Energy and Power

Hi

If you have data consisting of: time, voltage(rms), current(rms) and frequency, how do you calculate The Total Energy, Mean Power Consumption and Max Power Consumption?

An example of the data is:
time(sec), voltage(rms), current(rms), frequency
02.049824 231.401 0.022 49.738
02.668822 229.814 0.019 49.738
03.283705 228.528 2.599 49.726
03.898469 230.140 2.576 49.751

Power is the 'Rate of doing work".
This is given by "Volts x Amps". The result is voltamps.
However if the voltage is AC, then we must consider the phase relationship between the voltage waveform and the current waveform. So expended power is then Volts x Amps x cos of the phase angle between V and I. This has the units of watt.
The amount of work done is the product of the power multiplied by time (seconds). This is energy and is power x time. For mains electricity this has the units of kilowatt hours.
For your house supply, you buy energy and is 'so many kilowatthours'. (kWhr)

You can't. You need to know power factor.

I believe you can do Average power. remember there are apparent, real and reactive powers. The instantaneous average is the product of Vrms and I rms.
So, you have to do the Calculus thing to find the mean of the product.

For max, you can do the derivative thing or just multiply find the max.

Energy is the sum of (time * dv*dt ) or in your case delta v and delta t. Could be W-sec. W-min or Watt-hours etc. So, you do numerical integration.

But, you have to remember that for real AC power measurements you need instantaneous values of v(t) and i(t) much closer together.

The power factor is taken to be 1.

There is a formula where PF=cos(θ).
Taking PF to be 1, cos(θ) = 0 ;

Then P = Vrms * Irms * cos(θ)

Is this angle the same as the PF angle?

KeepItSimpleStupid said:

I believe you can do Average power. remember there are apparent, real and reactive powers. The instantaneous average is the product of Vrms and I rms.
So, you have to do the Calculus thing to find the mean of the product.
For max, you can do the derivative thing or just multiply find the max.
Energy is the sum of (time * dv*dt ) or in your case delta v and delta t. Could be W-sec. W-min or Watt-hours etc. So, you do numerical integration.
But, you have to remember that for real AC power measurements you need instantaneous values of v(t) and i(t) much closer together.

Click to expand...

I'm not sure I understand what you mean by the Calculus thing or Derivative thing.

The Calculus thing: https://www.wolframalpha.com/input/?i=average+sin(t)+from+0+to+PI

The integral of the function divided by the period. sin(t) was used as an example.

Numerical integration: https://en.wikipedia.org/wiki/Numerical_integration Rectangular or trapezoidal rule.

the OP said:

he power factor is taken to be 1.

There is a formula where PF=cos(θ).
Taking PF to be 1, cos(θ) = 0 ;

Then P = Vrms * Irms * cos(θ)

Is this angle the same as the PF angle?

Click to expand...

You mean cos(θ)=1, θ=0

The formula is only valid for sin waves.


Power factor is the ratio of true and apparent power. See: https://www.google.com/url?sa=t&rct...zeJreDmSd6hUyYA&bvm=bv.63934634,d.cWc&cad=rja

So your apparent power is your instantaneous power which is Vrms * Irms?
So to find the average power you say: Pavg = 1/T * [integral of P(t)]?

To calculate the average power it would be (V * I) for each line and them divided by 4 for this data set.
To find the maximum power you would choose the power calculated from the 4 values and choose the highest value. Must you multiply this value by root 2?

If you deem pf = 1, which is rarely true:

P[n] = V[n] x I[n] - that's your average power from the last n-th measurement period between times t[n] and t[n-1] (assuming V and I are true RMS for values for the period, is that so?)

E[t[n-1],t[n]] = P[n]*(t[n]-t[n-1]) - that's your energy used between t[n-1] and t[n]

You can sum up your energies over a period of time, say for an hour, or a day: E[t1,t2] = SUM(E[,])

P[t1,t2] = E[t1,t2]/(t2-t1) - that's your average power for period from t1 to t2

the OP said:

So your apparent power is your instantaneous power which is Vrms * Irms?

Click to expand...

Pretty much true except I'd use the definition of instantaneous power to be v(t)*i(t) when dealing with reactive waveforms.

Jason, In your post 5, you raise the question of power factor.
When the power factor is 1 ; then this is because the phase angle between the voltage maximum and the current maximum is zero. That is, the voltage and current waveforms are 'in phase' OR to put it another way, the phase angle is ZERO. The cosine (or cos) of zero is 1.
So in your questions and my post 2, you can calculate the wattage from the individual voltage and current values. But remember, you have ASSUMED or you KNOW that the power factor is 1 .
In post 8, the 'apparent' power is Vrms x I rms. The average power is not really as you put it. The average power is more taken to be the average over one mains cycle.
In post 11, the definition of 'instantaneous power' is given. If you calculated the instantaneous power at one degree intervals over a mains cycle, you would have 360 individual results. By adding these and dividing by 360 you would have the average power.
In post 10, the value of power factor was canvassed. For most heating and lighting, the power factor is one. For many electric motors etc, the power factor is around 0.8 . For overall household electricity consumption, the power factor is usually much better than this. The increasing use of electronic equipment in homes, has forced electronic equipment makers to consider the power factor of their appliances. Much equipment is now produced using power factor correction electronic devices in order that electricity consumers are using energy at a high power factor very close to 1.

Your post is very confusing.

rumpfy said:

he average power is more taken to be the average over one mains cycle.

Click to expand...

Since all of the data is a few volts of some average nominal value, you can throw the average over on mains cycle out the window for this problem.

You can compute average apparent power over the course of a line cycle, a day, a week or a month. So, someone was using a toaster, a curling iron, and an electric water heater etc.. No washers, garbage disposals, computers or light bulbs.

I guess, we really don't like the exercise, but that's life. What would be really useful and would separate the men from the boys would be given two v(t), i(t) waveforms compute all of the AC stuff. I'd make it easier and just give i(t) and v(t) for one line cycle. I might even give the value of Vrms and Irms for one cycle and make it extra credit to calculate those.

I think the real issue is understanding what RMS means. In the real world RMS has different meanings, You can say the value of a 12 VDC source is 12 V RMS, and ty can compute the RMS component of the AC component and the RMS of the AC+DC component with the right meter. When we use RMS casually, we tend to think of an AC waveform.

Suppose we had a 12 V power supply with ripple. We could say the ripple is maximum 10 mV p-p or 5 mV RMS. it's probably silly to say the 12 VDC source is 12.005 VRMS, but it could be technically correct.