If a shunt motor running with a load has its shunt field open, how would this effect the speed of the motor? We know its not going to slow down or stop suddenly so can rule those out fast. Stuck between unaffected or speed up. Im really thinking speeding up would be correct. Im about 99% sure its going to speed up

 

It can speed up to destruction in some cases, this is why DC shunt motor drives have a field loss detection circuit to shut the motor down.
This phenomenon also is seen on Series motors as a run-away condition because of field weakening as the rpm increases, and the reason they are operated with a load to restrict their maximum rpm.
An automotive starter motor works on this principle, one reason why you should not run one on the bench unloaded.
Max.

 

If the field developed a very high resistance, and thus very low current then I can agree that runaway and physical destruction would result. But if the field were opened completely then there would be no magnetic field for the armature to react against, and it is my prediction that the motor would coast to a stop and then catch fire. Or maybe catch fire before coming to a complete stop.

 

, and thus very low current then I can agree that runaway and physical destruction would result. But if the field were opened completely then there would be no magnetic field for the armature to react against,

Sorry that is not what happens in fact.
Every generator and motor (usually) possesses residual magnetism.
http://www.wisc-online.com/Objects/ViewObject.aspx?ID=IAU13808
And is borne out by many past disasters.
This is how a generator can come up to generation without any field voltage applied.
And then provide field voltage from the generated voltage, (self excitation).

Max.

 

Thanks for the correction Max.