Editing Diesel cycle (section)

== Idealized Diesel cycle ==
[[Image:DieselCycle PV.svg|thumb|upright=1.2|[[pressure–volume diagram|p–V diagram]] for the ideal '''Diesel cycle'''. The cycle follows the numbers 1–4 in clockwise direction.]]

The image shows a [[pressure–volume diagram|p–V diagram]] for the ideal Diesel cycle; where <math>p</math> is [[pressure]] and V the volume or <math>v</math> the [[specific volume]] if the process is placed on a unit mass basis. The ''idealized'' Diesel cycle assumes an [[ideal gas]] and ignores [[combustion]] chemistry, [[Exhaust gas|exhaust-]] and recharge procedures and simply follows four distinct processes:

* 1→2 : [[isentropic]] compression of the fluid (blue)
* 2→3 : constant pressure heating (red)
* 3→4 : isentropic expansion (yellow)
* 4→1 : constant volume cooling (green)<ref>Eastop & McConkey 1993, ''Applied Thermodynamics for Engineering Technologists'', Pearson Education Limited, Fifth Edition, p.137</ref>

The Diesel engine is a heat engine: it converts [[heat]] into [[Work (thermodynamics)|work]]. During the bottom isentropic processes (blue), energy is transferred into the system in the form of work <math>W_{in}</math>, but by definition (isentropic) no energy is transferred into or out of the system in the form of heat. During the constant pressure (red, [[isobaric process |isobaric]]) process, energy enters the system as heat <math>Q_{in}</math>. During the top isentropic processes (yellow), energy is transferred out of the system in the form of <math>W_{out}</math>, but by definition (isentropic) no energy is transferred into or out of the system in the form of heat. During the constant volume (green, [[isochoric process|isochoric]]) process, some of the energy flows out of the system as heat through the right depressurizing process <math>Q_{out}</math>. The work that leaves the system is equal to the work that enters the system plus the difference between the heat added to the system and the heat that leaves the system; in other words, net gain of work is equal to the difference between the heat added to the system and the heat that leaves the system.

* Work in (<math>W_{in}</math>) is done by the piston compressing the air (system)
* Heat in (<math>Q_{in}</math>) is done by the [[combustion]] of the fuel
* Work out (<math>W_{out}</math>) is done by the working fluid expanding and pushing a piston (this produces usable work)
* Heat out (<math>Q_{out}</math>) is done by venting the air
* Net work produced = <math>Q_{in}</math> - <math>Q_{out}</math>

The net work produced is also represented by the area enclosed by the cycle on the p–V diagram. The net work is produced per cycle and is also called the useful work, as it can be turned to other useful types of energy and propel a vehicle ([[kinetic energy]]) or produce electrical energy. The summation of many such cycles per unit of time is called the developed power. The <math>W_{out}</math> is also called the gross work, some of which is used in the next cycle of the engine to compress the next charge of air.

{{clear}}

=== Maximum thermal efficiency ===
The maximum thermal efficiency of a Diesel cycle is dependent on the compression ratio and the cut-off ratio. It has the following formula under cold [[Standard state|air standard]] analysis: 

<math>\eta_{th}=1-\frac{1}{r^{\gamma-1}}\left ( \frac{\alpha^{\gamma}-1}{\gamma(\alpha-1)} \right )</math>

where
:<math>\eta_{th} </math> is [[thermal efficiency]]
:<math>\alpha</math> is the cut-off ratio <math>\frac{V_3}{V_2}</math> (ratio between the end and start volume for the combustion phase)
:{{math|r}} is the [[compression ratio]] <math>\frac{V_1}{V_2}</math>
:<math>\gamma </math> is ratio of [[specific heat capacity|specific heats]] (C<sub>p</sub>/C<sub>v</sub>)<ref>{{cite web| url = http://230nsc1.phy-astr.gsu.edu/hbase/thermo/diesel.html| title = The Diesel Engine}}</ref>

The cut-off ratio can be expressed in terms of temperature as shown below:
:<math>\frac{T_2}{T_1} ={\left(\frac{V_1}{V_2}\right)^{\gamma-1}} = r^{\gamma-1}</math>

:<math> \displaystyle {T_2} ={T_1} r^{\gamma-1} </math>

:<math>\frac{V_3}{V_2} = \frac{T_3}{T_2}</math>

:<math>\alpha = \left(\frac{T_3}{T_1}\right)\left(\frac{1}{r^{\gamma-1}}\right)</math>

<math>T_3</math> can be approximated to the flame temperature of the fuel used. The flame temperature can be approximated to the [[adiabatic flame temperature]] of the fuel with corresponding air-to-fuel ratio and compression pressure, <math>p_3</math>.
<math>T_1</math> can be approximated to the inlet air temperature.

This formula only gives the ideal thermal efficiency. The actual thermal efficiency will be significantly lower due to heat and friction losses. The formula is more complex than the [[Otto cycle]] (petrol/gasoline engine) relation that has the following formula:

<math>\eta_{otto,th}=1-\frac{1}{r^{\gamma-1}}</math>

The additional complexity for the Diesel formula comes around since the heat addition is at constant pressure and the heat rejection is at constant volume. The Otto cycle by comparison has both the heat addition and rejection at constant volume.

=== Comparing efficiency to Otto cycle ===
Comparing the two formulae it can be seen that for a given compression ratio ({{math|r}}), the ''ideal'' [[Otto cycle]] will be more efficient. However, a ''real'' [[diesel engine]] will be more efficient overall since it will have the ability to operate at higher compression ratios.  If a petrol engine were to have the same compression ratio, then knocking (self-ignition) would occur and this would severely reduce the efficiency, whereas in a diesel engine, the self ignition is the desired behavior.  Additionally, both of these cycles are only idealizations, and the actual behavior does not divide as clearly or sharply. Furthermore, the ideal Otto cycle formula stated above does not include throttling losses, which do not apply to diesel engines.