Question 9.5.3: Energy Balance on an Adiabatic Reactor The dehydrogenation o......

Basis: 100 mol Feed
Material balances lead to the flowchart shown here.

Since only one reaction occurs, we could equally well choose the reactants and products [C_{2}H_{5}OH(v) , CH_{3}CHO(v) , H_{2}(g)] or their elemental constituents [C(s), H_{2}(g), O_{2}(g)] as references for enthalpy calculations. Let us choose the molecular species.

The open system energy balance neglecting kinetic and potential energy changes and shaft work and setting Q = 0 for this adiabatic reactor is

The expression for ΔH is that of Equation 9.5-1a, with the dots above H and n omitted since a molar amount and not a flow rate was chosen as a basis of calculation.
(a) Manual Calculation
Extent of Reaction:
We could use any reactant or product as a basis for calculating ξ. Let us use acetaldehyde. From Equation 9.1-3,

\xi = \left. \Large{\{} \right. \begin{matrix} Batch: \frac{(n_{i,out} – n_{i,in})(mol  i)}{\nu _{i}(mol  i)} = \frac{n_{ir}}{\left|\nu _{i}\right|} \\Continuous: \frac{(\dot{n}_{i,out} – \dot{n}_{i,in})(mol  i/ time)}{\dot{\nu} _{i}(mol   i/ time)} = \frac{\dot{n}_{ir}}{\left|\dot{\nu} _{i}\right|} \end{matrix}                 (9.1-3)

\xi = \frac{\left|(n_{CH_{3}CHO})_{out}  –  (n_{CH_{3}CHO})_{in} \right| }{\left|\nu_{CH_{3}CHO} \right| } = \frac{\left|30.0  mol –  0  mol\right| }{1   mol} =30.0

Standard Heat of Reaction:
From Equation 9.3-1 and Table B.1 (heats of formation),

Inlet Enthalpy:

\hat{H}_{1} = \int_{25°C}^{400°C}{(C_{p})_{C_{2}H_{5}OH}} \overset{C_{p}  from  Table  B.2 }{\left. \Large{\Longrightarrow } \right.} \hat{H}_{1} = 33.79  kJ/mol

Outlet Enthalpies:
The heat capacities of ethanol vapor and hydrogen are given in Table B.2. For acetaldehyde vapor, the heat capacity is given by Poling, Prausnitz, and O’Connell^{6} :

If the heat capacity formulas for the three species are substituted in this expression and the integrals are evaluated, the results are three fourth-order polynomial expressions for \hat{H}_{2}(T_{ad}) , \hat{H}_{3}(T_{ad}) , and \hat{H}_{4}(T_{ad}) :

Solve the Energy Balance for T_{ad}

This equation can be solved using an equation-solving program or a spreadsheet.^{7} The solution is

\boxed{T_{ad} = 185°C}

(b) Spreadsheet Calculation. Replicate the following solution for yourself in Excel with APEx to make sure you understand it. Each step matches a calculation in Part (a). The numbers and formulas entered in Column B—including an initially guessed temperature of 100°C—are shown below the spreadsheet.

[B3] 30
[B4] =DeltaHfg(“acetaldehyde”) – DeltaHfg(“ethanol”)
[B5] =Enthalpy(“ethanol”,25,400,,“g”)
[B6] 100
[B7] =Enthalpy(“ethanol”,25,B$6,,“g”)
[B8] =0.05048*(B$6 – 25) + 1.326e – 4*(B$6^2 – 25^2)/2

– 8.05e-8*(B$6^3 – 25^3)/3 + 2.38.e-11*(B$6^4 – 25^4)/4

[B9] =Enthalpy(“hydrogen”,25,B$6)
[B10] = B3*B4 + 70*B7 + 30*B8 + 30*B9 –  100*B5
Once the cell entries have been made, use Excel’s Solver or Goal Seek to set the value in Cell B10 (which equals ΔH) equal to 0 by varying the value in Cell B6 (the temperature). When you click “Solve” or “OK,” the value in Cell B6 changes to 185, so that the solution is \boxed{185°C} and the value in Cell B10 changes to – 2.4 × 10^{-7} , more than close enough to zero for our purposes.
Contrast the ease of this calculation with the cumbersome set of calculations in Part (a). This example would be even easier if the heat capacity formula for acetaldehyde were listed in Table B.2 so that the Enthalpy function of APEx could be used to determine \hat{H}_{3}. Since it isn’t, the formula for the integral of the given formula must be entered manually in [B8].

^{6}B. E. Poling, J. M. Prausnitz, and J. P. O’Connell, The Properties of Gases and Liquids, 5th Edition, McGraw-Hill, New York, 2001. The formula given has been derived from the one shown in this reference, which is for the heat capacity in J/(mol·K) with the temperature expressed in kelvin

^{7}To obtain the solution using a spreadsheet, put a guessed value of T_{ad} in one cell and the expression for ΔH in an adjacent cell, and use the Goal Seek tool to determine the value of T_{ad} for which the expression for ΔH equals zero. A first guess might be the value of T_{ad} obtained by dropping all higher-order terms in the expression, leaving 6.673T_{ad}  –  1134 = 0 \Longrightarrow T_{ad}\approx 170°C.