How to tell if a thermodynamic cycle is reversible without calculating entropy change? Consider the Carnot cycle, consisting of two reversible, isothermal processes and two isentropic processes. It is reversible, pretty much by definition.
Now consider the Lenoir cycle, consisting of an isochoric compression (heat addition), followed by an isentropic expansion, followed by an isobaric compression (heat loss). I calculated the entropy created by this cycle and found it to be strictly positive. 
However it's not clear intuitively why this cycle should be irreversible. Is heat change at constant volume or at constant pressure necessarily irreversible?
 A: If the isochoric and isobaric transformation are performed reversibly, i.e. quasistatically and without heat dissipation caused by friction or other effects, then your cycle will be reversible.
This is true for every thermodynamic cycle you can draw in the $PV$ plane: if every step is performed reversibly, then the cycle is reversible.
The peculiarity of the Carnot cycle is that it is the only reversible engine that operates between two heat sources only. 
You can easily see how many different heat sources you are using if you draw the cycle into the $TS$ diagram (picture from Wikipedia):

In this case, it is easy to verify that the change in entropy of the surroundings is
$$\Delta S_{surr} = -\frac{Q_H}{T_H}+\frac{Q_C}{T_C} =0$$
So that the engine is indeed reversible.
But now let's take your Lenoir cycle in the $TS$ diagram (picture from Wikipedia):

As you can see, during $1 \rightarrow 2 $ and $3 \rightarrow 1$ you are cutting infinitely many isotherms. The formula you have to use is in this case
$$\Delta S_{surr} = -\int_1^2 \frac{\delta Q}{T} + \int_3^1 \frac{\delta Q}{T}$$
But this time you cannot take out $T$ from the integral like you would do with a Carnot cycle, because it is not a constant.
What you can do is to assume that step $1 \rightarrow 2 $ and $3 \rightarrow 1$ are performed reversibly: in this case, $\Delta S_{surr}=0$ by definition.
