/*! 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 35 The rate of heat addition to an ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 35

The rate of heat addition to an air-standard Brayton cycle is \(800 \mathrm{~kW}\). The pressure ratio for the cycle is 12 and the minimum and maximum temperatures are \(298 \mathrm{~K}\) and \(1600 \mathrm{~K}\), respectively. Determine (a) the thermal efficiency of the cycle. (b) the mass flow rate of air, in \(\mathrm{kg} / \mathrm{s}\). (c) the net power developed by the cycle, in \(\mathrm{kJ} / \mathrm{s}\).

Short Answer

Expert verified
(a) 57.06%, (b) 0.613 kg/s, (c) 456.5 kW

Step by step solution

01

Determine the thermal efficiency

The thermal efficiency for an ideal Brayton cycle is given by \(畏 = 1 - (\frac{T_1}{T_2})^{\frac{纬 - 1}{纬}}\)where:\(纬 = 1.4\) (for air)\(T_1 = 298K\) (minimum temperature)\(T_2 = 1600K\) (maximum temperature)First, we need to calculate the intermediate temperature ratio:\[\frac{T_2}{T_1} = \frac{1600}{298} 鈮� 5.369\]Next, use the pressure ratio to find the temperature ratio for efficiency calculation:\[T_r = (\frac{T_2}{T_1})\]Now, compute efficiency:\[畏 = 1 - \frac{1}{(P_r)^{\frac{纬 - 1}{纬}}} = 1 - (\frac{1}{12})^{\frac{0.4}{1.4}}\approx 0.5706\]Therefore, the thermal efficiency is approximately 57.06%.
02

Calculate the mass flow rate of air

To find the mass flow rate, we use the heat addition rate and thermal efficiency.First convert efficiency to power terms:\[Q_{in} = 800 kW = 800 kJ/s\]It's known from thermodynamics:\[Q_{in} = 畏 \times \text{mass flow rate} \times c_p \times 螖T\]where:\[c_p = 1.005 \text{ kJ/kg.K} \text{ (specific heat at constant pressure for air)}\]Thus, rearrange to solve for mass flow rate:\[m = \frac{Q_{in}}{c_p \times 螖T}\]\[螖T = 1600 - 298 = 1302 K\]\[m = \frac{800}{1.005 \times 1302} 鈮� 0.613 \text{ kg/s}\]Therefore, the mass flow rate of air is approximately 0.613 kg/s.
03

Determine the net power developed by the cycle

The net power developed can be calculated from thermal efficiency and heat addition rate using:\[\text{Net Power} = 畏 \times Q_{in}\]Substitute the values:\[Net Power = 0.5706 \times 800 鈮� 456.5 \text{kJ/s or} \text{kW}\]Therefore, the net power developed by the cycle is approximately 456.5 kW.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Thermal Efficiency
In a Brayton cycle, thermal efficiency (\(畏\)) is a measure of how well the cycle converts heat into work. It's a key performance indicator. The formula for thermal efficiency in an ideal Brayton cycle is: \[畏 = 1 - (\frac{T_1}{T_2})^{\frac{纬 - 1}{纬}}\], where \(纬\) is the specific heat ratio, \(T_1\) is the lowest cycle temperature, and \(T_2\) is the highest cycle temperature. For this exercise:
  • \(纬 = 1.4\)
  • \(T_1 = 298 K\)
  • \(T_2 = 1600 K\)
After calculating the intermediate temperature ratio and using the pressure ratio 12, we determined that the thermal efficiency is approximately 57.06%. This means that 57.06% of the heat added to the cycle is converted into useful work.
Mass Flow Rate
The mass flow rate of air is symbolized as \(m\) and indicates how much air moves through the cycle per second. It's crucial for determining the amount of energy the cycle processes. We used the given heat addition rate (\(Q_{in} = 800 kW\)) and the specific heat at constant pressure for air (\(c_p = 1.005 kJ/kg路K\)) to calculate mass flow rate with the formula: \[m = \frac{Q_{in}}{c_p \times 螖T}\], where \(螖T\) is the temperature change within the cycle. For this exercise:
  • \(螖T = T_2 - T_1\)
  • \(螖T = 1600 K - 298 K = 1302 K\)
The mass flow rate was determined to be about 0.613 kg/s. This indicates the air's movement rate through the cycle is 0.613 kg every second.
Net Power Output
Net power output represents the work produced by the Brayton cycle after subtracting the work needed to compress the air. It's crucial for evaluating the cycle's practical performance. The net power developed by the cycle can be found using the thermal efficiency (\( 畏 \)) and heat addition rate (\( Q_{in} \)) with the formula: \[\text{Net Power} = 畏 \times Q_{in}\]. For this problem:
  • \(畏 = 57.06\%\)
  • \(Q_{in} = 800 kW\)
The net power output is calculated to be approximately 456.5 kW. This value reflects the cycle's efficiency in converting heat into useful work.
Heat Addition Rate
The heat addition rate (\(Q_{in}\)) is the total heat energy supplied to the system per second. This concept is essential for understanding how much energy enters the cycle. For this exercise:
  • \(Q_{in} = 800 kW\)
Knowing this value helps us understand the energy input into the Brayton cycle, which is crucial for subsequent calculations like thermal efficiency and net power output.
Pressure Ratio
The pressure ratio (\(P_r\)) in a Brayton cycle is the ratio of the pressure after compression to the pressure before compression. This ratio affects the cycle's efficiency and work output. A higher pressure ratio usually leads to higher thermal efficiency. In this exercise, the given pressure ratio is 12. Using this pressure ratio in our thermal efficiency calculation helps us gauge the performance improvements due to increased pressure.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A simple gas turbine is the topping cycle for a simple vapor power cycle (Fig. 9.23). Air enters the compressor of the gas turbine at \(15^{\circ} \mathrm{C}, 100 \mathrm{kPa}\), with a volumetric flow rate of \(20 \mathrm{~m}^{3} / \mathrm{s}\). The compressor pressure ratio is 12 and the turbine inlet temperature is \(1440 \mathrm{~K}\). The compressor and turbine each have isentropic efficiencies of \(88 \%\). The air leaves the interconnecting heat exchanger at \(460 \mathrm{~K}, 100 \mathrm{kPa}\). Steam enters the turbine of the vapor cycle at \(7000 \mathrm{kPa}, 480^{\circ} \mathrm{C}\), and expands to the condenser pressure of \(7 \mathrm{kPa}\). Water enters the pump as saturated liquid at \(7 \mathrm{kPa}\). The turbine and pump efficiencies are 90 and \(70 \%\), respectively. Cooling water passing through the condenser experiences a temperature rise from 15 to \(27^{\circ} \mathrm{C}\) with a negligible change in pressure. Determine (a) the mass flow rates of the air, steam, and cooling water, each in \(\mathrm{kg} / \mathrm{s}\). (b) the net power developed by the gas turbine cycle and the vapor cycle, respectively, each in \(\mathrm{kJ} / \mathrm{s}\). (c) the thermal efficiency of the combined cycle. (d) a full accounting of the net exergy increase of the air passing through the combustor of the gas turbine, \(\dot{m}_{\text {air }}\left[\mathrm{e}_{t 3}-\mathrm{e}_{42}\right]\), in \(\mathrm{kJ} / \mathrm{s}\). Discuss. Let \(T_{0}=300 \mathrm{~K}, p_{0}=100 \mathrm{kPa}\).

