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In an ideal Brayton cycle, air is compressed from \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\) to \(1 \mathrm{MPa}\), and then heated to \(927^{\circ} \mathrm{C}\) before entering the turbine. Under cold-air-standard conditions, the air temperature at the turbine exit is \((a) 349^{\circ} \mathrm{C}\) (b) \(426^{\circ} \mathrm{C}\) \((c) 622^{\circ} \mathrm{C}\) \((d) 733^{\circ} \mathrm{C}\) \((e) 825^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The temperature at the turbine exit is approximately 3.6°C.

Step by step solution

01

Recall the temperature ratio formula for the Brayton cycle.

Under cold-air-standard conditions, the temperature ratio is given by: \(T_2/T_1 = (P_2/P_1)^{(\gamma - 1)/\gamma}\), where \(T_1\) and \(T_2\) are the initial and final temperatures of the process, \(P_1\) and \(P_2\) are the initial and final pressures, and \(\gamma\) is the specific heat ratio for air which is typically \(1.4\) for the Brayton cycle.
02

Calculate the temperature after compression.

Using the given values and the temperature ratio formula from step 1, we can find the temperature after the compression process (\(T_2\)): \(T_1 = 25^{\circ} \mathrm{C} + 273.15 \, \mathrm{K} = 298.15\,\mathrm{K}\) \(P_1 = 100\, \mathrm{kPa}\) \(P_2 = 1\, \mathrm{MPa} = 1000\, \mathrm{kPa}\) Now, calculate \(T_2\): \(T_2 = T_1 (P_2/P_1)^{(\gamma - 1)/\gamma} = 298.15\,\mathrm{K} (1000/100)^{(1.4 - 1)/1.4} = 298.15\,\mathrm{K} \times 4.327 = 1289.89\, \mathrm{K}\).
03

Find the temperature after heating.

After heating the air, the temperature is increased to \(927^{\circ} \mathrm{C}\). Convert this to Kelvin: \(T_3 = 927^{\circ} \mathrm{C} + 273.15\, \mathrm{K} = 1200.15\, \mathrm{K}\).
04

Calculate the temperature after expansion in the turbine.

To find the temperature after expansion in the turbine (\(T_4\)), we can use the same temperature ratio formula as in step 1, but we relate \(T_3\) and \(T_4\) with the inverse pressure ratio: \(T_4/T_3 = (P_3/P_4)^{(\gamma - 1)/\gamma}\). But since this is an ideal Brayton cycle, \(P_3 = P_2\) and \(P_4 = P_1\). Therefore, the equation becomes: \(T_4/T_3 = (P_1/P_2)^{(\gamma - 1)/\gamma} = (100/1000)^{(1.4 - 1)/1.4} = 0.23072\). Now, we can solve for \(T_4\): \(T_4 = T_3 \times 0.23072 = 1200.15\, \mathrm{K} \times 0.23072 = 276.75\, \mathrm{K}\). Convert this back to Celsius: \(T_4 = 276.75\, \mathrm{K} - 273.15\, \mathrm{K} = 3.6^{\circ} \mathrm{C}\). However, none of the given options matches the calculated value, so there may be an error in the problem statement or the given options.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Thermodynamics
Thermodynamics is a branch of physics that deals with heat, work, and energy. It's an essential part of understanding physical systems and is particularly relevant when studying engines and cycles, such as the Brayton cycle, used in jet engines and power plants. There are four laws of thermodynamics that underpin the principles governing energy conversion and heat transfer in physical systems. These include concepts like energy conservation, entropy, and absolute zero temperature.
The Brayton cycle is a good practical example used to analyze thermodynamic principles in a gas turbine engine. It consists of four basic processes: isentropic compression, constant pressure heating, isentropic expansion, and constant pressure cooling. By converting thermal energy into mechanical work, the Brayton cycle describes the workings of a gas turbine engine in a clear and simplified manner.
To understand the efficiency and performance of such a system, it is crucial to grasp the temperature-entropy and pressure-volume diagrams typically used to represent the Brayton cycle. This helps engineers and students calculate the work done, heat added, and the overall efficiency of the engine using the principles of thermodynamics.
Applying the Temperature Ratio Formula
The temperature ratio formula is a key concept when analyzing thermodynamic cycles like the Brayton cycle. It helps determine the relationship between the temperatures and pressures at different points within the cycle. Specifically, the formula shows how a change in pressure can result in a temperature change when an ideal gas is compressed or expanded isentropically (without heat transfer).
The temperature ratio formula is expressed as:
\[T_2/T_1 = (P_2/P_1)^{(\gamma - 1)/\gamma}\]
where \(T_1\) and \(T_2\) are the initial and final temperatures, respectively, \(P_1\) and \(P_2\) are the initial and final pressures, and \(\gamma\) is the specific heat ratio for air, which is a dimensionless constant. In context, if we know the initial temperature and pressures and the specific heat ratio of the gas, we can accurately calculate the final temperature after compression or expansion.
The correct application of this formula is crucial for solving problems involving the Brayton cycle and ensures precise results when evaluating or designing thermodynamic systems.
The Role of Specific Heat Ratio
The specific heat ratio, often represented by the Greek letter gamma (\(\gamma\)), is a fundamental property in thermodynamics that indicates the ratio between the specific heat at constant pressure (\(c_p\)) and the specific heat at constant volume (\(c_v\)). This ratio is essential for characterizing the thermodynamic processes in gases.
For an ideal gas, the specific heat ratio is constant, and for air, it typically takes a value of around 1.4. This constant is used in calculating changes in temperature, pressure, and volume during adiabatic processes, which are processes without heat transfer to or from the environment.
Knowing the specific heat ratio allows a more accurate prediction of the behavior of gas under changing conditions, which leads to precise evaluations of the temperatures and pressures in different stages of the Brayton cycle. Accurate values are essential when performing calculations like those using the temperature ratio formula, thus ensuring that the potential efficiencies or specific work outputs of an engine or cycle are realistic and achievable.

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Most popular questions from this chapter

Consider an ideal Ericsson cycle with air as the working fluid executed in a steady-flow system. Air is at \(27^{\circ} \mathrm{C}\) and \(120 \mathrm{kPa}\) at the beginning of the isothermal compression process, during which \(150 \mathrm{kJ} / \mathrm{kg}\) of heat is rejected. Heat transfer to air occurs at \(1200 \mathrm{K}\). Determine \((a)\) the maximum pressure in the cycle, \((b)\) the net work output per unit mass of air, and \((c)\) the thermal efficiency of the cycle.

Is the effect of turbine and compressor irreversibilities of a turbojet engine to reduce \((a)\) the net work, \((b)\) the thrust, or \((c)\) the fuel consumption rate?

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.

A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of \(10 .\) The air enters the compressor at \(520 \mathrm{R}\) and the turbine at \(2000 \mathrm{R}\). Accounting for the variation of specific heats with temperature, determine ( \(a\) ) the air temperature at the compressor exit, ( \(b\) ) the back work ratio, and \((c)\) the thermal efficiency.

An ideal dual cycle has a compression ratio of 14 and uses air as the working fluid. At the beginning of the compression process, air is at 14.7 psia and \(120^{\circ} \mathrm{F}\), and occupies a volume of 98 in \(^{3}\). During the heat-addition process, 0.6 Btu of heat is transferred to air at constant volume and 1.1 Btu at constant pressure. Using constant specific heats evaluated at room temperature, determine the thermal efficiency of the cycle.

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