Measuring Flow

Hello all.

I have a flom meter wich measures in the range from 0 to 300 m3/h. I have to calculate how much gas was produced in an hour and throught the day.

I've made the program like so:
I add the curent value form the analog input, every second and increment a counter at the same time...
Then when the hour passes I divide, the value of the adding with the counter value and get the how much gas was made...
But my resoults are different from a Data logger that prints out data every hour and every 24 hours.
Is this the correct method that I have taken or do you guys do this differently?

I've read some specifications of these data loggers, that they take a reading every 125ms..., so basicly I'm takeing readings almoast 10 times slower, so I guess this could be the reason for the difference of the value that I get from the data logger. Correct?

Then when the hour passes I divide, the value of the adding with the counter value

Click to expand...

The cumulative sum is volume per unit time. The counter value is time. You should multiply them, not divide.

Check out some of these threads.

http://www.plctalk.net/qanda/search.php?searchid=546493

My first response was incorrect.

Each sample of flow (m3/hour) is taken at one-second intervals. When you accumulate 3600 samples, you have one hour's accumulated volume. That's because there are 3600 seconds per hour.

To get the volume at any time you multiply the sum of the samples by the number of samples and divide the result by 3600.

So basicly my method is correct. I'll have to find the problem elsewere... I think the logger isn't set correctly. Or at least the max value(at 20mA) is different than mine. So on monday I'll probably put 20mA on the logger for an hour an see what kind of resoult will the logger display.

Thanks for the input guys

So basicly my method is correct

Click to expand...

The way you defined your method and the method I described are not the same.

If you were using 125 mS sample times, then to calculate the volume at any time you would multiply the accumulated flow value by the number of samples and divide by 28,800.

I think the discrepancy between the datalogger and the plc stems from assuming (in the PLC) that the flow rate for the hour was uniform around the average value. The data logger probably averages the current and previous values and computes incremental volume.

In the PLC, if you take the average of the current flow and the previous flow (1 sec earlier) and divide by 3600, then add to an accumulator, I think the discrepancy will shrink.

Steve,
I don't understand your method:
multiply by 3600 then divide by 3600 ??

edit:
Since your measuring gas flow, the data logger may also be making adjustments for temperature and pressure.

Steve,
I don't understand your method:
multiply by 3600 then divide by 3600 ??

Click to expand...

NoName said he's taking sample once per second, adding the current sample value to a running total and also incrementing a counter each time he takes a sample. The counter accumulated value is the number of samples. At any given time, the running total of the sampled flow value times the number of samples divided by 3600 gives the current accumulated volume.

Steve Bailey said:

NoName said he's taking sample once per second, adding the current sample value to a running total and also incrementing a counter each time he takes a sample. The counter accumulated value is the number of samples. At any given time, the running total of the sampled flow value times the number of samples divided by 3600 gives the current accumulated volume.

Click to expand...

I believe the samples he's taking are flow, i.e. rate, not volume.

                flow(t) + flow(t-1) m3/hr            1 sec

delta volume = -----------------------------  * ---------------

                              2                   3600 sec/hr

NoName samples flow rate (m3/hour) and adds each sample to a running total. The units of the running total are m3/hour. He's also keeping track of the number of samples (N) in a counter. The cumulative number of samples has units of seconds.


Cum. Samples (m3/hour) * N seconds * (1 hour /3600 seconds) = m3

Averaging each sample as you've done will definitely improve the accuracy.