Editing Otto cycle (section)

==Cycle analysis==
In this process 1–2 the piston does work on the gas and in process 3–4 the gas does work on the piston during those isentropic compression and expansion processes, respectively. Processes 2–3 and 4–1 are isochoric processes; heat is transferred into the system from 2—3 and out of the system from 4–1 but no work is done on the system or extracted from the system during those processes. No work is done during an isochoric (constant volume) process because addition or removal of work from a system requires the movement of the boundaries of the system; hence, as the cylinder volume does not change, no shaft work is added to or removed from the system.

Four different equations are used to describe those four processes. A simplification is made by assuming changes of the kinetic and potential energy that take place in the system (mass of gas) can be neglected and then applying the first law of thermodynamics (energy conservation) to the mass of gas as it changes state as characterized by the gas's temperature, pressure, and volume.<ref name="Shapiro">{{Cite book |last=Moran |first=Michael J. |title=Fundamentals of engineering thermodynamics: SI version, SI units |last2=Shapiro |first2=Howard N. |date=2006 |publisher=Wiley |isbn=978-0-470-03037-0 |edition=5th |location=Chichester |pages=376}}</ref>{{page needed|date=July 2017}}<ref name="Fundamentals">Gupta, H. N. ''Fundamentals of Internal Combustion''. New Delhi: Prentice-Hall, 2006. Print.</ref>{{page needed|date=July 2017}}

During a complete cycle, the gas returns to its original state of temperature, pressure and volume, hence the net internal energy change of the system (gas) is zero. As a result, the energy (heat or work) added to the system must be offset by energy (heat or work) that leaves the system. In the analysis of thermodynamic systems, the convention is to account energy that enters the system as positive and energy that leaves the system is accounted as negative.

{{anchor |Equation 1a}}
Equation 1a.

During a complete cycle, the net change of energy of the system is zero:

:<math>\Delta E = E_\text{in} - E_\text{out} = 0</math>

The above states that the system (the mass of gas) returns to the original thermodynamic state it was in at the start of the cycle.

Where <math>E_\text{in}</math> is energy added to the system from 1–2–3 and <math>E_\text{out}</math> is energy removed from the system from 3–4–1. In terms of work and heat added to the system

{{anchor |Equation 1b}}
Equation 1b:

:<math>W_{1-2} + Q_{2-3} + W_{3-4} + Q_{4-1} = 0</math>

Each term of the equation can be expressed in terms of the internal energy of the gas at each point in the process:

:<math>W_{1-2} = U_2 - U_1</math>

:<math>Q_{2-3} = U_3 - U_2</math>

:<math>W_{3-4} = U_4 - U_3</math>

:<math>Q_{4-1} = U_1 - U_4</math>

The energy balance Equation 1b becomes

:<math>W_{1-2} + Q_{2-3} + W_{3-4} + Q_{4-1} = \left(U_2 - U_1\right) + \left(U_3 - U_2\right) + \left(U_4 - U_3\right) + \left(U_1 - U_4\right) = 0</math>

To illustrate the example we choose {{dubious|date=August 2022}} some values to the points in the illustration:
:<math>U_1  = 1</math>
:<math>U_2  = 5</math>
:<math>U_3  = 9</math>
:<math>U_4  = 4</math>
These values are arbitrarily but rationally {{dubious|date=August 2022}} selected. The work and heat terms can then be calculated.

The energy added to the system as work during the compression from 1 to 2 is

:<math>\left(U_2 - U_1\right) = \left(5 - 1\right) = 4</math>

The energy added to the system as heat from point 2 to 3 is

:<math>\left({U_3 - U_2}\right) = \left(9 - 5\right) = 4</math>

The energy removed from the system as work during the expansion from 3 to 4 is

:<math>\left(U_4 - U_3\right) = \left(4 - 9\right) = -5</math>

The energy removed from the system as heat from point 4 to 1 is

:<math>\left(U_1 - U_4\right) = \left(1 - 4\right) = -3</math>

The energy balance is

:<math>\Delta E = + 4 + 4 - 5 - 3 = 0</math>

Note that energy added to the system is counted as positive and energy leaving the system is counted as negative and the summation is zero as expected for a complete cycle that returns the system to its original state.

