DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING Urban Drainage: Hydraulics. Solutions to problem sheet 2: Flows in open channels

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1 DEPRTMENT OF CIVIL ND ENVIRONMENTL ENGINEERING Urban Drainage: Hydraulics Solutions to problem sheet 2: Flows in open channels 1. rectangular channel of 1 m width carries water at a rate 0.1 m 3 /s. Plot the corresponding specific energy diagram. Make necessary assumptions and use the diagram to solve the following problems: (a) The channel starts from a circular pipe of 60 cm in diameter. The cross-section gradually changes to the rectangle of 1 m width. Find the flow depth at the rectangular part of the channel if the entry pipe is half-full. [ ns: h 4 cm ] (b) The flow depth is 5 cm. The channel bed has a 7.5 cm high step. Find the water depth after the step. Find the critical step height when it starts influencing the flow upstream. [ ns: h 7 cm; z c 10 cm ] (c) What is the water depth after the step if it is 24 cm upstream the step? [ ns: h 15.5 cm ] (d) Water depth upstream a sluice gate is 24 cm. Find the depth after the gate. Expression for the specific energy E of the flow in an open channel is E = U 2 2 g + h = Q 2 2 g (h) 2 + h, [ ns: h 5 cm ] where (h) is the flow cross-section area, h flow depth, U mean flow velocity, Q flow rate and g gravity acceleration. For a rectangular channel of width b it becomes E = Q2 2 g b 2 + h. The graph of the function E = E(h) known as the specific energy diagram can now be plotted. n example of such a plot produced in EXCEL can be found in the provided file Q1.xls. See also figure Fig. Q1. (a) The specific energy for a half-full circular pipe with h = d/2 = 0.3 m is E 1 = Q 2 2 g (π /2) 2 + h = m 6 /s m/s 2 (π m m m. /2) 2 The corresponding point on the specific energy diagram is point (see Fig. Q1(a)). Note, that this point does not belong to the specific energy curve of the rectangular channel. For a properly designed transitional section between circular and rectangular parts of the channel we can neglect frictional losses. If the channel s bottom is horizontal the specific energy of the rectangular section is the same as the specific energy of the circular one: Z 2 = Z 1. The width of the channel increases during the transition. In such situations the flow tends to become shallower. Therefore the solution will be at the supercritical branch of the specific energy diagram for the rectangular channel (point, Fig. Q1(a)). The corresponding depth is approximately 4 cm within the available accuracy. 1

2 (a) (b) E 2 =E 1 c C (c) (d) E 2 =E 1 Fig. Q1 (b) Energy conservation for flow over the step is E 2 = E 1 z. The flow before the step is represented by the point of the specific energy diagram (Fig. Q(b)). ccording to the above equation the flow after the step is represented by the point. (c) nalogous to (b), see Fig. Q1(c). (d) The flow through a sluice gate is a rapidly varied flow. The flow upstream of the gate is subcritical and for a normally operating gate the flow becomes supercritical after the gate. lso, for a normally operating gate we can neglect frictional losses. Therefore the values of the specific energy before the gate (point, Fig. Q1(c)) and after the gate (point, Fig. Q1(c)) are the same. 2. Water depth in a 2 m wide rectangular channel is 50 cm. The bottom and walls of the channel are covered by concrete with Manning s roughness coefficient n = Calculate channel flow rate if its slope is Specify if the flow in the channel is sub- or supercritical (steep or mild slope)? [ ns: Q = 0.89 m 3 /s; mild ] 1 detailed table of Manning roughness coefficients can be found in: Mays, Larry W. Hydraulic Design Handbook. Chapter 3: Hydraulics of Open-Channel Flow. ppendix 3.. pp: McGraw-Hill. Online version available at: EXT KNOVEL DISPLY bookid=730 2

