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I'm trying to calculate the power of the Mississippi, using the section of the river as it passes through New Orleans. I have the flow rate of a river, turned into a mass flow rate of $16,992,000 \:\rm kg/s$. I have the area of a cross section of the river estimated at $10684 \:\rm m^2$.

I know the units need to end up as $\rm kg \: {m^2}/{s^3}$, but I can't quite fathom from where, after multiplying these two numbers together, to pull a $\rm s^{-2}$ from to multiply this by.

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The mass flow rate $Q_m$ and the cross-sectional area $A$ are almost enough to give you this. The only other thing that you need is the density of the water $\rho$.

Since $m=\rho V$, we have that $Q_m=\rho A v$, where $v$ is the flow velocity. To see this, think about the amount of water that passes through a square hole over a time period of one second. The water that passes through forms a rectangular box, with area the same as the hole and length equal to the distance that the water travels in one second (which, of course, is related to the flow velocity). Taking the volume of this box and multiplying by the density gives you the amount of mass that passed through that hole in one second, which means you can directly relate the flow velocity to the mass flow rate as long as you have the density and area.

With that in mind, what we want is the power, which is the same as the total amount of kinetic energy passing through the river per second. Given that kinetic energy is $K=\frac{1}{2}mv^2$, the power (i.e. the kinetic energy flow rate) is given by $P=\frac{1}{2}Q_m v^2$. But we know $v=\frac{Q_m}{\rho A}$ from the last paragraph, so, we have:

$$P=\frac{1}{2}Q_m\left(\frac{Q_m}{\rho A}\right)^2=\frac{Q_m^3}{2\rho^2A^2}$$

This is the amount of power you would extract if you stopped the flowing water. If you merely slow it down, you must take the difference between $P$ before extraction and $P$ after extraction.

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    $\begingroup$ Great explanation- thanks so much. I knew I needed to get the flow velocity somehow but couldn't quite think of it. Thanks for the visualization. $\endgroup$ – ROBINSONG Dec 6 '19 at 22:59
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I believe that you need to be more specific by what you mean by "power of the river". If you mean "how much power could be gained from the river by putting the whole thing through an electrical turbine" then the important quantities are the mass per second, the altitude above sea level and the Earth's gravitational field constant (little $g$).(Which give the right units multiplied).

If you want to calculate the kinetic energy flowing past a given point on the bank per second you want to first use the cross sectional area (which you have) the flow per second (which you have) and the density of water (which is basically $1000\:\rm kg/m^3$) to find the velocity:

$$ v = \text{flow} / (\text{cross section}\times\text{density}) $$ (which has units [$\rm s^{-1} \: m^1$], as it should).

Then the kinetic energy is the velocity squared times the mass,

$$E = m v^2.$$

This measure of "power" is the amount of kinetic energy per second, so multiplying by the mass passing per second instead of just the mass in the above equation will make it a power.

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