$$m \Delta u=m l \frac{\partial u}{\partial y}$$
The mass of fluid actually moving from one level $y_{0}$ to level $y_{0}+_{1}$ is proportional to $\rho\left|v^{\prime}\right|$, which leads to the expression for $\tau$
$$\tau=m \Delta u=\rho\left|v^{\prime}\right| l \frac{\partial u}{\partial y}$$

For continuity reasons we must have
$$v^{\prime} \sim u^{\prime} \sim|l| \frac{\partial u}{\partial y}$$
This means that
$$\tau=\rho l^{2}\left|\frac{\partial u}{\partial y}\right| \frac{\partial u}{\partial y}$$
which expresses $\tau$ in terms of the mean flow.

## PHYS97001 COURSE NOTES ：

The work done by the pressure is, per unit time
$$p u_{i} n_{i} d S=p u_{n} d S$$
so that the total energy flux through $d S$ is
$$d E_{f}\left(x_{i}, t\right)=\left(p+\rho g z+\frac{1}{2} \rho u_{i} u_{i}\right) u_{n} d S$$
and we can determine the total energy flux by integrating over the entire surface
$$E_{f}\left(x_{i}, t\right)=\int_{S}\left(p+\rho g z+\frac{1}{2} \rho u_{i} u_{i}\right) u_{i} n_{i} d S$$