Answer:
The answer is "".
Explanation:
Please find the correct question in the attachment file.
using formula:
A common process for increasing the moisture content of air is to bubble it through a column of water. The air bubbles are assum
ed to be spheres having a radius of 1.0mm, and are in thermal equilibrum with the surrounding water at 298K. The vapor pressure of water is 0.03 atm at 298K, and the total pressure of the gas inside the air bubble is 1.0 atm.
a) Draw a picture of the physical system and list at least five reasonable assumptions for the mass transfer aspects of the water evaporation process. State what coordinate system should be used.
b)What are the simplified differential forms of Fick's flux equation for water vapor (species A) and the general differential equation for mass transfer in terms of concentration cA ?
c) Propose reasonable boundary and initial conditions
a) Draw a picture of the physical system and list at least five reasonable assumptions for the mass transfer aspects of the water evaporation process. State what coordinate system should be used.
b)What are the simplified differential forms of Fick's flux equation for water vapor (species A) and the general differential equation for mass transfer in terms of concentration cA ?
c) Propose reasonable boundary and initial conditions
1 answer:
You might be interested in
Answer:
a) at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
b) daylight (d) = 0.50 μm
Incandescent ( i ) = 1 μm
Explanation:
To Calculate the band emission fractions we will apply the Wien's displacement Law
The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as
F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )
<em>Values are gotten from the table named: blackbody radiati</em>on functions
<u>a) Calculate the band emission fractions for the visible region</u>
at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
attached below is a detailed solution to the problem
<u>b)calculate wavelength corresponding to the maximum spectral intensity</u>
For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm
For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm
