# Confused about MW to MWh conversion

I have various datasets of energy production in MW.

I need to present everything in MWh. I need to know if my data is already by the hour (logging a value for each hour of each day of an entire year), if the sum of all those values in MW also equals the value in MWh.

To make the conversion, does that depend on whether the data is in 5-minute, 15-minute, or 1-hour increments? For example, if the data is registered in 5-minute increments, I think I would have to add all MW and then divide by 12 (because there are 12 5-minute periods in 1 hour).

Thus, if 1 MW produced every 5 minutes for a 2-hour period, then the calculation to obtain MWh is (1x2X12)/12 = 2 MWh? Is that correct? Add all MW and divide by number of periods in one hour?

If in 15 mins, it would be (1x24)/4 = 6 MWh.

If data is already by the hour, then (1x24)/1 = 24 MWh? So in this case, MW = MWh?

• MW is the rate of delivery, MWh is an absolute amount delivered. Your assumptions are correct; 1 MW for an hour delivers 1 MWh. 1MW for 5 minutes delivers 1/12 MWh. Commented Aug 18, 2022 at 21:25
• MW is a unit of power while MWh is a unit of energy... Commented Aug 18, 2022 at 21:25
• Think of it this way : if your speedo says 60mph (speed, rate) when you look at it every 5 minutes, it'll still take an hour to go 60 miles. 1MW is a rate; however often you measure it, it'll still take an hour to deliver 1 MWh. Measuring power more often just allows more accurate calculations if the power keeps changing.
– user16324
Commented Aug 18, 2022 at 22:18
• The watt is a "rate" that doesn't have something like "per second" as its unit, but it's already in the unit itself. The joule is more like you would expect (and 1 watt = 1 joule/second). It's like inventing a unit like "velocity", so you would drive "40 v" (say "40 velocities") which under the hood is say m/s. To go to meters you have to integrate over a time period. To go to joules you have to integrate likewise over time (and 1 joule = 1 wattsecond). Hopefully this helps and doesn't confuse more. :-) Commented Aug 19, 2022 at 10:01

I have various datasets of energy production in MW.

That's the source of your problem. That should read either,
"I have various datasets of power production in MW, or
"I have various datasets of energy production in MWh."

I'd go back to the source and get clarification.

For example, if the data is registered in 5-minute increments, I think I would have to add all MW and then divide by 12 .

Correct.

Thus, if 1 MW produced every 5 minutes for a 2-hour period, ...

That would be badly worded. "If I averaged 1 MW continuously **for each 5 minute perio ..." would be much better.

... then the calculation to obtain MWh is (1 x 2 x 12) / 12 = 2 MWh? Is that correct? Add all MW and divide by number of periods in one hour?

That's correct but it can be simplified to "I produced 1 MW continuously (or average) for two hours so MW × hours = 2 MWh.

If in 15 mins, it would be (1x24)/4 = 6 MWh.

No, you're mixing power and energy again. It's just average power × time. 1 MW for 2 hours = 2 MWh.

If data is already by the hour, then (1 x 24) / 1 = 24 MWh? So in this case, MW = MWh?

No. A MW is not the same as a MWh (same as a km is not the same as a km/h). The numeric value might work out the same but the concept is different.

simulate this circuit – Schematic created using CircuitLab

Figure 1. Average power vs time.

• In the first hour the energy used is (0.5 + 1 + 1 + 0.5) / 4 = 0.75 MWh.
• In the second hour the energy used is (1.5 + 2 + 1 + 1) / 4 = 1.375 MWh.
• Total for the two hour period is 2.125 MWh.
• The energy used is the integral of power with respect to time and is represented graphically by the area under the curve.
• Thank you!! I hate to trouble you again, but I'm running into this issue again. 😐 I'm not sure what to do if the dataset doesn't specify time increments. I have this dataset that is simply the nameplate capacity of different generators for a specific year in MW. I'm hoping to get a sum of all nameplate capacity for all solar generators for that year, but I would like to express it in MWh. Dataset is in link below... Any advice? view.officeapps.live.com/op/…
– Bex
Commented Sep 1, 2022 at 2:14
• First advice would be to remove the public access on that document! To get MWh for any particular solar panel you'd need the rating and the amount of insolation (sunshine) per year for each location. I don't know how you factor out excess insolation. (There's going to be some point at which additional sunshine doesn't produce additional output.) Commented Sep 1, 2022 at 8:54
• @Transistor: "I don't know how you factor out excess insolation." This loss is included, among others, in the performance ratio, which is typically around 90%. Commented Sep 1, 2022 at 12:26

Watts are a unit of power, energy consumed per unit time. Watt-hours are a unit of energy. Finding Watt-hours given time-discrete point readings depends on what assumptions you want to make about what happens between the readings. It's kind of like trying to figure out how far you've driven from a speedometer reading every 5, 15, 30 minutes etc. If you're cruising on the highway, assuming that the speed between readings is constant is probably going to get you a fairly accurate answer. If you're in stop-and-go traffic, it will probably be wildly off. You just need to decide what your assumption is going to be and state it clearly.

