Water In Steam Combustor Gas Out Gas In Generator Combustor exit v Steam out A m
ID: 503167 • Letter: W
Question
Water In Steam Combustor Gas Out Gas In Generator Combustor exit v Steam out A methane (CH4) and air mixture is burnt in the combustor then the heat is recovered to produce steam. The CH4/air mixture comes in at 25 oC and contains 10 kg/h of CH4 and 500 kg/h of air (assume 79 Nitrogen (N2) and 21 Oxygen (O2). The final outlet gas leaves at 100 oC. The combustion reaction follows the following chemical reaction: CH4 202 CO2 2H20 The heat of reaction is -802000 J/mol of CH4 reacted at 25 oC. The water enters the steam generator at 20 °C and leaves at 400 oC. Other useful information: Molecular masses (g/mol): N2 28, O 32, CH 16, H20 18, CO2 44 Specific heats (J/kgK) N2 1040, O2 935, CH4 2440, CO 920 H20 1890 (gas) 4200 (liquid) Heat of vaporisation J/kg) 2257000 Use a reference temperature of 25 oC.Explanation / Answer
2. The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it. The limiting reagent must be identified in order to calculate the percentage yield of a reaction. Here 2 moles of oxygen is required to react with 1 mole of methane.
3. The enthalpy of reaction
Let us examine the heat associated with reacting methane with oxygen gas to make carbon dioxide and water
CH4(g)+2O2(g)CO2(g)+2H2O(g)
The heat for a reaction is given the symbol Hrxn.It has units of energy per mole reaction (rxn). Per mole reaction refers to the balanced equation. For the given reaction 1 mole rxn would be the same as 1 mole of CH4(g) reacted, or 2 moles of O2(g) reacted, or 1 mole of CO2 formed, or 2 moles of H2O formed.
qp=H=n×Hrxn
where n is the number of "moles of reaction.
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