## Dynamic Viscosity

A certain fluid has a specific gravity of 1.375, and a kinematic viscosity of 0.00119 m^2/s. What is the absolute dynamic viscosity in kg/(m∙s)? Note the standard density of water is 1,000 kg/m^3.

Hint
$$\nu =\mu/\rho$$$where $$\nu$$ is the kinematic viscosity, $$\mu$$ is the absolute dynamic viscosity, and $$\rho$$ is the fluid’s density. Hint 2 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.
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 fluid’s 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 the absolute dynamic viscosity: $$\mu=\rho*\nu$$$
$$=(1,375\frac{kg}{m^{3}})(0.00119\frac{m^{2}}{s})=1.64\:\frac{kg}{m\cdot s}$$\$
1.64 kg/(m∙s)