Network, 2 loads, find complex power

[Color_of_Cyan]

Homework Statement

http://imageshack.us/a/img819/9791/sh5k.jpg

Given the network [above], find the complex power supplied by the source, the power factor of the source, and the voltage Vs(t). The frequency is 60 Hz.

Homework Equations

ω=1/f

Pf = cosø

P = IV

P_L = (V_rms)*(I_rms)*(Pf)

The Attempt at a Solution

Any hints on where to start would be helpful... not too sure about how to get & distinguish the different types of power

ω = 1/f = 0.0166

Apparent power = 12kVA ?

Z = (0.05 + 0.2j)Ω

cos^-10.7 = 45.5° for the first load and
cos^-10.9 = 25.8° for the other load

not sure what to do from here

 

[tiny-tim]

Hi Color_of_Cyan!

.. not too sure about how to get & distinguish the different types of power

Apparent power = S = V_r.m.sI_r.m.s

Average power = Scosφ

Reactive power = Ssinφ

Complex power = Se^jφ = S(cosφ + jsinφ)

 

[Color_of_Cyan]

Hi Color_of_Cyan! Apparent power = S = V_r.m.sI_r.m.s

Average power = Scosφ

Reactive power = Ssinφ

Complex power = Se^jφ = S(cosφ + jsinφ)

Ok, if i have

PL = (Vrms)*(Irms)*(Pf)

and the power factor is given for the load, can I find the current through that load first then?

(PL)/(Pf*V_rms) = (_Irms)

= 20000/(0.7*240)

is the current (119 ∠ 0)A for this load?

(and isn't VA the same as watts?) and your sig is missing the '∠ ' symbol

 

[tiny-tim]

Ok, if i have

PL = (Vrms)*(Irms)*(Pf)

and the power factor is given for the load, can I find the current through that load first then?

(PL)/(Pf*V_rms) = (_Irms)

= 20000/(0.7*240)

is the current (119 ∠ 0)A for this load?

no, 119 A is I_rms

I is I_maxcos(ωt + φ)

(φ is the phase, cosφ is the power factor)

I_rms = I_max/√2 … so knowing I_rms tells you nothing about φ

(and isn't VA the same as watts?)

yes

and your sig is missing the '∠ ' symbol

ooh, that's a good idea, i'll add it!

 

[rude man]

(and isn't VA the same as watts?)

No. VA is volts times amps. W is the product of volts times amps times the power factor.

Anyway, I don't understand your diagram. What's the "240V < 0 deg rms" mean? Is that the voltage on the right-hand side?

 

[Color_of_Cyan]

Anyway, I don't understand your diagram. What's the "240V < 0 deg rms" mean? Is that the voltage on the right-hand side?

Yes that's what it looks like (even though I could have copied it over better); I think it's the drop across both of the "loads" if they are in parallel like that.

no, 119 A is I_rms

I is I_maxcos(ωt + φ)

(φ is the phase, cosφ is the power factor)

I_rms = I_max/√2 … so knowing I_rms tells you nothing about φ

Thanks. I don't see what would be good about knowing that angle though. I am trying to find the total current first and then the "impedances" of the loads... wouldn't that help find Vs?

No. VA is volts times amps. W is the product of volts times amps times the power factor.

So instead of (PL)/(Pf*Vrms) = (Irms)

is it just (PL)/(Vrms) = (Irms)?

 

[tiny-tim]

Thanks. I don't see what would be good about knowing that angle though. I am trying to find the total current first and then the "impedances" of the loads... wouldn't that help find Vs?

"that angle" is the phase, the amount of lead or lag

the total complex current (is that what you mean by total current"? ) is:

I_max ∠ φ​

if you want impedance, you need to know phase

 

[rude man]

I would start by detrmining the actual components represented by the two blocks, then do a KVL or whatever to determine what the source has to be.

 

[Color_of_Cyan]

"that angle" is the phase, the amount of lead or lag the total complex current (is that what you mean by total current"? ) is:

I_max ∠ φ​

if you want impedance, you need to know phase

So would there be separate phases for each of the "blocks" then? I had cos^-10.7 = 45.5° and cos^-10.9 = 25.8° for the blocks, would these be the phases for them (be it for impedance & current)? Not that good with phases here (yet).

 

[tiny-tim]

So would there be separate phases for each of the "blocks" then?

yes (as the diagram shows)

I had cos^-10.7 = 45.5° and cos^-10.9 = 25.8° for the blocks, would these be the phases for them (be it for impedance & current)?

yes (except I'm not familiar with this decimal notation: i don't know whether it's cos^-1 or sin^-1)

and of course + for leading and - for lagging

 

[gneill]

If I may offer a suggestion... Look up "Power Triangle". It's a way to graphically represent the various power components (Real, Reactive, and Apparent). It also specifies their relationships to each other and the power factor.

