Question 8.4.2: Vaporization and Heating One hundred g-moles per hour of liq......
Vaporization and Heating
One hundred g-moles per hour of liquid n-hexane at 25°C and 7 bar is vaporized and heated to 300°C at constant pressure. Neglecting the effect of pressure on enthalpy, estimate the rate at which heat must be supplied.
An energy balance yields
\dot{Q} =\Delta \dot{H} (\dot{W}_{s} =\Delta \dot{E}_{p} =0, \Delta \dot{E}_{k} \approx 0)
Therefore an evaluation of Δ\dot{H} will yield the desired value of \dot{Q} .
From the Antoine equation (Table B.4) or the APEx function AntoineT (“n-hexane,” 5250), the temperature at which the vapor pressure of n-hexane is 7 bar (5250 mm Hg) is 146°C, and this is therefore the temperature at which the vaporization actually occurs. However, Table B.1 lists a value of Δ\hat{H}_{v} at the normal boiling point of n-hexane,
\Delta \hat{H}_{v} = 28.85 kJ/mol at 69°C
We must therefore find a path that takes hexane from a liquid to a vapor at 69°C, rather than at the true vaporization temperature of 146°C.
As noted previously, the change in enthalpy associated with a process may be determined from any convenient path as long as the initial and final points of the chosen path correspond to those of the process.
The diagram shown on the following page illustrates several possible paths from liquid hexane at 25°C to hexane vapor at 300°C.
If we knew Δ\hat{H}_{v} at 146°C, we would follow path ABEH (the true process path) to evaluate the overall Δ\hat{H} for the process, or if we knew Δ\hat{H}_{v} at 25°C, we would follow path CF, which would require only two calculations.
Since we have Δ\hat{H}_{v} at 69°C, we must follow path ADG, which includes vaporization at that temperature.
n-C_{6}H_{14} (l,25°C, 7 bar) \overset{\Delta \hat{H}_{A}}{\longrightarrow } n-C_{6}H_{14} (l,69°C, 1 atm) \\ \left. \Large{\Downarrow} \right. \Delta \hat{H}_{D} \\ n-C_{6}H_{14} (v,69°C, 1 atm) \overset{\Delta \hat{H}_{G}}{\longrightarrow } n-C_{6}H_{14}(v,300°C, 7 bar)
For the overall process
\dot{Q} =\Delta \dot{H} = \dot{n} (mol/h) \Delta \hat{H}(kJ/mol) \\ \left. \Large{\Downarrow} \right. \Delta \hat{H} = \Delta \hat{H}_{A} +\Delta \hat{H}_{D}+ \Delta \hat{H}_{G} = 85.5 kJ/mol \\ \dot{Q} = \begin{array}{c|c|c|c} 100 mol &85.5 kJ &1 h& 1 kW\\ \hline h &mol &3600 s& 1 kJ/s\end{array} = \boxed{2.38 kW}
Notice that the pressure change term in the first step (\hat{V}ΔP = – 0.0782 kJ/mol) accounts for less than 0.1% of the overall process enthalpy change. We will generally neglect the effects of pressure changes on Δ\hat{H} unless ΔP is on the order of 50 atm or more.
