Capter UNIFORM FLOW.. Introduction.. Basic equations in uniform open-cannel flow.3. Most economical cross-section.4. Cannel wit compound cross-section.5. Permissible velocity against erosion and sedimentation Summary Te capter on uniform flow in open cannels is basic knowledge required for all ydraulics students. In tis capter, we sall assume te flow to be uniform, unless specified oterwise. Tis capter guides students ow to determine te rate of discarge, te dept of flow, and te velocity. Te slope of te bed and te cross-sectional area remain constant over te given lengt of te cannel under te uniform-flow conditions. Te same olds for te computation of te most economical cross section wen designing te cannel. Te concept of permissible velocity against erosion and sedimentation is introduced. Key words Uniform flow; most economical cross-section; discarge; velocity; erosion; sedimentation.. INTRODUCTION... Definition Uniform flow relates to a flow condition over a certain lengt or reac of a stream and can occur only during steady flow conditions. Uniform flow may be also defined as te flow occurring in a cannel in wic equilibrium as been reaced between gravitational force and sear force. Many irrigation and drainage canals and oter artificial cannels are designed to carry water at uniform dept and cross section all along teir lengts. Natural cannels as rivers and creeks are seldom of uniform sape. Te design discarge is set by considerations of acceptable risk and frequency analysis, wereas te cannel slope and te cross-sectional sape are determined by topograpy, and soil and land conditions. Uniform equilibrium open-cannel flows are caracterized by a constant dept and a constant mean flow velocity: V 0 and 0 (-) s s were s is te coordinate in te flow direction, te flow dept and V te flow velocity. Uniform equilibrium open-cannel flows are commonly called uniform flows or normal flows. Note: Te velocity distribution in fully-developed turbulent open cannel flows is given approximately by Prandtl s power law (Fig..): V y N (-) Vmax were te exponent /N varies from ¼ down to ½ depending on te boundary friction and te cross-section sape. Te most commonly-used power law formulae are te one-sixt Capter : UNIFORM FLOW 5
power (/6) and te one-sevent power (/7) formulas. It sould be noted tat te velocity in open-cannel flow is assumed constant over te entire cross-section. V max y V v velocity distribution Fig... Velocity distribution profile in turbulent flow Suc flow conditions are represented scematically in Fig... Considering Bernoulli s teorem of te conservation of energy, between cross-sections and, leads to te expression: p V p V E z E L z L (-3) g g were and are te Corriolis-coefficients corresponding to te velocities V and V, respectively. Tey are also called te kinetic-energy correction coefficients. is equal to or larger tan but rarely exceeds.. (Li and Hager, 99). For a uniform velocity distribution, =. Te slope of te energy gradient line S is equal to te bed slope i of te cannel, or: L S = i (-4) L energy-gradient line E V g V S g p ydraulic-gradient line g p i g z L z Fig... Energy and ydraulic gradient in uniform-flow cannel L Datum E If te flow is uniform, te cross sections at points and must be constant. Consequently, te velocity and te dept of flow must also remain constant, or: V = V and = (-5) Te flow resistance in an open cannel is more difficult to quantify. Te importance of te resistance coefficient goes beyond its use in cannel design for uniform flow. Capter : UNIFORM FLOW 6
