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● Dry air at 15 oC and 85 kPa flows over a 2m long wet surface with a free stream velocity of 3 m/s. Determine the mass transfer coefficient. • Because of low mass flux conditions, we can use dry air properties for the mixture at the specified temperature of 15°C and 1 atm, for which v= 1.47 *105 m²/s. The mass diffusivity of water vapor in air at 288 K is determined from Eq. below to be: T2.072 DAB 1.87 x 10-5 × P Answer:0.0115 m/s ● ●

● Dry air at 15 oC and 85 kPa flows over a 2m long wet surface with a free stream velocity of 3 m/s. Determine the mass transfer coefficient. • Because of low mass flux conditions, we can use dry air properties for the mixture at the specified temperature of 15°C and 1 atm, for which v= 1.47 *105 m²/s. The mass diffusivity of water vapor in air at 288 K is determined from Eq. below to be: T2.072 DAB 1.87 x 10-5 × P Answer:0.0115 m/s ● ●

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Dry air at 15 oC and 85 kPa flows over a 2m long wet surface with a free stream velocity of 3 m/s. Determine the mass transfer coefficient. • Because of low mass flux conditions, we can use dry air properties for the mixture at the specified temperature of 15°C and 1 atm, for which v= 1.47 *105 m²/s. The mass diffusivity of water vapor in air at 288 K is determined from Eq. below to be: T2.072 DAB 1.87 x 10-5 x. P Answer:0.0115 m/s
Transcribed Image Text:Dry air at 15 oC and 85 kPa flows over a 2m long wet surface with a free stream velocity of 3 m/s. Determine the mass transfer coefficient. • Because of low mass flux conditions, we can use dry air properties for the mixture at the specified temperature of 15°C and 1 atm, for which v= 1.47 *105 m²/s. The mass diffusivity of water vapor in air at 288 K is determined from Eq. below to be: T2.072 DAB 1.87 x 10-5 x. P Answer:0.0115 m/s
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