Air at \(3.5\) bar, \(520 \mathrm{~K}\), and a Mach number of \(0.3\) enters a converging-diverging nozzle operating at steady state. A normal shock stands in the diverging section at a location where the Mach number is \(M_{x}=1.7\). The flow is isentropic, except where the shock stands. If the air behaves as an ideal gas with \(k=1.4\), determine (a) the stagnation temperature \(T_{\mathrm{ox}}\), in \(\mathrm{K}\). (b) the stagnation pressure \(p_{\mathrm{ox}}\), in bar. (c) the pressure \(p_{x}\), in bar. (d) the pressure \(p_{\mathrm{y}}\) in bar. (e) the stagnation pressure \(p_{\mathrm{oy}}\), in bar. (f) the stagnation temperature \(T_{\mathrm{oy}}\), in \(\mathrm{K}\). If the throat area is \(7.5 \times 10^{-4} \mathrm{~m}^{2}\), and the exit plane pressure is \(2.5\) bar, determine the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\), and the exit area, in \(\mathrm{m}^{2}\).

An air-standard Brayton cycle has a compressor pressure ratio of 10 . Air enters the compressor at \(p_{1}=100 \mathrm{kPa}\), \(T_{1}=20^{\circ} \mathrm{C}\) with a mass flow rate of \(11 \mathrm{~kg} / \mathrm{s}\). The turbine inlet temperature is \(1222 \mathrm{~K}\). Calculate the thermal efficiency and the net power developed, in \(\mathrm{kW}\), if (a) the turbine and compressor isentropic efficiencies are each \(100 \%\). (b) the turbine and compressor isentropic efficiencies are 88 and \(84 \%\), respectively. (c) the turbine and compressor isentropic efficiencies are 88 and \(84 \%\), respectively, and a regenerator with an effectiveness of \(80 \%\) is incorporated.

Air enters the first compressor stage of a cold air-standard Brayton cycle with regeneration and intercooling at \(100 \mathrm{kPa}\), \(300 \mathrm{~K}\), with a mass flow rate of \(6 \mathrm{~kg} / \mathrm{s}\). The overall compressor pressure ratio is 10 , and the pressure ratios are the same across each compressor stage. The temperature at the inlet to the second compressor stage is \(300 \mathrm{~K}\). The compressor stages and turbine 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 each compressor stage and the turbine stage as well as the regenerator, in \(\mathrm{kW}\), for \(T_{0}=300 \mathrm{~K}\)

Air enters the compressor of a simple gas turbine at \(p_{1}=96 \mathrm{kPa}, T_{1}=298 \mathrm{~K}\). The isentropic efficiencies of the compressor and turbine are 85 and \(89 \%\), respectively. The compressor pressure ratio is 13 and the temperature at the turbine inlet is \(1340 \mathrm{~K}\). The net power developed is \(1450 \mathrm{~kW}\). On the basis of an air-standard analysis, calculate (a) the volumetric flow rate of the air entering the compressor, in \(\mathrm{m}^{3} / \mathrm{s}\). (b) the temperatures at the compressor and turbine exits, each in \(\mathrm{K}\). (c) the thermal efficiency of the cycle.
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Radiation

Read Explanation

Fluids

Read Explanation

Classical Mechanics

Read Explanation

Fields in Physics

Read Explanation

Torque and Rotational Motion

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.