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How to model a storage
#1
Hi,

I am trying to model a storage device modifying the DEMO 4 provided by the package of Veda installation. I attached the file from DEMO 4 with the modifications. In this file, I hid some sheets that I was not working with. 

I created the mining process "MINMIC1" in the sheet "PRI_COA" for a commodity "MIC".  The mining process is able to produce 2000 PJ during his whole life. Per year, ACT_BND  =1500 PJ and for the year 2006, ACT_BND is 0. 

In the sheet "DemTechs_TPS", there is the demand process "DDEMMIC1" that needs the commodity "MIC" as input, with EFF = 1. This process has as output the commodity "DEMMIC1". The final demand for "DEMMIC1" is 1500 PJ in year 2005 and 500 PJ in year 2006. (sheet "Demands")

I would like to produce 2000 PJ of "MIC" in the mining process in the year 2005, supply 1500 PJ in year 2005, and store 500 J for the year 2006. 

In the results, I have:

Production of 1500 PJ by the mining process in the year 2005 and Imports of 500 PJ by a dummy process in year 2006. 

I have tried many parameters from TIMES documentation - Part II , section 4.3 without success.   

How could I obtain the storage being used?

Thank you,


Attached Files
.xlsx   VT_REG_PRI_V04.xlsx (Size: 172.27 KB / Downloads: 15)
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#2
Your test model had multiple serious issues.

First, your storage process (there was a single one defined, "STORAGE") is an inter-period storage. The documentation clearly states that the type of inter-period storage is STK, and for STK the commodity entering and leaving the storage (the charged and discharged commodity) should be a member of the PCG (and both should be, if they are different).

Problems I corrected in the attached:
 1.  Type of inter-period storage STORAGE  must be STK
 2.  Primary group must contain both input and output, so changed to NRG and added dummy NRG output
 3.  Added the dummy output flow of type NRG, and  a FLO_FUNC for defining the auxiliary DEMMIC1 output
 4.  VEDA by default prohibits new investments into B-Y technologies with RESID, so removed RESID=0
 5.  AFC only bounded the output flow, so added AFC(ACT)=1000 for defining max. storage level
 6.  Model was able to by-pass the demand technology DDEMMIC1 via storage without any INVCOST, so added equivalent INVCOST (321 * 500, because for storage one capacity unit was defined to deliver at most 500 PJ)

Normally, inter-period storage processes would have the same commodity in and out, and therefore normally there is no need for adding a dummy flow like I did, to circumvent the problem of different commodity types. One could also create an UD-CG in SysSettings for defining a PCG consisting of different types of commodities, but for "simplicity" I resorted to the dummy flow approach instead...


Attached Files
.xlsx   VT_REG_PRI_V04.xlsx (Size: 172.03 KB / Downloads: 38)
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#3
Dear Moderator,

Thank you a lot for the help. It really clarifies how to fill the excel file.

I forget to say there are maybe 15~20 days that I playing with TIMES and VEDA. Hence, considering the answers, I would like to make sure I understood all of it correctly.

Considering only the sheet "DemTechs_TPS":

"2. Primary group must contain both input and output, so changed to NRG and added dummy NRG output"
When you say "primary group", does it means "Primary Commodity Group" (PCG) - column Y in the excel sheet?
For every storage device, both input and output must belong to the same PCG and be explicitly defined in the PCG column. Is it correct?

"3. Added the dummy output flow of type NRG, and a FLO_FUNC for defining the auxiliary DEMMIC1 output"
If I understood well, there is two outputs now. How do I know which one is the auxiliary?

What is the difference between "Comm-OUT" and "CommGrp"?

About FLO_FUNC, accordingly to the documentation: "Establishes the relationship between storage charging / discharging and a related commodity flow (VAR_FLO) in the auxiliary storage flow equation (EQ_STGAUX)."
I am confused with the STG_EFF.
Let suppose that, for 1 PJ of input in the storage, the output of DEMMIC1 is 0.9 PJ. In this case, STG_EFF = 0.9 and FLO_FUNC = 1?

Is FLO_FUNC like an efficiency for the auxiliary storage process? Let suppose that are 3 commodities as output of the storage. For 2 of them, should I define a FLO_FUNC and the main commodity output is regulated via STG_EFF?

Instead of creating line 18 to set FLO_FUNC, could cell P15 be set to 1?

