OPEN CHANNEL FLOW Open channel flow is a flow of liquid, basically water in a conduit with a free surface. The open channel flows are driven by gravity alone, and the pressure gradient at the atmospheric interface is negligible. Closed duct flows are full of fluid, have no free surface within, and are driven by a pressure gradient along the duct axis. That is a surface on which pressure is equal to local atmospheric pressure. Open channel flows are characterized by the presence of a liquid-gas interface called the free surface. Some are natural flows such as rivers, creeks and floods, some are human made systems such as fresh water aquaducts, irrigation, sewers and drainage ditches.
An open channel always has two sides and a bottom, where the flow satisfies the no-slip condition. Therefore even a straight channel has a three-dimensional velocity distribution. Some measurements of straight channel velocity contours are shown below. The profiles are quite complex, with maximum velocity typically occurring in the midplane about 0% below the surface.
COMPARISON OF OPEN CHANNEL FLOW AND PIPE FLOW OPEN CHANNEL FLOW Open channel flow must have a free surface. A free surface is subject to atmospheric pressure. Flow area is determined by the geometry of the channel plus the level of free surface, which is likely to change along the flow direction and with as well as time. The cross section may be of any from circular to irregular form of natural streams, which may change along the flow direction and as well as with time. The depth of flow, discharge and the slopes of channel bottom and of the free surface are interdependent. PIPE FLOW No free surface in pipe flow. No direct atmospheric pressure, hydraulic pressure only. Flow area is fixed by the pipe dimensions. The cross section of a pipe is usually circular. The cross section of a pipe is usually circular. No such dependence.
TYPES OF OPEN CHANNEL FLOWS The most common method of classifying open channel flow is by the rate of change of the free surface depth. The simplest and most widely analyzed case in uniform flow. Type of flow Steady flow Uniform flow Uniform steady flow Varied steady flow Varied unsteady flow Rapidly varying flow Gradually varying flow Description When discharge (Q) does not change with time. When depth of fluid does not change for a selected length or section of the channel. When discharge does not change with time and depth remains constant for a selected section. Cross section should remain unchanged, referred to as a prismatic channel. When depth changes but discharge remains the same. When both depth and discharge change along a channel length of interest. Depth change is rapid. Depth change is gradual.
. Rapidly varying flow. Gradually varying flow. Hydraulic jump 4. Weir and waterfall 5. Gradually varying 6. Hydraulic drop due to change in channel slope
PARAMETER USED IN OPEN CHANNEL FLOW ANALYSIS Hydraulic radius, R of open channel flow R is a ratio of flow cross sectional area, A and wetted perimeter (WP) R = A WP R : Hydraulic radius A : Flow cross sectional area WP : Wetted perimeter
Reynolds number for open channel flow. Re = ρvr VR = μ υ Re < 500 laminar flow Re > 000 turbulent flow Reynolds number for pipe flow Re = ρvd VD = μ υ Re < 000 laminar flow Re > 4000 turbulent flow
Froude number, Fr The Froude number (Fr) is a dimensionless number defined as the ratio of channel velocity to the speed of propagation of a small disturbance wave in the channel. For a rectangular or very wide constant depth channel, Froude number can be defined as : Flow velocity Fr = = Surface wave speed where, V gy h V = Velocity g = gravity y h = Hydraulic depth y h = A T A = Area T = Top width of the channel Fr <.0 Sub-critical flow Fr =.0 or when V = gyh Critical flow Fr >.0 Super-critical flow A combination of both numbers is used to describe channel flow conditions.
THE CHEZY FORMULA Uniform flow can occur in long straight runs of constant slope and constant channel cross section. The water depth is constant at y = y n, and the velocity is constant at V = V 0. The slope be S = 0 tanθ, where θ is the angle the bottom makes with the horizontal, considered positive for downhill flow. From Bernoulli equation, the head loss becomes: h f = z z = S 0 L where L is the horizontal distance between section and. Head loss from Darcy-Weisbach is: h R f D D h h h L V = f Dh g A = 4 R h = 4 WP = Hydraulic diameter = Hydraulic radius 8g V0 = R h S f 0 For a given shape and bottom roughness, the quantity 8g C = f 8g f is constant and can be denoted by C.
Finally, the velocity V0 can be expressed as : V 0 8g = f R h S 0 = C ( R hs 0 ) Q = CA ( RhS 0 ) These are called Chezy formulas, first developed by the French engineer Antoine Chezy in conjuction with his experiments on the Saine River and the Courpalet Canal in 769. The quantity C, called the Chezy coefficient, varies about 60 ft ½ /s for small rough channels to 60 ft ½ /s for large smooth channels (0 m ½ /s to 90 m ½ /s in SI units).
