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Slide 1 - Lecture 15: Batch Distillation 1 Batch Distillation In differential distillation a feed mixture (an initial charge) of a given composition is placed in a single stage separator (a still pot, retort or flask) and heated to boiling. The vapor is collected and condensed to a distillate. The composition of the remaining liquid and the distillate are functions of time. There may be several reasons for running a batch process such as this: 1) Small capacity doesn’t warrant continuous operation 2) Separation is to be done only occasionally 3) Separation is preparative to produce a new product 4) Upstream operations are batchwise or feedstocks vary with time or from batch to batch 5) Feed materials are not appropriate for a continuous flow system. The Differential Distillation operation requires a much simpler apparatus, but is complicated because the process is now a function of time.
Slide 2 - Lecture 15: Batch Distillation 2 Batch Distillation To analyze this process we must perform component balances in the form of rates: The rate of depletion of the liquid is equal to the rate of distillate output The instantaneous rate of depletion of a component in the liquid is given by: Change in total amount of that component in the liquid Change in composition in the liquid Change in the total amount of the liquid The instantaneous rate of the component leaving in the distillate is: Conservation of species requires that these two rates be equal to each other:
Slide 3 - Lecture 15: Batch Distillation 3 Batch Distillation Multiplying the above equation by dt gives: But we know that the rate of total liquid depletion is equal to the flow rate of distillate: Which then gives: Rearranging to use separation of variables gives: The distillate composition and liquid composition are related through an equilibrium equation (y=kx). We can then integrate both sides: Rate of depletion equals the component flow rate in Distillate
Slide 4 - Lecture 15: Batch Distillation 4 Batch Distillation and then integrate both sides to obtain: If the mixture is a binary and the relative volatility constant we can substitute the relationship: Into The average composition of the liquid and vapor during the process is just: The average composition of the distillate is the amount of light key vaporized dividedby the amount of material vaporized.
Slide 5 - Lecture 15: Batch Distillation 5 Procedure: Given an amount and composition of a feed stock, the equilibrium data for the liquid-vapor equilibrium, Rate D, and pressure: 1) Determine W using the W(x) relationship derived above for x from x0 to xf in increments. 2) Relate y to time by using x as a function of time and 3) Relate W (and thus x) to time using the rate D: t=(W0-W)/D 4) Determine T as a function of t, x and W by using the equilibrium data (which relates T to y and x) Batch Distillation
Slide 6 - Lecture 15: Batch Distillation 6 1) Determine W using the W(x) relationship derived above for x from x0 to xf in increments. 2) Relate y to time by using x as a function of time and 3) Relate W (and thus x) to time using the rate D: t=(W0-W)/D 4) Determine T as a function of t, x and W by using the equilibrium data (which relates T to y and x) Batch Distillation Example: A batch still is loaded with 100 kmol 50% benzene in toluene with a relative volatility = 2.41. The boilup rate is constant at 10 kmol/hr.
Slide 7 - Lecture 15: Batch Distillation 7 Batch Rectification with Constant Reflux In Batch Rectification a feed mixture (an initial charge) of a given composition is placed in a multistage separator (a rectifier column) and heated. The vapor is collected and condensed to a distillate some of which is returned as reflux. The composition of the remaining liquid and the distillate are functions of time. Batch Rectification with constant reflux operation requires a simpler apparatus than a distillation column, but is complicated because the process is now a function of time. No boilup No reboiler No feed stream Composition xi(t) Distillate composition y(t) Distillate flow D(t) Total condenser N 2 1 Batch Distillation Reflux Distillate QB
Slide 8 - Lecture 15: Batch Distillation 8 Batch Rectification with Constant Reflux Total condenser N 2 1 Batch Rectification Reflux drum Reflux Distillate QB Equilibrium curve 45° line y yB yN Operating line at time 1 Slope=L/V=R/(R+1)<1 xD1 Operating line at time 0 Slope=L/V=R/(R+1)<1 xD0 For Batch Distillation we integrated the equation: because the relationship between y and x was simple. In this case y and x depends on the equilibrium relationship, the number of trays, and the ratio V/L and so the above equation cannot be integrated analytically. It can be analyzed graphically using a McCabe-Thiele procedure.
Slide 9 - Lecture 15: Batch Distillation 9 Batch Rectification with Constant Reflux Total condenser N 2 1 Batch Rectification Reflux drum Reflux Distillate QB Equilibrium curve 45° line y yB yN Operating line at time 1 Slope=L/V=R/(R+1)<1 xD1 Operating line at time 0 Slope=L/V=R/(R+1)<1 xD0 Batch Distillation of a two-stage column with R=1: x0 xW xD0 xD
Slide 10 - Lecture 15: Batch Distillation 10 Batch Rectification with Constant Reflux Equilibrium curve 45° line y yB yN xD1 xD0 Example: Batch Distillation with 3 stages, R=1 is initially Charged with 100 kmol of 20mol% n-hexane in n-octane. How many moles of charge must be distilled to produce an average distillate of 70mol% n-hexane? The boilup rate is 10kmol/hr. x0 xW xD0 xD 1) We start by drawing a series of operating lines. 2) We step off 3 stages on one of the operating lines. This gives us a Xw. 3) We determine From the the graph. 4) We construct a table of Xw and 1/(Y-X) andnumerically integrate 5) We set the integral to ln (W0/W) and solve for W. 6) We use To determine the average composition. This compositionwill typically be higher or lower than our target and we repeat the above steps for a different XW