Equivalence Ratio, Stoichiometric Ratio
4. Equivalence Ratio, Stoichiometric Ratio
The equivalence ratio φ = (F:A)/(F:A) stoich = (A:F) stoich /(A:F) = (O 2 :F) stoich /(O 2 :F)
(6) As an example, if φ = 0.5 for methane–air combustion, this implies that the excess is air sup-
plied for every kmole of fuel that is burned. In general, for methane–air combustion CH 4 + (2/ φ)(O 2 + 3.76 N 2 ) → CO 2 + 2H 2 O + 2((1/ φ)–1) O 2 + (2/ φ)×3.76N 2 , or (7)
Another term used to represent the fuel–air mixture composition is the stoichiometric ratio
SR = (actual air supplied)/(stoichiometric air demand of fuel) = 1/ φ.
For instance, if φ = 0.5, then SR = 2, i.e., the air supplied is two times as large as the stoichiometric or theoretical air demand of the fuel.
a. Example 1 Consider the metabolism of glucose in the human body. As we breathe in air, we transfer oxygen from our lungs into our bloodstream. That oxygen is transported to the cells of our tissues where it oxidizes glucose. Write down the stoichiometric reac-
tion for the consumption of glucose (s) C 6 H 12 O 6 by pure oxygen and by air. Deter- tion for the consumption of glucose (s) C 6 H 12 O 6 by pure oxygen and by air. Deter-
is consumed per minute? Solution The stoichiometric or theoretical reaction equation for this case is
(A) Applying an atom balance,
C 6 H 12 O 6 +aO 2 → b CO 2 +cH 2 O
Carbon C: 6 = b, (B) Hydrogen H: 12 = 2 c, i.e., c = 6, and
(C) Oxygen O: 6 + 2a = 2 b + c, i.e., a = 6.
(D) Therefore, the stoichiometric relation assumes the form
(E) The corresponding stoichiometric or theoretical reaction equation in the case of air is
C 6 H 12 O 6 +6O 2 → 6 CO 2 +6H 2 O.
C 6 H 12 O 6 + 6 (O 2 + 3.76 N 2 ) → 6 CO 2 +6H 2 O + 22.56 N 2 . (F) When excess air is supplied, 400% excess air = (supplied air–stoichiometric air) ×100/stoichiometric air
= (supplied air – (6 + 22.56)) × 100 ÷ (6+22.56), i.e., the supplied air = 142.8 kmole of air/kmole of glucose.
(G) Therefore, the excess air as a percentage of theoretical air equals (A:F/(A:F) stoich ) × 100 = 142.8÷28.56 = 500%.
(H) The equivalence ratio
φ = ((A:F) stoich /(A:F)) = 28.56 ÷142.8 = 0.2. (I) The actual reaction equation with 400% excess air or five times the theoretical air is
C 6 H 12 O 6 +5 ×6(O 2 + 3.76 N 2 ) → 6CO 2 +6H 2 O+aO 2 +bN 2 . (J) Applying the atom balance for Oxygen O: 5 ×6×2 = 6×2 + 6×1 + 2a, i.e., a = 24, and
(K) Nitrogen N: b = 30 ×2×3.76÷2 = 112.8, i.e.,
(L)
C 6 H 12 O 6 +5 ×6 (O 2 + 3.76 N 2 ) → 6CO 2 +6H 2 O + 24 O 2 + 112.8 N 2 . (M) The air consumption is specified as 360 L hr ––1 or 0.0147 kmole hr –1 . Therefore, the
glucose consumption per hour is ω = 0.0147÷142.8 = 0.0001029 kmole hr –1 .
(N)
The mass of one kmole of glucose is 180.2 kg, so that the glucose consumption per hour in terms of mass is
m = 00.0001029 kmole hr –1 × 180.2 kg kmole –1 = 18.6 g hr –1 or 0.31 g min –1 . (O) Remarks
The ratio of CO 2 to O 2 is called respiratory quotient (RQ) in the medical literature. For glucose, the stoichiometric reaction indicates that RQ =1. Fats (e.g., palmitic acid) are also used by the body for metabolism. The stoichiometric reaction for fat is given as
C 15 H 31 COOH + 23 O 2 → 16CO 2 + 16 H 2 O.
The RQ is given as 0.7. Since older persons have problems with excreting CO 2 , then fats are preferable compared to glucose due to lower RQ values.
Parts
» COMPUTATIONAL MECHANICS and APPLIED ANALYSIS
» Explicit and Implicit Functions and Total Differentiation
» Exact (Perfect) and Inexact (Imperfect) Differentials
» Intermolecular Forces and Potential Energy
» Internal Energy, Temperature, Collision Number and Mean Free Path
» Vector or Cross Product r The area A due to a vector product
» First Law for a Closed System
» First Law For an Open System
» STATEMENTS OF THE SECOND LAW
» Cyclical Integral for a Reversible Heat Engine
» Irreversibility and Entropy of an Isolated System
» Degradation and Quality of Energy
» SINGLE–COMPONENT INCOMPRESSIBLE FLUIDS
» Evaluation of Entropy for a Control Volume
» Internally Reversible Work for an Open System
» MAXIMUM ENTROPY AND MINIMUM ENERGY
» Generalized Derivation for a Single Phase
» LaGrange Multiplier Method for Equilibrium
» Absolute and Relative Availability Under Interactions with Ambient
» Irreversibility or Lost Work
» Applications of the Availability Balance Equation
» Closed System (Non–Flow Systems)
» Heat Pumps and Refrigerators
» Work Producing and Consumption Devices
» Graphical Illustration of Lost, Isentropic, and Optimum Work
» Flow Processes or Heat Exchangers
» CLASSICAL RATIONALE FOR POSTULATORY APPROACH
» Generalized Legendre Transform
» Van der Waals (VW) Equation of State
» Other Two–Parameter Equations of State
» Compressibility Charts (Principle of Corresponding States)
» Boyle Temperature and Boyle Curves
» Three Parameter Equations of State
» Empirical Equations Of State
» State Equations for Liquids/Solids
» Internal Energy (du) Relation
» EXPERIMENTS TO MEASURE (u O – u)
» Vapor Pressure and the Clapeyron Equation
» Saturation Relations with Surface Tension Effects
» Temperature Change During Throttling
» Throttling in Closed Systems
» Procedure for Determining Thermodynamic Properties
» Euler and Gibbs–Duhem Equations
» Relationship Between Molal and Pure Properties
» Relations between Partial Molal and Pure Properties
» Mixing Rules for Equations of State
» Partial Molal Properties Using Mixture State Equations
» Ideal Solution and Raoult’s Law
» Completely Miscible Mixtures
» DEVIATIONS FROM RAOULT’S LAW
» Mathematical Criterion for Stability
» APPLICATION TO BOILING AND CONDENSATION
» Physical Processes and Stability
» Constant Temperature and Volume
» Equivalence Ratio, Stoichiometric Ratio
» Entropy, Gibbs Function, and Gibbs Function of Formation
» Entropy Generated During an Adiabatic Chemical Reaction
» MASS CONSERVATION AND MOLE BALANCE EQUATIONS
» Evaluation of Properties During an Irreversible Chemical Reaction
» Criteria in Terms of Chemical Force Potential
» Generalized Relation for the Chemical Potential
» Nonideal Mixtures and Solutions
» Gas, Liquid and Solid Mixtures
» Availability Balance Equation
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