EXAMPLE A straight rectangular channel is 6 ft wide and ft deep and laid on a slope of. The friction factor is 0.0. SOLUTION C = 8g f = 8. 0.0 = 08 ft A = 6 = 8 ft A 8 R h = = =.5 ft WP + 6 + = tan S 0 s Q h = CA S 0 = ( R ) = ( 08)( 8) (.5 ( tan )) 450 ft /s For SI units, it can be used directly.
Uniform steady flow and Manning s equation When discharge remain the same and depth does not change, then we have uniform steady flow. In this condition, the surface of water is parallel to the bed of the channel. To make sure water (liquid) flow inside the channel, it must have certain angle of inclination, or the channel s slope. The slope of the channel (S) can be expressed as :- An angle = degree As percent = % As fraction = 0.0 or in 00
Manning s equation is used to estimate the velocity of flow in a channel. The SI units form of Manning s equation:- V. 0 = R n S V = Velocity of flow in a channel (m/s) n = Channel surface roughness. Values developed through experimentation. R = Hydraulic radius (A/WP) in meter S = Slope of the channel The English units form of Manning s equation:- V.49 = R n S V = Velocity of flow in a channel (ft/s) R = Hydraulic radius (A/WP) in feet
Example of the n values:-
The flowrate of a channel could be determined by :- Q = AV. 0 = AR n S where Q is in m /s. For uniform flow, Q is referred to as normal discharge. The above equation can also be re-arranged such that :- AR = nq S The left hand side equation is based on channel geometry.
Example #0 Determine normal discharge for a 00 mm inside diameter common clay drainage tile running half-full if the slope drops m over 000m.. 0 Q = AR n S n = 0.0 πd A = = 0.057 m 4 Wetted perimeter, WP = π D = 0.4 m A 0.057 R = = = 0.05 m WP 0.4 S = 0.00. 0 Q = AR S = 5.8 0 m /s n
Example #0 Calculate slope of channel if normal discharge is 50 ft /s. Channel is formed, unfinished concrete. English unit!!!.49 Q = AR S n A = ft WP = 9.66 ft A R = =.4 ft WP n = 0.07 S = Qn.49AR S = 0.0069 (Channel should drop.69 feet for every 000 feet length.
Example #0 Design a rectangular channel in formed, unfinished concrete with below mention specifications. Normal flowrate = 5.75 m /s S =.% Normal depth = half of the width of the channel. Since we have to design the channel, use this equation. AR = nq S B AR nq 0.07 5.75 = = = 0.89 S ( 0.0) Y=B/ B A = BY = B = WP = B + Y = B A B R = = WP 4 B B B AR = 4 B =.76 m Y = 0.88 m = 0.89
Example #04 In a rectangular channel as mention in Example #0, the final width was set at m and the maximum discharge is m /s. Find the normal depth for this maximum discharge. AR = nq S AR nq 0.07 = = =.8657 S A = Y WP = + Y A Y R = = WP + Y ( 0.0) Y=B/ AR = Y + Y ( Y ) =. 8657 Cannot solve directly. Use trial and error method. Can use MS Excel datasheet. From Excel, The normal depth must be.48 meter.
From MS Excel worksheet:
CONVEYANCE AND MOST EFFICIENT CHANNEL SHAPES. 0 Q = AR n S Other than the S term, all other terms are related to channel cross section and its features. These terms together are referred to as the Conveyance (K) of the channel. K. 0 = AR n Then, Q = K S R = A WP K is maximum when wetted perimeter (WP) is the least for a give area. This is also the most efficience cross section for conveying flow. For circular section, half full flow is the most efficient.
COMPOUND SECTIONS It is occurs when channel shape changes with flow depth. It is a typical idea in natural stream sections during flooding. At normal condition, water flows in the main channel. During floods, water spills over the flood plain. We need to know the flowrate, Q at various depths or vice-versa. So that we could design channels or determine channel safety for various flood magnitudes.
Example #05 Channel type: Natural channel with levees. Slope: 0.0005 Determine the normal discharge, Q for depth ft and 6 ft.
COMPOUND SECTION This is more realistic situation, where the channel roughness (value of n) may be different for floodplain than the main channel. In this case, we need to determine velocity for each sub-section, and then sum up the discharges for the sections. Example #06 Slope: 0.5% n for bank = 0.06 n for main channel = 0.0 Calculate discharge for depth of 8 feet?
