Mixtures, combustion and reacting-system balances

Describe the mixture basis

A mixture needs a composition as well as pressure and temperature. Mole fraction yi=ni/ntotal counts the share of amount in moles; mass fraction wi=mi/mtotal counts the mass share. They are generally different because species have different molar masses. For an ideal gas mixture, species partial pressure pi=yi p, and the partial pressures sum to total absolute pressure.

For instance, an invented mixture with 2 mol nitrogen and 1 mol oxygen has nitrogen mole fraction2/3. At300 kPa total pressure, its partial pressure is 200 kPa. Using rounded molar masses28 and 32 g/mol, its nitrogen mass share is 56/(56+32)=63.6%, rather than 66.7%. The chosen basis matters when converting a composition into a flow record.

Balance atoms before energy

In CH4, the 4 belongs to one methane molecule: it contains one carbon atom and four hydrogen atoms. In 2 H2O, the leading 2 counts two water molecules, or two moles of water molecules on a molar basis; each contains two hydrogen atoms and one oxygen atom. The total is therefore four hydrogen and two oxygen atoms per pair of molecules. Changing a leading coefficient changes the amount; changing a formula subscript changes the species.

Complete idealised oxidation of methane is CH4+2 O2→CO2+2 H2O. Coefficients count molecules or moles, not masses. Carbon, hydrogen and oxygen totals match on both sides. Real combustion can leave carbon monoxide, unburned fuel and other products; an atom-balanced proposed reaction does not establish actual completeness or reaction rate.

Stoichiometric oxidant is the amount required by the chosen complete reaction. Excess oxidant means more than that amount is supplied. Air also introduces nitrogen and other constituents; pure oxygen is a distinct model. Distinguish wet product composition, including water vapor, from a dry composition that excludes it. Condensed water changes both phase and the gas-mole denominator.

Keep chemical and sensible enthalpy on one reference

A reacting-flow energy balance can use species enthalpies built from consistent formation references plus temperature-dependent changes. The chemical contribution must be included once. If chemical energy is already present in those stream enthalpies, adding an extra release term for the same reaction double-counts it.

An adiabatic reacting control volume can become hot even with Q=0: incoming chemical energy is redistributed into product internal/flow energy or other outputs. Product temperature requires composition, phase and property information, not merely a balanced equation. Higher and lower heating values differ in their treatment of product water and stated conditions; compare fuels or efficiencies on the same basis. These balances do not prescribe a combustion operation.

Build a species-enthalpy ledger on one reference

A molar formation enthalpy hf° assigns the enthalpy change for making one mole of a specified species and phase from its constituent elements in their stated reference forms. Reference-element enthalpies are assigned zero at the chosen reference conditions; compounds need not be zero, and another phase has its own value. For a species at a different supplied state, write its molar enthalpy as h̄=hf°+Δhstate. The state increment accounts for the change from that same reference species/phase, including any phase change when present. Here the supplied increments below are sensible temperature changes within fixed phases.

For steady reacting flow with no shaft work and negligible storage, speed and elevation changes, heat into the control volume is the sum of product molar-flow×molar-enthalpy terms minus the corresponding reactant sum. A molar flow in mol/s times kJ/mol gives kJ/s, or kW. Formation and sensible terms can be displayed separately to check them, but their sum enters the balance once.

Follow excess oxygen through a constructed reaction

Supply1 mol methane and 3 mol oxygen to the complete-oxidation model. The reaction requires 2 mol oxygen, so one remains: CH4+3 O2→CO2+2 H2O+O2. Excess oxygen is(3−2)/2=50% of the stoichiometric requirement.

If all product water is vapor, products total4 mol and the oxygen wet mole fraction is 1/4=25%. Excluding water gives 2 mol dry gas and 50% oxygen on that different basis. Using rounded molar masses, inputs16+3×32=112 g and outputs44+2×18+32=112 g. No temperature or real exhaust composition follows from this bookkeeping alone.

For a separate fictional isomer conversion A→B, A and B have the same elemental composition. Use the table’s supplied phases, formation values and sensible increments; these are teaching numbers, not real chemical property data or a process prescription. Reactant enthalpy flow is 2×(−20+5)=−30 kW and product flow is 2×(−70+15)=−110 kW. Thus Q̇=−110−(−30)=−80 kW: 80 kW leaves the control volume under the stated assumptions. Negative reference enthalpies are permitted; the difference determines heat transfer.

Separating the same ledger gives a formation contribution 2×[−70−(−20)]=−100 kW and a sensible contribution 2×(15−5)=+20 kW, summing to −80 kW. The larger product sensible increment retains 20 kW more in the stream enthalpy account, reducing outward heat from 100 to 80 kW; the temperatures of the different species cannot be compared from these increments alone. Writing another −100 kW “chemical release” alongside the already complete −80 kW would count formation energy twice: it would falsely predict −180 kW and fail the original stream balance.

Try the reasoning

1. Must the number of gas molecules be conserved in a chemical reaction?

   Show the answer and reasoning

   No. Elements and ordinary total mass are conserved; molecules can be rearranged into different counts and species.

2. Why can wet and dry oxygen percentages differ for the same products?

   Show the answer and reasoning

   They use different denominators because the dry account excludes water.

3. Does adiabatic mean a reacting stream cannot become hotter?

   Show the answer and reasoning

   No. Chemical energy can become sensible energy without heat crossing the outside boundary.

4. In the fictional A→B ledger, change only the supplied B sensible increment from +15 to +30 kJ/mol, keeping its formation value, phase and both 2 mol/s flows. What heat rate now enters?

   Show the answer and reasoning

   B has h̄=−70+30=−40 kJ/mol, so its flow carries −80 kW. A still carries −30 kW; Q̇=−80−(−30)=−50 kW. Fifty kW leaves, 30 kW less than before because the product increment retained an additional 2×15=30 kW. The increment is supplied, not a temperature inferred without properties.

5. A calculation uses the original full enthalpy ledger and reports 180 kW of outward heat after adding a separate 100 kW reaction release. Identify the double count and check the signed balance.

   Show the answer and reasoning

   The formation difference is already in each total h̄. Correctly, −30+(−80)=−110 kW in the inlet-plus-heat-equals-outlet form. Replacing Q̇ by −180 gives −210 kW on the left, inconsistent with −110 kW out. One may separate formation and sensible contributions or use their totals, but must not use both twice.

What to carry forward

Declare mole or mass basis, conserve each element, and count chemical energy only once with consistent species states.