Trouble understanding this 'simple' circuit
The idea is that when Q10 is open, the valve turns off.  It does this by not providing a path to ground.  VALVE7- goes to +24V by leakage through C27 and D8 (assuming no valve is attached).  If the valve is attached, the entire thing is at +24V relative to ground.

The leakage through C27 and D8 shouldn't be much.  Enough to see the voltage, but if you shorted it, there would be so little current you probably couldn't measure it.