From the Clausius inequality we can derive that the efficiency of a Carnot (reversible) cycle is given by: $$e= 1 - \frac{T_c}{T_h}$$
Is this true for every reversible cycle? Is the efficiency of all reversible cycle equal to the efficiency of a Carnot Cycle? If not, what is responsible of this difference? (Is it due to the fact that the temperature are not the same throughout the whole cycle as they are during the isothermal transformation of the Carnot cycle?)
Is this true for every reversible cycle? Is the efficiency of all reversible cycle equal to the efficiency of a Carnot Cycle?
Yes. They are indeed.
The equality in Clausius' Inequality $$\oint \frac{đq_\textrm{sys}}{T_\textrm{source}}=0$$ is strictly valid for all reversible cycles.
Temperature of a reversible engine is at all times equal to the temperature of the heat sources that it is in contact with.
Thus, the entropy of the universe would be zero always when the cycles are reversible and that means efficiency of all reversible cycles would be the same viz.
$$e~=~ 1-\frac{T_\mathrm C}{T_\mathrm H}\;.$$
As summed up by Fermi in his lectures:
If there are several cyclic heat engines, some of which are reversible, operating around cycles between the same temperatures $t_1$ and $t_2,$ all the reversible ones have the same efficiency, while the nonreversible ones have efficiencies which can never exceed the efficiency of the reversible engines.
