Internal Combustion Engines – General Course Layout

SUBJECT: ME-448 INTERNAL COMBUSTION ENGINES

CREDIT HOURS: 3-0

CONTACT HOURS: 3 Hours per Week

TEXT BOOK:

• Internal Combustion Engines: Applied Thermo sciences, Colin R.Ferguson, Allan T. Kirkpatrick, 2nd Edition, Wiley

REFERENCE BOOKS:

• Edward F. Obert, Internal Combustion Engines and Air Pollution,Harper & Row NewYark.

• Internal Combustion Engines by V.Ganesan

• Internal Combustion Engine by Willard W. Pulkrabek

Thermochemistry and Fuel

𝐶𝐻4+ 2𝑂2 = 𝐶𝑂2 + 2𝐻2𝑂

𝐶8𝐻18+ 12.5𝑂2 = 8𝐶𝑂2 + 9𝐻2𝑂

Reactants Products

What is mole?

It signifies the amount of chemical.

01 mole = 6.022 * 1023 particles

01 mole = formula (molecular) mass (g)

Combustion

• Thus one mole of CH4 = 16.04 g

• One mole of O2 = 32.00 g

• Similarly One Kg mole of CH4 = 16.04 kg

• One kg mole of O2 = 32.00 kg

• And then there is the formulae m = NM

•Where N = No of moles

M = Formula or Molecular mass

Particles Moles Grams

Combustion

• Combustion with pure oxygen is ideal, but the cost factor prohibits its wide scale use.

• Air is the best and most in expensive alternate.

• Air is composed of

78 percent nitrogen

21 percent oxygen

01 percent argon

Traces of other gases like carbon dioxide, methane, helium etc.

• Nitrogen and argon are chemically neutral and are generallyconsidered as a combination, which adds up to 79 percent

• Oxygen is the remaining 21 percent.

Combustion

• Thus we can say that for every 0.21 moles of oxygen, there is 0.79 moles of nitrogen.

• For one mole of oxygen, there are 0.79/0.21 moles of nitrogen.

• In simple words, for every mole of oxygen needed for combustion we, need 4.76 moles of air.

• Thus the stoichiometric relation for methane can be written as;

𝐶𝐻4+ 2𝑂2 + 2(3.76)𝑁2 = 𝐶𝑂2 + 2𝐻2𝑂 + 2(3.76)𝑁2

And that for octane;

𝐶8𝐻18+ 12.5𝑂2 + 12.5 3.76 𝑁2 = 8𝐶𝑂2 + 9𝐻2𝑂 + 12.5(3.76)𝑁2

Combustion

• Remember to balance the equations for 01 kg mole of fuel (why?).

• Combustion can also occurs when the mixture is lean or rich.

• If methane is burned with 150 percent stoichiometric air then the

reaction is reshaped as under;

𝐶𝐻4 + 3𝑂2 + 3(3.76)𝑁2 = 𝐶𝑂2 + 2𝐻2𝑂 + 3(3.76)𝑁2 + 𝑂2

Thus there is excess oxygen in the product.

• When octane is burned with 80 percent stoichiometric air, there isn’t

enough oxygen to convert all the carbon molecules to CO2 and then CO

also ends up in the products.

𝐶8𝐻18 + 10𝑂2 + 10 3.76 𝑁2 = 3𝐶𝑂2 + 9𝐻2𝑂 + 10(3.76)𝑁2 + 5CO

Table for Molecular Weights

Combustion• Carbon Monoxide is a colorless and odorless gas which is

extremely dangerous.

• It is a product of incomplete combustion, and can beconverted into carbon dioxide if its burned further.

• Various terminologies are used to describe the amount of airor oxygen used in combustion.

• Remember equivalence ratio ∅.

Example Problem

Isooctane is burned with 120% theoretical air in a small three-cylinder turbocharged automobile engine :

Calculate

1. Air-fuel ratio

2. Fuel-air ratio

3. Equivalence ratio

𝐶8𝐻18 + 12.5𝑂2 + 12.5 3.76 𝑁2 = 8𝐶𝑂2 + 9𝐻2𝑂 + 12.5(3.76)𝑁2

𝐶8𝐻18 + 15𝑂2 + 15 3.76 𝑁2 = 8𝐶𝑂2 + 9𝐻2𝑂 + 15(3.76)𝑁2+2.5𝑂2

Chemical Equilibrium

1. Chemical equilibrium is the condition which occurs when the

concentration of reactants and products participating in a chemical

reaction exhibit no net change over time.

2. Chemical equilibrium may also be called a "steady state reaction."

This does not mean the chemical reaction has necessarily stopped

occurring, but that the consumption and formation of substances has

reached a balanced condition.

3. The quantities of reactants and products have achieved a constant

ratio, but they are almost never equal. There may be much more

product or much more reactant.

Reference: http://chemistry.about.com/

Chemical Equilibrium

• Here A and B represent the reactant species whereas C andD represent the products. The coefficients before them arethe stoichiometric coefficients of A,B, C and D respectively.

• Equilibrium composition for a reaction can be found out ifwe know a certain constant, commonly known as thechemical equilibrium constant.

