Consider a system of 6-in. ID vertical pipe with a falling film of water on the wall. Estimate he gas-pahse mass transfer coefficients for water vapor evaporating into air at 2 atm and 25°C, and a mass flow rate of 1570 Ib/hr. Take DAB=0.130 cm?/s. Use the following correlation for gases in a wetted-wall column:
Consider a system of 6-in. ID vertical pipe with a falling film of water on the wall. Estimate he gas-pahse mass transfer coefficients for water vapor evaporating into air at 2 atm and 25°C, and a mass flow rate of 1570 Ib/hr. Take DAB=0.130 cm?/s. Use the following correlation for gases in a wetted-wall column:
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
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![Consider a system of 6-in. ID vertical pipe with a falling film of water on the wall. Estimate
the gas-pahse mass transfer coefficients for water vapor evaporating into air at 2 atm and
25°C, and a mass flow rate of 1570 Ib/hr. Take DAB=0.130 cm?/s. Use the following
correlation for gases in a wetted-wall column:
0.83
0.44
Sc = 0.023 Re8 Sc°44 for Re > 2000
µ(g / cm·s)= 0.0716+4.94×10T- 3.71×10*T² where T(°C)
v(cm?/s) = 0.5[0.13269+8.73×10 T+9.8×10"T*]
where T(°C)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff432765d-5243-41e3-ac4a-5a9bf8e4d2a1%2F007e5b71-1b93-47e9-839e-1aa1fea6f544%2Fx49n2wp_processed.png&w=3840&q=75)
Transcribed Image Text:Consider a system of 6-in. ID vertical pipe with a falling film of water on the wall. Estimate
the gas-pahse mass transfer coefficients for water vapor evaporating into air at 2 atm and
25°C, and a mass flow rate of 1570 Ib/hr. Take DAB=0.130 cm?/s. Use the following
correlation for gases in a wetted-wall column:
0.83
0.44
Sc = 0.023 Re8 Sc°44 for Re > 2000
µ(g / cm·s)= 0.0716+4.94×10T- 3.71×10*T² where T(°C)
v(cm?/s) = 0.5[0.13269+8.73×10 T+9.8×10"T*]
where T(°C)
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