... Momentum analysis Consider a control volume of lengt L in uniform flow, as sown in Fig..3. L F P Wsin A o PL W = gal F P = F P P Fig..3. Force balance in uniform flow By definition, te ydrostatic forces, F p and F p, are equal and opposite. In addition, te mean velocity is invariant in te flow direction, so tat te cange in momentum flux is zero. Tus, te momentum equation reduces to a balance between te gravity force component in te flow direction and te resisting sear force: A L sin = o P L (-6) in wic = g = specific weigt of te fluid, A = cross-sectional area of flow, o = mean boundary sear stress, and P = wetted perimeter of te boundary on wic te sear stress acts. If Eq. (-6) is divided by PL, te ydraulic radius R = A/P appears as an intrinsic variable. Pysically, Eq. (-6) represents te ratio of flow volume to boundary surface area, or sear stress to unit weigt, in te flow direction. Eq. (-6) can be written as: o = R sin RS (-7) if we replace sin wit S = tan for small values of. Furtermore, if we solve Eq. (-7) for te bed slope, wic equals te slope of te energy grade line, L /L, and express te sear stress in terms of te friction factor f for uniform pipe flow according Darcy- Weisbac: o f (-8) V 8 we ave te Darcy-Weisbac equation (for uniform pipe flow): f o f..v f V i = S =. (-9) L R 8R 4R g from wic it is evident tat te appropriate lengt scale, wen applied to open-cannel flow, is 4R. It seems reasonable to use 4R as te lengt scale in te Reynolds-number and te relative rougness as well. Before applying uniform flow formulas to te design of open cannels, te background of Cezy s as well as Manning s formulas for steady, uniform in open cannels are presented in te next section. Capter : UNIFORM FLOW 7
.. Basic equations in uniform open-cannel flow... Cezy s formula Consider an open cannel of uniform cross-section and bed slope as sown in Fig..4: L Q i VA Q Fig..4. Sloping bed of a cannel Let L = lengt of te cannel; A = cross-sectional area of flow; V = velocity of water; P = wetted perimeter of te cross-section; f = friction coefficient according to Darcy-Weisbac; and i = uniform slope of te bed. It as been experimentally found, tat te total frictional resistance along te lengt L of te cannel, follows te law: Frictional resistance = f 8 contact area (velocity) = f 8 P.L Vn (-0) Te exponent n as been experimentally found to be nearly equal to. But for all practical purposes, its value is taken to be. Terefore, Frictional resistance = f 8 P.L V (-) Since te water moves over a distance V in second, terefore, te work done in overcoming te friction reads as: Frictional resistance distance V in second = f 8 PLV V = f 8 PL V3 (-) Te weigt of te water, W, in te cannel over a lengt of L is: W =.A.L (-3) Tis water falls vertically down over a distance V.i in second, so Loss of potential energy = Weigt of water Heigt =.A.L.V.i (-4) We know tat work done in overcoming friction = Loss of potential energy f i.e. 8 P.L V3 =.A.L.V.i (-5) Capter : UNIFORM FLOW 8
V = 8.A.i f..p or V = were C = 8 A i.f P (-6) 8 8g is known as Cezy s coefficeint and R = A.f f P as ydraulic radius. Te discarge of flow ten is Q = A V = AC Ri (-7) Note: Unlike te Darcy-Weisbac coefficient f, wic is dimensionless, te Cezy coefficient C as te dimension, [L / T - ], as mentioned in Capter. Cezy s coefficient C depends on te mean velocity V, te ydraulic radius R, te kinematic viscosity and te relative rougness. Tere is experimental evidence tat te value of te resistance coefficient does vary wit te sape of te cannel and terefore wit R and possibly also wit te bed slope i, wic for uniform flow will be equal to te slope of te energy-ead line i o, yielding a relationsip for te velocity of te form: V = K. R x y.i o were K, x and y are constants. (-8) Example.: A rectangular cannel is 4 m deep and 6 m wide. Find te discarge troug te cannel, wen it runs full. Take te slope of te bed as :000 and Cezy s coefficient as 50 m / s -. Solution: Given: Dept = 4 m, Widt b = 6 m, Bed slope i = /000 = 0.00, Cezy s coefficient C = 50 m / s - Q =? (m 3 /s) b Area of te rectangular cannel: A = b = 4 m Perimeter of te rectangular cannel: P = b + = 4 m Hydraulic radius of te flow: R = A P =.7 m Discarge troug te cannel: Q = AC Ri = 49.6 m 3 s - Ans. Example. : Water is flowing at te rate of 8.5 m 3 s - in an earten trapezoidal cannel wit a bed widt 9 m, a water dept. m and side slope :. Calculate te bed slope, if te value of C in Cezy s formula be 49.5 m / s -. Solution: Given: Discarge Q = 8,5 m 3 /s, Bed widt b = 9 m, Dept =. m, Side slope m =, Cezy s coefficient C = 49.5 m / s -, Bed slope =? / B b Surface widt of te trapezoidal cannel B = b + ( ) = 0. m Capter : UNIFORM FLOW 9
