Problem with AC Circuits and power

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HBK

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Two machines and a lighting load are connected in parallel to a 220 V, 60 Hz line. The first machine uses 40 kW of power at 0.81 power factor lagging. The second machine uses 27 kVA at 0.93 power factor leading. The lighting load consumes 13 kW. What is the current in the line feeding the three loads? What kVAR value of capacitance connected across the line is necessary to obtain a power factor of 0.98 lagging? With this capacitor installed, what will be the current flowing in the line feeding the four loads?

Could any one help me with this one?

Thanks!!
 
mivey

mivey

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Why don't you tell us what you got first? Most of us do not just provide test question answers without some input from the poster. How are you tackling the problem and what approaches have you tried? What answers did you get?
 
H

HBK

Member
jumper said:
The answer to all questions: 42.:)
Click to expand...

mivey said:
Why don't you tell us what you got first? Most of us do not just provide test question answers without some input from the poster. How are you tackling the problem and what approaches have you tried? What answers did you get?
Click to expand...

all of them are 42??

I have no idea how to even start the problem. I do know that the best way to solve this problem is by doing something with triangles, but thats about it. Any help would be appreciated.
 
david luchini

david luchini

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HBK said:
I have no idea how to even start the problem. I do know that the best way to solve this problem is by doing something with triangles, but thats about it. Any help would be appreciated.
Click to expand...

I'd suggest starting by figuring out the load currents for each of the three loads and adding them up to find the line current, and the power factor for that load.

Unless my math is wrong (and odds are it probably is) I get a line current of 365.44 Amps; adding a 3.168KVar capacitor to obtain a 0.98 lagging power factor; and a new line current of 362.31 Amps.
 
G

G._S._Ohm

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DC area
HBK said:
Two machines and a lighting load are connected in parallel to a 220 V, 60 Hz line. The first machine uses 40 kW of power at 0.81 power factor lagging.
Click to expand...
Lagging, so it's inductive?

I'd start this way.
Pavg = EI{Cos A}
PF = Cos A = 0.81
so I = Pavg/(E{0.81}) = 40,000/(220{0.81}) = 224.47 A.

BTW, arc cos 0.81 = 35.9 degrees so this is a Z of some magnitude at this angle.
 
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wiigelec

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Red Desert
The first thing you will want to do is draw a circuit diagram and then draw a phasor diagram representing the true and apparent power of the loads. From there you should be able to calculate the currents. Remember the currents will also be represented in a phasor diagram...
 
charlie b

charlie b

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jumper said:
The answer to all questions: 42.
Click to expand...
Not only is it the answer to the ultimate question of life, the universe, and everything, but it has also provided an answer to two other questions:
1. What do you get when you multiply six by nine?
and
2. How many roads must a man walk down?
 
kingpb

kingpb

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SE USA as far as you can go
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Engineer, Registered
HBK said:
Two machines and a lighting load are connected in parallel to a 220 V, 60 Hz line. The first machine uses 40 kW of power at 0.81 power factor lagging. The second machine uses 27 kVA at 0.93 power factor leading. The lighting load consumes 13 kW. What is the current in the line feeding the three loads? What kVAR value of capacitance connected across the line is necessary to obtain a power factor of 0.98 lagging? With this capacitor installed, what will be the current flowing in the line feeding the four loads?

Could any one help me with this one?

Thanks!!
Click to expand...

In any problem, first thing is to decipher what you are looking for. In this case the system current, NOT each load current.

Second thing to do is state your assumptions, i.e. 220V is line-line/single-phase, the lighting load is operating at 1.0 pf, and the loads always operate simultaneously.

Next, calculate your system current adding up the load KVA values, then divide by the voltage. Not so straight forward in your problem because you do have a leading pf for one load, so, lagging and leading come into play.

Determine the KVAR from each KVA. This is where it gets tricky because you need to understand the power triangle/trigonometry/complex numbers/vectors.

KVA = P + jQ where KVA is a vector

and,

KVAR = KVA * (sin(acos(pf)) (lagging)
KVAR = KVA * (sin(-acos(pf)) (leading)

To determine the "system" pf and current:
KVA = (40+25+13)KW + j(29-10-0)KVAR
KVA = 78KW + 19KVAR
|KVA| = sqrt(78KW^2 + 19KVAR^2)
|KVA| = 80.3KVA ( This is the magnitude)

I = 80.3KVA/220V
I = 365A (confirms david luchini post)

The system pf is therefore;

pf = KW/KVA
pf = 0.97

In the next step you need to improve your PF by going from 0.97 to 0.98

Take desired system pf and determine system KVAR, then subtract old KVAR from new KVAR. Another way to look at it is the amount of capacitance (-KVAR) that you need to add, noticing that the KVAR "added" is a negative (capacitance) value.

KVAR added = KVAR(new) - KVAR(old) = 16KVAR - 19KVAR = -3KVAR or 3KVAR of capacitance

Little play on words in the problem to trip up the unwary in saying determine the current in the four loads, as they are meaning the capacitor as the "fourth" load.

KVA = (40+25+13+0)KW + j(29-10-0-3)KVAR
KVA =78KW + 16KVAR
|KVA| = 79.6KVA

Current is calculated the same as earlier;

I = 79.6KVA/220V
I = 362A (confirms david luchini post)

That is a lot of cost to go from what would already be considered a very good PF, to a new very good PF, and only reduce amps by 3. But hey, nobody ever said book problems are practical..........
 
H

HBK

Member
kingpb said:
In any problem, first thing is to decipher what you are looking for. In this case the system current, NOT each load current.

Second thing to do is state your assumptions, i.e. 220V is line-line/single-phase, the lighting load is operating at 1.0 pf, and the loads always operate simultaneously.

Next, calculate your system current adding up the load KVA values, then divide by the voltage. Not so straight forward in your problem because you do have a leading pf for one load, so, lagging and leading come into play.

Determine the KVAR from each KVA. This is where it gets tricky because you need to understand the power triangle/trigonometry/complex numbers/vectors.

KVA = P + jQ where KVA is a vector

and,

KVAR = KVA * (sin(acos(pf)) (lagging)
KVAR = KVA * (sin(-acos(pf)) (leading)

To determine the "system" pf and current:
KVA = (40+25+13)KW + j(29-10-0)KVAR
KVA = 78KW + 19KVAR
|KVA| = sqrt(78KW^2 + 19KVAR^2)
|KVA| = 80.3KVA ( This is the magnitude)

I = 80.3KVA/220V
I = 365A (confirms david luchini post)

The system pf is therefore;

pf = KW/KVA
pf = 0.97

In the next step you need to improve your PF by going from 0.97 to 0.98

Take desired system pf and determine system KVAR, then subtract old KVAR from new KVAR. Another way to look at it is the amount of capacitance (-KVAR) that you need to add, noticing that the KVAR "added" is a negative (capacitance) value.

KVAR added = KVAR(new) - KVAR(old) = 16KVAR - 19KVAR = -3KVAR or 3KVAR of capacitance

Little play on words in the problem to trip up the unwary in saying determine the current in the four loads, as they are meaning the capacitor as the "fourth" load.

KVA = (40+25+13+0)KW + j(29-10-0-3)KVAR
KVA =78KW + 16KVAR
|KVA| = 79.6KVA

Current is calculated the same as earlier;

I = 79.6KVA/220V
I = 362A (confirms david luchini post)

That is a lot of cost to go from what would already be considered a very good PF, to a new very good PF, and only reduce amps by 3. But hey, nobody ever said book problems are practical..........
Click to expand...


dude you are beyond awesome!!

Thanks a lot bro!
saved my life!
 
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