# 9.4: Flood waves in a turbulent river

In homework exercise 29, we predict river speed using the stress-strain relation for a Newtonian fluid, and find that the speed is ∼ 8 km/s! The result is rendered reasonable only by imposing an eddy viscosity, much greater than the true viscosity, to represent the retarding effect of turbulent fluctuations. Here we will do that calculation again with turbulence represented in a more realistic fashion. We will also allow for a variable surface elevation $$\eta(x,t)$$ and hence the possibility of waves.

We begin with Cauchy’s equation Equation 6.3.18, with the stress tensor expanded into pressure and deviatoric parts as per Equation 6.3.27:

$\rho \frac{D u_{j}}{D t}=\rho g_{j}-\frac{\partial p}{\partial x_{j}}+\frac{\partial \sigma_{i j}}{\partial x_{i}}.\label{eqn:1}$

We assume that the flow is homogeneous ($$\rho=\rho_0$$) and two-dimensional ($$v=0$$, $$\partial/\partial y=0$$). We also assume that the stress varies primarily in $$z$$, and therefore neglect its $$x$$-derivatives. Finally, we assume that the motion is nearly hydrostatic and therefore $$\partial u/\partial z=0$$.

Writing out the components of Equation $$\ref{eqn:1}$$, we have

$\frac{D u}{D t}=g \sin \theta-\frac{1}{\rho_{0}} \frac{\partial p}{\partial x}+\frac{1}{\rho_{0}} \frac{\partial \sigma_{31}}{\partial z}\label{eqn:2}$

$\frac{D w}{D t}=-g \cos \theta-\frac{1}{\rho_{0}} \frac{\partial p}{\partial z}+\frac{1}{\rho_{0}} \frac{\partial \sigma_{33}}{\partial z}.\label{eqn:3}$

Assuming that the vertical acceleration $$Dw/Dt$$ is negligible, the vertical momentum equation gives us an altered form of hydrostatic balance:

$\frac{1}{\rho_{0}} \frac{\partial p}{\partial z}=-g \cos \theta+\frac{1}{\rho_{0}} \frac{\partial \sigma_{33}}{\partial z}.$

Integrating over $$z$$ and assuming $$p = 0$$ at $$z = \eta$$, we have

$p=-\rho_{0} g \cos \theta(\eta-z)+\left.\sigma_{33}\right|_{z} ^{\eta}.$

We now differentiate with respect to $$x$$ (remembering that stress is independent of $$x$$) to get

$\frac{\partial p}{\partial x}=\rho_{0} g \cos \theta \frac{\partial \eta}{\partial x}.$

Substituting this into the $$x$$-momentum equation, we have

$\frac{D u}{D t}=g \sin \theta-g \cos \theta \frac{\partial \eta}{\partial x}+\frac{1}{\rho_{0}} \frac{\partial \sigma_{31}}{\partial z}.$

In summary, the along-stream flow is driven by three forces: the downhill pull of gravity, pressure gradients due to surface deflection, and the stress divergence. The tangential stress $$\sigma_{31}$$ is exerted on the fluid by the solid bottom boundary and by the air above the surface, then transmitted through the fluid interior by turbulent eddies.

Assuming again that $$u$$ is independent of $$z$$ (the long-wave approximation), we can integrate in the vertical to obtain

$(H+\eta) \frac{D u}{D t}=(H+\eta)\left(g \sin \theta-g \cos \theta \frac{\partial \eta}{\partial x}\right)+\left.\frac{\sigma_{31}}{\rho_{0}}\right|_{-H} ^{\eta}\label{eqn:4}$

Here we see a major advantage of working directly from Cauchy’s equation: all we need to know about the stress tensor is the values of $$\sigma_{31}$$ at the surface and the bottom. The bottom stress is given by a very well-tested empirical relationship:

$\left.\sigma_{31}\right|_{z=-H}=\rho_{0} C_{D} u^{2},$

where $$C_D$$ is the drag coefficient. Typically, $$C_D$$ is in the range 10−3 −10−2. Here we treat $$C_D$$ as a constant. A similar relationship holds at the surface:

$\left.\sigma_{31}\right|_{z=\eta}=\rho_{A} C_{D}\left(u_{A}-u\right)^{2},$

where $$\rho_A$$ is the density of air and $$u_A$$ is the wind speed. For a typical river flow the effect of wind is negligible, so we will consider only the bottom stress. Substituting in Equation $$\ref{eqn:4}$$ and dividing by $$H +\eta$$, we have

$\frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}=\underbrace{g \sin \theta}_{\text {gravity }}-g \cos \theta \frac{\partial \eta}{\partial x}-\underbrace{\frac{C_{D} u^{2}}{H+\eta}}_{\text {friction }}.\label{eqn:5}$

We now assume that the terms involving $$\partial/\partial t$$ and $$\partial/\partial x$$ in Equation $$\ref{eqn:5}$$ are negligible. In other words, the dominant balance is between the downhill pull of gravity and the retarding effect of bottom friction. This assumption makes it easy to solve for the flow velocity1:

$u=\sqrt{\frac{g(H+\eta) \sin \theta}{C_{D}}}.\label{eqn:6}$

For example, consider the river described in homework exercise 29, with $$H$$ = 2 m, $$\eta$$ = 0, $$g$$ = 9.8 ms-2, $$\sin\theta$$ = 4.3 × 10−4 and set $$C_D$$ = 3×10−3 (Li et al. 2004). The result is u = 1.7 m/s, a reasonable river speed.

Exercise $$\PageIndex{1}$$

Derive an expression for the Froude number based on Equation $$\ref{eqn:6}$$. What is its value for the river parameters given above? Now derive a simple formula for the slope angle at which the flow becomes supercritical. If $$C_D$$ = 3×10−3, what is this critical slope angle in meters per kilometer? Based on this model, could a river become supercritical as a result of increased flow rate (in flood conditions, for example)?2

We now allow for small but nonzero variations in $$x$$ and $$t$$. Note that we have not yet had to invoke mass conservation: with such simple flow geometry mass is conserved automatically. To enforce mass conservation in the presence of variations in $$x$$ and $$t$$, we write Equation 9.2.6 with bottom topography omitted ($$h$$ = 0):

$\frac{\partial \eta}{\partial t}+\frac{\partial}{\partial x}[u(H+\eta)]=0.\label{eqn:7}$

Substituting for $$u$$ using Equation $$\ref{eqn:6}$$, this becomes3:

$\frac{\partial \eta}{\partial t}+\frac{\partial}{\partial x} \sqrt{\frac{g(H+\eta)^{3} \sin \theta}{C_{D}}}=0.$

After carrying out the differentiation with respect to $$x$$, this becomes

$\frac{\partial \eta}{\partial t}+\frac{3}{2} u \frac{\partial \eta}{\partial x}=0.$

What this tells us is that $$\eta(x,t)$$ is constant on trajectories $$x(t)$$ given by

$\left.\frac{d x}{d t}\right|_{\eta=\text { const. }}=-\frac{\partial \eta / \partial t}{\partial \eta / \partial x}=\frac{3}{2} u,\label{eqn:8}$

i.e., signals propagate 50% faster than the current itself. In the river described in exercise 29, the current travels 1.7 m/s. Suppose now that a rain event upstream causes a sudden increase in flow rate. Our model predicts that increase will propagate downstream at 2.6 m/s.

1This is called the Chézy formula for river speed. It was developed by the French engineer Antoine de Chézy (1718-1798) and tested using measurements of the River Seine.

2In fact, measurements show that the drag coefficient decreases with increasing depth, approximately as depth−1/3, so the Froude number is proportional to depth1/6 (e.g., White 2003).

3Does it bother you that we discarded terms involving $$\partial/\partial t$$ and $$\partial/\partial x$$ in Equation $$\ref{eqn:5}$$ but retain such terms in Equation $$\ref{eqn:7}$$? In Equation $$\ref{eqn:5}$$, we neglected the terms involving $$\partial/\partial t$$ and $$\partial/\partial x$$ not on the grounds that these terms are zero (i.e., that nothing varies in $$x$$ or $$t$$), but rather because the remaining terms are much bigger. In the mass equation Equation $$\ref{eqn:7}$$, both terms involve partial derivatives, so there are no larger terms to dominate the balance. It is therefore permissible to retain the partial derivatives in Equation $$\ref{eqn:7}$$.