Regulatory structures of the waterGeneral informationThese structures are used to distribute irrigation water within the system to a level and desired flow; so they used for desired control of water (level and flow) at the points of the system. A structure of regulation in an irrigation canal is used to control the flow, or raising the level of water or both. The type of structures that perform these functions are: Thomas or intakes, retention or retention dikes and retentions with jumps, Toms with hatches, structures distributing or parting.Some of these structures are structures, structures of flow distributing, making and delivery structures as retention.FunctionsThey must control the erosive action of the currents on the bottom and side of runways.Reguladorasno structures create reservoirs as a general rule, they act on the direction and magnitude of flow velocities.May belong to this group dykes, batteries of spurs, the dams, etc.Ensure the depths and conditions necessary for navigation and floating timber, create conditions for water catchment rivers, earn land to the sea, etc.LandfillWeir is a form of hydraulic control upstream. For establishing the level of the crest or a position, the flow through the structure can change only if the conditions of the water level change in the feeder canal. In the structures using landfills as control of water transfer from the channel of the Agency to the channels property, control of water is responsibility of the Agency and not of the users.Weir control is quite effective for the distribution of water among a group of users with minimal operational control channels, however this does not guarantee that there is a good fit between the water supply and the water required by the users.In by submerged hole hydraulic control, the flow is a function of the conditions both upstream as downstream. Although a gate hole can better represent the interests of both the users and the Agency. Flow can be easily changed, by regulating the opening of the hole in response to the demand of users change; However, it must be recognized that a greater degree of freedom of the regulacionya is wasted is by users as the Agency.Weir gives a precise measurement and flow regulation, but requires a greater difference in head that requires it submerged orifice. Alternatively, discharge or flow through sliding gates remain relatively constant even with slight variations in the surface of the water upstream, then the floodgates are preferable structures to control the flow in channels a constant flow.Landfills, for which the flow is very sensitive to the fluctuations of the water level upstream, they are preferable for the safe flow measurement.In its basic form a landfill is a vertical contrition that is placed normal to the direction of flow. For this type of hydraulic control, depth and flow rate are related by:Q = Cd Where Cd is a discharge coefficient, L is the width of the Weir, and H is the height above the crest of the Weir of water level. Note that for a specific location of the landfill, the flow is a function of the contributor canal water level or conditions upstream channel.Landfill of wide Ridge or diversion damWall heightTaking the maximum flow rate of the project, H(crt) is calculated by means of the formula:Q = r(s) B(t) H 3/2 (crt) Where: Q = flow rate in m 3 s;r(s) = degree of immersion;B(t) = width of the Weir perpendicular to the current;h (Crt) = height of the water surface waters above with respect to the diversion dam.The width of the wall of the structure is obtained by adding to H(crt) the height of the Weir on the sill and a proper topsides.Length of the StillingThe following formula is used to determine the length of the Stilling:LSB = 5 (H(J) 2 -H(J) 1) Where:l SB = length of the StillingH(j) 2 = theoretical height of water downstream from the hydraulic bossH(j) 1 = theoretical height of water upstream of the hydraulic boss.It adopts the coefficient "5" is the Stilling of trapezoidal section; However, there is still enough experience to demonstrate this. However, the dimensions chosen according to this formula have been satisfactory so far already built structures.The heights of the water, H(j) 1 and H(j) 2 are calculated by conventional methods, applying Bernoulli's equation.Dimensions of a tertiary outletThe diameter of the pipe of concrete is determined by assuming that the maximum speed of the water flowing through the pipe is 1 m/s, for the maximum flow rate of the tertiary canal project.Taking losses are determined according to the following formula:J(HR) = h(l, ent)+ h(l.f) + h(l.OL) =h(l.ent) = load loss in the inputh(l.f) = pressure drop in the pipeh(l.OL) = load loss in the outputJ(HR) = load loss in the work of Jack(IN) c = coefficient of Socket = 1.5(F) c = coefficient of friction = 0.023(P) v = velocity in a pipeExample: Is then exposed an example of project for a secondary channel that has a maximum capacity of 1000 l/s with an output of 250 l/s. The width of the bed of the secondary channel is of 1 m fall = 1.4 m.Project of Weir of regulation for the secondary channelIs projected a Creager section for a width of dam of 1 m and ana height of 0.40 m on the level of the bed or floor slab upstream.Height of the wallsBy applying the formula:
Data:Q = 1 m3/sr (s) = 2.00B (t) = 1,00 mH (crt) = 0.63 mWe get: Using an approximate topsides of 30 per cent of the depth of the water, we get:H (wall) = (0.40 + 0.63) * 1.3 = 1, 34mThis height of the walls of the structure applies also to the project of canal upstream, so the rise in load does not cause overflow.Length of the StillingFor a fall of 1, 40m in a depth of water in the channel and work upstream 1, 03 m, the speed of approximately upstream of the Weir will be:V1 = 0.97 m/s; And the total energy will be:HSB = 1,40 + 1.03 + 0, 05 = 2, 48mApplying Bernoulli's equation:
where:v(j) and H(j) are the conditions before the hydraulic boss.Now well
Substituting the value of v(j) 1 in equation
Applying the equation, H(j) 2 = 0, 5 H(j) 1get the depth of the water after the hydraulic boss.H(j) 2= 1, 10 mLength of the Stilling = (1.10 - 0.15) 5 = 4, 75mDraft decision on tertiary channelDownload of tertiary channel = 0.25 m 3 /sThe pipe, v(p) = 1, 00 m/sLength of the pipe, L(P) = 100, 00 mHydraulic surface = The selected pipe diameter = 0.62 m; (surface = 0, 29m2)Losses of load in the decision are the same a:J(HR) = h(l, ent)+ h(l.f) + h(l.OL)(ll.ent) h = 1.5 (loss of input)h(l.f) = (losses by reasoning)(L.out) h = 1.0 (output losses)J(HR) =This tells us that the water level upstream in the secondary channel must be 0, 08m above the level on the tertiary channel in order to derive 0, 25 m3s.Hydraulic design of retention In the hydraulic design of a retention it should be taken into account:a. The area of the central part of the body of retention must be such that the tie rod and speed remain roughly equal in the channel and retention in order to avoid lossesb. The total length of the measuring jugs Crest (the central part more side dumps) must be sized in such a way that invading part of the free edge will pass a percentage of the cost of design in order to ensure a proper functioning of the structure when failures occur in the operation of the system. We will set up that spending to pass is 20-40% of the cost of design and we'll pour it invading 80% free in particular edge.c. Pouring Crest speed should not be greater than 1.10 m/s since the operation of the boards is difficult.d. Transitions must be designed to prevent excessive height in order to maintain the water level as horizontal as possible.The steps to follow in the design are as follows:a. Determination of the width of the central part of the "B" check so that your area is equal to or greater than the area of the section of the canal
being: width B: of the central part of the retentionA: area of the section of the canald height of water retention and the channel
Definition of characteristic dimensions in retention
b. determination of expenditure which passes over the crest measuring jugs,
being: Qvert: spending that passes over the crest measuring jugsL: average measuring jugs Crest width is: H: 80% of free specifically rimC: coefficient of expenditure, its approximate value can be taken in the #1 table;usually you can take a value of 1.84It must be met to Qvert > or = Qnorma = 40% Qdiseno. (B) comply with this standard must be increased otherwise.c. determination of the velocity over the crest measuring jugs,
Being: Vvert: speed over the crest measuring jugsQvert: expenditure that it passes over the crest measuring jugsAvert: area slopeIt must be met to Vvert 1.10 m/s. (B) comply with this standard must be increased otherwise.d. determination of the lengths of the transitions in and out. These transitions, as you can see in the chapter on transitions, must have a length that the maximum angle of the surface of the water with the axis of the channel is 12 30 '.Example of calculation Determination of the width of the central part:
schematic form of withholding.
