What is the expected maximum voltage for the pulse from this flow meter circuit?

The input resistance may be nonlinear - something like an optocoupler LED, maybe with some additional parts. However, I would still expect a voltage more like you calculated, or perhaps even a bit higher.

Maybe your resistor is actually 47kΩ .. I suggest measuring it to be sure. Also if it's actually delivering pulses, you could be reading the average of hi and lo.

I would like to know the expected maximum amplitude for the pulse from this circuit if using 24V as the power supply.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Equivalent circuit.

• The output voltage when switch is open will be 24 V (provided R2 << R_PLC).
• The output voltage when SW1 is closed will be 0 V. With an NPN transistor turned hard-on the output voltage will be about 0.7 V (according to the specification you posted).

The input impedance at the PLC input is 3kΩ. The PLC seems to pick up the signal, but the voltage measured 2.58VDC maximum when measured between PLC Common and PLC Input when the circuit is connected to the PLC and flow meter is running. Is the voltage seen by the PLC input not supposed to be based on a voltage divider? 24V × (3000 Ω / (4700 Ω + 3000 Ω)) = 9.35 V.

Your maths looks right but that's when the switch is open. The specification says that the flowmeter has a duty cycle of 50% when running, meaning that half the time it will be at 0.7 V. That gets your average down to 9.35 / 2 = 4.7 V.

So where's the error? Let's look more closely at the specification of the PLC input:

• Input impedance: 3 kΩ.
• Input current: 6 mA @ 24 V DC.

From Ohm's law, \$ R = \frac V I = \frac {24}{6m} = 4\ \text k \Omega \$. It doesn't help us much as it pushes the voltage away from what we expected.

My guess is that the 50% duty cycle isn't accurate and that the switch is closed less than half the time. An oscilloscope would be useful to diagnose further.