/*! 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 24 A Diesel engine has a compressio... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A Diesel engine has a compression ratio of 14 and cut-off takes place at \(7 \%\) of the stroke. Determine the air standard efficiency.

Short Answer

Expert verified
The air standard efficiency is approximately 64.96%.

Step by step solution

01

Understand Relevant Formula

The efficiency of a Diesel engine can be determined using the air-standard efficiency formula for a Diesel cycle: \[ \text{Efficiency} = 1 - \frac{1}{r^{\frac{\text{k}-1}{\text{k}}}} \times \frac{(r_c^{k} - 1)}{k(r_c - 1)} \] where \( r \) is the compression ratio, \( r_c \) is the cut-off ratio, and \( k \) is the adiabatic index (ratio of specific heats, typically 1.4 for air).
02

Substitute Given Values

Given: Compression ratio \( r = 14 \) Cut-off ratio \( r_c = 1.07 \) (as 7% of stroke means an increase of 7% after compression). Adiabatic index \( k = 1.4 \). Substituting these values into the formula gives: \[ \text{Efficiency} = 1 - \frac{1}{14^{\frac{0.4}{1.4}}} \times \frac{(1.07^{1.4} - 1)}{1.4(1.07 - 1)} \]
03

Simplify the Expression

Calculate the term \( 14^{\frac{0.4}{1.4}} \): \[ 14^{\frac{0.4}{1.4}} = 14^{0.2857} \ \text{(approx. 2.942)} \] Calculate the term \( 1.07^{1.4} \): \[ 1.07^{1.4} \ \text{(approx. 1.101)} \] The expression simplifies to: \[ 1 - \frac{1}{2.942} \times \frac{(1.101 - 1)}{1.4 \times 0.07} \]
04

Calculate Individual Components

Calculate the denominator and numerator separately: \[ \frac{1}{2.942} = 0.34 \] \[ 1.101 - 1 = 0.101 \] \[ 1.4 \times 0.07 = 0.098 \] Thus, the expression becomes: \[ 1 - 0.34 \times \frac{0.101}{0.098} \]
05

Final Efficiency Calculation

Calculate the fraction: \[ \frac{0.101}{0.098} \ \text{(approx. 1.0306)} \] Multiply and subtract from 1: \[ 1 - 0.34 \times 1.0306 = 1 - 0.3504 = 0.6496 \] Therefore, the air standard efficiency is approximately \( 64.96\text{\%} \)

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 91Ó°ÊÓ!

Key Concepts

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

Air-standard efficiency
Air-standard efficiency is a theoretical measure of how efficiently an engine converts the heat energy from fuel into useful work.
The concept comes from the idealized assumptions about the working fluid, often air, and the cycle's processes, which are simplified for analysis. It provides an upper bound that real engines aim to approach but never quite reach.

The efficiency of a Diesel cycle can be calculated using the formula:
\[ \text{Efficiency} = 1 - \frac{1}{r^{k-1/k}} \times \frac{(r_c^k - 1)}{k(r_c - 1)} \]

Here, the variables are defined as:
  • \(r\): Compression ratio
  • \(r_c\): Cut-off ratio
  • \(k\): Adiabatic index (ratio of specific heats)
This formula considers the idealized behavior of the gas during compression and expansion within the engine. While actual efficiency in practical engines tends to be lower due to real-world factors like friction and heat losses, understanding air-standard efficiency offers insights into potential performance improvements.
Compression Ratio
The compression ratio of a Diesel engine is crucial for determining its efficiency.
It is defined as the ratio of the volume of the cylinder and combustion chamber when the piston is at the bottom of its stroke to the volume when the piston is at the top.

Mathematically, the compression ratio (\(r\)) is represented as:
\[ r = \frac{V_{\text{max}}}{V_{\text{min}}} \]
Where:
  • \(V_{\text{max}}\): Volume when the piston is at the bottom (BDC)
  • \(V_{\text{min}}\): Volume when the piston is at the top (TDC)
Higher compression ratios generally lead to higher efficiencies because compressing the air-fuel mixture more effectively leads to better combustion.
In our problem, the compression ratio is given as 14, which means the volume is reduced by 14 times during compression before ignition takes place.
Cut-off Ratio
The cut-off ratio is another vital factor in the efficiency of a Diesel engine.
It is the volume ratio at which the fuel injection is stopped relative to the cylinder's volume after compression.