"5. AFC only bounded the output flow, so added AFC(ACT)=1000 for defining max. storage level"
I got confused here. ACT is written in the column of outputs, but it seems to be an attribute. I do not have realized, in the documentation-Parte IV, the syntaxe that allows write attributes in lines instead of write it in columns. Could you point me some documentation about it?
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#4
Please find some quick answers from me below:

> When you say "primary group", does it means "Primary Commodity Group" (PCG)

Yes.  Both PG and PCG are used to refer to the same thing, primary commodity group, interchangeably.

> For every storage device, both input and output must belong to the same PCG.

As I said, for the charging/discharging flows of inter-period storage and regular timeslice storage that is indeed the case. But never for auxiliary input / output flows.

> …. and be explicitly defined in the PCG column. Is it correct?

No.  If you choose to rely on the defaults, you don't need to define the PG explicitly.  VEDA always defines the PG anyway even if not defined. If you don't know how, you should maybe ask the VEDA developers, or you can also explicitly define the PG if either unsure or you know that the default will be wrong. And some additional default QA adjustments are made by TIMES. But in your case, there was no way to make it right by default.

>  If I understood well, there is two outputs now. How do I know which one is the auxiliary?

I defined the PCG as NRG.  Only one of the outputs was NRG.  So, surely you should know which one is an auxiliary, no?

>  Let suppose that, for 1 PJ of input in the storage, the output of DEMMIC1 is 0.9 PJ. In this case, STG_EFF = 0.9 and FLO_FUNC = 1?

You defined STG_EFF=1, so it had no effect originally. With this dummy approach, it will never have any efficiency effect (it would only have a capacity effect if AFC is defined on the discharge flow).  Instead, if you would want a 90% efficiency, you should adjust FLO_FUNC accordingly when using this dummy approach. But as I said already, 1) it is rather unusual to model inter-period storage with different IN/OUT commodities (and even more so with different commodity types), and 2) you don't need to use the dummy approach even if you do have such.

> Is FLO_FUNC like an efficiency for the auxiliary storage process?

Yes, in this specific case it is, as only the auxiliary commodity actually provides the useful output. But as said, it is just one way of modelling it.  If you don’t like about any such dummy flow approaches, you can always just define the PCG as {MIC,DEMMIC1}.

> Instead of creating line 18 to set FLO_FUNC, could cell P15 be set to 1?

No, of course not.  FLO_FUNC expects two commodity groups, and under VEDA the first one must be specified under Other_Indexes / CommGrp.  Please refer to the VEDA Attributes documentation!

> ACT is written in the column of outputs, but it seems to be an attribute.

No, ACT is not written in the column of outputs.  It is in the CommGrp column.  It is a reserved name for the pre-defined activity-defining group.
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#5
Thank you for the help. I really appreciate it. 

>  If I understood well, there is two outputs now. How do I know which one is the auxiliary?

>>I defined the PCG as NRG.  Only one of the outputs was NRG.  So, surely you should know which one is an auxiliary, no?

Accordingly with Documentation Part II :"The main flows of timeslice and inter-period storage processes are the flows of the charged and discharged commodities included in the set primary commodity group PCG of the process." 

What happens if all the commodities are in NRG? Ex: Let suppose I have a storage system with 2 commodities as input and also as output: ELEC1 and DEMMIC1. Both belongs to NRG now.
In this case, should the dummy be in another "Commodity Set Membership", for  exemple MAT? Must the PCG in cell Y17 be MAT and  cell E18 be "DEMMIC1, ELEC1" ?

> Instead of creating line 18 to set FLO_FUNC, could cell P15 be set to 1?

>>No, of course not.  FLO_FUNC expects two commodity groups, and under VEDA the first one must be specified under Other_Indexes / CommGrp.  Please refer to the VEDA Attributes documentation!

I have looked for "FLO_FUNC" with control+F in adobe through the "Documentation for the TIMES Model PART IV: VEDA2.0". There is no results about it.
Anyway, in the excel file, DEMMIC1 is in Comm-OUT (cell D16), DEMMIC1 is in CommGrp (cell E18), DUMNRG is in Comm-OUT (cell D18), DUMNRG is in CommGrp (cell E16). Hence, both commodities are in the Comm-OUT and in CommGrp. Could you be more clear about "the first one must be specified under Other_Indexes / CommGrp"?

> ACT is written in the column of outputs, but it seems to be an attribute.

