How to Calculate for Energy? Newbie

what if it gets cloudy from 11:30 to 1?

Sir, I can't assume that the power is constant for every 30min.

Here's a snapshot, sir. If you could tell me exactly how to calculate for the total energy, sir. I would be so happy.

Your graph is drawn using straight lines to connect data points. This is at best an approximation, because the actual graph (as drawn by a continuously connected recorder) would look smooth.

Your question indicates that you can't do this using calculus, which is how we would do it.

That being said, there is a way to get a "close" approximation using your graph.

Using data point 1 and 2 (segment 1), average the two points then multiply by 1/2 hour (1+2=x.. ..x/2=average watts.. ..average watts x 1/2 = watt hours for segment 1).

The next average would be segment 2 using data points 2 and 3.

That is the closest you will get with your methods.

Yes, see #8

Didn't read down that far; stopped at the graph.

JohnAnthony:

Use a power usage integrator on continuous bases and look at its value in watts after 1 hour it will be exact watts.

CR4 Admin Modified Post: Changed "watts/hour" to "watts" per user request.

Sorry, but all of those units are wrong.

You are right.

Watt is a unit of power, energy per unit time, joule/sec in SI units. Watt-hour is a unit of energy, joule or watt-hour or kWh etc. That does seem to be going round in circles, but it works OK. Joule is small for many purposes and electric energy supplied is traditionally measured in kWh = 3.6*10⁶ joule.

Are you really recording power (watt), at 30 min intervals, or energy on a watt-hour meter? If the latter you have the answer to your question. If the former, do as Tornado says, but to get an accurate figure need the average power over each 1/2 hour.

Thanks for the reply sir.

Yes sir I am recording power (watt) every 30min, not energy. That's what I want to get sir the energy. I attached a photo above sir but I don't how to calculate for energy given that graph.

I hope you can help me, sir. I really appreciate your time.

The energy is the integral of power over time. In your graph it would be the area of your graph between the 0 line and the power line that you drew. If you never took calculus in school, check the definition of Integral on the web

I think other posts have covered it, but if you need more detail, I would take the average power P_ave in each 1/2 hour section of your graph, multiply by 1/2 hour, and that gives 1/2*P_ave watt-hour. Add all the 1/2 hour sections to get total. This assumes the power varies linearly between the measurement points, which won't be exact, but it's all we have.

So 1st bit P_ave = 24 watt approx, giving 12 watt-hour, 2nd bit P_ave = 38 watt, giving 19 watt-hour, and so on.

I must have my stupid head on.

You're recording watts per half hour, what are you using to record this?

This may come as a silly idea, pulsed output KWHr meters are cheap. Just count the pulses over your recording period. Even cheaper, read the damn meter!

As far as I can see, he's taking a power reading, presumably by reading a meter, every 1/2 hour. He then has to do some maths to find what he wants, energy over the 6 hours. It would be more accurate, but more work, to use shorter intervals.

As you say, easier (and more accurate) to use a kWh meter, assuming readily available for readings small by usual power supply standards. Only about 0.18 kWh over the 6 hours.

Without using calculus or Simpson's Rule, you can get a fair estimate of the total energy by cutting out the area under the graph with scissors and weighing it on some very sensitive laboratory scales. You then compare the weight of the graph with that of a rectangle of known area (say 60 watts X 6 hours) and obtain your integrated energy quantity that way. You are, in effect, substituting weight of paper for area. Sounds crude, but it works.

Ha!

Or draw it on a really big piece of paper, then cut it out and weigh it on your bathroom scale.

Excellent idea! Ratiometrics without calculus. Sometimes you just have to use what is available!

Get one of these:

KILL A WATTRealtime pow...$19.00

Watt is a unit if power and watt hour is that of energy. Power in watt multiplied by perod in hours will get you watt hour, larger unit is kilowatt hour (KWH ) commonly known as unit in electrical terms. As suggested by others calculate area below your graph or why don't use kilowatt meters available in market which can get you the power straight without any calcs. For calculating from data mathematically you are bound to take the power constant between two consecutive readings. So more frequent readings mean better accuracy and continuous readings mean exactness to the extant of measuring accuracy and human errors..

Haven't done this since I left school! (1953) So there well may be mistakes!

Total the vertical readings (which is approx. 538w) divide this number by the

total number of readings (13) taken, gives an average wattage of 41.38w

which, over 6 hours, gives a total of 248.38 wh.

Cannot get much simpler than this, or am I missing something?

jt.

Is it true that an apple a day keeps the doctor away?
Or is it just one of Granny's myths?

Power and energy for the calculus challenged. Analogy: Power is to speed in your car, as energy is to distance traveled. Since power is the rate, multiply the rate x time (interval, i.e. 30 mins = 0.5 hours), for each "30" minute interval, and add them up and you have total energy produced (and either consumed or stored less efficiency of storage battery). Power is Watts, Kilowatts, Megawatts, or Gigawatts in normal usage. Learn what the prefixes mean, i.e. 1000, 1000000, 1000000000.

Energy is measured in power-time: Watt-hour, kW-hr, MW-hr, and so on. 1000 watt-hr is equivalent to: 1 kW-hr (duh), and is also equivalent to 3412.141633 BTU (mechanical-thermal unit equivalency), and is also equivalent to 3,600,000 Joules.

1 Joule of energy is 1 Newton-meter, also 1 Kg-m²/sec² , (thus energy is equivalent to mechanical (or other) force moving something through a distance. It is also equivalent to Pressure moving a boundary (piston for example) through a volume.