Re: Change 1 PQ to PV -> infeasible problem

[Jose Luis Marín]

Chris,

Iterative methods in general (whether they are N-R, FD, G-S, etc.) are
prone to failure; you can sometimes mitigate this problem, but you can
never eliminate it 100%. It sounds like you may have hit one of those cases
where NR iteration fails and it's hard to find "the right seed".  Our
company provides tools that help precisely in cases like this, but I don't
want to spam the list.  We can continue this conversation off-list.

-- 
Jose L. Marin
Grupo AIA


2016-05-10 10:40 GMT+02:00 Chris Prokop :

> Dear Jose and Ray,
>
> thanks for your replies, I've tested another few things:
>
> I deactivated the other two PV-buses, so there is only one reference bus,
> many PQ-buses and the problematic PQ/PV-bus.
>
> With the PQ (result is ok):
> - Slack inserts: -60 Mvar
> - Problematic bus: 0.118 Mvar, V=1.0335 pu
>
> With PV without limits (result is ok)
> - Slack inserts: -61 Mvar
> - Problematic bus: 0.851 Mvar, V=1.05 pu (as set as Vg)
>
> With PV with limits (not ok):
> - Slack inserts: +84 Mvar (145 Mvar difference)
> - Problematic bus: 0.7 Mvar, V=0.15 pu (generator has reached its upper
> limit, hence it was converted to PQ)
>
> But the result of V of the problematic bus should be somewhere in between
> 1.0335 and 1.05 p.u.. All other P and Q of the generators and buses remain
> constant within all variants. For me it looks like the result is in the
> unstable area of the QV-curve.
>
> @ Jose: when setting Vg=1.0335 than the result is ok again. The problem
> starts at the point the generator reaches it's Q-Limit and gets converted
> to PQ again.
>
> I don't understand why the result is in the unstable area of the QV-curve,
> because the reactive power is capacitive and not inductive, hence I
> wouldn't expect a stability Problem (I'm increasing the voltage).
>
> Interesting point: with the fast-decoupled solver the power flow with
> Q-limit converges to V=1.0465 p.u. (result is ok), with Gauss-Seidel it
> doesn't converge. Maybe there is a problem with Newton's method?!
>
> Thanks for your help/explanations, nice regards,
> Chris
>
>
> 2016-05-06 7:33 GMT+02:00 Jose Luis Marín :
>
>>
>> Sorry for the confusion, Ray:  where I said, *"...solvable with the bus
>> running as PV, but unfeasible when the bus is running as PQ"*, I meant
>> to say:
>>
>>"*a valid powerflow solution* with the bus running as PV, but an
>> *unstable* one when the bus is running as PQ"
>>(it is one of the multiple low-voltage solutions of the full
>> mathematical problem--these solutions are unstable form the point of view
>> of real operations)
>>
>> So yes, in that situation the solution obtained with the bus type being
>> PV is (of course) still a mathematical solution to the same problem,
>> regardless of the bus type being reinterpreted as PQ.  But the point is
>> that, only if you look at the problem in this second case (bus type
>> switched to PQ) it is then possible to see that the solution is a low
>> voltage one, and therefore not valid.  One way to detect this is to look at
>> the slope dQ / dV of the Q-V curve, because this situation corresponds to
>> being "on the wrong side" of the curve.  It may happen either if the
>> setpoint V was too low, or if the P output is too high.
>>
>> What the Newton method may do in that case is hard to say, since N-R does
>> not "see" the kind of instability that we are talking about here.  It may
>> converge to this unstable solution, or to the operationally correct one, or
>> diverge.  It all depends on the complex landscape of the respective basins
>> of attraction for each solution, which as you know is fractal.  But unless
>> the case is right at the bifurcation point (here that would be the minimum
>> of the Q-V curve), the basin of attraction around this unstable solution
>> will be finite, so there's a good chance that if you provide the original
>> solution (obtained when the bus was type PV) and use it as the seed to the
>> new problem (bus type now switched to PQ), then NR will converge to the
>> same solution.  You'd probably need to start from a flat start in order to
>> find the high-voltage solution instead.
>>
>> --
>> Jose L. Marin
>> Grupo AIA
>>
>>
>> 2016-05-05 22:17 GMT+02:00 Ray Zimmerman :
>>
>>> Thanks Jose. I’ve added that to the manual as you suggested.
>>>
>>> I also agree with your suggestions for Chris. However, I’m curious about
>>> your [*] note. In the situation you describe, it’s just that the Newton
>>> method will diverge, correct? The original solution will still be a
>>> solution of the power flow equations won’t they?
>>>
>>> Ray
>>>
>>>
>>> On Apr 30, 2016, at 2:01 PM, Jose Luis Marin 
>>> wrote:
>>>
>>>
>>> Interesting, I didn't know that enforcing Q-limits also affects
>>> generators when their bus-type is set to PQ.  Ray, I suggest documenting
>>> this behavior in the manual, probably at the end of the last paragraph in
>>> Section 4.1.  