>>No, ACT is not written in the column of outputs.  It is in the CommGrp column.  It is a reserved name for the pre-defined activity-defining group.

ACT is written in cell E17. Accordingly to cell E12, this is a "Output Commodity". Could you make it clear?  About the syntaxe: if one uses a reserved name in lines, is this used to set constraints?
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#6
Ok, some more answers below:

> What happens if all the commodities are in NRG?

If you define the PCG to include all NRG comodities in the process topology, that's what you will get.  If you want to exclude some, please just define the PCG accordingly. 

> Ex: Let suppose I have a storage system with 2 commodities as input and also as output: ELEC1 and DEMMIC1. Both belongs to NRG now. In this case, should the dummy be in another "Commodity Set Membership", for  exemple MAT?

I am not sure I follow, but if you define DEMMIC1 to be of type NRG, you don't need any dummy flow or auxiliary flow.  Why would you then define another output ELEC1?

> I have looked for "FLO_FUNC" with control+F in adobe through the "Documentation for the TIMES Model PART IV: VEDA2.0". There is no results about it.

I suggested that you should look at the "on-line" VEDA Attributes documentation when in any doubt about using attributes under VEDA: Information → TIMES Attributes.  It makes it clear that two commodity groups must be specified. If you would put 1 into cell P15 as you suggested, only one CG would be specified.  However, I admit now that the VEDA summary does not make it clear which one of the two CGs should go into CommGrp (Other_Indexes), but you can at least check from the DD files where each one goes... (I guess that's how I learned to know it myself).

Please also read section 4.3 Storage processes in the documentation. Especially sections 4.3.5 Auxiliary storage flows and 4.3.7 Availability factors for storage processes.

> Could you be more clear about "the first one must be specified under Other_Indexes / CommGrp"?

From the documentation, you can see that the FLO_FUNC for the STORAGE process (as it was modified by me) must be defined as FLO_FUNC(r,y,'STORAGE','DEMMIC1','DUMNRG','ANNUAL') = Value.  And so it should be obvious that DEMMIC1 must go under CommGrp and DUMNRG must be in a commodity column.

> ACT is written in cell E17. Accordingly to cell E12, this is a "Output Commodity". Could you make it clear?

The comment in Cell E12 ("Output Commodity") was written by you, not by me.  Sorry that I inadvertently copied your comment as well when making a new column for CommGrp and copying the column formats...  I am very sorry!  But note that comments like that have no impact on the resulting model.  The Comm-IN and Comm-OUT columns define process topology and they must hold commodities only; the CommGrp column does not define any topology entries and it can hold any commodity group, whenever needed for a TIMES attribute to be defined.

> About the syntaxe: if one uses a reserved name in lines, is this used to set constraints?

As a reserved commodity group label, 'ACT' cannot be defined by the user as a commodity in the RES topology.  There are basically no other reserved commodity group / commodity names other than ACT and the commodity types. 'ACT' can be used as a CG in the following TIMES attributes: ACT_EFF, FLO_EMIS, FLO_EFF, FLO_SHAR, FLO_CUM, IRE_FLOSUM, NCAP_AFC, PRC_ACTFLO, STG_SIFT (and FLO_FUNCX). For example ACT_EFF(r,y,p,'ACT',ts) defines a process efficiency from the shadow primary group (SPG) to the activity. And FLO_EMIS(r,y,p,'ACT',com,ts) defines an emission factor for emissions com in proportion to the process activity.
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#7
(19-06-2021, 08:46 PM)Antti-L Wrote: Ok, some more answers below:

> What happens if all the commodities are in NRG?

If you define the PCG to include all NRG comodities in the process topology, that's what you will get.  If you want to exclude some, please just define the PCG accordingly. 

> Ex: Let suppose I have a storage system with 2 commodities as input and also as output: ELEC1 and DEMMIC1. Both belongs to NRG now. In this case, should the dummy be in another "Commodity Set Membership", for  exemple MAT?

I am not sure I follow, but if you define DEMMIC1 to be of type NRG, you don't need any dummy flow or auxiliary flow.  Why would you then define another output ELEC1?

> I have looked for "FLO_FUNC" with control+F in adobe through the "Documentation for the TIMES Model PART IV: VEDA2.0". There is no results about it.

