Homework #10 GEOL 4880 Humphrey Fall 2022

The first 2 questions are basic to river studies.  The 3rd question looks complex, but is just solving a cubic equation.  Don’t panic, there are cubic equation solver apps out there in web-land.  There are calculations in this homework, take your time and think a little.  And underline or box your answers!

1)    a.   In class we did a rough estimate of the hydraulic radius (the R in the Mannings eqn.).  Let’s do a better job: estimate the Hydraulic Radius of the Laramie River at minor flood, 1m deep and 4m wide.  Assume a trapezoidal channel shape, 4m wide at the surface, with straight 45 degree sloping banks, so that the middle (flat) bed section is 2m wide

b.   Search the web for an appropriate value of Mannings ‘n’ for the Laramie river. You will have to look at pictures of other rivers and find one that looks similar.  The USGS has a good web site for this. 

c.  Back-calculate the required value of manning’s n for the actual Laramie river, using the hydraulic radius from question a, and (for the Laramie river in minor flood) use a depth of 1.0m, flow velocity of 1m/s, slope of 4x10-4.  Compare this n with that in b.

 

2) In class we have reduced the logarithmic profile, along with Manning’s equation, Shield’s criterion and the Meyer_Peter_Mueller equation to get the somewhat ‘over simplified’ alluvial river relation Qsed ~ S Qwater, which is sometimes written as an equation: C Qsed = S Qwater, where S is the slope, and the C mainly comes from Manning’s n or  the critical shear stress due to the sediment size, and therefore can be considered a roughness parameter, or even more crudely, a function of the D84 of the coarse sediment on river bed.  This relationship for how a river that is approximately in “steady state” or quasi-equilibrium with the need to transport Qsed and Qwater is surprisingly useful when thinking about big-picture river behavior.  It is the basis for the ‘LANE’ graph I showed in class (the plot is on our web pages).

A  Some results are obvious: What would tend to happen to Qsed , if water discharge increases, but nothing else changes?

B  How would the river try to adjust the slope, if Qwater increases, but everything else stays the same?

C Some results are more difficult: What happens if the size of the sediment increases (for example a landslide introduces larger bedload)?  Remember that it takes a long time for slope to adjust.  What is the expected long time response?  For the shorter term response we should have to look at non-equilibrium behavior, but we can probably make a guess by assuming S doesn’t change.  What is your guess at short response?

D In most rivers the water discharge and sediment flux increases downstream, while the Slope decreases.  How does the Lane eqn predict the sediment size will change as you go downstream in such a river system?

 

3  (this weeks geomorphic puzzle, it is not hard, but it requires being careful!) In class I quickly sketched the energy argument for the difference in behavior of super-critical and sub-critical flow (Froude number greater or less than 1).  The development in class was a sketch of the tradeoff between potential energy and kinetic energy.  I want you to understand this important concept: that water only has so much energy, and the trade off from kinetic to potential (or vice versa) controls a lot of the river’s behavior.  I would like you to follow the logic outlined here and find solutions to illustrate this behavior. 

Consider a rectangular cross section channel of constant width and steady discharge, which has a small bump of height ‘h’ which the flow must cross.  Assume the slope of the flow is so low that we can ignore it and set the elevation of the bed of the incoming flow to be 0, while the bed of the flow over the bump is just h. We will call the initial location 1 and the location at the bump 2.

The incoming depth is d1 and the incoming velocity is v1, since the width is constant, we just look at a 1m wide section of this flow.  The flow over the bump will have depth d2 and velocity v2, and the water surface height over the bump is d2 + h.  Although we don’t yet know values for d2 or v2: they will tell us if the water surface goes up or down, and that is what we solve will for.  The mass flux of water per time incoming is r*v1*d1 and the mass over the bump is the same (or equivalently r*v2*d2).

The kinetic energy of the water is the mass times the square of the velocity divided by 2, while the potential energy is the mass times gravity times the height (d1/2).  Since the flow velocity is changing, it is important to multiply by the local velocity (v), to get a total energy flux per unit width per time at each of the two locations.  For example the total kinetic energy per meter width per time, is the energy per cubic meter r*v1 / 2, times the depth d1, and times the velocity v1. We assume no energy is lost between points 1 and 2, and claim that the total energy flux at the 2 points will be equal.  We write the energy balance for K.E. and P.E. as a statement that the energy flux per time at the two places is equal: [equation A]:

r*v13*d1/2 + r*g*v1*d1 *d1/2   =    r*v23*d2/2 + r*g*v2*d2 *(d2/2 + h)  ,             …….  A

where we have been careful to include the extra height of the bump (h) in the potential energy at location 2.  This simplifies considerably since r*v1*d1 is equal to r*v2*d2 since the discharge is constant and the width does not change.  Further simplification occurs by eliminating v2 (using constant water flux and constant channel width), if we write v2 = v1*d1/d2. We can multiply the whole equation by d22, to get [equation B]:

0 = [g*v1*d1] *d23 + [2*g*d1*v1*h - v13*d1 -  g*v1*d12] *d22 + [v13*d13 ],            …….. B

where all terms in ‘[ ]’ brackets are known if we know  v1*d1 (the water flux).  This is a cubic equation in the only unknown, d2 .  (e.g. the equation is simply 0 = a*d3 + b*d2 + 0*d + c)

      a. Write equation A and produce equation B by the indicated steps. Show your work.

b. To see what the Laramie river might do in encountering a bump in the bed, try using eqn B with an input velocity of 1m/s, a depth of 1m, and a bump height of 0.1m.  Report on how much does the water surface drop over the bump?  [don’t forget to add the height h of the bump to get the water surface above 0.  You can directly solve the cubic equation or use a web app. [hint, you are looking for the largest positive real root] Or it is a fairly quick iteration to get a solution: try a d2 of 0.6m to start, improve on that if you can]

c. If you have made it this far, in this question, and just to see the opposite behavior for high Fr number flow, find how far the surface (d2 + h) rises using an (unrealistic) flow velocity of 5m/s, depth of .5m and bump of 0.1m? Typically in super-critical flow, the water rises so high above bed obstacles that a standing breaking wave is formed.