Hydraulics OverviewPeter WilcockJuly 30, 2019

• General observation & some terms• Conservation of mass & force balance• Flow resistance – what sets the depth?• Using all the tools at once• Energy in open channel flow

Rapidly varied & gradually varied flow• Water surface, or backwater modeling

The open channel toolbox(for flow and sediment transport)

• Conservation RelationsWater Mass  (aka continuity)Momentum  (aka F = ma, Newton’s 2nd Law)

Energy• Constitutive Relations

Flow resistance   (we’ll use Manning eq.)

Sediment transport  (Wednesday)

• Morphodynamics – add conservation of sediment mass (Wednesday/Thursday)

• Distorted view (channels are much wider than they are deep;

b/h = 20 is a narrow channel)

• Width b, area A, wetted perimeter P• Hydraulic radius R = A/P• Mean depth h = A/b• Often, b P, so R h

e.g. for a rectangular channel

(b)

(hmax)

2bh bhR h

b h b

3

1:20

Continuity

• Conservation of water mass• Just accounting, like your 

checkbook• Rate of change of water 

mass in reach = net rate of input and output

• Inputs and outputs: • Constant discharge: 

input = output, no change in water mass in reach

Channel cross section 2

Channel cross section 1

1 1 1 2 2 2

2 11 2

1 2

and

For ,

Q U A Q U AU AQ QU A

Q is water discharge  (e.g. ft3/s or m3/s); U is mean velocity (e.g. ft/s or m/s)

Continuity for unsteady and nonuniform 1d flow –

Now depth h varies in both time and space

( )

( )For constant , .

( )or

( ) 0

If / t = 0, we recover constant

x x xUhBUhB UhB x

t xh x UhB x

t xh Uht x

h Uht x

UhB Q

S = I - 0

Qx

Qx+x

B x

hx hx+x

For a control volume,

The rate of change of mass = the net rate of mass flowing through the sides

slope top width depth wetted perimeter x/s area

SBhPA

oBoundary stress: (stresss is force/area)Boundary force: PL

Volume of water: Weight of water: Downslope component of weight of water:

ALgAL

gALS

L

o

o

PL gALSgRS

The ‘depth‐slope’ product

Momentum: Newton’s 2nd Law, F = ma Normal flowno accelerations, so F = 0

is the boundary shear stress - the flow force per unit area - it drives the sediment transport

o

o ≈ hS (h in cm, S in %, o in Pa)

1b. Boundary shear stress — Unsteady, Nonuniform Flow

We still use Newton’s second law,

But the flow is accelerating in (1) space and (2) time. These accelerations are balanced by the sum of forces, to which we now must add a third force, due to the difference in flow depth down the reach

maF

Px

B x

Px+xg

0g

gsin( gS P gh / 2

hx hx+x

oU U hU gS gt x x R

If the flow is predominantly one-dimensional (no rapid changes in width or depth that give rise to vertical or lateral accelerations), we get the

1d St. Venant Eqn:

“local” acceleration

“convective” acceleration

bodyforce

pressure force

boundary force

An approximate derivation is in the primer

0 1o

y V V VSgR x g x g t

Steady, uniform flow

Steady, nonuniform flow

Unsteady, nonuniform flow

Rearrange to solve for o

m a F

tU

gxU

gU

xhSgR 1

0

We can rearrange the St. Venant eqn, to solve for shear stress

1. For steady, uniform flow, we recover the depth-slope product

2. But flow is never perfectly steady and nonuniform!

3. So, how do we know whether the other terms are important or not?

4. Are we making an assumption or an approximation?

flood.xls

0 fh U UgR S gRSx g x

2

2f bd US z hdx g

NOTE:

Backwater programs (HEC-RAS) compute:

10

Flow from a tributary at point a increases rapidly from Q1 to Q2 over t. At time t, any increase in flow has not yet reached a distance x downstream, creating a nonuniform flow ‘wedge’. The flow is unsteady (it is increasing with time) and it is nonuniform (it is increasing upstream). Question: do the unsteady and nonuniform terms contribute significantly to o?

flood.xls

Given a rectangular channel: The discharge at pt a increases Channel width b 15 (m) from Q1 to Q2 over t

Channel slope So 0.001 Q1 10 (m^3/s)Hydraulic roughness n 0.035 Q2 15 (m^3/s)

t 3600 (s)Enter Data only in green cellsFlow is unsteady and nonuniform, so we ask. which terms in the St Venant Eqn are significant?How does the shear stress change, approximately, over this reach of nonuniform flow?>>> we need to estimate the magnitude of the different terms in the St Venant Eqn.The length of the flood wave x is found from Uf = x/t, where Uf is the speed of the flood wave and is approximately 1.5 times mean velocity in the rea

x 4470 (m)We need to determine depth and velocity at the upstream and downstream end of the reach

