Dear Sir,
In order to determine the total amount of a substance in a cell/node I would like to ask what volume is atributed during the calculation to a cell/node?
I use HP1 for simulating the transport and decay in a clay column. I want the contaminant to be ‘injected’ for in fixed period of time into the lowest cell/node of the system and I use a trick in this purpose: I consider a reaction in which I define a negative stoechiometric coefficient for my contaminant and it just ‘appears’ in my system from nothing.
I use the following Basic statement:
Source
-start
10 REM parm(1)=the rate at which the contaminant is injected into the system
20 REM parm(2)=time interval of injection
40 rate = parm(1)
50 IF (TOTAL_TIME > parm(2)) then rate = 0
60 moles = - rate * TIME
70 save moles
-end
A:
KINETICS 50-51 @Layer 1@
Source
-formula Cont -1
-m0 0.0
-parms parm(1) parm(2)
B:
KINETICS 51 @Layer 1@
Source
-formula Cont -1
-m0 0.0
-parms parm(1) parm(2)
I have saturated conditions, no vertical flow, clay, 50 cells of 1 meter each. The source is situated in the lowest cell. In general it works, I have tested, but it is not quite clear what volumes are attributed to one cell (A) and to one node (B)? While specifying one or more cells (KINETICS 50-51 for example), this statement places the total specified amount in every liter clay. To determine the total amount injected into the system I have to know the volume attributed to the cel? Is it precisely the volume between the two specified nodes (in this case 1m3) or is it (1m3+the volume atributed to the upper node, in this case node 50)
On the othet hand, while specifying one node (KINETICS 51) it places 1/125,4 of the total amount in each liter. What volume is used in this case?
Hope You can help me whis this.
Sincerely,
Ecaterina
Volume atribited to a cell and to a node
If you want to know the amount of mass added to the system, you ask for the kinetic phase in the output (kin("Source")). This change in the amount of kin("Source") will give you the amount of Source added to the solution. However, for the lowest node, the actual amount will be lower because part of the node is outside the domain (see e.g. equation 8.52 in the HYDRUS manual).