Non-uniform Flows in Open ChannelFaculty of Engineering & TechnologyINTI University College

Open Channel FlowsSteady flow velocity, pressure and cross-section do not change with timeUnsteady flow velocity, pressure and cross-section change with time

Uniform flow velocity or depth is same at every sectionNon-uniform flow velocity or depth is not same at every section

Open Channel FlowsSteady uniform flow Steady uniform flow will occur in long channels (of constant cross-section and slope) over that portion (which is far enough from entry or exit) for the flow to have reached its terminal velocity. Under these conditions, the depth is constant and known as the normal depth.

Steady non-uniform flow At entry and exit where the depth is varying and wherever the cross-section is changing as would be the case in most rivers and natural channels, steady non-uniform flow will occur if conditions do not change with time. This is also referred to as varied flow.

Uniform flows in open channels

Chezys formula Chezy Formula: (m/s) where v = mean velocity (m/s)C = Chezy's coefficient (L1/2T-1) m = hydraulic radius = Area/Wetted perimeter = A/P i = slope of energy lineFor steady uniform flow, the slope of the energy line, i is equal to the bed slope, S (m/s) where C = Chezy's coefficient (L1/2T-1) R = hydraulic radius = Area/Wetted perimeter = A/PS = bed slope

Discharge Q = ACmi (m3 /s)

Mannings formulaMannings formula: (m/s)

where v = mean velocity (m/s) n = Mannings roughness coefficient (L-1/3T) m = hydraulic radius = Area/Wetted perimeter i = slope of energy lineFor steady uniform flow, the slope of the energy line, I is equal to the bed slope, S (m/s) where n = Mannings roughness coefficient (L-1/3T) R = hydraulic radius = Area/Wetted perimeterS = bed slope Discharge (m3 /s)

Table 15.2: Values of n in Mannings formulaSurface of channelRoughness n Neat cement0.010 to 0.013Cement mortar0.011 to 0.015Concrete, in situ0.012 to 0.018Concrete, precast0.011 to 0.013Brick with cement mortar0.012 to 0.017Canals, earth, straight and uniform0.017 to 0.025Natural streams, clean, smooth and straight0.025 to 0.035Natural streams, rough0.045 to 0.060

Uniform flow formulaeContinuity equation: Q = AV

Chezy equation:

Mannings equation:

Geometry of Channel Cross-Sections

Geometry of Channel Cross-Sections

1 1

ZX-sectional Area A = (B + zD)DWetted Perimeter P = B + 2D1+z2 Hydraulic radius R = A/P

Channel Geometric PropertiesDepth of water = D Wetted Parameter = PCross-sectional area = A

Hydraulic radius R = A/P

Top width (or) Surface width = T (or Bs) Hydraulic Mean depth = Dm = A/T or Dm = A/Bs T (or) BsDArea = A

Example 1A rectangular open channel has a width of 4.5 m and a slope of 1 vertical to 400 horizontal. Find the mean velocity of flow v and the discharge Q when the depth D of water is 1.2 m, if C in the Chezy formula is 49 in SI units. A = BD = 4.5 x 1.2 = 5.4P = B + 2D = 4.5 + 2 x 1.2 = 6.9 DR = A/P = 0.784 BChezy formula:Mean velocity V = 49 x (0.784 x 1/400) = 2.164 m/sDischarge Q = AV = 5.4 x 2.164 = 11.718 m3/s

Example 2A trapezoidal canal has a bottom width B of 3 m and sides with a slope of 1 vertical to 2 horizontal. The slope of the bed is 1 in 5000 and the depth of water D is 1.2 m. Using the Manning formula with n = 0.025, calculate the rate of discharge in m3/s.

A = (B + DZ) D = (3 + 1.2 x 2) x 1.2 = 6.48 m2P = B + 2D (1 + Z2) = 3 + 2 x 1.2 x (1 + 22) = 8.37 m R = A/P = 0.775 1 D 2Mannings formula: BMean velocity V = 1/0.025 x (0.775)2/3 (1/5000)1/2 = 0.476 m/sDischarge Q = AV = 6.48 x 0.476 = 3.09 m3/s

B + 2zD

Example 3A rectangular channel has a width B of 2.4 m and a slope i of 1 in 400. What will be the depth of water D if the rate of flow Q is 8.5 m3/s and the coefficient C in the Chezy formula is 51 in SI units?