Area of te trapezoidal cannel: A = b B Wetted perimeter: P = b =.5 m =.68 m Hydraulic radius: R = A P = 0.986 m Now using te relation: Q = AC Ri i Q = R(AC) = 4440 Ans.... Manning s formula Manning, after carrying out a series of experiments, deduced te following relation for te value of C in Cezy s formula: C = R 6 n (-9) were n is te Manning constant in metric units, n = [m -/3 s]. n is expressing te cannel s relative rougness properties and values are given in Table. Now we see tat te velocity: V = C Ri = R 6 Ri n = R 6 R i n V 3 = R i n (-0) 3 Now, te discarge is: Q = AV = A R i n (-) Table.: Values of Manning coefficient n [m -/3 s] Wetted perimeter n Wetted perimeter n A. Natural cannel D. Artificially lined cannel Clean and straigt Glass 0.00 Sluggis wit deep pools Brass Major rivers Steel, smoot 0.0 B. Flood plain Steel, painted Pasture, farmland Steel, riveted Ligt brus Cast iron Heavy brus 0.075 Concrete, finised 0.0 Trees 0.50 Concrete, unfinised C. Excavated eart cannels Planned wood 0.0 Clean 0.0 Clay tile Gravelly 0.05 Brickwork Weedy Aspalt 0.06 Stony, cobbles Corrugated metal 0.0 Rubble masonry 0.05 For a more detailed description, we can take te value of Manning s n from te Table at te end of tis capter. Capter : UNIFORM FLOW 30
Example.3: An earten trapezoidal cannel wit a 3 m wide base and side slopes : carries water wit a dept of m. Te bed slope is /600. Estimate te discarge. Take te value of n in Manning s formula as 0.04 m -/3 s. Solution: Given: Base widt b = 3 m, Side slope = :, Water dept = m, Bed slope /600, Manning s coefficient n = 0.04 m -/3 s Discarge Q =? (m 3 /s) B b Surface widt of te trapezoidal cannel B = b + = 5 m Area of te trapezoidal cannel: A = b B Wetted perimeter: P = Hydraulic radius: R = A P b = 5.88 m = 4 m = 0.686 m Now using te relation: Q = 3 A R i =.94 m 3 /s Ans. n Example.4 : Water at te rate of 0. m 3 /s flows troug a vitrified sewer wit a diameter of m wit te sewer pipe alf full. Find te slope of te water surface, if Manning s n is m -/3 s. Solution: Given: Discarge Q = 0. m 3 /s, Diameter of pipe D = m, Manning s n = m -/3 s Sewer slope i =? D Area of te flow: A = 4 D Wetted perimeter: P = Hydraulic radius: R = A P = 0.393 m =.57 m = 0.5 m D Using Manning s formula: Q = A R i n 3 Qn Water surface slope: i = = A R 3 430 Ans. Capter : UNIFORM FLOW 3
..3. Discussion of factors affecting f and n Te dependence of f on te relative rougness in open cannel flow is not as well known as in pipe flow, because it is difficult to assign an equivalent sand-grain rougness to te large values of te absolute rougness eigt typically found in open cannels. Te dependence of flow resistance on te cross-sectional sape occurs as a result of canges of bot te cannel ydraulic radius, R, and te cross-sectional distribution of velocity and sear. Tere is no substitute for experience in te selection of Manning s n for natural cannels. Table.. (at te end of tis capter) from Ven Te Cow (959) gives an idea of te variability to be expected in Manning s n..3. MOST ECONOMICAL CROSS-SECTION.3.. Concept A typical uniform flow problem in te design of an artificial canal is te economical proportioning of te cross-section. A canal, aving a given Manning coefficient n and a slope i, is to carry a certain discarge Q, and te designer s aim is to minimize te cross-sectional area. Clearly, if A is to be a minimum, te velocity V is to be a maximum. Te Cezy and Manning formulas indicate, terefore, tat te ydraulic radius R = A/P must be a maximum. It can be sown tat te problem is equivalent to tat of minimizing P for a given constant value of A. Tis concept as a practical application in estimating te cost for a canal excavation and /or lining. From economic considerations of minimizing te flow cross-sectional area for a given design discarge, a teoretically optimum cross-section will be introduced..3.. Conditions for maximum discarge Te conditions for maximum discarge for te following cross-sections will be dealt wit: (a). Rectangular cross-section, and (b). Trapezoidal cross-section. (a). Cannel wit rectangular cross-section Consider a cannel of rectangular cross-section as sown in Fig..5. Let b = breadt of te cannel, and = dept of te cannel. b Area of flow: A = b b = A (-0) Fig..5. A rectangular cannel Discarge: Q = A V = A AC Ri AC i (-) P Keeping A, C and i constant in te above equation, te discarge will be maximum wen A/P is maximum or te wetted perimeter P is minimum. Or in oter words, wen: dp 0 d (-3) We know tat P = b + = A (-4) Capter : UNIFORM FLOW 3
Differentiating te above equation wit respect to and equating to zero yields: dp A 0 (-5) d A = = b or b = i.e. te breadt is two times te dept (-6) In tis case, te ydraulic radius is: A b R P b 4 (-7) Hence, we can say tat for te maximum discarge or te maximum velocity, tese two conditions (i.e. b = and R = /) sould be used for solving te problem of optimizing cannels of rectangular cross-section..00 Q Q max 0.95 0.90 0.85 b A = b = constant 0 3 4 5 b Fig..6. Experimental relationsip between Q Q max and b As can be seen in Fig..6, te maximum represented by tis optimal configuration is a rater weak one. For example, for aspect ratios, b, between and 4, te flow rate is witin 96% of te maximum flow rate obtained wit te same area and by b/ =. Example.5.: Find te most economical cross-section of a rectangular cannel to carry 0.3 m 3 /s of water, wen te bed slope is /000. Assume Cezy s C = 60 m -/3 s -. Solution: Given: Discarge Q = 0.3 m 3 /s, Bed slope i = /000, Cezy coefficient C = 60 m -/3 s - Breadt of cannel b =? (m) and dept of te cannel =? (m) We know tat for te most economical rectangular section: b = Area: A = b = = and ydraulic radius: R = / = 0.5 Using te relation: Q = AC Ri Capter : UNIFORM FLOW 33
and squaring bot sides yields: 5 5 0.09 m 7. 5 5 or m Dept of te cannel: = 0.46 m Ans. And breadt: b = d = 0.83 m Ans. (a). Cannel wit trapezoidal cross-section Consider a cannel of trapezoidal cross-section ABCD as sown in Fig..7. D B = b + n C n A b B n Fig..7. A trapezoidal cannel Let b = breadt of te cannel at te bottom, = dept of te cannel, and = side slope (i.e. vertical to n orizontal) n Area of flow: A = (b + n) A or = b + n b = A - n (-8) A Discarge: Q = A V AC Ri AC i (-9) P Keeping A, C and i constant in te above equation, te discarge will be maximum, wen A/P is maximum or te wetted perimeter P is minimum. Or in oter words: dp 0 d We know tat: P = b n b n (-30) Substituting te value of b from equation (-8) yields: A P = n n (-3) Differentiating te above equation wit respect to d and equating to zero results into: dp A n n 0 (-3) d A n n Capter : UNIFORM FLOW 34
or (b n) n n [A = (b + n)] b n n n b n n (-33) We see tat b + n = B is te top widt of te cannel and n is te lengt of te sloping side, i.e. te lengt of te sloping side is equal to alf te top widt In tis case, te ydraulic radius: A (b n) (b n) (b n) R = P b n b (b n) (b n) (-34) Hence, we can say tat for te maximum discarge or te maximum velocity, tese two b n conditions (i.e. n and/or R = ) sould be used for solving te problems in te case of cannels of trapezoidal cross-section. Example.6: A canal of trapezoidal cross-section as to be excavated troug ard clay at te least cost. Determine te dimensions of te cannel for a discarge equal to 4 m 3 /s, a side slope for ard clay n = :, a bed slope :500 and Manning s n = 0.0 m -/3 s. Solution: Given: Discarge Q = 4 m 3 /s; Bed slope i = /500; Side slope n = : Manning s n = 0.0 m -/3 s Breat b =? (m) Dept =? (m) b We know tat for te least cost: alf of te top widt = lengt of sloping side b n n wit n = b.83 b = 0.83 Area of flow: A = (b + n) =.83 Using Manning s formula: Q = A R i n yields: 8 3 3.4 m 8/3.55 m Ans. And b = 0.83 =. m Ans. Capter : UNIFORM FLOW 35
