HYDRAULIC PRINCIPLES FOR FLOOD CONTROL CHANNELS

FLOOD CONTROL CHANNELS ARE CONSTRUCTED FOR THE PURPOSE OF CONVEYING HEAVY STORM WATER FLOWS THROUGH AND FROM AREAS WHICH WOULD OTHERWISE BE INUNDATED, USUALLY RESULTING IN PROPERTY DAMAGESAND LOSS OF LIVES

DESIGN RATIONALE:

The hydraulic design of a flood control channel must result in a safe, efficient, reliable and cost effective project with appropriate consideration of environmental and social aspects.

SAFETY:Potential hazards to humans and property, creation of a false of security, consequences of flows exceeding the improved channel capacity.

EFFICIENCY:Channel cross section, plan and bottom profile configurationto optimize conveyance and operation and maintenance.

RELIABILITY:Ability to achieve project purposes throughout the project economic life; proper functioning of facilities such as pumps, gates, trash fenders, etc.

COST EFFECTIVENESS:Initial, operational, maintenance and replacement costs optimized on an annual costs basis.

ENVIRONMENTAL AND SOCIAL:Fish and wildlife, beautification, recreational opportunities,handicap access and mitigation of adverse impacts.

OPEN CHANNEL FLOW & its CLASSIFICATION

A. STEADY FLOW1) UNIFORM FLOW

2) VARIED FLOWa. Gradually varied flowb. Rapidly varied flow

B. UNSTEADY FLOW1) UNSTEADY UNIFORM FLOW (rare)

2) UNSTEADY FLOW (i.e., unsteady varied flow)a. Gradually varied unsteady flowb. Rapidly varied unsteady flow

Steady flowUnsteady flowE.L.E.L.E.L.E.L.E.L.W.S.W.S.W.S.W.S.W.S.UNIFORMGRADUALLY VARIEDRAPIDLY VARIEDGRADUALLY VARIEDRAPIDLY VARIEDVVSOME TYPES OF OPEN CHANNEL FLOW

STEADY UNIFORM FLOW

FLOW THAT IMPLIES THAT THE DEPTH, WATER AREA, VELOCITY AND DISCHARGE DONT CHANGE WITH DISTANCE ALONG THE CHANNEL. IT ALSO IMPLIES THAT THE ENERGY GRADE LINE, WATER SURFACE AND CHANNEL BOTTOM ARE PARALLEL FOR UNIFORM FLOW.

IT IS THE FUNDAMENTAL TYPE OF FLOW TREATED IN OPEN CHANNEL HYDRAULICS.

STEADY VARIED FLOW

FLOW IS VARIED IF THE DEPTH OF FLOW CHANGES ALONG THE LENGTH OF THE CHANNEL

THE FLOW IS RAPIDLY VARIED IF THE CHANGES TO THE FLOW (depth and/or velocity) OCCUR ABRUPTLY OVER A COMPARATIVELY SHORT DISTANCE (also known as local phenomenon)

STEADY NON UNIFORM FLOW(gradually varied flow)

RESULTS FROM THE CHANGES OCCURING ALONG THE CHANNEL BOUNDARIES (e.g., channel geometry changes), FROM LATERAL INFLOWS TO THE CHANNEL OR BOTH. THIS TYPE OF FLOW OCCURS MOSTLY ON NATURAL RIVERS AND CHANNELS.

KINDS OF OPEN CHANNEL

1) NATURAL and

2) ARTIFICIAL

NATURAL CHANNEL

INCLUDE ALL WATERCOURSES THAT EXIST NATURALLY ON THE EARTH, VARYING IN SIZE FROM TINY HILLSIDE RIVULETS, THROUGH BROOKS, STREAMS SMALL AND LARGE RIVERS TO TIDAL ESTUARIES.

- VERY IRREGULAR IN SECTION, USUALLY FROM AN APPROXIMATE PARABOLA TO AN APPROXIMATE TRAPEZIOD.

