Sample Problem: AC Power 4

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Del curso dictado por Georgia Institute of Technology

Linear Circuits 2: AC Analysis

41 calificaciones

Georgia Institute of Technology

Linear Circuits 2: AC Analysis

41 calificaciones

This course explains how to analyze circuits that have alternating current (AC) voltage or current sources. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.

De la lección

Module 4: Power

The description goes here

4.1 Power Module Overview1:18

4.2 Root Mean Square8:29

4.3 Power Factor12:06

4.4 Power Triangles10:10

4.5 AC Power Transfer13:59

Sample Problem: AC Power 15:12

Sample Problem: AC Power 214:30

Sample Problem: AC Power 37:41

Sample Problem: AC Power 49:38

Sample Problem: AC Power 54:42

Conoce a los instructores

Dr. Bonnie H. Ferri
Professor
Electrical and Computer Engineering
Dr. Joyelle Harris
Director of the Engineering for Social Innovation Center
School of Electrical and Computer Engineering

The topic of this problem is AC Steady State Power.

The problem is to determine the power supplied by the power company to

the given 480 V load.

So we have a load, which is external to our company,

which has a voltage of 480 V at the load wall outlet, if you will.

And the power associated with that load is 88 KW.

That's the average power.

It's the real power associated with the AC steady state load.

We also have a power factor, and the power factor is 0.707 lagging.

Lagging means that the current lags a voltage.

The phase angle of the current

lags the phase angle of the voltage by a certain amount.

Remember that the power factor definition is, power factor is equal to cosine of

the angle of the voltage minus the angle of the current.

So we can use this information to determine what that difference in the two

phase angles are.

Our voltage is supplied by the power company and

a certain current is supplied by the power company.

There's a line voltage of 0.08 ohms associated

with the path from the power company to the customer,

and we'll take that into the count as well.

So, the first thing we want to find if we're looking for

the power supplied by the power company to the given 480 V load,

is we need to find the current Is.

And we're going to work in RMS terms.

Remember, the RMS, in the case of current,

if we have a sinusoidal current with an amplitude of I sub m,

then our RMS value is going to be I sub m divided by squared root of 2.

So it gives us our RMS value.

We can write a similar expression for

V sub m, and VRMS is going to be equal to

V sub m divided by square root of 2.

So now if we want to find IRMS, we're going to take information that we know,

namely that first of all, if we want to go back to our equation for average power.

We know that average power is 1/2 V sub m

I sub m cosine ( theta V- theta sub i).

And so, if we plug in RMS values for V and

I, and this VRMS IRMS times the power of factor.

Again, the power of factor consign of theta V minus theta I.

So we can use this information to find IRMS.

Therefore, IRMS is going to be equal

to power average for a load,

which is (88 KW) divided by

our VRMS, which is 480 V for

a load time times our power factor,

which is 0.707.

So if we take that expressions and solve for

IRMS then we have IRMS which is 259.3 A.

And again, this is RMS current, root mean square current.

So the power.

Supplied is going to be a combination

Of the power that's absorbed in the source,

Resistance, plus the power and the load.

So the power of the source from the lead

lines that attach the power company to

its customers is 0.08 times I squared RMS.

And that's simply using our equation that we have up above.

We know the power factor for the source, which is a resistance,

the power factor is 0.

The angle between the voltage and the current is 0.

We know that IRMS is IRMS.

And we know VRMS is IRMS times R,

which is the 0.08.

So we get this first term from our expression for the average power.

In this case, it's for our source resistance.

Then we're going to add to it the power to the load.

And the power of the load is given to us, it's 88 KW.

So, the total power that's

supplied is going to be equal to 93.38 KW.

Now, if we take the same problem and we change the power factor,

right now, our power factor is 0.707.

Let's say, that we improve the power factor and

we bring it closer to an ideal factor of one.

Ideally, our power company would be delivering its power to a purely resistive

load, meaning the community.

But in fact, that's not the way it works.

In practice we know that there is some capacitive and inductive,

which in totality end up shifting our

load away from a purely resistive load to an impedance base load.

But the closer we can get to a purely resistive network,

the more efficient our power company can be with delivering power to the customer.

Because now we're just delivering real power, or average power, and

we're not going to be delivering any of the reactive power that is

ultimately lost in heat or electromagnetic radiation or other lost mechanisms.

So let's say that we increase our power factor, we improve our power factor.

And we're going to raise it to 0.9 lagging.

It means, the current lags the voltage, and

the cosine of the angle between those two is 0.9.

So in this case, IRMS using the same

approach that we had above to finding IRMS.

Instead of 0.707 for it,

putting in 0.9 for it and solving,

we end up with IRMS which is 203.7 A.

And again, this is RMS.

So first of all, if we look at this we first of all increase or

improve our power factor from 0.7 to 0.9.

And what's that do to the amount of current that the power company has

supplied to the supplier?

It goes down, and was originally 259 A required.

Now it's only 203 A required to satisfy the same

load at the output, the same average power load.

We've improved the power factor once again.

So, what is the power that has to be supplied in this case?

So if we look at the power supplied.

The power supplied is going to be the 88 KW from the load again, because we know

that that's part of what's being absorbed at the far end of the circuit.

Plus we have, again, our 0.08 line resistance,

we haven't improved our lines.

And we're only using 203.7 A of current now.

And so we end up supplying 91.32 KW.

So the power company has a reduction in the amount of power that

they have to supply at their end in order to satisfy the needs of the load

in terms of real power if the power factory increases from 0.707 to 0.9.

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