RIVERS & STREAMS HYDRAULICS

Benoit Cushman-Roisin Dartmouth College

River flow is 3D and unsteady (turbulent).

But, length of river >> width & depth

As a result, the downstream velocity, aligned with the channel, dominates the flow.

The 1D assumption may be made, with h(x,t) = water depth u(x,t) = water velocity with x = downstream distance and t = time.

Another assumption may be made: Incompressibility density  = constant = 1000 kg/m3 (freshwater)

1 The reduction to two flow variables [h(x,t) and u(x,t)] necessitates only two physical statements.

These are:

1. Conservation of massvolume (what goes in, goes out)

2. Downstream momentum budget (with 3 forces: pressure, gravity and friction).

Budget for a stretch dx of river

A = Area W = Width h = max depth P = Wetted perimeter S = Slope = sin 

2 Conservation of mass:

Amount stored in stretch dx = what goes in – what goes out.

Adxtdt  Au dt Au dt t xxdx dx0, dt 0  AAu0 tx Then   constant  AAu 0 tx where A(h) is a known function of the water depth h (channel profile given)

This equation is attributed to Leonardo da Vinci (1452-1519), which he wrote in an algebraic way instead of using derivatives as we do now.

Case of a rectangular channel with constant width:

A = Wh, with W = constant

P = h + W + h

    A  Au  0  h  hu  0 t x t x

3 Momentum Budget:

d Momentum inside stretch Momentum entering at x dt Momentum exiting at xdx  Pushing pressure force in rear  Braking pressure force ahead  Downslope gravitational force  Braking frictional force along bottom

h Pressure force  F pdA  p ( z ) w ( z ) dz p  0 z  height from deepest point at bottom pz( ) hydrostatic (gage) pressure  gh ( z ) wz( ) channel width at level z (0 z h )

h Fp  g(h  z)w(z)dz a function of h only 0

dFp h  [g(h  z)w(z)]zh  gw(z)dz dh 0 h  g w(z)dz  gA 0

F dF h h For later: p  p  gA x dh x x

4 Gravitational force  (mg )sin  () Adx gS  gASdx

Frictional force  bottom stress bottom area

 b ()Pdx 2  ()CuPdxD  2 with bD bottom stress Cu and P  wetted perimeter

Putting it altogether:

[Audx]at tdt [Audx]at t 2 2  Au  Au Momentum in and out dt at x at xdx Pressure force, rear and front  Fp |at x  Fp |at xdx  gASdx Downslope gravity 2 Bottom friction  CD Pu dx

or, in differential form: Fp h  gA from earlier x x   F Au  Au 2   p   gAS  C  Pu 2 t x x D

and after some simplifications and use of volume conservation:

u u h P  u  g  gS  C u 2 t x x D A friction inertia gravity

5 For convenience, we define the hydraulic radius:

A cross -sectional area R   P wetted perimeter so that the momentum equation becomes:

uu  h u2 uggSC tx  xD R

For a broad flat channel, which is a good approximation for most rivers:

Wh Wh u u h u 2 R    h   u  g  gS  C W  2h W t x x D h

This equation is attributed to Adhemar de Saint-Venant (1797-1886).

Together, this momentum equation and the mass-conservation equation form a 2 x 2 nonlinear system for the flow variables h and u.

1. Uniform frictional flow:

   0 and  0 t x

Only the momentum equation remains and it becomes:

u 2 ghS 0  gS  CD  u  h CD

The balance is between the forward force of gravity and the retarding force of bottom friction. The formula is due to Antoine de Chézy (1718-1798).

The Chézy formula specifies one relation between the velocity u and the water depth h. How can these quantities be determined separately?

6 Answer: We need to know the volumetric flow (discharge) of the river!

When Q  Au  Whu is given, then 1  C Q2 3 h   D   2   gSW  1  gSQ 3 u     CDW 

Note that both water depth and velocity increase with the discharge. This explains why the water level rises and the current increases simultaneously when the river discharge rises.