ARTIFICIAL CHANNEL

CONSTRUCTED OR DEVELOPED BY HUMAN EFFORTS; NAVIGATION CHANNELS, POWER CANALS, IRRIGATION CANAL AND FLUMES, DRAINAGE DITCHES, TROUGH SPILLWAYS, FLOODWAYS, LOG CHUTES, ROADSIDE GUTTERS.

- USUALLY DESIGNED WITH SECTIONS OF REGULAR GEOMETRY.

CHANNEL GEOMETRY

1) PRISMATIC CHANNEL

2) NON PRISMATIC CHANNEL

PRISMATIC CHANNEL

CHARACTERIZED BY UNVARYING CROSS SECTION, CONSTANT BOTTOM SLOPE, AND RELATIVELY STRAIGHT ALIGNMENT.

NON PRISMATIC CHANNEL

WHEN THE CROSS SECTION, ALIGNMENT AND OR BOTTOM SLOPE CHANGES ALONG THE CHANNEL.

GEOMETRIC PROPERTIES OF COMMON OPEN CHANNEL SHAPES

SHAPE SECTION FLOW WETTED HYDRAULIC AREA, A PERIMETER, P RADIUS, R

Trapezoidal y(b+ycot) b+ 2y y(b+ycot)

Triangular y2cot 2y ycos sin 2

Rectangular by b+2y by y b + 2y

Wide flat by b y

sin b+(2y/sin) ybybyb>>y

COMMON TYPES OF FLOOD CONTROL CHANNEL CROSS SECTION

1. TRAPEZIODAL CROSS SECTION- have sloped sides and are formed by excavating in situ material- usually the most economical channel when ROW is available

2. RECTANGULAR CROSS SECTION- have vertical or near vertical sides- may be required for channels located in urban areas where the ROW is severely restricted.

UNIFORM FLOW IN A CHANNEL ENERGY GRADE LINE

HYDRAULIC GRADE LINEy

V22g

UNIFORM FLOWTWO (2) UNIFORM FLOW EQUATIONS:

1.) CHEZY FORMULA and2) MANNINGS EQUATION

VARIABLES:A = cross sectional area of the channelV = channel velocityP = wetted perimeter of the channelR = hydraulic radiusS = energy slope (equal to the slope of the bed)C or n = roughness coefficient

1) CHEZY FORMULAV=C ( RS)1/2

2) MANNINGS EQUATION

V=(1/n) R2/3 S1/2 V=(1.49/n) R2/3 S1/2

FOR THE DISCHARGE, Q

Q=A V

CHEZYS C

IMPORTANT FORMULAS

1. THE G. K. FORMULA:

C = [41.65 + (0.00281/S) + (1.811/n)] [1 + (41.65 + 0.00281/S)(n/R1/2)]

where:C=Chezys resistant factorS=slopeR=hydraulic radiusn=Kutters coefficient of roughness

2. THE BAZIN FORMULA:

C = 157.6 1 + m/(R1/2)where:C=Chezys resistant factorR=hydraulic radiusm=Bazins coefficient of roughness

PROPOSED VALUE OF BAZINS mDESCRIPTION OF CHANNELBAZINS m1. Very smooth cement of planed wood------------------------------0.112. Unplaned wood, concrete or brick---------------------------------0.213. Ashlar, rubble masonry, or poor brickwork--------------------0.834. Earth channels in perfect condition-------------------------------1.545. Earth channels in ordinary condition-----------------------------2.366. Earth channels in rough condition---------------------------------3.17

VALUES OF MANNINGS COEFFICIENT, n

SURFACE/DESCRIPTION RANGEMIN.MAX.1. Natural stream channels (top flood width < 30 m.)(i) Fairly regular sectiona. Some grass and weeds, little or no brush0.0300.035b. Dense growth of weeds, depth of flow materially greater than the weed height0.0350.05c. Some weeds, light brush on banks0.0350.05d. Some weeds, heavy brush on banks0.050.07e. Some weeds, dense trees0.060.08f. For trees within channel, with branches submerged at high flood, increase all above values by0.010.02