• It is given by;

Chemical Equilibrium

• Ke is very dependent on temperature, changing many orders of

magnitude over the temperature range experienced in an IC engine.

• As Ke gets larger, equilibrium is more towards the right (products).

This is the maximizing of entropy.

• For hydrocarbon fuels reacting with oxygen (air) at high engine

temperatures, the equilibrium constant is very large, meaning that

there are very few reactants (fuel and air) left at final equilibrium.

• However, at these high temperatures another chemical phenomenon

takes place that affects the overall combustion process in the engine

and what ends up in the engine exhaust.

Chemical Equilibrium

The important question to answer is what actually happens.

• Dissociation of normally stable components will occur atthese high engine temperatures

𝐶𝑂2yields

CO + O

𝑂2yields

O + O (monoatomic oxygen)

𝑁2yields

N+ N (monoatomic nitrogen)

So what!! What's the big deal

• Nitrogen in the diatomic form does not react with other

substances, but when it dissociates to monatomic nitrogen at high

temperature it readily reacts with oxygen to form nitrogen oxides,

NO and NO2, a major pollutant.

• To avoid generating large amounts of nitrogen oxides, combustion

temperatures in automobile engines are lowered, which reduces

the dissociation of N2.

• Unfortunately, this also lowers the thermal efficiency of the

engine.

Chemical Equilibrium

Exhaust Dew Point Temperature

• When exhaust gases of an IC engine are cooled below their dew

point temperature, water vapor in the exhaust starts to condense

to liquid.

• It is common to see water droplets come out of an automobile

exhaust pipe when the engine is first started and the pipe is cold.

• Very quickly the pipe is heated above the dew point temperature,

and condensing water is then seen only as vapor when the hot

exhaust is cooled by the surrounding air, much more noticeable in

the cold wintertime.

Combustion Temperature

• Heat liberated in a combustion reaction – in terms of enthalpy?

Ni = number of moles of component i

hi = (h°𝑓)i + ∆hi

(h°𝑓)i = enthalpy of formation, the enthalpy needed to form one mole

of that component at standard conditions of 25°C and 1 atm

∆hi = change of enthalpy from standard temperature for component i

Thermodynamic Tables

Combustion Temperature

• Q will be negative, meaning that heat is given up by the reactinggases.

• Heating value QHV is the negative of the heat of reaction for oneunit of fuel, and thus is a positive number.

• It is calculated assuming both the reactants and the products areat 25°C.

• Care must be used when using heating values, which almostalways are given in mass units (kJ/kg), whereas heats of reactionare obtained using molar quantities.

• Two values of heating value are given in any thermodynamic table.

Higher Heating Value 𝑄𝐻𝐻𝑉 is used when water in theexhaust products is in the liquid state

Lower Heating Value 𝑄𝐿𝐻𝑉 is used when water in the productsis vapor.

The difference is the heat of vaporization of the water

Combustion Temperature

• Now we already know, how to use 𝑄𝐿𝐻𝑉 to find out theoutput work.

• All we need to do is to find the heat in and the problem canbe moved forward from there;

• And how do we find the heat in;

𝑄𝑖𝑛 = 𝜂𝑐𝑚𝑓𝑄𝐿𝐻𝑉

Now the question arises as to why we are studying all these laws of thermodynamics and chemistry.

The answer lies in the heading of the slide.

Combustion Temperature

• We want to formulate a method through which we can predict themaximum possible temperature that can be reached in an ICEengine.

• In more technical words, we want to calculate the Adiabatic FlameTemperature of the input air fuel mixture.

• Assuming that inlet conditions of the reactants are known, it isnecessary to find the temperature of the products such that thisequation will be satisfied. This is the adiabatic flame temperature.

Combustion Temperature

• Adiabatic flame temperature is the ideal theoretical maximumtemperature that can be obtained for a given fuel and airmixture.

• The actual peak temperature in an engine cycle will be severalhundred degrees less than this. (WHY?)

• There is some heat loss even in the very short time of one cycle.

• Combustion efficiency is less than 100% so a small amount offuel does not get burned.

• Some components dissociate at the high engine temperatures.All these factors contribute to making the actual peak enginetemperature somewhat less than adiabatic flame temperature.

Example Problems

𝐶8𝐻18 + 15𝑂2 + 15 3.76 𝑁2 = 8𝐶𝑂2 + 9𝐻2𝑂 + 15(3.76)𝑁2+2.5𝑂2

Q No 1. An SI engine operating on stoichiometric propane fuel burns0.00005 kg of fuel in each cylinder during each cycle with a combustionefficiency of 95 percent. When combustion starts at end of the compressionstroke, the temperature and pressure in the cylinder are 700 K and 2000KPa. Exhaust temperature is 1200 K.

Find combustion heat input (heat of reaction) and combustion heat inputusing QLHV.

𝐶3𝐻8 + 5𝑂2 + 5 3.76 𝑁2 3 C𝑂2 + 4 𝐻2𝑂 + 5 3.76 𝑁2

Q No 2. Find the adiabatic flame temperature of iso-octane burned with anequivalence ratio of 0.833 in dry air. It can be assumed that the reactants areat a temperature of 700K after the compression stroke.