Quickly Estimate Batch Distillation Time

Figure 1. Equation assumes liquid continues to contact the straight sides of the vessel.

DERIVING THE EQUATIONFor simplification, let’s assume the liquid is in a jacketed tank and remains on the straight side of the vessel as it is concentrated (Figure 1). For purposes of integration, let’s also assume the heat of vaporization (ΔH), liquid density (ρ), overall heat transfer coefficient (U) and the temperature differential between the steam or other heat-transfer medium in the jacket and the liquid in the tank (ΔT) remain constant. First, let’s consider the geometry of the batch as it remains on the tank straight side:The wetted area is:A_S = πDh     (1)    The volume relationship is:V_S = πD²h/4     (2)Dividing Eq. 1 by Eq. 2 provides the area-to-volume relationship:A_S/V_S = 4/D    (3)So, the total wetted area is:    A_T = A_H + 4(V_T-V_H)/D     (4)Differentiating Eq. 4 with respect to time gives: dA_T/dt = (4/D)dV_T/dt    (5)Next, let’s write an alternative expression for dV_T/dt using the heat transfer rate and fluid properties: dV_T/dt = -Q/(ρΔH)    (6)Substituting the standard expression for heat transfer yields:dV_T/dt = -UATΔT/ρΔH    (7)Using Eq. 5 to eliminate volume in the relationship leads to:(D/4)dA_T/dt = -UA_TΔT/ρΔH    (8)Solving for dA_T/dt gives:dA_T/dt = -4UA_TΔT/ρDΔH    (9)At this point, it is convenient to define a characteristic time constant: Θ = ρDΔH/4UΔT    (10)Substituting Eq. 10 into Eq. 9 gives:dA_T/dt = -A_T/Θ    (11)Rearrangement yields:dA_T/A_T = -dt/Θ    (12)We then can solve Eq.12 per standard procedure:dln(A_T ) = -1/Θdt     (13)Integrating to the indefinite solution gives:ln(A_T) = -t/Θ     (14)Then, substituting the integration limits (A_t at time t, A₀ at time 0), we obtain:ln(A_t/A₀) = <-t/Θ     (15)Rewriting Eq. 15 provides the following simple equation: A_t/A₀ = e^–t/Θ    (16)where Θ is given by Eq. 10.PROCESS EXAMPLEWe want to concentrate a batch from 735 gal to 617 gal in a 5-ft-diameter tank. The bottom head holds 74 gal and provides 23 ft² of wetted area, in addition to the jacket. The solvent has a heat of vaporization of 1,036 Btu/lb and a density of 62.3 lb/ft³. We can assume the overall heat transfer coefficient is 50 Btu/(hr∙ft²∙°F). What is the time to evaporate the batch if the ΔT across the jacket is maintained at 165°F?First, we must convert volumes to cubic feet: V₀ = 735/7.481 = 98.25 ft³V_t = 617/7.481 = 82.48 ft³V_H = 74/7.481 = 9.89 ft³

Then, we calculate wetted areas per Eq. 4:
A₀ = 23 + 4(98.25 - 9.89)/5 = 93.7 ft²
A_t = 23 + 4(82.48 - 9.89)/5 = 81.1 ft²

From Eq. 10, we determine characteristic time:
Θ = (62.3 × 5 × 1,036)/(4 × 50 × 165) = 9.78 hr

Using a rearrangement of Eq. 15 we get:

t = -Θ ln(A_t/A₀) = - 9.78 ln(81.1/93.7) = -9.78 ln(0.8655) = 1.4 hr
So, the time needed to concentrate the batch from 735 gallons to 617 gallons is 1.4 hr.
 

MICHAEL J. GENTILCORE is Hazelwood, Mo.-based chemical process engineering manager for Mallinckrodt, a business unit of Covidien that will become a standalone company in 2013. E-mail him at [email protected]

NOMENCLATURE
A    Area, ft²
D    Tank diameter, ft
h    Liquid height above head, ft
ΔH    Heat of vaporization, Btu/lb
Q    Heat transfer rate, Btu/hr
ΔT    Temperature difference between jacket and process, °F
t    Time, hr
U    Overall heat transfer coefficient, Btu/(hr∙ft²∙°F)
V    Volume, ft³
ρ    Liquid density, lb/ft³
Θ    Time constant (per Eq. 10), hr

Subscripts
0    Time zero
H    Bottom head
S    Straight side of vessel
T    Total
t    Time t

REFERENCES
1. Kern, D. Q., “Process Heat Transfer,” p. 624, McGraw-Hill, New York (1950) [still available in McGraw-Hill Classic Textbook Reissue Series].
2. Green, D. W. and Perry, R. H., “Perry’s Chemical Engineers’ Handbook,” 6th Ed., p. 10-38, McGraw Hill, New York (1997).
3. Gentilcore, M. J., “Estimate Heating and Cooling Times for Batch Reactors,” p. 41, Chem. Eng. Progress (March 2000).