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16 Cards in this Set
- Front
- Back
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Cards by adac - please rate and favorite.
Flip this card for the source of information and a guide on what certain terms mean. I recommend studying these cards in order for the first few times. |
Notes from Easterbrook, D.J., (1969) Principles of Geomorphology - Part 3: Fluvial Morphology - Chapter 6: Stream Processes
My cards are designed so you describe the subject, rather than naming something after seeing the description. My method will better prepare you for short answer and essay questions. Equations use . for multiplication (not x) |
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How dominant is running water at sculpting the Earth's surface? (Tip: Locally and worldwide)
Give an example of how powerful this process is. |
Running water is the most dominant process worldwide, but locally other processes MAY dominate.
Eg. The Mississippi drainage system sees ~22 trillion cubic feet of water flow annually from land to sea. ~517 million tons of rock material removed form land surface (cite). |
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How is water impelled forward downstream?
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Generally: By the force of gravity
Specifically: By the component of the gravitational force F which is parallel to the stream floor (F sin slope angle). Hint: Try to visualize this force parallel to the stream floor. |
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What variable most affects the stream's ability to erode its channel and transport sediment?
(Remember this for later when talking about slope) |
Velocity
(Remember this for later when talking about slope) |
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What impedes the flow of water?
How is the acceleration of water related? |
Frictional forces.
If there was no friction - water would accelerate to very high velocities. No acceleration means a balance between the downslope component of gravitational force and the internal and external frictional forces. |
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How do frictional forces affect the distribution of velocities within a stream channel?
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They tend to reduce velocity near the bed and walls of a stream.
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What is the value of velocity on the channel floor?
At what distance from the surface of the stream to the bed do mean velocities occur? |
v = 0 at channel floor
mean velocity occurs at 0.6d - that is, 0.6 of the total depth (Leopold, Wolman, and Miller, 1964). |
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Give the relationship between velocity and discharge (Q)
Tip: The equation for discharge |
Q = w . v . d
where Q = discharge (m^3/s) w = width (m) d = mean depth (m) v = mean velocity (m/s) |
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Give the equations for:
- Stream cross-sectional area - Wetted perimeter If Q increases, these values also increase. Which of the two increases more rapidly? What are the implications of this rate of change on frictional retardation? Tip: Simply substitute some values for w and d and see what happens. |
Cross-sectional area: w . d
Wetted perimeter: w + 2d w . d increases more rapidly. This results in a relative decrease in frictional retardation by the floor and banks of the channel. Important: For this reason, velocity increases with discharge. Hint: Draw a cross-section of a stream and change w and d values to visualize the changes. |
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Read Leopold and Maddock, 1953 ... seriously, it will help you a lot.
Write the equations for w, d, and v. What do the exponents represent? |
w = a . Q ^ b
d = c . Q ^ f v = k . Q ^ m The exponents represent the slope of the lines shown in figures 4A and 4B. |
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We are told by our lecturers that Q = w.v.d and also that Q = a.Q^b . c.Q^f . k.Q^m
We are then told, that if this is true, then the sum of the exponents b+f+m must equal 1, while the product of a.c.k must also equal 1. I found this to be pretty confusing. Try and re-arrange the equation above to help you understand why these conditions are true. |
Q = w.d.v = a.Q^b . c.Q^f . k.Q^m
Q = a.c.k (Q^b+f+m) Try and remember the rules of algebra. In order for Q = Q, the exponent of Q must be equal to 1 (Q^1 = Q). Also, the only way for Q to equal Q when being multiplied by another number is for that number to be equal to 1. So, for (a.c.k) . (Q), a.c.k must equal 1. Thus Q = 1 . (Q^1) = Q |
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How is velocity related to slope?
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Velocity is directly related to slope and increases as slope increases.
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With regard to transporting power, why does the mean velocity have little meaning?
Tip: Think about how we got the mean velocity in a previous card... |
Two reasons mean velocity is not useful regarding transport power:
1. Only the velocity near the floor of the channel is usable for transportation of bed load (mean velocity is at 0.6 of the total depth). 2. Much of the bed load is transported during flood events. |
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Both slope and discharge affect velocity. What is their correlation (positive or negative)?
If slope and discharge increase, which is more effective at increasing transportation of bed load? More importantly: Why? |
Positive - if slope and/or discharge increases, velocity increases (and vice-versa).
An increase in slope will be more effective at increasing transportation of bed load in comparison to increased discharge (Gilbert, 1914). The reason is because an increase in slope (at constant discharge) will increase velocity (see card 4) and decrease depth whereas an increase in discharge will increase depth. |
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We mentioned an increase in slope increases velocity while decreasing depth. Explain why this would result in increased transport of bed load.
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The maximum velocity that can be developed without causing increased erosion of the channel floor is a function of depth;
i.e. with the same velocity a deep channel will not erode its channel as much as a shallow channel, since, as depth increases, the water above the bed won't affect particles on the channel floor. Hint: It really helps to view the idealized velocity curves in Gilbert (1914). |
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Name four ways sediment can be transported by a stream.
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- In solution
- Suspension - Saltation - Traction |
