1. Lecture no 15 &16 The Continuity Equation
Prepared by
Engr. Sarafarz Khan Turk
Lecturer at IBT LUMHS Jamshoro

2. The Continuity Equation
A continuity equation in physics is an equation that describes the transport of a conserved quantity. Since mass, energy, momentum, electric charge and other natural quantities are conserved under their respective appropriate conditions, a variety of physical phenomena may be described using continuity equations.

3. The Continuity Equation
Continuity equations more generally can include "source" and "sink" terms, which allow them to describe quantities that are often but not always conserved, such as the density of a molecular species which can be created or destroyed by chemical reactions. In an everyday example, there is a continuity equation for the number of living humans; it has a "source term" to account for people giving birth, and a "sink term" to account for people dying

4. 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.

5. Cross Section of a Floodplain
Qin
Qout t3
Terre-Firme t2 t1
Floodplain
River Channel
Q is Discharge
t is time

6. A “Glass” of Water
Glass is empty at t1

7. A “Glass” of Water Qout < Qin Qin Qout t2 ΔS increases
Glass is filling at time t2

8. A “Glass” of Water Qout < Qin Qin Qout t3 ΔS increases t2
Glass is nearly full at t3

9. A “Glass” of Water Qout > Qin Qin Qout t3 ΔS decreases t4
Glass is emptying at t4

10. The Basic Form of the Continuity Equation
ΔS = Qin - Qout
Qout
Qin t1 t2 t3 t4
Qout<Qin
Qout<Qin
Qout>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.

11. 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.

12. River Discharge Q = 3m * 10 m * 1 m/s = 30 m3/s

13. 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

14. 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.

15. A “Cube” of Water Qin Qout Qin Qout 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.

16. A “Cube” of Water Vin along z axis Vout along y axis Vin along y axis
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.

17. 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 * Here, multiplying ΔY/ΔY is the same as multiplying by 1,
but ΔX * ΔY * ΔZ is the volume of the cube, lets call the
volume, “vol”
ΔVy * Using this same approach, there are similar expressions for the change in discharge within the cube for the X and Z axis.
ΔVx * ΔVz * ΔY
= ΔQy
vol ΔY
= ΔQy
vol ΔX
vol ΔZ
= ΔQx
= ΔQz

18. 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 ΔVy ΔVz
vol ΔX
vol ΔY
vol ΔZ
These are a series of simple algebraic steps replacing variables with their equivalent expressions.
ΔVx ΔX
ΔVy ΔY
ΔVz ΔZ
ΔQ = vol ( )
ΔVx ΔX
ΔVy ΔY
ΔVz ΔZ
ΔS = vol ( )
This is the basic form of the continuity equation.
We want this equation in terms of h, the thing we can easily measure.

19. A Basic Physics Fact for Water
Water flows downhill!
An expression for this is:
Applying this expression to the algebra:
What is Δ(Δh)?
How do we re-write ΔS as something with h, the thing that is measurable?
Velocity = Constant * Slope Δh ΔX Δh ΔY Δh ΔZ
Vx = K *
Vy = K *
Vz = K *
ΔS = vol ( )
ΔVz ΔZ
ΔVx ΔX
ΔVy ΔY
K * Δ(Δh)
ΔX * ΔX
K * Δ(Δh)
ΔY * ΔY
K * Δ(Δh)
ΔZ * ΔZ
ΔS = vol ( )

20. 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.
Δ(Δh)
ΔX * ΔX
Δ(Δh)
ΔY * ΔY
Δ(Δh)
ΔZ * ΔZ
Δh/Δt = vol*K ( )

21. 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.
Δ(Δh)
ΔX * ΔX
Δh/Δt = h*K ( )

22. Applying the Continuity Equation to a Floodplain
Terre-Firme t1 t2
Floodplain
River Channel
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?

23. Applying the Continuity Equation to a Floodplainx
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.

24. Applying the Continuity Equation to a Floodplain
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. t1
This is the run, or Δx, it’s the difference in x values. Both lines have the same Δx. t2 x
Slope is rise over run. The t1 line has a shallow slope compared to the t2 line, which has a steeper slope.

25. Applying the Continuity Equation to a Floodplainx 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?

26. Applying the Continuity Equation to a Floodplainx
This is the change in x, the run.
Δh/Δx
These are the change in slope. Note that the change in slope is negative. t1 t2
Change in slope is found the same way as finding slope, using the rise over run method.

27. Applying the Continuity Equation to a Floodplainx
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.

28. 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?
Δ(Δh)
ΔX * ΔX
Δh/Δt = h*K ( )

29. Applying the Continuity Equation to a Floodplainx
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.

30. Applying the Continuity Equation to a Floodplainx
This is Δh calculated from times t1 and t2.

31. Applying the Continuity Equation to a Floodplain
ΔX * ΔX
Δh/Δt = h*K ( )
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: t1 h t2 x Δh x x Δh ΔX
Vx = K * t1
Δ(Δh)/(ΔxΔx) t2

32. 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 cross-section 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 “cross-sections” 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.