Re: A question about transformer modeling with uk, ur

[Jose Luis Marín]

Certainly, testing with a small system is the only way to find out what the
internal model *really* is (since you can't inspect the code in this
case).  Anyway, I would add another suggestion: export from Netomac to
PSS/E RAW format, and inspect the transformer records very carefully.  Read
the specs of the RAW format (the specific version of the format used), and
have a look at the internal model used by PSS/E (I'd assume that Netomac
uses the same, but who knows).  You can find it here:


http://w3.usa.siemens.com/datapool/us/SmartGrid/docs/pti/2009July/PDFs/Modeling_of_two_winding_voltage_regulating_transformers.pdf

Compare this model with the one use by MATPOWER (page 22 of the PDF
manual).  You'll see they're not exactly the same.  You said you're not
using magnetizing branches (Bs), which simplifies things a lot.  But pay
attention to the tap ratios (both sides!), just in case.

-- 
Jose L. Marin
Grupo AIA



2016-05-10 13:50 GMT+02:00 Chris Prokop :

> Dear Shuo,
>
> I'd try to simulate a simply network with 1 reference bus with 1
> transformer and 1 subsequent load to test the behaviour between Matpower
> and netomac.
>
> I'm not used to netomac, so can't help here directly.
>
> Nice regards,
> Chris
>
> 2016-05-10 13:22 GMT+02:00 Shuo Chen :
>
>> Dear Chris,
>>
>> thanks for your reply, actually we are getting the same transformer data.
>> In my power system  uk is far more larger than ur (more than 50:1), so it
>> can be assumed that ux = uk.
>>
>> i'm wondering whether there is any little difference between netomac and
>> matpower by modeling the transformer, so that even if I give them the same
>> input, PF results could still differ.
>>
>> best regards
>> Shuo
>>
>> Zitat von Chris Prokop ：
>>
>> Dear Shuo Chen,
>>>
>>> I'm used to similar data and calculate r and x as (ur, uk in p.u.,
>>> base_MVA
>>> and S_transformer in MVA):
>>> - r = ur * base_MVA / S_transformer
>>> - x = sqrt(uk^2 - ur^2) * base_MVA / S_transformer
>>> - ratio = 1 (in your case, where rated = nominal voltage)
>>> - angle = 0 (in your case, without phase shifting)
>>>
>>> Maybe one of your problems is using uk instead of ux (uk^2 = ux^2 +
>>> ur^2)?
>>>
>>> Nice regards,
>>> Chris
>>>
>>> 2016-05-10 0:33 GMT+02:00 Shuo Chen :
>>>
>>>
 Dear Ray and matpower users,

 i'm writing my thesis about a data-converter for two power system
 simulation softwares: PSSE@Netomac from Siemens and matpower. Here is a
 problem with transformer modeling, i've read a lot in this archive but
 still can't solve it, so i decide to post my question here.

 My goal is to get the same or a simular power flow convergency using the
 Newton's method after converting the net topology from one to the other.
 However, I'm stucked when I try to convert a Netomac net into a matpower
 case. Comparing the PF results of the 2 softwares, there is always a big
 deviation (by bus voltage magnitude, bus voltage angle and branch P/Q
 injection). More specifically, the bus voltages in pu of matpower are
 all
 lower than those of Netomac, the deviation could be up to 5%, like 0.977
 instead of 0.997 in Netomac.

