/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 An ideal air-standard Brayton cy... [FREE SOLUTION] | 91Ó°ÊÓ

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An ideal air-standard Brayton cycle operating at steady state produces \(10 \mathrm{MW}\) of power. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.9. Sketch the \(T-5\) diagram for the cycle and determine (a) the mass flow rate of air, in \(\mathrm{kg} / \mathrm{s}\) (b) the rate of heat transfer, in \(\mathrm{kW}\), to the working fluid passing through the heat exchanger. (c) the thermal efficiency. $$ \begin{array}{crcr} \text { State } & p(\mathrm{kPa}) & T(\mathrm{~K}) & h(\mathrm{k}] / \mathrm{kg}) \\ \hline 1 & 100 & 300 & 300.19 \\ 2 & 1200 & 603.5 & 610.65 \\ 3 & 1200 & 1450 & 1575.57 \\ 4 & 100 & 780.7 & 800.78 \end{array} $$

Short Answer

Expert verified
The mass flow rate is calculated using the power and enthalpies. The heat transfer rate is mass flow rate times enthalpy difference between states 3 and 2. The efficiency is derived from enthalpy differences.

Step by step solution

01

- Sketch the T-s diagram

Draw the Temperature vs. Entropy (T-s) diagram for the Brayton cycle. The cycle consists of four processes: 1-2 isentropic compression, 2-3 constant pressure heat addition, 3-4 isentropic expansion, and 4-1 constant pressure heat rejection. Be sure to label each state (1, 2, 3, 4) on the diagram.
02

- Calculate the mass flow rate

Using the given power output of the cycle (10 MW) and the enthalpies at each state, apply the first law of thermodynamics. The power output can be represented as: \( \text{Power} = \text{mass flow rate} \times (h_3 - h_4 - h_2 + h_1) \) Rearrange it to find the mass flow rate, \( \text{mass flow rate} = \frac{\text{Power}}{(h_3 - h_4 - h_2 + h_1)} \). Use values from the table for enthalpies.
03

- Calculate the rate of heat transfer

The rate of heat transfer into the working fluid (between state 2 and 3) in the heat exchanger is given by: \( \text{Rate of Heat Transfer} = \text{mass flow rate} \times (h_3 - h_2) \). Use the mass flow rate found in Step 2 and enthalpy values from the table.
04

- Determine thermal efficiency

Thermal efficiency of the Brayton cycle is given by: \( \text{Thermal Efficiency} = \frac{\text{Net Work Output}}{\text{Heat Input}} = 1 - \frac{h_4 - h_1}{h_3 - h_2} \). Use the enthalpy values from the table to calculate the efficiency.

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Key Concepts

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

Thermodynamics
Thermodynamics is the study of energy, heat, and work. It helps us understand how heat energy is converted to mechanical energy and vice versa. In thermodynamics, systems and surroundings are analyzed using laws. The **First Law of Thermodynamics** is all about energy conservation. It's expressed mathematically as \( \text{dU} = \text{dQ} - \text{dW} \), where \( \text{dU} \) represents the change in internal energy, \( \text{dQ} \) is the heat added to the system, and \( \text{dW} \) is the work done by the system.

The **Second Law of Thermodynamics** introduces the concept of entropy, a measure of system disorder. It states that entropy in an isolated system always increases. This law indicates natural processes are irreversible. These laws are crucial in analyzing cycles like the Brayton cycle.
Ideal gas cycles
Ideal gas cycles are theoretical models used to simplify and analyze real gas behavior in thermodynamic processes. One prominent example is the Brayton cycle, commonly used in jet engines and power plants. The Brayton cycle operates with the following stages:
  • Isentropic compression (1-2): Air is compressed adiabatically in a compressor.
  • Constant pressure heat addition (2-3): Heat is added to the compressed air at constant pressure.
  • Isentropic expansion (3-4): The heated air expands adiabatically in a turbine.
  • Constant pressure heat rejection (4-1): Heat is rejected at constant pressure, resetting the cycle.
Understanding these stages helps in analyzing the cycle's performance.
Thermodynamic efficiency
Thermodynamic efficiency measures a cycle's ability to convert heat into work. For the Brayton cycle, thermal efficiency is calculated using the expression:
\[ \text{Thermal Efficiency} = \frac{\text{Net Work Output}}{\text{Heat Input}} = 1 - \frac{h_4 - h_1}{h_3 - h_2} \]
Here, \( h_1, h_2, h_3, \) and \( h_4 \) refer to the specific enthalpies at states 1, 2, 3, and 4, respectively.

