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Questions Related to Power System Design

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meyra31

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Right.. There are 2 different questions which I couldn't answer..

There is an interconnector which is a transmission line used to t4ransfer large amounts of power from one section to another part of the system.. Short circuit test has been carried out and the question asks the effect of introducing resistance into the interconnector? How could we minimse the effect of this resistance when transferring bulk power between the generator and the consumer?

Second question is on three phase power transformer.. we were asked to calculate short circuit impedance and resistance and then we found short circuit reactance by using this equation: Z^2=X^2+R^2

and then we are given that the nominal rating of the transformer is 3kVa at 400 V line to line. Whats the percentage reactance? Assuming this iis star configuration, how would we solve this question?

Thank you in advance friends.
 

Hi,

Power loss ina a series resistor...

P = I * I * R

To lower power dissipation it is obvious, isn't it?

Either you lower resistance, or you lower current.
And since I is squared i'd prefer it. Half the current gives a quarter of power dissipation.

Klaus
 

Thanks bro..

any help with the second part of the question?
 

The more copper you use on the interconnection the less resistance.
You need to know the resistance and the inductance of primary winding of the 3KVA transformer, with these you can calculate reactance.
 

INductance values are not given but resistance is 1.76 Ohms. I have no idea how to calculate percentage reactance..
 

Sounds like a incomplete problem specification. There must be separate numbers for series resistance and reactance given in some way (may be it's said they are equal).
 

Reactive power is the power on the series resistance and core loses. I guess you don't count core loses.
I=3KVA/400V=7.5A
P=I*I*R=7.5*7.5*1.76 ohm=99W
99W/3KVA*100=3.3%

Don't click 'helped me' before you passed your exam.
 

Reactive power is the power on the series resistance and core loses.
No. Series resistance voltage drop is active power. Leakage inductance voltage drop is reactive.

Respectively you distinguish Rs (or %R) and Xs (or %X) components of transformer impedance. If Rs is 1.76 Ohm (%R = 3.3%), the transformer will have a typically larger (e.g. 5 or 10%) %X.
 

Reactive power is the power on the series resistance and core loses. I guess you don't count core loses.
I=3KVA/400V=7.5A
P=I*I*R=7.5*7.5*1.76 ohm=99W
99W/3KVA*100=3.3%

Don't click 'helped me' before you passed your exam.

No. Series resistance voltage drop is active power. Leakage inductance voltage drop is reactive.

Respectively you distinguish Rs (or %R) and Xs (or %X) components of transformer impedance. If Rs is 1.76 Ohm (%R = 3.3%), the transformer will have a typically larger (e.g. 5 or 10%) %X.

Thanks for answers but I guys isnt 3kva the apparent power (S) ? And I thought in three phase transformers, S= root3 x vL x Il where Il = Iphase in star connections and Vl= root 3 x Vphase.. So cant we just use S=root 3 VL x IL and find Il from here
 

Hi,

It's a long time ago .... i hope i remember right.

Isn't it so that 3kVA in a 400V 3Ph system means:
* Each line carries 1kVA
* and to calculate the line current you need to use star voltage = 400V / sqrt(3) = 231V.
* it is independent if load is connected in star or delta connection
(For sure star or delta connection matters when you want to calculate transformer winding current.
Imagine the power plant needs to supply 3kVA, and thus the line current must be the same independent of load connection)
Then the line current is 3kVA / 231V = 12.98A

Klaus
 

Hi,

It's a long time ago .... i hope i remember right.

Isn't it so that 3kVA in a 400V 3Ph system means:
* Each line carries 1kVA
* and to calculate the line current you need to use star voltage = 400V / sqrt(3) = 231V.
* it is independent if load is connected in star or delta connection
(For sure star or delta connection matters when you want to calculate transformer winding current.
Imagine the power plant needs to supply 3kVA, and thus the line current must be the same independent of load connection)
Then the line current is 3kVA / 231V = 12.98A

Klaus

I really dunno mate..
I mean this is probably an easy question but i just forgot the main principles in these cases.. How are we supposed to find reactancc by knowing current and rated power though? can we use Q=I^2 R ?
 