Determination of the expenditure side
> It is okDetermination of speed above the measuring jugs Crest:
Determination of the length of transition of input and output.Assuming the characteristics of the canal waters above and waters down the same retention is:
When it comes to deductions in prefabricated channels the width of the central part is identical to the of precast and will only be necessary to determine the total length of the measuring jugs crest to allow it to spend 20-40% of the cost of design invading 80% of the free edge.Example of calculation:
a width of the central portion (B)
b. determination of the measuring jugs Crest length (L)
clearing L is:L = (Qvert) / (C H3/2 ))Qvert = 40% Qdiseno = 0.40 * 0 085 = 0.034 m3/sH = 80% B L = 0.80 * 0 10 = 0.08 mH 3/2 = 0.0226C = 1. 92tomado of table 1.L = 0.034 / (1.92 * 0 0226) = 0.78 m = 0.80 m.c.Calculo of the speed above the crest measuring jugs.Vvert = Qvert / AvertQvert = 0.034 m3/sAvert = L * H = 0.80 * 0 08 = 0.064 m2Vvert = 0.034 / 0.064 = 0.53 m/s < 1.10 m/sd. determination of the transitions in and out.Assuming that the sections of the channel before and after the retention are the same.LTE = Lts (L - b) /(2*tg 12 30) =(0.8 0.43) / (2 * 0 22169) = 0.90 m.Table 3.5. Values of 'C' (expense ratio) for the calculation of deductions.H
D0.060.150.300.460.610.911.221.52
0.151.972.082.262.442.452.592.712.81
0.30.1.921.942.042.152.252.422.442.51
0.611.911.891.931.992.042.142.232.32
0.911.901.871.891.921.962.032.102.17
1.521.901.861.861.871.891.941.982.03
3.051.901.851.841.841.841.861.881.91
1.901.851.831.821.821.811.811.81
Definition of the relationship H versus d
Structure of control and pipeFor the design of these outlets we must bear in mind that the same be derived perpendicular to the feeder canal to facilitate both calculation and construction, therefore channels must derive normal feeder Canal.In the calculation, we will work with the norms established by the United States Bureau of Reclamation.We will usually use Gates that will be flattened to circular ducts, the dimensions of the floodgates being equal to the diameter of the duct, it will have a variable length for each specific case, except for special cases such as when the socket has to cross a road or any other structure we can set that length in 5 meters to allow a width envelope of the berm of the channel on the site for reasons of operation.Then the formula for the calculation of the losses is exposed through the floodgates of outlet and ductwork.E1 = E2 + loss (3.8)Energy input e1 =Y1 + Power output e2 = Y2 + Assuming enough approach thatY2 = D +V1 v2Lost through the circular slitQ = C.A0Where:Q = flow rate (m s)C = coefficient of discharge = 0.75 for inlet Louvreg = acceleration of gravity (m/s2 ))h = charge needed for to pass the flow rate Q (m)0 = Area of the pipe (m2 ))The speed of the flow through the gate"0V" is given by the expression:
Being = load speed0.78 = load loss to defeat or loss of load in the gateLoss by friction in the pipeDuct friction loss is directly proportional to "f" friction factor, length and load speed; and inversely proportional to the diameter.The expression is as follows:
Where: Vt = velocity in the pipeL = length of ductD = diameter of the ductf = friction factorHF = duct friction lossLoss of output transitionIn the transition from output losses are usually negligible in reason to that the speed in the duct and channel are similar.In the case of little refined transitions its value is 0.25 of the difference of loads of speed at the beginning and at the end of the transition.So the formula is as follows:HS = 0.25 Where:Vt = speed in the duct (m/s)V2 = speed in the channel's output (m/s)g = acceleration of gravity (m/s)This term is negligible and therefore not taken into account.The total head loss will be:Losses = 0.78 Applying continuity is0 0 V = en Vt clearing V0is:V0=But Cc; where DC = contraction coefficientV0 =And the equation (3.14) is transformed into:Loss = Substituting in the equation (a) must:
If the speed of approach, V1 = 0 then it is despicable and equation becomes:
It will be the formula to be used for the calculation of the socket. We will then see how to obtain each of the factors involved therein.Calculation of Vt in ductAs mentioned above we use circular ducts of diameter D equal to the dimensions of the inlet Louvre, and that Dlo will choose in such way that speed in the tube is similar to the branch canal to avoid any appreciable loss in output.Thus Vt = where att =Calculation of coefficient of contraction CcFor Gates vertical flat and over 10 h/d ratios normally may be a value of DC. = 0,644 without a significant mistake in the calculation.BeingD = pipe diameter (m)H = height of water before the gate, as you will see more later in the submergenceis equal to twice the diameter of the pipe (m)Calculation of friction factorThe friction factor is a function of the number of Reynolds "r" and the factorR = = Viscosity kinematics of water = 1, 12 x 10 -6 m 3 /s Vt = speed of the duct (m/s)D = diameter of the duct (m)e = thickness of the roughness = 0.001 foot concrete = 0, 0003mOnce obtained these values we get the "f" factor according to the formula of Swamme (1993). By substituting the values of Vt Cc and f in the equation (3.16) Gets the value of the loss of charge H.
As we have seen previously in channel outlets, the requirement in the regulation of the flow is more rigorous that in the plot, in which we will take care of the delivery of water in plots, and where we have been using simple structures that are usually placed steel pipes, iron ductile or galvanized iron and when to make use of stops always be preferred to the concrete Sockets.For hydraulic calculation we will take into consideration the following loss of loadby friction(b) by inputFor the calculation of friction losses, we will work with the modified by Darcy Chzy formula.