For example, if you have a power reading of 1MW at noon and a power reading of 1.1MW at 12:05, you might assume that from the first reading to the second one, the power was constant at 1MW and for that 5-minute period, the energy produced was 0.083MWh. For the next 5-minute period, the energy produced was 0.092MWh, and so forth.

Watts (W) or in your case megawatts (MW) is a measure of power (P) (heat flow). Watt-hour (Wh) is a measure of energy (E) delivered over a period of time.

If the power MW is constant over 2 hours, then the total energy consumed or delivered over the 2 hours is$$E_{2hours} = P\Delta t$$ If the power is 1MW then$$E_{2hours}=2\text{ MWh}$$

If the power changes over the total measuring period and is measured say every 5 minutes or 1/12 of an hour, then you must average the power measured for 2 consecutive periods, then multiply by the time of the period. Over two hours there would be 24, 5 minute periods.$$E_{2hours}= \Sigma \frac {P_{i}-P_{i-1}}{2}\Delta t$$$$E_{2hours}= \sum_{i=1}^{24} \frac {P_{i}-P_{i-1}}{2}\frac {1}{12}\text{ MWh}$$

In order to avoid confusion between power and energy, or between 5min values and 1h values, I like to explicitly write measurements as $$\frac{\mathrm{energy}}{\mathrm{time}}$$ for example : $$10 \frac{\mathrm{MWh}}{\mathrm{h}}$$

This seems redundant at first, but it makes it clear that it's an energy amount, which has been converted/delivered/used in an hour. If you simplify the unit by removing both h, you notice that it's actually an average power : $$10 \mathrm{MW}$$

The advantage becomes clear with other time ranges, for example:

$$2 \frac{\mathrm{MWh}}{\mathrm{15min}}$$

An energy amount of 2MWh, delivered in 15 minutes. If you simplify, you notice that it represents an average power of 8MW.

And if you have four consecutive measurements (e.g. 2, 3, 1 and 4 MWh/(15min)), you can simply add the energy amounts in order to get MWh (10 in this example), and convert them back to average power (in MWh/h) if you want.

• Thanks! My issue was more that the data is presented in 5-minute intervals, over a whole year, in MW, and I needed to get a result in MWh for the whole year. So I added all the MW values over this year, and divided by 12, since it is in 5-minute increments, to get the total value in MWh for the entire year. Hopefully that is still correct.
– Bex
Commented Aug 19, 2022 at 17:30
• @bex, thanks for the info. My advice should still work in this case. You can convert your MW values to MWh/(5min). This includes the factor 12 you're talking about (e.g. 12MW is 1MWh/(5min)). You can add all the MWh amounts, and you get MWh for the year. Yet another alternative would be to calculate the average power, in MW, and multiply by 8760h in order to get a yearly energy value. Commented Aug 19, 2022 at 17:36
• Thank you!! I hate to trouble you again, but I'm running into this issue again. 😐 I'm not sure what to do if the dataset doesn't specify time increments. I have this dataset that is simply the nameplate capacity of different generators for a specific year in MW. I'm hoping to get a sum of all nameplate capacity for all solar generators for that year, but I would like to express it in MWh. Dataset is in link below... Any advice? view.officeapps.live.com/op/…
– Bex
Commented Sep 1, 2022 at 2:13
• @Bex: That should be another question, possibly on sustainability.stackexchange.com . To keep it simple, you could use the yearly insolation (in kWh/(m²*a), e.g. based on nrel.gov/gis/assets/images/…), multiply it by a typical performance ratio (~85%), divide by lab irradiance (1kW/m²), and you'd get the specific yield (kWh/(kW*a)). Multiply it by nominal capacity, and you get MWh/a. Commented Sep 1, 2022 at 12:25

MW is a unit of power, it's instantaneous, it's the rate at which energy is being delivered. MWh is a unit of energy, and is absolute.

Think of it like speed and distance, where power (W) is speed and energy (Wh) is distance.

if 1 MW produced every 5 minutes

would be like saying

if I go 1km/h every 5 minutes

it doesn't make much sense. Our speed/distance example should read

if I go 1km/h for 5 minutes (then I would cover 1km/h * 5 minutes * (1/60 minutes/hour) = 1/12 km)

or for power/energy

if 1 MW is produced for 5 minutes (there would be 1MW * 5 minutes * (1/60 minutes/hour) = 1/12 MWh)

If all your units are in MW and you want to convert to MWh, you can add them all together and multiply by the time to get total MWh.

e.g. if you have 3 sources generating 250W, 100W, and 150W and you want to know how many MWh this is equivalent to, multiply their sum (500W) by the amount of time they will be running for.

in 1 hour: 500W*1h=500Wh

in 1 day 500W*24h=12000Wh

So you can't say that 1W = 1Wh, because it depends on the amount of time that the power is being delivered for. (in the special case that it is being delivered for 1h, the W and Wh will be equal, but that is not a general rule)

• Merci! My main difficulty was the fact that I need to present the data in MWh, but the dataset is in MW, and in 5-minute increments. So I had to determine how many MWh of energy were being put out during a certain month, or year. So I was adding all the MW over a year, which is logged in 5-min intervals, and trying to convert this MW amount to MWh. But yes for the last equivalence I just meant that only in this case, the numerical values would be the same!
– Bex
Commented Aug 19, 2022 at 17:25