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I have a system thet consists of a pump to move the fluid and a solenoid valve to allow the liquid to pass through.

I have a flow sensor that has the following characteristics:

F = (2.5* Q), Q=L/Min

Where Q is the flow rate in liters per minute, and F is the frequency of the signal generated by the sensor in Hz.

According to this, the flow would be:

Q = L / 60[s].

So:

F = 2.5*L/60[s].

The frequency would be the square signal pulses (P) that occur in 1 second, therefore:

P/[s] = L / 24[s] ( 2.5/60 = 1/24)

Therefore:

L = 24P.

Using a pulse counter, I count 24 pulses to obtain one liter (obviously activating and deactivating the pump and the solenoid valve), but the volume obtained is very small.

So, in my calculations, something is wrong.

Any comment or suggestion is welcome.

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    \$\begingroup\$ Q = L / 60[s]. - what does it mean? Q is expressed in liters per minute. So you get Q by dividing F by 2.5. Then integrate/multiply it over the time in minutes getting the volume. That is Volume[Liters] = F/2.5 * T[min] assuming F is constant. You seem to consistently mix units with amounts. \$\endgroup\$ Commented Feb 16, 2024 at 15:24
  • \$\begingroup\$ @EugeneSh. But if I want to work in seconds, it is 60, since a minute has 60 seconds \$\endgroup\$ Commented Feb 16, 2024 at 15:29
  • \$\begingroup\$ Then if you have T[sec], convert it to minutes by dividing it by 60. That is Volume[Liters] = F/2.5 * T[sec] / 60 \$\endgroup\$ Commented Feb 16, 2024 at 15:31
  • \$\begingroup\$ Here F*T will give you the number of pulses, so if you are interested in number of pulses in one liter you can say 1[Liter] = [Pulses]/150. That is each liter is 150 pulses. Unless I missed something too. \$\endgroup\$ Commented Feb 16, 2024 at 15:39
  • \$\begingroup\$ @EugeneSh. Thanks for the suggestion. With 24 pulses, I get to a tenth of a liter, maybe less, with 150, maybe I get to about a half liter, maybe a little more. I was reviewing some tutorials, and it seems that I should not measure pulses, what I should do is measure pulses in a given time, for example 1 second, that is, I should measure the frequency and use the sensor equations \$\endgroup\$ Commented Feb 16, 2024 at 15:44

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\$ f = 2.5 \times Q\$ where \$f\$ is the frequency and \$Q\$ is the flow rate in L/min.

So if Q = 1 L/min, f = 2.5 Hz which results in 150 pulses for a 1 L flow.

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  • \$\begingroup\$ Yes, you are right. It has worked. At first, as I mentally determined, it was going to be about half a liter. With the tests I got exactly half a liter, after analyzing the situation, it must be 2 times 150. The reason is that I use the timer of a microcontroller as a counter. I thought that the increment was due to a rising or falling edge, however, I see that it works by state, that is, any transition from 0 to 1, or vice versa, produces the increment of the counter. \$\endgroup\$ Commented Feb 16, 2024 at 19:02
  • \$\begingroup\$ Usually you just count the number of times it goes high. If you count both highs and lows you might get twice the resolution (but remember that it's then 300 pulses per L). \$\endgroup\$ Commented Feb 16, 2024 at 20:11
  • \$\begingroup\$ You didn't give a part number, but be careful if the meter requires the flow to be above a minimum rate to register correctly. You could imagine an impeller style allowing liquid to flow through it very slowly without turning the impeller. \$\endgroup\$ Commented Feb 16, 2024 at 20:13
  • \$\begingroup\$ I'm afraid to add a serial number or indicate what device it is. It happened to me once that after receiving several responses, I indicated what device it was, and the question was blocked, supposedly for asking for suggestions on a particular device. I didn't understand what the sin was committed. \$\endgroup\$ Commented Feb 16, 2024 at 21:12
  • \$\begingroup\$ The sin was probably that you were looking for help on the use of a commercial part. The site tries to filter out non-design questions. You were fortunate with this one as it was really a maths question (until you started to mention rising and falling edges in the comments to my answer). Have fun! \$\endgroup\$ Commented Feb 16, 2024 at 21:45

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