Entropy of loops in the PV plane The change in entropy of the Carnot and reversible cycles is said to be 0. Several other loops are supposed to have a non-negative change in entropy.
This presents 2 problems which I cannot reconcile. 1) Entropy is supposed to be a state function so shouldn't any loop on the PV plane bring the system to its initial state and thus its initial entropy? 2) Any loop on the PV plane is reversible given enough heat sources. Shouldn't all reversible cycles, not just the Carnot, have a zero change in entropy. Wouldn't this mean that any loop on the PV plane would have a zero change of entropy? 
 A: 
1) Entropy is supposed to be a state function so shouldn't any loop on the PV plane bring the system to its initial state and thus its initial entropy?

Pair of values $P,V$ may not be enough to specify the state of the system, there may be other thermodynamics state variables. If so, it is possible system returns to the same values $P,V$, but it is in a different thermodynamic state because the additional quantities have different value.
But when the pair of values $P,V$ is sufficient to describe thermodynamic state, like it is for a simple homogeneous system such as gas far from condensation, its entropy is function of $P,V$ only and the system indeed gets the same entropy after the values of $P,V$ are restored.
This is true also for any irreversible cycle that has a point where the state is thermodynamic equilibrium state with definite $P,V$. If the the system gets back to such state, it does not matter whether this happened reversibly or not - the entropy returns to the same value characteristic for the equilibrium state $P,V$.
The claim that entropy increases when irreversible cycle is performed means that the entropy of (system + its environment (heat reservoir)) increases. That way, the entropy of the system may return to its original value, while the total entropy still increases.
A: 
1) Entropy is supposed to be a state function so shouldn't any loop on the PV plane bring the system to its initial state and thus its initial entropy?

Yes, for a reversible process. For an irreversible process where heat has been lost along the way you do not reach the exact same state (less internal energy).

2) Any loop on the PV plane is reversible given enough heat sources.

While that is true for a part of the system, you need to consider the full closed system to see the total entropy change. A part of a system can easily have entropy changes that are zero or negative.
When ice is melting in a water bath (assuming isolated system):


*

*The entropy of the water is decreasing, $\Delta S_{water} < 0$.

*The entropy of the ice block is increasing, $\Delta S_{ice} > 0$.

*The whole system will all in all experience an increase in entropy, $\Delta S_{system} = \Delta S_{water} +\Delta S_{ice} >0 $.

A: 
1) Entropy is supposed to be a state function so shouldn't any loop on the PV plane bring the system to its initial state and thus its initial entropy?

It is, but this is the PV plane, it is a projection of the total state space, so every PV point can correspond to different entropies. A reversible cycle will be closed in the PV plane, but not all closed cycles PV cycles are reversible.

Shouldn't all reversible cycles, not just the Carnot, have a zero change in entropy. Wouldn't this mean that any loop on the PV plane would have a zero change of entropy?

It would if you were in the SV or SP plane, but not in the PV plane.
