# Continuity Equation Presentation ```The Continuity Equation
This equation is pervasive, appearing in many
different aspects of science and engineering.
We’ll investigate its form and usage for water
flow across a floodplain.
Cross Section of a Floodplain
Qin
Qout
t3
TerreFirme
t2
t1
River
Channel
Floodplain
Q is Discharge
t is time
A “Glass” of Water
Glass is empty at t1
A “Glass” of Water
Qout &lt; Qin
Qout
t2
Qin
ΔS increases
Glass is filling at time t2
A “Glass” of Water
Qout &lt; Qin
Qout
Qin
t3
ΔS increases
t2
Glass is nearly full at t3
A “Glass” of Water
Qout &gt; Qin
Qout
Qin
t3
ΔS decreases
t4
Glass is emptying at t4
The Basic Form of the Continuity Equation
ΔS = Qin - Qout
t1
t2
t3
Qout&lt;Qin Qout&lt;Qin
t4
Qout&gt;Qin
The water flowing into the glass (Qin) minus
the water flowing out of the glass (Qout)
equals the amount of water that was added
to or taken away from the glass (ΔS) during
the time step (e.g., t2 to t3).
For example, say a sink faucet is filling your
glass at 2 gal/min and you have a straw
sucking out 0.5 gal/min. And, you do this for
one minute. Then the glass will gain 1.5
gallons of water.
Qout
Qin
What is “Discharge”
• Q is discharge, which we typically think-of
as the flow of water down a river.
• This flow, or Q, is estimated by measuring
the river cross-section and measuring the
velocity of the water, then multiplying
these together.
River Discharge
Q = 3m * 10 m * 1 m/s = 30 m3/s
Q = 9m * 10 m * 2 m/s = 180 m3/s
How to Apply Continuity to a
Floodplain
• Why is it difficult to know Q on a floodplain? We can
estimate Q in a river channel, but measuring flow
velocities on a floodplain or measuring cross sectional
area is very difficult (without satellites).
• Instead, need to describe the continuity equation with
things that are measurable, so… what is easily
measured?
– the water surface elevation, call that “h”
– h on a number of different days, or every hour, or every week,
etc., call that “Δh/Δt”, the change in h with the change in time
– h at a number of different locations across the floodplain, call
that “Δh/Δx”, the change in h with space, ie, the slope
The Amazon Floodplain
Its difficult to know the flow velocities and depths in
these environments. Characterizing the entire
location would take too many surveys. This
situation is unlike a typical river channel. However,
it should be straightforward to measure the water
surface elevation, h.
A “Cube” of Water
Qin
Qout
Qout
Qin
Qout
Imagine the floodplain to consist of a entire series of small cubes of water. Our
goal is to describe the flows into and out-of each cube using h as the primary,
measurable item. Then, summing each cube will give the Q for the floodplain.
A “Cube” of Water
Vin along z axis
ΔY
ΔZ
Vout along y axis
Vin along y axis
X1
ΔX
ΔX = X1 – X2
X2
Vout along z axis
Because Q = velocity * area
we can use Vin Vout ΔX, ΔY, and ΔZ instead of Q
Note: V along Y axis is part of this schematic, but not shown for clarity of the drawing.
Some Algebra
Vin(y) – Vout(y) = ΔVy
this is the difference in velocity flowing into and out of the cube
along the y axis, or you can think of it as the change in velocity
that occurs inside the cube.
ΔVy * ΔX * ΔZ = ΔQy Multiplying the change in velocity by the cross sectional area of
the cube gives the change in discharge that occurs inside the
cube along the y axis.
ΔVy * ΔX * ΔZ * ΔY = ΔQy Here, multiplying ΔY/ΔY is the same as multiplying by 1,
ΔY
but ΔX * ΔY * ΔZ is the volume of the cube, lets call the
volume, “vol”
ΔVy *
ΔVx *
vol
= ΔQy
ΔY
vol
= ΔQx
ΔX
Using this same approach, there are similar expressions for
the change in discharge within the cube for the X and Z axis.
ΔVz *
vol
= ΔQz
ΔZ
A Little More Algebra
Qin – Qout = ΔQ
This is an expression that says that all of the flow into the
cube minus all of the flow out of the cube is equal to the
flow change that occurs inside the cube.
ΔQ = ΔQx + ΔQy + ΔQz The total flow change the occurs inside the cube is equal to
the sum of the three parts.
ΔQ = ΔVx
vol
vol
vol
+ ΔVy
+ ΔVz
ΔX
ΔY
ΔZ
ΔQ = vol (
ΔVx
ΔVy
ΔVz
+
+
ΔX
ΔY
ΔZ
)
ΔS = vol (
ΔVx
ΔVy
ΔVz
+
+
ΔX
ΔY
ΔZ
)
These are a series of simple
algebraic steps replacing
variables with their equivalent
expressions.