I suggested that you should look at the "on-line" VEDA Attributes documentation when in any doubt about using attributes under VEDA: Information → TIMES Attributes.  It makes it clear that two commodity groups must be specified. If you would put 1 into cell P15 as you suggested, only one CG would be specified.  However, I admit now that the VEDA summary does not make it clear which one of the two CGs should go into CommGrp (Other_Indexes), but you can at least check from the DD files where each one goes... (I guess that's how I learned to know it myself).

Please also read section 4.3 Storage processes in the documentation. Especially sections 4.3.5 Auxiliary storage flows and 4.3.7 Availability factors for storage processes.

> Could you be more clear about "the first one must be specified under Other_Indexes / CommGrp"?

From the documentation, you can see that the FLO_FUNC for the STORAGE process (as it was modified by me) must be defined as FLO_FUNC(r,y,'STORAGE','DEMMIC1','DUMNRG','ANNUAL') = Value.  And so it should be obvious that DEMMIC1 must go under CommGrp and DUMNRG must be in a commodity column.

> ACT is written in cell E17. Accordingly to cell E12, this is a "Output Commodity". Could you make it clear?

The comment in Cell E12 ("Output Commodity") was written by you, not by me.  Sorry that I inadvertently copied your comment as well when making a new column for CommGrp and copying the column formats...  I am very sorry!  But note that comments like that have no impact on the resulting model.  The Comm-IN and Comm-OUT columns define process topology and they must hold commodities only; the CommGrp column does not define any topology entries and it can hold any commodity group, whenever needed for a TIMES attribute to be defined.

> About the syntaxe: if one uses a reserved name in lines, is this used to set constraints?

As a reserved commodity group label, 'ACT' cannot be defined by the user as a commodity in the RES topology.  There are basically no other reserved commodity group / commodity names other than ACT and the commodity types. 'ACT' can be used as a CG in the following TIMES attributes: ACT_EFF, FLO_EMIS, FLO_EFF, FLO_SHAR, FLO_CUM, IRE_FLOSUM, NCAP_AFC, PRC_ACTFLO, STG_SIFT (and FLO_FUNCX). For example ACT_EFF(r,y,p,'ACT',ts) defines a process efficiency from the shadow primary group (SPG) to the activity. And FLO_EMIS(r,y,p,'ACT',com,ts) defines an emission factor for emissions com in proportion to the process activity.



Thank you for the feedback.

Some of the points are hard for me yet.

> What happens if all the commodities are in NRG?

>>If you define the PCG to include all NRG commodities in the process topology, that's what you will get.  If you want to exclude some, please just define the PCG accordingly.

> Ex: Let suppose I have a storage system with 2 commodities as input and also as output: ELEC1 and DEMMIC1. Both belongs to NRG now. In this case, should the dummy be in another "Commodity Set Membership", for  exemple MAT?

>>I am not sure I follow, but if you define DEMMIC1 to be of type NRG, you don't need any dummy flow or auxiliary flow.  Why would you then define another output ELEC1?

I'm trying to get the syntaxe of this excel file. 
Let try with 2 other examples. I would like to represent a compressed air energy storage system (CAEX).

Example 1: There are one converter (CONV1) before the storage device (CAEXSTORAGE) and one converter after the storage device (CONV2). Focusing on the flow of commodities, CONV1 needs, as input, 5kg of air (AIR) and 1kWh of electricity (ELEC1) for each kg of air stored (AIRSTOCK).  This AIRSTOCK enters in the storage as input and it is also the only output. I would like to suppose there are no losses in the storage (STG_EFF=1 , correct?) The output of  CAEXSTORAGE will be input for CONV2. Beyond AIRSTOCK, CONV2 needs COA as input. For each 3kg of AIRSTOCK and 10 GJ of COA, the output of CONV2 1 kWh of ELEC1.

I tried to implement this example in file "VT_REG_PRI_V04-corrected.xlsx". Mining process for AIR and ELEC1 are defined in sheet "PRI_COA". They can produce only in 2005. There are demand for ELEC1 in 2005 and 2006 (sheet "Demands" in the excel file), so for 2006 the storage must be used. Converters and the CAEXSTORAGE are defined at sheet "DemTechs_TPS" in the excel file.  I considered only one unit (PJ) to make easy. Results gave me the mining of 2 AIR and 1 ELEC1. CONV1 has no usage of ELEC1 (ELEC1 must be produced in such a way to have AIRSTOCK - so how could I correct it?). CONV2 is not taking COA as input, but it produces ELEC1 (Figure R1.png)

