Overview

 

Topic 1 - Engine cycles

We will make the following Assumptions for analysis of a Standard Engine Cycle:

1. The gas mixture is assumed to be air for the entire cycle, even though after combustion the gases are changed.
2. The system is treated as a closed cycle, i.e. the same gas is heated, cooled, and reheated, through the cycle, and never exchanged.
3. Combustion is represented more simply as a heat addition process.
4. Because of the closed cycle assumption, the open exhaust is treated as a closed, heat rejection process.
5. All processes in the cycle are represented as ideal processes
   1. The intake and exhaust strokes are considered to be constant pressure. The intake stroke, at WOT, is assumed to be at ambient pressure of 1 atmosphere.
   2. Compression stroke and compression stroke are treated as isentropic (constant entropy)
   3. Combustion is idealized by a constant-volume process for SI engines and constant pressure for CI engines.
   4. Exhaust stroke is considered constant volume
   5. All processes are considered to be reversible.
6. The air is treated as an ideal gas and thus the ideal gas equation applies: PV = mRT
7. Standard air conditions will be used:
   1. c_p = 1.108 kJ/kg-K = 0.265 BTU/lbm-R
   2. c_v = 0.821 kJ/kg-K = 0.196 BTU/lbm-R
   3. k = c_p /c_v = 1.35
   4. R = 0.287 kJ/kg-K

The Otto cycle is used to model the relationship between pressure and volume in a 4 stroke spark ignition engine.

Topic 1 - Process 6-1

Constant pressure intake of ambient air at P₀ intake valve open. Exhaust valve closed.

Topic 1 - Process 1-2

Isentropic compression stroke. All valves closed.

Topic 1 - Process 2-3

Combustion - constant volume heat input.  All valves closed.

Topic 1 - Process 3-4

Isentropic power stroke. All valves closed.

Topic 1 - Process 4-5

Exhaust blowdown - constant volume heat rejection. Exhaust valve open.

Topic 1 - Process 5-6

Topic 1 - Process 5-6 cont.

Topic 1 - Diesel cycles

The diesel engine causes combustion of its fuel-air mixture by compression rather than the spark from a sparkplug.  It operates on a different cycle.

Topic 2 - Thermochemistry

As we move on to considering how we extract energy from our fuel, we need to define some more terms:

Maximum energy is released from the fuel when it reacts with a Stoichiometric amount of oxygen.  Stoichiometric conditions are those which supply precisely enough oxygen to combust all of the available fuel.

We need to use either the Fuel-to-Air Ratio (FA) or the Air-to-Fuel Ratio (AF)

Equivalence ratio  $= \text{Φ}$

$= \text{F}{\text{A}}_{\text{a}\text{c}\text{t}\text{u}\text{a}\text{l}}/\text{F}{\text{A}}_{\text{s}\text{t}\text{o}\text{i}\text{c}\text{i}\text{o}\text{m}\text{e}\text{t}\text{r}\text{c}}$

$= \text{A}{\text{F}}_{\text{s}\text{t}\text{o}\text{i}\text{c}\text{i}\text{o}\text{m}\text{e}\text{t}\text{r}\text{c}}/\text{F}{\text{A}}_{\text{a}\text{c}\text{t}\text{u}\text{a}\text{l}}$

When $\text{Φ}$ = 1 the mixture is referred to as lean, with inadequate fuel to use all of the available Oxygen, resulting in excess Oxygen in the exhaust gasses.

When $\text{Φ}$  > 1 the mixture is referred to as rich, with inadequate Oxygen to combust all of the fuel, resulting in unburned fuel in the exhaust gasses.

When $\text{Φ}$ = 1  combustion is stoichiometric.

Topic 3 - Combustion

Methane is a viable fuel.  Let us consider how it would combust in our engine.

