Lecture 2.-sewer-hydraulics1
- 1. L2. Sewer Hydraulics The Islamic University of Gaza- Civil Engineering Department Sanitary Engineering- ECIV 4325 Based on Dr. Fahid Rabah lecture notes
- 2. HGL HGL HGL h L Gravity flow: full flow Gravity flow: Partial flow Pressure flow: full flow S= HGL = slope of the sewer R = Af/Pf = D/4 S= HGL = slope of the sewer R =Ap/Pp S= h/L = slope of the HGL R=D/4 Sewer Hydraulics
- 3. 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: 2 1 3 2 1 SR n V = …………………………………………………. …...(1) 2 1 3 8 312.0 SD n Q = ………………………………………………...(2) Note: Equation 1 is used for calculating the velocity in pipes either flowing full or partially full. Equation 2 same for flow rate. R = Hydraulic radius (Area/ wetted parameter) S = slope. n = manning coefficient. D = pipe diameter. Q = flow rate. 2 1 3 21 S f R nf V = … … … … … … … … … … … … … … … … … … … … … … ..… ..(3) 2 1 3 21 SpR npV = … … … … … … … … … … … … … … … … … … … … … ..… … .(4) 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: Manning equation
- 4. 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: −= 2 cos1 2 1 θ D d ……………………… (5) −= π θθ 2360 Sin Af Ap …………………… (6) −= πθ θ 2 360 1 Sin Rf Rp ……………………..(7) 3 2 = f p f p R R V V …………………………… (8) = ff pp f p VA VA Q Q ………………………….(9) 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 D d D Ө
- 5. 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 1. Find the flow angle : ? −= 2 cos1 2 1 θ D d = 0.67 From this relation ? = 219.75o 2. Find Qp/Qƒ : −= πθ θ 2 360 1 Sin Rf Rp = ( ) − 75.2192 75.219360 1 π Sin = 1.167 −= π θθ 2360 Sin Af Ap = − π2 75.219 360 75.219 Sin = 0.712 3 2 = f p f p R R V V = (1.167)2/3 = 1.1088 = ff pp f p VA VA Q Q = 0.712*1.1088 = 0.789 d D Ө
- 6. 4. Find the diameter of the pipe (D): 2 1 3 8 312.0 SD nf Q = = ( ) 2 1 004.0 8 013.0 312.0 3D = 0.2355 D = (0.15517)3/8 = 0.497 ? 0.50m ? 20" (design pipe diameter) 5. Find the partial flow velocity (VP): 2 1 3 21 S f R nf V = = ( ) 2 1 3 2 004.0 4 497.0 013.0 1 = f V = 1.203 m/s f V f Q f A = = 203.1 2355.0 = 0.196 m2 (or 4 2D f A π = ) Ap = Aƒ *0.712 = 0.196* 0.712 = 0.139 m2 Vp = Vƒ *1.109 = 1.203*1.109= 1.33 m/s 3. Calculate Qƒ: Qƒ = 789.0 P Q = 789.0 186.0 = 0.2355 m3 /s 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.
- 7. 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 From partial flow curves: • Start from the Y-axis with 67.0=⇒ D d , and draw a horizontal line until you intersect the Q curve (for n = constant, the dashed line), then draw a vertical line to intersect the X- axis at .78.0=⇒ f Q P Q • Extend the horizontal line until it intersects the velocity curve, then draw a vertical line to intersect the X-axis (for n = constant, the dashed line) at 12.1=⇒ f V P V . •Calculate Qƒ : 78.0 P Q f Q =⇒ = 238.0 78.0 186.0 = m3 /s
- 8. Use the nomographs for pipes flowing full to find D: • Locate the slope ( 0.004) on the “S” axis. • Locate the manning coefficient “n” ( 0.013) on the “n” axis. • Draw a line connecting “S” and “n” and extended it until it intersects the Turning Line. • Locate the Qƒ ( 0.238 m3 /s) on the “Q” axis. • Draw a line connecting “Qƒ” and the point of intersection on the Turning Line and find the diameter “D” by reading the value that this line intersect the D axis at. D = 500 mm = 20”. • find the Velocity “Vƒ” by reading the value that this line intersect the V axis at. Vƒ = 1.2 m/s. • Calculate Vp = Vƒ * 1.12 = 1.2*1.12= 1.34 m/s.
- 10. 21.1 Partial Flow Curves Q &V d / D