A = BD = 2.4 x DP = B + 2D = 2.4 + 2DR = A/P DChezy formula: or B8.5 = 2.4 D x 51 x (2.4D/(2.4 + 2D) x (1/400))1.388 = (2.4 D/(2.4 + 2D))1/2DBy trial and error,D =1 R.H.S. = 0.7385 R.H.S L.H.SD = 2= 1.732D = 1.65= 1.325D = 1.66= 1.385 R.H.S = L.H.SD = 1.66 m

Example 15.1

Example 15.1

Non-uniform flows in open channels

Specific EnergySpecific energy is defined as the energy per unit weight of water at a cross section measured above bed level at that point. If D is the depth and v is the velocity, E = D + V2/2g (1) Consider a rectangular channel, width B, cross-sectional area A, through which there is a volume rate of flow Q, V = Q/A = Q/BDE = D + q2/2gD2 where q = Q/B (2)

D3 ED2 + q2/2g = 0 (3)

This equation has three roots of which two are positive and real and the other is negative and unreal. BD

Alternative depths of flowFor a constant value of E, there are 2 alternative depths for a given value of qFor a constant value of q, there are 2 alternative depths for a given value of EThe larger depths correspond to deep slow flow (or) subcritical flowThe smaller depths correspond to shallow fast flow (or) supercritical flow

Alternative depths of flowThere is a critical depth Dc at which the 2 roots coincide, when the discharge q for a given specific energy E is maximum (qmax) and the energy E required for a given discharge q is a minimum (Emin) .

Alternative depths of flow

Critical depth Dc in Rectangular ChannelsAssume q is constant: E = D + V2/2g E = D + q2/2gD2dE/dD = 1 2q2/2gD3 = 0q2 = gD3 (or) Dc= (Q2/gB2)(1/3)

E = Dc + gDc3/2gDc2 at critical flowE = Dc + Dc/2 = 3/2Dc Emin= (3/2)DcThus, critical depth of flow Dc in a rectangular channel will be 2/3 EBDDv2/2gv

Critical depth Dc in Rectangular ChannelsAssume E is constant: E = D + q2/2gD2 D3 ED2 + q2/2g = 0 q2/2g = ED2 - D3q = [2g(ED2 - D3)] = D[2g(E - D)] = (2g) [(E D) D] dq/dD = 2g[(E - D) + D(E - D)- ()(-1)]For maximum discharge, when D = Dc , dq/dD = 0(E D) = D/(E D)(E Dc) = Dc Emin = 3/2 Dc

The same result have been obtained.

BD

Maximum discharge qmax at critical depth DcE = D + V2/2gE = D + q2/2gD2q2 = 2gD2(E - D)q = D[2g(E - D)]

At critical flow, E = (3/2) Dq = D[2g(3/2D D)] = D[2g(D)] = [gD3] = g1/2 D3/2 BD

Critical velocity Vc in Rectangular channelsVelocity of flow corresponding to critical depth Dc is known as critical velocity VcE = D + V2/2g3/2Dc = Dc + Vc2/2gVc2/2g = Dc/2

Froude No:

At critical flow, Vc /gDc = 1 = Frwhen Fr = 1 Vc = gDc critical flowwhen Fr < 1 V < gD subcritical flowwhen Fr > 1 V > gD supercritical flowBD

Example on rectangular channelA rectangular channel 3.6 m wide carries 2.5 m3/s. What will be the critical depth, critical velocity and critical slope? Use Mannings n = 0.02

Q = 2.5 m3/s, B = 3.6 m, q = Q/B = 2.5/3.6 = 0.694m3/s/m (or) 0.694m2/s

= (0.6942/9.81)1/3 = 0.366 m

= (9.81x0.3661/2 = 1.896 m/s

Mannings eqn. R = A/P = BD/(B+2D) = 3.6x0.366/(3.6+2x0.366) = 0.304 m

1.896 = (1/0.02)(0.304)2/3S1/2 S1/2 = 0.0838 Sc = 0.007

BD

Channel Geometric PropertiesDepth of water = D Wetted Parameter = PCross-sectional area = A