Example 2: Same example as before (actually, example 1 is better because is more near to reality). However, I would like to model only CAEXSTORAGE and use FLO_FUNC. I tried to implement this example in file "VT_REG_PRI_V04.xlsx".  Mining process for AIR and ELEC1 are defined in sheet "PRI_COA". They can produce only in 2005. There are demand for ELEC1 in 2005 and 2006 (sheet "Demands" in the excel file), so for 2006 the storage must be used. The CAEXSTORAGE are defined at sheet "DemTechs_TPS" in the excel file. Now, CONV1 and CONV2 are aggregated in CAEXSTORAGE.
Inputs, for CAEXSTORAGE, are AIR, COA and ELEC1. Output is only ELEC1. There are the need of 5kg of air (AIR), 10 GJ of COA and 1kWh of electricity (ELEC1) as input for each kWh of ELEC1 as output (Note that the aggregated input is same as in the example 1).  I considered only one unit (PJ) to make easy. In the results, the inputs of CAEXSTORAGE is 3 AIR, no COA and no ELEC1 in 2005. I was expecting 5 AIR, 10 COA and 1 ELEC1 as input in 2005 in such a way 1 ELEC1 can be released in 2006. (Figure R2.png) How could I correct it?

About FLO_FUNC(r,y,'STORAGE','DEMMIC1','DUMNRG','ANNUAL') = Value:  ---  Must the order of the columns in excel correspond to the order of the parameters in the function? Ex.: In the excel file, Comm-OUT appears before the CommGrp. So, line 19 (range B19 to P19)  in "VT_REG_PRI_V04-corrected.xlsx" is created to match the parameters order of the function FLO_FUNC. Is it right?


Attached Files Thumbnail(s)
       

.xlsx   VT_REG_PRI_V04-corrected.xlsx (Size: 176.87 KB / Downloads: 6)
.xlsx   VT_REG_PRI_V04.xlsx (Size: 175.17 KB / Downloads: 4)
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#8
Ok, seems like you need still some more help.  Shy

I am willing to help through these (tomorrow), but could you first clarify some details:

For example1 you say:
> CONV1 needs, as input, 5kg of air (AIR) and 1kWh of electricity (ELEC1) for each kg of air stored (AIRSTOCK).  This AIRSTOCK enters in the storage as input and it is also the only output. I would like to suppose there are no losses in the storage. The output of  CAEXSTORAGE will be input for CONV2. Beyond AIRSTOCK, CONV2 needs COA as input. For each 3kg of AIRSTOCK and 10 GJ of COA, the output of CONV2 1 kWh of ELEC1.

In my mind, those 3kg of AIRSTOCK needed per 1 kWh of ELEC1 produced (in CONV2) would seem to result in 3*5 = 15 kg of AIR input needed per 1 kWh of ELEC1 produced, because you say 5 kg of AIR is needed per 1 kg of AIRSTOCK stored.  Similarly, I think it would result in 3*1 kWh of ELEC1 consumed per 1 kWh of ELEC1 produced.(??)

However, for example2 you say: There are the need of 5kg of air (AIR), 10 GJ of COA and 1kWh of electricity (ELEC1) as input for each kWh of ELEC1 as output (Note that the aggregated input is same as in the example 1).

But for me it strikes that the aggregate balances look quite different between example 1 and 2.  Can you point out where my blind spot is:  Why are my calculations for Example1 invalid?
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#9
(21-06-2021, 01:03 AM)Antti-L Wrote: Ok, seems like you need still some more help.  Shy

I am willing to help through these (tomorrow), but could you first clarify some details:

For example1 you say:
> CONV1 needs, as input, 5kg of air (AIR) and 1kWh of electricity (ELEC1) for each kg of air stored (AIRSTOCK).  This AIRSTOCK enters in the storage as input and it is also the only output. I would like to suppose there are no losses in the storage. The output of  CAEXSTORAGE will be input for CONV2. Beyond AIRSTOCK, CONV2 needs COA as input. For each 3kg of AIRSTOCK and 10 GJ of COA, the output of CONV2 1 kWh of ELEC1.

In my mind, those 3kg of AIRSTOCK needed per 1 kWh of ELEC1 produced (in CONV2) would seem to result in 3*5 = 15 kg of AIR input needed per 1 kWh of ELEC1 produced, because you say 5 kg of AIR is needed per 1 kg of AIRSTOCK stored.  Similarly, I think it would result in 3*1 kWh of ELEC1 consumed per 1 kWh of ELEC1 produced.(??)