(ii) Irregular sections, with pools, slight channel meander; increase values given above about0.010.02

VALUES OF MANNINGS COEFFICIENT, n

SURFACE/DESCRIPTION RANGEMIN.MAX.(iii) Mountain streams, no vegetation in channel,banks usually steep, trees and brush alongbanks submerged at high flood:a. Bottom of gravel, cobbles and few boulders0.0400.05b. Bottom of cobbles, with large boulders0.0500.072) Larger stream channels (top flood width >30m)Reduce smaller stream coefficient by 0.010

3) Flood plains (adjacent to stream beds)Pasture, short grass, no brush0.0300.035Pasture, tall grass, no brush0.0350.050Cultivated land- no crop0.0300.040Cultivated land- nature field crop0.0450.055Scrub and scattered bush0.0500.070Wooded0.120.16

VALUES OF MANNINGS COEFFICIENT, n

SURFACE/DESCRIPTION RANGEMIN.MAX.4) Man made channels and ditchesEarth, straight and uniform0.0170.025Grass covered0.0350.050Dredged0.0250.033Stone lined and rock cuts, smooth and uniform0.0250.035Stone lined and rock cuts, rough and irregular0.0350.045Lined- metal corrugated0.0210.024Lined- smooth concrete0.0120.018Lined- grouted riprap0.0170.030

OPTIMUM CHANNEL CROSS SECTION

- A CROSS SECTION THAT REQUIRES A MINIMUM FLOW AREA IN WHICH THE HYDRAULIC RADIUS, R IS A MAXIMUM AND THE WETTED PERIMETER, P IS A MINIMUM

- ALSO KNOWN AS THE BEST HYDRAULIC SECTION

PROPERTIES OF OPTIMUM CHANNEL SECTIONS

SHAPE SECTION OPTIMUM NORMAL X-SECTIONAL GEOMETRY DEPTH, yn AREA, A = 600Trapezoidal b = 2/3(yn) 0.968 Qn 1.622 Qn Sb1/2 Sb1/2 Triangular = 450 1.297 Qn 1.682 Qn Sb1/2 Sb1/2

Rectangular b = 2yn 0.917 Qn 1.682 Qn Sb1/2 Sb1/2

Wide flat none 1.000 (Q/b)n ------------ Sb1/2

ynbynbynb>>yyn3/83/43/83/43/83/43/8

NON UNIFORM FLOW IN A CHANNEL ENERGY GRADE LINE

HYDRAULIC GRADE LINE

CHANNEL BOTTOMDATUM1V12

2gFLOWhl2V222gy1z1z2y2L

SPECIAL CONSIDERATION

1) FREEBOARD

- THE VERTICAL DISTANCE MEASURED FROM THE DESIGN WATER SURFACE TO THE TOP OF THE FLOOD CONTROL CHANNEL WALL.

CRITERIA:

a) A freeboard of 0.80 m. for velocities of 10 m/s or less. For curved alignments, the wall height must be at least 0.30 m above the superelevated water surface.

b) A one (1) m freeboard for velocities higher than 10.0 m/s. For curved alignments, the wall height must be at least 0.60 m above the superelevated water surface.

M1M2

c) If the flow is supercritical, the wall height shall be equal to the sequent depth but not less than the heights required under items 1 and 2.

2. SLOPE

- THE SIDE SLOPE OF THE FLOOD CONTROL CHANNEL DEPENDS ON THE KIND OF STREAM BANK MATERIALS AS SHOWN BELOW.SIDE SLOPESTREAM BANK MATERIALS

nearly verticalrock1:4mud and peat soils1:2 to 1:1stiff clay or earth with concrete lining1:1earth with stone lining, earth for large channel3:2firm clay or earth for small ditches2:1loose sandy earth3:1sandy loam in porous clay

- Provision of revetments along the riverbanks subjected to direct attack of the river flow for protection against scouring and wave wash. It should be located along the meander belts of the river, at the downstream and upstream sections of the hydraulic structures where turbulent flow occurs among others.

