60 likes | 164 Views
EXAMPLE 1. SECTION 1. SECTION 2. TOTAL. k = 4 ft/hour = 6,67x10 -2 ft/min. EXAMPLE 2. Determination of h. Section 1. Section 2. h = 38.33 cm. Flow Nets: an example. Flow Nets: an example. The flow net is drawn with: m = 5 n = 17. Flow Nets: the solution.
E N D
EXAMPLE 1 SECTION 1 SECTION 2 TOTAL k = 4 ft/hour = 6,67x10-2 ft/min
EXAMPLE 2 Determination of h Section 1 Section 2 h = 38.33 cm
Flow Nets: an example The flow net is drawn with: m = 5 n = 17
Flow Nets: the solution • Solve for the flow per unit width: q = (m/n) K h = (5/17)(150)(35) = 1544 ft3/day per ft
From the flow net, the total head loss, H, is 6.2 -2.2 = 4.0 meters. • There are 6 flow channels (m) and 21 head drops along each flow path (n): • Q = (KmH/n) x dam length = (0.527 m/day x 6 x 4m / 21) x (dam length) = 0.60 m3/day per m of dam • = 43.4 m3/day for the entire 72-meter length of the dam