Need Help with this Question; Power Factor

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schicco

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New York
I am taking electrical classes. I need help with this problem.

A balanced 3 phase, 60 Hz load consists of machinery of 500 kva at 0.5 PF lagging, and a 3 phase resistive load 300 Kw. The two loads are supplied by a 4160 volts feeder.

a. What is the magnitude of the feeder current?
b. Specify the number and ratings of capacitors required if it is desired to improve the feeder power factor to 0.8. Assume caps will be connected in Y.

Thanks, Steve
 
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Smart $

Smart $

Esteemed Member
Location
Ohio
I'll attempt to get you headed in the right direction. The key to answering the questions is to know how to determine combined current through vector math. Being a balanced load makes the problem less complex than it could be. You'll only have to determine the current of one line, as the magnitude of the other two will be the same , ?120?.

First determine the current of each load. For the resistive load kW = kVA. The formula is simple I = kVA ? E. The resistive load has a power factor of 1.0, so it neither leads nor lags, so assign it an angle of 0? (Line 1 or A). For the machinery load, current angle is directly related to power factor: theta = arccos(PF).

Vector add the two currents:
_A @ 0?
_A @ -[arccos(PF)]?​
+_______________
_A @ _?​
 
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Smart $

Smart $

Esteemed Member
Location
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Here's a rough plot of the vectors...

vectors.gif
 
Sahib

Sahib

Senior Member
Location
India
schicco said:
Thanks a lot Smart $ for the help.
Click to expand...
Complex notation of vectors dispenses with graphical solution and problem can be solved easily algebraically then. Here is the way. Find the KW and KVAR of the load: 250KW and 433KVR. So the load KVA in complex notation=250+j433. For resistors:300+j0. So the total KVA=550+J433. Its magnitude=9010. So the current is 9010/1.732*4.16=1.25KA.
Find the theta of cos0.5. Let it be theta1. Find the theta of cos0.8. Let it be theta2.
The KVAR of the capacitor bank=550*[tan(theta1)-tan(theta2)].
 
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Smart $

Smart $

Esteemed Member
Location
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Sahib said:
Complex notation of vectors dispenses with graphical solution and problem can be solved easily algebraically then. Here is the way. Find the KW and KVAR of the load: 250KW and 433KVR. So the load KVA in complex notation=250+j433. For resistors:300+j0. So the total KVA=550+J433. Its magnitude=9010. So the current is 9010/1.732*4.16=1.25KA.
Find the theta of cos0.5. Let it be theta1. Find the theta of cos0.8. Let it be theta2.
The KVAR of the capacitor bank=550*[tan(theta1)-tan(theta2)].
Click to expand...
But the power factor of the feeder is not 0.5. The 0.5 power factor is only for one out the two loads. Need the power factor with both loads: cos[arctan(433/550)]... so tan(theta1) is simply 433/550.
 
Smart $

Smart $

Esteemed Member
Location
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Sahib said:
Complex notation of vectors dispenses with graphical solution and problem can be solved easily algebraically then. Here is the way. Find the KW and KVAR of the load: 250KW and 433KVR. So the load KVA in complex notation=250+j433. For resistors:300+j0. So the total KVA=550+J433. Its magnitude=9010. So the current is 9010/1.732*4.16=1.25KA.
Find the theta of cos0.5. Let it be theta1. Find the theta of cos0.8. Let it be theta2.
The KVAR of the capacitor bank=550*[tan(theta1)-tan(theta2)].
Click to expand...
Magnitude is 700kVA. Current is 700kVA/4.16kV/sqrt(3)=97.2A
 
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G

GoldDigger

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Retired PV System Designer
VA is never a vector, so magnitude is not a correct term.
VA is itself just a scalar (number).
You can get a scalar by multiplying two magnitudes (VA) or by taking the dot product of two vectors (giving actual watts.)
If you multiply the magnitude of the voltage by the vector current you get an expression which is more confusing than helpful.
Sent from my XT1080 using Tapatalk
 
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Smart $

Smart $

Esteemed Member
Location
Ohio
GoldDigger said:
VA is never a vector, so magnitude is not a correct term.
Click to expand...
I agree VA is not a vector, but the term magnitude isn't always associated with a vector. One of its definitions is simply "number". So it really depends on the vernacular in use.

That said, Sahib was demonstrating a shortcut method to determine kW and kVAR without extracting I from each value, working the equations, then converting back to VA/W values Essentially it is just doing calculations with I scaled by E.
 
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schicco

Member
Location
New York
Sahib said:
Complex notation of vectors dispenses with graphical solution and problem can be solved easily algebraically then. Here is the way. Find the KW and KVAR of the load: 250KW and 433KVR. So the load KVA in complex notation=250+j433. For resistors:300+j0. So the total KVA=550+J433. Its magnitude=9010. So the current is 9010/1.732*4.16=1.25KA.
Find the theta of cos0.5. Let it be theta1. Find the theta of cos0.8. Let it be theta2.
The KVAR of the capacitor bank=550*[tan(theta1)-tan(theta2)].
Click to expand...

Sahib
How did you get 9010? I thought 550+J433 should give us (sqrt of)550?+433? = 699.99 KVA
 
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