In a state of thermodynamic equilibrium there are no internal gradients of mechanical, thermal or chemical potential; consequently the macroscopic properties of the system remain steady in time. If a disturbance (for example a change in external pressure, a temperature difference or a change in composition) is imposed on such a system, the system departs from its initial equilibrium and evolves toward a new equilibrium. Whether this evolution is reversible or irreversible depends on whether the process can be reversed so that both the system and its surroundings are returned exactly to their initial states through the same sequence of intermediate equilibrium states.
Consider a simple closed system consisting of a gas confined by a frictionless piston. Initially the gas pressure equals the external pressure due to the piston weight and its temperature equals that of the surroundings; hence the system is at equilibrium (state A). If a small mass is placed on the piston, the external pressure becomes slightly larger than the gas pressure and the piston moves downward until a new equilibrium (state B) is reached. Repeating this operation with more masses in small increments moves the system through a sequence of states from A to X.
If each mass is removed in the reverse order and in the same incremental manner, and if at every intermediate stage the system and surroundings return through exactly the same states as in the forward process, the overall process A → X → A is reversible. If the reverse path does not coincide with the forward path, or if the surroundings are not restored to their initial condition, the process is irreversible.
A quasi-static process is one that proceeds infinitely slowly so that the system at each instant remains arbitrarily close to equilibrium. Quasi-staticity is necessary but not sufficient for reversibility: all reversible processes must be quasi-static, but a quasi-static process can still be irreversible if dissipative effects (friction, viscosity, finite temperature differences, etc.) are present.
When the piston is loaded by infinitesimally small masses added one after another, the motion can be made arbitrarily slow and the system stays almost in equilibrium throughout. In this limiting case the forward and reverse paths coincide and the process is reversible. If, however, a finite mass is dropped suddenly on the piston, the piston will accelerate and oscillate; viscous dissipation and internal friction convert mechanical energy into heat and the oscillation dies out, leaving the system in a new equilibrium. That sudden process is irreversible.
A process is thermodynamically reversible if, after reversing the direction of all driving forces, the system and surroundings can be returned to their exact initial states without any net changes in the rest of the universe. Equivalently, the process must produce zero entropy generation. This last statement leads naturally to the entropy criterion below.
Define the total entropy differential of an isolated universe (system + surroundings) as dStotal. Then the second law gives the inequality
dStotal = dSsystem + dSsurroundings ≥ 0
For a reversible process dStotal = 0. For an irreversible process dStotal > 0. Thus entropy generation is the quantitative measure of irreversibility.
Only for processes that proceed through equilibrium states can the instantaneous intensive variables (pressure, temperature) be assigned uniquely to the system and used to evaluate work and heat in the familiar integral form.
For a reversible mechanical process in a closed system the work exchanged is
Wrev = ∫ Pint dV
where Pint is the internal (equilibrium) pressure of the system at each instant. If the process is irreversible, internal pressure is not uniquely defined during transient non-equilibrium states and the simple integral above referring to system pressure is not valid. In many irreversible practical cases the work can be computed using the external pressure (for example, when expansion is opposed by a known constant external pressure),
W = ∫ Pext dV
but this does not represent the maximum work extractable between the same end states; reversible work is an upper bound.
Heat transfer is reversible only in the limiting case of an infinitesimal temperature difference between system and surroundings. If heat flows across a finite temperature difference ΔT, the transfer is irreversible because entropy is generated. In the reversible limit the exchanged heat δQ_rev is related to entropy change as
δQrev = T dS
Quasi-static means the system passes through a continuous sequence of equilibrium states. Reversible requires, additionally, absence of dissipative effects (friction, viscosity, finite ΔT, etc.). Thus:
Although strictly reversible processes are idealisations that cannot be realised in finite time with real substances, the concept is central to engineering because reversible processes set upper bounds on performance. For instance, the maximum work obtainable or the maximum thermal efficiency of a heat engine operating between two temperatures is achieved only by reversible (Carnot) processes; any irreversibility reduces efficiency. In civil engineering practice the ideas are relevant in the design and analysis of HVAC and refrigeration systems, power plants used for building services, thermal insulation and heat-flow problems in structures, and energy efficiency studies.
Recognising sources of irreversibility permits targeted improvements: reducing flow resistance to lower viscous losses, increasing heat exchange area or reducing ΔT to reduce entropy generation in heat transfer, and avoiding abrupt expansions or compressions where possible.
29 videos|153 docs|36 tests |
| 1. What is a reversible thermodynamic process? | ![]() |
| 2. What is an irreversible thermodynamic process? | ![]() |
| 3. What are some examples of reversible thermodynamic processes? | ![]() |
| 4. What are some examples of irreversible thermodynamic processes? | ![]() |
| 5. Why are reversible processes considered ideal in thermodynamics? | ![]() |