# Should I multiply by time to determine Watt-hours?

I have done an experiment with solar panels over the course of the day to determine the total power output in Wh. I measured voltage and current through my solar panel circuit using a multimeter every half an hour. I multiplied these together to find the power being generated each half hour of the day. Assuming the power rises and falls linearly between each measurement, I determined the energy for each half hour is equal to 30 times the power at the mid-mark (either 15 or 45 minutes into the hour). Now, to determine the total energy in Wh the panel produced during the day, do I multiply the result of the previous operation by 0.5 hrs and then add all the interval's power together or can I just add them all together without multiplying by 0.5 hrs?

• Watt-hours measures energy, not power. – The Photon Mar 20 '16 at 3:54
• I understand how watt-hours measure energy, not power. I never meant to say that. I also want to note that I do know calculus but not coding so sadly that will not be the greatest. I also do NOT want to average them. I would like to do the process I described, I just don't know if I need to multiply by 0.5 or not. – solarstudent Mar 21 '16 at 2:16
• You can edit your question at any time to make it more clear (in your first sentence you say you want to "determine the total power output in Wh"). – The Photon Mar 21 '16 at 2:17
• See seethraman's answer...he shows (by slightly different reasoning) why you should multiply by 0.5. – The Photon Mar 21 '16 at 2:19

The term "watt-hour" (Wh) is often not correctly understood. Watt (W) is a unit of power. Energy is measured in Joules (J) and is related to power by $$E = {P_\text{avg}}\times \Delta t,$$ where $\Delta t$ is time in question and $P_\text{avg}$ is the average power spent during that time.

With only two measurements, one at the start and one at the end, the best guess is that the average power was $$P_\text{avg} = \frac{P_\text{start} + P_\text{end}}{2}$$

Notice that "watt-hours" is power multiplied with time, so it is a measure of energy. This is what power companies really sell you, and what they should charge you for; though that's for some reason not always the case.

Joules is the same as a "watt-second", so $$1\;\text{Wh} = (1\;\text{W})\times(1\;\text{h}) = (1\;\text{W})\times(3,600\;\text{s}) = 3,600\;\text{J}=3.6\;\text{kJ}$$

### Sidenote

If you know calculus, then a more accurate formula for the relationship between power and energy is $$E = \int_{t_\text{start}}^{t_\text{end}} P(t)\;dt,$$ or even $$P(t) = \frac{dE(t)}{dt},$$ which says that power is the rate-of-change of energy consumption. You can make use of this, if you know the sample-rate of your measurements, for instance by using an ADC from a micro-controller. If the time between samples is dt, then the total energy can be calculated by:

loop() {
time = time_now();
dt = time - time_last;
power = current * voltage;
energy += power * dt;
time_last = time;
}


Let's say you have 10 readings that is captured with an interval of 30 minutes over the period of 5 hours.

• Take an average of all the 10 readings.(there's no point in finding the 15 min interval power, by averaging).

• Multiply it by 5(no of hours in this case), you'll get Wh.

• This is just the sum of the measurements, divided by 10 and multiplied by 5. So it's the same as taking the sum and multiplying by 0.5 as proposed by OP. – The Photon Mar 20 '16 at 15:40

Yes, you need to multiply average power by time (that the power was averaged over) in hours to get Wh

$$P_{AVG} = \frac {P_2 + P_1}{2}$$ where $P_1$ and $P_2$ are the power at the start ($T_1$) and end ($T_2$) of the period.

$$Energy = P_{AVG} \cdot (T_2 - T_1) = \frac {P_2 + P_1}{2} \cdot (T_2-T_1)$$

So if we got two readings 17 W and 21 W half an hour apart we calculate

$$Energy = \frac {P_2 + P_1}{2} \cdot (T_2-T_1) = \frac {17W + 21W}{2} \cdot (0.5h) = 9.5~Wh$$

Graph all the VA measurements you took on an x-y plot. Make your Volt Amps your y-axis, and the time your x-axis. You should have a bunch of dots forming a sort of dome in the middle of your graph. Connect them. The area under the curve is the energy generated.

Are you using a passive (resistive) load, or do we need to be considering power factor?