A = 80 4 = 0 A = 50 8 = 400 A = 00 5 = 500 P = 80 + 4 = 84 P = 4 + 50 + = 57 P = 00 + 5 = 05 V i Q = Q =.49 Ai = ni Pi n i= A V i i S (.49)( 0.005) Q = 900 ft /s ( 0 / 84) 0 ( 400 / 57) 400 ( 500 /05) 0.06 + 0.0 + 0.06 500
ENERGY PRINCIPLES FOR OPEN CHANNEL FLOW Energy at particular point in the channel is potential energy and kinetic energy. Specific energy: V Q E = y + = y + g ga y = Depth of flow V = Velocity Q = Discharge A = Cross sectional flow area Total energy: V Q E = y + z + = y + z + g ga z = Height of the channel bottom from the datum
Example: Rectangular channel width = m Depth = m Q = 4.0 m/s Height above datum = m Determine the specific energy and total energy. Specific energy: m m m E = V y + g = + 4 ( )( 9.8)( ) =.0 m Datum Total energy: E = datum height + specific energy =.0 +. =. m
Specific energy diagram The specific energy can be plotted graphically as a function of depth of flow. V Q E = y + = y + g ga y = Static energy, E Q ga = Kinetic energy, E s (potential energy) k Relationship between y and static energy, E s
Relationship between y and kinetic energy, E k E k = Q ga For a rectangular channel; Substitute Q with specific discharge (discharge per unit width), Substitute area, A with, A = B y Q q = B Q E k = = q ga gy
EXAMPLE A rectangular channel, width is 4 m, flowrate is m /s and depth of flow is.5 m. Draw specific energy diagram Find critical and alternate depth? Q E = y + ga q E = y + gy Q q = = = m /s B 4
4.5 4.5 static E = y kinetic E total E.5 y.5 0.5 0 0 4 5 6 Energy
Meaning from the graph:
. The diagram applies for a given cross section and discharge.. As the depth of flow increases, the static energy increases, and the kinetic energy decreases. The total energy curve approaches the static energy curve for high depths and the kinetic energy curve for small depths 4. The specific energy is minimum (Emin) for a particular depth this depth happens to be the critical depth Depth for which the Froude s number =.0. velocity = Vc. 5. Emin only energy value with a singular depth! 6. Depths less than the critical depths supercritical flow. Froude Number >.0. V > Vc. 7. Depths greater than the critical depths subcritical flow. Froude Number <.0. V < Vc. 8. For all other energy values there are two depth associated one greater than the critical depth and one less than the critical depth. 9. The two depths associated with the same energy values are referred to as Alternate depths 0. As discharge increases, the specific energy curves move to the upper right portion of the chart.
E = E = Q y + ga q y + gy There is a minimum value of E at a certain value of y called the critical depth. de dy q = 0 = gy It shows that E min occurs at y c y c q = g Q = b g The associated minimum energy is ; E min = E = ( y c ) y c The depth, y c corresponds to channel velocity equal to the shallow-water wave propagation speed, C 0. Fr = V gy ( gy ) y V y q = gyc = c c = c c
By comparison it follows that the critical channel velocity is: V c = = ( gyc ) C0 Fr = For For E < E min, no solution exists. E > E min, two solutions are possible: () Large depth with V < V c, called sub-critical. () Small depth with V > V c, called super-critical. In sub-critical flow, disturbances can propagate upstream because wave speed C 0 > V. In super-critical flow, waves are swept downstream: Upstream is a zone of silence, and a small obstruction in the flow will create a wedge-shape wave exactly analogous to the Mach waves. The angle of these waves must be: μ = sin C0 V = sin ( gy) V The wave angle and the depth can thus be used as a simple measurement of super-critical flow velocity.
Critical depth, y c y c q = g = 9.8 = 0.97 m Minimum specific energy, E min E y = ( 0.97).457 m min = c = Since given depth was.5 m > 0.97 m, the given depth is sub-critical and the other depth should be super-critical. Now, determining alternate depths (Energy at.5 m) E = q y + gy =. 5 + 9.8.5 =.57 m This energy value is the same for the other alternate (super-critical) depth, so;.57 = y + 9.8 y Determine value of y by trial and error method. (use Excel ok)
super-critical alternate depth, y = 0.467 m
HYDRAULIC JUMP In open channel flow a supercritical flow can change quickly back to a subcritical flow by passing through a hydraulic jump The upstream flow is fast and shallow. The downstream flow is slow and deep. The hydraulic jump is quite thick, ranging in length from 4 to 6 time the downstream depth. It is very important that such jumps be located on specially designed aprons; otherwise the channel bottom will be badly scoured by the agitation. Jumps also mixed fluids very effectively and have application to sewage and water treatment designs.