However, for example2 you say: There are the need of 5kg of air (AIR), 10 GJ of COA and 1kWh of electricity (ELEC1) as input for each kWh of ELEC1 as output (Note that the aggregated input is same as in the example 1).

But for me it strikes that the aggregate balances look quite different between example 1 and 2.  Can you point out where my blind spot is:  Why are my calculations for Example1 invalid?

Thank you for the replay. 

Actually, there is a mistake in my example, one number missing.  So sorry about it. Let me correct it: 

CONV1 needs, as input, 5kg of air (AIR) and 1kWh of electricity (ELEC1) for each kg of air stored (AIRSTOCK).  This AIRSTOCK enters in the storage as input and it is also the only output. I would like to suppose there are no losses in the storage. The output of  CAEXSTORAGE will be input for CONV2. Beyond AIRSTOCK, CONV2 needs COA as input. For each 3kg of AIRSTOCK and 10 GJ of COA, the output of CONV2 1 kWh of ELEC1.
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#10
Ok, thanks!

These are interesting examples, for which I hope my suggestions will help you further (tomorrow). But probably they have nothing to do with reality, as the total energy efficiency is so low?

The aggregate energy inputs thus appear to be 10 GJ + 1 kWh = 10.0036 GJ, and the energy output  is 1 kWh = 0.0036 GJ, and so the overall efficiency of both configurations appears to be only 0.036%, correct?  Shy
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#11
(21-06-2021, 02:37 AM)Antti-L Wrote: Ok, thanks!

These are interesting examples, for which I hope my suggestions will help you further (tomorrow). But probably they have nothing to do with reality, as the total energy efficiency is so low?

The aggregate energy inputs thus appear to be 10 GJ + 1 kWh = 10.0036 GJ, and the energy output  is 1 kWh = 0.0036 GJ, and so the overall efficiency of both configurations appears to be only 0.036%, correct?  Shy

Thank you. I will wait your answer then. About the numbers, they are just numbers and not reality. Let us say everything is in PJ. I think I can use the table ~UnitConversion in the "Defaults" sheet of file SysSettings to make those conversions (I still using the DEMO4 with modifications for these examples). Am I right?
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#12
Very well, please find attached my suggested modifications to make the two CAEX examples working.

I integrated both examples in the same model, and defined the demand as 3.6 PJ (so 1 TWh) for both DEMELEC1
an DEMELEC2.  I also expressed all energy flows in PJ, and assumed MWh where you said kWh (to make it look more sensible).  Anyway, as you said, the parameters are just numbers.


.xlsx   VT_REG_PRI_V04-CAEX.xlsx (Size: 174.43 KB / Downloads: 16)
Reply
#13
(21-06-2021, 05:02 PM)Antti-L Wrote: Very well, please find attached my suggested modifications to make the two CAEX examples working.

I integrated both examples in the same model, and defined the demand as 3.6 PJ (so 1 TWh) for both DEMELEC1
an DEMELEC2.  I also expressed all energy flows in PJ, and assumed MWh where you said kWh (to make it look more sensible).  Anyway, as you said, the parameters are just numbers.

Thank you for the file. 

I have some questions about: the results, a questions that was lost before, and some news. 

Q1) 

Let us forget the units, because in only interested in the flow for now. If you multiply the ELEC by 3.6, so we have the following:

CONV1 needs, as input, 5 (AIR) and 3.6 electricity (ELEC1) for each 3 air stored (AIRSTOCK).  This AIRSTOCK enters in the storage as input and it is also the only output. The output of  CAEXSTORAGE will be input for CONV2. Beyond AIRSTOCK, CONV2 needs COA as input. For each 3 of AIRSTOCK and 10 of COA, the output of CONV2 is 3.6 ELEC1.

For example 1:
The expected result: 
The demand of ELEC1 is now 3.6 in 2005 and also in 2006. 
In 2005, the demand is supplied by the mining process. So, 3.6 is produced and delivered directly. 
In 2006, there is a need of storage in 2005 for the supply in 2006. 
Since 3.6 of ELEC1 must be delivered by CONV2, that means CONV2 needs, as input, 3 AIRSTOCK and 10 COA in 2006. Hence, from CAEXSTORAGE there is a need of  3 AIRSTOCK as output in 2006, so 3 AIRSTOCK as input in 2005. That means, CONV1 needs to make 3 AIRSTOCK as output in 2005, which means a need of 3.6 of ELEC1 and 5 AIR as input in 2005. 