How are we supposed to find reactancc by knowing current and rated power though?
You can't. %R and %X are independent transformer parameters that can be measured or read from the type plate. You are proabably overlooking a sentence in your homework problem.
 

Hi,

Then the line current is 3kVA / 231V = 12.98A

Did i really write this?
I meant: line current =line power / line voltage = 1kVA / 231V = 4.33A.

Sorry for that.

Klaus
 

the problem seems like thyristor controlled series braking resistor.
you can find every detail in understanding FACTS by: Hingorani.

S=P+jQ=V* conjugate(I)=V*(Ia-jIr)
Ssc=Psc+JQsc=V2/Zsc==>R=Zsc cos(theta) = E^2/Ssc cos(theta)

also you can find R and X using your transformer voltage regulation factor.

good luck
 

the problem seems like thyristor controlled series braking resistor.
you can find every detail in understanding FACTS by: Hingorani.

S=P+jQ=V* conjugate(I)=V*(Ia-jIr)
Ssc=Psc+JQsc=V2/Zsc==>R=Zsc cos(theta) = E^2/Ssc cos(theta)

also you can find R and X using your transformer voltage regulation factor.

good luck

I dont get what u mean by R=Zsc cos theta.. ??

Basically isnt reactance ''X'' ??
so If I use the equation S=p+jq, i know S and i know line voltage and short circuit resistance, i can find the current right? Which voltage am i supposed to use in this case? Line voltage must be converted into phase voltage, is that right as well?
And once we get the line current, cant we just use Q = I^2 * X(reactance) and to find percentage reactance, i just do Zsc-Xsc/Xsc *100.. Anyone? can u double check what I just said please.. I am not sure
 

Hi,

It's a long time ago .... i hope i remember right.

Isn't it so that 3kVA in a 400V 3Ph system means:
* Each line carries 1kVA
* and to calculate the line current you need to use star voltage = 400V / sqrt(3) = 231V.
* it is independent if load is connected in star or delta connection
(For sure star or delta connection matters when you want to calculate transformer winding current.
Imagine the power plant needs to supply 3kVA, and thus the line current must be the same independent of load connection)
Then the line current is 3kVA / 231V = 12.98A

Klaus

ok but how can we calculate the reactive power? I ll try to use Q= I^2 * X and I found I but to find Q qw have to know angle ??? we arent given any in this case.. how can we calculate Q anyone??
 

Hi,

. how can we calculate Q anyone??

As said before you need additional information.

Inductance, current, cos(phi), L, xL, active Power, apparent power...
Some of the values above...


Klaus
 

Zsc is short circuit impedance, Theta is phase angle.

so If I use the equation S=p+jq, i know S and i know line voltage and short circuit resistance, i can find the current right?
you also need reactance .
And once we get the line current, cant we just use Q = I^2 * X(reactance) and to find percentage reactance, i just do Zsc-Xsc/Xsc *100.. Anyone? can u double check what I just said please.. I am not sure

Doesn't seem correct. you cannot find current by ur data. you need short circuit data or voltage regulation data of your transformer.

you have missing data.

please straighten your question.
 

Zsc is short circuit impedance, Theta is phase angle.


you also need reactance .


Doesn't seem correct. you cannot find current by ur data. you need short circuit data or voltage regulation data of your transformer.

you have missing data.

please straighten your question.

the thing is next question asks us to calculate voltage regulation from the per- unit reactance we obtained.. So as I said short circuit data is Rsc is 1.759 and Zsc is 2.644 S is 3kVa at 400 V line to line and its star connection.. these are the only data we got
 

Sounds like a incomplete problem specification. There must be separate numbers for series resistance and reactance given in some way (may be it's said they are equal).

Hi,

Power loss ina a series resistor...

P = I * I * R

To lower power dissipation it is obvious, isn't it?

Either you lower resistance, or you lower current.
And since I is squared i'd prefer it. Half the current gives a quarter of power dissipation.

Klaus

Look at post #19
I admire your patient asking again and again for all the data.
I guess this one is just another example: We told Meyra31 that the resistance and inductance are needed, we got the resistance immediately but it took another 2 days to get Meyra31 to disclose the inductance data.
 

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