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Fluvial geomorphologists describe stream power, $\Omega$, as the 'rate of energy dissipation against the bed and banks of a river per unit downstream length'.

It is expressed as a function of water density $\rho$, channel slope $S$, discharge $Q$, and the gravitational constant $g$:

$$\Omega = \rho g Q S$$

I'm aware geomorphologists are usually interested in a local measure of power for predicting things like sediment transport. I'd like to estimate the total power dissipated by a river network above a certain elevation (i.e., for a watershed, above a gauging station, etc...)

[Note: It's for fun. I'm curious about the energy dissipated for an entire mountain range.]


Here's what I've thought about so far.

Given the following:

With that I can imagine writing

$$ \begin{align} \int \Omega \ dx & = \rho g \int Q \ \frac{d h}{d x} \ dx \\ & = \alpha \rho g \int A^{\beta} \ dh \end{align} $$

Computing the change in drainage area with height should be fairly easy with GIS, but I'm left with the messy empirically-determined constants, $\alpha$ and $\beta$. Overall, I expect I'm missing a simpler principle in my approach.

Is there a simpler or more elegant way to estimate the total power dissipated by a river?

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  • $\begingroup$ By "power relationship" do you just mean that it's a power law. If so, isn't the implication that $\alpha$ and $\beta$ should be functions of $h$ or something? $\endgroup$ – Mike Feb 20 '18 at 1:59
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Looking at the definition you give for "stream power", it is evident that the assumption that allows you to derive the expression for $\Omega$ is that the kinetic energy does not change significantly — or maybe even more simply that you can just ignore the kinetic energy entirely.

On that basis, and using that assumption, I would say that fundamentally the way to compute the power dissipated is to simply use energy conservation. Compute the total potential energy of the water when it entered the watershed (fell from the sky or melted out of a glacier, I guess), subtract the potential energy that that water has at the bottom (and maybe the kinetic energy it has at the bottom), and the difference gives you the amount of energy that must have been dissipated in between. Of course, to get power, you just replace the mass used to compute the potential energy with the rate at which mass enters the stream or passes the lower station.

So if $h$ is the height at which a certain quantity of water enters the watershed, $p$ is the precipitation rate at that point, and $h_0$ is the height of the bottom reference point, the total power looks like \begin{equation} P = \int \rho\, g\, (h-h_0)\, p\, dA = \rho\, g\, \int (h-h_0)\, p\, dA, \end{equation} where $dA$ is the infinitesimal area element of that point.

I want to emphasize that this equation is sort of the fundamental calculation in this scenario; I don't think you can get around the problem without either knowing or estimating the things in this expression. So that point is that you fundamentally need some way of knowing the precipitation rate $p$ in any given place (along with the elevation of that place).

Now, comparing to your expression, it looks like you're trying to just parametrize this as $\alpha A^\beta$. I don't know if that's the easiest way, if $\alpha$ and $\beta$ are well known among fluvial morphologists, for example. I would think that if this really is just for fun, you might be able to make good progress by simply assuming some dependence of $p$ on $h$, and knowing that $\int p\, dA$ gives you the total rate of water flow out of the watershed (which is presumably relatively well known), you could normalize $p$ by doing this integral with your GIS magic. Then the integral for $P$ should be comparably easy.

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  • $\begingroup$ Thanks Mike, makes sense! As I understand it you're taking the surface integral of precipitation rate at each point (x, y) over the entire drainage to calculate the overall gravitational potential difference. $p$ will be difficult to estimate, I think, but there might be enough information on snowpack/rainfall to make enjoyable comparisons of different mountain ranges. My only follow-up question: does it matter what units $p$ has? (i.e., volume versus mass) $\endgroup$ – D. Betchkal Feb 20 '18 at 3:27
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    $\begingroup$ @D.Betchkal The dimensions of $p$ matter, and they should be length over time — it can be measured in inches per hour, feet per year, etc. This way, the integral $\int p\, dA$ will give you a volume per unit time and the integral for $P$ will have dimensions of power. But be mindful of what comes out at the end. You might get power in units of pound-inches per year, or something. But a straightforward conversion at the end will get you whatever units you want. $\endgroup$ – Mike Feb 20 '18 at 15:02
  • $\begingroup$ Also be careful that you use consistent units. For example, $\rho$ might have units of kilograms per cubic meter. But if $h$ and $dA$ are measured in feet and square-feet, they won't cancel the cubic meter correctly. Again, a simple conversion factor will set you right. $\endgroup$ – Mike Feb 20 '18 at 15:04

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