Steady state process/systems and entropy generation?

Question

As far as I know in a steady-state process, each point in the control volume (system) does not undergo a change in state with time, which means that its properties don't change with time. However, How can we have a steady-state system with entropy generation, since entropy generation tends to increase the entropy?

Answer 1

In a steady state process all internal parameters of the system are constant in time but only the parameters of the system. Outside the system the entropy increases as it should in any real process. For example, if you take a metal rod and fix the temperatures at each end while isolating the cylindrical side of the rod and if the temperatures at the ends are different but kept constant then, after a short transient, steady state heat will flow at a constant rate from the higher temperature end toward the lower temperature end.

If the conductivity is $\kappa$ then the heat flux through a temperature gradient is $$\mathbf q = -\kappa \nabla T$$ and the internally generated entropy production rate is $$\dot \sigma = \mathbf q \cdot \nabla \frac{1}{T}$$ or $$\dot \sigma = (-{\kappa \nabla T})\cdot({\nabla \frac{1}{T}})=\frac{\kappa|\nabla T|^2}{T^2}>0.$$ Given $\dot\sigma$ is the entropy produced per unit volume in the rod and being positive it would grow unbounded were it not dumped to the environment, in this case to the lower temperature reservoir (thermostat) into which the excess heat flows. That is there are two sources of entropy into dumping entropy in the lower temperature reservoir: there is a steady transfer of entropy from the higher temperature reservoir into the lower temperature reservoir $and$ there is the internally (irreversibly) produced entropy by the conduction is also dumped to the lower temperature reservoir.

Answer 2

There is a difference between equilibrium, which, as you say, imply tha tall state variables are constant and there is no net flow of anything (heat, mass, etc.). In this case, there can be no entropy production because the system is frozen.

In the steady state most variables are still constant but there is a (constant) flow. For example, some heat is entering the system somewhere and leaving it somewhere else: one reaches a state where the heat entering is the same as the heat leaving (constant flow), but there still is a net flow of heat in the system which can lead to entropy production. The heat, by flowing in the system, can be used to set the state variables to a new constant value which is constant - but it is not the equilibrium one.

In mathematical terms, if you have a function $f$ describing your state $x$ over time $t$, it evolves over time because of its current $J$ according to the continuity equation (in 1D)

$$\dot{f}=-A\partial_x J$$

i.e. the time derivative of $f$ is proportional ($A$ is a constant) to the space derivative of the current $J$. The current "transports" $x$ around and makes it change state.

A steady state is a state in which $$\dot{f}=0$$ i.e. the state does not change over time. However, this does not imply that the current is zero but simply $$\partial_x J=0$$ i.e. that the current is constant, $J(x)=j_0$ so that $-\partial_x j_0 = 0$.

This current is the source of the entropy production.

At equilibrium, to the condition of steady state $\dot{f}=0$ one has to add the stronger condition $$J=0$$ which is due to detailed balance, time-reversal, etc i.e. general equilibrium physics.

Thus the difference lays in the presence of a current $J$.

Answer 3

At each given location within the control volume, the entropy per unit mass of fluid is not changing with time, but it is a function of spatial position within the control volume (just as temperature, internal energy per unit mass, enthalpy per unit mass, etc. can be functions of spatial position). So as the parcels of fluid move through the control volume, their temperature, internal energy per unit mass, entropy per unit mass, etc. change along the path. But at any given position, they are constant. So if you integrate over the total control volume, the average values of all these parameters does not change with time.

However, for the parcels of fluid exiting the control volume at any time, their temperatures, internal energies per unit mass, entropies per unit mass, etc. are different than those of the new parcels of fluid entering the control volume at that time. Thus, the entropy change is not for the fluid within the control volume (which is constant in time) but for the fluid leaving minus the fluid entering the control volume.

Answer 4

You are right, entropy of the system is a state function, so if the state is steady, then system entropy is constant. However, to be in a nonequilibrium steady state, the system must be open, which usually means there will be heat flow out to the surroundings. That increases the entropy of the surroundings. If the energy dissipated to the surroundings came in as heat, that process will simultaneously decrease the entropy of the surroundings, so the 2nd law says the increase due to the outflow must be greater than the decrease due to the inflow, that is, the intake temperature must be higher than the exhaust temperature. Alternatively, the energy can come in as work being done on the system at a steady rate (e.g. a pressure gradient driving flow through a pipe). That may happen with no decrease in the surroundings' entropy. In that case the steady state means that all the work in is dissipated to the surroundings as heat, and again the entropy of the surroundings increases.

Answer 5

When we use the word process, we indicate that we want to analyze what happens when a change occurs. The absence of the term process is an indication that we are not analyzing a change. We are analyzing a static state.

When we attach the term unsteady state to the term process, we indicate that the change we want to consider occurs at a fixed position as a function of time. When we attach the term steady state to the term process, we indicate that the change we want to consider is not a function of time but is a function of some other parameter. The easiest other parameter to apply is spatial position.

A control volume is a defined region in spatial coordinates. We are at liberty to choose whatever control volume we want, wherever we want, and as big or small as we want. When we want to analyze a steady state process, we should not stand inside a static control volume to try to analyze change as a function of position. Instead, we should pick a control volume that has is as small as is needed to have a uniform state throughout, we should stand outside of that control volume, and we should move it in spatial position throughout the system of interest.

In order for something in a control volume to change during a process (steady state or unsteady state), the entropy of the universe must increase. In an unsteady state process, the entropy of the universe increases with time, the state of the contents in the control volume changes with time while remaining constant throughout the entire control volume, and the control volume remains fixed in its coordinates. In a steady state process, the entropy of the universe increases with time, the state of the substance inside the control volume changes as it is moved from one spatial location to the other, and the rate for the change in state is constant with time.

In summary, for a steady state process, we should use a moving control volume that is as small as is needed to have a uniform state throughout. When the control volume moves its spatial location through the system, the state of the contents in the control volume changes and the entropy of the universe increases for whatever reason. Regardless of when or how we view the movement of the control volume in spatial position (regardless of how long we view the process), the change in the state of the substance in the control volumes occurs at the same rate. This is how best to analyze a steady state process to clear the confusion.