Determining power utilization...

[AmazingTrans]

Hi there, I am trying to determine the power that is utilized by my dc system. I plot a active power curve of my system in 24 seconds. It fluctuates between 88kW to 52kw. I have a graph that looks like a sinusoidal shape.

I also have plot the RMS value of it and the RMS is 74 kW.

I am trying to determine how much kw is my system using to calculate the kwH.

Hope someone can guide me?

 

[Smart $]

74kW*24secs*1hr/3600secs =0.49kWh

 

[AmazingTrans]

Since this signal is DC sinusoidal, can I use RMS to determine the values?
Here is a sample of the graph:
http://www.tiikoni.com/tis/view/?id=42fb257

 

[gar]

140325-0037 EDT

If the measurement is power, and I believe from your description that is the case, then you simply average the power curve over a time period of interest, and multiply that average power by the time period to get the energy used over that time period.

An RMS value of a power curve is not what you want.

Note: instantaneous power = v * i . To obtain energy you integrate instantaneous power over a desired time period. In other words the area under the instantaneous power curve of the desired time period is energy.

.

 

[AmazingTrans]

140325-0037 EDT

If the measurement is power, and I believe from your description that is the case, then you simply average the power curve over a time period of interest, and multiply that average power by the time period to get the energy used over that time period.

An RMS value of a power curve is not what you want.

Note: instantaneous power = v * i . To obtain energy you integrate instantaneous power over a desired time period. In other words the area under the instantaneous power curve of the desired time period is energy.

.

Thanks everyone.so with that said I can determine the average power by summing up: p=kW ; t=ms
{P(I+1) + p(I) /2}? t(I+1) -t(I)}

Then to get kWh I just: (sum kwms /1000)/3600

Does that sound right?

 

[gar]

140325-0908 EDT

I don't quite understand your equation. What is meant by capital P vs lower case p. Is +1 = to 1 millisec? Does the instantaneous power follow a straight line between t and t+1?


Assume you have a device that provides you with the average power in kw at the end of each 1 millisec sample period, then the energy per 1 millisec sample in kWh is:

....... E(kWh average per millisec) = P(kW average per millisec)*1 / 3600

Thus a constant 1 kW load consumes 1/3600 kWh in one millisecond. If the power remains constant at 1 kW for 1 hour, then the sum of the 3600 one millisecond samples is 1 or 1 kWh. If for 1800 samples the power is 1 kW, for 900 samples 0.5 kW, and for another 900 samples 0.1 kW, then the total energy used in that 1 hour period is 0.5 + 0.125 + 0.025 = 0.65 kWh. It matters not how those samples are distributed thru the hour.

The total energy over some longer time is the sum of each of the contiguous E values in that longer time period.

.

 

[AmazingTrans]

140325-0908 EDT

I don't quite understand your equation. What is meant by capital P vs lower case p. Is +1 = to 1 millisec? Does the instantaneous power follow a straight line between t and t+1?


Assume you have a device that provides you with the average power in kw at the end of each 1 millisec sample period, then the energy per 1 millisec sample in kWh is:

....... E(kWh average per millisec) = P(kW average per millisec)*1 / 3600

Thus a constant 1 kW load consumes 1/3600 kWh in one millisecond. If the power remains constant at 1 kW for 1 hour, then the sum of the 3600 one millisecond samples is 1 or 1 kWh. If for 1800 samples the power is 1 kW, for 900 samples 0.5 kW, and for another 900 samples 0.1 kW, then the total energy used in that 1 hour period is 0.5 + 0.125 + 0.025 = 0.65 kWh. It matters not how those samples are distributed thru the hour.

The total energy over some longer time is the sum of each of the contiguous E values in that longer time period.

.

gar,

Not sure if you can view this link. http://www.tiikoni.com/tis/view/?id=42fb257
I have a trace every 2ms for 3600 points.

for i =1 to 3600
sumPower = {P(i+1) + P(i) /2}? t(i+1) -t(i)}

 

[iceworm]

Amazing -
With a couple of assumptions this gets really simple.

The plot looks like a startup followed by steady state. The curve does not have to be sinusoidal - just the same shape upper and lower. If the energy part you are interested in is the steady state section, then:

The "average power" (as gar noted - this is the one you want, not the rms) is right down the middle of the sine wave, equal between the upper peak and lower peak.

ice

 

[gar]

140325-1150 EDT

AmazingTrans:

In response to your PM go to http://beta-a2.com/energy.html and at the bottom of the page is my contact information.

I can eyeball your average power at about 65 to 70 kW from your curve. Energy used in kWh for any time period where this estimate is valid is the time in hours of the time period times 67.

To do a moderately accurate estimate of your energy use assume that your 2 mS measurement is in of itself an accurate average, then take your 2 mS samples and multiply each by 1/1800 to get the energy in kWh per sample. Add all the 2 mS energy values in the desired time period and that is your energy used in that time period.

.

 

[AmazingTrans]

140325-1150 EDT

AmazingTrans:

In response to your PM go to http://beta-a2.com/energy.html and at the bottom of the page is my contact information.

I can eyeball your average power at about 65 to 70 kW from your curve. Energy used in kWh for any time period where this estimate is valid is the time in hours of the time period times 67.

To do a moderately accurate estimate of your energy use assume that your 2 mS measurement is in of itself an accurate average, then take your 2 mS samples and multiply each by 1/1800 to get the energy in kWh per sample. Add all the 2 mS energy values in the desired time period and that is your energy used in that time period.

.

Thanks for the info.,