Question

Problem: Carbon tetrachloride at 20 0C, has an absolute viscosity of 9.67×10-4Pa.s and a kinematic viscosity of 6.08×10-7m2/s. Calculate its density and specific weight.

Answer 1

Answer on Question #53008 - Physics - Mechanics - Kinematics - Dynamics

Carbon tetrachloride at $20{}^{\circ}\mathrm{C}$, has an absolute viscosity of $\mu = 9.67 \times 10^{-4} \mathrm{~Pa} \cdot \mathrm{s}$ and a kinematic viscosity of $v = 6.08 \times 10^{-7} \frac{\mathrm{m}^2}{\mathrm{s}}$. Calculate its density and specific weight.

Solution:

The connection between absolute and dynamic viscosities is given by

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

where $\rho$ – is the density of the fluid. Since $\mu = 9.67 \times 10^{-4} \, \mathrm{Pa} \cdot \mathrm{s}$ and $v = 6.08 \times 10^{-7} \frac{\mathrm{m}^2}{\mathrm{s}}$, we obtain

$$\rho = \frac{\mu}{v} = \frac{9.67 \times 10^{-4} \, \mathrm{Pa} \cdot \mathrm{s}}{6.08 \times 10^{-7} \, \frac{\mathrm{m}^2}{\mathrm{s}}} = 1590.5 \, \frac{\mathrm{kg}}{\mathrm{m}^3}$$

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

$$\gamma = \rho \cdot g = 1590.5 \, \frac{\mathrm{kg}}{\mathrm{m}^3} \cdot 9.8 \, \frac{\mathrm{m}}{\mathrm{s}^2} = 15587 \, \frac{\mathrm{N}}{\mathrm{m}^3}$$

Answer:

$\rho = 1590.5 \, \frac{\mathrm{kg}}{\mathrm{m}^3}, \gamma = 15587 \, \frac{\mathrm{N}}{\mathrm{m}^3}.$

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