**Example Problems**

Physical and transport properties, where required, are calculated from empirical correlations given by Yaws *(16).*

*Example I. *Assume 50 gal of methanol spills onto a level surface outdoors. A local thermometer read *T*_{A}*= *59 ° F, and a local anemometer gives an average wind speed of __u__ = 5 mi/h. Estimate the greatest depth of the spill (*h*) and the time it will take the spill to evaporate (*t*_{2}-t_{1}).

Summarize the known conditions and the physical properties of methanol: *P*_{vp}* *= 69 .0 58 mmHg, *M *= 32.044 lb/lbmol, *W *= 332.24 lb, *R *= 555 mmHg-ft^{3}/lbmol^{o}R, µ= 0.61 9 cP, cr = 24.869 dyne/cm, p = 49.707 lb/ft^{3}, and *D*_{AB} = lb/ft.

As a first pass, assume that the evaporative flux is independent of the dimensions of the spill *( i.e., E *remains constant during the evaporation process). A preview of the calculation s reveals that the EPA method yields the shortest evaporation time, while the Stiver-MacKay method

yields the longest evaporation time. Therefore, for a conservative estimate, the Stiver-MacKay method will be used.

Calculate the initial spreading time, t_{vs} using Eq. I:

t_{vs} = 0.023462[(32.I 74 ft/s^{2})(50 gal x 7.48 gal/fl^{3}) (49.707 l b/ft^{3})(0.6l9 lb/ft-s)/24.869 dyne/ cm] = 6.24 s.

Calculate the pool radius at t_{vs} using Eq. 23:

*a*_{0} = 1 .413 1 42 [ (24.869 dyne/cm)(6.684 ft3)(6.243 s)/

(0.619cP)]^{1/4 }= 9.04 ft.

In this calculation, the unit conversion factors for µ and cr have been worked into the coefficient. The liquid pool is assumed to take the form of a spherical cap, due to the effects of surface tension. Given the volume and the radius at time zero, solve Eq. 5 for the maximum depth of the pool at its center:

1 2.77 ft^{3}; therefore, *h* = 0.0S2 ft

*B* is found by rearranging g Eq. 9 and using a spreadsheet solver function:

Tan^{3} P + 3tan*B* = (1/_{ao}^{3})(6V_{o}). Thus , tan*B* = 0.005755 rad, and *B* = 0.00576.

Per Eqs. 43 and 45:

*E *= 0.1 758(5 mi/h)(69.058 mm Hg)(32.044 lb/lbmol)/(555 m mHg-ft^{3} lb/mo1°R)(59+ 453.49°R)) = 6.76 x 10^{-3}l lb/ft^{2}min.

Use this result in Eq. 37 to find the evaporation time, t_{2} - t_{1}. Solve for t_{2} with t_{1} = 0 and *W*_{2} = 0. This leads to:

t_{2} = 3W_{1 }^{1/3 }/2bc£ where:

*b *= (6W/((1 + 3cot^{2}*B*))) ^{1/3} = 7.51 x 10^{1/3} ft/lb^{1/3}

and

*c *= 3(1 + 3cot^{2}*B*)^{2/4}p)^{1/3} + (6/(p(1 + 3cot^{2}P)))^{1/3}

= 113.42 ft/lb^{1/3}.

Thus:

t_{2} = 3(332.24lb)^{1/3}/(2 x 3.14 x (7.51 x 10^{-3} ft/lb^{1/3}) x

(113.42 ft/lb^{1/3} )(6.76 x 10^{-3} lb/ft^{2}min))= 574.1 7 min.

**Example 2: **Repeat Example 1, but this time, assume that the evaporative flux is a function of the pool radius *(a0) *under conditions of forced convection. Since the flux varies throughout the evaporation process, one needs an integrated mass balance that accounts for the effect of the pool's shrinkage on the flux. The Stiver-MacKay method is the only one that includes an explicit term for *k*, and will be used to perform the calculations. All of the physical properties and constants *(e.g., b* and *c*) are consistent with those cited in Example 1.

First, determine whether convection is turbulent or laminar using Eq. 46:

*Re = *(5 mi/h)(5,280 ft/mi)(0.076 lb/ft^{3})(2X 9.04 ft)/ ((0.018 cP)(2.419 lb/ft-h)/cP)= 8.33 x 10^{5}.

Since *Re *is greater than 3 x 10^{5}, now is turbulent and Eq. 64 should be used. This equation requires the determination of several constants.

*D*_{AB} is determined using physical properly estimation methods described in Ref. 1 2, 11-4.4

and Table 11 - 1 10 be 0.160 cm2/s = 0.010 ft2/min . The central half- angle is calculated as P = 0 .0 0576 rad. Also

per Eq. 66:

*x *= 0.0365((0.010 ft2/min ) ln )((5 mi/h **x 88**

fi/mi n/(mi/h))413(0.619 cP(0 .04032 lb/fl

mi n/cP))"10(2(7.5 I x JQ- l ft/lb1il )co 1(0.00576 rad))-113

11.62 fl-lb111S/ min.

And, per Eq. 65:

*q *= *2rcbcxP"l' _sMI R'TA *= 2rc(7.5 l x lQ- l ftllb til) x (113 .4 2 ft/lb1il)(I 1.62 ft-lb1l15/min )(69.058 mmHg) x (32.0422 lb/lbmol)/((555 mmHg-ft3/lbmol0 R)(59 +

453.49°R)) = 0.484 lb215/min.

Assume *W2 *= 0 and 11 = 0 , a od solve for r2

using Eq. 64, which is rearranged as:

12 = *5W .2'512q *= *5 *x (332 .24 l b)215/(2 x 0.484) =

52.68 min .

As may be expected, the predicted time required to evaporate the entire spill decreases significantly when one ac count s for a change in the evaporative flux with the decreasing size of the pool.

*Example 3. *Consider a smaller spill *( V*_{0} = *5 *gal) of methanol. Once again, assume that the evaporative flux varies during the evaporation process. Assume that 11 = 0 ft/s and thus, only free convection takes place. Also, as

sum e that the air above the spill contains a negligible concentration of vapor. Calculate the amount of time ii will take to evaporate the entire pill.

The Stiver-MacKay method will be use d because it includes an explicit term for k. This case exhibits turbulent flow free convec1ion, since *ScGr *= - 3.9 x 10^{9}. Thus, Eq. 76 is used with W_{2} = 0 and t1 = 0 to calculate /2:

where:

*11 *= *2rcbcmP*_{vp}^{S}MIR'T ,. = 0.0887 (lb/min )11l

*m *= 0. 14 ( D,._{8})2f.l(g 1'16/ µ)1il = 7.93 ft/mi n. Thus, 12 = 63 .6 min .

*Example 4. *Building o n Example 3, in which the evaporative flux varies, calculate lhc mass of liquid remaining, along wi1h the volume and radius of the spill, a evaporation progresses, until all of the liquid is evaporated.

Us e Eq. 76 10 solve for W2 with 11 = 0 and 12 varying from I min to 60 min. T o solve for *a, *find *V *us in g W/ p.

T he n, us i ng Eq. 2, solve for *a. *The results are shown in

t he Table and Figu re 2.