AC power - Simulation and hand calculated values different

[eehelp150]

Homework Statement

Homework Equations

P=I*(I*)*R/2

The Attempt at a Solution

R1 = 12
R2 = 20
L = 0.0531H
C = 265.26*10^(-6)F
W = 2*pi*60
L -> jwL = 20j
C = 1/(jwc) = -j10

Mesh equations:
Mesh 1 (left mesh, clockwise)
-Vs +I1(R1+L+C)-I2(C)=0

Mesh2 (right mesh, clockwise)
I2(C+R2)-I1(C)=0

I1=6.78-j5.09
I2=-0.68-j3.73

Hand calculated:
PowerR1 = (I1 * I1conjugate *R1)/2 = 431W
PowerR2 = (I2*I2conjugate*R2)/2 = 143W

PowerL = (I1 * I1conjugate *L)/2=j719.87VAR
PowerC = -j297.79VARWhen I simulate in MATLAB, I get different results for reactive power of L and C:
Matlab provides these following RMS Voltages:
VL = 119
VC = 53.53

PowerL = $\frac{VL^2}{L}$ = -j719.87VAR
PowerC = $\frac{V_C^2}{C}$=j297.79VAR

Is my MATLAB simulation wrong or my hand worked calculations wrong?

 

[gneill]

What value are you using for the source voltage?

 

[eehelp150]

What value are you using for the source voltage?

Completely forgot that, sorry.
Voltage Source is $120\sqrt2cos(2\pi60)$

 

[gneill]

Completely forgot that, sorry.
Voltage Source is $120\sqrt2cos(2\pi60)$

Okay, so it's a 120 V rms source. You'll want to use rms if you're doing power calculations.

 

[eehelp150]

Okay, so it's a 120 V rms source. You'll want to use rms if you're doing power calculations.

The formula I'm using accounts for RMS, doesn't it? The divided by 2 should account for that.

 

[gneill]

The formula I'm using accounts for RMS, doesn't it? The divided by 2 should account for that.

I suppose it would. Personally I find it simpler to just use 120V in the calculations and stick to P = VI*, where P is the complex power.

 

[eehelp150]

I suppose it would. Personally I find it simpler to just use 120V in the calculations and stick to P = VI*, where P is the complex power.

Is there any reason I'm getting 90degrees (+i) when I calculate it by hand and -90degrees(-i) when I use the measured values from MATLAB?
Note: For "measured" values of power, I am squaring the RMS voltage given by Matlab and dividing by the impedance.

 

[gneill]

From what I can see, your RMS voltages from Matlab don't include their phase angles. So only the impedance angle is contributing to the calculation.

 

[eehelp150]

From what I can see, your RMS voltages from Matlab don't include their phase angles. So only the impedance angle is contributing to the calculation.

Would it be 'acceptable' to exclude the phase angles? If I recall correctly they are irrelevant without context, right?

 

[gneill]

Would it be 'acceptable' to exclude the phase angles? If I recall correctly they are irrelevant without context, right?

I think it would depend upon what you want to do with the result. If you were determining the power factor for a circuit you'd probably want to know if it were leading or lagging. If you just want to know the magnitude of the reactive power in a component, losing the angle wouldn't matter.

 

[eehelp150]

I think it would depend upon what you want to do with the result. If you were determining the power factor for a circuit you'd probably want to know if it were leading or lagging. If you just want to know the magnitude of the reactive power in a component, losing the angle wouldn't matter.

For percent error I could ignore it right?
The assignment is to solve the circuit by hand and get the real/reactive power for each element.
Then use MATLAB to solve the circuit and get the real/reactive power for each element
Then calculate percent error

 

[gneill]

For percent error I could ignore it right?

Sure, since I presume you'll be comparing the magnitudes.

 

[eehelp150]

Sure, since I presume you'll be comparing the magnitudes.

If I want the total reactive power, I just add Q1 and Q2 right?
Do I ignore the phases as well for that?

 

[gneill]

If I want the total reactive power, I just add Q1 and Q2 right?
Do I ignore the phases as well for that?

I wouldn't ignore the phases for the total. The reactive powers being out of phase will at least partially cancel.

Instead, why not use the complex current supplied by the source and determine the total complex power? Then you can pull out the real and reactive components.

 

[eehelp150]

I wouldn't ignore the phases for the total. The reactive powers being out of phase will at least partially cancel.

Instead, why not use the complex current supplied by the source and determine the total complex power? Then you can pull out the real and reactive components.

How would I get total complex power using MATLAB? I can get the RMS Current of the source, but I wouldn't have any degree, only magnitude.

 

[gneill]

How would I get total complex power using MATLAB? I can get the RMS Current of the source, but I wouldn't have any degree, only magnitude.

You used mesh analysis did you not? That gives you the complex current in the first loop...

 

[eehelp150]

You used mesh analysis did you not? That gives you the complex current in the first loop...

Yes, but the professor wants the Total Real/Reactive power measured with MATLAB as well as by hand. I already have the by hand part.

 

[gneill]

I don't have any idea how the Matlab portion was implemented, so I can't really advise. Did you write equations or run some kind of simulation?

 

[eehelp150]

I don't have any idea how the Matlab portion was implemented, so I can't really advise. Did you write equations or run some kind of simulation?

I am using Simulink + SimPowerSystems. I think I am only able to get the RMS values.
Edit:
Nevermind. I found a power measurement tool.

 

[gneill]

I am using Simulink + SimPowerSystems. I think I am only able to get the RMS values.
Edit:
Nevermind. I found a power measurement tool.

