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What does the area enclosed by the cycle represent on a \(P\) - \(V\) diagram? How about on a \(T\) -s diagram?

Short Answer

Expert verified
In a \(P\) - \(V\) diagram (pressure-volume diagram), the area enclosed by a cycle represents the total work done by the system during the thermodynamic cycle. On the other hand, in a \(T\) -s diagram (temperature-entropy diagram), the area enclosed by a cycle represents the total heat absorbed by the system during the entire thermodynamic cycle.

Step by step solution

01

Understand the concept of a thermodynamic cycle

A thermodynamic cycle is a series of thermodynamic processes, during which a system (such as a gas or working fluid) undergoes various changes and ultimately returns to its initial state. In an ideal cycle, the working fluid exchanges heat and work with its surroundings, and some of this work can be extracted as useful output (such as running an engine). Examples of thermodynamic cycles include the Carnot cycle, Otto cycle, and Rankine cycle.
02

The significance of the area enclosed in a \(P\) - \(V\) diagram

A \(P\) - \(V\) diagram (also known as a pressure-volume diagram) is a graphical representation of the changes in pressure and volume of a system during a thermodynamic cycle. The x-axis represents the volume and the y-axis represents the pressure. As the system goes through the cycle, the path traced out on the \(P\) - \(V\) diagram represents the relationships between pressure and volume for each process in the cycle. The area enclosed by the cycle on a \(P\) - \(V\) diagram has a specific meaning: it represents the total amount of work done by the system during the entire thermodynamic cycle. This can be deduced from the fact that the work done by the system during a specific process is given by: $$W = -\int_{V_1}^{V_2}P(V) \, dV$$ where \(W\) represents the work done, \(P\) is the pressure of the system, \(V\) is the volume, and the integration limits (\(V_1\) and \(V_2\)) represents the volume at the initial and final states, respectively. Since the work done during the whole cycle is the sum of work done during each process, the total area under each process on the \(P\) - \(V\) diagram becomes the total work done during the cycle.
03

The significance of the area enclosed in a \(T\) -s diagram

A \(T\) -s diagram (also known as a temperature-entropy diagram) is another way to visualize the thermodynamic processes in a cycle. In this case, the x-axis represents the entropy and the y-axis represents the temperature. As the system goes through the cycle, the path traced out on the \(T\) -s diagram represents the relationships between temperature and entropy for each process in the cycle. The area enclosed by the cycle on a \(T\) -s diagram represents the total amount of heat absorbed by the system during the entire thermodynamic cycle. To understand this, recall that heat transfer, denoted as \(Q\), during a particular process in a cycle can be expressed as: $$Q = T \int_{s_1}^{s_2} dS$$ Where \(T\) is the temperature of the system, and the integration limits (\(s_1\) and \(s_2\)) represent the entropy at the initial and final states, respectively. The change in entropy during a process is proportional to the heat absorbed or rejected by the system (since entropy is a measure of energy dispersal), so the area enclosed by the cycle on a \(T\) -s diagram represents the net heat absorbed during the entire cycle.
04

Conclusion

To summarize, the area enclosed by a cycle on a \(P\) - \(V\) diagram represents the total work done by a system during the thermodynamic cycle, while the area enclosed by a cycle on a \(T\) -s diagram represents the total heat absorbed by the system during that cycle. Both diagrams provide useful insights into the performance and efficiency of thermodynamic cycles and allow engineers to design and analyze complex systems such as engines and refrigeration systems.

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Most popular questions from this chapter

Air in an ideal Diesel cycle is compressed from 2 to \(0.13 \mathrm{L},\) and then it expands during the constant pressure heat addition process to 0.30 L. Under cold air standard conditions, the thermal efficiency of this cycle is (a) 41 percent (b) 59 percent \((c) 66\) percent \((d) 70\) percent \((e) 78\) percent

Consider an ideal gas-turbine cycle with two stages of compression and two stages of expansion. The pressure ratio across each stage of the compressor and turbine is 3 The air enters each stage of the compressor at \(300 \mathrm{K}\) and each stage of the turbine at \(1200 \mathrm{K}\). Determine the back work ratio and the thermal efficiency of the cycle, assuming \((a)\) no regenerator is used and \((b)\) a regenerator with 75 percent effectiveness is used. Use variable specific heats.

Consider an aircraft powered by a turbojet engine that has a pressure ratio of \(9 .\) The aircraft is stationary on the ground, held in position by its brakes. The ambient air is at \(7^{\circ} \mathrm{C}\) and \(95 \mathrm{kPa}\) and enters the engine at a rate of \(20 \mathrm{kg} / \mathrm{s}\) The jet fuel has a heating value of \(42,700 \mathrm{kJ} / \mathrm{kg},\) and it is burned completely at a rate of \(0.5 \mathrm{kg} / \mathrm{s}\). Neglecting the effect of the diffuser and disregarding the slight increase in mass at the engine exit as well as the inefficiencies of engine components, determine the force that must be applied on the brakes to hold the plane stationary.

A simple ideal Brayton cycle without regeneration is modified to incorporate multistage compression with intercooling and multistage expansion with reheating, without changing the pressure or temperature limits of the cycle. As a result of these two modifications, (a) Does the net work output increase, decrease, or remain the same? (b) Does the back work ratio increase, decrease, or remain the same? \((c) \quad\) Does the thermal efficiency increase, decrease, or remain the same? (d) Does the heat rejected increase, decrease, or remain the same?

An ideal Brayton cycle has a net work output of \(150 \mathrm{kJ} / \mathrm{kg}\) and a back work ratio of \(0.4 .\) If both the turbine and the compressor had an isentropic efficiency of 85 percent, the net work output of the cycle would be \((a) 74 \mathrm{kJ} / \mathrm{kg}\) \((b) 95 \mathrm{kJ} / \mathrm{kg}\) \((c) 109 \mathrm{kJ} / \mathrm{kg}\) \((d) 128 \mathrm{kJ} / \mathrm{kg}\) \((e) 177 \mathrm{kJ} / \mathrm{kg}\)

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