UPSTREAM DOWNSTREAMhu 1.12 (m) hd 0.87 (m)Ru 0.98 (m) Rd 0.78 (m)Uu 0.89 (m/s) Ud 0.77 (m/s)

We also need the mean velocity and hydraulic radius for the reachR 0.88 (m)U 0.83 (m/s)h -0.253 (m)U -0.12 (m/s)

Now we can compute all of the components in the brackets of the St Venant eqnChannel slope So 0.00100

h/x -0.00006(U/g)*(U/x) 0.00000(1/g)*(U/t) 0.00000

0.00106

Mean o in nonuniform reach, approximated from St Venant eqn.o 9.16 Pa

o in steady uniform flow before flood7.65 Pa

Flood Water Normal flow before flood

a

drop if any of these terms are more than an order of magnitude smaller 

Note that a number of cells are namedand that those names are used in formulas

To make approximation, express in difference form

0 2/3

2/30

2/3

3/52/5

0

and

so, for a rectangular channel ( , 2 )

or 2

1 2

which can be solved iteratively for

SU R Q UA

nA bh P b h

S bhQ bh

n b h

nQh b hb S

h

0 01dh U dU dUgR S

dx g dx g dt

0 0

1h U U UgR Sx g x g t

01h V V VgR S

x g x g t

Use St. Venant to consider sediment transport for a steady river flow entering a reservoir.  Assume a wide channel, so R = h.  As flow enters the reservoir, it deepens and slows (by continuity!).  Which effect wins out?

0h V Vgh Sx g x

+ ‐

Consider for the case of constant width (for simplicity)

( ) such that and 0 such that

. Replace in the first equation above, givi

h U U bx g x

q Uh h Uq Uh const U hdx dx dx dx

U U h Udx h dx dx

2 2

ng

1 . Because 1 in a reservoir,

the entire quantity is negative. Because is itself positive

for flow entering a reservoir, we see tha

h U U h h U Ux g h dx dx gh gh

h U U hx g x x

t is reduced from . A smaller value of means that transport capacity decreases and sediment deposits.

o

o

ghS

Does boundary stress oincrease or decrease as we enter the reservoir? 

The answer

11

Consider a typical problem: Specify a channel, with known geometry

(fluid and acceleration of gravity g are also known).For a specified discharge Q, what is the flow depth h for steady uniform flow?Continuity gives us U (when A(h) is known). Momentum gives o (when R(h) is known). Energy gives Sf = So for steady uniform flow.We need one more relation to find depth h!

This is a flow resistance relation (a constitutive relation).

slope top width depth wetted perimeter x/s area

SBhPA

Flow Resistance

• A relation between velocity, flow depth, boundary stress, and boundary roughness 

• We’ll use Manning’s eqn.• Prefer an eqn. with constant roughness (n)

• Using continuity:

2 / 3SU Rn

2 / 3SQ UA ARn

S is the slope of the ‘energy grade line’, which is also the slope of the channel for uniform flow

• For a useful approximation, consider a wide (R  h) rectangular channel

• Solve for depth• In field, relation between Q, A, & R

more complex, though not more complicated

• In practice, a flow resistance relation is often used to find the flow depth for a specified flow in a given channel

3/ 5nQhb S

2 / 3SQ ARn

5 / 3b ShnQ

5 / 3b ShQ

n

Manning’s Eq., 3 ways:

14

2 / 3( )SQ bh hn

5 / 3SQ bhn

The hard part: determining roughness ‘n’• How to measure:

survey cross section and reach, which gives S and A, R as a function of water surface elevation (aka stage or WSEL)

• Measure discharge at a known stage, solve for n

2 / 3Sn R AQ

2 / 3SQ ARn

• If you don’t measure, you are guessing• Picture books, tables• Formulas n = fn(D, R, veg., …)

2/3

5/3

15/3 1

/

so

or

o

Q BhU

B Qh gS

SU hn

SQ QgS n

SQn gS

Relating Q to  in a simple, prismatic, wide channel

Combining the tools

This is how we link transport rates (a function of ) to water discharge Q

Discharge = width * depth * velocity     continuity

Width increases with discharge           channel geometry

The ‘depth‐slope’ product                     momentum

Manning’s eqn flow resistance

Combine. It’s only algebra!

Boundary stress is related to discharge, slope, roughness, channel shape

This solves the FLOW problem

2/3

5/3

15/3 1

15/3 1

/

so

or

cc

Q BhU

B Qh gS

SU hn

SQ QgS n

SQn gS

SQn gS

Why?

