                                                    Alternate Calculation Basis for Piping System Friction
                                                                                   Head Loss: Liquids

             then, h  = Kv"2g   =  16.941 (4'292 )  = 4.84 ft kerosene   Pressure  loss  in  a  piping  system  (not including  the
                                 2(32.2)
                                                                   tanks, heat exchangers, distillation columns, etc.) is usu-
               Total friction loss for discharge  side pump  due to fiic-   ally expressed in units of feet offlowingfluid, or the equiva-
             tion:                                                 lent  converted  to pounds per  square  inch.  Some published
                                                                   pressure loss data is expressed as per 100 equivalentfeet of
                                                                   the size pipe being used or estimated.
                h  = 11.64 + 4.84 = 16.48 ft fluid kerosene
               hf = Ap  = 16.48/[(2.31)/(0.81)]  = 5.77 psi
                                                                   Equivalent Feet Concept for Valves, Fittings, Etc.
             Example 2-2: Laminar Flow Through Piping System
                                                                     With pipe  of  any specified size as the basis, the  total
                                                                   footage  of  straight  pipe in a system is just the measured
               A heavy weight oil, No.  5 fuel  oil, is  to  be  pumped
             through 350 ft of  existing 4in. Schedule 40 pipe at 350   length  (totaled).
             gpm. Oil data:                                          For fittings, valves, etc., in the same system, these can
                                                                   be expressed as equivalent straight pipe, then added to the
               Temperature = 100°F                                 straight pipe described above, to arrive at a total equiva-
                   Viscosity = 150 cp                              lent straight length of pipe of the specific size in question.
                     Sp Gr  = 0.78 = 48.6 lb/cu ft                   Figure 2-20 presents equivalent lengths of straight pipe
                   Pipe I.D.  = 4.026 in. = 0.3355 ft              (feet) for various pipe system components. For example,
                                                                   a  standard  threaded  6-inch  90" elbow  is  equivalent  to
             Reynolds number = 50.6 Qp/(dp)                        adding 17 feet of straight pipe to the system. This 17 feet
                            = 50.6 [ (350) (48.6)/(4.026) (150)l   is additive to the lengths of  nominal 6-inch straight pipe
                            = 1425                         (2-50)   in the system (dotted line). However, there is an impor-
                                                                   tant consideration in the use of this chart, i.e., use only for
               Flow is <2000, therefore, flow of viscous or laminar sys-   threaded  or  screwed  pipe/fittings,  and  only  for  sizes
             tem  consists of friction factor, fT, for 4in. pipe  = 0.017   under 2-inch nominal  size. It is not practical in  current
             (Table 2-2).                                          industry  practice  to  thread  a  process  or  utility  system
                                                                   much greater in nominal diameter than 2 inches. For spe-
                1 gate valve = K = 8 fT = 8 (0.017) = 0.136, (Table 2-2)   cial situations, the  larger  sizes can  be  used,  but from  a
                3 90" = K = 20 fT = 20  (0.017) = 0.345            handling standpoint, sizes greater than 3 inches or 4 inch-
                1 90" = R/D  = 5; 5/D  = 0.00045 (Figure 2-11); K  =   es are not practical.
             0.1 (Figure 2-13A)                                      For pipe sizes greater than 2 inches nominal, industry
                1 pipe  entrance  to  tank projecting inward, K  = 0.78   practice is to weld the pipe and fittings into one continu-
             (Table 2-2)                                           ous system, and then use flanged or special bolted  con-
                For 350 ft pipe, K = f (L/D)  = 0.0449 (350/0.3355)  =   nections  for  attaching  the  valves,  orifices, and  connec-
             51.12                                                 tions  to  vessels or  other  equipment.  For  special lethal,
                [For f, see calculations below]                    high  pressure,  and  steam  power  plant  high  tempera-
                                                                   ture/high  pressure  utility  systems, even  the  valves  and
                f = 64/&                                           connections  to  vessels  are welded  into  the  system  (See
                f = 64/1425  = 0.0449                      (2-1 7)   ASME and ANSI Codes). For these situations of about 1%-
                                                                   inch to 2-inch nominal pipe size and larger, use Figure 2-

                Total Kvalues =51.12 + 0.7'8 + 0.1 + 0.136 + 0.345 =   21 to determine the equivalent pipe lengths for these fit-
             52.48                                                 tings, valves, etc. For example, a 45" welding elbow, or an
                                                                   open 6-inch gate valve  (see line on chart) have an equiva-
                Velocityv,  = 0.408 Q/d2 = 0.408 (350)/(4.026)2    lent length of 6-inch pipe of four feet (straight), which is
                        = 8.8 ft/sec                       (2-51)   an addition to the actual straight pipe in the system. In
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