The results (Figure R1.png):
20 AIR and 10.8 of ELEC1 are produced (by mining) in 2005.  Since the mining has a positive production cost, the production should not be higher than the total demand, which is 7.2 ELEC1. 

The flow: 
In 2005: the demand is supplied by the mining process. So, 3.6 is produced and delivered directly
Storage in 2005:
  In CONV1: 10 AIR and 7.2 ELEC1 as input. 6 AIRSTOCK as output. This is twice as needed.
  In CAEXSTORAGE : 6 AIRSTOCK  as input. 

In 2006: 
   In CAEXSTORAGE : AIRSTOCK  as output. This means 3  AIRSTOCK  remains in storage. The storage should be empty. 
   In CONV2:  AIRSTOCK and 10 COA  as input. 3.6 ELEC1 as output. 

For Example 2: 

+ How can I see the CAEXSTORAG2 in the list of process? I cannot see the this process neither ELEC2 (Figure R2.png)
+ The results (Figure R3.png):

The output of CAEXSTORAG2 in 2006 is 3.6 ELEC2. In this configuration, I believe that COA cannot be consumed in 2006. Thus, the input in 2005 should be 5 AIR and 3.6 ELEC2 and 10 COA. The results are not inline with the expected result, since the actual results has the input as twice the expected result. 

Where are the problems about that difference between the results and the expected result?

Q2) About FLO_FUNC(r,y,'STORAGE','DEMMIC1','DUMNRG','ANNUAL') = Value:  ---  Must the order of the columns in excel correspond to the order of the parameters in the function? Ex.: In the excel file, Comm-OUT appears before the CommGrp. So, line 19 (range B19 to P19)  in "VT_REG_PRI_V04-corrected.xlsx" is created to match the parameters order of the function FLO_FUNC. Is it right?


In the sheet "DemTechs_TPS": 
Q3) In cell O25, the storage capacity was set to 2. But, there was 6 units of AIRSTOCK in the flow. Could you give a little word about it?

Q4) CAEXSTORAGE has no auxiliary commodities. Why cell O24 empty is not empty?

Q5) When should I use the column "Output"? If the Output is empty, does that mean a default equals to 1?
I used the column "INPUT" and "OUTPUT" based on the explanations in https://forum.kanors-emr.org/showthread.php?tid=965&highlight=two+inputs. I am not sure if I understood it well. I have understood that the numbers describing the amount of inputs and outputs should be specified.  Documentation does not help a lot about it (Figure OUTPUT.png)


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#14
You have some serious misunderstandings about your own model:

> 20 AIR and 10.8 of ELEC1 are produced (by mining) in 2005.  Since the mining has a positive production cost, the production should not be higher than the total demand, which is 7.2 ELEC1.

You are perhaps missing the fact the the mining process of AIR now supplies both examples, 10 for each.  And that is fully correct.  Also 10.8 of ELEC1 is fully correct and exactly according to you specs: 3.6 PJ goes directly to the demand and 7.2 PJ is needed for the storage (the demand for two years is stored!). So, 3.6 + 2*3.6 = 10.8, which is thus fully correct. For AIR, the consumption was 5 kg per 3 kg of AIRSTOCK, and so we get 10 kg for both examples (for the two years' demand to be stored). So 2*5 = 10 of AIR for each example, making 20 in total (for both examples).

> In CONV1: 10 AIR and 7.2 ELEC1 as input. 6 AIRSTOCK as output. This is twice as needed.

AIRSTOCK must be stored in 2005 to supply the demand for two years, 2006 and 2007, which makes 6. So it is fully correct again.

>  In 2006:    In CAEXSTORAGE : 6 AIRSTOCK  as output. This means 3  AIRSTOCK  remains in storage. The storage should be empty.

You are wrong.  In 2006 the CAEXSTORAGE has 3 AIRSTOCK  as output. I have verified that, and it is fully correct for satisfying one year's demand (the milestone year 2006).

> The output of CAEXSTORAG2 in 2006 is 3.6 ELEC2. In this configuration, I believe that COA cannot be consumed in 2006. Thus, the input in 2005 should be 5 AIR and 3.6 ELEC2 and 10 COA. The results are not inline with the expected result, since the actual results has the input as twice the expected result.

