3/16 Temperature Effects Example

Overview of Temporary Effects in Industrial Reactions

This section focuses on the concept of temporary effects as they pertain to industrial reactions, particularly in the context of a group project due on the 27th. A review of prior materials is necessary, especially for those who missed the previous class.

Importance of Understanding Temporary Effects

  • Definition of Temporary Effects: These are changes in reaction conditions that temporarily affect the outcome of chemical processes.
  • Typical Scenarios in Industrial Reactions:
    • Reactions may not reach completion.
    • The process may not be isothermal, meaning temperature varies.
    • The presence of both innate species and multiple concurrent reactions complicates the analysis.

Exercise Introduction

  • The upcoming exercise will illustrate two complications in reaction analysis:
    • The presence of nitrogen from air used in combustion (reacting with hydrocarbon fuels).
    • The reactants are not present in their main proportions, impacting the balance calculations.

Molar Balance

  • Understanding Molar Balance: A foundational concept for approaching temperature effect problems, particularly vital because various reactions may influence reactants and products.
  • Typical Strategy in Steady-State Industrial Reactions:
    1. Set accumulation term to zero because there is no change in the amount of reactants/products over time at steady state.
    2. Focus on inert outlet flows and the production/consumption terms from reactions.

Molar Balance Equation

  • Molar balance is typically expressed as:

    ext{No accumulation} = ext{Molar Out} - ext{Molar In} + ext{Production} - ext{Consumption}
  • During an industrial operation, this means that:
    • Accumulation = 0
    • Molar Out = Molar In + Production - Consumption

Example Reaction: Reverse Water Gas Shift Reaction

  • Chemical Equation:
    extCO<em>2+extH</em>2→extCO+extH2extOext{CO}<em>2 + ext{H}</em>2 \rightarrow ext{CO} + ext{H}_2 ext{O}
  • Key Concepts for Calculating Stoichiometric Amounts:
    • Stoichiometric coefficients help quantify amounts of products and reactants in equations.
  • Consumption/Production Relations:
    • For CO2:
    • extMolarBalanceforCO2=extMolarOut(CO2)−extMolarIn(CO2)=−ζext{Molar Balance for CO2} = ext{Molar Out (CO2)} - ext{Molar In (CO2)} = -\zeta
    • For CO and Water:
    • extMolarBalanceforCO=ζext{Molar Balance for CO} = \zeta
    • extMolarBalanceforH2O=0.54+2ζext{Molar Balance for H2O} = 0.54 + 2\zeta

Enthalpy Calculation Methods

  • Direct Way:
    • ΔH=extEnthalpyIn−extEnthalpyOut\Delta H = ext{Enthalpy In} - ext{Enthalpy Out}
  • Indirect Way: More favored due to ease.
    • Combining sensible heat contributions and heat of reaction:
    • extSensibleHeatGain=C<em>p(T</em>f−Ti)ext{Sensible Heat Gain} = C<em>p (T</em>f - T_i)
    • Two distinct pathways:
    • Temperature increase/decrease of products vs. reactants.

Class Exercise Follow-Up

  • The today’s in-class exercise focuses on a problem based on earlier demonstrations, aimed at reinforcing concepts discussed prior.
  • Key points raised during setup include:
    • Reactants burned completely.
    • Use of 30% excess air in the reaction, indicating more oxidant than necessary for complete combustion.
    • The presence of water vapor in combustion air.
    • The inert nitrogen present alongside oxygen in the air.

Setting up the Equation for Molar Balance in the Exercise

  1. Write the balanced equation for combustion:
    extCH<em>4+2extO</em>2→ext2CO+2extH2extOext{CH}<em>4 + 2 ext{O}</em>2 \rightarrow ext{2CO} + 2 ext{H}_2 ext{O}
  2. Establish moles based on one mole of methane:
    • Moles of O2 required: 2 moles are typical; with 30% excess:
    • m=1×(2+0.3⋅2)=2.6m=1 \times (2 + 0.3 \cdot 2) = 2.6 moles of O2
  3. For nitrogen in air (79% N2, 21% O2):
    • Compute nitrogen based on total O2 present:
    • N<em>2=2.6moles O</em>20.21≈9.78N<em>{2} = \frac{2.6 \text{moles O}</em>2}{0.21} \approx 9.78 moles of nitrogen.
  4. Consider water vapor present at 4.2 mole percent, leading to:
    • H<em>2O</em>in=4.2%⋅(moles O<em>2+moles N</em>2)H<em>2O</em>{in} = 4.2 \% \cdot (moles \ O<em>2 + moles \ N</em>2) leading to needed adjustments for molar balance formulation.

Spreadsheet Illustrations and Techniques

  • Use of Name Boxes in Excel for clarity in formula applications.
    • Define a name for specific data cells to reference clearly across calculations.
  • Formula implementation illustrated with reversible reactions and direct output relationships in spreadsheet setups.
  • Usage of functions like SUM to total enthalpy across components for both input and output phases.

Final Remarks and Homework

  • Students encouraged to replicate processes learned in class exercises, with additional practice problems provided in course materials.
  • Emphasis on iterative learning, revising concepts through practical spreadsheet scenarios and preparation for upcoming evaluations.