Note: Te semi-circular section (te semi-circle aving its center in te surface) is te best ydraulic section. Te best ydraulic cross-section for oter sapes can be drawn as presented in Fig..8. D R R D 90 circular cannel rectangular cannel trapezoidal triangular cannel cannel Fig..8. Cross-sections of maximum flow rate: i.e. optimum design Students sould try to proof te conditions for circular and triangular cannels for te best ydraulic cross-section based on te below relationsip: 3 3 5 A 3 3 Q A Q.n V.R.i..i = constant (-5) A n n P P i Table..: Te best design summary for several cannel cross-sections Cross-section Optimum widt B Optimum cross-section A Semi-circle D D 8 Rectangular D D Trapezoidal D 3 3 D 4 Triangular D D D R Optimum wetted perimeter P D D 3D D 60 D Optimum ydraulic radius R D 4 D 4 D 4 D 4 R.3.3. Problems of uniform-flow computation Te computation of uniform flow may be performed by te use of two equations: te continuity equation and a uniform-flow formula. Wen te Manning formula is used as te uniform-flow formula, te computation will involve te following six variables: () te normal discarge Q (4) te mean velocity of flow V () te normal dept (5) te coefficient of rougness n (3) te cannel slope i (6) te geometric elements tat depend on te sape of te cannel section, suc as A,R, etc, Capter : UNIFORM FLOW 36
Wen any four of te above six variables are given, te remaining two unknowns can be determined by te two equations. Te following represents some types of problems of uniform flow computation: A. to compute te normal discarge: In practical applications, tis computation is required for te determination of te capacity of a given cannel or for te construction of a sysntetic rating curve of te cannel. B. to determine te velocity of te flow: Tis computation as many applications. For example, it is often required for te study of scouring and silting effects in a given cannel. C. to compute te normal dept: Tis computation is required for te determination of te stage of flow in a given cannel. D. to determine te cannel rougness: Tis computation is used to ascertain te rougness coefficient in a given cannel; te coefficient tus determined may be used in oter similar cannels. E. to compute te cannel slope: Tis computation is required for adjusting te slope of a given cannel. F. to determine te dimensions of te cannel section: Tis computation is required mainly for design purposes. Table.3. lists te known and unknown variables involved in eac of te six types of problems mentioned above. Type of problem A B C D E F Notes: Discarge Q? - Table.3: Problems of uniform-flow computation Velocity V -? - - - - Dept d? Rougness n? Slope i? Geometric elements Te known variables are indicated by te ceck mark () and te unknown required in te problems by te question mark (?). Te unknown variable(s) tat can be determined from te known variables is(are) indicated by a das (-). Table.3. does not include all types of problems. By varying combinations of various known and unknown variables, more types of problems can be formed. In design problems, te use of te best ydraulic section and of empirical rules is generally introduced and tus new types of problems are created.? Capter : UNIFORM FLOW 37
