I feel dubious about the use of “integral” for this - you have Watts, measured

once per minute. That’s either going to be an instantaneous measurement (such

as “7.8 Watts right now, no idea what it was 5 seconds ago”), or it’s going to

be an average over the previous minute (such as “468 Joules used in the past

60 seconds = an average of 7.8 Watts”) - which it is depends on the instrument

making the measurements.

If you want to convert Watts (measured once per minute) back into Joules (or

kiloWatt-hours; same thing except for a multiplication factor of 3.6 x 10^6)

then I think you need a simple “sum” operator:

Minute 1: 7.8 Watts

Minute 2: 10.2 Watts

Minute 3: 13.3 Watts

Minute 4: 6.4 Watts

Minute 5: 9.6 Watts

assume this pattern repeats (mysteriously) 12 times to create 1 hour’s data.

The sum of the above 5 values is 47.3; multiply this by 60 (seconds) to get

2838 Joules (or Watt-seconds). Since the pattern repeats 12 times in an hour,

multiply by 12 to get 34056 Joules of energy used in that hour.

34056 Jouleṡ ÷ 3.6 x 10^6 = 9.46 x 10-³ kWh

So, I think you should be using “sum” to add up the Watt power values, and

then dividing by ( 3.6 x 10^6 ) ÷ ( 60 x 12 ) = 2500 to convert into kWh.

Note that 2500 is approximately twice 1440 (which is what you’re dividing by

at present), hence explaining the values you’re currently getting as being

about twice what you expected.

Antony.