 I simply grab all the bus/gen/branch parameters from Netomac except for
 the transformer impedance, so I guess there might be a mistake when the
 transformer model is built in matpower.
 The transformers have 2-windings and no tap changer

 The parameters I could get from Netomac are:

 - Rated voltage of HV side winding (Un1)
 - Nominal network voltage HV side (UB1? the value is equal to Un1)
 - Rated voltage of LV side winding (Un2)
 - Nominal network voltage LV side (UB2? the value is equal to Un2)
 - Rated apparent power (equal to baseMVA)
 - ur in % (may not be zero)
 - uk in % (must be larger than ur)
 - P0 and I0 are not given
 - vector group YY0
 in matpower a transformer is treated as a transmission line, where
 r(p.u.)
 and x(p.u.) are needed for the power flow calculation. (b is omitted in
 my
 case)
 According to the formels:

 z_pu = uk% / 100
 x_pu = uk% *(Un/UB).^2 * (SB/Sn) / 100
 r_pu = sqrt(z_pu.^2 - x_pu.^2) (here r_pu = ur%/ 100)

 Since in my case Un/UB = 1, SB/Sn = 1, so x = uk%/100, r = ur%/100, b =
 0
 I set ratio = 1 and angle = 0, status = 1, angmin/max = +-360
 other values are set to 0

 Netomac can also export a .raw file for PSSE, the conversion above will
 generate equivalent r and x values as the exported .raw file.
 Theoretically, with the same net topology and P/Q accuracy, the PF
 convergency of the two simulation tools should be almost the same.

 Here is one more hint, as far as I know, the transformer impedance in
 matpower is modeled at "to" side, while in Netomac it's modeled at "HV"
 side 

Re: A question about transformer modeling with uk, ur

[Chris Prokop]

Dear Shuo,

I'd try to simulate a simply network with 1 reference bus with 1
transformer and 1 subsequent load to test the behaviour between Matpower
and netomac.

I'm not used to netomac, so can't help here directly.

Nice regards,
Chris

2016-05-10 13:22 GMT+02:00 Shuo Chen :

> Dear Chris,
>
> thanks for your reply, actually we are getting the same transformer data.
> In my power system  uk is far more larger than ur (more than 50:1), so it
> can be assumed that ux = uk.
>
> i'm wondering whether there is any little difference between netomac and
> matpower by modeling the transformer, so that even if I give them the same
> input, PF results could still differ.
>
> best regards
> Shuo
>
> Zitat von Chris Prokop ：
>
> Dear Shuo Chen,
>>
>> I'm used to similar data and calculate r and x as (ur, uk in p.u.,
>> base_MVA
>> and S_transformer in MVA):
>> - r = ur * base_MVA / S_transformer
>> - x = sqrt(uk^2 - ur^2) * base_MVA / S_transformer
>> - ratio = 1 (in your case, where rated = nominal voltage)
>> - angle = 0 (in your case, without phase shifting)
>>
>> Maybe one of your problems is using uk instead of ux (uk^2 = ux^2 + ur^2)?
>>
>> Nice regards,
>> Chris
>>
>> 2016-05-10 0:33 GMT+02:00 Shuo Chen :
>>
>>
>>> Dear Ray and matpower users,
>>>
>>> i'm writing my thesis about a data-converter for two power system
>>> simulation softwares: PSSE@Netomac from Siemens and matpower. Here is a
>>> problem with transformer modeling, i've read a lot in this archive but
>>> still can't solve it, so i decide to post my question here.
>>>
>>> My goal is to get the same or a simular power flow convergency using the
>>> Newton's method after converting the net topology from one to the other.
>>> However, I'm stucked when I try to convert a Netomac net into a matpower
>>> case. Comparing the PF results of the 2 softwares, there is always a big
>>> deviation (by bus voltage magnitude, bus voltage angle and branch P/Q
>>> injection). More specifically, the bus voltages in pu of matpower are all
>>> lower than those of Netomac, the deviation could be up to 5%, like 0.977
>>> instead of 0.997 in Netomac.
>>>
>>> I simply grab all the bus/gen/branch parameters from Netomac except for
>>> the transformer impedance, so I guess there might be a mistake when the
>>> transformer model is built in matpower.
>>> The transformers have 2-windings and no tap changer
>>>
>>> The parameters I could get from Netomac are:
>>>
>>> - Rated voltage of HV side winding (Un1)
>>> - Nominal network voltage HV side (UB1? the value is equal to Un1)
>>> - Rated voltage of LV side winding (Un2)
>>> - Nominal network voltage LV side (UB2? the value is equal to Un2)
>>> - Rated apparent power (equal to baseMVA)
>>> - ur in % (may not be zero)
>>> - uk in % (must be larger than ur)
>>> - P0 and I0 are not given
>>> - vector group YY0
>>> in matpower a transformer is treated as a transmission line, where
>>> r(p.u.)
>>> and x(p.u.) are needed for the power flow calculation. (b is omitted in
>>> my
>>> case)
>>> According to the formels:
>>>
>>> z_pu = uk% / 100
>>> x_pu = uk% *(Un/UB).^2 * (SB/Sn) / 100
>>> r_pu = sqrt(z_pu.^2 - x_pu.^2) (here r_pu = ur%/ 100)
>>>
>>> Since in my case Un/UB = 1, SB/Sn = 1, so x = uk%/100, r = ur%/100, b = 0
>>> I set ratio = 1 and angle = 0, status = 1, angmin/max = +-360
>>> other values are set to 0
>>>
>>> Netomac can also export a .raw file for PSSE, the conversion above will
>>> generate equivalent r and x values as the exported .raw file.
>>> Theoretically, with the same net topology and P/Q accuracy, the PF
>>> convergency of the two simulation tools should be almost the same.
>>>
>>> Here is one more hint, as far as I know, the transformer impedance in
>>> matpower is modeled at "to" side, while in Netomac it's modeled at "HV"
>>> side (which is the "from" side), so there could be a difference between
>>> the
>>> two models but i got no clue how to unify them.
>>>
>>>
>>> Has anyone ever met this kind of problem before? Or the transformer is
>>> converted in a right way, but there could be something wrong in other
>>> parts?
>>> I know this is a tough one, hopefully I could get some hints here. Many
>>> thanks for all!
>>>
>>> Best Regards
>>> Shuo
>>>
>>>
>>>
>>>
>>>
>
>
>
>

