Am I allowed to ask for PE advice on this forum? (Power EE)

[firstoption]

i know students aren't allowed, not sure if test takers are not allowed either. (tried to find this in FAQs)


Anyways, i am having an issue regarding a problem i saw. It says to calculate what capacitor must be added in parallel to a load in order to counter-act the line impedance and keep the source voltage and load voltage equal. Problem says to have it so there is zero voltage drop across the line impedance (R + X[sub]L[/sub])

The answer to the question doesn't even add to make source and load voltages equal (not even in magnitude).


So my questions are:
- can a current/impedance combination (with both R and X) exist so there is zero voltage drop?
- is it considered a voltage drop if the angle of voltage shifts, but magnitude stays the same? (eg. 277<10deg to 277<72deg)
- if the last question is true, then is there basically an infinite values of I given Z that would simply shift the voltage while keeping voltage magnitude equal?

thanks

 

[firstoption]

Also, i suppose that whether jX goes up or down, it'd still be considered a "drop," yes?

or are drops only for resistance?

 

[waross]

Yes capacitors may be used at the end of a power line to raise the voltage.
The voltage drop depends on the current.
The voltage rise depends on the capacitance.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[firstoption]

Yes capacitors may be used at the end of a power line to raise the voltage.
The voltage drop depends on the current.
The voltage rise depends on the capacitance.

none of these answer my questions

thanks

 

[Parchie]

The answers are there. You're just not being attentive. Voltage drop exists because all conductors have impedances. You can't improve on the resistive portion but you can correct the line inductance by offsetting the inductance with capacitance.

 

[HamburgerHelper]

The magnitude of the voltage drop or rise will be impacted by the power factor of the load. For example, if your load is full compensated the current to the load will be in phase with the voltage. The voltage drop will be I*Z. Just for this example, let's imagine Z is full reactive. (Z<90 deg). The voltage drop for a fully compensated load will be at 90 degrees to your source voltage. The affect on the magnitude of the voltage at the load will be small due to voltage drop being 90 degrees out of phase with the source voltage. If you overly compensate your load so that the current leads the voltage, the voltage drop will be in the opposite direction to your source voltage and cause the voltage at your load to be higher in magnitude than your source voltage. If you draw out the phasors for a overly compensated, fully compensated, and under compensated load and look at the voltage drops it will all be fairly clear.

 

[wroggent]

- can a current/impedance combination (with both R and X) exist so there is zero voltage drop? Only if I=0.
- is it considered a voltage drop if the angle of voltage shifts, but magnitude stays the same? (eg. 277<10deg to 277<72deg) Yes, in this example the voltage drop is 187.2-j215.3 V
- if the last question is true, then is there basically an infinite values of I given Z that would simply shift the voltage while keeping voltage magnitude equal? I guess so

 

[magoo2]

Start off in a simple manner. Assume the load is a combination of resistance and inductance elements. In other words, it is a lagging power factor.

Vs is the sending voltage.

The voltage drop magnitude is I*R* cos theta + I*X* sine theta

I is the load current and R and X are the series resistance and reactance of the circuit.

From the magnitude of Vs and the voltage drop, you know the load voltage.

Next go back to the same circuit without load. Now you want to add a capacitor at the load end. I in this case is the capacitor current.

The voltage drop is now - I*X. Looking at the voltage drop equation, since it is capacitive current, angle is -90 degrees so the first term drops out and the sine of -90 is -1. The sign is negative, indicating that it is a voltage rise.

You now can solve for the capacitor rating that gives you a voltage rise equal to the drop.

You will actually have to iterate because the capacitor current will vary depending on the load voltage. That is the gist of it.

 

[firstoption]

The answers are there. You're just not being attentive. Voltage drop exists because all conductors have impedances. You can't improve on the resistive portion but you can correct the line inductance by offsetting the inductance with capacitance.

ironic you say that seeing that you weren't being attentive to what exactly my questions were asking

now the questions have been mostly answered.



Thank you for everyone for the help. I think the problem was simply presented wrong (in a video) because what everyone says here makes more sense than assuming no voltage drop.


I attached the problem statement in case anyone is curious
I figure that its easier to find an "I" that works and work backwards to find the capacitor. The solution shows something completely different

 

[stevenal]

"- is it considered a voltage drop if the angle of voltage shifts, but magnitude stays the same?"

Since the problem is stated to look at magnitude, no. In general, it is the magnitude that is of interest.

Yes, you may ask.

 

[jghrist]

See thread238-314544