Re: [ferret_users] query regarding the calculation of zonally averaged ocean heat content in joules rather than joules/m^2

Thank you Ryo,

thats works fine for me. it is zonally integrated heat content.

Apart from this i am have one more small doubt (just for the sake of confirmation) that while making zonal/meridional/box averages of ocean data lets say SST for the region x=20E:140E and y=90S:90N so does the ferret considering land region also for averaging or it is just averaging the ocean region in that box area. Because if ferret is making the averages considering the land part then the zonal/meridional/box averages of ocean data is wrongly calculated and interpreted.

i hope the land values are automatically taken care by the ferret while making zonal/meridional/box averages. is it so?

regards,saurabh

On Wed, Nov 23, 2016 at 4:18 PM, Ryo Furue <furue@xxxxxxxxxx> wrote:

Hi Saurabh,

Before talking about Ferret, "zonally averaged heat content
in joules" doesn't make sense. I suppose you mean either

zonally integrated heat content in joules per meter

or

zonally averaged heat content in joules per meter squared,

because you get the total amount of heat only after integrating
the former in the meridional direction or after multiplying
the latter by the zonal width and integrating it the meridional direction.

Without looking at your script, the zonally integrated heat content is

heat_per_unit_volume[z=<depthrange>@DIN, x=<xrange>@DIN]

if heat_per_unit_volume has the right units. If it's joules per meter cubed
and x and y are longitudes and latitudes, you should get the right result.

Ryo

i want to calculate the total energy from the ocean heat content analysis. suppose i want to calculate for indian ocean (20E:120E, 30N:65S) in the following way

1) for heat content calculation i am having temperature and density data so i calculated the heat content from 0m-700m

use ptemp_new.nc
use pden_new.nc

sh da

let temp1=ptemp[d=1,y=63N:64S]
let dens1=pden[d=2,y=63N:64S]

let cp=3992
let heat=dens1*cp*temp1
let ohcg=heat[z=0:700@din]

repeat/l=1:142:1 (save/append/file=ohc_globe.nc ohcg;set mem/size=700)

2) i use this heat content file for the total energy calculation

use ohc_globe.nc

let ohc2=ohcg/10^22 !!!!!!!! for scaling !!!

use climatological_axes
cancel data climatological_axes
let ohc_cli=ohc2[gt=month_`ohc2,return=cal`@mod] !<<<<<<<<<<<forurisional file<<<<<<<<<<<<<<<<<<<<!
let ohc_anm1=ohc2-ohc_cli[gt=ohc2@asn]

define axis/edges/t/units=months/t0=15-dec-2004/t=15-jan-2005:15-nov-2016:1/calendar=360_days tmonthly
let ohc_new=ohc_anm1[gt=tmonthly@asn]

define axis/edges/t/units=years/t=15-jan-2005:15-oct-2016:1/calendar=360_days t_yearly
let ohc_year=ohc_new[gt=t_yearly@ave]

let ohc_diff=ohc_year[l=12]-ohc_year[l=1]

let ohc_diff_zonal_indi=(ohc_year[l=12,x=20E:120E@ave]-ohc_year[l=1,x=20E:120E@ave])*(120-20)*(63+64)*1000000*110*110

!!!!!!!!!! here total longitudinal distance is (120-20)*110*1000 and similarly for latitudinal distance (63+64)*110*10000 !!!!!!!!!!!!!!!!!!!

is this the correct way to calculate the zonally averaged total heat content in joules as we already averaged in longitudes. But if i calculate the ocean heat content in joules/m^2 then results vary significantly i am attaching both the figures.

please note :- both the figures are having different units joules and joules/m^2

--

REGARDS

Saurabh Rathore
Research Scholar (PhD.)
Centre For Oceans, Rivers, Atmosphere & Land Science Technology
Indian Institute Of Technology, Kharagpur
contact :- 91- 8345984434

REGARDS

Saurabh Rathore

Research Scholar (PhD.)

Centre For Oceans, Rivers, Atmosphere & Land Science Technology

Indian Institute Of Technology, Kharagpur

contact :- 91- 8345984434