1 mole of CH₄ weights  12 + 4(1) = 16kg   (16 lb in imperial system)

1 mole of O₂ weights 2(16)  = 32 kg  (32 lb in SI imperial system)

Let us combust some iso-octane, C₈H₁₈

___C₈H₁₈     +     ___O₂         >>>>>>>            ___CO₂      + ___H₂0

We must balance the chemical equation

1 C₈H₁₈     +     12.5 O₂          >>>>>>>             8  CO₂       + 9 H₂0

But we do not burn pure oxygen…….we use air.  Air is 79% inert gasses (mostly Nitrogen, so that is what we usually call it) along with 21% Oxygen.  That means that there will be .79/.21 = 3.76 times as much Nitrogen as Oxygen.

Let’s re-combust our iso-octane, using air not oxygen

_a_C₈H₁₈   +  _b_O₂   +  _b_(3.76) N₂       >>>>>>             _c_CO₂    + _d_H₂0  + _b_ (3.76) N₂

Now we must balance the equation again.

1 C₈H₁₈   +  12.5 O₂   +  12.5 (3.76) N₂      >>>>>>             8 CO₂    +  9 H₂0  +  12.5 (3.76) N₂

This is really the same as before, but with the Nitrogen included.  But these have all been Stoichiometric……just enough Oxygen to combust all of the fuel.

What if that is not the case?

Let us re-combust our iso-octane, using 120% stoichiometric air   (i.e. too much Oxygen).  The O₂ on the left side of the equation was previously was 12.5.  So at120% too much would air it would be 12.5(1.2)  = 15

_a_C₈H₁₈   +  _15_O₂   +  _b_(3.76) N₂      >>>>>       _c_CO₂    + _d_H₂0  + _b_ (3.76) N₂           

Now Balance the equation

1 C₈H₁₈   +  15 O₂   +  15 (3.76) N₂        >>>>>>           8 CO₂    + 9 H₂0  + 15 (3.76) N₂  +  2.5 O₂

Now let us look at the fuel-to-air ratio

$\text{F}\text{A}=\frac{8(12.01+18 (1.01)}{15(1+3.76) (28.97)}=0.055$

At Stoichiometric it was

$\text{FA}=\frac{8(12.01+18 (1.01)}{12.5(1+3.76) (28.97)}=0.066$

So the equivalence ratio is $\Phi$= 0.055/0.066    =   0.833 

Now let us re-combust our iso-octane, using 80% stoichiometric air   (i.e. too little Oxygen)

The O₂ on the left side of the equation was was 12.5.  So 80% Oxygen would be 12.5(0.8)  = 10.

_a_C₈H₁₈  + _10_O₂  +_b_(3.76) N₂     >>>>>      _c_CO₂  + _d_H₂0 +  _b_ (3.76) N₂  + _e_CO

Now when we balance the equation, we end up with CO, Carbon Monoxide.

1 C₈H₁₈   +  10 O₂ +  10 (3.76) N₂             3 CO₂  + 9 H₂0  +10 (3.76) N₂ + 5 CO

$\text{F}\text{A}=\frac{8(12.01+18 (1.01)}{10(1+3.76) (28.97)}=0.082$

At Stoichiometric

$\text{F}\text{A}=\frac{8(12.01+18 (1.01)}{12.5(1+3.76) (28.97)}=0.066$

Resulting in an Equivalence Ratio $\text{Φ}$   = 0.082/0.066    =   1.24

Note that if you keep reducing the Oxygen, and eventually you will have raw C₈H₁₈ in the exhaust

Topic 4 - Exhaust gas analysis

The main emissions of a car engine are:

• Nitrogen gas (N₂) - Air is 78-percent nitrogen gas, and most of this passes right through the car engine.
• Carbon dioxide (CO₂) - This is one product of combustion. The carbon in the fuel bonds with the oxygen in the air.
• Water vapor (H₂O) - This is another product of combustion. The hydrogen in the fuel bonds with the oxygen in the air.
• The emissions above are mostly benign, although carbon dioxide emissions are believed to contribute to global warming.

