A field sample of an aquifer is packed in a test cylinder. The cylinder has a length of 120 cm and a diameter of 12 cm. The field sample with a porosity of 0.24 is tested under a constant head difference with water at 10°C. If the estimated hydraulic conductivity of the sample is 40 m/day and the hydraulic gradient is 1.5, calculate the average flow velocity (pore velocity). Arca Sand colume Datum plane O a. u = 166.63 m/d O b.u = 250.00 m/d Oc.u = 200.00 m/d O d. u = 48.00 m/d
A field sample of an aquifer is packed in a test cylinder. The cylinder has a length of 120 cm and a diameter of 12 cm. The field sample with a porosity of 0.24 is tested under a constant head difference with water at 10°C. If the estimated hydraulic conductivity of the sample is 40 m/day and the hydraulic gradient is 1.5, calculate the average flow velocity (pore velocity). Arca Sand colume Datum plane O a. u = 166.63 m/d O b.u = 250.00 m/d Oc.u = 200.00 m/d O d. u = 48.00 m/d
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![### Calculating Average Flow Velocity in a Test Cylinder
**Problem Statement:**
A field sample of an aquifer is packed in a test cylinder. The cylinder has a length of 120 cm and a diameter of 12 cm. The field sample with a porosity of 0.24 is tested under a constant head difference with water at 10°C. If the estimated hydraulic conductivity of the sample is 40 m/day and the hydraulic gradient is 1.5, calculate the average flow velocity (pore velocity).
**Diagram Explanation:**
Below the problem statement, there is a diagram depicting the test cylinder setup:
- The test cylinder contains a sand column positioned between two pipes facilitating the water flow.
- Distances \( l \) (length of the sand column) and \( r \) (radius of the cylinder) are indicated.
- The datum plane is marked for reference.
**Options for the Average Flow Velocity (u):**
- **a. u = 166.63 m/d**
- **b. u = 250.00 m/d**
- **c. u = 200.00 m/d**
- **d. u = 48.00 m/d**
**Solution Approach:**
To solve for the average flow velocity \( u \), we can use the formula:
\[ u = \frac{K \cdot i}{n} \]
Where:
- \( K \) = Hydraulic conductivity = 40 m/day
- \( i \) = Hydraulic gradient = 1.5
- \( n \) = Porosity = 0.24
Plugging in the values:
\[ u = \frac{40 \, \text{m/day} \times 1.5}{0.24} \]
\[ u = \frac{60}{0.24} \]
\[ u = 250 \, \text{m/day} \]
**Answer:** b. \( \text{u} = 250.00 \, \text{m/day} \)
The calculation concludes that the average flow velocity (pore velocity) for the given field sample and conditions is 250.00 meters per day.
For more detailed explanations and additional practice problems, please proceed to the subsequent modules.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2d7e9b1-7208-47e7-acf1-5a4d04c3ff3b%2Fa120b3f9-b1b7-4008-8fff-0085a325a8c8%2F9kufkl9_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculating Average Flow Velocity in a Test Cylinder
**Problem Statement:**
A field sample of an aquifer is packed in a test cylinder. The cylinder has a length of 120 cm and a diameter of 12 cm. The field sample with a porosity of 0.24 is tested under a constant head difference with water at 10°C. If the estimated hydraulic conductivity of the sample is 40 m/day and the hydraulic gradient is 1.5, calculate the average flow velocity (pore velocity).
**Diagram Explanation:**
Below the problem statement, there is a diagram depicting the test cylinder setup:
- The test cylinder contains a sand column positioned between two pipes facilitating the water flow.
- Distances \( l \) (length of the sand column) and \( r \) (radius of the cylinder) are indicated.
- The datum plane is marked for reference.
**Options for the Average Flow Velocity (u):**
- **a. u = 166.63 m/d**
- **b. u = 250.00 m/d**
- **c. u = 200.00 m/d**
- **d. u = 48.00 m/d**
**Solution Approach:**
To solve for the average flow velocity \( u \), we can use the formula:
\[ u = \frac{K \cdot i}{n} \]
Where:
- \( K \) = Hydraulic conductivity = 40 m/day
- \( i \) = Hydraulic gradient = 1.5
- \( n \) = Porosity = 0.24
Plugging in the values:
\[ u = \frac{40 \, \text{m/day} \times 1.5}{0.24} \]
\[ u = \frac{60}{0.24} \]
\[ u = 250 \, \text{m/day} \]
**Answer:** b. \( \text{u} = 250.00 \, \text{m/day} \)
The calculation concludes that the average flow velocity (pore velocity) for the given field sample and conditions is 250.00 meters per day.
For more detailed explanations and additional practice problems, please proceed to the subsequent modules.
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