A PECULIER PROBLEM, Please Help!

Thread Starter

b.shahvir

Joined Jan 6, 2009
457
Hi Guys,

:D Firstly, I apologize to all my friends on AAC for adding one more thread to the already existing clutter of nonsensical threads started by me. I tried my level best to solve this doubt pertaining to physics of ‘Load Sharing’ on my own with whatever study material I have, but in vain (browsing the net didn’t help me either). As a result, I had no option but to put it up as an extra thread on this forum. I hope you guys would be kind enough to bear with me!
But before I start my never ending rant, I want to clarify that this problem is theoretical in nature and not related to the practical conditions pertaining to large generators hooked up to large power grids (Infinite bus).

:confused: Consider two power sources A and B (DC generators/batteries……. I have considered DC power for clarity), connected in parallel and sharing a common load. Suppose I increase the EMF of say, source A (EMF of source B remains un-altered). On applying ‘Superposition Theorem’ I found that the terminal voltage Vt across the load would obviously rise also resulting in an increase in the total load current I(load)…….so far, so good.

But then I tried comparing the above results with the ‘Terminal voltage Vt’ versus ‘Load current I(load)’ characteristics of two DC generators operating in parallel and sharing a common load. I observed that if the EMF of say Gen A is increased, Gen A takes up more load with a significant increase in the bus terminal voltage Vt across load.

However, according to the Vt – I(load) characteristics, in spite of the increase in Vt, the load current and hence output power is shown to remain constant.
This, in my opinion, directly defies Ohm’s Law…..which in turn is the cause of my doubt.

A voltmeter joined parallel with the connected load would then never indicate the rise in Vt as the ‘voltmeter current Iv’(however small) is now part of the total load current I(load), but which remains constant {according to the Vt - I(load) characteristics} as mentioned above.

:( Any kind of help/suggestion in this regard will be greatly appreciated……..or else, this thread too will be destined for (like several others before it), a pre-mature demise!

Thanks & Kind Regards,
Shahvir


 

thingmaker3

Joined May 16, 2005
5,083
Have you considered the internal resistance of the supplies? If the supplies are not equal, the lower supply must be considered part of the load.
 

Thread Starter

b.shahvir

Joined Jan 6, 2009
457
Have you considered the internal resistance of the supplies? If the supplies are not equal, the lower supply must be considered part of the load.
:) Yes, i have done my calculations accounting for the internal resistances of the supplies as well. Also, I have considered batteries as power sources A and B, and all remaining circuit components as resistances to avoid complexity and maintain clarity.

I have applied Superposition Theorem considering 4 different cases, including 3 with unequal voltages. In all the 4 cases the values of internal resistances are considered to be constant. In each case i have increased the EMF of source A by a certain amount (EMF of source B assumed to remain un-altered).
The results obtained are as below;

CASE 1:- With both the EMFs Ea and Eb equal at 6VDC
Terminal voltage Vt = 5.8 Volts
Load current I(load) = 0.5 Amps
Ia = 0.28 Amps ; Ib = 0.28 Amps
Both currents Ia and Ib are equal and positive (discharging).

CASE 2 :- With Ea = 6.2VDC and Eb = 6VDC
Terminal voltage Vt = 5.9 Volts
Load current I(load) = 0.59 Amps
Ia = 0.38 Amps ; Ib = 0.20 Amps
Ia is more and Ib is less but both currents are positive.
This indicates that source B is still supplyinga small part of the total load current I(load). Ia and Ib are both discharge currents.

CASE 3 :- With Ea = 7VDC and Eb = 6VDC
Terminal voltage Vt = 6.2 Volts
Load current I(load) = 0.62 Amps
Ia = 0.80 Amps ; Ib = - 0.19 Amps
Ia is positive and Ib is negative.
This indicates that source B now becomes part of the load. Ia is discharging { Ia = Ib + I(load)} while Ib is charging.

CASE 4 :- With Ea = 12VDC and Eb = 6VDC
Terminal voltage Vt = 8.6 Volts
Load current I(load) = 0.86 Amps
Ia = 3.42 Amps ; Ib = - 2.57 Amps
Ia is positive and Ib is negative.
Again, this indicates that source B is part of the load. Again, Ia is discharging { Ia = Ib + I(load)} while Ib is charging.

From above, in CASE 2, Ib is still a discharge current (though < Ia) and contributing to total load current I(load) in a small way.

In CASE 3 and 4, Ib is negative and now becomes part of the load….albeit with source A now charging source B and supplying current to the connected load as well.

In all the 4 cases, an increase in Vt has resulted in a proportionate increase in I(load) as expected. This is in accordance with Ohm’s Laws. The problem is when I tried comparing this with Vt – I(load) characteristics for DC machines in parallel. If Ea is increased, Vt also increases but I(load) and hence output power is depicted as remaining constant…….albeit with load shifting between the individual machines.

Thanks & Kind Regards,
Shahvir
 

t_n_k

Joined Mar 6, 2009
5,455
CASE 1:- With both the EMFs Ea and Eb equal at 6VDC
Terminal voltage Vt = 5.8 Volts
Load current I(load) = 0.5 Amps
Ia = 0.28 Amps ; Ib = 0.28 Amps
Both currents Ia and Ib are equal and positive (discharging).

Did these values come from actual measurements or calculations?
 

Thread Starter

b.shahvir

Joined Jan 6, 2009
457
CASE 1:- With both the EMFs Ea and Eb equal at 6VDC
Terminal voltage Vt = 5.8 Volts
Load current I(load) = 0.5 Amps
Ia = 0.28 Amps ; Ib = 0.28 Amps
Both currents Ia and Ib are equal and positive (discharging).