20

Because we generally relate changes in the bed and banks –erosion and deposition – to the bed shear stress  and we will want to be able to relate  to the water dischargee.g. at what Q will  be sufficient to move river bed material?

For example, if the grain size of the river bed is used to find the critical shear stress for incipient motion c, this relation gives us the critical discharge  Qc needed to move the river bed.

Conservation of Energy• Units: Length• Head ‘loss’• H, EGL• Components

21

2U

g22

2U

g1h

2h1z

2z

1

2 1EEnergy grade line

2H

2 21 2

1 1 2 2

1 2

2 2or

f

f

U Uz h z h hg g

H H h

• Gradually varied flowEstimate hf, use energy eqn. to estimate depth at one place given depth at another place (next)

• Rapidly varied flow• (gates, weirs, rapid changes in 

width or bottom elevation)• flow often complex, but can be 

approximated by assuming negligible energy loss (if flow converging)

• Over short distances, with small energy loss,

1 22 21 2

1 1 2 2

2 21 2

1 2

1 2

and 0, so

2 2

becomes 2 2

or

f

f

z z h

U Uz h z h hg g

U Uh hg g

E E

ch

Weir Gate

2

2UE h

g

22

The Specific Energy E2

2UE h

g

2

2

For a rect. channel Define

/ so / and

2

Q UAQ Ubhq Q b Uh

U q h

qE hgh

flow energy relative to the bed

The extraordinary properties of the Specific Energy E

2

22qE hgh

E

hSubcritical

Supercritical

F = 1

• 2 flow states• Change in h with step 

or constriction• Properties of Min(E)

F = 1

31U q

gh gh F q

2

2

2

3

2 22

3

3

To find the minimuim , set 0.

02

1 0

1

so 1 at minimum .

Note, too, that where a subscript 'c' is included

to indicate that this holds at critical flow

c

dEEdh

d qhdh gh

qgh

q Ughgh

E

q gh

F

F

.

ch

Weir

Basis for critical flow meters!

What happens when flow goes over a smooth step of height z, or through a contraction in width, with negligible head loss (H const)?

1 2

1 1 2 2

1 2

and 0,

or

fz z z h

z E z EE z E

Flow Over Step

Flow Through Contraction E1 = E2

Diagrams from Roberson & Crowe, Fluid Mechanics, Wiley

Diagrams use y for flow depth and V for velocity

2

22qE hgh

1Ugh

F

1Ugh

F

Gradually Varied Flow: Modeling Water Surface Profilesaka backwater or step‐backwater modeling

21

2U

g22

2U

g1h

2h1z

2z

1E Energy grade line

2H

fh

Long distances: Z1 > Z2 and hf > 0

Conservation of energy in open channel flow2 21 2

1 1 2 22 2 fU Uz h z h h

g g

26

Switching back depth from y to h and mean velocity from V to U  ….   Sorry!

The rate of head loss is defined as SfMust estimate head loss, so we need

length of stream over which the head loss is occurring x = (x2 – x1) 

ff

hS

x

2 22 1

2 2 1 12 2 f fU U

z h z h h S xg g

2 22 1

2 2 1 1

2 1

2 2f

U Uz h z h

g g H Sx x x

2 21 2

1 1 2 22 2 fU Uz h z h h

g g

Just restating the definition of energy in open channel flow

2 1

2 1o

z zS

x x

2 22 1

2 1

2 1

2 2o f

U Uh h

g gS S

x x

2 22 1

2 2 1 1

2 1

2 2f

U Uz h z h

g g H Sx x x

Using

becomes

An ordinary differential equation that tells us the rate (h/x) at which h changes alongstream.  This relation only helps if we know the depth at some place to start with.  If we do, then the equation can be used to calculate depth at some other location and we are off to the races.

2 22 1

2 1

2 1

2 2o f

U Uh h

g gS S

x x

2 22 1

2 1

2 12 2

o f

U Uh h

g gx x

S S

21

2U

g

1h

1z

1x

2x ? ? 1 1 2 2Given at , at what will occur?h x x h Direct Step

Specify Q, So, and h1.  Continuity gives U1Pick h2, continuity gives U2

Solve for x2.

Not very convenient, except for prismatic canals!

2 22 1

2 1

2 1

2 2o f

U Uh h

g gS S

x x

Specify Q, So, h1 and xContinuity gives U1Guess h2, continuity gives U2

Iterate.

? 21

2U

g

1h

1z

1x

2x

?

