problems and solutions2

Chapter 6

6.1. Carbon dioxide is diffusing through nitrogen in one direction at atmospheric pressure and 0°C. The mole fraction of

CO2 at point A is 0.2; at point B, 3 m away, in the direction of diffusion, it is 0.02. Diffusivity D is 0.144 cm2/s. The gas

phase as a whole is stationary; that is, nitrogen is diffusing at the same rate as the carbon dioxide, but in the opposite

direction. (a) What is the molal flux of CO2, in kilogram moles per square meter per hour? (b) What is the net mass flux, in

kilograms per square meter per hour? (c) At what speed, in meters per second, would an observer have to move from one

point to the other so that the net mass flux, relative to him or her, would be zero? (d) At what speed would the observer have

to move so that, relative to him or her, the nitrogen is stationary? (e) What would be the molal flux of carbon dioxide

relative to the observer under condition (d)?

solution:

(a)from(6.1-8)

27308206.01RTPcccMBA

from equation(6.1-19)

hmkmolyyzDcJNAAiMAA244/10388.102.02.027308206.03360010144.0

(b) net mass flux

for carbon dioxide (molecular weight=44)

mass flux of CO2= 44×1.388×10-4 kg/m2h

for nitrogen (molecular weight=28)

mass flux of N2= 28×1.388×10-4 kg/m2h

so the net mass flux in the direction of CO2 diffusion

m=(44-28) ×1.388×10-4 =2.221×10-3 kg/m2h

(c) Here JA=NA=NB, since the diffusion is equimolal. The concentration at any point depends on position due to the

concentration profile of the equimolal diffusion, so does velocity based on equations (6.1-3a) and (6.1-3b). To select two

points, yA=0.2 and 0.02, respectively, to calculate the positions of observer

for yA=0.2, CA=Cm× yA=00893.027308206.02.0

from equation (6.1-3a)

410388.1AAAAuCNJ

the diffusing velocity of A: 01555.000893.010388.14Au

for B: CB=Cm× (1-yA)=03571.027308206.02.01

the diffusing velocity of B: 003887.003571.010388.14BABCJu

let uo is the velocity of the observer moving in the direction of CO2 diffusion, then net velocity (uA-uo) gives a mass transfer

rate mA equal to that in opposite direction mB corresponding to (uB+uo)

from equations (6.1-4) and (6.1-5)

)(44oAAAAAuuCJMm

)(28oBBABBuuCJMm for mA = mB, and rearranging two equations above gives

hmCCNCCCuuCuBAABABBAAo/00159.099988.039292.00022208.003571.02800893.0441610388.12844)2844(284428444 at yA=0.2,

It is similar to calculating uo at the point of yA=0.02

(d)

When the velocity of observer moving is equal to that of nitrogen diffusing, the nitrogen is stationary

uo=m/h003887.003571.010388.14BABCJu

(e) When the velocity of observer moving is equal to that of nitrogen diffusing, the molal flux of carbon dioxide diffusing is

indicated by

hkmoluuCuuCJAAoAAA/10736.1)003887.001555.0(00893.0)()(4B

Chapter7

7.1

solution:

The data from third column in the table are used as calculating

Mole fraction x of ammonia in water:

01048.018100170.1170.1x

molar ratio of ammonia to water X in liquid

01059.001048.0101048.01xxX

molar ratio of ammonia to inert gas 00796.08.03.1018.0pPpY

the results of calculation are list in the table

p/kPa 0 0.4 0.8 1.2 1.6 2.0 2.43 3.32 4.23 6.67 9.28

x 0 0.00527 0.01048 0.01563 0.02074 0.0258 0.03079 0.04063 0.05028 0.07357

0.09574

X 0 0.00530 0.01059 0.01588 0.02118 0.0265 0.03177 0.04235 0.05294 0.07940 0.1059

Y 0 0.00396 0.00796 0.01198 0.01600 0.0201 0.02453 0.03387 0.04353 0.07040 0.1008

The data in the table are used to plot the mole fraction x versus partial pressure p diagram and molar ratio X-Y diagram

0.000.020.040.060.080.100246810p /kPax / mole fraction B

0.000.020.040.060.080.100.120.000.020.040.060.080.10Y / molar ratioX / molar ratio B

7.3 Vapor-pressure data for a mixture of pentane (C5H12) and hexane (C6H14) are given by the table. Calculate the vapor and

liquid composition in equilibrium for the pentane-hexane at 13.3kpa pressure on assuming that the vapor of mixture 虚线范围表示符合

Herry’s law

虚线范围表示符合

Herry’s law approaches ideal behavior and liquid follows Raoult’s low.

t,K 260.6 265 270 275 280 285 289

PA,kPa 13.3 17.3 21.9 26.5 34.5 42.5

48.9

PB, kPa 2.83 3.5 4.26 5.0 8.53 11.2

13.3

Solution: From Raoult’s low （equations (7.1-4) and (7.1-5) ）

And the total pressure of system is equal to sum of the partial pressures of two substances

)1(ABAABBAABAxPxPxPxPppP

rearranging equation above

BABAPPPPx 1

and

AAAAxPpyp

rearranging equation above gives

PxPyAAA 2

substituting the data for the table into the equations 1 and 2 gives the results in the following table