Riemann surface
matlab

Show that the kinematic viscosity has the primary dimensions of $\mathrm{L}^2 \mathrm{~T}^1$.

Solution:
The kinematic viscosity is defined as the ratio of the dynamic viscosity by the density of the fluid.
The density has units of mass $(\mathrm{kg})$ divided by volume $\left(\mathrm{m}^3\right)$; whereas the dynamic viscosity has the units of mass ( $\mathrm{kg})$ per meter $(\mathrm{m})$ per time (s).
Hence:
$$v=\frac{\mu}{\rho}=\frac{M L^{-1} T^{-1}}{M L^{-3}}=L^2 T^{-1}$$

2.2 Water at $36 \mathrm{~m}$ above sea level has a velocity of $18 \mathrm{~m} / \mathrm{s}$ and a pressure of $350 \mathrm{kN} / \mathrm{m}^2$. Determine the potential, kinetic and pressure energy of the water in metres of head. Also determine the total head.

Ans $(35.68 \mathrm{~m}, 16.5 \mathrm{~m}, 36 \mathrm{~m}, 88.2 \mathrm{~m})$

Solution:
Take each term separately
\begin{aligned} & \frac{p_l}{\rho g}=\frac{350 \times 10^3}{1000 \times 9.81}=35.678 \mathrm{~m} \ & \frac{V_l^2}{2 g}=\frac{18^2}{2 x 9.81}=16.514 \mathrm{~m} \ & z_1=36 \mathrm{~m} \end{aligned}
The total head $=35.678+16.514+36=88.192 \mathrm{~m}$

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