EXAMPLE 19.1-4. Evaporation of a Naphthalene Sphere A sphere of naphthalene having a radius of 2.0 mm is suspended in a large volume of still air at 318 K and 1.01325 × 105 Pa (1 atm). The surface temperature of the naphthalene can be assumed to be at 318 K and its vapor pressure at 318 K is 0.555 mmHg. The DAB of naphthalene in air at 318 K is 6.92 × 10-6 m²/s. Calculate the rate of evaporation of naphthalene from the surface. 789 Solution: The flow diagram is similar to Fig. 19.1-3a. DAB = 6.92 × 10-6m²/s, PA1 = (0.555/760) (1.01325 × 105) = 74.0 Pa, PA2 = 0, r1 = 2/1000 m, R = 8314 m³ Pa/kg mol K, PB₁ = P - PA1 = 1.01325 × 105-74.0 = 1.01251 × 105 Pa, and PB2 = P - PA2 = 1.01325 × 105- 0. Since the values of PB1 and PB2 are close to each other, PBM = PB1+PB2 2 (1.0125 +1.01325) × 105 = 1.0129 x 105 Pa 2 ΝΑ Substituting into Eq. (19.1-23), DABP (PAL-PA2) RTr₁PBM m² 6.92 x 10-6 (1.01325 × 105 Pa)(74.0 Pa - 0 Pa) N₁ = m³ Pa 8314. S kgmole. K (318 K)(2/1000 m)(1.0129 × 105 Pa) NA = 9.68 × 10-8 kgmole m.s Consider a sphere shaped naphthalene (CH) having a radius of 20 mm suspended into a large volume of still air at 50 °C and 1 atm absolute pressure. At 50 °C, the vapor pressure of naphthalene is 0.9 mm Hg, its density is about 1.1 g/cm³ and its diffusivity in air is 7.02 X106 m²/s. (a) Calculate the rate of evaporation of naphthalene into the still air in kgmol/s.m² (b) Calculate the time for the complete evaporation of the naphthalene into the still air. Hint: See example 6.2-4 of the 4th edition or example 19.1-4 of the 5th edition of the class text.
EXAMPLE 19.1-4. Evaporation of a Naphthalene Sphere A sphere of naphthalene having a radius of 2.0 mm is suspended in a large volume of still air at 318 K and 1.01325 × 105 Pa (1 atm). The surface temperature of the naphthalene can be assumed to be at 318 K and its vapor pressure at 318 K is 0.555 mmHg. The DAB of naphthalene in air at 318 K is 6.92 × 10-6 m²/s. Calculate the rate of evaporation of naphthalene from the surface. 789 Solution: The flow diagram is similar to Fig. 19.1-3a. DAB = 6.92 × 10-6m²/s, PA1 = (0.555/760) (1.01325 × 105) = 74.0 Pa, PA2 = 0, r1 = 2/1000 m, R = 8314 m³ Pa/kg mol K, PB₁ = P - PA1 = 1.01325 × 105-74.0 = 1.01251 × 105 Pa, and PB2 = P - PA2 = 1.01325 × 105- 0. Since the values of PB1 and PB2 are close to each other, PBM = PB1+PB2 2 (1.0125 +1.01325) × 105 = 1.0129 x 105 Pa 2 ΝΑ Substituting into Eq. (19.1-23), DABP (PAL-PA2) RTr₁PBM m² 6.92 x 10-6 (1.01325 × 105 Pa)(74.0 Pa - 0 Pa) N₁ = m³ Pa 8314. S kgmole. K (318 K)(2/1000 m)(1.0129 × 105 Pa) NA = 9.68 × 10-8 kgmole m.s Consider a sphere shaped naphthalene (CH) having a radius of 20 mm suspended into a large volume of still air at 50 °C and 1 atm absolute pressure. At 50 °C, the vapor pressure of naphthalene is 0.9 mm Hg, its density is about 1.1 g/cm³ and its diffusivity in air is 7.02 X106 m²/s. (a) Calculate the rate of evaporation of naphthalene into the still air in kgmol/s.m² (b) Calculate the time for the complete evaporation of the naphthalene into the still air. Hint: See example 6.2-4 of the 4th edition or example 19.1-4 of the 5th edition of the class text.
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
Related questions
Question
Solve for an and b. draw the flow diagram with include values like example below. I also attach Example 19.1-4.
Transcribed Image Text:EXAMPLE 19.1-4. Evaporation of a Naphthalene Sphere
A sphere of naphthalene having a radius of 2.0 mm is suspended in a
large volume of still air at 318 K and 1.01325 × 105 Pa (1 atm). The
surface temperature of the naphthalene can be assumed to be at 318 K
and its vapor pressure at 318 K is 0.555 mmHg. The DAB of
naphthalene in air at 318 K is 6.92 × 10-6 m²/s. Calculate the rate of
evaporation of naphthalene from the surface.
789
Solution: The flow diagram is similar to Fig. 19.1-3a. DAB = 6.92 ×
10-6m²/s, PA1 = (0.555/760) (1.01325 × 105) = 74.0 Pa, PA2 = 0, r1 =
2/1000 m, R = 8314 m³ Pa/kg mol K, PB₁ = P - PA1 = 1.01325 ×
105-74.0 = 1.01251 × 105 Pa, and PB2 = P - PA2 = 1.01325 × 105-
0. Since the values of PB1 and PB2 are close to each other,
PBM =
PB1+PB2
2
(1.0125 +1.01325) × 105
= 1.0129 x 105 Pa
2
ΝΑ
Substituting into Eq. (19.1-23),
DABP (PAL-PA2)
RTr₁PBM
m²
6.92 x 10-6
(1.01325 × 105 Pa)(74.0 Pa - 0 Pa)
N₁ =
m³ Pa
8314.
S
kgmole. K
(318 K)(2/1000 m)(1.0129 × 105 Pa)
NA
= 9.68 × 10-8
kgmole
m.s
Transcribed Image Text:Consider a sphere shaped naphthalene (CH) having a radius of
20 mm suspended into a large volume of still air at 50 °C and 1
atm absolute pressure. At 50 °C, the vapor pressure of
naphthalene is 0.9 mm Hg, its density is about 1.1 g/cm³ and its
diffusivity in air is 7.02 X106 m²/s.
(a) Calculate the rate of evaporation of naphthalene into the
still air in kgmol/s.m²
(b) Calculate the time for the complete evaporation of the
naphthalene into the still air.
Hint: See example 6.2-4 of the 4th edition or example 19.1-4
of the 5th edition of the class text.
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