Hydraulic radius R = A/P

Top width (or) Surface width = T (or Bs) Hydraulic Mean depth = Dm = A/T or Dm = A/Bs T (or) BsDArea = A

Critical flow in Non-rectangular ChannelsE = D + V2/2g = D + Q2/2gA2For critical flow, dE/dD = 0 for a given value of Q dE/dD = (or) From figure, dA = Bs dD(or) dA/dD = Bs

(or)

where Bs and A are the surface breadth and area of the flow at critical flow.Critical velocity Vc = Q/A = (g A3 /Bs) (1/A) = (gA/Bs) where Dmc = A/Bs = Area/Surface width (at critical flow). Dmc = mean depth at critical flow

Critical depth Dc in Non-rectangular ChannelsCritical velocitywhere Dmc = mean depth at critical flow Dmc = Ac/Bs = Area/Surface width (at critical flow) Ac = function of critical depth (Dc)

Froude No: where Dm = mean depth

when Fr = 1v = gDm critical flowwhen Fr < 1v < gDm subcritical flowwhen Fr > 1 v > gDm supercritical flow

Example on trapezoidal channelA channel of trapezoidal cross-section, with a base width of 0.6 m and side slopes 45 carries 0.34 m3/s. Determine the critical depth.

Side slope 45 z = 1Critical velocity

Q = A V = AcVc = = Ac(gAc /Bs) where Bs is the surface width or top widthQ2 = Ac2g Ac/Bs = Ac3g/Bs

Ac = (B + zDc)Dc = BDc + zDc2 and Bs = B + 2zDc Q2 = Ac3g/Bs Q2Bs = Ac3g Q2(B + 2zDc) = (BDc + zDc2)3g

BD1z

Example on trapezoidal channelQ2(B + 2zDc) = (BDc + zDc2)3 g(0.34)2(0.6 + 2Dc) = (0.6Dc+ Dc2 )3x 9.81

= 84.8616 i.e., f(Dc) = 84.8616

By trial & error,If Dc = 0.3, f(Dc) = 60.966= 0.2 = 244.14= 0.26 = 100.18= 0.27 = 87.95= 0.273 = 84.65

critical depth Dc = 0.273 m

Problem: Non-uniform flows in a channelWater flows down a rectangular channel 4.0 m wide at a depth of 1.8 m. The discharge is 18.0 m3/s. (i) Determine whether the flow is subcritical or supercritical. (ii) Calculate the alternative depth of flow that could occur in the channel at the same discharge. (iii) Determine the critical slope of the channel assuming that it is lined with concrete with a Manning n of 0.012 s/m1/3.

Problem: Non-uniform flows in a channelRect. channel: B = 4.0 m, D = 1.8 m, Q = 18 m3/s, n = 0.012; q = Q/B = 18/4 = 4.5 m2/s(i) To find whether the flow is subcritical or supercritical Dc = (q2/g)1/3 = (4.52/9.81)1/3 = 1.273 m i.e. D > Dc subcritical flow

(or) = 2.5/(9.81x1.8) = 0.59 < 1 i.e. Fr < 1 subcritical flow

(ii) To find the alternative depth of flow: E1 = E2 = E = constant = energy line 1.8 + = D2 + 1.8 + = D2 + 1.8 + 0.3186 = D2 +

Problem: Non-uniform flows in a channel2.1186 = D2 +

By trial and error,D2 = 1 m, RHS = 2.0321 = 0.9 m, = 2.174 = 0.95 m, = 2.0936 = 0.94 m, = 2.108 Alternative depth (second depth) = 0.94 m

(iii) Critical slope Sc

Problem: Non-uniform flows in a channel

Ac = BDc = 4x1.273 = 5.092 m2, P = B + 2Dc = 4 + 2x1.273 = 6.546 mR =A/P = 5.092/6.546 = 0.778 m

= 0.0502

Sc = 0.0025 (or) Bed slope = 1 in 400

Occurrence of Critical Flow ConditionsSince at the critical depth, the discharge is a maximum for a given specific energy, cross-sections at which the flow passes through the critical depth are known as control sections.

