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What is the second-law efficiency? How does it differ from the first-law efficiency?

Short Answer

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Answer: The main difference between the Second-law efficiency and the First-law efficiency is that the Second-law efficiency takes into account the irreversibilities and wasted energy present in a system, while the First-law efficiency is based on the conservation of energy principle and does not consider the irreversibilities.

Step by step solution

01

Introduction to Second-law Efficiency

The Second-law efficiency, also called the exergy efficiency, takes into account the inefficiencies or irreversibilities present in a system due to factors such as friction, heat losses, and chemical reactions. It is defined as the ratio of the useful exergy (the maximum available work from a system) obtained from a process, to the exergy supplied to the process. It can be mathematically represented as: \[ \eta_{II} = \frac{Exergy\: Obtained}{Exergy\: Supplied} \]
02

Introduction to First-Law Efficiency

The First-law efficiency, also known as the energy efficiency, is based on the conservation of energy principle, and is a measure of how effectively a system converts energy from one form to another. It is defined as the ratio of the useful energy (work or heat) output of a process to the energy input to the process. Mathematically, it can be represented as: \[ \eta_{I} = \frac{Useful\: Energy\: Output}{Energy\: Input} \]
03

Difference between Second-law and First-law Efficiencies

The main differences between Second-law and First-law efficiencies are: 1. Second-law efficiency takes into account the irreversibilities present in a system, while the First-law efficiency considers only the energy conversion and does not account for irreversibilities. 2. Second-law efficiency is based on the maximum available work (exergy) from a process, while the First-law efficiency is based on the conservation of energy principle. 3. First-law efficiency may approach 100% for an idealized system without irreversibilities, whereas Second-law efficiency can never reach 100% due to the presence of irreversibilities in real-life processes. 4. Second-law efficiency provides a more meaningful evaluation of the system's performance, as it not only takes into account the energy conversion but also the wasted energy due to inefficiencies. In summary, the Second-law efficiency is a more comprehensive performance measure that takes into account the irreversibilities and wasted energy present in a system, while the First-law efficiency is a simpler measure based on the conservation of energy principle and does not consider the irreversibilities.

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Most popular questions from this chapter

Argon gas enters an adiabatic compressor at \(120 \mathrm{kPa}\) and \(30^{\circ} \mathrm{C}\) with a velocity of \(20 \mathrm{m} / \mathrm{s}\) and exits at \(1.2 \mathrm{MPa}\) \(530^{\circ} \mathrm{C},\) and \(80 \mathrm{m} / \mathrm{s}\). The inlet area of the compressor is \(130 \mathrm{cm}^{2} .\) Assuming the surroundings to be at \(25^{\circ} \mathrm{C}\), determine the reversible power input and exergy destroyed.

An insulated vertical piston-cylinder device initially contains \(15 \mathrm{kg}\) of water, \(13 \mathrm{kg}\) of which is in the vapor phase. The mass of the piston is such that it maintains a constant pressure of \(300 \mathrm{kPa}\) inside the cylinder. Now steam at \(2 \mathrm{MPa}\) and \(400^{\circ} \mathrm{C}\) is allowed to enter the cylinder from a supply line until all the liquid in the cylinder is vaporized. Assuming the surroundings to be at \(25^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\), determine \((a)\) the amount of steam that has entered and ( \(b\) ) the exergy destroyed during this process.

Obtain the following information about a power plant that is closest to your town: the net power output; the type and amount of fuel used; the power consumed by the pumps, fans, and other auxiliary equipment; stack gas losses; temperatures at several locations; and the rate of heat rejection at the condenser. Using these and other relevant data, determine the rate of irreversibility in that power plant.

Writing the first- and second-law relations and \(\operatorname{sim}-\) plifying, obtain the reversible work relation for a uniformflow system that exchanges heat with the surrounding medium at \(T_{0}\) in the amount of \(Q_{0}\) as well as a heat reservoir at \(T_{R}\) in the amount of \(Q_{R^{*}}\) (Hint: Eliminate \(Q_{0}\) between the two equations.)

Combustion gases enter a gas turbine at \(627^{\circ} \mathrm{C}\) and \(1.2 \mathrm{MPa}\) at a rate of \(2.5 \mathrm{kg} / \mathrm{s}\) and leave at \(527^{\circ} \mathrm{C}\) and \(500 \mathrm{kPa} .\) It is estimated that heat is lost from the turbine at a rate of \(20 \mathrm{kW}\). Using air properties for the combustion gases and assuming the surroundings to be at \(25^{\circ} \mathrm{C}\) and 100 kPa, determine \((a)\) the actual and reversible power outputs of the turbine, (b) the exergy destroyed within the turbine, and \((c)\) the second-law efficiency of the turbine.

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