- #1
andyskin
- 5
- 2
Please can you have a look at the file at point me in the right direction if I'm wrong.
Thank you
Thank you
A 50 kW load operates from a 60 Hz 10 kV rms line with a power factor of 60% lagging -- Determine the capacitance that must be placed in parallel
- Thread starter andyskin
- Start date
In summary, In order to correct the Power Factor from 0.60 to 0.90, you will need to find the Power Factor Correction.pdf file attached. This file has the derivation for how to equate the two equations of "XC".
- #2
anorlunda
Staff Emeritus
- 11,308
- 8,732
Welcome to PF.
I moved your post to the homework help forum. Normally, we don't allow posts in the form of all pictures in a PDF file, but your pictures are easy to read. I hope you attract some hints from our helpers.
Next time, please type in the text parts of your question and use copy/paste or the ATTACH button to include pictures of the graphics.
I moved your post to the homework help forum. Normally, we don't allow posts in the form of all pictures in a PDF file, but your pictures are easy to read. I hope you attract some hints from our helpers.
Next time, please type in the text parts of your question and use copy/paste or the ATTACH button to include pictures of the graphics.
- #3
andyskin
- 5
- 2
Sorry didn’t know there was a homework section
- #4
ammarb32
- 26
- 9
Hi,
I'm not sure what you're trying to ask in the question. Determine the capacitance that must be placed in parallel to...? What are you trying to do? In the PDF file you attached, it seems as though you're trying to correct the Power Factor from 0.60 to 0.90. However, you're also solving for the Power that needs to be supplied?
Is there a part of the question you might have forgotten to add?
If you're trying to correct the Power Factor from 0.60 to 0.90, please find the Power Factor Correction.pdf file attached. This is the method that should lead you to the correct answer if you want to determine the value of the capacitor required in order to correct the Power Factor.
I'm not sure what you're trying to ask in the question. Determine the capacitance that must be placed in parallel to...? What are you trying to do? In the PDF file you attached, it seems as though you're trying to correct the Power Factor from 0.60 to 0.90. However, you're also solving for the Power that needs to be supplied?
Is there a part of the question you might have forgotten to add?
If you're trying to correct the Power Factor from 0.60 to 0.90, please find the Power Factor Correction.pdf file attached. This is the method that should lead you to the correct answer if you want to determine the value of the capacitor required in order to correct the Power Factor.
Last edited:
- #5
andyskin
- 5
- 2
the full question meant to read:
A 50 kW load operates from a 60 Hz 10 kV rms line with a power factor of 60% lagging. Determine the capacitance that must be placed in parallel with the load to achieve a 90% lagging power factor.
A 50 kW load operates from a 60 Hz 10 kV rms line with a power factor of 60% lagging. Determine the capacitance that must be placed in parallel with the load to achieve a 90% lagging power factor.
- #6
ammarb32
- 26
- 9
I see. Then, in that case, you just need to change the ending of your method. You need to equate the two equations of "XC" (Finding the reactance that's coming from the capacitor). I attached the derivation of that in the Equations.PNG file. The reason for this is because your "Active Power" (P) is not changing when you change the Power Factor. However, your "Reactive Power" (Q) is changing with the different Power Factors.andyskin said:
Capacitors supply reactive power "Q" to the circuit. Therefore, you need to use the "change in reactive power" to calculate the capacitance value needed to supply that amount of "Q".
Attachments
Last edited:
- #7
andyskin
- 5
- 2
thank you for your help, i'll have a look
1. What is the purpose of determining the capacitance in this scenario?
The purpose of determining the capacitance is to improve the power factor of the load, which is a measure of how efficiently the load uses the electricity from the power source. A low power factor can lead to higher energy costs and can also cause strain on the power grid.
2. How is the power factor calculated in this situation?
The power factor is calculated by dividing the real power (in this case, 50 kW) by the apparent power (which is the product of the voltage and current, or 10 kV and the square root of 3). This gives a power factor of 60% lagging, which means the load is using 60% of the available power and the remaining 40% is being wasted.
3. Why is the power factor lagging in this scenario?
The power factor is lagging because the load is inductive, meaning it requires more reactive power (in the form of a magnetic field) to operate. This causes the current to lag behind the voltage, resulting in a lower power factor.
4. How does adding capacitance improve the power factor?
Adding capacitance in parallel with the load helps to counteract the inductive effect and bring the power factor closer to 1 (or 100%). This is because capacitance produces a leading current, which balances out the lagging current from the inductive load.
5. What is the formula for calculating the required capacitance in this scenario?
The formula for calculating the required capacitance is: C = (Q x tan φ) / (2πfV2), where Q is the reactive power (in this case, 40 kVAr), φ is the power factor angle (60 degrees), f is the frequency (60 Hz), and V is the voltage (10 kV). Plugging in these values gives a required capacitance of approximately 26.7 microfarads.
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