Such sections are a limiting factor in the design of a channel and can be expected to occur under the following circumstances.

Transition from subcritical to supercritical flow (Channel Transitions)Entrance to a steep-slope channel from a reservoirFree outfall from a channel with a mild-slope Change of bed level or channel width

Occurrence of Critical Flow ConditionsTransition from subcritical to supercritical flow (Channel Transitions)

This may occur where there is a change of bed slope S. Upstream, the slope is mild and S < Sc. Downstream, the slope is steep and S > Sc. The depth will change smoothly from D1 to D2 and at the break in the slope, the water depth will pass through the critical depth forming a control section which regulates the depth upstream.

The reverse transition from supercritical to subcritical flow occurs abruptly by means of a hydraulic jump.

Occurrence of Critical Flow ConditionsEntrance to a steep-slope channel from a reservoir

Reservoir

If the depth of flow in the channel is less than the critical depth for the channel, the water surface must pass through the critical depth in the vicinity of the entrance, since conditions in the reservoir correspond to subcritical flow.

Occurrence of Critical Flow ConditionsFree outfall from a channel with a mild-slope

If the channel bed slope S < Sc , the upstream flow will be subcritical.At the outfall, there is no resistance to flow, so that, theoretically, it will be a maximum and the depth should be critical.In practice, the critical depth occurs at a distance of 3Dc to 10Dc from the brink.

If the channel bed slope S > Sc , the upstream flow will be supercritical and the depth will everywhere be less than the critical depth.

Occurrence of Critical Flow ConditionsChange of bed level or channel widthUnder certain circumstances, flow will occur at critical depth if a hump is formed in the bed of the channel or if the width of the channel is reduced.

Change of bed level Broad-crested Weir

Change of channel width Venturi Flume

Broad-crested Weir

Flow over a Broad-crested Weir

Case (a): Critical flow over the weirFor a rectangular channel,

=

z

Flow over a Broad-crested WeirAlso

Q = 1.705BE3/2 where E = H + V2/2g

If the depth upstream is large compared to the depth over the weir, V2/2g is negligible and the above equation can be written as, Q = 1.705BH3/2 where H = head above the crest level.

If the level of the flow downstream is raised, the surface level will draw down over the hump, but the depth may not fall to the critical depth. [See Case (b)]

Flow over a Broad-crested Weir

Case (b): Water level do not fall to the critical depthRate of flow (Q) can be calculated by applying energy equation and the continuity equation and will depend upon the difference in surface level upstream and over the weir.

Applying energy equation between 1 and 2, assuming no losses,

Flow over a Broad-crested WeirContinuity equation, BD1V1= BD2V2

where h = difference in surface level at upstream and over the weir.Discharge

Problem: Flow over a Broad-crested WeirWater flows across a broad-crested weir in a rectangular channel 400 mm wide. The depth of the water just upstream of the weir is 70 mm and the crest of weir is 40 mm above the channel bed. Considering the velocity of approach, calculate the fall of the surface level and the corresponding discharge.

Flow over a Broad-crested weir (using formula)

Assume a free fall over the weir:Q = 1.705 BE3/2 where E = h1 + v12/2g h1 = y1 z = 0.07 0.04 = 0.03 m

Assume V1 = 0 for 1st trial:Q = 1.705 Bh13/2 = 1.705(0.4)(0.03) 3/2 = 0.00354 m3/s v1 = Q/A = 0.00354/(0.07x0.4) = 0.126 m/s

Assume V1 = 0.126 m/s for 2nd trialE = h1 + v12/2g = 0.03 + 0.00082 = 0.03082 mQ = 1.705 BE3/2 = 1.705(0.4)(0.03082) 3/2 = 0.00368 m3/s

h1V12/2gZ = 0.04ycEy1 = 0.07h

Flow over a Broad-crested weir (using formula)

Q = 1.705 BE3/2 = 1.705(0.4)(0.03082) 3/2 = 0.00368 m3/s

= = (0.003682 /9.81x0.42) 1/3 = 0.0205 m

h = y1 z yc = 0.07 0.04 0.0205 = 0.0095 m = 9.5 mm

h1V12/2gZ = 0.04ycEy1 = 0.07h

Effect of lateral contraction of a channelWhen the width of a channel is reduced while the bed remains flat, the discharge/width , q increases.