3. PREVENTION OF SCOUR

- EROSION OF SANDS IS ESPECIALLY CRITICAL AS THE SAND PARTICLE WEIGHT IS SMALL AND THERE IS NO COHESION BETWEEN GRAINS AS WELL AS THERE IS USUALLY LITTLE VEGETATION ALONG THE CHANNEL. ON THE OTHER HAND, WHILE CLAY AND SILT ARE FAIRLY RESISTANT TO SCOUR, ESPECIALLY IF COVERED WITH THICK VEGETATION, IT IS LIKEWISE NECESSARY TO PROVIDE CHANNEL SCOUR PROTECTION WHEN THE FLOW IS SUFFICIENTLY HIGH TO CAUSE EROSION OF FINE MATERIALS IN CHANNELS.

SUGGESTED MAXIMUM PERMISSIBLE MEAN CHANNEL VELOCITIES:

CHANNEL MATERIALMEAN CHANNEL VELOCITY (ft/sec)FINE SAND 2.0COARSE SAND 4.0FINE GRAVEL 6.0EARTHSandy silt 2.0silty clay 3.5clay 6.0 GRASS LINED EARTHSandy silt 5.0-6.0 Silty clay 7.0-8.0POOR ROCK (usually sedimentary) 10.0 Soft sandstone 8.0Soft shale 3.5GOOD ROCK (usually igneous and hard metamorphic) 20.0

4. SEDIMENT TRANSPORT

- THE DESIGNED SLOPE AND CROSS SECTION FOR A FLOOD CONTROL CHANNELS MUST BE CAPABLE OF MAINTAINING TRANSPORT OF INCOMING BED MATERIAL OTHERWISE DEPOSITION AND LOSS OF HYDRAULIC CONVEYANCE WILL OCCUR. HOWEVER, SEDIMENT TRANSPORT HAVE DIFFERENT EFFECTS ON STABILITY: INCREASING BED MATERIAL TRANSPORT TENDS TO INCREASE INSTABILITY, BUT HEAVY WASH LOADS OF FINE MATERIALS MAY PROMOTE STABILITY BY DEPOSITING COHESIVE LAYERS ON BANKS AND ENCOURAGING VEGETATION.

SAMPLE PROCEDURE FOR THE DESIGN OF FLOOD CONTROL CHANNEL PROJECT

I. ESTABLISH PROJECT OBJECTIVESA. Flood damage reduction ------ level of protection desiredB. Environmental1. Water quality2. Recreation3. Fish and wildlife4. Historic preservation5. Aesthetics

II. IDENTIFY ALTERNATIVES FOR ACHIEVING THE PROJECT OBJECTIVES

A. Non structural

B. Structural

1. Reservoirs2. Levees3. Flood control channels

III. EVALUATE ALTERNATIVES

IV. DETAILED PROJECT DESIGN FOR FLOOD CONTROL CHANNELS

A. Data collection and analysis, existing conditions1. Watershed conditionsa. climateb. topographyc. soils/geologyd. sediment yield

e. land use/cover(existing and recent changes)f. hydrology

2. Stream and floodplain (each reach)a. Hydrologyi. generate flood frequency series ii. determine corresponding stage data iii. calculate flow duration curves (hydrographs)

b. Hydraulicsi. identify resistance components and determine existing n values at various stages ii. Determine amount and size distribution of bed and suspended load

c. Geomorphologyi. survey cross section and existing channel grade ii. establish relationship of cross section geometry to discharge i.i.i. measure pool-riffle spacing and meander geometry and relate to discharge and channel width i.v. evaluate stability of bed and banks v. measure size distribution of bed and bank material v.i. measure cohesiveness of banks v.i.i. Identify and map locations of hard points in bed or banks

