Question 8.4.4: Partial Vaporization of a Mixture An equimolar liquid mixtur......
Partial Vaporization of a Mixture
An equimolar liquid mixture of benzene (B) and toluene (T) at 10°C is fed continuously to a vessel in which the mixture is heated to 50°C. The liquid product is 40.0 mole% B, and the vapor product is 68.4 mole% B.
How much heat must be transferred to the mixture per g-mole of feed?
Basis: 1 mol Feed
We start with a degree-of-freedom analysis:
\begin{matrix} 3 unknown variables (n_{V},n_{L},Q) \\ – 2 material balances \\ – 1 energy balance \\ \hline =0 degrees of freedom\end{matrix}
We could count each specific enthalpy to be determined as an unknown variable, but then we would also count the equations for each of them in terms of heat capacities and latent heats, leaving the number of degrees of freedom unchanged.
We next determine n_{V} and n_{L} from material balances, and then Q from an energy balance.
The energy balance for this process has the form Q = ΔH. An enthalpy table for the process appears as follows:
The values of n_{out} were determined from the known mole fractions of benzene and toluene in the outlet streams and the calculated values of n_{V} and n_{L}. We do not know the feed-stream pressure and so we assume that ΔH for the change from 1 atm to P_{feed} is negligible, and since the process is not running at an unusually low temperature or high pressure, we neglect the effects of pressure on enthalpy in the calculations of \hat{H}_{1} through \hat{H}_{4}. (The pressure can be estimated from the Antoine equation and Raoult’s law to be 164 mm Hg.)
The heat capacity and latent heat data needed to calculate the outlet enthalpies are obtained from Tables B.1 and B.2.
The formulas (including the APEx formulas) and values of the unknown specific enthalpies are given below. Convince yourself that the formulas represent Δ\hat{H} for the transitions from the reference states to the process states.
[=Enthalpy(“benzene”,10,80.1,“C”,;“l”) + Hv(“benzene”)
+ Enthalpy(“benzene”,80.1,50,“C”,“g”)] = 37.53 kJ/mol
\hat{H}_{4}=\int_{10°C}^{110.62°C}{(C_{p})_{C_{7}H_{8}(l)}} dT + (\Delta \hat{H}_{v} )_{C_{7}H_{8}} (110.62°C)+ \int_{110.62°C}^{50°C}{(C_{p})_{C_{7}H_{8}(v)}} dT
[=Enthalpy(“toluene”,10,110.62,“C”,“l”) + Hv(“toluene”)
+ Enthalpy(“toluene”,110.62,50,“C”,“g”)] = 42.94 kJ/mol
The energy balance is
Q=\Delta H = \sum\limits_{out}{n_{i}\hat{H}_{i}} – \sum\limits_{in}{n_{i}\hat{H}_{i}} \Longrightarrow \boxed{Q= 17.7 kJ}
References: B(l, 10°C, 1 atm), T(l, 10°C, 1 atm)
| Substance | n_{in}
mol |
\hat{H}_{in}
(kJ/mol) |
n_{out}
(mol) |
\hat{H}_{out}
(kJ/mol) |
| B(l) | 0.500 | 0 | 0.259 | \hat{H}_{1} |
| T(l) | 0.500 | 0 | 0.389 | \hat{H}_{2} |
| B(v) | — | — | 0.241 | \hat{H}_{3} |
| T(v) | — | — | 0.111 | \hat{H}_{4} |