This is the basic form of the
continuity equation.
We want this equation in terms of h, the thing we can easily measure.
A Basic Physics Fact for Water
• Water flows downhill!
• An expression for this is:
Velocity = Constant * Slope
• Applying this expression to the algebra:
Δh
Δh
Vz = K *
ΔY
ΔZ
ΔVx
ΔVy
ΔVz
ΔS = vol (
+
+
)
ΔX
ΔY
ΔZ
K * Δ(Δh)
K * Δ(Δh)
K * Δ(Δh)
ΔS = vol (
+
+
)
ΔX * ΔX
ΔY * ΔY
ΔZ * ΔZ
Vx = K *
Δh
ΔX
Vy = K *
• What is Δ(Δh)?
• How do we re-write ΔS as something with h, the
thing that is measurable?
The Continuity Equation for a
Floodplain
• Δ(Δh)/(ΔxΔx) is the change in water slope.
• Recall that ΔS is a measure of the time of
Qin and Qout. In the algebra of the cube
(i.e., the glass), the glass doesn’t change
its size with time, only the water level in
the glass changes. So ΔS during the time
interval is Δh/Δt.
Δh/Δt = vol*K (
Δ(Δh)
+
ΔX * ΔX
Δ(Δh)
+
ΔY * ΔY
Δ(Δh)
ΔZ * ΔZ
)
Applying the Continuity Equation to
a Floodplain
• All of the proceeding was to determine K,
it is unknown.
• To determine K, we’ll use the continuity
equation from a number of locations and
times.
• We’ll simplify the equation to just the x
dimension.
Δh/Δt = h*K (
Δ(Δh)
ΔX * ΔX
)
Applying the Continuity Equation to a Floodplain
TerreFirme
t1
t2
River
Channel
Floodplain
Q is Discharge t is time
Here’s the floodplain water level at two different times. The simple form of the
continuity equation needs h and x. What are these in this schematic?
Applying the Continuity Equation to a Floodplain
h
t1
t2
x
This is h at two different times for all of the x values. To solve the continuity
equation, we need Δ(Δh)/(ΔxΔx) which is the change in slope. Next slides
give the slope and then the change in slope.
Applying the Continuity Equation to a Floodplain
h
t1
t2
This is the run, or Δx, it’s the
difference in x values. Both lines
have the same Δx.
x
These are the
rise, or Δh, it’s
the difference
in h values.
Lines have
different Δh
values, thus
different
slopes. Note
that Δh is
negative.
Slope is rise over run. The t1 line has a shallow slope compared to the t2 line,
which has a steeper slope.
Applying the Continuity Equation to a Floodplain
x
t1
Δh/Δx
t2
Because line t2 has a steeper slope than line t1 and because both slopes are
negative, line t2 plots below line t1. Now, what is the change in slope?
Applying the Continuity Equation to a Floodplain
x
This is the change in x, the run.
t1
t2
Δh/Δx
These are the
change in
slope. Note
that the
change in
slope is
negative.
Change in slope is found the same way as finding slope, using the rise over
run method.
Applying the Continuity Equation to a Floodplain
x
Δ(Δh)/(ΔxΔx)
t1
t2
Because line t2 has a steeper change in slope than line t1 and because both
change in slopes are negative, line t2 plots below line t1.
Applying the Continuity Equation to
a Floodplain
• Now, we’ve got all of the values for the
right hand side of the equation (except K,
which we’ll solve for).
• What is the left hand side of the equation?
Δh/Δt = h*K (
Δ(Δh)
ΔX * ΔX
)
Applying the Continuity Equation to a Floodplain
h
t1
t2
x
This is h at two different times for all of the x values. So Δh is the difference in
h values for the two times.
Applying the Continuity Equation to a Floodplain
Δh
x
This is Δh calculated from times t1 and t2.
Applying the Continuity Equation to a Floodplain
Δh/Δt = h*K (
t1
t2
h
x
Δh
x
x
Δ(Δh)
ΔX * ΔX
)
• To determine K, input h,
Δh/Δt, and Δ(Δh)/(ΔxΔx)
from a given x location.
• With K, we can determine
velocity anywhere on the
floodplain. Recall that:
Vx = K *
t1
t2
Δ(Δh)/(ΔxΔx)
Δh
ΔX
Applying the Continuity Equation to a Floodplain
• With velocity determined anywhere on the floodplain, we can
determine discharge anywhere on the floodplain.
• Essentially, discharge in a river channel requires just one crosssection because water is confined to flow between the banks of a
river.
• Water on a floodplain can move in many different directions. It is not
confined by the banks of the river.
• Discharge on the floodplain requires knowledge of all “crosssections” where flow might go. So, if we know the floodplain
topography and the h surface, we can calculate the “cross-sections”,
these are just the difference between h and topography.
• The next slide presentation shows the two-dimensional version of
this equation.
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