d. Stratigraphy (from test borings)i. determine stratigraphic sequence

ii. describe stratigraphic units in detail i.i.i. establish average depth to seasonal water tables e. Existing structures i. types i.i. location i.i.i. design i.v. scour and deposition patterns

f. Ecologyi. map riparian vegetation and locate and identify unique or valuable trees i.i. evaluate terrestrial ecology i.i.i. evaluate aquatic ecology

g. Water qualityi. physical

ii. chemical i.i.i. biological h. Aesthetic resources - identify, describe and photograph major components i. Historical and recreational resources - identify and describe major resources, with particular attention to historical and archeological components

B. Flood control channel design

1. Fix exact location and alignment geometry of channels2. Hydraulic designa. Rapid flow channels - used lined channels; choice of environmental features limited

b. Tranquil flow channelsi. Select best combination of channel cross section alignments and construction techniques to meet flood control and environmental objectives i.i. Select additional features to meet environmental objectives i.i.i. Establish downstream water surface elevation and the water surface control line including freeboard i.v. select n values for eachv. size channel v.i. Check channel stability for anticipated flows (if unstable, stabilize by one or more of the following: adjust cross section, adjust grade, line, or armor channel, provide grade control, provide bank protection)

3. Review design for maintenance consideration; adjust design if necessary.

4. Review design for aesthetics; adjust if needed

C. Design environmental features of project that are not part of the flood channel proper

D. Develop detailed cost estimates; if cost too high, modify project design beginning at step IVB.

SOME NATURAL STREAM CHANNEL TYPES, THEIR CHARACTERISTICS AND STABILITY PROBLEMS

CHANNEL TYPICAL STABILITY TYPESFEATURESPROBLEMS

1) MOUNTAIN- STEEP SLOPES- BED SCOUR AND TORRENTS- BOULDERS DEGRADATION- DROPS & CHUTES- DEBRIS FLOW2) ALLUVIAL FANS- MULTIPLE CHANNELS- SUDDEN CHANNEL - COARSE DEPOSITS- DEGRADATION- DEPOSITION

3) BRAIDED- INTERLACING CHANNELS- FREQUENT SHIFTS CHANNELS - COARSE SEDIMENTS OF MAIN CHANNEL- HIGH BED LOAD-SCOUR/DEPOSITION

4) MEANDERING- ALTERNATING BENDS- BANK EROSION RIVERS- FLAT SLOPES- MEANDER- WIDE FLOODPLAINS MIGRATION-SCOUR/DEPOSITION

5) MODIFIED RIVERS- PREVIOUSLY - MEANDER CHANNELIZED DEVELOPMENT - ALTERED BASE LEVELS- DEGRADATION & AGGRADATION - BANK EROSION

6) REGULATED - UPSTREAM RESERVOIRS- DEGRADATION RIVERS- IRRIGATION DIVERSION BELOW DAMS - LOWERED BASE LEVEL FOR TRIBUTARIES -AGGRADATION IN TRIBUTARY MOUTH

7) DELTAS - MULTIPLE CHANNELS- CHANNEL SHIFTS - FINE DEPOSITS- DEPOSITION ANDEXTENSION

STABILITY PROBLEMS FOR FLOOD CONTROL MODIFICATION

FORM OF CHANNEL MODIFICATION

1. Clearing and snagging- bank erosion and bed scour within reach directly affected, headcutting on the upstream section and sedimentation on the downstream.2. Cleanout or enlargement- bank erosion and sedimentation within reach directly affected, and headcutting on the upstream section.

3. Realignment- bank erosion, bed scour and meandering within reach directly affected, headcutting on the upstream section and sedimentation on the downstream.4. Levees- meander encroachment on setback within reach directly affected and increase flood peaks on the downstream section.

