# Quickly Estimate Batch Distillation Time

## A simple equation suffices in many situations

Batches often are concentrated by applying heat from steam in a jacket (or similar means). Common references such as Kern’s “Process Heat Transfer” [1] and “Perry’s Chemical Engineers’ Handbook” [2] cover transient heating and cooling, as does an earlier paper of mine [3]. However, their unsteady-state analyses treat wetted area as a constant, which clearly isn’t the case when a batch is boiled down.

Equally unhelpful, batch distillation literature primarily focuses on predicting component splits via vapor/liquid equilibrium and the Rayleigh Equation. No publication appears to offer a quick way to predict the time needed to distill or evaporate a batch to a desired concentration. This article introduces just such an equation.

**DERIVING THE EQUATION**

For 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*(1)

_{S}= πDhThe volume relationship is:

*V*(2)

_{S}= πD^{2}h/4Dividing Eq. 1 by Eq. 2 provides the area-to-volume relationship:

*A*(3)

_{S}/V_{S}= 4/DSo, the total wetted area is:

*A*(4)

_{T}= A_{H}+ 4(V_{T}-V_{H})/DDifferentiating Eq. 4 with respect to time gives:

*dA*(5)

_{T}/dt = (4/D)dV_{T}/dtNext, let’s write an alternative expression for

*dV*using the heat transfer rate and fluid properties:

_{T}/dt*dV*(6)

_{T}/dt = -Q/(ρΔH)Substituting the standard expression for heat transfer yields:

*dV*(7)

_{T}/dt = -UATΔT/ρΔHUsing Eq. 5 to eliminate volume in the relationship leads to:

*(D/4)dA*(8)

_{T}/dt = -UA*ΔT/ρΔH*_{T}Solving for dA

_{T}/dt gives:

*dA*(9)

_{T}/dt = -4UA*ΔT/ρDΔH*_{T}At this point, it is convenient to define a characteristic time constant:

*Θ = ρDΔH/4UΔT*(10)

Substituting Eq. 10 into Eq. 9 gives:

*dA*(11)

_{T}/dt = -A_{T}/ΘRearrangement yields:

*dA*(12)

_{T}/A_{T}= -dt/ΘWe then can solve Eq.12 per standard procedure:

dln(A

_{T}) = -1/Θdt (13)

Integrating to the indefinite solution gives:

*ln(A*(14)

_{T}) = -t/ΘThen, substituting the integration limits (A

_{t}at time t, A

_{0}at time 0), we obtain:

*ln(A*(15)

_{t}/A_{0}) = <-t/ΘRewriting Eq. 15 provides the following simple equation:

*A*(16)

_{t}/A_{0}= e^{–t/Θ}where Θ is given by Eq. 10.

**PROCESS EXAMPLE**

We 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 ft

^{2}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/ft

^{3}. We can assume the overall heat transfer coefficient is 50 Btu/(hr∙ft

^{2}∙°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:

*V*

_{0}= 735/7.481 = 98.25 ft^{3}*V*

_{t}= 617/7.481 = 82.48 ft^{3}*V*

_{H}= 74/7.481 = 9.89 ft^{3}Then, we calculate wetted areas per Eq. 4:*A _{0} = 23 + 4(98.25 - 9.89)/5 = 93.7 ft^{2}*

*A*

_{t}= 23 + 4(82.48 - 9.89)/5 = 81.1 ft^{2}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}/A_{0}) = - 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 Mike.Gentilcore@Covidien.com**NOMENCLATURE***A* Area, ft^{2}*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∙ft^{2}∙°F)*V* Volume, ft^{3}*ρ* Liquid density, lb/ft^{3}

*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).

## What are your comments?

You cannot post comments until you have logged in. Login Here.

## Comments

No one has commented on this page yet.

RSS feed for comments on this page | RSS feed for all comments