Formally, the cut-off ratio (\(r_c\)) is:
\[ r_c = \frac{V_{\text{cut-off}}}{V_{\text{min}}} \]
In our problem, the cut-off ratio provided is 1.07, which signifies that after the air is compressed, the volume increases by 7% due to the fuel injection.
The smaller the cut-off ratio, the higher the efficiency because less time is spent in the less efficient constant pressure heat addition process.
Adiabatic Index
The adiabatic index, often denoted as \(k\), is the ratio of the specific heats of a gas at constant pressure (\(C_p\)) and constant volume (\(C_v\)).
In formula form, it is given as:
\[ k = \frac{C_p}{C_v} \]
For air, this ratio is typically around 1.4. The adiabatic index is important because it affects how the gas behaves when it is compressed or expanded without heat transfer.

In the context of Diesel engines, the adiabatic index affects the air-standard efficiency formula and how well the engine can convert thermal energy into mechanical energy.
By understanding and applying the value of the adiabatic index properly, we can determine more accurate theoretical efficiencies.

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

An engine working on the air standard Otto cycle is supplied with air at \(0.1 \mathrm{MPa}, 27^{\circ} \mathrm{C}\). The compression ratio is 8 . The heat supplied is \(1400 \mathrm{~kJ} / \mathrm{kg}\). Calculate the maximum pressure and temperature of the cycle, the cycle efficiency, and the mean effective pressure.

A four-cylinder, four-stroke internal combustion engine has a bore of \(66 \mathrm{~mm}\) and a stroke of \(54 \mathrm{~mm}\). The clearance volume is \(13 \%\) of the cylinder volume at bottom dead center and the crankshaft rotates at 3500 RPM. The processes within each cylinder are modeled as an air-standard Otto cycle with a pressure of \(100 \mathrm{kPa}\) and a temperature of \(27^{\circ} \mathrm{C}\) at the beginning of compression. The maximum temperature in the cycle is \(2250 \mathrm{~K}\). Based on this model, calculate the net work per cycle, in \(\mathrm{kJ}\), and the power developed by the engine, in \(\mathrm{kW}\).

Air enters the compressor of a simple gas turbine at \(100 \mathrm{kPa}, 298 \mathrm{~K}\), and exits at \(700 \mathrm{kPa}, 550 \mathrm{~K}\). The air enters the turbine at \(700 \mathrm{kPa}, 1120 \mathrm{~K}\) and expands to \(100 \mathrm{kPa}, 770 \mathrm{~K}\). The compressor and turbine operate adiabatically, and kinetic and potential energy effects are negligible. On the basis of an air-standard analysis, (a) develop a full accounting of the net exergy increase of the air passing through the gas turbine combustor, in \(\mathrm{kJ} / \mathrm{kg}\). (b) devise and evaluate an exergetic efficiency for the gas turbine cycle. Let \(T_{0}=298 \mathrm{~K}, p_{0}=100 \mathrm{kPa}\).

The term regeneration is used to describe the use of regenerative feedwater heaters in vapor power plants and regenerative heat exchangers in gas turbines. In what ways are the purposes of these devices similar? How do they differ?

Air at \(30 \mathrm{kPa}, 240 \mathrm{~K}\), and \(200 \mathrm{~m} / \mathrm{s}\) enters a turbojet engine in flight. The air mass flow rate is \(26 \mathrm{~kg} / \mathrm{s}\). The compressor pressure ratio is 11 , the turbine inlet temperature is \(1360 \mathrm{~K}\), and air exits the nozzle at \(30 \mathrm{kPa}\). The diffuser and nozzle processes are isentropic, the compressor and turbine have isentropic efficiencies of \(85 \%\) and \(88 \%\), respectively, and there is no pressure drop for flow through the combustor. Kinetic energy is negligible everywhere except at the diffuser inlet and the nozzle exit. On the basis of air-standard analysis, determine (a) the pressures, in \(\mathrm{kPa}\), and temperatures, in \(\mathrm{K}\), at each principal state. (b) the rate of heat addition to the air passing through the combustor, in \(\mathrm{kJ} / \mathrm{s}\). (c) the velocity at the nozzle exit, in \(\mathrm{m} / \mathrm{s}\).

See all solutions

Recommended explanations on Physics Textbooks

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.