Cool. Sorry I can't be of more help there, I've never used Matlab.

 

[eehelp150]

Cool. Sorry I can't be of more help there, I've never used Matlab.

No worries. I really appreciate all the help you've given me this semester. You are literally (figuratively) a life saver.

 

[eehelp150]

Cool. Sorry I can't be of more help there, I've never used Matlab.

Quick question: To reduce reactive power consumption, we would add a larger capacitor, right?

 

[gneill]

Quick question: To reduce reactive power consumption, we would add a larger capacitor, right?

Not necessarily. It depends on how the impedances combine in the circuit, and which side of the natural frequency the circuit is operating. You might find that there's a "sweet spot" for the capacitance that minimizes the reactive impedance that's not where you'd expect from the usual power factor correction approach. Usually there the idea is to stick a new capacitor in parallel with the original circuit, not change a capacitor that's already buried in the circuitry.

I presume that Matlab will let you plot a function fairly easily. Why not write a function that gives you the complex impedance as a function of the value of C in microfarads. Plot the imaginary component versus C. I think you'll find a minimum reactance when C is smaller than the original value (I estimate somewhere around 50 μF).

 

[eehelp150]

Not necessarily. It depends on how the impedances combine in the circuit, and which side of the natural frequency the circuit is operating. You might find that there's a "sweet spot" for the capacitance that minimizes the reactive impedance that's not where you'd expect from the usual power factor correction approach. Usually there the idea is to stick a new capacitor in parallel with the original circuit, not change a capacitor that's already buried in the circuitry.

I want to reduce reactive power by 50%
This is my work:
$Q_{original}=430.8$
$Q_{desired}=215.4$
$Q_{delta}=215.4$
$X_C=\frac{V_{rms}^2}{Q_{delta}}=\frac{120^2}{215.4}$
$C=\frac{1}{X_C*2*pi*60}=3.967e-5$
Where would I stick this capacitor in the circuit to get my desired reactive power?

I presume that Matlab will let you plot a function fairly easily. Why not write a function that gives you the complex impedance as a function of the value of C in microfarads. Plot the imaginary component versus C. I think you'll find a minimum reactance when C is smaller than the original value (I estimate somewhere around 50 μF).

Don't really know how to plot with Simulink...

 

[gneill]

Where would I stick this capacitor in the circuit to get my desired reactive power?

Presumably across the source.

Don't really know how to plot with Simulink...

Matlab itself should let you write and plot functions, no? Your total circuit impedance is just:

$Z = R1 + ZL + \frac{R2*ZC}{R2+ZC}$

where: $ZC = \frac{1}{jω C}$

 

[eehelp150]

Presumably across the source.

Matlab itself should let you write and plot functions, no? Your total circuit impedance is just:

$Z = R1 + ZL + \frac{R2*ZC}{R2+ZC}$

where: $ZC = \frac{1}{jω C}$

For some reason the simulator will not let me put a capacitor in parallel with the source.

 

[gneill]

For some reason the simulator will not let me put a capacitor in parallel with the source.

Strange. Try putting a very small resistance in series with the capacitor. Something like 10^-6 Ohms.

 

[eehelp150]

Strange. Try putting a very small resistance in series with the capacitor. Something like 10^-6 Ohms.

That works! Would there be any way to calculate a different value for the original capacitor instead of adding a capacitor?

 

[gneill]

That works! Would there be any way to calculate a different value for the original capacitor instead of adding a capacitor?

Yes, but I imagine that the algebra could get tedious. You want to find the reactive power as a function of C, so that means finding the imaginary part of the complex power. It would be easier to plot the reactive power and pick out the solution on the plot.

 

[eehelp150]

Yes, but I imagine that the algebra could get tedious. You want to find the reactive power as a function of C, so that means finding the imaginary part of the complex power. It would be easier to plot the reactive power and pick out the solution on the plot.

Ok, thanks!
Going back to this:
$Z = R1 + ZL + \frac{R2*ZC}{R2+ZC}$
Total complex power would be V^2/Z, where Z is the above right?

How would I solve this using algebra?

 

[eehelp150]

Yes, but I imagine that the algebra could get tedious. You want to find the reactive power as a function of C, so that means finding the imaginary part of the complex power. It would be easier to plot the reactive power and pick out the solution on the plot.

I am plugging in the original total real power and total reactive power. I cannot get the original value of capacitance.

Edit:
I used 576-430.8i instead and got the original capacitance value. How do I solve for C if I don't have a "defined" real power to solve for? Do I simply divide that by 2 (50%) as well?

 

[gneill]

I used 576-430.8i instead and got the original capacitance value. How do I solve for C if I don't have a "defined" real power to solve for? Do I simply divide that by 2 (50%) as well?

No, the real and imaginary parts of the power will have some complicated relationship.

Since you want to reduce the reactive power by 50% you can simplify things a bit by setting Vs = 1. Then ignoring units the current will be $I = 1/Z$ and the power will be $P = V_s I^* = I^*$.

Isolating the imaginary part of $I^*$ might be tedious. You can write your impedance as:

$Z = 12 + j20 + \frac{1}{\frac{1}{20} + jωC}$

Might as well define $x = ωC$, then
$$I = \frac{1}{12 + j20 + \frac{1}{\frac{1}{20} + jx}}$$
Normalizing that to pull out the complex component is not complicated, just labor intensive. Then find its value using the original capacitor value in x = ωC as the "benchmark" value for the imaginary component. You'll want to find a new value for x that yields half of that power.