5. Floodways and bypasses- sedimentation of original channel within reach directly affected.

References:

1. Chow, V. T. 1973. Open Channel Hydraulics, International Edition, McGraw Hill Company, Auckland Bogota Guatemala Johannesburg Lisbon London Madrid Mexico New Delhi Panama San Juan Sao Paulo Singapore Sydney Tokyo

2. Bedient, P. and Huber, W. 1952. Hydrology and Floodplain Analysis, Second Edition, Addison-Wesley Publishing Company, Reading, Massachusetts. Menlo Park, California. New York. Don Mills, Ontario. Wokingham, England. Amsterdam. Bonn. Sydney. Singapore. Tokyo. Madrid. San Juan. Milan. Paris

3. Roberson, J., Cassidy, J. and Chaudhry, M.H. 1998. Hydraulic Engineering, Second Edition, John Wiley & Sons, Inc., New York, Chichester, Weinheim, Brisbane, Singapore, Toronto

References:

4. US Army Corps of Engineers 1993. Engineering and Design. River Hydraulics. Engineer Manual 1110-2-1416, Department of the Army, US Army Corps of Engineers, Washington, DC 20314-1000

5. US Army Corps of Engineers 1994. Engineering and Design. Hydraulic Design of Flood Control Channels. Engineers Manual 1110-2-1601, Department of the Army, US Army Corps of Engineers, Washington, DC 20314-1000

6. US Army Corps of Engineers 1982. Engineering and Design. Hydraulic Design for Local Flood Protection Projects. Engineer Regulation 1110-2-1405, Department of the Army, US Army Corps of Engineers, Washington, DC 20314-1000

References:

7. US Army Corps of Engineers 1989. Engineering and Design. Environmental Engineering for Flood Control Channels. Engineer Manual 1110-2-1205, Department of the Army, US Army Corps of Engineers, Washington, DC 20314-1000

8. US Army Corps of Engineers 1994. Engineering and Design. Channel Stability Assessment for Flood Control Projects. Engineers Manual 1110-2-1418, Department of the Army, US Army Corps of Engineers, Washington, DC 20314-1000

9. Department of Public Works and Highways. Design Guidelines Criteria and Standards for Public Works and Highways. Volume II

EXAMPLE PROBLEM NO. 1

The flow is uniform in a trapezoidal channel at a depth of 1.5 m. The bottom width is 15.0 m., side slopes of 1:1, bottom slope is 0.0001 and the Manning's n is 0.02. Determine the discharge capacity of the channel.

Figure:

d = 1.5 m

b = 15 m11

SolutionGiven:b =15 md =1.5 ms =0.0001n =0.02

From the eq.V = 1/n(R^2/3)(s^1/2) for metric systemIt is known that Q = AV, soQ = (1/n)A(R^2/3)(s^1/2)

Knowing that R = A/P and referring to the Figure,A = (0.5)(15+18)(1.5) = 24.75 sq. m.P = (1.5)(2^0.5)(2)+(15) =19.24264 m.

soR = A/P = 1.286206 m

thereforeQ = (1/0.02)(24.75)(1.29^2/3)(0.0001^0.5) =14.66465 cu. m./sec

EXAMPLE PROBLEM NO. 2If the channel shown below has the following properties, n(a,b), n(b,c), n(g,h) = 0.03 and n(c,d), n(d,e), n(e,f) = 0.025; slope = 1/2000. Determine the discharge in the channel.

Figure:

d1d230 m38 m44 m1 2 1 1

SolutionGiven:b1 = 30m.b2 = 38m. b3 = 44m.d1 = 3 m. = d2 s = 0.0005From the eq., V = 1/n(R^2/3)(s^1/2) for metric systemIt is known that Q = AV, so Q = (1/n)A(R^2/3)(s^1/2)Knowing that R = A/P and referring to the Figure,A1 = say the area of the high water channelA1 = (0.5)(30+36)(3) + (0.5)(44+50)(3) = 240 sq.m.P1 = [(3)(5^0.5)+(30)] + [(44)+ (3)(5^0.5)] =87.41641msoR1 = A1/P1 =2.74548 m, thereforeQ1 = (1/0.03)(240)(2.74548^2/3)(0.0005^0.5) = 350.7415 cu. m./sec

