PC-Progress Discussion Forums

Hello
<u><i>my second question is: </i></u>
When I put pulse input as a boundary condition, I require also adjusting the application time of the pulse input,
Because if i don’t do that, the model underestimate the BTC observed.
I send to u my project folder


My experiment is as follow:

I applied an average of 0.61mm of water of 73g/l of bromide (45g/m2); the period of the experiment was 57days; the average transport velocity was about 0.5cm/day
My question is:

Pcp_File_Version=1
*** BLOCK A: Model Description ***************************************
Welcome to CXTFIT
J.Simunek
LUnit TUnit MUnit (indicated units are obligatory for all input data)
cm
days
g
Inverse Mode NRedu
1 1 1
MODC ZL
3
*** BLOCK B: Parameters for Inverse problem **************************
MIT ILMT MASS
100 0 1
*** BLOCK C: Parameter Names *****************************************
v
D
R
Mu
Parameters ...
0.49 10 1 0
1 1 0 0
*** BLOCK D: Boundary Value Problem **********************************
MODB - Boundary Value Mode
3
73 0.1
0 1
*** BLOCK E: Initial Value Problem **********************************
MODI - Initial Value Mode
0
*** BLOCK F: Production Value Problem **********************************
MODP - Production Value Mode
0
*** BLOCK G: Observed Data for Inverse Problem *************************
INPUTM - Input data code
2 15
57
Z(I) C(I)
2.5 0.0493005
7.5 0.116744
12.5 0.165457
17.5 0.21076
22.5 0.284233
27.5 0.290358
32.5 0.155127
37.5 0.0966828
45 0.0491646
55 0.0363105
65 0.0253317
75 0.0119924
85 0
95 0.00907649
105 0
*** BLOCK H: Position and time for direct problem ********************
NZ DZ ZI NT DT TI OutputCode
120 1 0 1 0 57 2
*** End of the Input File ********************************************


Thanks a lot

Meha:

As stated in the previous topic with Tina, it is necessary to have proper mass recovery in the concentration measurements (BTC or rofile) for a proper parameter estimation. In your case, total amount of solutes observed at 57d is 8.16 mg/cm2 based on the area of the solute profile.

---------------

z conc g/l area
0 0
2.5 0.0493005 0.061625625
7.5 0.116744 0.41511125
12.5 0.165457 0.7055025
17.5 0.21076 0.9405425
22.5 0.284233 1.2374825
27.5 0.290358 1.4364775
32.5 0.155127 1.1137125
37.5 0.0966828 0.6295245
45 0.0491646 0.54692775
55 0.0363105 0.4273755
65 0.0253317 0.308211
75 0.0119924 0.1866205
85 0 0.059962
95 0.00907649 0.04538245
105 0 0.04538245

sum 8.159840525 (g/l)*cm
-> 8.16 mg/cm2
---------------

The input mass is 73 g/l*0.061cm=4.459mg/cm2 (<8.16). Since the observed mass is greater than the input mass, it would be difficult to have a proper estimation.

However, I tried to include the total mass in the estimation for a delta input and a pulse input.

For a delta input, the initial estimate for m_b/v was (4.459 mg/cm2) / (0.5 cm/d).

For a pulse input, the initial estimate for the duration time was 0.061cm/ 0.5cm/d = 0.122 d.

When I estimated D & v for the CDE, the results were almost identical in both cases and the adjusted mass in the soil profile was 7.38 mg/cm2.

Although I am very skeptical to the estimated parameter values, I confirm the program works well at least for the mass estimation.

If you want, I will send you the CXTFIT projects.

Nobuo

Dear

can u explain to me please me also how have u integrated the observed mass in the profile?
doesn't we need to multiply the concentration by the volumetric water content of each depth to integrate the observed mass in the profile (conc*height of each depth* Ovol)

depth(cm)g/l g/dm2
2,5 0,0493005 0,0098601[=g/l*0.5(dm)*0.37(Ovol)]
7,5 0,11674406 0,02334881
12,5 0,16545653 0,03309131
17,5 0,21075985 0,04215197
22,5 0,28423297 0,05684659
27,5 0,29035815 0,05807163
32,5 0,15512673 0,03102535
37,5 0,09668282 0,01933656
45 0,04916457 0,01966583 [=g/l*1(dm)*0.37(Ovol)]
55 0,03631055 0,01452422
65 0,02533172 0,01013269
75 0,01199244 0,00479698
85 0 0
95 0,00907649 0,0036306
105 0 0

sum of observed profile
(0,32648263g/dm2)




Thanks again
maha

excuse me sir
i have used an average volumetric water content 0.4 instead of 0.37
thanks again

Maha: I use the trapezoidal rule. For example. for 2.5<z<7.5,
(0.0493005+0.116744)*(7.5-2.5)/2=0.41511125
Note that since g/l = mg/cm3, the unit of the area is
mg/cm3*cm= mg/cm2.

z conc g/l area
2.5 0.0493005
7.5 0.116744 0.41511125

The input mass is c*water flux*t =c*v*theta*t. The mass in the soil profile is integral c*theta dz. When theta is constant, theta can be canceled out, i.e., c*v*t = Integral c dz. Nobuo