Where:hf = friction head loss (L)F = Factor that depends on the material and the State of the pipelineL = length of pipe (L)d = diameter of the pipe (L)V = velocity of the water in the tube (LT- 1 ))g = gravity Acceleration (LT- 2 ))The values of "f" will be available from the formula of Swame (1993) shown above.b. loss of charge per ticket will appoint it by I, and we will work with the following formula:
In which "ke" is a coefficient that depends on the degree of flaring from the entrance, for that purpose, the following values are given:For pipe incoming ke = 0.78For entry with edge angle ke = 0.50For entries with edges ke = 0.23For input flare ke = 0.04Where V = velocity in the duct (m/s)In our projects we will work with ke = 0.50 for that being the case of our structures. Also when using tap must be addressed head of the same losses. With these formulas can draughtsman with base to the flow necessary to derive and the required minimum load on site, calculate structure.We will use the formula presented above, which will give us the loss of charge in decision and thus we can outline the structure.Suppose that a trapezoidal channel it will derive a flow rate of 185 litres by segundo(0.185 m3/s)The characteristics of the feeder canal in site are the followingLower base or template (b) = 0.40 mWater depth (d) = 0.63 mArea (A) = 0.79 m 2 Slope of the Fund (s) = 0.0005Speed (v) = 0.742 m/sFlow rate (Q) = 0.625 m 3 /sWe know that speed in the duct of the structure must be similar to the branch canal, so let's assume that the speed on that channel is 0.80 m/s.Therefore;
Then scores are made to choose the gate that conforms to a pipe to let flow (0.185 m3s) without any appreciable loss. For example, we are looking for in a Handbook of hatches and select a sliding gate model 5.0 18 '' x 18 '' and a tube of diameter (D) of 18 "= 0.4572 m.
WhereT = duct Area where is Vt the speed in the ductAs speed is high (greater than 0.8 m/s), then choose one gate higher within the commercial catalogue.The gate which follows is a 24 "x 24" and 24 "tube = (0. 61m) in diameter.
as this speed is somewhat less than the branch canal can be passed to the calculation of H, recalling that:
Calculation of CcFor this you can use tables that are in specialized books; However, let's assume that it is 0.644Calculation of fThe number of Reynolds = R = where:Vt= tube = 0.633 m/sD = pipe diameter = 0.61 m= kinematic viscosity of the fluid = 1.12 x 10 -6 m/sR = Then we need to find the relationship where is the roughness of the material as concrete = 0.0003 = 0.3 x 10 -3 m and D = 0.61
Now we calculate fwhere f = 0. 01708o f = 0.017We will use a tube of 5 m lengthThis would be the loss of cargo due to the flow to pass through the structure, in order to ensure that it passes a little more flow and which is controlling the tail-lift approach up to a value of H = 0.07 m or 7 cm; then the loss of load that we would have in making is 7 cm; now moves on to outlining it.N 1 = flush of the feeder canal in site. Is assumed that N1 is 99.669, know the strap of the feeder canal which is 0.63 m; Thus N10 = 99.669 + 0.63 = 100,299 mN10 = 100,299 m (altitude level water on the feeder canal)Then assuming that the strap of the derived channel of 0. 40m; We can calculate:N2 = N10 - 0.40 - H = 100,299 - 0.40 - 0.07 = 99.829 (altitude background derived channel).For the calculation of N8 , we must comply with the rule of submergence (s); Like thisn 8= N10 - 2D = 100,299 - 2 x 0.61 = 99,079 m (elevation of the floor of the tube at the entrance)n 4 = N8 - 0.10 (for installation of the gate) = 99.079 - 0.10 = 99.069 m(dimension fund the recess of the gate)N3= N8 - L x 0.02 = 99,079 - 5x0.02 = 98.979 (altitude Fund at the end of the tube)The value of 0.02 corresponds to the slope of the pipen 5= N1 + d + BLc + BLT where:d = strap of the feeder canal = 0.63 mBLc = free in particular edge = 0.20 mBLt = free ground edge = 0.10 mn 5 = 99.669 + 0.63 + 0.20 + 0.10 = 100.599 (dimension of the top berm).n 6 = height of berm in derived channeln 7 = N2 + height of the derived channelAssuming the height of the canal at 0.54 mn 7= 99.829 + 0.54 = 100.369 (altitude derived channel berm)n 11= N2 + d (derived channel) = 99.829 + 0.40 = 100.229 (altitude level water in the)derived channel)n 9 = N + D + Hc = 99.079 + (0.5 * 0.61) + 1.42 = 100.804 (base head dimensionthe gate)For Hc will refer to table the manufacturer of sluice gates that for this example of a gate sliding model 5.0 24 "x 24" is 1.42 m.The hydraulic design is accompanied of a planted and tables with all the features of construction, for the found data. An outline of this Jack is shown in Figure 3.8.
take in diagrammatic form structure
http://www.virtual.unal.edu.co/cursos/sedes/palmira/5000117/docs_curso/contenido.html