Did these values come from actual measurements or calculations?
:) These values have come from calculations. I had applied Superposition Theorem to the network. Then, Studiot suggested i apply Kirchoff's Laws (as per Thingmaker3's advice) in my earlier thread on 'Two PM Alternators in Parallel Query' in the 'Physics section' of this forum and obtained exactly the same results in both the cases!

Actually this thread is an offshoot of my earlier thread on 'Two PM Alternators in Parallel' as mentioned above. This was done to avoid mixing of concepts and for clarity.

Thanks & Kind Regards,
Shahvir
 

Thread Starter

b.shahvir

Joined Jan 6, 2009
457
I don't think .28 + .28 = .5 :confused:
Ok, 0.28 + 0.28 = 0.56 to be accurate……I rounded off on the lower side. :p

:) Thanx very much Thingmaker3 and t_n_k for trying to help me out. Ultimately, the point of all this mathematical exercise (as suggested by a very good friend of mine) is to re-affirm an already evident point about voltage sources in parallel……. that if EMF of one of the two power sources connected in parallel to each other is increased, the terminal voltage Vt across the common load shared by them will rise causing load current I(load) to increase. This is not my doubt, rather I agree with it in totality.

The problem arose when I tried comparing these results with Vt – I(load) characteristics pertaining to DC generators (AC has been left out to prevent complexity)……..wherein increasing the EMF of say, Gen A increases Vt, but I(load) and output power are shown to remain constant as per the graphical characteristics. I myself plotted the co-ordinates and verified it graphically (again as suggested by a very good friend of mine) and found the same results. This has stumped me! (for details, one can refer the last pages of my thread on ‘Two PM Alternators in Parallel query’ in the ‘Physics’ section of this forum as this thread is an off-shoot of the same).

I fail to understand as to how during the load shifting process between the two power sources, I(load) remains un-altered in spite of an increase in Vt…..that too since the circuit in question is linear (even though it might consist of rotating machines at some point) and should obey Ohm’s laws!
But all this has been said before. I will be grateful if you guys can guide me thru this problem, which in turn would be a considerable learning experience for me and hopefully for others too…..whosoever might be interested.

Kind Regards,
Shahvir
 

Thread Starter

b.shahvir

Joined Jan 6, 2009
457
Suggestion: Use a bigger font or a sans serif font. I'm trying squint with a pretty big LCD. Pitchers would help as well. :D
SORRY! I DO NOT KNOW WHAT IS SANS SERIF FONT :confused:

P.S. I WOULD HAVE GONE FOR A LARGER FONT BUT I DO NOT WANT TO WASTE PRECIOUS FORUM INK IN THESE TIMES OF ECONOMIC RECESSION!! :D :D
 

thatoneguy

Joined Feb 19, 2009
6,359
Greenpeace thanks you for your effort to save electrons.

Sans Serif fonts do not have the 'ticks' at the edges of letters, Like this font.

This is a serif font (Garamond), and is hard to read. Especially in italics, or when small italics.

End of lithography lesson for today. Next week, we will cover what those little 'ticks' are called, and their origin. :D
 

Thread Starter

b.shahvir

Joined Jan 6, 2009
457
Greenpeace thanks you for your effort to save electrons.
Thanx...i am proud to have contributed in my little way to make mother earth a better place to inhabit.

Sans Serif fonts do not have the 'ticks' at the edges of letters, Like this font.

This is a serif font (Garamond), and is hard to read. Especially in italics, or when small italics
End of lithography lesson for today. Next week, we will cover what those little 'ticks' are called, and their origin. :D
Thanx for the enlightenment, but i wish you would have been able to enlighten me in my query! ;)
 

thatoneguy

Joined Feb 19, 2009
6,359
Try using a larger difference in voltage, see if results are linear.

Measure Ia and Ib seperately, then measure when both are connected.

Does adding a diode have any effect on the outcome?
 

Thread Starter

b.shahvir

Joined Jan 6, 2009
457
Try using a larger difference in voltage, see if results are linear.

Measure Ia and Ib seperately, then measure when both are connected.

Does adding a diode have any effect on the outcome?
I have already used larger differences in voltages, the results are still linear.

Superposition theorem measures Ia and Ib seperately and then combines them.

Adding a diode is irrelevant in this case as i have considered DC supply.

P.S. I have presented the results in the earlier posts, u can refer them if interested.
 

t_n_k

Joined Mar 6, 2009
5,455
I'd like to return to your statement about parallel operation of generators and their VI characteristic.

"The problem arose when I tried comparing these results with Vt – I(load) characteristics pertaining to DC generators (AC has been left out to prevent complexity)……..wherein increasing the EMF of say, Gen A increases Vt, but I(load) and output power are shown to remain constant as per the graphical characteristics."

Do you mean absolutely constant? No change at all? This seems to be a pivotal point for your arguments.

I'd suggest there will always be a change and the degree of change depends on the EMF variation due (say) to changes in the individual generator's driving speed or DC field excitation. One can consider many examples of this and the outcome will be consistent.

The following website covers several examples of interest :

http://books.google.com.au/books?id...X&oi=book_result&resnum=3&ct=result#PPA129,M1

Under practical operating conditions there would be speed / field feedback controls etc. to govern the load power - per system requirements. Under these circumstances one would expect to obtain very tight control over load power.
 

Thread Starter

b.shahvir

Joined Jan 6, 2009
457
Dear t_n_k, :)

Thanx for reply.......I humbly request you to refer my thread on 'Two PM Alternators in Parallel Query' in the 'Physics section' of this forum Page no.9; post # 83. In it i have explained the cause of my doubt, it will give you a clearer picture as to what I'm after.

Kind Regards,
Shahvir
 
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