If losses are neglected, the specific energy E remains constant and therefore, for subcritical flow, the depth will decrease, whilefor supercritical flow, the depth will increase as the channel narrows.

1 2

Q

q1 q2 B1,D1 B2 ,D2 Plan view of channel

Venturi Flume

Venturi Flume

Venturi FlumeA lateral contraction followed by an expansion can be used for flow measurement as an alternative to the broad-created weir.

If the conditions are such that the free surface does not pass through the critical depth, the arrangement forms a venturi flume.

Venturi FlumeIf the degree of contraction and the flow conditions are such that the upstream flow is subcritical, then the free surface passes through the critical depth in the throat. The supercritical flow beyond the throat will then revert to subcritical flow downstream by means of an hydraulic jump or a standing wave. A venturi flume operating in this mode is known as a standing wave flume.

Venturi Flume

Venturi FlumeFigure (a) Venturi Flume For continuity of flow, B1D1v1 = B2D2v2.(1)Applying energy equation to the upstream and throat sections and ignoring losses, D1 + v12/2g = D2 + v22/2g(2)Substituting for v1 from equation (1),

Discharge, Q = B2D2V2 =

Venturi Flume

Discharge, Q =

Actual discharge,

where Cd is a coefficient of discharge ( 0.95 to 0.99 ).

Venturi FlumeFigure (b) Standing wave FlumeIf the degree of contraction and the flow conditions are such that the upstream flow is subcritical, then the free surface passes through the critical depth in the throat.

Specific energyE = Dc + Vc2/2g (or)Vc = [2g(E-Dc)]

Discharge,Q = B2Dcvc = B2Dc[2g(E-Dc)]

where E is the specific energy measured above the bed level at the throat Critical depth,

Discharge, Q = B2 x (2/3)E(2gx(1/3)E)

Q = 1.705BE3/2

Venturi FlumeAssuming if the upstream velocity head is negligible,Q = 1.705BH3/2

where H is the height of the upstream face above the bed level at the throat.

Venturi Flume with a humpIn some cases, in addition to the lateral contraction, a hump is formed in the bed (as shown in the figure below), in which case H = D1 - Z.DischargeQ = 1.705BH3/2 where H = D1 Z

Ex 16.2: Venturi flumeA venture flume is formed in a horizontal channel of rectangular cross-section 1.4 m wide by constricting the width to 0.9 m and raising the floor level in the constricted section by 0.25 m above that of the channel. If the difference in levels of the free surface between the throat and upstream is 30 mm and both the upstream and downstream depths are 0.6 m, calculate the volume rate of flow. If the downstream conditions are changed so that a standing wave forms clear of the constriction, what will be the volume rate of flow if the upstream depth is maintained at 0.6 m?

Ex 16.2: Venturi flumeVenturi flume:Discharge

= 0.2352 m3/s

Standing wave flume:Discharge Q = 1.705BE3/2 If it is assumed that E = H = D1 Z = 0.35 m, then Q = 1.705x0.9x0.353/2 = 0.3177 m3/s

Effect of neglecting upstream velocity V1 :V1 = Q /B1D1 = 0.3177/(1.4x0.6) = 0.3782 m/s V12 /2g = 0.0073 mE = H + V12 /2g = 0.35 + 0.0073 = 0.3573 Q = 1.705x0.9x0.35733/2 = 0.3277 m3/s

% error in Q = Q/Q = (0.3277 0.3177)/ 0.3177 = 0.03 or 3%

Crump Weir

Crump Weir1:21:5ychEnergy linevv2 /2gVc2 /2g

Crump Weir

ycVc2 /2gP

Crump Weir

Sluice gatesUse to control the flow in rivers and man-made channelsSometimes referred to as underflow gatesAlso use to measure the discharge

Sluice gates

Problem: Broad-crested weir and Venturi flumeUniform flow occurs at a depth of 1.5 m in a long rectangular channel 3 m wide and laid to a slope of 0.0009. If Mannings n = 0.015, calculate (i) maximum height of hump on the floor to produce critical depth (ii) the width of contraction which will produce critical depth without increasing the upstream depth of flow.

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