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Liquid \(n\) -pentane at \(25^{\circ} \mathrm{C}\) is burned with \(30 \%\) excess oxygen (not air) fed at \(75^{\circ} \mathrm{C}\). The adiabatic flame temperature is \(T_{\mathrm{ad}}\left(^{\circ} \mathrm{C}\right)\) (a) Take as a basis of calculation \(1.00 \mathrm{mol} \mathrm{C}_{5} \mathrm{H}_{12}(1)\) burned and use an energy balance on the adiabatic reactor to derive an equation of the form \(f\left(T_{\mathrm{ad}}\right)=0,\) where \(f\left(T_{\mathrm{ad}}\right)\) is a fourth-order polynomial \(\left[f\left(T_{\mathrm{ad}}\right)=c_{0}+c_{1} T_{\mathrm{ad}}+c_{2} T_{\mathrm{ad}}^{2}+c_{3} T_{\mathrm{ad}}^{3}+c_{4} T_{\mathrm{ad} \mathrm{d}}^{4}\right]\). If your derivation is correct, the ratio \(c_{0} / c_{4}\) should equal \(-6.892 \times 10^{14} .\) Use a spreadsheet program to determine \(T_{\mathrm{ad}}\) (b) Repeat the calculation of Part (a) using successively the first two terms, the first three terms, and the first four terms of the fourth-order polynomial equation. If the solution of Part (a) is taken to be exact, what percentage errors are associated with the linear (two-term), quadratic (three-term), and cubic (four-term) approximations? (c) Why is the fourth-order solution at best an approximation and quite possibly a poor one? (Hint: Examine the conditions of applicability of the heat capacity formulas in Table B.2.)

Step by step solution

01

Write Complete Combustion Reaction

Write down the combustion reaction \(C_{5}H_{12}(l) + aO_{2}(g) + bN_{2}(g) \rightarrow xCO_{2}(g) + yH_{2}O(l) + cN_{2}(g) + Heat\). Balancing this equation gives \(C_{5}H_{12}(l) + (30/100)* 15/2 O_{2} \rightarrow 5CO_{2} + 6H_{2}O+ Heat\). Assume complete combustion, so \(a=19.5\) and \(b=0\) (no nitrogen as excess oxygen is used not air) and \(x=5\), \(y=6\) and \(c=0\).

02

Establish Energy Balance Equation

Establish the energy balance equation for the adiabatic reactor, which is \(\displaystyle∑H_{reactants}(T_{ref}) -∑H_{products}(T_{ad}) = 0\). Assume temperature of reactants is \(T_{ref} = 25^{\circ} C\) and Flame temperature is \(T_{ad}\). Substitute the enthalpy terms by their respective heat capacities to convert them to temperature dependant.

03

Identify Polynomial

This equation transforms into a fourth-order polynomial \(f(T_{ad}) = c_{0}+c_{1}T_{ad}+c_{2}T_{ad}^{2}+c_{3}T_{ad}^{3}+c_{4}T_{ad}^{4}=0\). Substitute data from Table B.2 to find the coefficients \(c_{0}, c_{1}, c_{2}, c_{3}, c_{4}\). The ratio \(c_{0}/c_{4}\) should be \(-6.892x10^{14}\)

04

Solve Polynomial to find Flame Temperature

Use a spreadsheet program or any polynomial solver to determine \(T_{ad}\) by solving \(f(T_{ad}) = 0\).

05

Repeat Calculation to Determine Errors

Repeat the calculation for only first two terms, then three terms, and then four terms of the polynomial equation. Calculate the percentage of errors associated with the linear (two-term), quadratic (three-term), and cubic (four-term) approximations by considering the result from part(a) with all terms as the exact solution.

06

Analyze Fourth Order Solution

Analyze why the fourth-order solution at best is an approximation and possibly a poor one. This can be done by examining the conditions of applicability of the heat capacity formulas in Table B.2 or understanding that such polynomial equations give only approximations and any initial conditions or assumptions made affect the accuracy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combustion Reaction

When understanding adiabatic flame temperature, it’s crucial to begin with the concept of a combustion reaction. Combustion reactions entail the burning of a substance, usually a hydrocarbon, in the presence of an oxidant, generally oxygen. For pentane, a hydrocarbon, the reaction follows:

\[ C_{5}H_{12}(l) + ext{oxygen} ightarrow ext{carbon dioxide} + ext{water} + ext{heat} \]In a perfect or complete combustion, all fuel combines with the oxidant without leaving any residual unburned fuel or oxygen. This yields carbon dioxide (\(CO_2\)) and water (\(H_2O\)), as illustrated by the simplified reaction:

• \(C_{5}H_{12}(l) + aO_{2}(g) \rightarrow 5CO_{2}(g) + 6H_{2}O(l) + ext{Heat}\)
• Here, \(aO_{2}\) is calculated based on stoichiometry and the condition that there's 30% excess oxygen.

Balancing the reaction is pivotal to ensure the stoichiometric ratios of fuel to oxidant are correct, influencing the energy calculations and ultimately the adiabatic flame temperature.

Energy Balance Equation

The crux of determining the adiabatic flame temperature is the energy balance equation. For an adiabatic process, where no heat is lost to the surroundings, the principle of conservation of energy applies. The total enthalpy of reactants must equal the total enthalpy of products:\[ \ ∑H_{reactants}(T_{ref}) = ∑H_{products}(T_{ad}) \ \]Where:

• \(∑H_{reactants}(T_{ref})\) is the initial enthalpy of reactants at the reference temperature, typically seen as \(25^{\circ} C\).
• \(T_{ad}\) represents the adiabatic flame temperature.

For practical calculations, we express enthalpy changes with heat capacity formulas, making adjustments as temperatures change, thus making them temperature-dependent equations. This transformation leads naturally to a polynomial that you can solve for \(T_{ad}\). By maintaining an energy balance, we ensure that combustion efficiency and energy conversions are correctly assessed, lending accuracy to adiabatic flame temperature estimation.

Fourth-Order Polynomial

The polynomial equation representing the energy balance is usually a fourth-order polynomial as seen in combustion calculations:

• \(f(T_{ad}) = c_{0} + c_{1}T_{ad} + c_{2}T_{ad}^{2} + c_{3}T_{ad}^{3} + c_{4}T_{ad}^{4} = 0 \)

This mathematical approach mirrors how changes in temperature affect reaction thermodynamics. Coefficients \(c_{0}\) through \(c_{4}\) are calculated using data, like calorific capacity samples from reference materials (e.g., Table B.2 in our exercise). By setting up this polynomial, we're using both the known quantities and molecular energy estimations to solve for \(T_{ad}\), where the root of the polynomial gives the flame temperature.However, approximating through such polynomials always introduces slight inaccuracies because real-world conditions rarely align perfectly with theoretical calculations.

Heat Capacity Formulas

The role of heat capacity formulas in the energy balance cannot be overstated. Heat capacity indicates how much heat a substance can store as it changes temperature. As temperatures in reactions vary, heat capacity helps to assess energy changes over that range.For instance, to convert enthalpies into temperature related values, formulas for specific heat capacities \( C_{p} \) are used:

• These formulas typically vary with temperature, often expressed in the polynomial form \(C_{p} = A + BT + CT^{2} + ...\).
• Dependence of \(C_{p}\) on temperature means these coefficients must be precise to ensure accurate energy calculation.

Neglecting these variabilities or misapplying conditions can cause significant errors. Subtle temperature changes or assumptions in heat capacity's temperature dependency might make a fourth-order polynomial less responsive to abrupt or irregular conditions, and this is why calculated results, though close, remain approximations.