Questions: Problem 4. The approximate profile of a stream channel is given in the graph below. The depth is modeled using the equation d(w) = 1/8(w^4 - 20w^3 + 141w^2 - 406w) + 51 (a) The stream is deep, but not so wide. 1. At pre-flood stage, the water depth is 20 ft (the green line), and the edge of the stream spans from w=0.8 ft to w=9 ft. What is the cross-section of the river at this stage? 2. Calculate the cubic feet per second if the water is moving at 9 feet per second at pre-flood stage. 3. Find the cubic feet per second of water passing by for "typical" early summer flows (10 ft depth marked in purple) when the river spans from w=1.3 ft to w=8.5 ft, and the water moving at 5 feet per second. 4. (Bonus) You can see from the profile that the stream depth is not uniform. Calculate the average depth at "typical" early summer flow. Draw a picture showing the simplified stream profile. Use your simplified stream depth model to create an equation that gives the cubic feet per second which relies only on the variable for stream speed v. What are the limitations of your model?

Problem 4. The approximate profile of a stream channel is given in the graph below. The depth is modeled using the equation d(w) = 1/8(w^4 - 20w^3 + 141w^2 - 406w) + 51 (a) The stream is deep, but not so wide. 1. At pre-flood stage, the water depth is 20 ft (the green line), and the edge of the stream spans from w=0.8 ft to w=9 ft. What is the cross-section of the river at this stage? 2. Calculate the cubic feet per second if the water is moving at 9 feet per second at pre-flood stage. 3. Find the cubic feet per second of water passing by for "typical" early summer flows (10 ft depth marked in purple) when the river spans from w=1.3 ft to w=8.5 ft, and the water moving at 5 feet per second. 4. (Bonus) You can see from the profile that the stream depth is not uniform. Calculate the average depth at "typical" early summer flow. Draw a picture showing the simplified stream profile. Use your simplified stream depth model to create an equation that gives the cubic feet per second which relies only on the variable for stream speed v. What are the limitations of your model?
Transcript text: Problem 4. The approximate profile of a stream channel is given in the graph below. The depth is modeled using the equation \[ d(w)=\frac{1}{8}\left(w^{4}-20 w^{3}+141 w^{2}-406 w\right)+51 \] (a) The stream is deep, but not so wide. 1. $\left.{ }^{* * *}\right)$ At pre-flood stage, the water depth is 20 ft (the green line), and the edge of the stream spans from $w=.8 \mathrm{ft}$ to $w=9 \mathrm{ft}$. What is the cross-section of the river at this stage? 2. (**) Calculate the cubic feet per second if the water is moving at 9 feet per second at pre-flood stage. 3. (*) Find the cubic feet per second of water passing by for "typical" early summer flows ( 10 ft depth marked in purple) when the river spans from $w=1.3 \mathrm{ft}$ to $w=8.5 \mathrm{ft}$, and the water moving at 5 feet per second. 4. (Bonus) You can see from the profile that the stream depth is not uniform. Calculate the average depth at "typical" early summer flow. Draw a picture showing the simplified stream profile. Use your simplified stream depth model to create an equation that gives the cubic feet per second which relies only on the variable for stream speed $v$. What are the limitations of your model?
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Solution

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Solution Steps

Step 1: Calculate the cross-sectional area at pre-flood stage.

The cross-sectional area is a rectangle with width (9 ft - 0.8 ft) = 8.2 ft and depth 20 ft. Area = 8.2 ft * 20 ft = 164 ft²

Step 2: Calculate the flow rate at pre-flood stage.

Flow rate = cross-sectional area * velocity = 164 ft² * 9 ft/s = 1476 ft³/s

Step 3: Calculate the flow rate at typical early summer flow.

The cross-sectional area is a rectangle with width (8.5 ft - 1.3 ft) = 7.2 ft and depth 10 ft. Area = 7.2 ft * 10 ft = 72 ft² Flow rate = cross-sectional area * velocity = 72 ft² * 5 ft/s = 360 ft³/s

Final Answer

  1. 164 ft²
  2. 1476 ft³/s
  3. 360 ft³/s
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