Because the combustion process is never perfect, some smaller amounts of more harmful emissions are also produced in car engines:

• Carbon monoxide (CO) is a colorless, odorless, deadly gas. It reduces the flow of oxygen in the bloodstream and can impair mental functions and visual perception. In urban areas, motor vehicles are responsible for as much as 90 percent of carbon monoxide in the air.
• At high temperatures (i.e. like our combustion) CO₂ can disassociate to form CO and O

Topic 4 - Catalytic converter

Topic 4 - Application 1

Oxides of Nitrogen, NO_x, results when high temperatures allow the N₂ molecules to break into two separate N molecules. Same with O₂.

This can the allow the following reactions in the combustion processes:

O + N₂    >>>>   NO + N

N + O₂    >>>>   NO + O

N + OH   >>>>   NO + H

NO₂ can then result from:

NO + H₂O     >>>>       NO₂ + H₂

NO + O₂       >>>>     NO₂ + O

NO₂ can then react with energy from sunlight to create photochemical smog

NO₂ + sunlight    >>>>     NO + O + smog

And the loose Oxygen can react

O + O₂    >>>>     O₃ which is ozone which is harmful to lung tissues

Many fuels used in CI engines contain small amounts of sulfur

O₂ + S     >>>>    SO₂

This combines with oxygen in the air

2 SO₂ + O₂      >>>>          2 SO₃

These combines with water vapor

SO₃ + H₂O     >>>>          H₂SO₄  sulfuric acid

SO₂ + H₂O     >>>>          H₂SO₃  sulurous acid

Catalytic Converters promote a thermo-chemical reaction that converts NOx and CO to less harmful combinations

NO + CO     >>>>      ½ N₂ + CO₂

2 NO + 5 CO + 3 H₂O   >>>>        2 NH₃ + 5 CO₂

2 NO + CO   >>>>         N₂O + CO₂

NO + H₂      >>>>       ½ N₂ + H₂0

2 NO + 5 H₂   >>>>        2NH₃ + 2H₂O

2 NO + H₂     >>>>       N₂O + H₂O

CO + H₂O     >>>>    CO₂ + H₂

Topic 4 - Application 2

An engine has a have a fuel rich mixture and a dry gas exhaust analysis is performed, yielding the following results:

CO₂ = 15%, C₈H₁₈ = 1 %, CO = 0% H₂ = 0%, O₂ = 0% and the rest is Nitrogen.

Determine $\text{F}\text{A}$ and $\mathrm{\Phi .}$

x C₈H₁₈ + y O₂ + y (3.76) N₂ >>>>> 15CO₂ + 1 C₈H₁₈ + z H₂O + 84 N₂

Balance Nitrogen: y(3.76) = 84, so y = 22.3

Balance Oxygen: y(2) = z +15(2), so z = 14.6

Balance Hydrogen: x(18) = 1(18) +z(2) , so x = 2.6

2.6 C₈H₁₈ + 22.3 O₂ + 22.3 (3.76) N₂ >>>>> 15CO₂ + 1 C₈H₁₈ + 14.6H₂O + 84 N₂

$\text{F}\text{A}=\frac{(2.6)8(12.01) + (2.6)18 (1.01) }{22.3(1+3.76) (28.97)}=0.096$

At Stoichiometric

$\text{F}\text{A}=\frac{8(12.01) + 18 (1.01) }{12.5(1+3.76) (28.97)}=0.066$

$\text{Φ}$ = 0.096/0.066 = 1.46

Topic 5 - Heat energy from combustion

${\text{Q}}_{\text{i}\text{n}}={\text{h}}_{\text{c}}{\text{m}}_{\text{i}} {\text{Q}}_{\text{L}\text{H}\text{V}}$

Q_LHV is the Low Heating Value (see table below for various fuels).  It assumes exhaust water is a gas.

Q_HHV  is the High Heating Value, and assumes exhaust water is liquid.