. 4. CHANNEL WITH COMPOUND CROSS-SECTION A compound cannel consists of a main cannel, wic carries te base flow (frequently running off up to bank-full conditions), and a floodplain on one or bot sides tat carries over-bank flow during te time of flooding as sketced in Fig..9. Te compound cross-section of a cannel may be composed of several distinct subsections wit eac subsection different in rougness form te oters. canal bank ig water level low water level dyke (): A, P, n (): A, P, n (3): A 3, P 3, n 3 Fig..9. Over-bank flow in a compound cannel Te rougness of te side cannels will be different (generally rouger) from tat of te main cannel and te metod of analysis is to consider te total discarge to be te sum of te component discarges computed by te Manning equation. Te mean velocity for te wole cannel section is equal to te total discarge divided by te total water area. Te classical metod of computation of discarge, as presented by Cow in 959, consisted in subdividing te composite cross-section into sub-areas wit vertical interfaces in wic te sear stresses are neglected. Te discarge for eac sub-area is calculated by assuming a common friction slope i for te wole cannel. Tus in te cannel, as sown in Fig..9, assuming tat te bed slope is te same for te tree sub-areas, it olds: m m 3 AiR i Q Qi i (-6) n i i i Te division of te cannel by tese artificial vertical boundaries assumes implicity tat te sear stress on tese interfaces is relatively small wit respect to te sear stress acting on te wetted perimeter of te cannel. Note: It as been found by more recent experimentation tat tis ypotesis is incorrect and tat it leads to a considerable over-estimation of te discarge in te compound cannel. Example.7: Water flows along a drainage canal aving te properties sown in te figure below. If te bottom slope i = /500=0.00, estimate te discarge. 3 m m 3 m 0.6 m n = 0.00 () () n 3 = (3) n = 0.8 m [n i ] = m -/3 s Capter : UNIFORM FLOW 38
Solution: We divide te cross-section into tree subsections as is indicated in te figure and write te discarge as Q = Q + Q + Q 3, were for eac section, it olds: 3 Q i =A i V i = Ai Ri i ni Te appropriate values of A i, P i, R i and n i are listed in te table below: i A i (m ) P i (m) R i (m) n i ( m -/3 s) () ()* (3).8.8.8 3.6 3.6 3.6 0.500 0.778 0.500 0.00 0.00 Note tat te imaginary portions of te wetted perimeter between te sections (denoted by te dased lines in te figure) are not implemented in P i. Tat is, for section (): A = (0.8 + 0.6) m =.8 m P = { + (0.8)} m = 3.6 m So tat A.8 R m P 3.6 = 0.778 m Tus te total discarge is: Q = Q + Q + Q 3 = Q = 3 3.8 0.500.8 0.778.8 0.500 0.00 0.00 3 m 3 s - Q =.75 m 3 /s Ans. If te entire cannel cross-section were considered as one flow area, ten : A = A + A + A 3 = 6.4 m P = P + P + P 3 = 0.8 m Ten R = A = 0.593 m P Te total discarge can be written as 3 Q = A R i n n eff were n eff is te effective value of n for te wole compound cannel. Wit Q =.75 m 3 /s, as determined above, te value of n eff is found to be: 3 A R i n eff = 0.068 m -/3 s Q As expected, te effective rougness (Manning s n) is between te minimum (n = m -/3 s) and maximum (n 3 = m -/3 s) values for te individual subsections. Capter : UNIFORM FLOW 39
.5. PERMISSIBLE VELOCITY AGAINST EROSION AND SEDIMENTATION Te excavation and lining cost of open cannels or conduits varies wit teir size. Wit respect to water-resources-system economics, erosion of and sedimentation in cannels are problems in ydraulic engineering. Erosion and sedimentation must be predicted because tey can cange te bed slope, te cannel widt and terefore te flow conditions. So, if te available slope permits, te cost of te initial construction may be reduced by using te igest velocity. However, if te velocity is becoming too ig, te cannel may be damaged or destroyed by erosion. Tis must be avoided by limiting te velocities according to te boundary materials. For clear water in ard-surfaced water conductors, te limiting velocity is beyond practical requirements. Velocities above 0 m/s for clear water in concrete cannels ave been found to do no arm. If te water carries abrasive material, damage may occur at lower velocities. No definite relation as been establised