Question 9.5.4: Simultaneous Material and Energy Balances The ethanol dehydr......
Simultaneous Material and Energy Balances
The ethanol dehydrogenation reaction of Example 9.5-3 is carried out with the feed entering at 300°C. The feed contains 90.0 mole% ethanol and the balance acetaldehyde and enters the reactor at a rate of 150 mol/s.
To keep the temperature from dropping too much and thereby decreasing the reaction rate to an unacceptably low level, heat is transferred to the reactor. When the heat addition rate is 2440 kW, the outlet temperature is 253°C. Calculate the fractional conversion of ethanol achieved in the reactor.
A degree-of-freedom analysis based on atomic species balances (see Section 4.7) is as follows:
(Convince yourself that there are only two independent atomic balances by writing the C and O balances and observing that they yield the same equation.)
Balance on C
Balance on H
Energy Balance
In the last example we used molecular species as references for specific enthalpy calculations. This time we will use elemental species [C(s), H_{2}(g), O_{2}(g)] at 25°C and 1 atm. (For a single reaction both choices require about the same computational effort.) The energy balance neglecting shaft work and kinetic and potential energy changes becomes
\dot{Q}=\Delta \dot{H}= \sum{\dot{n}_{out}\hat{H}_{out}} – \sum {\dot{n}_{in}\hat{H}_{in}}
\Delta \dot{H}= \sum{\dot{n}_{out}\hat{H}_{out}} – \sum{ \dot{n}_{in} \hat{H}_{in}} (9.5-2)
The value of \dot{Q} is 2440 kJ/s and the expression forΔ\dot{H} is that of Equation 9.5-2. The specific enthalpies of the species at the inlet and outlet of the process relative to their elemental constituents are calculated as
\hat{H}_{i} = \Delta \hat{H}^{°}_{fi} + \int_{25°C}^{T}{C_{pi}(T) dT}
where T is 300°C at the inlet and 253°C at the outlet. Taking standard heats of formation from Table B.1 (or APEx function DeltaHfg), formulas for C_{p} of ethanol vapor and hydrogen from Table B.2 (or APEx function Enthalpy for the integrals of those formulas), and the formula for C_{p} of acetaldehyde vapor from Example 9.5-3, we calculate the values of \hat{H}_{i} shown in the inlet–outlet enthalpy table.
The energy balance ( \dot{Q}= \sum{\dot{n}_{out} \hat{H}_{out}} – \sum {\dot{n}_{in}\hat{H}_{in}} ) becomes
Solving Equations 1 through 3 simultaneously (e.g., with Excel’s Solver) yields
\dot{n}_{1}= 92.0 mol C_{2}H_{5}OH/s\\ \dot{n}_{2} = 58.0 mol CH_{3}CHO/s\\ \dot{n}_{3} = 43.0 mol H_{2}/s
The fractional conversion of ethanol is
x= \frac{(\dot{n}_{C_{2}H_{5}OH})_{in} – (\dot{n}_{C_{2}H_{5}OH})_{out}}{(\dot{n}_{C_{2}H_{5}OH})_{in}} = \frac{(135 – 92.0) mol/s}{135 mol/s} =\boxed{0.319}
References: C(s), H_{2}(g), O_{2}(g) at 25°C and 1 atm
| Substance | \dot{n}_{in}
(mol/s) |
\hat{H}_{in}
(kJ/mol) |
\dot{n}_{out}
(mol/s) |
\hat{H}_{out}
(kJ/mol) |
| C_{2}H_{5}OH | 135 | -212.19 | \dot{n}_{1} | – 216.81 |
| CH_{3}CHO | 15 | -147.07 | \dot{n}_{2} | – 150.90 |
| H_{2} | — | — | \dot{n}_{3} | 6.595 |
