diameter of 0.20 millime = 1.6) with a Reynolds n
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![**Problem Statement:**
A particle has a diameter of 0.20 millimeters (Specific gravity = 2.5) falling in a liquid (Specific gravity= 1.6) with a Reynolds number of 10,000, determine its terminal velocity in m/s.
**Detailed Explanation:**
To determine the terminal velocity of a particle falling in a liquid, we can use the following fundamental fluid mechanics concepts:
1. **Particle Diameter (d):**
- Diameter \( d = 0.20 \) millimeters = \( 0.20 \times 10^{-3} \) meters
2. **Specific Gravity (SG):**
- Specific Gravity of particle, \( SG_p = 2.5 \)
- Specific Gravity of liquid, \( SG_l = 1.6 \)
3. **Reynolds Number (Re):**
- Given \( Re = 10,000 \)
**Specific Gravity to Density Conversion:**
\[ \rho_{object} = SG \times \rho_{water} \]
\[ \rho_{water} \approx 1000 \; \text{kg/m}^3 \]
Thus,
- Density of particle \( \rho_p = 2.5 \times 1000 \) kg/m³ = \( 2500 \) kg/m³
- Density of liquid \( \rho_l = 1.6 \times 1000 \) kg/m³ = \( 1600 \) kg/m³
**Reynolds Number Definition:**
\[ Re = \frac{\rho_l \cdot V \cdot d}{\mu} \]
Where:
- \( \rho_l \) is the density of liquid (1600 kg/m³)
- \( V \) is the velocity to be determined (m/s)
- \( d \) is the diameter of the particle (0.20 x 10^-3 meters)
- \( \mu \) is the dynamic viscosity of the liquid
The expression can rearrange to solve for terminal velocity \( V \):
\[ V = \frac{Re \cdot \mu}{\rho_l \cdot d} \]
Note: Without the dynamic viscosity \( \mu \), the problem cannot be precisely solved. For a typical liquid, this value should be provided, or an assumption must be made based on standard liquids, such as water having an approximate dynamic viscosity of \( 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2c07220d-ca17-4b4a-9c07-9176cc38cf32%2F532a8b53-ddcc-4ebe-be96-b44c8b5f25f9%2F1oll0zp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A particle has a diameter of 0.20 millimeters (Specific gravity = 2.5) falling in a liquid (Specific gravity= 1.6) with a Reynolds number of 10,000, determine its terminal velocity in m/s.
**Detailed Explanation:**
To determine the terminal velocity of a particle falling in a liquid, we can use the following fundamental fluid mechanics concepts:
1. **Particle Diameter (d):**
- Diameter \( d = 0.20 \) millimeters = \( 0.20 \times 10^{-3} \) meters
2. **Specific Gravity (SG):**
- Specific Gravity of particle, \( SG_p = 2.5 \)
- Specific Gravity of liquid, \( SG_l = 1.6 \)
3. **Reynolds Number (Re):**
- Given \( Re = 10,000 \)
**Specific Gravity to Density Conversion:**
\[ \rho_{object} = SG \times \rho_{water} \]
\[ \rho_{water} \approx 1000 \; \text{kg/m}^3 \]
Thus,
- Density of particle \( \rho_p = 2.5 \times 1000 \) kg/m³ = \( 2500 \) kg/m³
- Density of liquid \( \rho_l = 1.6 \times 1000 \) kg/m³ = \( 1600 \) kg/m³
**Reynolds Number Definition:**
\[ Re = \frac{\rho_l \cdot V \cdot d}{\mu} \]
Where:
- \( \rho_l \) is the density of liquid (1600 kg/m³)
- \( V \) is the velocity to be determined (m/s)
- \( d \) is the diameter of the particle (0.20 x 10^-3 meters)
- \( \mu \) is the dynamic viscosity of the liquid
The expression can rearrange to solve for terminal velocity \( V \):
\[ V = \frac{Re \cdot \mu}{\rho_l \cdot d} \]
Note: Without the dynamic viscosity \( \mu \), the problem cannot be precisely solved. For a typical liquid, this value should be provided, or an assumption must be made based on standard liquids, such as water having an approximate dynamic viscosity of \( 1
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