Extending our equations to multicomponent systems is straightforward. For a bubble calculation we have
For a dew calculation we have
These equations may be used for bubble- or dew-pressure calculations without iterations. For bubble- or dew-point temperatures, iteration is required. A first guess may be obtained from one of the following formulas:
But these are somewhat inaccurate guesses which require subsequent iteration.7
For flash calculations, the general formula is:
to find L/F and xi and yi are then found using Eqns. 10.15 and 10.16.
Use the next example to understand how to apply the strategies described in Section 10.2. Note how the column conditions are used to decide which routine to use. Note also that developing skill to determine which routine to run is as important as proficiency with the routines.
Example 10.1. Bubble and dew temperatures and isothermal flash of ideal solutions
The overhead from a distillation column is to have the following composition:
A schematic of the top of a distillation column is shown below. The overhead stream in relation to the column and condenser is shown where Vprod represents vapor flow and Dprod represents liquid flow. In an ideal column, the vapor leaving each tray (going up) is in phase equilibrium with the liquid leaving the same tray (going down). If the cooling water to the condenser is turned off, then only vapor product will be obtained, but this is not typical because the column works better with some liquid L returning to the column. To obtain liquid product only, cooling water is provided and the vapor product stream is turned off, and the condenser is known as a total condenser. If cooling water is provided to partially condense the vapor stream, the liquid product stream is typically turned off. Then the condenser provides additional separation, operating as a partial condenser. In an ideal partial condenser, the exiting vapor and liquid leave in phase equilibrium with each other.
a. Using the shortcut K-ratio, calculate the temperature at which the condenser must operate in order to condense the overhead product completely at 8 bar.
b. Using the shortcut K-ratio, and assuming the overhead product vapors are in equilibrium with the liquid on the top plate of the column, calculate the temperature of the overhead vapors and the composition of the liquid on the top plate when operating at the pressure of part a.
c. The vapors are condensed by a partial condenser operating at 8 bar and 320 K. Using the shortcut K-ratio, what fraction of liquid is condensed?
Solution
Use the shortcut estimates of the K-ratios. Use of a solver tool is recommended after developing an understanding of the manual iterations summarized below.
a. To totally condense the overhead product, the mixture must be at the bubble-point temperature or lower. The maximum temperature is the bubble-point temperature. To find the bubble-point temperature for the ternary system, we apply Eqn. 10.20 extended to three components. The calculation requires trial-and-error iteration on temperature as summarized in Table 10.1.
Tabulated below, the shortcut K-ratio is calculated using Eqn. 10.7 at each temperature guess at 8 bar, and the y values from Eqn. 10.9 are summed to check for convergence following the procedure set forth in Table 10.1. Iterations are repeated until the y’s sum to 1.
It is easy to create an Excel spreadsheet to do these calculations quickly (e.g., Flshr.xlsx can be modified).
The temperature has been bracketed, interpolating,
The temperature has been bracketed, interpolating,
b. The saturated vapor leaving the tray is in equilibrium with the liquid and is at its dew-point temperature at 8 bar. Eqn. 10.21 is used. The calculation requires iteration on temperature. Calculating the Ki ratios as in part (a), the liquid phase compositions are calculated at each iteration using Eqn. 10.12 until the values sum to 1.
Repeating the procedure at this temperature a final time results in the liquid compositions,
c. Recognize that the solution involves an isothermal flash calculation because P and T are both specified. Begin by noting that the specified condenser temperature is between the bubble temperature, 317 K, and the dew temperature, 324 K, so vapor-liquid equilibrium is indeed possible. Because T and P are already set, Eqn. 10.7 is used to calculate the K-ratio for each component. Then, we seek a solution for Eqn 10.23 at 320 K and 8 bar. In the general flash routine, Fflash, Vflash, and Lflash are used to denote the flow rates for the flash drum and we must adapt the flash variables to the column stream names. We add “flash” and “column” subscript descriptors for increased clarity. In the solution, Fflash, Vflash, and Lflash represent the flow rates for the partial condenser, Fflash = V1,column, Vflash = Vprod, and Lflash = Dprod + Lcolumn. In the summarized calculations below, Vflash/Fflash = Vprod/V1,column and Di in the table is the objective variable for the flash calculation as used in Eqn. 10.23, not the column liquid product flow rate Dprod. Each table column summarizes a guess for flash ratio Vflash/Fflash and the resultant flash objective variable Di. The composition of the feed is given by zi as conventional for a flash calculation, which is the composition of V1,column.
Flshr.xlsx or Chap10/Flshr.m can be modified to work this example.
