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Flood Routing 2 Review of Energy Conservation Energy Grade Line Gradually Varied Flows, Head Loss Runge-Kutta Routing

Flood Routing 2 Review of Energy Conservation Energy Grade Line Gradually Varied Flows, Head Loss Runge-Kutta Routing. Review of Conservation of Energy.

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Flood Routing 2 Review of Energy Conservation Energy Grade Line Gradually Varied Flows, Head Loss Runge-Kutta Routing

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  1. Flood Routing 2Review of Energy ConservationEnergy Grade LineGradually Varied Flows, Head Loss Runge-Kutta Routing

  2. Review of Conservation of Energy • Conservation of energy says you can’t create energy or destroy it. You can only shuffle it around into various places. You can store energy as: • --mechanical energy (energy of motion) • --potential energy (energy of position) • --molecular energy (pressure) • --chemical energy • --heat

  3. Components for incompressible, frictionless water • Movement (kinetic) energy: • Potential (height)energy: • where h is the height above some reference elevation and M is the mass. • Pressure (P) has units of force per unit area. It is a measure of the molecular energy. To give it units of energy , multiply by a unit volume, v. Discussion: work

  4. Bernoulli Equation • So, if we ignore heat (friction), we can make an expression for conservation of energy: • The total energy doesn’t change from place 1 to place 2, although it may change form. • We assumed no heat changes due to friction, and water is incompressible

  5. Alternate forms of Bernoulli • Mass is an extensive property, its value depends on the amount of matter. Instead divide mass M by the unit volume v to get the density r of the fluid: • Density is an intensive property, i.e. it doesn’t depend on the amount of matter present. • Now we have Cons. of Energy per unit volume.

  6. Alternative forms of Bernoulli 2 • Water is incompressible, so the fluid density ρ is a constant, and gravity acceleration g is a constant. Define another constant gamma, as γ = ρg . Divide by γ = ρg and rewrite the equation: Units are now depth, and each term is called head, a height.

  7. Energy Head • In fluid dynamics, energy head is a concept that relates the energy in an incompressible fluid to the height of an equivalent static column of that fluid. From Bernoulli's Principle, the total energy at a given point in a fluid is the energy associated with the movement of the fluid, plus energy from pressure in the fluid, plus energy from the height of the fluid relative to an arbitrary datum.

  8. Energy heights in Hydrology - Pressure • Energy head is expressed in units of height such as meters or feet. What is the physical meaning of this height? For the pressure term, It’s the height to which pressurized water would rise in a well. Remember, the water isn’t moving. • Pressure as weight (a force) of water above per unit area

  9. Energy lengths in Hydrology – movement • For the Kinetic Energy velocity head, it’s the height the water would rise to — if it hit a vertical wall—the faster the water is moving, the higher it will rise. 1/2 rV2 P Pitot (left)and static pressure gages

  10. Energy lengths in Hydrology height • The Potential Energy elevation head is the difference in height from one place to some datum. • We often divide the elevation head into two pieces, the elevation of the channel bottom above some datum (maybe sea level, for example) z, and the depth of the water, d. h just equals z + d.

  11. Open Channels • In unconfined flows (open channel flows), water is open to the sky. • There can be no large pressure differences between one section of the stream and another, so we can be rid of the pressure term. With potential energy head h divided into depth d and height above datum z, Bernoulli becomes:

  12. Energy Grade Line • The constant “height” that all of these things reach, a measure of the total energy in the system is called the energy grade line (EGL). Up to now, it has been horizontal, meaning that no energy has left the system. However, we haven’t dealt with another form of energy—heat. Example: h = d + z decreases, velocity increases

  13. Frictional Head Losses • Energy is being lost as heat because the flowing water comes in contact with the channel sides, causing friction. This lost energy is called frictional head loss hf. It results in the energy grade line having a slight (always negative) slope. upstream downstream

  14. Detention Basin Routing • The amount of outflow from a detention basin depends on the height of the water, as you recall from Homework 1 and your test. The greater the depth portion of elevation head, the faster the water flows out of the spillway with velocity V2. • Potential Energy at the reservoir surface is converted to Kinetic Energy of motion in water released to the atmosphere at the outlet spillway. Notice the pressure is the same, atmospheric, at 1 & 2 Also the mass dropping at 1 is the mass leaving at 2 So V2 is square root of 2gh

  15. Detention Basin Routing • Consider some reservoir or lake. We’d like to know how inflow flood water will be attenuated during its passage through the reservoir. The method starts by stating the same continuity equation we’re used to: where H is the head in the reservoir. For standing water (like in a reservoir) H is just the depth.

  16. solve top equation for dS, subs in bottom left, divide both sides by A(H) The Storage Function and STAGE • The storage S in the reservoir is some measured function of the depth H and the area A, and the change in storage with depth is: Combining this with yields This is a differential equation

  17. Mult both sides by Dt, subs D for d Finite Difference Methods • From here, we can do the same finite differencing technique we did for the Muskingum method, and: where A(Hn) is the surface area for a particular depth H. This is the form we use when we convert from flow rate, say cfs, to depth, say inches.

  18. Calculate deriv=slope project line to t n+1 First Order Method Because each differential equation for reservoir storage is made of different curve fits, we usually couldn’t solve the actual differential equation exactly. Instead we use Numerical Methods. • In a first-order (Euler) solution, a finite increment Dt is chosen, a slope is projected, • and Hn is known so • This method has large error if Dt is large.

  19. Second Order Method • Step 1: Calculate First order estimate DH1, • Step 2: use DH1 to estimate DH2 (eqn. lower left) • Step 3: use average of DH1 andDH2 for DH • Step 4: Hn+1 = Hn + DH Notice Step 1 uses the slope at tN The Step 2 line to get DH2 is the slope at t N+1 projected back to tN

  20. Fourth-order Runge-Kutta Methods • Runge-Kutta methods follow the same reasoning with more steps. The equation is: • and • The constants k are iterative evaluations of f(t,H). We look up I(t) in our input hydrograph, and we calculate O(H) and A(H)from a curve fit I at this time, O at this H, A at this H

  21. An Example • As usual we’ll go through an example, then you will do a similar homework problem. • For simplicity, we will look at problems with straight sides, so Area doesn’t change with height

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