First A Course in Thermodynamics 2006 F. A. Kulacki
Description of chemically reacting systems Reaction chemistry
Stoichiometry
A First Course in Thermodynamics 2006 F. A. Kulacki
deniifed smetsys gniitcaer yllllaciimehC den fed smetsys gn tcaer y ac mehC
Definition: A multi-component system in which multienergy transfer is accompanied by a change in composition.
Reactants
Reaction takes place at a finite rate.
A First Course in Thermodynamics 2006 F. A. Kulacki M od-51 page-3 Reacting Systems-1
1 -- smetsyS 1 smetsyS gniitcaeR gn tcaeR
Basic Concepts Reaction Chemistry
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Products
+
Energy Transfer
denifed smetsys gnitcaer yllacimehC
Systems: Open and closed are considered
Heat Transfer Reactants Products
A First Course in Thermodynamics 2006 F. A. Kulacki M od-51 page-4 Reacting Systems-1
deniifed smetsys gniitcaer yllllaciimehC den fed smetsys gn tcaer y ac mehC
Three Criteria: (1) Conservation of mass (stoichiometry) (2) Conservation of energy (1st Law) (3) Increase of Entropy, or general equilibrium considerations
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smellborp llatnemadnuF sme borp atnemadnuF
A First Course in Thermodynamics 2006 F. A. Kulacki
Is energy needed or released when a systems undergoes a chemical reaction? (Is the reaction endothermic or exothermic?) exothermic?) How much energy is involved? How with the composition of the system change? How can the equilibrium state for the system be described?
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stcudorP & ,,reziidiixO ,,stnatcaeR & rez d xO stnatcaeR
Consider a closed system.
Products: CO22,, H22O, CO, N22,, O22 Products: CO H O, CO, N O
Reactants
Products
+
Energy transfer
Reactants --usually a fuel Reactants usually a fuel Oxidizer --oxygen for carbon based fuels Oxidizer oxygen for carbon based fuels
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riia noiitsubmoC r a no tsubmoC
riiA rA
Air is a mixture of oxygen, nitrogen, argon, carbon dioxide, and other trace gases. The molar composition of air at 1 atm is usually represented in terms of O2 and N2 in most combustion situations. The O2:N2 ratio by moles is
1 O2 = N 2 3.76
A First Course in Thermodynamics 2006 F. A. Kulacki M od-51 page-8 Reacting Systems-1
Fuel & Air
Products
+
Heat
Products of combustion Un-reacted O2 & all of the original N2
Air:Fuel Ratio AF = (Mass of Air)/(Mass of Fuel)
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xC4H10 + y(O2 +3.76N2 ) 7.8CO +1.1CO+ 8.2O2 +82.9N2 + zH2O 2
Note: The chemical equation is written on a molar basis. Note: The chemical equation is written on a molar basis.
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xC4H10 + y(O2 +3.76N2 ) 7.8CO +1.1CO+ 8.2O2 +82.9N2 + zH2O 2
Assumes 100 moles of products to keep the algebra simple
(The chemical equation conserves all moles of each element. Total moles most likely will change.)
A First Course in Thermodynamics 2006 F. A. Kulacki M od-51 page-12 Reacting Systems-1
1 ellpmaxE -- 1 e pmaxE
Given: Find:
A First Course in Thermodynamics 2006 F. A. Kulacki
Recant is butane, C4H10 Combustion takes place so that the products on a dry basis are: CO2 7.8% CO 1.1% 8.2% O2 82.9% N2 The percent theoretical air used in this reaction.
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Solution: Solution: The term theoretical air is introduced air here. So the first step is to determine the air needed for complete combustion. The chemical balance by species is written as
Molar Balance: Nitrogen: Carbon: Hydrogen:
3.76 y = 82 .9
4 x = 7.8+ 1.1 = 8.9
10 x = 2 z 2 z 22
Oxygen:
Moles = 44
y = 22, x = 2.2, and z = 11
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The theoretical air is given by a chemical balance that assumes just enough air present to oxidize all of the carbon:
xC4H10 + y(O2 +3.76Ns )= aCO +bH2O+3.76yN2 2
Results: a = 4, b = 5 , and y = 6.5 Moles of theoretical air = 6.5 (1+3.76)
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Moles of theoretical air = 6.5 (1+3.76) Moles actual air = (y/x)(1+3.76) = 10(4.76)
Percent excess air = 10/6.5 = 1.54, or 54% excess air, or 154% theoretical air.
A First Course in Thermodynamics 2006 F. A. Kulacki
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This is the data as given. The overall mole percent of O2 is 10% in the outlet stream. The actual moles of O2 present is determined by stoichiometry.
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10CO 2 + 13H 2 O + 2 CO + 3O 2 + 72 N2 + x(O 2 + 3.76 N 2 ) (10 + A)CO 2 + (13 + B) H2 O + (3 + C)O 2 + (72 + x3.76) N2
2 ellpmaxE 2 e pmaxE
1 2
A First Course in Thermodynamics 2006 F. A. Kulacki
Given: Given: Two streams of gases react in an open system. The reactant compositions are known, and the products contain a total of 10% oxygen. Find: Find: the moles of oxygen present in the products.
3
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1 3 2
1 CO2 H2 O CO O2 N2 10% 13% 2% 3% 72%
2
3
0% 21% 79% 10%
The chemical equation is written on the basis of 100 moles of stream 1 for which the composition is known.
Based on the requirement that O2 is % of the exhaust:
3+ C = 0.10 (10 + A) + (13 + B) + (3 + C) + (72 + 3.76x)
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10CO 2 + 13H 2 O + 2CO + 3O 2 + 72 N2 + x(O2 + 3.76N 2 ) (10 + A)CO 2 + (13 + B) H2 O + (3+ C)O 2 + (72 + x3.76) N2
Carbon: 10 + 2 = 10 + A A=2 Oxygen: 20+13+2+6+2x = 2(10+A)+(13+B) +2(3+C) 41 + 2x = 43 + B + 2C Hydrogen: 26 = 2(13 + B) B=0
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10CO2 + 13H 2O + 2CO + 3O2 + 72 N2 + x(O2 + 3.76 N2 ) (10 + A)CO2 + (13 + B)H 2 O + ( 3 + C)O2 + (72 + x3.76 )N2
With B = 0, x = 15.05, and C = 14.05 Thus, 15.05 moles of O2 are needed to produce the required 10% of O2 in the exhaust on a 100 mole basis. This means that 71.8 moles of air are needed in Stream 2 per 100 moles of exhaust.
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15 doM fo dnE
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