Sept_29_12_08_PM_KS_HW4_solutions

Sept_29_12_08_PM_KS_HW4_solutions - Chemical Engineering...

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Page 1 of 1 Chemical Engineering 150B- Fall 2005 Problem Set #4 Solutions Problem 1. (15 points) Consider liquid A evaporating into initially pure gaseous B in a closed vessel shown below: This vessel initially has 0.4 liter of A with 75 cm 2 of surface area in a total volume of 20 liters. After 4 minutes, B is five percent saturated with A. (a) What is the mass transfer coefficient of A? Note that the vessel is isothermal at 298 K, and the equilibrium vapor pressure of A at 298K is 24 torr. Both A and B are ideal in the gas phase. The flux at 4 minutes can be found directly from the values given: ( )( ) () ( ) 1 vapor concentration air volume N liquid area time = 2 1 2 1 24 273 005 196 760 22 4 298 75 240 mol .. .L Nm o l / c m s cm sec ⎛⎞ ⎜⎟ ⎝⎠ == -8 7.0 x 10 The concentration difference is that at the interface between A and B minus that in the solution. That at the surface is the value at saturation; that in the bulk at short times is essentially zero. Thus, ( ) 1 A ,i A Nk c c =− 2 33 1 24 273 0 760 298 22 4 10 mol mol/cm s k .* cm k = = -8 7.0 x 10 0.054 cm/s This value is lower than that commonly found for transfer in gases. (b) How long will it take to reach ninety percent saturation of A in B? The time required for 90 percent saturation can be found from a mass balance: accumulation in gas phase = evaporation rate B ( g ) A ( l ) 2 pt 2 pt 3 pt
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Page 2 of 2 [] 11 1 d Vc AN dt d Vc Ak c ( sat ) c dt = = B is initially pure, so t=0, c 1 =0. We use this condition to integrate the mass balance: 1 1 1 (kA/V)t c e c(sat) =− () 1 1 33 2 1 20 0 4 10 10 9 0 054 75 11052 31 c V tl n kA c ( sat ) .c m n . m / s * c m ts t. h r ⎛⎞ ⎜⎟ ⎝⎠ = = [ Note that credit will be given to answers that use vapor concentration of A at 4 minutes as 5% of total pressure (1 atm), only because the statement “five percent A” may be interpreted in that manner. Answers for that situation are as follows (a) k=1.72 cm/s and (b) t=349 s. Since the mol fraction of A at saturation is y A = 24/760=0.03 < 0.05 , that interpretation of 5% is not physically possible.] Problem 2. (15 points) A 1-m square, thin plate of napththalene is oriented parallel to an air stream flowing at 30 m/s. The air is at 310 K and 1 atm. The plate is at 300 K. Determine the rate of sublimation from the top of the plate. [The following properties for naphthalene are known: Diffusivity of naphthalene is at 298 K = 6.11 x 10 -6 m 2 /s ; vapor pressure at 300 K = 25 Pa.] Note that diffusivity changes with temperature, so we must first use equation 24-41 at the average temperature to calculate the appropriate diffusivity. Next we check whether this flow is laminar or turbulent: 3 pt 3 pt 2 pt 2 pt
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Page 3 of 3 Then we use the derivation in example 3 on p562 of W3R to find the average mass transfer coefficient taking care of both turbulent and laminar portions of the plate. Finally we evaluate the flux based on the definition of the average mass transfer coefficient: 3 pt 3 pt 3 pt 2 pt
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Page 4 of 4 Problem 3. (20 points) The boundary layer solution for a flat plate predicts the following equations: For laminar flow: 1/2 1/3 c AB kx 0.332Re D x Sc = For turbulent flow: 4/5 c AB 0.0292Re D x Sc =
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This homework help was uploaded on 04/02/2008 for the course CHEM 150b taught by Professor Bell during the Spring '08 term at University of California, Berkeley.

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Sept_29_12_08_PM_KS_HW4_solutions - Chemical Engineering...

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