Re: A question about transformer modeling with uk, ur

[Shuo Chen]

Dear Chris,

thanks for your reply, actually we are getting the same transformer  
data. In my power system  uk is far more larger than ur (more than  
50:1), so it can be assumed that ux = uk.


i'm wondering whether there is any little difference between netomac  
and matpower by modeling the transformer, so that even if I give them  
the same input, PF results could still differ.


best regards
Shuo

Zitat von Chris Prokop ：


Dear Shuo Chen,

I'm used to similar data and calculate r and x as (ur, uk in p.u., base_MVA
and S_transformer in MVA):
- r = ur * base_MVA / S_transformer
- x = sqrt(uk^2 - ur^2) * base_MVA / S_transformer
- ratio = 1 (in your case, where rated = nominal voltage)
- angle = 0 (in your case, without phase shifting)

Maybe one of your problems is using uk instead of ux (uk^2 = ux^2 + ur^2)?

Nice regards,
Chris

2016-05-10 0:33 GMT+02:00 Shuo Chen :



Dear Ray and matpower users,

i'm writing my thesis about a data-converter for two power system
simulation softwares: PSSE@Netomac from Siemens and matpower. Here is a
problem with transformer modeling, i've read a lot in this archive but
still can't solve it, so i decide to post my question here.

My goal is to get the same or a simular power flow convergency using the
Newton's method after converting the net topology from one to the other.
However, I'm stucked when I try to convert a Netomac net into a matpower
case. Comparing the PF results of the 2 softwares, there is always a big
deviation (by bus voltage magnitude, bus voltage angle and branch P/Q
injection). More specifically, the bus voltages in pu of matpower are all
lower than those of Netomac, the deviation could be up to 5%, like 0.977
instead of 0.997 in Netomac.

I simply grab all the bus/gen/branch parameters from Netomac except for
the transformer impedance, so I guess there might be a mistake when the
transformer model is built in matpower.
The transformers have 2-windings and no tap changer

The parameters I could get from Netomac are:

- Rated voltage of HV side winding (Un1)
- Nominal network voltage HV side (UB1? the value is equal to Un1)
- Rated voltage of LV side winding (Un2)
- Nominal network voltage LV side (UB2? the value is equal to Un2)
- Rated apparent power (equal to baseMVA)
- ur in % (may not be zero)
- uk in % (must be larger than ur)
- P0 and I0 are not given
- vector group YY0
in matpower a transformer is treated as a transmission line, where r(p.u.)
and x(p.u.) are needed for the power flow calculation. (b is omitted in my
case)
According to the formels:

z_pu = uk% / 100
x_pu = uk% *(Un/UB).^2 * (SB/Sn) / 100
r_pu = sqrt(z_pu.^2 - x_pu.^2) (here r_pu = ur%/ 100)

Since in my case Un/UB = 1, SB/Sn = 1, so x = uk%/100, r = ur%/100, b = 0
I set ratio = 1 and angle = 0, status = 1, angmin/max = +-360
other values are set to 0

Netomac can also export a .raw file for PSSE, the conversion above will
generate equivalent r and x values as the exported .raw file.
Theoretically, with the same net topology and P/Q accuracy, the PF
convergency of the two simulation tools should be almost the same.

