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Exploring Engineering

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Exploring Engineering

Chapter 5

Chemical energy

- Chemical formulae
- Atomic and molecular masses
- The mole/mol and Avogadro’s number
- Basic combustion reactions
- Stoichiometry
- Tabular method of determining stoichiometric cfts.
- Tabular method of solving combustion problems

- Atoms are the building blocks of matter and irreducible by chemical methods.
- A molecule is the smallest possible piece of a chemical compound.
- Molecules are made of two or more atoms.
- Since molecules are extremely small entities, and so we use large units called “mole or mol” and “kilo-mole or kmol”
- The unit* is the mole and its symbol is the mol – in practice used synonymously

- Same concept as force in newtons or N

- In chemical reactions, the identity and number of atoms, but not molecules, remains constant
- E.g., S + O2 SO2
- For engineering purposes, per atom or molecule basis is way too small
- Use either g mole (abbreviated mole or mol) or kg mole (abbreviated kmole or kmol )
- We’ll usually use kmol

- The “gram mole” or “mol” is defined to be the amount of substance containing as many "elementary entities" as there are atoms in exactly 0.012 kg of pure carbon‑12. (The “kmol” is a factor 103 larger)
- The relative masses of atoms is called “molecular mass” and commonly but incorrectly “molecular weight”

- All to 3 significant figures

- The “elementary entities” in pure C-12 are atoms of carbon defined as exactly mass 12.
- In exactly 1 mol any “elementary entity” (atoms, molecules, ions, electrons, etc.) there are 6.0221367 x 1023 atoms. This is Avogadro’s number
- A kmol is 103 larger. These are sizes interesting to engineers.
- Avogadro’s number is to mols as a dozen eggs is to 12 eggs - a shorthand notation.
- Just as a dozen eggs may include 5 turkey and 6 chicken eggs and 1 dinosaur egg, the “elementary entities” can include mixtures of molecules, atoms, electrons, neutrons, etc.

- How many kg are there in 1.00 kmol of CO2? How many kmols, mols, and lbm moles are there in 11.0 lbm of CO2?
- Need: m (mass) and n (moles) in specified CO2
- Know: Atomic masses of C = 12.0, O = 16.0; also 11.0 lbm = 5.00 kg. (Courtesy of Convert.exe)
- How: n = m/M where M = molecular mass of CO2

- Solve: M for CO2 = 12.0 + 2 x16.0 = 44.0 kg/kmol or 44.0 lbm/lbm mole, or 44.0 tons/ton mole etc.
- Thus 1.00 kg CO2 = 1.00/44.0 [kg][kmol/kg] = 2.27 10-2 kmol
- 5.00 kg (from 11.0 lbm) into kmols, n = 5.00/44.0 [kg][kmol/kg] = 0.114 kmol or 114 mols
- 11.0 lbm mols, n = 11.0/44.0 [lbm][lbm mole/lbm] = 0.251 lbm mole

- How much of “A” reacts with “B” according to a stipulated chemical combustion reaction?
- E.g., C2H5OH + aO2 bCO2 + cH2O
- C2H5OH is called “ethanol” or “ethyl alcohol” and popularly just plain “alcohol”.
- We are completely combusting the alcohol to carbon dioxide and to water (vapor) in this example.

- We need to find unique values of a,b & c that satisfy this equation and preserve the number of C, H and O atoms
- This leads to three equations and three unknowns

- C2H5OH + aO2 bCO2 + cH2O. To solve, the following tabular method is used in this course:

- C2H5OH + 3O2 2CO2 + 3H2O
- Visually check that this is correct!

- A stoichiometric method is an essential starting place to calculate fuel-to-air ratios in combustion applications
- Air is considered to be 79% N2 and 21% O2 in most combustion problems. This works out to 3.76 kmols of N2 per kmol of O2
- The mixture of 3.76 kmols of N2 and 1.00 kmol of O2 is 4.76 kmols of air containing 4.76 NAvg (Avogadro’snumber)

- How much energy released when a fuel (such as a hydrocarbon) is burned?

- For most simple hydrocarbons with 3 or more carbon atoms/molecule assume heat of combustion is 45,500 J/mol
- Example: If 1.00 kg of gasoline is burned, what is 1) the stoichiometric equation, how many lbm of CO2 is produced and how much heat is generated?

- Need: For 1 lbm of gasoline combusted what’s stoichiometry. How much CO2 produced and how much thermal energy?
- Know: Gasoline modeled by isooctane C8H18 and it produces 45.5 kJ/mol heat
- How: C8H18 + aO2 = bCO2 + cH2O

C8H18 + 12.5O2 = 8CO2 + 9H2O

CO2 = 8 44.0 = 352 lbm.

Heating Value = Energy in fuel and oxidizer

- Energy in the products of combustion

Heating value = 45.5 kJ

- Because we asserted that the heat of combustion was 45.5 kJ/kg the energetics table was mostly filled with zeros
- But many fuels do not obey this rule and instead use heats of formation

The values in the last column are not uniform in sign.

The use of the matrix table in the last example is then

Invaluable.

- “Elementary entities” means atoms, molecules, even sub-nuclear particles
- Every mol has the same enormous number of such particles (Avogadro’s number)
- Includes mixtures of molecules such as air

- Stoichiometry based on conservation of atoms in a reacting mixture
- The tabular matrix method allows a solution of stoichiometric coefficients.
- A related tabular matrix method allows a solution of thermal energetics