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# If the kinematic viscosity of glycerin is $v=1.15\left(10^{-3}\right) \mathrm{m}^2 / \mathrm{s}$, find its viscosity in FPS units. At the temperature considered, glycerin's specific gravity is $S_g=1.26$

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First, let us convert the kinematic viscosity of glycerin from SI to FPS units. Since $\nu_g=1.15\cdot 10^{-3}\ \mathrm{m^2/s}$, we have to convert meters to feet. As we know, it holds that $1\ \mathrm{m}=3.28\ \mathrm{ft}$. So, $1\ \mathrm{m^2}=10.8\ \mathrm{ft^2}$, and we obtain:

$$$\nu_g=1.15\cdot 10^{-3}\cdot 10.8\ \mathrm{ft^2/s}=12.4\cdot 10^{-3}\ \mathrm{ft^2/s}.$$$

Now, since is the kinematic viscosity of glycerin is, by definition, equal to:

$$$\nu_g=\dfrac{\mu_g}{\rho_g},$$$

it follows that the viscosity of glycerin $\mu_g$ is given by:

$$$\mu_g=\nu_g\rho_g.$$$

The density is related to the specific gravity by the following equation:

$$$S_g=\dfrac{\rho_g}{\rho_w},$$$

where $\rho_w=1940\ \mathrm{slug/ft^3}$ is the density of water (in FPS units). Therefore, $\rho_g=S_g\rho_w$, which, since $S_g=1.26$, can be plugged into equation (1) to obtain:

$$$\mu_g=\nu_gS_g\rho_w=12.4\cdot 10^{-3}\ \mathrm{ft^2/s}\cdot 1.26\cdot 1940\ \mathrm{slug/ft^3}=\boxed{30.3\ \mathrm{lb\cdot s/ft^2}},$$$

where we have used that $1\ \mathrm{slug\cdot ft/s^2}=1\ \mathrm{lb}$.

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