This problem specifies two loads: one via real power and power factor, the other via apparent power (kVA rather than Watts) and power factor. You can drop the known values onto a power triangle and determine the related values via Pythagoras.

 

[Color_of_Cyan]

yes (as the diagram shows) yes (except I'm not familiar with this decimal notation: i don't know whether it's cos^-1 or sin^-1) and of course + for leading and - for lagging

It was for power factor & the angle. For the complex current of the leftmost block then I got

mag = 119 * √2 = 168.2

then the angle = cos^-10.7 = 45.5°

so would the complex current for it be (168,2∠45.5°)A ?

for the other block would it be

I rms = 12000/240

then 50 * √2 = 70.7

then cos^-10.9 = 25.8

so would the complex current for the other block be (70.7 ∠ 25.8)A ?

 

[NascentOxygen]

It was for power factor & the angle. For the complex current of the leftmost block then I got

mag = 119 * √2 = 168.2

then the angle = cos^-10.7 = 45.5°

so would the complex current for it be (168,2∠45.5°)A ?

for the other block would it be

I rms = 12000/240

then 50 * √2 = 70.7

then cos^-10.9 = 25.8

so would the complex current for the other block be (70.7 ∠ 25.8)A ?

Almost.

How are you going to designate the angle to show 'lagging'?

It is probably the convention that voltages and currents are denoted by their RMS values, so there is no need to convert to peak values (since you invariably are going to convert back to RMS values in the end).

Now that you know both currents, sum them to give the current through the L and R elements representing the transmission line.

 

[Color_of_Cyan]

Nice, thanks. I still don't know how to even tell between lead or lag though.

Converting to rect. form they are:

(117.9 + 120j)A

and

(63.6 + 30.77j)A

so I_tot = 181.5 + 150.77j

I_tot = (236 ∠ 39.7)AThe impedance of the other 2 elements is (0.05 + 0.2j)Ω = ( 0.206 ∠ 75.96)Ω

so V = IZ

= ( 0.206 ∠ 75.96)Ω * (236 ∠ 39.7)A

= (48.61 ∠ 115.6)V

= (-21 + 43.8j) - why is the real component negative?

Adding the two voltages now (240V being the other one) I get:

Vtotal = 219 + 43.8j

=(223 ∠ 11.3)V

Would this be the same as Vs?

 

[NascentOxygen]

Nice, thanks. I still don't know how to even tell between lead or lag though.

A lagging power factor indicates the load is inductive. It is a property of an inductor that the current lags the voltage. The angle of current relative to voltage will lie between 0° and -90°.

The converse applies to a leading power factor, and is an indication of the load being capacitive. The angle of current relative to voltage now lies between 0° and +90°.

Converting to rect. form they are:

(117.9 + 120j)A

and

(63.6 + 30.77j)A

One of the above won't be quite right. You need to treat one as leading and the other lagging. You diagram in the first post indicates this difference.

 

[Color_of_Cyan]

A lagging power factor indicates the load is inductive. It is a property of an inductor that the current lags the voltage. The angle of current relative to voltage will lie between 0° and -90°.

The converse applies to a leading power factor, and is an indication of the load being capacitive. The angle of current relative to voltage now lies between 0° and +90°.


One of the above won't be quite right. You need to treat one as leading and the other lagging. You diagram in the first post indicates this difference.

But the currents both have positive angles there in the current phasors... how do you know if the angle is negative then (if that is what you were implying) ?

 

[tiny-tim]

But the currents both have positive angles there in the current phasors... how do you know if the angle is negative then (if that is what you were implying) ?

Because (as NascentOxygen says) the diagram says that one is leading and the other is lagging …

it's like a diagram that says that two velocities are 30 mph north and 70 mph south … even though they're both positive, you know that they're opposite because it says so!

 

[Color_of_Cyan]

Okay okay, so I guess the current goes from + to - (on the diagram for the 240V drop) where it "leads" then?

 

[NascentOxygen]

Okay okay, so I guess the current goes from + to - (on the diagram for the 240V drop) where it "leads" then?

You "guess"?! Who guesses when you can google?

On the phasor diagram, it's the Imaginary component of the current which is drawn pointing in the negative direction to show lagging current. The Real component is not drawn any differently, whether for leading or lagging conditions. It's solely the Imaginary component that determines the lead or lag. http://www.labsanywhere.net/circuit/lectures/lect12/lecture12.php

 

[Color_of_Cyan]

Well that helped to see it, so the currents would then be

0.7 leading pf block: (168,2∠45.5°)A = (117.9 + 120j)A

and now for the lagging block:

0.9 lagging pf block: (70.7 ∠ -25.8)A = (63.6 -[/color] 30.77j)A
which would mean

I_tot = (181.5 - 89.23j)A

I_tot = (202 ∠ -26.1°)Aimpedance of the other 2 elements is still (0.05 + 0.2j)Ω = ( 0.206 ∠ 75.96)Ω

so V = IZ

= ( 0.206 ∠ 75.96)Ω * (202 ∠ -26.1°)A= (41.612 ∠ 49.86)V = (26.82 + 31.8j)V

so V_tot = Vs(t) = (266.82 + 31.8j)V then?Then apparent power is finding V_{rms * Irms. Would that just be the magnitudes of Vtot and Itot multiplied out?}

 

[NascentOxygen]

That's looking a tiny bit better. I haven't checked all your maths, but where your leading current exceeds the lagging component, shouldn't the sum of them be still leading?

