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HGL

Sewer Hydraulics. Gravity flow: full flow. Gravity flow: Partial flow. Pressure flow: full flow. L. h. HGL. HGL. HGL. S= HGL = slope of the sewer R =A p /P p. S= h/L = slope of the HGL R=D/4. S= HGL = slope of the sewer R = A f /P f = D/4. Manning equation.

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HGL

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  1. Sewer Hydraulics Gravity flow: full flow Gravity flow: Partial flow Pressure flow: full flow L h HGL HGL HGL S= HGL = slope of the sewer R =Ap/Pp S= h/L = slope of the HGL R=D/4 S= HGL = slope of the sewer R = Af/Pf = D/4

  2. Manning equation Many formulas are used to solve the flow parameters in sewers were discussed in the hydraulic course. The most used formula for sanitary sewers is Manning equation: R = Hydraulic radius (Area/ wetted parameter) S = slope. n = manning coefficient. D = pipe diameter. Q = flow rate. Note:Equation 1 is used for calculating the velocity in pipes either flowing full or partially full. Equation 2 same for flow rate. Vƒ = velocity flowing full VP = velocity flowing partially Note:Equations 3 and 4 are the same as Equation 1, but they are written using the subscript (ƒ) and (P), to indicate flowing full and partially full, respectively:

  3. In sanitary sewers the flow is not constant; consequently the depth of flow is varying as mentioned above. In this case it is difficult to find the hydraulic radius to apply Manning’s equation. For partially full pipe the following relations are applied: D Ө d D • d = partial flow depth. • R = Hydraulic radius (P = partial, ƒ = full) • Ө = flow angle in degrees. • Maximum capacity of the pipe when d/D = 0.95 • A = Flow area. • Maximum velocity in the pipe occurs at d/D=0.81 Ө d

  4. D Ө d Example 1 Find the diameter of the pipe required to carry a design flow of 0.186 m3/s when flowing partially, d/D = 0.67, slope = 0.4%, n = 0.013 use the relations of partial flow. Solution

  5. It is noticed that it is quite long procedure to go through the above calculations for each pipe in the system of large numbers of pipes. The alternative procedure is to use the nomographs of Mannings equation and the partial flow curves.

  6. Example 2 Find the diameter of the pipe required to carry a design flow of 0.186 m3/s when flowing partially, d/D = 0.67, slope = 0.4%, n = 0.013 using the nomographs and partial flow curves . Solution

  7. D d Partial Flow Curves d/D d/D Qp/Qf Vp/Vf

  8. Partial Flow Curves Q &V d / D 1.1 2

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