A2 = say the area of the low water channelA2 = (0.5)(38+44)(3) + (44)(3) = 255 sq.m.P2 = [(3)(2^0.5)(2)+(38)] = 46.48528 m.

soR 2= A2/P2 = 5.485607 m.

thereforeQ2 = (1/0.025)(255)(5.485607^2/3)(0.0005^0.5) = 709.4171 cu. m./sec

Total Q = Q1 + Q2Total Q = 1060.159 cu.m./sec

EXAMPLE PROBLEM NO. 3Determine the design flood level of the channel (trapezoidal section) from the given data below:upstream elev = 58.612 mdownstream elev = 58.598 mlength of stream, L = 50.0 mFigure:

DFLElev. 70.453Elev. 60.3d22.0 m11

SolutionGiven:Q = 137 cum./secs = 0.00028 b = 22 m.ss = 1H:1Vn = 0.03 arb = el. 60.3

Try depth of flow, d1 = 4.0 mFrom the eq.V = 1/n(R^2/3)(s^1/2) for metric systemIt is known that Q = AV, so, Q = (1/n)A(R^2/3)(s^1/2)Knowing that R = A/P and referring to the Figure,A1 = (0.5)(22+30)(4) = 104 sq.m.P1 = (4)(2^0.5)(2)+(22) = 33.31371 m.soR1 = A1/P1 = 3.121838 m.

thereforeQ1 = (1/0.03)(104)(3.121838^2/3)(0.00028^0.5) = 123.9076 cu. m./sec < 137 cu. m./sec Try depth of flow, d2 = 4.5 mA2 = (0.5)(22+31)(4.5) = 119.25 sq. m.P2 = [(4.5)(2^0.5)(2)+(22)] = 34.72792 m.soR 2= A2/P2 = 3.433836 m.

thereforeQ2 = (1/0.03)(119.25)(3.433836^2/3)(0.00028^0.5) = 151.39186 cu. m./sec > 137 cu. m./sec

By interpolation:y/(Q2-Q) = (d2-d1)/(Q2-Q1)y/(151.39-137) = (4.5-4.0)/(151.39-123.91)y = 0.261827 m.

therefored = d1 + yd = 4.261827 m.

DFL = 60.3+ 4.26187DFL = 64.56187 m

EXAMPLE PROBLEM NO. 4Given the properties of the earth canal shown, find the discharge.Figure:

d1b1b2b3b4d3d2n = 0.023n = 0.030s = 0.00042s = 0.00042

SolutionGiven:b1 = 6 m.b2 =20 m. b3 = 4 m. b4 = 150 m.d1 = 4 m.d2 = 3 m.d3 = 1 m.n1 = 0.023n2 =0.03From the eq.V = 1/n(R^2/3)(s^1/2) for metric systemIt is known that Q = AV, soQ = (1/n)A(R^2/3)(s^1/2)Knowing that R = A/P and referring to the Figure,A1 = say the area of the main sectionA1 = (0.5)(4)(6) + (20)(4) + (0.5)(1+4)(4) = 102 sq.m.P1 = ((6^2+4^2)^0.5)+(20)+((4^2+3^2)^0.5) =32.2111 m

soR1 = A2/P2 = 3.16661 m.thereforeQ1 = (1/0.023)(102)(3.16661^2/3)(0.00042^0.5) = 213.8391 cu. m./sec

A2 = say the area of the overflow sectionA2 = (0.5)(38+44)(3) + (44)(3) = 255 sq.m.P2 = [(3)(2^0.5)(2)+(38)] = 46.48528 m.soR2= A2/P2 = 5.485607 m.Q2 = (1/0.025)(255)(5.485607^2/3)(0.0005^0.5) = 709.4171 cu. m./sec

Total Q = Q1 + Q2Total Q = 923.2561 cu.m./sec