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                                                   Water Hydraulics, Transmission, and Appurtenances
                                          Chapter 5
                                      EXAMPLE 5.29 EFFECT OF A FLOATING OBJECT ON WATER DEPTH
                                      A floating cylinder 8 cm in diameter and weighing 960 g is placed in a cylindrical container 26 cm in diameter partially full of water.
                                          Determine the increase in the depth of water in the container due to placing the float in it.
                                      Solution:
                                                                                              3
                                      A 960 g cylinder will displace 960 g of water. Since 1 g of water occupies 1 cm volume, the cylinder will displace 960 cm of water.
                                          The change in total volume beneath the water surface ΔV equals the area of the cylindrical container A, times the change in
                                      water level Δh,or
                                                                           ΔV = AΔh
                                                                           Δh =ΔV∕A
                                                                                            2
                                                                                     3
                                                                                              2
                                                                              = 960 cm ∕[  (26) cm ]                         3
                                                                              = 1.81 cm.
                                    5.2.2 Exponential Equation for Surface             will give the following Manning formula using US customary
                                    Resistance                                         units:
                                    Because of practical shortcomings of the Weisbach formula,  v = (1.486∕n)(r) 0.67 (s) 0.5  (US customary units) (5.25)
                                    engineers have resorted to the so-called exponential equa-
                                                                                       where v = velocity, ft/s; n =coefficient of roughness, dimen-
                                    tions in flow calculations. Among them the Chezy formula
                                                                                       sionless; r = hydraulic radius, ft; and s = slope of energy
                                    is the basic for all:
                                                                                       grade line, ft/ft.
                                                             √
                                                        v = C rs              (5.18a)     Equation (5.26) is the equivalent Manning formula using
                                                                                       SI units:
                                    where v = average velocity, ft/s; C = coefficient; r = hydraulic
                                    radius, which is defined as the cross-section area divided by  v = (1∕n)(r) 0.67 (s) 0.5  (SI units)  (5.26)
                                    the wetted perimeter, ft; and s = slope of water surface or
                                                                                       where v = velocity, m/s; n = coefficient of roughness, dimen-
                                    energy gradient:
                                                                                       sionless; r = hydraulic radius, m; and s = slope of energy
                                                        r = A∕P w              (5.19)  grade line, m/m.
                                                                                          For a pipe flowing full, the hydraulic radius, Eq. (5.19),
                                                        s = h ∕L               (5.20)  becomes
                                                             f
                                    where r = hydraulic radius, ft or m; A = cross-section area,   r = [(  ∕4)D ]∕(  D) = D∕4     (5.27)
                                                                                                             2
                                            2
                                    ft 2  or m ; and P = wetted perimeter, ft or m; s =
                                                    w
                                    slope of water surface, dimensionless; and L = pipe length,  where r = hydraulic radius, ft or m; and D = pipe diameter,
                                    ft or m.                                           ft or m.
                                        The coefficient C can be obtained by using one of the  Substituting for r into Eqs. (5.25) and (5.26), the follow-
                                    following expressions:                             ing Manning equations are obtained for practical engineering
                                                                                       designs for circular pipes flowing full:
                                      Chezy expression: C = (8g∕f) 0.5         (5.21)
                                                                                         v = (0.59∕n)(D) 0.67 (s) 0.5  (US customary units) (5.28)
                                      Manning expression: C = (1.486∕n)(r) 1∕6  (5.22)
                                                                                         Q = (0.46∕n)(D) 2.67 (s) 0.5  (US customary units) (5.29)
                                      Bazin expression: C = 157.6∕(1 + mr −0.5 )  (5.23)
                                                                                                        3
                                                                                       where Q = flow rate, ft /s; v = velocity, ft/s; D = pipe diame-
                                      Kutter expression: C = (41.65 + 0.00281∕s + 1.811∕n)∕  ter, ft; s = slope of energy grade line, ft/ft; and n = roughness
                                                         [1 + (n∕r 0.5 )(41.65 + 0.00281∕s)]
                                                                                       coefficient, dimensionless. For SI measurements:
                                                                               (5.24)                    0.67  0.5
                                                                                             v = (0.40∕n)(D)  (s)   (SI units)    (5.30)
                                        In the preceding expressions, f, n, and m are the friction       2.67  0.5
                                                                                            Q = (0.31∕n)(D)  (s)    (SI units)    (5.31)
                                    or roughness factors determined by hydraulic experiments
                                                                                                         3
                                    using water only. Of the above hydraulic equations, Robert  where Q = flow rate, m /s; v = velocity, m/s; D = pipe diam-
                                    Manning’s expression is commonly used for both open chan-  eter, m; s = slope of energy grade line, m/m; and n = rough-
                                    nels and closed conduits. Combining Eqs. (5.18a) and (5.22)  ness coefficient, dimensionless.
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