The viscosity of glycerin at 20 C is µ = 1.414 Pa.s and the mass density is = 1264 kg/m3 . Calculate the kinematic viscosity.
The Correct Answer and Explanation is :
Solution:
The kinematic viscosity (( \nu )) is calculated using the formula:
[
\nu = \frac{\mu}{\rho}
]
Where:
- ( \mu ) is the dynamic viscosity (( 1.414 \, \text{Pa·s} )),
- ( \rho ) is the mass density (( 1264 \, \text{kg/m}^3 )).
Substituting the given values:
[
\nu = \frac{1.414}{1264}
]
Performing the calculation:
[
\nu = 0.001119 \, \text{m}^2/\text{s}
]
Final Answer:
The kinematic viscosity of glycerin at ( 20^\circ \text{C} ) is:
[
\nu = 1.119 \times 10^{-3} \, \text{m}^2/\text{s}
]
Explanation:
1. Kinematic Viscosity Definition:
Kinematic viscosity (( \nu )) is a measure of a fluid’s resistance to flow under the influence of gravity, expressed as the ratio of dynamic viscosity (( \mu )) to the fluid’s mass density (( \rho )). Its SI unit is ( \text{m}^2/\text{s} ).
2. Relationship Between ( \mu ) and ( \nu ):
Dynamic viscosity (( \mu )) represents the internal resistance to flow when an external force is applied. However, kinematic viscosity takes into account the fluid’s density, which affects how the fluid’s inertia influences its flow behavior.
3. Calculation Process:
In this problem, we directly substitute the given values into the formula. The dynamic viscosity is ( 1.414 \, \text{Pa·s} ), which is equivalent to ( 1.414 \, \text{kg·m}^{-1}\text{s}^{-1} ) (since ( 1 \, \text{Pa} = 1 \, \text{kg·m}^{-1}\text{s}^{-2} )). The mass density of ( 1264 \, \text{kg/m}^3 ) tells us how concentrated the mass is in the fluid’s volume. Dividing ( \mu ) by ( \rho ) gives ( \nu ).
4. Practical Meaning:
The result (( 0.001119 \, \text{m}^2/\text{s} )) shows that glycerin flows relatively slowly due to its high viscosity and density. This property is crucial in industries where glycerin is used, such as in lubricants, pharmaceuticals, and food production, as it influences pumping, mixing, and thermal transport.