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Why does a nonquasi-equilibrium compression process require a larger work input than the corresponding quasiequilibrium one?

Short Answer

Expert verified
Answer: A nonquasi-equilibrium compression process requires a larger work input because the rapid changes in pressure and volume do not allow the system to maintain a near-equilibrium state throughout the process. This leads to increased resisting forces (pressure) from the system, requiring more work to compress the system. In contrast, a quasi-equilibrium process, where the system remains close to the equilibrium state, results in minimal resisting forces and less work required for compression.

Step by step solution

01

Quasi-equilibrium processes definition

A quasi-equilibrium process is a process whereby the system remains infinitesimally close to the equilibrium throughout the entire process, which allows us to apply the equilibrium state properties at every stage. In a quasi-equilibrium process, the pressure and volume of the system change slowly, and the system can adjust its internal distribution near the equilibrium state at all times.
02

Nonquasi-equilibrium processes definition

A nonquasi-equilibrium process is a process in which the system does not remain close to the equilibrium state during the process. In this case, the changes in pressure, volume, or temperature occur rapidly, and the system cannot maintain a near-equilibrium state throughout the process.
03

Work done during compression processes

The work done on a system during a compression process is given by \(W = -\int_{V_1}^{V_2} P\,\mathrm{d}V\), where \(W\) is the work done, \(V_1\) and \(V_2\) are the initial and final volumes respectively, and \(P\) is the pressure. In a quasi-equilibrium process, the pressure is well-defined at every stage, and the integral can be easily calculated. In a nonquasi-equilibrium process, the pressure is not well-defined at all stages, and we have to consider the worst-case scenario in our calculations.
04

Comparison of work input between quasi-equilibrium and nonquasi-equilibrium processes

Due to their nature, quasi-equilibrium processes always produce the least amount of work required for a process to occur. As the system remains close to the equilibrium state, the resisting force (pressure) is always adjusted to a minimum value, resulting in less work done by the external agent to compress the system. On the other hand, in a nonquasi-equilibrium process, the system does not have an opportunity to adjust its internal distribution properly due to the rapid changes in the pressure and volume. This leads to higher resisting forces (pressure) from the system and, therefore, requires a greater work input to compress the gas. In summary, a nonquasi-equilibrium compression process requires more work input than the corresponding quasiequilibrium one because the pressure and volume change rapidly and the system cannot maintain a near-equilibrium state throughout the process, which results in increased resisting forces, requiring more work to compress the system.

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Most popular questions from this chapter

A heat pump cycle is executed with \(R-134 a\) under the saturation dome between the pressure limits of 1.4 and \(0.16 \mathrm{MPa}\) The maximum coefficient of performance of this heat pump is \((a) 1.1\) (b) 3.8 \((c) 4.8\) \((d) 5.3\) \((e) 2.9\)

A Carnot heat engine operates between a source at \(1000 \mathrm{K}\) and a sink at \(300 \mathrm{K}\). If the heat engine is supplied with heat at a rate of \(800 \mathrm{kJ} / \mathrm{min}\), determine \((a)\) the thermal efficiency and \((b)\) the power output of this heat engine.

A commercial refrigerator with refrigerant- 134 a as the working fluid is used to keep the refrigerated space at \(-35^{\circ} \mathrm{C}\) by rejecting waste heat to cooling water that enters the condenser at \(18^{\circ} \mathrm{C}\) at a rate of \(0.25 \mathrm{kg} / \mathrm{s}\) and leaves at \(26^{\circ} \mathrm{C}\). The refrigerant enters the condenser at \(1.2 \mathrm{MPa}\) and \(50^{\circ} \mathrm{C}\) and leaves at the same pressure subcooled by \(5^{\circ} \mathrm{C}\). If the compressor consumes \(3.3 \mathrm{kW}\) of power, determine \((a)\) the mass flow rate of the refrigerant, \((b)\) the refrigeration load, \((c)\) the \(C O P,\) and \((d)\) the minimum power input to the compressor for the same refrigeration load.

Consider a Carnot heat-pump cycle executed in a steady-flow system in the saturated liquid-vapor mixture region using refrigerant-134a flowing at a rate of \(0.22 \mathrm{kg} / \mathrm{s}\) as the working fluid. It is known that the maximum absolute temperature in the cycle is 1.2 times the minimum absolute temperature, and the net power input to the cycle is \(5 \mathrm{kW}\). If the refrigerant changes from saturated vapor to saturated liquid during the heat rejection process, determine the ratio of the maximum to minimum pressures in the cycle.

A Carnot heat engine receives heat from a reservoir at \(900^{\circ} \mathrm{C}\) at a rate of \(800 \mathrm{kJ} / \mathrm{min}\) and rejects the waste heat to the ambient air at \(27^{\circ} \mathrm{C}\). The entire work output of the heat engine is used to drive a refrigerator that removes heat from the refrigerated space at \(-5^{\circ} \mathrm{C}\) and transfers it to the same ambient air at \(27^{\circ} \mathrm{C}\). Determine \((a)\) the maximum rate of heat removal from the refrigerated space and ( \(b\) ) the total rate of heat rejection to the ambient air.

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