Higher efficiency means better performance of the cycle. Engineers aim to optimize these cycles to achieve maximum efficiency by improving component design, increasing compressor and turbine efficiencies, and reducing temperature losses.
Heat transfer calculations
Heat transfer calculations are essential in evaluating the performance of thermodynamic cycles. In the Brayton cycle, heat transfer mainly occurs during two processes:
  • Heat addition (2-3): Calculated using \[ Q_{\text{in}} = \text{mass flow rate} \times (h_3 - h_2) \]
  • Heat rejection (4-1): Similarly, calculated using \[ Q_{\text{out}} = \text{mass flow rate} \times (h_4 - h_1) \]
The mass flow rate can be determined from the power output and enthalpy changes, as described in the step-by-step solution. Accurately calculating these heat transfers ensures efficient thermal management and optimal cycle performance.

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Most popular questions from this chapter

Steam expands isentropically through a converging nozzle operating at steady state from a large tank at \(1.83\) bar, \(280^{\circ} \mathrm{C}\). The mass flow rate is \(2 \mathrm{~kg} / \mathrm{s}\), the flow is choked, and the exit plane pressure is 1 bar. Using steam table data as needed, determine the diameter of the nozzle, in cm, at locations where the pressure is \(1.5\) bar, and 1 bar, respectively.

Air enters the compressor of a regenerative gas turbine at \(14.5 \mathrm{lbf} / \mathrm{in}^{2}, 77^{\circ} \mathrm{F}\), and is compressed to \(60 \mathrm{lbf} / \mathrm{in}^{2}\) The air then passes through the regenerator and exits at \(1120^{\circ} \mathrm{R}\). The temperature at the turbine inlet is \(1700^{\circ} \mathrm{R}\). The compressor and turbine each have an isentropic efficiency of \(84 \%\). The net power developed is \(1000 \mathrm{hp}\). Using an air-standard analysis, calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the regenerator effectiveness (d) the mass flow rate of the air, in lb/s.

Air enters the compressor of a regenerative air-standard Brayton cycle at \(14 \mathrm{lbf} / \mathrm{in}^{2}, 520^{\circ} \mathrm{R}\). The compressor pressure ratio is 14 and the turbine inlet temperature is \(2500^{\circ} \mathrm{R}\). The compressor and turbine have isentropic efficiencies of 83 and \(87 \%\), respectively. The net power developed is \(5 \times 10^{6}\) Btu/h. For regenerator effectiveness values ranging from 0 to \(100 \%\) plot (a) the thermal efficiency. (b) the percent decrease in heat addition to the air.

Air enters the compressor of a cold air-standard Brayton cycle with regeneration, intercooling, and reheat at \(100 \mathrm{kPa}\), \(300 \mathrm{~K}\), with a mass flow rate of \(6 \mathrm{~kg} / \mathrm{s}\). The compressor pressure ratio is 10 , and the pressure ratios are the same across each compressor stage. The intercooler and reheater both operate at the same pressure. The temperature at the inlet to the second compressor stage is \(300 \mathrm{~K}\), and the inlet temperature for each turbine stage is \(1400 \mathrm{~K}\). The compressor and turbine stages each have isentropic efficiencies of \(80 \%\) and the regenerator effectiveness is \(80 \%\). For \(k=1.4\), calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in \(\mathrm{kW}\). (d) the rates of exergy destruction in the compressor and turbine stages as well as the regenerator, in \(\mathrm{kW}\), for \(T_{0}=300 \mathrm{~K}\).

A four-cylinder, four-stroke internal combustion engine operates at 2800 RPM. The processes within each cylinder are modeled as an air-standard Otto cycle with a pressure of \(14.7 \mathrm{lbf} / \mathrm{in} .^{2}\), a temperature of \(80^{\circ} \mathrm{F}\), and a volume of \(0.0196 \mathrm{ft}^{3}\) at the beginning of compression. The compression ratio is 10 , and maximum pressure in the cycle is \(1080 \mathrm{lbf} / \mathrm{in}^{2}\) Determine, using a cold air-standard analysis with \(k=1.4\), the power developed by the engine, in horsepower, and the mean effective pressure, in lbf/in. \({ }^{2}\)

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