Note: As the discharge increases, the water depth (÷Q2/3) increases faster than the velocity (÷Q1/3).

Manning’s formula

River data show that the drag coefficient CD is not a constant but depends on depth.

If we use the logarithmic velocity profile of wall turbulence, we obtain:  2 CD  2 [ln(h / z0 ) 1]

and the Chezy formula becomes

ghS ghS  h  u   2 ln 1 CD   z0 

Using abundant data, Robert Manning (1816-1897) determined that a power of h was adequate, with the 2/3 power giving the best fit, and he wrote: 1 u  R2 / 3 S1/ 2 n in which the coefficient n is now called the Manning Coefficient.

Note that this expression is not dimensionally correct. So, care must be taken to use metric units.

7 Examples of Manning Coefficients:

Smooth cement canal Clark Fork at St. Regis, Montana Clark Fork above Missoula, Montana n = 0.012 n = 0.028 n = 0.030

Middle Fork Flathead River South Fork Clearwater River Rock Creek near Darby, Montana near Essex, Montana near Grangeville, Idaho n = 0.041 n = 0.051 n = 0.075

See also: https://wwwrcamnl.wr.usgs.gov/sws/fieldmethods/Indirects/nvalues/index.htm For details, consult: https://pubs.usgs.gov/wsp/2339/report.pdf

Numerical values of the Manning Coefficient:

CHANNEL TYPE n Artificial channels finished cement 0.012 unfinished cement 0.014 brick work 0.015 rubble masonry 0.025 smooth dirt 0.022 gravel 0.025 with weeds 0.030 cobbles 0.035 Natural channels mountain streams 0.045 clean and straight 0.030 clean and winding 0.040 with weeds and stones 0.045 most rivers 0.035 with deep pools 0.040 irregular sides 0.045 dense side growth 0.080 Flood plains farmland 0.035 small brushes 0.125 with trees 0.150

8 The enigma of Roman water engineers

Roman engineers had no conception of time at the scale of the minute and second. So, they could not quantify the water velocity and dealt only with water depths.

So, how were they able to build properly designed aqueducts and sewage drains? Segovia, Spain

Answer:

The Romans were lucky because velocity is directly related to water depth, and water depth could then be used as the only variable.

It also helped that water depth happens to be the more sensitive function of discharge among the two variables.

9 A nice exercise:

Subject the Chezy solution to small, time-dependent fluctuations, to find that it is stable only as long as

Fr  2  S  4CD

If this condition is not met, waves grow to finite amplitude.

These are so-called “roll waves”

Photo credit: Alden Adolph, Thayer School student, January 2013

10 2. Steady frictionless flow:

 Now, take  0 (steadiness) and C  0 (no friction) t D The pair of governing equations become:

d Q Mass : (hu)  0  hu  constant  dx W du dh Momentum : u  g  gS dx dx

If we define the bottom elevation b(x) above a reference datum (sea level) db then S   and the momentum equation can be cast as : dx

d  u 2    gh  gb  0 dx  2 

Therefore, the expression

u 2 B   gh  gb 2

is conserved along the flow.

This is relation is due to Daniel Bernoulli (1700 -1782), and is known as the Bernoulli principle.

The essence of the Bernoulli principle is conservation of energy :

u 2  kinetic energy, g(b  h)  potential energy. 2

11 3. Linear waves:

Now, ignore friction (CD = 0) and bottom slope (S = 0).

Then, linearize the equation around a basic state of no flow and a uniform water depth H:

hxt(,) H axt (,)

Governing equations reduce to: au  u  a Hg0,  tx  t  x

A solution is the shallow-water gravity wave: axt(,) A sin( kx ct ) cA uxt(,) sin( kx ct ) H with cgH

Summary of particular solutions and cases to be considered later:

  S CD t x Shallow-  water wave Bernoulli 

Chézy  Steady non-   uniform flow Roll waves    Flooding

12