OPEN CHANNELS I CHAPTER 4 (1) - ROBERSON ET AL., WITH ADDITIONS OPEN CHANNEL FLOW: FLOW OF A LIQUID IN A CONDUIT IN WHICH THE UPPER SURFACE OF THE LIQUID (THE FREE SURFACE) IS IN CONTACT WITH THE ATMOSPHERE. EXAMPLE: WATER FLOW IN RIVERS, IRRIGATION CANALS, SEWER LINES FLOWING PARTIALLY FULL, STORM DRAINS, STREET GUTTERS. Cabana irrigation canal, Peru. IN A NATURAL CHANNEL, THE PROBLEM IS OF PREDICTING THE WATER SURFACE PROFILE, OR TO ESTIMATE VELOCITY AND DEPTH. STEADY UNIFORM FLOW: WHEN DISCHARGE, VELOCITY AND DEPTH ARE CONSTANT. STEADY GRADUALLY VARIED FLOW: WHEN VELOCITY AND DEPTH ARE A FUNCTION OF DISTANCE. UNSTEADY FLOW: WHEN DISCHARGE, VELOCITY, AND DEPTH ARE A FUNCTION OF TIME AND DISTANCE. ALL ASPECTS OF THE DESIGN PROBLEM MUST BE CONSIDERED, INCLUDING WATER AND SEDIMENT TRANSPORT. UNUSUAL HAZARDS: -- LANDSLIDES -- SEDIMENTATION -- ACCIDENTAL GATE OPERATION -- BREACH OF EMBANKMENT IN STEEP CHANNELS, THERE IS THE POSSIBILITY OF ROLL WAVES. Roll waves in Cabana lateral canal, Peru. ROLL WAVES CAN BE AVOIDED WITH SEQUENTIAL DROPS. Drops in steep La Joya canal, Peru. STEADY UNIFORM FLOW: ARTIFICIAL CHANNELS: THERE IS NO CHANGE IN VELOCITY AND FLOW DEPTH. VELOCITY HEAD IS CONSTANT. EGL AND W.S. WILL HAVE SAME SLOPE AS CHANNEL BOTTOM. PRISMATIC CHANNEL: CHANNEL OF CONSTANT CROSS SECTION. NORMAL DEPTH: FLOW IN UNIFORM CHANNEL. NET FORCE ACTING ON FLOW IS ZERO. ALL FORCES ARE BALANCED. GRAVITATIONAL FORCE EQUALS RESISTANCE FORCE. THE FORCE BALANCE EQUATION IS: W sin θ = τo P Δx τo = SHEAR STRESS ALONG THE BOTTOM θ = ANGLE OF INCLINATION OF CHANNEL BOTTOM P = WETTED PERIMETER OF CHANNEL CROSS SECTION Gravitational and resistance forces under uniform flow. SINCE: W = γ A Δx γ A Δx sin θ = τo P Δx τo = γ (A/P) sin θ BECAUSE θ IS SMALL: sin θ ≅ tan θ = So τo = γ (A/P) So τo = γ R So R = HYDRAULIC RADIUS THE SHEAR STRESS CAN ALSO BE EXPRESSED AS THE QUADRATIC LAW: τo = Cd ρ V2 = Cd (γ/g) V2 Cd IS A DRAG OR FRICTION COEFFICIENT, WHICH IS DIMENSIONLESS N/M2 = Cd [(N/M3)/(M/S2) ] (M2/S2) Cd ρ V2 = γ R So V = (g/Cd) 0.5 (R So) 0.5 V = C (R So) 0.5 C = CHEZY COEFFICIENT Cd = f/8 IF f = 0.024, Cd = 0.003 V = (8g/f) 0.5 (R So) 0.5 THIS IS A FORM OF THE DARCY-WEISBACH EQUATION APPLIED TO OPEN CHANNELS V2= (8g/f) R So V2= (8g/f) R (hf/L) V2= (2g/f) 4R (hf/L) hf = f [L/(4R)] (V2/ 2g) THIS IS THE SAME AS DARCY-WEISBACH, BUT WITH (4R) INSTEAD OF D. SINCE 4R = D IN A PIPE, THE HYDRAULIC RADIUS IS AN EQUIVALENT MEASURE OF CROSS SECTION IN AN OPEN CHANNEL R = A/P = [πD2/4]/(πD) = D/4 V = (8g/f)0.5 (R So)0.5 Q = A V = (8g/f) 0.5 A (R So)0.5 IN ROCK-BEDDED STREAMS: f = [1.2 + 2.3 log (R/d84) ]-2 d84 = diameter for which 84% by weight is smaller. Rock Creek near Darby, Montana. HYDRAULIC DEPTH IN A CHANNEL: D = A/T HYDRAULIC RADIUS: R = A/P IN A WIDE CHANNEL: T ≅ P THEREFORE: R ≅ D V = (8g/f)0.5 (D So)0.5 V2 = (8g/f) D So So = (f/8) [V2/(gD)] So = (f/8) F2 F IS THE FROUDE NUMBER = V /(gD)0.5 For F = 1:   So = Sc = f/8 So = Sc F2 EXAMPLE 4-1 DETERMINE THE DISCHARGE IN A LONG, RECTANGULAR CONCRETE CHANNEL THAT IS 5-FT WIDE, SLOPE = 0.002, AND WITH SECTION DEPTH = 2 FT. ASSUME ROUGHNESS OF CONCRETE ks = 0.005 ft. (range for concrete 0.001-0.01) A = (2) (5) = 10 R = A/P = 10 /( 2 + 2 + 5) = 1.11 FT. ks / (4R) = (0.005) /(4 X 1.11) = 0.00113 ASSUME REYNOLDS NUMBER Re = 106 FROM MOODY DIAGRAM: f ≅ 0.02 FIRST TRIAL OF Q: Q = (8g/f) 0.5 A (R So)0.5 Q = [(8)(32.2)/(0.02) ]0.5 10 [(1.11)(0.002)]0.5 Q = 53.5 CFS V = Q/A = 5.35 FPS ASSUME TEMPERATURE = 60 oF KINEMATIC VISCOSITY ν = 1.22 X 10-5 FT2/SEC Re = V (4R)/ν = 5.35 (4) (1.11) / 0.0000122 = 1.95 X 106 FROM MOODY DIAGRAM: f ≅ 0.02 THEN: Q = 53.5 CFS. THE MANNING EQUATION WHEN THE CHEZY COEFFICIENT IS NOT A CONSTANT, BUT RATHER A FUNCTION OF THE HYDRAULIC RADIUS: C = (1.49/n) R1/6 CHEZY EQUATION: V = C (R So)1/2 MANNING EQUATION: V = (1.49/n) R1/6 (R So)1/2 MANNING EQUATION: V = (1.49/n) R 2/3 So1/2 DISCHARGE: Q = (1.49/n) A R 2/3 So1/2 CALCULATION OF EXAMPLE ASSUMING n = 0.013 RESULT: Q = 54.83 CFS (COMPARE WITH PREVIOUS RESULT). THE RESISTANCE FACTOR IN OPEN CHANNELS IS A FUNCTION OF THE CHANNEL ROUGHNESS, AMONG OTHER FACTORS. THE RESISTANCE FACTOR IN FLOOD PLAINS. FLOW IN CIRCULAR CONDUITS HIGHWAY CULVERTS AND URBAN SEWERS ARE EXAMPLES OF CIRCULAR OPEN CHANNELS. THE DISCHARGE IS PROPORTIONAL TO A R2/3 THE VELOCITY IS PROPORTIONAL TO R2/3 THE RATIO Q/Qo = (A R2/3)/(Ao Ro2/3) IS INDEPENDENT OF THE ROUGHNESS OR SLOPE, WHERE Q IS DISCHARGE FOR A GIVEN DEPTH AND Qo IS THE DISCHARGE WHEN THE CONDUIT FLOW FULL (CONSTANT n ASSUMED). THE MAXIMUM DISCHARGE OCCURS AT A DEPTH LESS THAN FULL. AS THE CONDUIT GETS FULL, THE WETTED PERIMETER INCREASES MUCH FASTER THAN THE AREA, THUS DECREASING THE HYDRAULIC RADIUS. EXAMPLE 4-2 DETERMINE THE DISCHARGE IN A 3-FT DIAMETER SEWER PIPE IF THE DEPTH OF FLOW IS 1 FT, SLOPE = 0.0019, n = 0.012. USE FIG. 4-6, WITH n = 0.012 and S= 0.0019. FROM THIS FIGURE: Qo = 30 CFS. y/do = 0.3333 FROM FIG. 4-5: Q/Qo = 0.2 Q = 0.2 Qo = 0.2 (30) = 6 CFS. COMPARE