A summary of the isothermal flash calculation is given below.:
Interpolating between the last two results that bracket the answer,
Vflash/Fflash = 0.25, applying Eqns. 10.15 and 10.16,
⇒ {xi} = {0.1829, 0.7053, 0.1117} and {yi} = {0.3713, 0.5642, 0.0648}
The compositions can be confirmed to be converged. The outlet composition of the vapor, Vprod, is given by {yi} and it is clearly more enriched in the volatile components than the inlet from the top of the column V1,column.
Note: The flash problem converges more slowly than the bubble- and dew-point calculations, so the third iteration is necessary.
Ethanol + methanol form a nearly ideal solution as shown in Fig. 10.2. An equimolar feed at 760 mmHg is subjected to an adiabatic flash operating at 200 mmHg. Feed enters at 70°C and 43 mol/min. Find the exiting stream temperatures, flow rates, and compositions. Assume ideal solutions and use the Antoine equation for vapor pressures.
Solution
This is a direct application of a procedure, so it is clear which VLE routine to use: the FA row of Table 10.1. We must combine the VLE procedure with energy balance. A bubble-pressure calculation at 70°C (not shown) shows that the feed is all liquid. Two solutions are provided using different pathways for the enthalpy calculations. Both solutions will use the same flash calculation procedures and the Antoine equation is used with {methanol, ethanol}:
A = {8.081, 8.1122}, B = {1582.3, 1592.9}, C = {239.73, 226.18}
Solution 1
This solution method calculates component enthalpies using a reference state of liquid at 25°C for all species where we set HR = 0. The enthalpy calculations use the pathway of Fig. 2.6(a). The pathway is taken through the boiling point of each component, as in Example 3.3. To compute stream enthalpies, we use ideal solutions as shown in Example 3.3, ignoring heat of mixing. Heat capacity constants and heats of vaporization are taken from Appendix E.
The solution requires a guess of T resulting in 0 ≤ V/F ≤ 1 that provides two phases, and then a check of the energy balance. Due to the complexity of the calculation, we iterate on the T guess manually, and automate the tedious parts of solving for V/F and checking the energy balance. The solution is provided in MATLAB file Ex10_02.m. Some intermediate results are tabulated.
Because a computer is used, we skip preliminary bubble and dew calculations. Note that we do not tabulate all values until 0 ≤ V/F ≤ 1. The first guess of 45°C is above the dew temperature. The second guess of 35°C is below the bubble temperature. The next guess happens to give a condition close to the bubble temperature, so we raise the temperature guess slightly. The column OBJEB = (Eqn. 10.19)/1000. The compositions and enthalpies are shown below and the last row is converged. The exiting flow rates are V = V/F(43) = 0.09(43) = 3.87 mol/min, and L = 43 – 3.87 = 39.13 mol/min. About 9% (molar basis) of the inlet is flashed, and the outlet temperature is 40.2°C compared to an inlet of 70°C.
Solution 2
This solution method calculates component enthalpies using the pathway of Fig. 2.6(c), the reference state of the elements at 25°C, the heats of formation of ideal gases, the generalized correlation for heat of vaporization in Eqn. 2.45, and the Cpig(25°C) from the back flap. The results are slightly different from Solution 1, owing to the imprecision of Eqn. 2.45 and differences between the heat capacities. Process simulation software typically uses this enthalpy path and reference state.
We begin by finding the enthalpy of the feed relative to the elements at 25°C, noting that it is a liquid ideal solution. HF = HL(70°C) = Σ xi(ΔH°Cf,i + Cpig(T–TR) – ΔHivap) = 0.5(–200,940 + 5.28(8.314)(70 – 25) – 35,976) + 0.5(–234,950 + 7.88(8.314)(70 – 25) – 38,595) = –252,769 J/mol. This takes care of the first term in Eqn. 10.19. Noting that the feed is liquid, we might suspect the flash outlet to be mostly liquid. Performing a bubble-temperature calculation at 200 mmHg gives T = 40.00°C and HL(40°C) = –256,901 and with no vapor stream results in Q = –4132 J/mol.a The temperature must be slightly higher to move Q toward zero. Suppose we “guessed” that the temperature is 40.23°C.b Then the flash calculation gives xEtOH = 0.5173, yEtOH = 0.3515, V/F = 0.1042, HL(40.23°C) = –257,502. The formula for HV is similar to that for HL but omits the ΔHivap contribution and replaces xi with yi, so HV(40.23°C) = –212,088. Following Eqn. 10.19, Q = (1 – 0.1042)(–257,502) + 0.1042(–212,088) + 252,769 = 0.1 J/mol. We may assume that 0.1 J/mol is sufficiently close to zero.
a. Note that the heats of vaporization must be recomputed at the new temperature.
b. Obviously, this was not our first guess. Alternatively, you could call the solver with the added constraint that Q = 0 while Σ Di = 0 by changing T and V/F simultaneously.
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