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                                                                                              5.2 Fluid Transport  161

                                             3. For rough pipes, the relative roughness e>r takes command and
                                                                   1> 1f = 2 log r>e + 1.74                  (5.14)
                                                where r is the radius of the pipe (ft or m) and e>r is the relative roughness,
                                                dimensionless.
                                             4. In the critical zone between R   2,000 and R   4,000, both R and e>r make their
                                                appearance in the semiempirical equation of Colebrook and White (1937–1938):
                                                             1> 1f = 1.74 - 2 log(e>r + 18.7>R1f )           (5.15)

                                             The magnitudes of absolute roughness e in the f:R and f:(e>r) relationships depend on
                                          the angularity, height, and other geometrical properties of the roughness element and its
                                          distribution. Common magnitudes of e>r have been evaluated for large pipes by Bradley
                                          and Thompson (1951).
                                             Despite the logic and inherent conceptual simplicity of the combination of friction-factor
                                          diagram and Weisbach formulation, there are important reasons why water engineers do not
                                          make use of them for the routine solution of fluid transport problems encountered in water
                                          transmission lines and pipe networks. Among the reasons are the following:
                                             1. Because the relative roughness e>r is a key to f, it is not possible to find r (or d) di-
                                                rectly when h , v or Q, e, and water temperature (or  ) are given. A trial-and-error
                                                           f
                                                solution is required.
                                             2. Because transmission lines may include noncircular as well as circular conduits,
                                                additional f:R diagrams are needed.
                                             3. Because entry 2 also often applies to partially filled sections, additional diagrams
                                                are also necessary for them. For such sections, moreover, trial-and-error solutions
                                                must be performed whenever the depth of flow is unknown.


                     EXAMPLE 5.2   HEAD LOSS USING DARCY-WEISBACH FORMULA
                                                                       2
                                                                                2
                                         Oil of absolute viscosity 0.0021 lb-s/ft (0.1 N-s/m ) and specific gravity 0.851 flows through
                                                                                                    3
                                                                                                             3
                                         10,000 ft (3048 m) of 12-in. (300-mm) stainless steel pipe at a flow rate of 1.57 ft /s (0.0445 m /s).
                                         Determine the Reynolds number, the friction factor, and the head loss in the pipe.
                                         Solution 1 (U.S. Customary System):
                                             Velocity, v   Q>A
                                                            3
                                                                    2
                                                       (1.57 ft /s)>[(1 ft)   0.785]
                                                       2 ft/s
                                             Reynolds number, R   vd >    vd > g
                                                                                                          2
                                                                                                  2
                                                                                      3
                                                               (2 ft/s)(1 ft)(0.851   62.4 lb/ft )>[(0.0021 lb-s/ft )(32.2 ft/s )]
                                                               1,570   2,000, which indicates a laminar flow
                                             Friction factor,  f   64>R
                                                             64>1,570
                                                             0.041
                                                               2
                                             Head loss, h f   f(L>d)(v >2g)
                                                                              2
                                                                                         2
                                                         (0.041)(10,000 ft/1 ft)(2ft/s) >(2   32.2 ft/s )
                                                         25.5 ft
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