between te nature of abrasive materials, te material of cannel bank and bed, and a permissible velocity. In unlined earten cannels, te limiting velocity involves many factors. Generally, a fine soil is eroded more easily tan a coarse one, but te effect of te grain size may be obscured by te presence or absence of a cementing or binding material. Te tendency to erode is reduced by seasoning. Groundwater conditions can exert an important influence. Seepage out of te cannel, particularly if te water is turbid, tends to tougen te banks, wereas infiltration reduces te resistance to erosion. Erosion can be reduced or avoided by designing for low velocities. If te water carries an appreciable amount of silt in suspension, too low a velocity will cause te canal to fill up until its capacity is impaired. It is necessary to coose a velocity tat will keep te silt in motion but tat will not erode te bank or bottom of te canal. Te margin of permissible velocities depends on te amount and nature of te silt in te water, te nature of te bank material, te size and sape of te canal, and many oter factors. Te silt content of most turbid water varies wit te season, as does also te demand for water and te resultant velocity of te flow. Te determination of non-scouring, non-silting velocities for earten canals as attracted te attention of many investigators over a long period of time, and a considerable mass of data and formulas ave been accumulated. However, for preliminary purposes, and for design in many cases, use may be made of te approximate values purposed by Fortie and Scobey, in 96, as sown in Table.4. Were te silt is important, it is better to make te slope a little too steep rater tan a little too flat. A gradient tat proves to be too steep can be controlled by cecks. In ard-surfaced cannels, silting is easily controlled if fall for scouring velocity is available. Capter : UNIFORM FLOW 40
Table.4: Permissible canal velocities (Fortier and Scobey, 96) Velocity, m/s, after aging, of canal carrying: Original material excavated for canal Clear water, no detritus Water transporting colloidal silts Water transporting non-colloidal silts, sands, gravels, or rock fragments () () (3) (4) Fine sand (non-colloidal) Sandy loam (non-colloidal) Sandy loam (non-colloidal) Alluvial silts wen non-colloidal Ordinary firm loam Volcanic as Fine gravel Stiff clay (very colloidal) Graded, loam to cobbles, wen non-colloidal Alluvial silts wen colloidal Graded, loam to cobbles, wen colloidal Coarse gravel (non-colloidal) Cobbles and singles Sales and ardpans 0.46 0.53 0.6 0.6 0.76 0.76 0.76.4.4.4...5.83 0.76 0.76 0.9.07.07.07.5.5.5.5.68.83.68.98 0.46 0.6 0.6 0.6 0.69 0.6.4 0.9.5 0.9.5.98.98.5 Capter : UNIFORM FLOW 4
Table.5: Value of Manning s Rougness Coefficient n [m -/3 s] (Ven Te Cow, 973) Type of cannel and description Minimum Normal Maximum A. Closed Conduits Flowing Party Full A.. Metal a. Brass, smoot b. Steel. Lock bar and welded. Riveted and spiral c. Cast iron. Coated. Uncoated d. Wrougt iron. Black. Galvanized e. Corrugated metal. Subdrain. Storm drain A.. Nonmetal a. Lucite b. Glass c. Cement. Neat surface. Mortar d. Concrete. Culvert, straigt and free of debris. Culvert wit bends, connections, and some debris 3. Finised 4. Sewer wit manoles, inlet, etc., straigt 5. Unfinised, steel form 6. Unfinised, smoot wood form 7. Unfinised, roug wood form e. Wood. Stave. Laminated, treated f. Clay. Common drainage tile. Vitrified sewer 3. Vitrified sewer wit manoles, inlet, etc. 4. Vitrified subdrain wit open joint g. Brickwork. Glazed. Lined wit cement mortar. Sanitary sewers coated wit sewage slimes, wit bends and connections i. Paved invert, sewer, smoot bottom j. Rubble masonry, cemented 0.009 0.00 0.00 0.0 0.0 0.008 0.009 0.00 0.00 0.0 0.0 0.00 0.0 0.0 0.06 0.08 0.00 0.0 0.06 0.06 0.09 0.04 0.009 0.00 0.0 0.0 0.06 0.09 0.05 0.06 0.0 0.00 0.06 0.00 0.00 0.08 0.06 0.00 Capter : UNIFORM FLOW 4
