Stream power

Stream power is the rate of energy dissipation against the bed and banks of a river or stream per unit downstream length. It is given by the equation:

$\Omega=\rho g Q S$

where Ω is the stream power, ρ is the density of water (1000 kg/m³), g is acceleration due to gravity (9.8 m/s²), Q is discharge (m³/s), and S is the channel slope.

It can be derived by the fact that if the water is not accelerating and the river cross-section stays constant (generally good assumptions for an averaged reach of a stream over a modest distance), all of the potential energy lost as the water flows downstream must be used up in friction or work against the bed: none can be added to kinetic energy. Therefore, the potential energy drop is equal to the work done to the bed and banks, which is the stream power.

We know that change in potential energy over change in time is given by the equation:

$\frac{\Delta PE}{\Delta t} = m g \frac{\Delta z}{\Delta t}$

where water mass and gravitational acceleration are constant. We can use the channel slope and the stream velocity as a stand-in for ${\Delta z}/{\Delta t}$: the water will lose elevation at a rate given by the downward component of velocity $u_z$. For an channel slope (as measured from the horizontal) of $\alpha$:

$\frac{\Delta z}{\Delta t} = u_z = u \sin(\alpha) \approx u S$

where $u$ is the downstream flow velocity. It is noted that for small angles, $\sin(\alpha) \approx \tan(\alpha) = S$. Rewriting the first equation, we now have:

$\frac{\Delta PE}{\Delta t} = m g u S$

Remembering that power is energy per time and using the equivalence between work against the bed and loss in potential energy, we can write:

$\Omega = \frac{\Delta PE}{\Delta t}$

Finally, we know that mass is equal to density times volume. From this, we can rewrite the mass on the right hand side

$m = \rho L b h$

where $L$ is the channel length, $b$ is the channel width (breadth), and $h$ is the channel depth (height). We use the definition of discharge

$Q = u b h$

where $A$ is the cross-sectional area, which can often be reasonably approximated as a rectangle with the characteristic width and depth. This absorbs velocity, width, and depth. We define stream power per unit channel length, so that term goes to 1, and the derivation is complete.

$\Omega = \rho g Q \cancelto{1}{L} S$

Unit stream power is stream power per unit channel width, and is given by the equation:

$\omega=\frac{\rho g Q S}{b}$

where ω is the unit stream power, and b is the width of the channel.

Stream power is used extensively in models of landscape evolution and river incision. Unit stream power is often used for this, because simple models use and evolve a 1-dimensional downstream profile of the river channel. It is also used with relation to river channel migration, and in some cases is applied to sediment transport.^[1]

References

1. Bagnold, R. A. (1966). An approach to the sediment transport problem from general physics (Geological Survey professional paper). US Geological Survey, U. S. Govt. Print. Off.