Here is one more hint, as far as I know, the transformer impedance in
matpower is modeled at "to" side, while in Netomac it's modeled at "HV"
side (which is the "from" side), so there could be a difference between the
two models but i got no clue how to unify them.


Has anyone ever met this kind of problem before? Or the transformer is
converted in a right way, but there could be something wrong in other parts?
I know this is a tough one, hopefully I could get some hints here. Many
thanks for all!

Best Regards
Shuo

Re: A question about transformer modeling with uk, ur

[Chris Prokop]

Dear Shuo Chen,

I'm used to similar data and calculate r and x as (ur, uk in p.u., base_MVA
and S_transformer in MVA):
- r = ur * base_MVA / S_transformer
- x = sqrt(uk^2 - ur^2) * base_MVA / S_transformer
- ratio = 1 (in your case, where rated = nominal voltage)
- angle = 0 (in your case, without phase shifting)

Maybe one of your problems is using uk instead of ux (uk^2 = ux^2 + ur^2)?

Nice regards,
Chris

2016-05-10 0:33 GMT+02:00 Shuo Chen :

>
> Dear Ray and matpower users,
>
> i'm writing my thesis about a data-converter for two power system
> simulation softwares: PSSE@Netomac from Siemens and matpower. Here is a
> problem with transformer modeling, i've read a lot in this archive but
> still can't solve it, so i decide to post my question here.
>
> My goal is to get the same or a simular power flow convergency using the
> Newton's method after converting the net topology from one to the other.
> However, I'm stucked when I try to convert a Netomac net into a matpower
> case. Comparing the PF results of the 2 softwares, there is always a big
> deviation (by bus voltage magnitude, bus voltage angle and branch P/Q
> injection). More specifically, the bus voltages in pu of matpower are all
> lower than those of Netomac, the deviation could be up to 5%, like 0.977
> instead of 0.997 in Netomac.
>
> I simply grab all the bus/gen/branch parameters from Netomac except for
> the transformer impedance, so I guess there might be a mistake when the
> transformer model is built in matpower.
> The transformers have 2-windings and no tap changer
>
> The parameters I could get from Netomac are:
>
> - Rated voltage of HV side winding (Un1)
> - Nominal network voltage HV side (UB1? the value is equal to Un1)
> - Rated voltage of LV side winding (Un2)
> - Nominal network voltage LV side (UB2? the value is equal to Un2)
> - Rated apparent power (equal to baseMVA)
> - ur in % (may not be zero)
> - uk in % (must be larger than ur)
> - P0 and I0 are not given
> - vector group YY0
> in matpower a transformer is treated as a transmission line, where r(p.u.)
> and x(p.u.) are needed for the power flow calculation. (b is omitted in my
> case)
> According to the formels:
>
> z_pu = uk% / 100
> x_pu = uk% *(Un/UB).^2 * (SB/Sn) / 100
> r_pu = sqrt(z_pu.^2 - x_pu.^2) (here r_pu = ur%/ 100)
>
> Since in my case Un/UB = 1, SB/Sn = 1, so x = uk%/100, r = ur%/100, b = 0
> I set ratio = 1 and angle = 0, status = 1, angmin/max = +-360
> other values are set to 0
>
> Netomac can also export a .raw file for PSSE, the conversion above will
> generate equivalent r and x values as the exported .raw file.
> Theoretically, with the same net topology and P/Q accuracy, the PF
> convergency of the two simulation tools should be almost the same.
>
> Here is one more hint, as far as I know, the transformer impedance in
> matpower is modeled at "to" side, while in Netomac it's modeled at "HV"
> side (which is the "from" side), so there could be a difference between the
> two models but i got no clue how to unify them.
>
>
> Has anyone ever met this kind of problem before? Or the transformer is
> converted in a right way, but there could be something wrong in other parts?
> I know this is a tough one, hopefully I could get some hints here. Many
> thanks for all!
>
> Best Regards
> Shuo
>
>
>
>

Re: Change 1 PQ to PV -> infeasible problem

[Chris Prokop]

Dear Jose and Ray,

thanks for your replies, I've tested another few things:

I deactivated the other two PV-buses, so there is only one reference bus,
many PQ-buses and the problematic PQ/PV-bus.