3 The mean velocity U of a steady uniform flow in an open channel of slope S is specified by the Chezy equation U = C R S, where the Chezy coefficient C can be calculated using Manning s equation C = R 1/6 /n. The corresponding flow rate is then Q = R1/6 R S. n For a rectangular channel of the depth h and width b the hydraulic radius R = /P is R = The value of the flow rate can now be found as b h b + 2 h. Q = (b h) 1 n ( ) b /3 S 1/ m 3 /s. b + 2 h The Froude number of the flow = Q g h is smaller than 1. The flow is therefore subcritical and the slope is mild. 3. wide (the depth is much smaller than the width) rectangular channel has a slope S = The bed and walls of the channel are paved with concrete. The value of Manning s roughness coefficient is n = and the discharge per unit width of the channel is q = 3 m 2 /s. Find the depth and velocity of the flow. Is the slope steep or mild? [ ns: h = 0.48 m; U = 6.25 m/s; steep ] The Chezy equation for velocity of a steady uniform flow on a slope S is: U = C R S. The Chezy coefficient can be found using Manning s equation C = R1/6 n, where the hydraulic radius R for a rectangular channel of width b and depth h R = b h 2 h + b can be reduced to R h for a wide channel since b h. Substitution into the Chezy equation gives the following equation for the steady uniform flow rate q = h U per unit of the width of the channel q = 1 n h5/6 S. The depth h of the steady uniform flow on a slope is therefore ( ) q n 3/5 h =. S 3

4 Substituting the values we have the value of the flow depth h = ( ) / m 0.02 and the corresponding flow speed The Froude number of the flow is U = q/h 3/0.48 = 6.25 m/s and since the value of the Froude number is grater than 1 the flow is supercritical and the slope is steep. 4. Water from a reservoir enters an overflow spillway via a broad-crested weir (Fig. Q4). The flow in the spillway is supercritical and it can be shown that the transition from subcritical to supercritical flow occurs near the weir crest. (a) Make necessary assumptions and derive the relation between the spillway flow rate and the level of water in the reservoir H if the width of the weir is 46 m. (b) The gauging station downstream the spillway showed the peak flow of 97.2 m 3 /s. Calculate the corresponding level of water in the reservoir. [ ns: H 1.15 m ] H h c Fig. Q4 Critical condition occurs when Froude number is 1. Therefore on the crest we have = Q b h 3/2 g 1/2 = 1, which gives Q = b g 1/2 h 3/2, where h = h c. The corresponding specific energy E = E c = U 2 2 g + h = 1 ( ) U 2 2 h g + h = 1 h F r2 + h = 3 2 h since 1. Let us neglect energy losses in the flow inside the reservoir before the spillway crest and let chose the crest as the datum for measuring the energy. This means that the total 4

5 energy at the crest equal to the corresponding specific energy and the total energy inside the reservoir is H since we can neglect kinetic energy there. Therefore H = E c = 3 2 h c and our final equation which relates Q and H is Q = b g 1/2 ( 2 3 H ) 3/2. If the measured flow rate is Q = 97.2 m 3 /s then the corresponding water level in the reservoir is H = 3 2 ( ) Q 2/3 b g 1/2 = 3 ( 97.2 m 3 ) 2/3 /s 2 46 m (9.82 m/s 2 ) 1/2 = 1.15 m. 5. Figure Q5 shows the cross section of a road. The road width W is 10 m, the height of the kerb h is 15 cm and the angle between the kerb and the road surface β is 85. Find the required distance along the road between storm drain inlets to prevent storm water overflow for the rainfall of 100 mm per hour if the road slope S = and the value of Manning s roughness coefficient is n = [ ns: L 289 m ] h β W Fig. Q5 Solution outline: The flow demand from the rainfall I is Q D = I L W 2. The maximal depth of the flow in the gutter is h. The corresponding cross-section area, wetted perimeter and hydraulic radius are = 1 ( 2 h2 tan β; P = + 1 ) ; R = h ( ) sin β cos β 2 cos β + 1 pplying the Chezy equation with Manning s coefficient we can find the channel flow capacity as Q C = 1 n R2/3 S 1/2. The distance L between drain inlets must be chosen from the condition that the demand does nor exceed the capacity Q D Q C. Therefore L 2 Q C I W. 5