# Evaluation of incremental sensor

Evaluating a volume flow incremental sensor does not deliver expected results.

The pump of which the volume flow is to be measured has a hand-calculated flow rate of

$$Q = \dfrac{n}{ 60 \;\mathrm{s/min}} e \pi\left(\frac{d}{2}\right)^2 \approx 2 \;\mathrm{cm^3/s}$$

where $n$ is the rotational speed of $2000 \;\mathrm{1/min}$, $e$ the excentricity of the pump of $1.1 \;\mathrm{mm}$, and $\pi\left(\frac{d}{2}\right)^2$ is the area of the piston with $d = 8 \;\mathrm{mm}$.

Now the sensor (VSE, Type: VS 0,1) delivers 10000 impulses per liter.

The calculation counts the "Delta Impulses", call it dI, and divides by the time delta dt in milliseconds.

$$\dfrac{dI}{dt} = \text{impulses/ms}$$

Now I say that $10$ impulses mean $1 \;\mathrm{cm^3}$ as the spec says ($1 \;\mathrm{liter} = 1000 \;\mathrm{cm^3}$). So $$1 \;\text{impulse} = 0.1 \;\mathrm{cm^3/ms}$$

Further with $1000 \;\mathrm{ms} = 1\;\mathrm{s}$, it is $100 \;\mathrm{cm^3/s}$. I can multiply the $\frac{dI}{dt}$ with $100$ and get the flow in $\mathrm{cm^3/s}$.

$$1 \frac{dI}{dt} = 100 \;\mathrm{cm^3/s}$$

Is that correct?

Now this calculation is done within the program. The factor $100$ is given as a parameter that I have set to this value to translate the flow into $\mathrm{cm^3/s}$. The program always uses milliseconds (PLC-Control with $1\;\mathrm{ms}$ cycle, increments time ticker each time +1).

I don't know the values of the used impuls count, but the result in $\mathrm{cm^3/s}$ is about $10$ times higher than the hand-calculated flow rate. I can't find a logical mistake. Do you?