 # Pertemuan 19 - 22 Open Channel 2. Bina Nusantara VARIED FLOW IN OPEN CHANNELS.

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Pertemuan 19 - 22 Open Channel 2

Bina Nusantara VARIED FLOW IN OPEN CHANNELS

Bina Nusantara In a closed conduit there can be a pressure gradient that drives the flow. An open channel has atmospheric pressure at the surface. The HGL is thus the same as the fluid surface. Sketch of downhill flow in an open channel Open Channels vs. Closed Conduits

Bina Nusantara In chapter before, we looked at some uniform open channel flows. Now we deal with varied flow which is steady but nonuniform. (Flow is constant in time, but velocity and depth may vary along the flow). We will only deal with two very simple cases here (there’s much more in chapter 15), but these do illustrate the main points of open channel flow.

Bina Nusantara Energy equation applied to open channel:

Bina Nusantara We make the following simplifications: 1. Assume turbulent flow (  = 1). 2. Assume the slope is zero locally, so that z 1 = z 2. 3. Write pressure in terms of depth (y = p /  ). 4. Assume friction is negligible (h L = 0).

Bina Nusantara Specific Energy: The combination is called the specific energy. For our example (no slope, turbulent, …)

Bina Nusantara The specific energy can be written in terms of discharge Q = V A (from continuity): For a channel with rectangular cross-section, A = b y, (where b is the width): For a given Q, we now have E in terms of y alone.

Bina Nusantara Thus, for flat slope (+ other assumptions…) we can graph y against E: (Recall for given flow, E 1 = E 2 ) Curve for different, higher Q. For given Q and E, usually have 2 allowed depths: Subcritical and supercritical flow.

Bina Nusantara Subcritical vs. Supercritical Flow These 2 different types of flow are in fact observed: Example: Flow past a sluice gate. Subcritical: Calm, tranquil flow. Supercritical: Rapid flow, “whitewater”. (Examples a and b above have different specific energy E)

Bina Nusantara Critical Depth and the Froude Number At the turning point (the left- most point of the blue curve), there is just one value of y(E). This point can be found from It can easily be shown (but we won’t do it here) that at

Bina Nusantara Define the Froude number (Recall that the Reynolds number is the ratio of acceleration to viscous forces). The Froude number is the ratio of acceleration to gravity. Perhaps more illustrative is the fact that surface (gravity) waves move at a speed of Flows with Fr < 1 thus move slower than gravity waves. Flows with Fr > 1 move faster than gravity waves. Flows with Fr = 1 move at the same speed as gravity waves.

Bina Nusantara Flows sometimes switch from supercritical to subcritical: (The switch depends on upstream and downstream velocities; our theory is not sufficient to determine which type of flow the fluid chooses) Gravity waves: If you throw a rock into the water, the entire circular wave will travel downstream in supercritical flow. In subcritical flow, the part of the wave trying to travel upstream will in fact move upstream (against the flow of the current).

Bina Nusantara Flow over a Bump Which will it be? or As it turns out: Left = subcriticalRight = supercritical We’ll derive this using the Bernoulli equation for frictionless, steady, incompressible flow along a streamline:

Bina Nusantara Apply Bernoulli equation along free surface streamline (p=0): For a channel of rectangular cross-section, again we have

Bina Nusantara Substitute Q = V yb into Bernoulli equation: To find the shape of the free surface, take the x-derivative: Solve for dz / dx:

Bina Nusantara (from last page) Since subcritical: Fr < 1 supercritical: Fr > 1 Subcritical flow with dh / dx > 1 dy / dx < 1 Supercritical flow with dh / dx > 1 dy / dx > 1 if flow is subcriticalif flow is supercritical

Bina Nusantara The Hydraulic Jump Example: May want to know: 1. How does water depth change? 2. Where does jump occur?

Bina Nusantara A look at the hydraulic jump in greater detail: Note that there is a lot of viscous dissipation ( = head loss ) within the hydraulic jump. So our previous analysis does not apply to the jump (and unless we know V 1, V 2, y 1, y 2, and Q, we cannot determine h L ).

Bina Nusantara It turns out that it is more useful to apply the momentum eqn.: Why? Because there is an unknown loss of energy (where mechanical energy is converted to heat). But as long as there is no friction along the base of the flow, there is no loss of momentum involved.

Bina Nusantara Momentum balance: The forces are hydrostatic forces on each end: (where and are the pressures at centroids of A 1 and A 2 )

Bina Nusantara … and that’s actually all for this problem: For example, if y 1 and Q are given, then for rectangular channel is the pressure at mid-depth. So entire left-hand side is known, and we also know the first term on the right-hand side. So we can find V 2.

Bina Nusantara There are obviously many more applications. For example, now that we have V 2 we could find h L (by using the energy equation)… But this is enough for this course.

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