## 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**

**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**

**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)