VARIED FLOW IN OPEN CHANNELS

Transcription

1 Chapter 15 Open Channels vs. Closed Conduits VARIED FLOW IN OPEN CHANNELS Fluid Mechanics, Spring Term 2011 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 Energy equation applied to open channel: In chapter 10, 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 (thereʼs much more in chapter 15), but these do illustrate the main points of open channel flow.

2 Specific Energy: We make the following simplifications: The combination energy. is called the specific 1. Assume turbulent flow (! = 1). 2. Assume the slope is zero locally, so that z 1 = z Write pressure in terms of depth (y = p / "). 4. Assume friction is negligible (h L = 0). For our example (no slope, turbulent, ) The specific energy can be written in terms of discharge Q = V A (from continuity): 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 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. For given Q and E, usually have 2 allowed depths: Subcritical and supercritical flow.

3 Subcritical vs. Supercritical Flow Critical Depth and the Froude Number These 2 different types of flow are in fact observed: Example: Flow past a sluice gate. At the turning point (the leftmost point of the blue curve), there is just one value of y(e). This point can be found from Subcritical: Calm, tranquil flow. Supercritical: Rapid flow, whitewater. (Examples a and b above have different specific energy E) It can easily be shown (but we wonʼt do it here) that at Define the Froude number Flows sometimes switch from supercritical to subcritical: (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. (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).

4 Flow over a Bump Apply Bernoulli equation along free surface streamline (p=0): Which will it be? or As it turns out: Left = subcritical Right = supercritical Weʼll derive this using the Bernoulli equation for frictionless, steady, incompressible flow along a streamline: For a channel of rectangular cross-section, again we have Substitute Q = V z b into Bernoulli equation: (from last page) To find the shape of the free surface, take the x-derivative: Since subcritical: Fr < 1 supercritical: Fr > 1 Subcritical flow with dh / dx > 1 dy / dx < 0 Supercritical flow with dh / dx > 1 dy / dx > 0 Solve for dz / dx: if flow is subcritical if flow is supercritical

5 The Hydraulic Jump A look at the hydraulic jump in greater detail: Example: May want to know: 1. How does water depth change? 2. Where does jump occur? 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 ). It turns out that it is more useful to apply the momentum eqn.: Momentum balance: 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. The forces are hydrostatic forces on each end: (where and are the pressures at centroids of A 1 and A 2 )

6 and thatʼs actually all for this problem: For example, if y 1 and Q are given, then for rectangular channel 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. 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. The End