The sound of a waterfall can add a sense of calm to a space. The owner of the pictured waterfall wanted to increase the sound effect by causing the water to develop turbulent flow 0.4 m from the top. i)If the kinematic viscosity of the water in the garden at 20°C is 1.004 x 10$ m²/s , what flow rate would the owner need to use. Show your calculations. ii) What would happen to the point of turbulence in the winter when the temperature dops a few degrees, but flow rate is maintained at the summer value, explain your answer?
The sound of a waterfall can add a sense of calm to a space. The owner of the pictured waterfall wanted to increase the sound effect by causing the water to develop turbulent flow 0.4 m from the top. i)If the kinematic viscosity of the water in the garden at 20°C is 1.004 x 10$ m²/s , what flow rate would the owner need to use. Show your calculations. ii) What would happen to the point of turbulence in the winter when the temperature dops a few degrees, but flow rate is maintained at the summer value, explain your answer?
Related questions
Question
![**Understanding Waterfall Sound and Turbulent Flow Creation**
The sound of a waterfall can add a sense of calm to a space. The owner of the pictured waterfall wanted to increase the sound effect by causing the water to develop turbulent flow 0.4 m from the top.
**Part i)** If the kinematic viscosity of the water in the garden at 20°C is \(1.004 \times 10^{-5} \, m^2/s\), what flow rate would the owner need to use? **Show your calculations.**
**Part ii)** What would happen to the point of turbulence in the winter when the temperature drops a few degrees, but the flow rate is maintained at the summer value? Explain your answer.
---
**Detailed Explanation:**
In this problem, we aim to find the necessary flow rate to achieve turbulent flow at a specified distance from the top of the waterfall, and then analyze what happens under different temperature conditions.
### Part i) Calculation of Flow Rate for Turbulent Flow
To achieve turbulent flow, we must consider the Reynolds number (Re), defined as:
\[ \text{Re} = \frac{\rho \cdot v \cdot L}{\mu} \]
However, for a waterfall, the flow along a vertical surface, the Reynolds number can be written in terms of kinematic viscosity (\( \nu \)):
\[ \text{Re} = \frac{v \cdot L}{\nu} \]
Where:
- \( \nu \) is the kinematic viscosity.
- \( D \) is the characteristic length (0.4 m in our case).
We assume turbulent flow occurs when Re > 4000. We solve for the flow rate.
Given that \( \nu = 1.004 \times 10^{-5} \, m^2/s \), and \( L = 0.4 \, m \):
\[ \text{Re} = \frac{Q \cdot L}{\nu} \]
where \( v \) is the velocity, which can be obtained from the flow rate \( Q \) (consider cross-sectional area as \( A \)):
\[ v = \frac{Q}{A} \]
Combining the equations:
\[ 4000 = \frac{Q \cdot 0.4}{1.004 \times 10^{-5}} \]
Solving for \( Q \):
\[ Q =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbd229fe5-a61f-4584-a09a-f2898fdc7c7e%2F1f490476-f58f-450f-91a7-475f23ad6237%2Fscv3n5f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Understanding Waterfall Sound and Turbulent Flow Creation**
The sound of a waterfall can add a sense of calm to a space. The owner of the pictured waterfall wanted to increase the sound effect by causing the water to develop turbulent flow 0.4 m from the top.
**Part i)** If the kinematic viscosity of the water in the garden at 20°C is \(1.004 \times 10^{-5} \, m^2/s\), what flow rate would the owner need to use? **Show your calculations.**
**Part ii)** What would happen to the point of turbulence in the winter when the temperature drops a few degrees, but the flow rate is maintained at the summer value? Explain your answer.
---
**Detailed Explanation:**
In this problem, we aim to find the necessary flow rate to achieve turbulent flow at a specified distance from the top of the waterfall, and then analyze what happens under different temperature conditions.
### Part i) Calculation of Flow Rate for Turbulent Flow
To achieve turbulent flow, we must consider the Reynolds number (Re), defined as:
\[ \text{Re} = \frac{\rho \cdot v \cdot L}{\mu} \]
However, for a waterfall, the flow along a vertical surface, the Reynolds number can be written in terms of kinematic viscosity (\( \nu \)):
\[ \text{Re} = \frac{v \cdot L}{\nu} \]
Where:
- \( \nu \) is the kinematic viscosity.
- \( D \) is the characteristic length (0.4 m in our case).
We assume turbulent flow occurs when Re > 4000. We solve for the flow rate.
Given that \( \nu = 1.004 \times 10^{-5} \, m^2/s \), and \( L = 0.4 \, m \):
\[ \text{Re} = \frac{Q \cdot L}{\nu} \]
where \( v \) is the velocity, which can be obtained from the flow rate \( Q \) (consider cross-sectional area as \( A \)):
\[ v = \frac{Q}{A} \]
Combining the equations:
\[ 4000 = \frac{Q \cdot 0.4}{1.004 \times 10^{-5}} \]
Solving for \( Q \):
\[ Q =
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 2 images
