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# Steady free-surface flows - PowerPoint PPT Presentation

Steady free-surface flows. Conservation of Mass. Simply it means that the mass of the system is constant.

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Presentation Transcript

Conservation of Mass
• Simply it means that the mass of the system is constant.
• The law of conservation of mass between different channel cross sections implies that the volumetric flow rates at these sections are equal to each other provided there is no lateral inflow or outflow.
• Continuity Equation

Liquid is assumed incompressible

(Chaudhry, 2008)

Water level rise in a reservoir
• A river discharge into a reservoir at a rate of 400,000 ft3/sec, and the outflow rate from the reservoir through the flow passage in the dam is 250,000 cfs. If the reservoir surface area is 40 mi2, what is the rate of rise of water in the reservoir?
Conservation of Momentum

The change in the linear momentum of the flow can be expressed as:

= sum of horizontal components of average velocities of flow at section 1 and 2, respectively

and = horizontal components of pressures acting at section 1 and 2, respectively

= horizontal component of W sin ϴ

= horizontal component of unknown force acting between sections 1 and 2

= weight of fluid between sections 1 and 2

= specific weight of fluid

momentum correction coefficients

= channel slope angle

(French, 1985)

With the assumption that, is small,, and then:

distances to centroids of respective flow areas and from free surface

Momentum function

• The lower limb AC asymptotically approaches the horizontal axis while the upper limb BC rises upward and extends indefinitely to the right.
• The M-y curve predicts two possible depths of flow (sequent depths of a hydraulic jump)

(French, 1985)

Force to Open an Elliptical Gate
• Q. An elliptical gate covers the end of a pipe 4 m in diameter. If the gate is hinged at the top, what normal force is required to open the gate when water is 8 m deep above the top of the pipe and the pipe is open to the atmosphere on the other side? Neglect the weight of the gate. Assume water temperature (10 °C, ν = 9810 N/m3

Example 3.10 from (Crowe, 2009)

Example Solution
• Solution Plan
• Calculate resultant hydrostatic force using
• Find the location of the center or pressure
• Draw a free body diagram of the gate
• Apply moment equilibrium about the hinge.
Example Solution
• Hydrostatic (resultant) force
• pressure at depth of the centroid
• area of elliptical panel (A = ∏ a b)
• Resultant force
Solution Conti’
• Center of pressure
• , where is the slant distance from the water surface to the centroid.
• Area moment of inertia of an elliptical panel
• Finding the center of pressure
• Moment equilibrium
Euler’s Equation of Motion
Specific Energy
• A concept introduced by Bakhmeteff (1912).
• Useful for application of Bernoulli equation
• Specific energy in a channel section is defined as the energy per pound of water at any section of a channel measured with respect to the channel bottom.

For a specific q

Specific Energy curves

Fig. 3-2 (Chow, 1959)

Application of Momentum and Energy Equations
• The momentum and energy equations should yield the same results if properly applied.
• Energy equation provides computational ease and conceptual simplicity as compared to momentum equation. (energy is scalar quantity while moment is a vector quantity)
• Channel transition, hydraulic jump and hydraulic jump at a sluice gate outlet applications
Channel Transition
• A change in the channel cross section (width or channel bottom slope)
• Channel transition is usually designed so that losses in the transition are small (H1 = H2)
• In example shown subcritical flow remains subcritical downstream of the transition
Example of channel transition
• A rectangular channel 10 ft wide is narrowed down to 8 ft by a contraction 50 ft long, built of straight walls and a horizontal floor. If the discharge is 100 cfs and the depth of flow is 5 ft on the upstream side of the transition section, determine the flow surface profile in the contraction (a) allowing no gradual hydraulic drop in the contraction, and (b) allowing a gradual hydraulic drop having its point of inflection at the mid-section of the contraction. The frictional loss through the contraction is negligable.
Solution (Part a)
• Total energy in the approaching flow measured above the channel bottom is
• This energy is kept constant throughout the contraction since energy losses are negligible.
• The alternate depths for the given total energy can be computed by

or

This is a cubic equation in which b is channel width.

• At the entrance where b=10 ft its solution gives two positive roots: a low stage which is the alternate depth; and a high stage , which is the depth of flow.
• At the exit section where b = 8 ft, this equation gives a low stage , and a high stage .
• With no gradual hydraulic drop is allowed in the contraction the depth of flow at the exit section should be kept at the high stage as shown.
• High stages for other intermediate sections are then computed by the above equation which gives the flow surface profile. Similarly, the low stages are computed by the above procedure and indicated by the alternate depth line.
Solution (Part b)
• When a gradual hydraulic drop is desired in the contraction, the depth of flow at the exit section should be at the low stage.
• Since the point of inflection of the drop or a critical section is maintained at the mid section of the contraction, the critical depth at this section is equal to the total head divided by 1.5 or
• The critical velocity is equal to , hence the width of this critical section should be
• With the size of the mid-section determined, the side walls of the contraction can be drawn in with straight lines.
• The critical depth line is shown to separate the high from low stage. The critical depth can be computed from the equation
• In reality the flow near the drop is more or less curvilinear and the actual profile would deviate from the theoretical one.
• The designer may fit any type of contraction walls he/she desires to suit a given flow profile, or vice versa.