WITH ONLINE CALCULATION. RESULT: Q = 7.5 CFS (THERE IS A SLIGHT DIFFERENCE). FLOW IN CONDUITS OF TRAPEZOIDAL CROSS SECTION TO ASSIST IN SOLVING PROBLEMS INVOLVING THE FLOW IN TRAPEZOIDAL CHANNELS: THE DIMENSIONLESS FACTOR AR2/3/b8/3 IS PLOTTED VS y/b (Fig. 4-7). EXAMPLE 4-3 DETERMINE THE NORMAL DEPTH FOR A TRAPEZOIDAL CHANNEL WITH THE FOLLOWING CHARACTERISTICS: SIDE SLOPES OF 1 VERTICAL TO 2 HORIZONTAL (z = 2); b = 8 ft, Q = 200 cfs, So= 0.001, n = 0.012. THE MANNING EQUATION (EQ. 4-7) CAN BE WRITTEN AS: A R2/3 = Q n / (1.49 So1/2 ) A R 2/3/b8/3 = Q n / (1.49 So1/2b8/3) SINCE: Q = 200 CFS, n= 0.012; So= 0.001; b= 8 ft EVALUATE THE RIGHT-HAND SIDE: A R2/3/b8/3 = 200 (0.012) / [1.49 (0.001)1/2(8)8/3] A R 2/3/b8/3 = 0.199 FROM FIG. 4-7:   y/b = 0.33. THEN: y = 0.33 (8) = 2.64 FT. COMPARE WITH ONLINE CALCULATION. RESULT: y = 2.63 FT (SAME RESULT). DESIGN OF ERODIBLE CHANNELS THE CHANNEL WILL ERODE IF THE VELOCITY IS TOO LARGE. TWO METHODS: -- PERMISSIBLE VELOCITY METHOD -- PERMISSIBLE TRACTIVE FORCE METHOD. DESIGN PROCEDURE: -- CHOOSE SIDE SLOPE z THAT WOULD BE STABLE (TABLE 4.2). -- CHOOSE MAXIMUM PERMISSIBLE VELOCITY V (TABLE 4.3). -- WITH Q, V, n, S, and z, CALCULATE b AND y. All-American Canal, near Yuma, California. EXAMPLE 4-4 AN UNLINED IRRIGATION CANAL IS TO BE CONSTRUCTED IN A FIRM LOAM SOIL. THE SLOPE IS 0.0006, AND Q = 100 CFS. DETERMINE THE CHANNEL DIMENSIONS z, b AND y. CHOOSE SIDE SLOPE z = 1.5 H : 1 V (TABLE 4-2) CHOOSE V= 2.5 FPS (TABLE 4-3) CHOOSE n= 0.020. CALCULATE HYDRAULIC RADIUS FROM MANNING EQUATION: V = (1.49/n) R2/3S 1/2 R = [nV/(1.49S1/2)]3/2 = R = [(0.02)(2.50) /(1.49 X 0.00061/2)] 3/2 = 1.6 A = Q/V = 100/2.5 = 40 sq ft. P = A/R = 40/1.6 = 25. A = y ( b + 1.5 y) = by + 1.5 y2 = 40 P = b + 2 (y2 + 1.52y2 ) 1/2 = b + 3.61y = 25 b = 18.1; y = 1.91; say b = 18.0 and y = 2.0 COMPARE WITH ONLINE CALCULATION, ASSUMING B = 18 FT. RESULT: V = 2.5 FPS; y = 1.92 FT (SAME RESULT). PROJECT SCOPE WATER RESOURCES PROJECTS INCLUDES OTHER STRUCTURES IN ADDITION TO CHANNELS INTAKE WORKS, FLUMES, CHECKS, DROPS, AND TRANSITIONS. Crossing of Tinajones Feeder Canal with Chiriquipe Creek, Peru. Creek crossing with Wellton-Mohawk Feeder Canal, Arizona. Drops in Cabana lateral canal, Peru. DIVERSION DAM IS CONSTRUCTED SO AS TO MAINTAIN A WATER LEVEL HIGH ENOUGH IN THE RIVER TO BE ABLE TO ALWAYS DIVERT THE REQUIRED FLOW INTO THE MAIN IRRIGATION CANAL. THE INTAKE TO THE MAIN CANAL CONSISTS OF A CANAL ENTRANCE STRUCTURE, INCLUDING GATES FOR CONTROLLING