EXAMPLE #0 Data:. Triangular channel with side slopes having ratio of :.5. Flowrate is 0.68 ft /s. Channel type clean, excavated earth Calculate:. Critical depth. Minimum specific energy. Plot specific energy curve 4. Determine energy for 0.45 ft and alternate depth 5. Velocity of flow and Froude number 6. Calculate require slope for given flow
SOLUTION FOR EXAMPLE #0 Used Excel. z =.5 Q = 0.68 ft /s n = 0.0 y = independence number Area, A = ( zy)( y) = zy Flow velocity, Q V = A Free surface width, T = zy Hydraulic depth, y h = A T
We cannot solve this question directly because the dimension of the triangular weir is not given. Trial and error method need to be applied. First step, create excel worksheet and calculate all the important parameter.
() Critical depth Critical depth occur at minimum energy, or at Froude number, Fr =.0 Find in the excel worksheet. Critical depth is 0.5 ft () Minimum specific energy, E min Minimum specific energy occur at critical depth. E min = 0.6675 ft
(c) Plot the graph y-axis is E (energy) and x-axis is depth (y) Specific energy.4. Depth (ft) 0.8 0.6 0.4 0. 0 0 0.5.5 Energy (ft)
(d) depth 0.45 ft, Energy = 0.70548 ft Alternate depth = 0.645 ft (which has same specific energy)
(e) Velocity for y = 0.450 ft, V =.4 ft/m, Fr =.5, super-critical flow for y = 0.645 ft, V =.09 ft/m, Fr = 0.6, sub-critical flow (f) slope? Take y = 0.45 ft For triangle, Cross section area, A = zy ( ) y ( )= zy = 0.075 Wetted perimeter, A = y + z =.65 Hydraulic radius, R = A WP = 0.075.65 = 0.87 Flowrate, Q =.49 n AR So, slope, S = S Qn (.49)AR = 0.0 Or foot drop for every 00 feets.
EXAMPLE #0 (0.) Water at 0 C flows in a 0 cm wide rectangular channel at a depth of 0 cm and a flowrate of 80,000 cm /s. Estimate (a) the Froude number and (b) the Reynolds number. EXAMPLE #0 (0.8) An earthquake near the Kenai peninsula, Alaska, creates a single tidal wave called tsunami that propagates southward across the Pacific Ocean. If the average ocean depth is 4 km and seawater density is 05 kg/m, estimate the time of arrival of this tsunami in Hilo, Hawaii. EXAMPLE #04 (0.) A rectangular channel is m wide and contains water m deep. If the slope is 0.85 and the lining is corrugated metal, estimate the discharge for uniform flow.
EXAMPLE #05 (0.7-0.8-0.9) The trapezoidal channel as shown below is made of brickwork and slopes at :500.. Determine the flowrate if the normal depth is 80 cm.. Determine the normal depth for which the flowrate will be 8 m /s.. Let the surface be clean earth, which erodes if V exceeds.5 m/s. What is the maximum depth to avoid erosion?
EXAMPLE #06 (0.84) Consider the flow under the sluice gate. If y =0 ft and all losses are neglected except the dissipation in the jump, calculate y and y and the percentage of dissipation, and sketch the flow to scale with the EGL included. The channel is horizontal and wide.
EXAMPLE #07 (0.89) Water 0cm deep is in uniform flow down a unfinished concrete slope when a hydraulic jump occurs. If the channel is very wide, estimate the water depth y downstream of the jump.
EXAMPLE #08 (0.9) Water in horizontal channel accelerates smoothly over a bump and then undergoes a hydraulic jump. If y = m and y = 40cm, estimate (a) V, (b) V, (c) y 4 and (d) the bump height, h.
EXAMPLE #09 (0.95) A 0 cm high bump in a wide horizontal water channel creates a hydraulic jump just upstream. Neglecting losses except in the jump, for case y = 0cm, estimate (a) V 4, (b) y 4, (c) V and (d) y.
EXAMPLE #0 (C0.6) Figure shows a horizontal flow of water through a sluice gate, a hydraulic jump, and over 6 ft sharp crested weir. Channel, gate, jump and weir are all 8 ft wide unfinished concrete. Determine (a) the flow rate, (b) the normal depth, (c) y, (d) y and (e) y 4.