Electronic – Current draw of IC

In the steady-state case, it's a simple linear interpolation:

\$I_S\$ is the stall current.

\$\omega_{nl}\$ is the no-load speed.

\$I_{nl}\$ is the no-load current.

For a motor loaded such that it spins with a speed \$\omega_L\$, the current draw is:

$$I_L = (I_S- I_{nl}) (1-\frac{\omega_L}{\omega_{nl}}) + I_{nl} $$

For example, you have a motor with a 2.2A stall current, 0.2A no-load current, and let's assume the no-load speed is 1000 rpm. Technically \$\omega\$ usually refers to rads/s, but here the units cancel out; the only important factor is what the relative speed of the motor compared to the no-load speed is.

At a 250 rpm load speed, the current draw is:

$$I_L = (2.2A - 0.2A) (1-\frac{250}{1000}) + 0.2A = 1.7A$$

Keep in mind that even though the steady-state current of the motor is 1.7A, the peak current could be much more. Starting up the motor would draw close to the stall current of 2.2A, decreasing until the motor reaches steady state. Depending on how quickly the motor is able to get to steady state, this may be significant or not.

You haven't given any reason you need deliberate current limiting at all. Just skip it.

It sounds like the transformer itself will limit the total delivered power. I don't get what exactly you are trying to protect against. The 7809 will protect itself because it has a thermal shutoff built in. You didn't give specs on the transformer, but there's a good chance that it can handle a shorted secondary indefinitely.

And no, the 7809 does not do current limiting. You are confusing the current capability of different 7809 variants with current limiting. The current capability is the max current the device is guaranteed to be able to output indefinitely, assuming you have ensured that all other parameters are met. Note that this almost certainly includes a heat sink or some kind of cooling.

The 78xx series usually needs 2 to 3 V of input headroom. At 2.5 V input headroom and 500 mA out, it will dissipate 1.25 W. Check the junction to ambient thermal resistance, but that probably means just sticking up a TO-220 package in free air isn't good enough.

Your real challenge will be designing the thermal system to deal with the heat produced in the 7809. A transformer followed by a full wave bridge and a cap isn't going to keep the 7809 input voltage right at the minimum threshold. The dissipation will therefore be higher than in the example above.