THE DISCHARGE INTO THE CANAL. WASTEWAYS ARE OFTEN PROVIDED AT INTERVALS ALONG THE MAIN CANAL IF AN EMERGENCY DEVELOPS WHERE DOWNSTREAM WATER USE IS STOPPED. A FLUME IS USED TO CONVEY WATER ACROSS A DEPRESSION. Dulzura flume (conduit), downstream of Barrett Dam, San Diego County (Note that the flume was overflowing on March 8, 2005). AN INVERTED SIPHON COULD BE USED FOR THIS PURPOSE. Inverted siphon, Cabana irrigation canal, Peru. TRANSITIONS ARE REQUIRED FOR SMOOTH PASSAGE OF WATER FROM CANAL TO FLUME. CHECK STRUCTURE CONTROLS WATER SURFACE LEVEL ON A CANAL TO MAINTAIN HIGH WATER SO THAT WATER CAN BE DIVERTED INTO SECONDARY CANAL. DROP STRUCTURE IS A VERY STEEP CHANNEL. Tinajones Feeder canal drop, Peru. BAFFLED DROP STRUCTURE IS SHOWN IN FIG. 4-10. SECONDARY CANAL DRAWS WATER FROM THE CANAL THROUGH THEIR OWN INTAKE STRUCTURES. FARM LATERALS TAKE WATER FROM THE SECONDARY CANALS. STEADY-NONUNIFORM FLOW IN OPEN CHANNELS THE ENERGY EQUATION IS: p1/γ + α1V12/(2g) + z1 = p2/γ + α2V22/(2g) + z2 + hL p1/γ = pressure head at point 1 α1V12/(2g) = velocity head at point 1 z1 = elevation (head) at point 1 hL = head loss between points 1 and 2 p1/γ + z1 = y1 + So Δx p2/γ + z2 = y2 ASSUMING α1 = α2 = 1: y1 + V12/(2g) + So Δx = y2 + V22/(2g) + hL FOR THE SPECIAL CASE WHEN CHANNEL IS HORIZONTAL (So = 0), AND THE HEAD LOSS IS ZERO (hL = 0): y1 + V12/(2g) = y2 + V22/(2g) THIS MAY BE THE CASE OF A HORIZONTAL AND SHORT CHANNEL. SPECIFIC ENERGY: E = y + V2/(2g) THE CONTINUITY EQUATION IS: Q = A1V1 = A2V2 y1 + Q2/(2gA12) = y2 + Q2/(2g A22) BECAUSE A = f (y), FOR A GIVEN DISCHARGE, THE SPECIFIC ENERGY IS SOLELY A FUNCTION OF FLOW DEPTH. FUNCTION y vs E IS SHOWN BELOW. FOR A GIVEN SPECIFIC ENERGY, THE DEPTH CAN BE LARGE OR SMALL. FOR LOW DEPTH, VELOCITY IS HIGH; FOR HIGH DEPTH, VELOCITY IS LOW. POTENTIAL ENERGY (DEPTH) AND KINETIC ENERGY (VELOCITY). FLOW UNDER A SLUICE GATE IS AN EXAMPLE OF FLOW WITH TWO DEPTHS. THE LARGE DEPTH AND LOW KINETIC ENERGY OCCURS UPSTREAM OF THE GATE. THE LOW DEPTH AND HIGH KINETIC ENERGY OCCURS DOWNSTREAM OF THE SLUICE GATE. THE TWO DEPTHS AT WHICH SAME FLOW CAN OCCUR ARE CALLED ALTERNATE DEPTHS. IF WE MAINTAIN SAME RATE OF FLOW, BUT SET THE GATE WITH A LARGER OPENING, THE U/S DEPTH WILL DROP AND THE D/S DEPTH WILL RISE. THUS, WE HAVE DIFFERENT ALTERNATE DEPTHS AND A SMALLER VALUE OF SPECIFIC ENERGY. A POINT IS REACHED WHEN THE SPECIFIC ENERGY IS A MINIMUM, AND ONLY A SINGLE DEPTH OCCURS. AT THIS POINT, THE FLOW IS TERMED CRITICAL. 090913