How to understand the difference between standard entropy and generated entropy

Theory

A process changes the system and its surroundings. If the process is reversible the entropy changes cancel out exactly, which is to say that the total entropy before and after the process are the same. If the process is irreversible, then the net change is positive and we call it entropy generation, or $sigma$ in your notation. We can combine both cases into as single equation: $$ tag{1} Delta S_text{sys}+Delta S_text{surr} = sigma geq 0 $$ with the equal sign for reversible process. This is the same as the equation that you wrote if we set $$ tag{2} Delta S_text{surr} = -int frac{dQ_text{rev}}{T} $$ (we use $-dQ_text{rev}$ because heat that is added to the system is removed from the surroundings and vice versa). Equation (2), however, is a special case. It implies that the surroundings are the same temperature as the system and that there are no mechanically irreversible processes in the surroundings. It is an assumption we make we we do not know the actual temperature of the surroundings and by invoking this assumption we calculate a minimum bound on the entropy generation (if the heat is transferred between two different temperatures or if there are other irreversibilities in the surroundings, $sigma$ will be even larger).

I recommend that you always start with Eq. (1) and go from there based on what you know in the particular problem.

Application to your example

In this case there is no entropy change in the surroundings and Eq. 1 becomes $$ sigma = Delta S_text{sys} = S_2 - S_1 $$ We need to go to the steam tables to find $S_1$ and $S_2$ of the system given that the internal energy of the initial state has increased by $W$.

As a demonstration, assume that the box is rigid, it contains only vapor, and the vapor is an ideal gas. Then: $$ sigma = Delta S_text{sys} = m c_Vlnfrac{T_2}{T_1} $$ with $T_2$ calculated from the energy balance: $$ T_2 = T_1 + frac{W}{m c_V} $$ Notice that since $T_>T_1$ we will have $sigma>0$.

If the box contains a liquid/vapor mixture we need a couple interpolations in the steam tables (more than a couple, actually).

Part 1, the integral ... is entropy due to heat transfer.

Yes, but not just any heat transfer. Only reversible heat transfer. Reversible heat transfer occurs across an infinitesimally small difference in temperature. A reversible transfer of heat does not generate entropy.

Therefore first equation should technically read

$$Delta S_{sys}=int_1^2frac{δQ_{rev}}{T}+σ$$

Notice that I have added the subscript "rev" to show that the first term on the right side only applies to a reversible transfer of heat. I have also added the subscript "sys" to indicate it is the change in entropy of the system.

Part 2, generated entropy ? --> entropy coming from where?

It comes from any irreversible process, be it irreversible work or irreversible heat transfer. An irreversible transfer of heat occurs if the transfer occurs across a finite temperature difference (thermal disequilibrium). Irreversible work occurs mainly as a result of friction and/or finite pressure differences (mechanical disequilibrium).

If heat is involved, then part 1 is responsible for an entropy change in the form of lost heat.

As indicated above, only a reversible transfer of heat is responsible for entropy change of part 1. Irreversible heat transfer generates entropy (part 2). The change can be in the form of loss or gain of heat, where the change in entropy is either negative (heat transfers out of the system) or positive (heat transfers into the system), respectively.

If work is involved, then part 2 is involved, in the form of irreversibilities.

Part 2 represents entropy generated and it occurs for any irreversible process, not just work. If work is involved it must be irreversible work. Reversible work does not generate entropy. But irreversible heat transfer also generates entropy. The difference between part 1 and part 2 regarding heat is that the temperature difference for part 1 is infinitesimally small whereas it is finite for part 2.

It's important to understand that in reality all real heat and work processes are irreversible. Heat always requires a finite temperature difference (thermal disequilibrium). Work always involves some form of mechanical disequilibrium as well as some friction. This is why it is said that the entropy of the universe is always increasing, due to the entropy generation that results from all real processes. The reversible process is an idealization where the total entropy change is zero.

How can I know the value of the generated entropy, other than measuring the difference between state 1 and 2 in a table? Being entropy a state function, we don't care about the "path" to reach state 2.

You can't determine the generated entropy without information on states 1 and 2, regardless of where you obtain it (table or otherwise). The difference in entropy between state 1 and 2 is the total change in entropy of the system, meaning the sum of reversible entropy (part 1) and generated entropy (part 2).

You can determine the part that is generated entropy by devising any reversible path that returns the system to its original state. When you return the system to its original state, the entropy that was generated in the system is transferred to the surroundings in the form of heat. That quantity of entropy equals the entropy generated.

In your example since there is no heat transfer with the surroundings, all of the change in entropy will necessarily be generated entropy due to irreversible work. With sufficient data on the initial state, and assuming you can ignore any heat transfer to the container and stirrer blade masses (discussed below), you can determine the entropy generated directly from the steam tables. Or you can determine the enthalpy change from the tables (which is equivalent to the effect of heat transfer) and divide it by an appropriate temperature to get the entropy change.

But... would it be different if my propeller is bigger or heavier or has more or less frictions with the container?

Friction heating of the water is automatically included in the entropy generated because it contributes to the phase change. But the propeller and container can make a difference.

Since the process is carried out in a rigid container, the change from a liquid vapor mix to total vapor will not occur at constant temperature and pressure (both will rise). I did a rough calculation assuming 1 kg of a 50-50 liquid-vapor mix and an initial temperature of 100 C, and found the final temperature would be between 120 C and 125 C. If the system before stirring was in internal equilibrium, the temperatures of all the contents would be initially the same. But when the stirring increases the temperature of the water, there will be heat transfer inside the container between the water and the container and propeller materials, raising their temperatures and increasing their entropies. As a result, the increase in entropy of the water as determined by the steam tables, will be less than the total entropy generated by the stirrer work.

Including the container and stirrer blade as part of the system obviously complicates matters. In these types of problems the mass of the materials involved in confining or doing work on the system are generally ignored. For example, in piston/cylinder problems the heat capacities of the piston and cylinder are generally ignored.

Hope this help