Type of cannel and description Minimum Normal Maximum B. Lined or built-up cannels B.. Metal a. Smoot steel surface. Unpainted. Painted b. Corrugated B.. Nonmetal a. Cement. Neat surface. Mortar b. Wood 3. Planed, untreated 4. Planed, creosoted 5. Unplanted 6. Plank wit battens 7. Lined wit roofing paper c. Concrete. Trowel finis. Float finis 3. Finised, wit gravel on bottom 4. Unfinised 5. Gunite, good section 6. Gunite, navy section 7. On good excavated rock 8. On irregular excavated rock d. Concrete bottom float finised wit sides of. Dressed stone in mortar. Random stone in mortar 3. Cement rubble masonry, plastered 4. Cement rubble masonry 5. Dry rubble or riprap e. Gravel bottom wit sides of. Formed concrete. Random stone in mortar 3. Dry rubble or riprap f. Brick. Glazed. In cement mortar g. Masonry. Cemented rubble. Dry rubble. Dressed aslar i. Aspalt. Smoot. Roug j. Vegetal lining 0.0 0.0 0.00 0.00 0.0 0.00 0.06 0.08 0.0 0.06 0.00 0.00 0.00 0.03 0.0 0.03 0.06 0.0 0.05 0.0 0.0 0.09 0.0 0.00 0.07 0.00 0.00 0.05 0.00 0.03 0.033 0.05 0.03 0.06. 0.08 0.06 0.00 0.00 0.03 0.05 0.00 0.04 0.04 0.05 0.06 0.036 0.08 0.500 Capter : UNIFORM FLOW 43
Type of cannel and description Minimum Normal Maximum C. Excavated or dredged a. Eart, straigt and uniform. Clean, recently completed. Clean, after weatering 3. Gravel, uniform section, clean 4. Wit sort grass, few weeds b. Eart, winding and sluggis. No vegetation. Grass, some weeds 3. Dense weeds or aquatic plants in deep cannels 4. Eart bottom and rubble sides 5. Stony bottom and weedy banks 6. Cobble bottom and clean sides c. Dragline-excavated or dredged. No vegetation. Ligt brus on banks d. Rock cuts. Smoot and uniform. Jagged and irregular e. Cannels not maintained, weeds and brus uncut. Dense weeds, ig as flow dept. Clean bottom, brus on sides 3. Same, igest stage of flow 4. Dense brus, ig stage 0.06 0.08 0.0 0.0 0.03 0.05 0.08 0.05 0.05 0.05 0.045 0.080 0.08 0.0 0.05 0.07 0.05 0.08 0.080 0.070 0.00 0.00 0.05 0.033 0.033 0.033 0.060 0.0 0.080 0.0 0.40 D. Natural streams D.. Minor stream (top widt at flood stage < 00 ft) a. Streams on plain. Clean, straigt, full stage, no rifts or deep pools. Same as above, but no more stones and weeds 3. Clean, winding, some pools and soals 4. Same as above, but some weeds and stones 5. Same as above, lower stages, more ineffective slopes and sections 6. Same as 4, but more stones 7. Sluggis reaces, weedy, deep pools 8. Very weedy reaces, deep pools, or floodways wit eavy stand of timber and underbrus b. Mountain stream, no vegetation in cannel, banks usually steep, trees and brus along banks submerged at ig stages. Bottom: gravels, cobbles, and few boulders. Bottom: cobbles wit large boulders 0.05 0.033 0.045 0.075 0.045 0.048 0.070 0.00 0.033 0.045 0.055 0.060 0.080 0.50 0.070 Capter : UNIFORM FLOW 44
Type of cannel and description Minimum Normal Maximum D. Natural streams D.. Flood plains a. Pasture, no brus. Sort grass. Hig grass b. Cultivated areas. No crop. Mature row crops 3. Mature field crops c. Brus. Scattered brus, eavy weeds. Ligt brus and trees, in winter 3. Ligt brus and trees, in summer 4. Medium to dense brus, in winter 5. Medium to dense brus, in summer d. Trees. Dense willows, summer, straigt. Cleared land wit tree stumps, no sprouts 3. Same as above, but wit eavy growt of sprouts 4. Heavy stand of timber, a few down trees, little undergrowt, flood stage below brances 5. Same as above, but wit flood stage reacing brances 0.05 0.00 0.05 0.045 0.070 0.0 0.080 0.00 0.060 0.070 0.00 0.50 0.060 0.00 0.0 0.045 0.070 0.060 0.080 0.0 0.60 0.00 0.080 0.0 0.60 D.3. Major streams (top widt at flood stage > 00 ft). Te n value is less tan tat for minor streams of similar description, because banks offer less effective resistance. a. Regular section wit no boulders or brus b. Irregular and roug section 0.05. 0.060 0.00 Capter : UNIFORM FLOW 45