With the PQ (result is ok):
- Slack inserts: -60 Mvar
- Problematic bus: 0.118 Mvar, V=1.0335 pu

With PV without limits (result is ok)
- Slack inserts: -61 Mvar
- Problematic bus: 0.851 Mvar, V=1.05 pu (as set as Vg)

With PV with limits (not ok):
- Slack inserts: +84 Mvar (145 Mvar difference)
- Problematic bus: 0.7 Mvar, V=0.15 pu (generator has reached its upper
limit, hence it was converted to PQ)

But the result of V of the problematic bus should be somewhere in between
1.0335 and 1.05 p.u.. All other P and Q of the generators and buses remain
constant within all variants. For me it looks like the result is in the
unstable area of the QV-curve.

@ Jose: when setting Vg=1.0335 than the result is ok again. The problem
starts at the point the generator reaches it's Q-Limit and gets converted
to PQ again.

I don't understand why the result is in the unstable area of the QV-curve,
because the reactive power is capacitive and not inductive, hence I
wouldn't expect a stability Problem (I'm increasing the voltage).

Interesting point: with the fast-decoupled solver the power flow with
Q-limit converges to V=1.0465 p.u. (result is ok), with Gauss-Seidel it
doesn't converge. Maybe there is a problem with Newton's method?!

Thanks for your help/explanations, nice regards,
Chris


2016-05-06 7:33 GMT+02:00 Jose Luis Marín :

>
> Sorry for the confusion, Ray:  where I said, *"...solvable with the bus
> running as PV, but unfeasible when the bus is running as PQ"*, I meant to
> say:
>
>"*a valid powerflow solution* with the bus running as PV, but an
> *unstable* one when the bus is running as PQ"
>(it is one of the multiple low-voltage solutions of the full
> mathematical problem--these solutions are unstable form the point of view
> of real operations)
>
> So yes, in that situation the solution obtained with the bus type being PV
> is (of course) still a mathematical solution to the same problem,
> regardless of the bus type being reinterpreted as PQ.  But the point is
> that, only if you look at the problem in this second case (bus type
> switched to PQ) it is then possible to see that the solution is a low
> voltage one, and therefore not valid.  One way to detect this is to look at
> the slope dQ / dV of the Q-V curve, because this situation corresponds to
> being "on the wrong side" of the curve.  It may happen either if the
> setpoint V was too low, or if the P output is too high.
>
> What the Newton method may do in that case is hard to say, since N-R does
> not "see" the kind of instability that we are talking about here.  It may
> converge to this unstable solution, or to the operationally correct one, or
> diverge.  It all depends on the complex landscape of the respective basins
> of attraction for each solution, which as you know is fractal.  But unless
> the case is right at the bifurcation point (here that would be the minimum
> of the Q-V curve), the basin of attraction around this unstable solution
> will be finite, so there's a good chance that if you provide the original
> solution (obtained when the bus was type PV) and use it as the seed to the
> new problem (bus type now switched to PQ), then NR will converge to the
> same solution.  You'd probably need to start from a flat start in order to
> find the high-voltage solution instead.
>
> --
> Jose L. Marin
> Grupo AIA
>
>
> 2016-05-05 22:17 GMT+02:00 Ray Zimmerman :
>
>> Thanks Jose. I’ve added that to the manual as you suggested.
>>
>> I also agree with your suggestions for Chris. However, I’m curious about
>> your [*] note. In the situation you describe, it’s just that the Newton
>> method will diverge, correct? The original solution will still be a
>> solution of the power flow equations won’t they?
>>
>> Ray
>>
>>
>> On Apr 30, 2016, at 2:01 PM, Jose Luis Marin 
>> wrote:
>>
>>
>> Interesting, I didn't know that enforcing Q-limits also affects
>> generators when their bus-type is set to PQ.  Ray, I suggest documenting
>> this behavior in the manual, probably at the end of the last paragraph in
>> Section 4.1.  Something to the effect of *"... Note also that this
>> option affects generators even if the bus they are attached to is already
>> of type PQ."*
>>
>> Going back to Chris's problem, I suggest you approach this as consisting
>> of two orthogonal issues:
>>
>>- Solving with Q-limits enforced vs. solving without
>>- Solving with a given generator running as PQ-type vs. running as
>>PV-type (by manually switching it)
>>
>> With regards to the first issue, I suggest to start by analyzing the
>> behavior of your case *without* enforcing Q limits: in particular, pay
>> close attention to the QG injections obtained *in the solution* for those
>> gens operating as