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# newsolutions - Chemical Engineering 150B Fall 2005 Problem...

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Page 1 of 14 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 0 05 19 6 760 22 4 298 75 240 mol . . . L N mol / cm 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 N k c c = - 2 3 3 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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[ ] 1 1 1 1 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 3 3 2 1 20 0 4 10 1 0 9 0 054 75 11052 3 1 c V t ln kA c ( sat ) . cm t ln . . cm / s* cm t s t . hr = - - - = - - = = [ 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:
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newsolutions - Chemical Engineering 150B Fall 2005 Problem...

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