1 1 2 2Given at , what is at ?h x h x Standard Step

2h

2z

2 22 1 1 2 2 1

1 ( )( )2 o fh h U U S S x x

g

Defining Sf.  We used Manning’s eqn to describe flow resistance

2 / 3SU Rn

2

2 / 3fnUS

R

For steady, uniform flow, So = Sf and we did not have to specify which S was used in Manning’s eq.

We now precisely define that slope as Sfand we use Manning’s eq. to define Sf.

Having chosen a value of h2, we calculate R2, U2, and Sf.  Inserting U2 and Sf. on the right side, we then calculate a new, improved guess for h2. 

This is step‐backwater modeling.  This is HEC‐RAS.

2 22 1 1 2 2 1

1 ( )( )2 o fh h U U S S x x

g

Big advantages of a canned program:

(a) Iterative solution to the governing equation is not always stable

(b) The world is messy (irregular cross‐sections, variable roughness, compound channels, bridges, culverts, ….)

Don’t believe a RAS solution unless you can sketch it yourself!  How?

(1) identify controls   (known depth for a specified discharge)critical depth, normal flow

(2) which way to go (start D/S in subcritical flow and U/S in supercritical flow

(3) Sketch!

In its most basic application, to build a HEC‐RAS model you (a) survey cross sections in your study reach and enter these into RAS, (b) assign roughness to each section (or portions of sections), (c) specify a discharge, and (d) step back and let it run.  

ROUGHNESS:  If you don’t calibrate, you are only guessing. 

3/ 5n

nqhS

1/ 32 /ch q gaka F = 1

aka Manning’s eq.

Types of gradually varied flow profiles

Mild Slope

Steep Slope

M1 M2M3

S3

S1S2hc

hn

hc

hn

hc

hn

hc

hn hc

hnhc

hn

Backwater

Backwater

HydraulicJump

HydraulicJump

Drawdown

Subcritical Flow – Downstream Control!

Supercritical Flow – Upstream Control!

Some Key PointsHow often does sediment move? What is the Sediment Balance?Flow COMPETENCE of bed material    Sediment Supply v. Transport CAPACITYThreshold Channel, Incipient motion  Alluvial channel problemShields curve + flood frequency  Transport relations give capacity. Supply?

All sediment transport problems have a FLOW PARTFlow problem hydraulics  boundary stress oSediment problem  sediment response to boundary stress o

All relations are approximate. You can quantitatively evaluate the error associated with approximations (e.g.  = gRS)

Channel hydraulics and transport: uncertainty is in the input!can estimate the magnitude and importance of uncertaintyneed actual measurements to obtain accuracy

The tools are general, their application is local.

Why do I have to learn RAS?  Provides useful basis for estimating flow and stressAccounts (coarsely) for nonuniform flow conditions

2 / 3 5 / 3

2 / 3

(1a) Finding Normal Depth - Rectangular Channel

and

where

, 2 ,2

ho o

h

R AV S Q Sn nP

bhA bh P b h Rb h

5 / 3

2 / 3

1

Given , , Find from Given , , Find from Given , , Find from Given , , To find , have to iterate!

Try arranged as

1 2

Solve for on top:

h o

h o

h o

o

o

no

R n Q SR n S QR Q S nS n Q h

b hQ Sn h

b

nQy hb S

3/ 5 2 / 51 2

where and 1 subscripts indicate successive approximations

nhb

n n

(1) Finding normal depth

5 / 3

2 / 32

2 2

(1b) Finding Normal Depth - Trapezoidal Channel

( )Manning's eqn:

2 1

where

2 2 1

oo

o

o o o o

h B shSQ

nB h s

B B sh A B h sh h B sh P B h s

2 / 523/ 5

1

Given , , Find from Given , , Find from Given , , Find from Given , , To find , have to iterate!

2 1Try arranged as

where and 1 subs

o

o

o

o

o nn

o no

h n Q Sh n S Qh S Q nS n Q h

B h snQhB shS

n n

cripts indicate successive approximations

h

Bosh sh

s1

3/ 2

2 / 33/ 23/ 2

(2) Finding critical depth

/Finding Critical Depth 1 or/

where is top width.

(2a) Rectangular Channel or

(2b) Trapezoidal Channel

2

c

o o

V Q A Q Ag Bg A Bgh

B

BhQ A Qhg B B gB

B B sh A B h sh

F

333/ 22

33

2

2

oo

o

o

o

h B shAh B shB shB

y B shQB shg

33

1/32

3

3 32

Given , Find 2

2Given , Find

Given , Find 2

c o cc o

o c

o co c

o c

cc o o c c

gh B shh B Q

B sh

B shQQ B yg B sh

ghh Q B B sh shQ

h

Bosh sh

s1