The magnitude of the supply voltage is of major interest, as well as its power factor. The VA is just supply voltage x current magnitude.

 

[Color_of_Cyan]

Ugh, my mistake.

I total is really

I_tot = (181.5 + 89.23j)A

I_tot = (202 ∠ 26.1°)A

and again impedance of the other 2 elements is still (0.05 + 0.2j)Ω = ( 0.206 ∠ 75.96)Ω

so V = IZ, now

= ( 0.206 ∠ 75.96)Ω * (202 ∠ 26.1°)A= (41.612 ∠ 102)V = (-8.65 + 40.7j)V

so V_tot = Vs(t) = (231.35 + 40.7j)V = (235 ∠ 9.97)V thenDo you mean the apparent power comes out as the product of magnitudes of the phasors as in

= (235V * 202A)? (= 47470 VA) ?

 

[gneill]

Unless specifically asked for peak values or time domain equations for the voltage or current ( for example: $V_{max} cos(\omega t + \phi)$ ) you should be keeping all your voltages and currents as RMS values when doing power calculations. There's no need to switch back and forth between peak and RMS as you go along, and you run the risk of mixing peak and RMS values in calculations.

In your calculation for Vs, you've used a peak current to calculate a potential drop across the series impedance, then added that result to the given 240 Vrms...

 

[NascentOxygen]

That's looking better. Yes, apparent power is without regard to power fsctor. So you have 47.47 kVA there.

 

[Color_of_Cyan]

Unless specifically asked for peak values or time domain equations for the voltage or current ( for example: $V_{max} cos(\omega t + \phi)$ ) you should be keeping all your voltages and currents as RMS values when doing power calculations. There's no need to switch back and forth between peak and RMS as you go along, and you run the risk of mixing peak and RMS values in calculations.

In your calculation for Vs, you've used a peak current to calculate a potential drop across the series impedance, then added that result to the given 240 Vrms...

So you can also convert that 240V to max value too then?

It would be 240 * √2 = 339V

You could still add the magnitudes of the drops that way too then?

So the drop of the parallel impedance was still

= (41.612 ∠ 102)V = (-8.65 + 40.7j)V

but you could still add the max value in magnitude to get

V_tot = Vs(t) = (330 + 40.7j)V = (332 ∠ 7)V


Now from my last post the apparent power from phasor magnitudes would be

(332V * 202A) = 67k VA


Sorry I keep making all these mistakes..

 

[gneill]

So you can also convert that 240V to max value too then?

It would be 240 * √2 = 339V

You could still add the magnitudes of the drops that way too then?

So the drop of the parallel impedance was still

= (41.612 ∠ 102)V = (-8.65 + 40.7j)V

but you could still add the max value in magnitude to get

V_tot = Vs(t) = (330 + 40.7j)V = (332 ∠ 7)V


Now from my last post the apparent power from phasor magnitudes would be

(332V * 202A) = 67k VA


Sorry I keep making all these mistakes..

No! Don't use max values at all. Use RMS values throughout. Power calculations demand RMS values unless you are specifically told to find instantaneous peak values.

 

[Color_of_Cyan]

Okay, thanksSo the drop of the parallel impedance was still

V_drop rms = (41.612 ∠ 102)V * √2

V_drop rms = (58.84 ∠ 102)V = (-12.23 + 57.56j)VV_tot rms = (227.77 + 57.56j)V = (122 ∠ 192)V

So you have to display Vs(t) (total voltage) as the peak value then?

 

[gneill]

I think your Vdrop calculation is twice what it should be. Could be you've still got some peak values mixed into the current and impedance calculations, and the √2's are combining to make a 2x multiplier.

State your two load currents in RMS to begin with. What are the values?

 

[Color_of_Cyan]

Ok, trying again

(PL)/(Pf*Vrms) = (Irms)

I_rms = 20000W/(0.7*240V)

I_rms = (119 ∠ 0)A for the 0.7 leading power factor

and

I_rms = (12000VA/240V)

I_rms = (50 ∠ 0)A for the other oneI'll get to the rest of the problem again later

 

[NascentOxygen]

The currents won't have an angle 0°. That would mean they are exactly in phase with the voltage (which I assume is being taken as the reference).