From the energy balance the work out of the system is:

:<math>\sum \text{Work} = W_{1-2} + W_{3-4} = \left(U_2 - U_1\right) + \left(U_4 - U_3\right) = 4 - 5 = -1</math>

The net energy out of the system as work is -1, meaning the system has produced one net unit of energy that leaves the system in the form of work.

The net heat out of the system is:

:<math>\sum \text{Heat} = Q_{2-3} + Q_{4-1} = \left(U_3 - U_2\right) + \left(U_1 - U_4\right) = 4 -3 = 1</math>

As energy added to the system as heat is positive. From the above it appears as if the system gained one unit of heat. This matches the energy produced by the system as work out of the system.

{{anchor |Thermal efficiency}}
[[Thermal efficiency]] is the quotient of the net work from the system, to the heat added to system.
Equation 2:

:<math>\eta = \frac{- \left(W_{1-2} + W_{3-4}\right) }{Q_{2-3}} = \frac{\left(U_1 - U_2\right) + \left(U_3 - U_4\right)}{ \left(U_3 - U_2\right)}</math>

:<math>\eta =1+\frac{U_1 - U_4 }{  \left(U_3 - U_2\right)} = 1+\frac{(1-4)}{ (9-5)} = 0.25  </math>

Alternatively, thermal efficiency can be derived by strictly heat added and heat rejected.

:<math>\eta=\frac{Q_{2-3} + Q_{4-1}}{Q_{2-3}}
=1+\frac{\left(U_1-U_4\right) }{ \left(U_3-U_2\right)} </math>

Supplying the fictitious values

<math>\eta=1+\frac{1-4}{9-5}=1+\frac{-3}{4}=0.25</math>

In the Otto cycle, there is no heat transfer during the process 1–2 and 3–4 as they are isentropic processes. Heat is supplied only during the constant volume processes 2–3 and heat is rejected only during the constant volume processes 4–1.

{{anchor |The above values}}
The above values are absolute values that might, for instance {{dubious|date=August 2022}}, have units of joules (assuming the MKS system of units are to be used) and would be of use for a particular engine with particular dimensions. In the study of thermodynamic systems the extensive quantities such as energy, volume, or entropy (versus intensive quantities of temperature and pressure) are placed on a unit mass basis, and so too are the calculations, making those more general and therefore of more general use.  Hence, each term involving an extensive quantity could be divided by the mass, giving the terms units of joules/kg ([[specific energy]]), meters<sup>3</sup>/kg ([[specific volume]]), or joules/(kelvin·kg) (specific entropy, heat capacity) etc. and would be represented using lower case letters, u, v, s, etc.

{{anchor |Equation 1}}
Equation 1 can now be related to the specific heat equation for constant volume. The [[Heat capacity|specific heats]] are particularly useful for thermodynamic calculations involving the [[ideal gas]] model.

:<math>C_\text{v} = \left(\frac{\delta u}{\delta T}\right)_\text{v}</math>

Rearranging yields:

:<math>\delta u = (C_\text{v})(\delta T)</math>

Inserting the specific heat equation into the thermal efficiency equation (Equation 2) yields.

