3
$\begingroup$

I have temperature, u wind, v wind and relative humidity. I am wondering to calculate water vapor flux divergence and convergence.

Would anybody kindly help to find our a solution for this?

$\endgroup$
6
$\begingroup$

The formula for moisture flux is $$q\vec{V} $$ where $q$ is the water vapor mixing ratio, which can be found using mixhum_ptr and $\vec{V}$ is the velocity.

Therefore the divergence of the moisture flux must be $$\nabla{\dot{}q\vec{V}}=\frac{\partial(qu)}{\partial x}+\frac{\partial(qv)}{\partial y}+\frac{\partial(qw)}{\partial z} $$ $$\approx\frac{\delta(qu)}{\delta x}+\frac{\delta(qv)}{\delta y}+\frac{\delta(qw)}{\delta z}$$

which is computable if you have gridded data.

For example, you could compute this by setting $$qu=q\ast u$$ $$qv=q\ast v$$ $$qfluxDiv=uv2dv\_cfd(qu,qv,lat,lon,opt) $$ Use uv2dv_cfd per http://www.ncl.ucar.edu/Document/Functions/Built-in/uv2dv_cfd.shtml

Note: If you wanted the turbulent moisture flux, you would need to subtract $\nabla{\dot{}\bar{\vec{V}}\bar{q}}$ from your answer.

$\endgroup$
  • 1
    $\begingroup$ Please suggest in NCL or grads. Thank you $\endgroup$ – Kay Jul 22 '16 at 1:30
  • 1
    $\begingroup$ For horizontal water vapor flux, divergence try this NCL code: qfluxDiv=uv2dv_cfd(qu,qv,lat,lon,opt) $\endgroup$ – BarocliniCplusplus Jul 22 '16 at 3:55
  • 1
    $\begingroup$ I added that to the answer, but without data and code, it is hard to give an example. I provided a link to the NCL function, so that you may know how to use the function. There is an example there on how to use the function. $\endgroup$ – BarocliniCplusplus Jul 22 '16 at 13:43
  • 1
    $\begingroup$ @BarocliniCplusplus - can you add in your answer what q is ? $\endgroup$ – gansub Oct 31 '16 at 9:32
  • 2
    $\begingroup$ @BarocliniCplusplus - Thank you. The only other suggestion is that his input is relative humidity. You may want to link how to convert water vapor mixing ratio to relative humidity. So this becomes a complete answer. $\endgroup$ – gansub Oct 31 '16 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.