## Kinematic Viscosity

A certain fluid has a specific gravity of 1.375, and an absolute dynamic viscosity of 1.64 kg/(m∙s). What is the kinematic viscosity in m^2/s? Note the standard density of water is 1,000 kg/m^3.

Hint
The units for the absolute dynamic viscosity ( $$\mu$$ ) are $$kg/(m\cdot s)$$ . The units for the kinematic viscosity ( $$\nu$$ ) are $$m_{}^{2}/s$$
Hint 2
$$\nu =\mu/\rho$$$where $$\nu$$ is the kinematic viscosity, $$\mu$$ is the absolute dynamic viscosity, and $$\rho$$ is the density The units for the absolute dynamic viscosity ( $$\mu$$ ) are $$kg/(m\cdot s)$$ . The units for the kinematic viscosity ( $$\nu$$ ) are $$m_{}^{2}/s$$ . The relationship between the the two viscosities is: $$\nu =\mu/\rho$$$
where $$\rho$$ is the density in $$kg/m^{3}$$ .

To find the fluid’s density, recall that Specific Gravity is:
$$SG=\frac{\rho }{\rho_w}$$$where $$\rho$$ is the fluid density and $$\rho_w$$ is the density of water at standard conditions. $$\rho=SG\cdot \rho_w= (1.375)(1,000\frac{kg}{m^3})=1,375\:\frac{kg}{m^3}$$$
Solving for kinematic viscosity:
$$\nu =\frac{1.64\frac{kg}{m\cdot s}}{(1,375\frac{kg}{m^3})}=0.00119\:\frac{m^2}{s}$$$$$0.00119\:\frac{m^2}{s}$$$