:<math>\eta = 1-\left(\frac{C_\text{v}(T_4 - T_1)}{ C_\text{v}(T_3 - T_2)}\right)</math>

Upon rearrangement:

:<math>\eta = 1-\left(\frac{T_1}{T_2}\right)\left(\frac{T_4/T_1-1}{T_3/T_2-1}\right)</math>

Next, noting from the diagrams <math>T_4/T_1 = T_3/T_2</math> (see [[Isentropic process|isentropic relations for an ideal gas]]), thus both of these can be omitted. The equation then reduces to:

{{anchor |Equation 2}}
Equation 2:
:<math>\eta = 1-\left(\frac{T_1}{T_2}\right)</math>

Since the Otto cycle uses isentropic processes during the compression (process 1 to 2) and expansion (process 3 to 4) the [[Isentropic process|isentropic equations]] of ideal gases and the constant pressure/volume relations can be used to yield Equations 3 & 4.<ref name=Reynolds>{{cite book|last=Reynolds & Perkins|title=Engineering Thermodynamics|url=https://archive.org/details/engineeringtherm00reyn|url-access=registration|year=1977|publisher=McGraw-Hill|isbn=978-0-07-052046-2|pages=[https://archive.org/details/engineeringtherm00reyn/page/249 249]}}</ref>

{{anchor |Equation 3}}
Equation 3:
:<math>\left(\frac{T_2}{T_1}\right)=\left(\frac{p_2}{p_1}\right)^{(\gamma-1)/{\gamma}}</math>

{{anchor |Equation 4}}
Equation 4:
:<math>\left(\frac{T_2}{T_1}\right)=\left(\frac{V_1}{V_2}\right)^{(\gamma-1)}</math>

::where

::<math>{\gamma} = \left(\frac{C_\text{p}}{C_\text{v}}\right)</math>

::<math>{\gamma}</math> is the specific heat ratio

::::The derivation of the previous equations are found by solving these four equations respectively (where <math>R</math> is the [[specific gas constant]]):

::::<math>C_\text{p} \ln\left(\frac{V_1}{V_2}\right) - R \ln \left(\frac{p_2}{p_1}\right) = 0</math>

::::<math>C_\text{v} \ln\left(\frac{T_2}{T_1}\right) - R \ln \left(\frac{V_2}{V_1}\right) = 0</math>

::::<math>C_\text{p} = \left(\frac{\gamma R}{\gamma-1}\right)</math>

::::<math>C_\text{v} = \left(\frac{R}{\gamma-1}\right)</math>

Further simplifying Equation 4, where <math>r</math> is the compression ratio <math>(V_1/V_2)</math>:

{{anchor |Equation 5}}
Equation 5:
:<math>\left(\frac{T_2}{T_1}\right) = \left(\frac{V_1}{V_2}\right)^{(\gamma-1)} = r^{(\gamma-1)}</math>

From inverting Equation 4 and inserting it into Equation 2 the final thermal efficiency can be expressed as:{{page needed|date=July 2017}}<ref name="Fundamentals" />{{page needed|date=July 2017}}

{{anchor |Equation 6}}
Equation 6:
:<math>\eta = 1 - \left(\frac{1}{r^{(\gamma-1)}}\right)</math>

From analyzing equation 6 it is evident that the Otto cycle efficiency depends directly upon the compression ratio <math>r</math>. Since the <math>\gamma</math> for air is 1.4, an increase in <math>r</math> will produce an increase in <math>\eta</math>. However, the <math>\gamma </math> for combustion products of the fuel/air mixture is often taken at approximately 1.3.
The foregoing discussion implies that it is more efficient to have a high compression ratio. The standard ratio is approximately 10:1 for typical automobiles. Usually this does not increase much because of the possibility of autoignition, or "[[Engine knocking|knock]]", which places an upper limit on the compression ratio.<ref name="Shapiro" />{{page needed|date=July 2017}} During the compression process 1–2 the temperature rises, therefore an increase in the compression ratio causes an increase in temperature. Autoignition occurs when the temperature of the fuel/air mixture becomes too high before it is ignited by the flame front. The compression stroke is intended to compress the products before the flame ignites the mixture. If the compression ratio is increased, the mixture may auto-ignite before the compression stroke is complete, leading to "engine knocking". This can damage engine components and will decrease the brake horsepower of the engine.