Applied ODE's: How high will it flood in a lake?

1. Applied ODE's: How high will it flood in a lake? How does lake level change with time? Why does Rt. 108 flooded? Was the water highest when the rain was strongest, or when it ended? How do the answers to these questions depend on the shape of dams?

2. What is our problem? • Qin, m3/s, is forcing and is given • Qout, m3/s, depends on lake level h • V, volume of lake, depends on lake level h • h, lake level, relative to whatever constriction sets Qout

3. So how do we make this into an ODE? • What is change of lake volume V with time? • So what controls this? • What are units of Qin and Qout? • on board, find

4. But Q depends on h, the lake level! • and h depends on V, the lake volume. • Lets assume a steep sided lake: • often true of reservoirs and rivers in valleys • Less true of many ponds, V increases more rapidly than h as area increases; Think of an ice-cream cone!

5. But we need Qout! How does it depend on h? • Why do we believe lake outflow is only influenced by conditions at one place in the outflow? • If outflow is through "V" shaped valley, Q depends on h5/2

6. Much more detail in:

7. So the equation is • Write as a matlab function, assume a two day rainstorm • Qin=0 for t<1 day (in seconds! Why?) • Qin=Qflood for 1<t<3 days • Qin=0 for t>3 days

8. What do we use as parameters? For Lamprey river at Newmarket. Get area.

9. Get flood Qin from USGS guage. Note units in CFM! • Heck of a flood, that!

10. What do we use as parameters? Get Newmarket dam width.

11. We write this function: • Data from floods March 13, 2010. Max discharge of about 700 m3/s for 2 days. function dVdt=dV_dt(t,V); %return the time rate of change of lake level as function of time W=10; % dam width, meters A=150000; %surface area of resevoir, m^2 Qmax=700; %max inflow in flood, m^3/s g=9.8; %gravitational acceleration, m/s^2 if (t>1*8.64e4) & (t<3*8.64e4) Qin=Qmax; else Qin=0; end dVdt=Qin-sqrt(g)*W/A^(3/2)*V^(3/2);

12. Euler routine is same as before: %very simple example of Euler's method to solve % lake level problem, ininital condition of V=0 at t=0 clf() clear all t=0; %initial time V=0; %intial condition dt=60; %time step tvec=[t]; %store time history Vvec=[V]; %store T history %integrate with Euler's method to time t>2pi while (t<=5*24*60*60) V=V+dt*dV_dt(t,V); t=t+dt; tvec=[tvec t]; Vvec=[Vvec V]; end plot(tvec/8.64e4,Vvec/150000,'r*-','linewidth',2) xlabel('time, days') ylabel('river height above dam') title('Lake level')

13. Results vrs. data: • Some error because gauge at Packers Falls. • But over all, good for a quick model. • Length of flood is set by time it takes water to reach "lake," Not time to get over dam.

14. Length of flood is set by time it takes water to reach "lake," Not time to get over dam. • Why does this matter? • Less important to change dam than to stretch out time it takes water to reach dam. • More delay ponds • More beaver ponds, wetlands, etc • Fewer parking lots • If we build parking lots, we need to buffer the inflow into the parking lot somehow. • How? Drainage percolators, holding ponds, etc

15. Why do beaver dams survive these big floods?

16. Where are they? And how wide? And what is the drainage basin that sets Qin?

17. Where are they? And how wide? And what is the drainage basin that sets Qin? • Qin much less • W is much greater • Thus more Qout for given change in h! • Of course, A is also much less. • Overall, change in height is much less • Beaver dams are much more secure