integration to get the total volumetric transport in Sv

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I'm a little lost on how to integrate the Sverdrup transport to get the total volumetric transport. the answer should be
The zonal wind stress contribution to the total Sverdrup transport: -22.5 Sv (negative sign means southward)
meridional wind stress contribution to the total Sverdrup transport: -3.4 Sv
Both zonal and meridional wind stress contribution: -25.9 Sv.
% sverdrup.m
% Reads and plots wind stress data in the subtropical North Atlantic Ocean.
% The wind stress data are defined at 2-degree intervals in longitude and
% latitude. Originally coded by Raymond G. Najjar 2004
% Modified by PRB in 2005, 2007
clear
% Read in monthly wind stress (dyne cm^2) fields at 25N and 27N in the Atlantic
load taux25n.txt
load tauy25n.txt
load taux27n.txt
load tauy27n.txt
% Compute annual averages and convert to N m^-2. (dyne is a cgs unit for force = g*cm s^-2)
tx25 = mean(taux25n)*0.1;
ty25 = mean(tauy25n)*0.1;
tx27 = mean(taux27n)*0.1;
ty27 = mean(tauy27n)*0.1;
% The wind stress data extend from 81W to 15W; that's from the east coast of Florida
% to the northwest coast of Africa.
% Plot the stresses as a function of distance in km from Florida.
imax = size(tx25,2);
i = 1:imax;
xkm = (i-1)*222*cos(26.*pi/180);
yo = i*0;
% Part I: The integrated Ekman and Sverdrup transports
% Computing Ekman Transport
omega = (7.29).*10.^-5;
f = 2.*omega.*sin((26./180).*3.14);
tx26 = (tx25+tx27);
ty26 = (ty25+ty27);
Mx26 = (ty26./f);
My26 = (-tx26./f);
% Computing wind stress curl and Sverdrup Transport
dx = 111.*cos((26./180).*3.14);
dy = 111.*1000;
a = 6.4.*10.^6;
B = (1./a).*2.*omega.*cos(26);
for i =1:imax-1
curl(i+1) = (1./B).*((ty26(i)-ty26(i))/(2.*dx) - (tx27(i)-tx25(i))/(2.*dy));
end
figure(1)
subplot (2,1,1)
plot(xkm,tx25,'-b',xkm,ty25,'-r',xkm,tx27,'-.b',xkm,ty27,'-.r',xkm,yo,'k')
legend('\tau_x at 25\circN','\tau_y at 25\circN','\tau_x at 27\circN','\tau_y at 27\circN')
title('Annual Mean Wind Stress at 25\circN and 27\circN in the Atlantic')
xlabel('distance east of Florida (km)')
ylabel('stress (N m^-^2)')
subplot (2,1,2)
plot(xkm,My26,'-.k',xkm,curl,'k')
title('Ekman and Sverdrup mass transports at 26°N');
xlabel('distance east of Florida (km)');
ylabel('transport (kg/ms)');
  1 个评论
darova
darova 2021-3-24
There are too much explanations. Can you please simplify your question? DO you have any picture or something?

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回答(1 个)

Abhimenyu
Abhimenyu 2024-11-8,6:54
Hi Hannah,
I understand that you want to integrate the Sverdrup transport to get the total volumetric transport. This can be done by first finding the the gradient of the zonal wind stress in the north-south direction (dtx_dy) and the gradient of the meridional wind stress in the east-west direction(dty_dx). Then they are updated in a loop according to the Sverdrup balance equations.
Please find the example MATLAB code below:
dx_m = 222000; % Convert dx to meters
rho = 1025; % Density of seawater in kg/m^3
% Initialize transport variables
zonal_transport = 0;
merid_transport = 0;
The negative sign in the zonal transport formula indicates the direction of the transport, while the positive sign in the meridional transport formula indicates the contribution to the total transport.
% Calculate transports
for i = 1:imax-1
% Zonal wind stress contribution
dtx_dy = (tx27(i) - tx25(i))/(2*dy);
zonal_transport = zonal_transport - (1/(B*rho))*dtx_dy*dx_m;
% Meridional wind stress contribution
dty_dx = (ty26(i+1) - ty26(i))/dx_m;
merid_transport = merid_transport + (1/(B*rho))*dty_dx*dx_m;
end
% Convert to Sverdrups (1 Sv = 10^6 m^3/s)
zonal_transport_sv = zonal_transport/1e6;
merid_transport_sv = merid_transport/1e6;
total_transport_sv = zonal_transport_sv + merid_transport_sv;
This should give the desired wind stress contributions.
I hope this helps!

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