${\text{Q}}_{\text{H}\text{H}\text{V}}={\text{Q}}_{\text{L}\text{H}\text{V}} + ∆{\text{h}}_{\text{v}\text{a}\text{p}\text{o}\text{r}\text{i}\text{a}\text{t}\text{i}\text{o}\text{n}}$

$\text{∆}{\text{h}}_{\text{v}\text{a}\text{p}\text{o}\text{r}\text{i}\text{a}\text{t}\text{i}\text{o}\text{n}}$ = (kg of water) (2442000 J/kg) / (kg of fuel)

                                   (at 25 C)

Pulkrabek, W. W, 'Engineering fundamentals of the internal combustion engine', (Pulkrabek 2015:380)

If our isooctane is burned in an engine with 90% combustion efficiency and .00005 kg of fuel is burned in each cylinder per each cycle, then the heat into the engine cycle is

${\text{Q}}_{\text{in}}$=   (.90) (.00005 kg) (44300 kJ/kg) = 2 kJ

Topic 6 - Self-ignition and octane

The self-ignition temperature is the temperature at which a fuel will self-combust, without a spark, assuming that oxygen is present.  This can result in a phenomena commonly known as “knock.”

The Octane number of gasoline (petrol) is a measure of how well a fuel self-ignites.  The higher the octane number, the less susceptible the engine will be to knock.  There are actually two octane numbers, depending on the test conditions.

MON = Motor Octane Number

RON = Research Octane Number       

Anti-Knock Index (AKI) is the average of the two  =   (MON + RON)/2

To truly understand the thermochemistry of the IC engine combustion process, one must understand the fuel.  Gasoline is a mixture of different hydrocarbons, with as many as 25,000 available in crude oil.  Some common ones are:

Topic 6 - Application

A high performance fuel consists of 20% isooctane, 20% triptane, 20% isodecane, and 40% toluene by moles.  Find its research octane mumber.

Write the chemical reaction equation for the fuel.

$0.2{\text{C}}_8{\text{H}}_{18} + 0.2{\text{C}}_7{\text{H}}_{16} + 0.2{\text{C}}_{10}{\text{H}}_{22}$$+ 0.4{\text{C}}_7{\text{H}}_8 + \underline{\text{a}}{\text{O}}_2 + \underline{a}(3.76){\text{N}}_2$  >>>>>>  $\underline{\text{b}}\text{C}{\text{O}}_2+ \underline{\text{d}}{\text{H}}_{2}0 + \underline{\text{a}}(3.76){\text{N}}_2$

Now Balance the equation

$0.2{\text{C}}_8{\text{H}}_{18} + 0.2{\text{C}}_7{\text{H}}_{16} + 0.2{\text{C}}_{10}{\text{H}}_{22}$$+ 0.4{\text{C}}_7{\text{H}}_8 + 11.4{\text{O}}_2 + 11.4(3.76){\text{N}}_2$  >>>>>> $7.8\text{C}{\text{O}}_2+ 7.2{\text{H}}_{2}0 + 11.4(3.76){\text{N}}_2$

Estimate the Q_LHV of the fuel (you will need the previous table).

${\text{Q}}_{\text{L}\text{H}\text{V}} = \frac{22.8\left(44300\right) + 20\left(44440\right) + 28.4\left(44220\right) + 36.8\left(40600\right)}{108}=43044\text{k}\text{J}/\text{k}\text{g}$

Estimate the RON of the fuel

$\text{R}\text{O}\text{N}=(.211)\left(100\right)+(.185)\left(112\right)+(.263)\left(113\right)+(.341)\left(120\right)=112$

Summary

After completing this section, you should be comfortable with:

• The Otto cycle as applied to the 4 stroke spark ignition engine
• How to balance the chemical equation for the combustion of fuel in an IC engine
• Stoichiometric, lean and rich fuel mixtures
• How to determine Air-to-Fuel ratio and Equivalence Ratio
• The gasses that comprise automotive exhaust
• Pollutants in the exhaust gas mixture
• How to determine heat energy from the combustion process
• What causes self-ignition or knock
• How octane rating can be determined, and its impact on knock

Reference

Heywood, J. (2018). Fundamentals of Internal Combustion Engines. New York, USA: McGraw-Hill.

Pulkrabek, W. (2015). Engineering Fundamentals of the Internal Combustion Engine.  New York, USA: Pearson.