CHE 493: FLUID MECHANICSChapter 9: Flow in Open Channels

INTRODUCTIONOpen channel flow implies flow of liquids in channels open to the atmosphere or in partially filled conduitsCharacterized by the presence of a liquid-gas interface called the free surfaceMost of natural flows encountered in practice are open-channels flowEg: Rivers, floods, draining of rainwater through roofs, highways

UNIFORM/VARIED FLOWUniform flow - if the flow depth (average velocity) remains constantEncountered in long straight sections of channels with constant slope and cross section the liquid accelerates until the head loss due equals the elevation drop reaches terminal velocity uniform flow is establishedRemains uniform as long as the slope, cross section and surface roughness of the channel remain unchangedFlow depth is called the normal depth important characteristic for open-channel flowsNon-uniform/varied flow - Flow depth varies with distance in the flow direction

Uniform Flow in Channel

Head loss = Elevation Loss

Flow depth = yAverage flow velocity = VBottom slope = S0 = tan

During open channel, Head Loss = Elevation Drop

--------------- (1)

Since hL = S0L and Dh = 4Rh ------------------------- (2)

Sub (2) in (1):------------------------ (3)

Rearrange (3), uniform flow velocity:

where

Flow rate:Chezy CoefficientAntoine Chezy (1718-1798)Note: Determine using Moody chart, open channel typically is turbulent flow and fully develop.

Gauckler and Manning made recommendations:

Where: a = dimensional constant = 1 m1/3/sn = Manning coefficient (depends on roughness of the channel surface)

For uniform flow velocity & flow rate: and

Mean value for Manning coefficient

Types of ChannelsHydraulic radius

Circular channel

Rectangular channel

Trapezoidal channel

Example 1Water is flowing in a weedy excavated earth channel of trapezoidal cross-section with a bottom width of 0.8m, trapezoid angle of 60 and a bottom slope angle of 0.3. If the flow depth is measured to be 0.52 m, determine the flow rate of water through the channel. (Given n = 0.030)

SPECIFIC ENERGYConsider flow of a liquid in a channel

Where:y - flow depthV - average velocityZ elevation of the bottom of channel at that location relative to some reference datum

Total mechanical energy in terms of head:

Not realistic representing true energyIt can be realistic if the reference datum is taken to be the bottom of the channel so Z = 0Then, the total mechanical energy = Pressure + Dynamic HeadThis term is called specific energy, Es

------------------------ (1)

Consider flow in an open channel of constant width, b. Volume flowrate: .

So, the average flow velocity --------------- (2)

Sub (2) into (1)

There is minimum specific energy Es,min required to support specific flow rate, Q

Therefore, Es cannot be below Es,min for a given QSo,

Critical flow depth

Critical velocity

To find character and flow, using Froude Number

Lc = Critical LengthFr < 1 = Subcritical or tranquil flowFr = 1 = Critical flowFr > 1 = Supercritical or rapid flow

Example 2Water is flowing steadily in a 0.65 m wide rectangular open channel at a rate of 0.25 m3/s. If the flow depth is 0.15 m, determine(a) The flow velocity and type of flow(b) The alternate flow depth (Es1=Es2 ) if the character of flow were to change

HYDRAULIC JUMPIt called rapidly varied flow (RVF) if the flow depth changes markedly over a relatively short distance in the flow direction.Occur when there is a sudden change in flow, such as an abrupt change in cross section.RVF is complicatedsince there will be affect of backflow and flow separation.In compressible flow, a liquid can accelerate from subcritical to supercritical flowIt can also decelerate from supercritical to subcritical flow by undergoing a shock which is known as hydraulic jumpHydraulic jump involves considerable mixing and agitation and thus significant amount of mechanical energy dissipation

Consider steady flow through a control volume that encloses the hydraulic jump

Assumption from figure:Velocity is nearly constant across the channel at section 1 & 2 therefore the momentum flux correction factors 1 = 2Pressure in the liquid varies hydrostatically, we consider gage pressure only since atmospheric pressure acts on all surfaces and its effect cancel out.The wall shear stress and associated losses negligible relative to the losses that occur during the hydraulic jump due to intense agitation.The channel is wide and horizontalNo external or body forces

From momentum equation

For channel width b

Substituting and simplifying:

Eliminating V2 by from the continuity eqn gives:

Canceling factor y1 y2 from both side and rearranging gives:

where

Therefore, depth ratio:

The energy equation for this horizontal flow section can be expressed as:

Noting that; and

The head loss associated with hydraulic jump is expressed as:

The specific energy of the liquid before the hydraulic jump is

Then , the energy dissipation ratio:

Example 3Water is discharged into a 8 m wide rectangular horizontal channel from a sluice gate is observed to have undergone a hydraulic jump. The flow depth and velocity before the jump are 0.8 m and 7 m/s respectively. Determine:The flow depth and the Froude number after the jumpThe head loss and the dissipation ratioThe wasted power production potential due to the hydraulic jump