Wrong again. The input in 2005 should be 10 AIR and 7.2 ELEC2 and 20 COA.  And that is exactly what the model results are!


Could you please first confirm that you understand the above, and agree that the results are fully as expected, before continuing...    Big Grin
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#15
(21-06-2021, 08:32 PM)Antti-L Wrote: You have some serious misunderstandings about your own model:

> 20 AIR and 10.8 of ELEC1 are produced (by mining) in 2005.  Since the mining has a positive production cost, the production should not be higher than the total demand, which is 7.2 ELEC1.

You are perhaps missing the fact the the mining process of AIR now supplies both examples, 10 for each.  And that is fully correct.  Also 10.8 of ELEC1 is fully correct and exactly according to you specs: 3.6 PJ goes directly to the demand and 7.2 PJ is needed for the storage (the demand for two years is stored!). So, 3.6 + 2*3.6 = 10.8, which is thus fully correct.  For AIR, the consumption was 5 kg per 3 kg of AIRSTOCK, and so we get 10 kg for both examples (for the two years' demand to be stored).  So 2*5 = 10 of AIR for each example, making 20 in total (for both examples).

> In CONV1: 10 AIR and 7.2 ELEC1 as input. 6 AIRSTOCK as output. This is twice as needed.

AIRSTOCK must be stored in 2005 to supply the demand for two years, 2006 and 2007, which makes 6. So it is fully correct again.

>  In 2006:    In CAEXSTORAGE : 6 AIRSTOCK  as output. This means 3  AIRSTOCK  remains in storage. The storage should be empty.

You are wrong.  In 2006 the CAEXSTORAGE has 3 AIRSTOCK  as output. I have verified that, and it is fully correct for satisfying one year's demand (the milestone year 2006).

> The output of CAEXSTORAG2 in 2006 is 3.6 ELEC2. In this configuration, I believe that COA cannot be consumed in 2006. Thus, the input in 2005 should be 5 AIR and 3.6 ELEC2 and 10 COA. The results are not inline with the expected result, since the actual results has the input as twice the expected result.

Wrong again. The input in 2005 should be 10 AIR and 7.2 ELEC2 and 20 COA.  And that is exactly what the model results are!


Could you please first confirm that you understand the above, and agree that the results are fully as expected, before continuing...    Big Grin

You have some serious misunderstandings about your own model:
Actually, it is not my model rsrs - I'm not even able to construct one without know the syntax of the excel files.  This model comes from DEMO 4, which is provided by VEDA, with some modifications. Using it, I'd like to understand some points that is not clear, at least for me, in the documentation. For sure, I have misunderstandings about that (a lot, not few). 

You are perhaps missing the fact the the mining process of AIR now supplies both examples, 10 for each.  And that is fully correct.  
Good for me. No problem with AIR. 

Also 10.8 of ELEC1 is fully correct and exactly according to you specs: 3.6 PJ goes directly to the demand and 7.2 PJ is needed for the storage (the demand for two years is stored!). 
I do not understand this point. The storage must be done in 2005 to be used in 2006 only. Only one year of storage. Why 2 years is stored? (Actually, this may be the main point. If we assume to have 3.6 of demand for 2007 and storage is sent to 2007, this will be good, but 2007 does not exist in the demand table. 2007 does not even appear in the results.)

AIRSTOCK must be stored in 2005 to supply the demand for two years, 2006 and 2007, which makes 6. So it is fully correct again.
I do not understand this point. Why 2007? Only in 2005 and in 2006 there are some demand. Accordingly to the Demands sheet (Figure Demand.png), 2007 is not even in the table. This is quite close to the point above. 

You are wrong.  In 2006 the CAEXSTORAGE has 3 AIRSTOCK  as output. I have verified that, and it is fully correct for satisfying one year's demand (the milestone year 2006).
I understand parts of this. I agree that In 2006 the CAEXSTORAGE has 3 AIRSTOCK  as output. But, where are the other 3 AIRSTOCK at the end of 2006?

Wrong again. The input in 2005 should be 10 AIR and 7.2 ELEC2 and 20 COA.  And that is exactly what the model results are!
Well, rsrs, there is really some stuff not clear in our communication. After 2005, there is only demand in 2006. The only demand is 3.6 ELEC2. However, the amount of inputs are enough to have 7.2 ELEC2 as output, twice of the final output. If 3.6 is the output in 2006, where are the other 3.6 of ELEC2?


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