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.
Expand 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)
Time Analysis
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