Determine the flow area of storm water, Civil Engineering

Storm water flows at a depth of 4 ft in the natural channel as shown in the figure. The channel has a slope of 0.0l ft/ft and n = 0.025.

487_Determine the flow area of Storm water.png

1. Determine the flow area:

    Area 1: Tan σ1 = opp/adj      adj = opp/Tan σ1       adj = 4/1.19 = 3.36 ft

               (1/2)(b)(h) = (1/2)(3.36)(4) = 6.7 ft2

     Area 2 = (b)(h) = (9)(4) = 36 ft2

    Area 3 = Tan σ2 = opp/adj      adj = opp/Tan σ2     adj = 4/0.84 = 4.76 ft

                 (1/2)(b)(h) = (1/2)( 4.76)( 4) = 9.5 ft2  

    Total Area= Area 1 +Area 2 +Area 3 = 6.7 ft2 + 36 ft2 + 9.5 ft2 = 52 ft22. 

2. Determine the flow velocity:  

v = (1.49/n) (R)0.67 (S)0.5       R = A/P      Ptotal = PArea 1 + PArea 2 + PArea 3

PArea 1 = hypArea 1    sin σ1 = oppArea 1 /hypArea 1      hypArea 1 = 4/sin σ1 = 5.2 ft

PArea 3= hypArea 3    sin σ2 = oppArea 3 /hypArea 3      hypArea 3 = 4/ sin σ1 = 6.2 ft

    P = PArea 1 + PArea 2  PArea 3 = 5.2 ft + 9 ft + 6.2 ft = 20.4 ft

    R = A/P = 52 ft2 / 20.4 ft = 2.5 ft

    v = (1.49/0.025)(2.5)0.67 (0.01)0.5 = (59.6)(1.88)(0.1) = 11.2 ft/sec

3. Determine the flow rate:    Q = vA = (11.2 ft/sec)(52 ft2) = 582 ft3/sec  

4. Determine the hydraulic grade line (HGL) at this channel section:  

    The hydraulic grade line is the depth of flow HGL = 4 ft  

5. Determine the energy grade line (EGL) at this channel section:  

    P/γ = 4ft    Z = 0 ft (bottom)    v2/2g = (11.2)2/64.4 = 1.95 ft

    EGL = 4ft + 0 ft + 1.95 ft = 5.95 ft

Posted Date: 5/22/2013 1:37:18 AM | Location : United States







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