Question

Problem: The density of a fluid is 1257.5kg/m3 and its absolute viscosity is 1.50Pa.s. Calculate its specific weight and kinematic viscosity.

Answer 1

Answer on Question #53007 - Physics - Mechanics - Kinematics - Dynamics

The density of a fluid is $\rho = 1257.5\frac{\mathrm{kg}}{\mathrm{m}^3}$ and its absolute viscosity is $\mu = 1.50\mathrm{Pa}\cdot \mathrm{s}$. Calculate its specific weight and kinematic viscosity.

Solution:

The connection between absolute and dynamic viscosities is given by

$$v = \frac {\mu}{\rho},$$

where $v$ – is the kinematic viscosity of the fluid. Since $\mu = 1.50 \, \text{Pa} \cdot \text{s}$ and $\rho = 1257.5 \frac{\text{kg}}{\text{m}^3}$, we obtain

$$v = \frac {\mu}{v} = \frac {1.50 \, \text{Pa} \cdot \text{s}}{1257.5 \frac {\text{kg}}{\text{m} ^ {3}}} = 1.19 \times 10 ^ {- 3} \frac {\text{m} ^ {2}}{\text{s}}$$

The specific weight is given by

$$\gamma = \rho \cdot g,$$

where $g$ – is the acceleration due to gravity. Since $g = 9.8\frac{\mathrm{m}}{\mathrm{s}^2}$, we obtain

$$\gamma = \rho \cdot g = 1257.5 \frac {\mathrm{kg}}{\mathrm{m} ^ {3}} \cdot 9.8 \frac {\mathrm{m}}{\mathrm{s} ^ {2}} = 12323.5 \frac {\mathrm{N}}{\mathrm{m} ^ {3}}$$

Answer: $v = 1.19 \times 10^{-3} \frac{\mathrm{m}^2}{\mathrm{s}}, \gamma = 12323.5 \frac{\mathrm{N}}{\mathrm{m}^3}$.

http://www.AssignmentExpert.com/