tutorial4

CENG 241 Heat and Mass Transfer

Tutorial Example 4

n

1a. Show that J

i 1

i 0



1b. Show that DAB = DBA

A

2a. For the gas, the mass fraction of A, wA, is defined as wA  , where i is the



mass density

Show that

1. M = yAMA + yBMB, where yA and yB are the molar fraction of A and B

2. the molar fraction, yA,of A is

wA wA

MA M

yA   A

wA w 1

 B

MA MB M

where MA, MB and M are the molar mass of A, B and the mixture respectively.

dwA

2b. Show that dy A  2

 1 

M AM A  

M 

2c. Show that j A   DAB A and cv A  vB    AB A are equivalent.

dw cD dy

dz y A y B dz

3. Show that

a. j A  jB   V *  V 

b. J A  J B  cV  V * 

where V = molar average velocity

*

V = mass average velocity

ji = mass flux of i relative to molar average velocity

Ji = molar flux of i relative to mass average velocity



4. A gas mixture at a total pressure of 1.5 x 105 Pa and 295K contains 20% of

hydrogen, 40% of oxygen and 40% of water vapour by volume. The absolute

velocities of each species are –10 m/s, -2 m/s and 12 m/s respectively, all in the

direction of the z-axis.



a. Determine the mass average velocity and the molar average velocity for the

mixture.

b. Determine the molar flux of oxygen relative to the molar average velocity.

c. Determine the molar flux of oxygen relative to the stationary axis.

(Ans. a. 2.784 m/s, 2 m/s, b. –97.85 mol/m2s, c. –48.927 mol/m2s)





Edward Chan

CENG 241 Heat and Mass Transfer

Solution for Tutorial 4

1a. Ji = ci (vi – V)

  J i  ci vi   ciV

  J i  ci vi  V  ci , as the V is independent of ci

c v c

  J i  ci vi  i i



c

i

i



  J  c v   c v  0

i i i i i



dc A dc

1b. J A   DAB and J B   DBA B

dz dz

dc dc

c  c A  cB  A   B

dz dz

Using Part 1a., JA + JB = 0

dc A dc

 DAB  DBA B  0

dz dz

dcA dc

 DAB  DBA A  0  DAB  DBA

dz dz



2a.1 Recalling the Ideal Gas Law

W RT

PV  nRT  PV  RT  M  W

M PV

WA RT

PAV  n A RT  y A PAV  RT  y A M A  WA

MA PV

W RT

PBV  nB RT  y B PBV  B RT  y B M B  WB

MB PV



WA  WB  W or  A   B  

 M = yAMA + yBMB







 AV wA

nA MA MA

2a.2 yA    , where V is the total volume

n A  nB  AV   BV wA



wB

MA MB MA MB

wA wB 1

M   , as wA  wB  1

wA wB wA wB wA wB

  

MA MB MA MB MA MB

wA

M

 yA  A

1

M

 wA wB  dwA wA  dwA dwB 



M  M  M  M  M  M 

  

2b. dy A   A B  A A  A B 

2

 wA wB 



M  M  

 A B 



wA + wB = 1  dwA + dwB = 0  dwA = - dwB

dy A 

1  wA dwA  wA dwA 

dwA

2 2

 1   1 

M 

M AM B   M 

M AM B  

 A  A



dwA

2c. j A   DAB

dz

2





  A vA  V *

  1  dy A

  DAB M A M B   , where V* is mass average velocity

 M  dz

2

 1  dy A

  A wA  wB v A  wAv A  wAvB    DAB M A M B   ,  wA  wB  1

 M  dz

 1  dy A   

2



 wA wB v A  vB    DAB M A M B   ,  wA  A 



 M  dz   



2

 1  dy A

 cv A  vB   cDAB

M AM B

 

wA wB  M  dz

2

 1  dy A

 cv A  vB   cD AB

M AM B

 

M A y A M B y B  M  dz

M M

 cDAB  1  dy A

2



 cv A  vB    

y A y B  M  dz



3a. jA + jB = A(vA - V) + B(vB - V)

= AvA + BvB - V(A + B)

  v  BvB 

 (  A   B ) A A

    V 



 A B 

= (V* - V)



3b. JA + JB = cA(vA - V*) + cB(vB - V*)

= cAvA + cBvB - V*(cA + cB)

c v c V 

 c A  cB  A A B B  V * 

 c c 

 A B 

= c(V - V*)



mi

4a. In this case, the Ideal Gas Law is modified to PV  yi  ni RT  RT ,

Mi

since they have the same pressure but occupy with different portion, yi.

PVyi M i

 mi vi

RT

vi  yM v

V 

*

  i i i



 mi  PVyi M i yM i i

RT

0.2  2   10  0.4  32   2  0.4  18  12

V*   2.784 m/s

0.2  2  0.4  32  0.4  18



PVyi

 ni vi

RT

vi  yv yv

V   i i



 ni  PVyi y

i i

i

RT

V = 0.2 (-10) + 0.4 (-2) + 0.4 (12) = 2 m/s



b. Molar flux of oxygen relative to the molar average velocity

= cO2 (vO2 - V)

PV  yO 2 1.5  10 5  0.4

 vO 2  V    2  2  97 .85 mol/m 2s

RT 8.314  295

c. Molar flux of oxygen relative to the stationary axis

= cO2 vO2

PV  yO 2 1.5  10 5  0.4

 vO 2   2  48 .927 mol/m 2s

RT 8.314  295









Edward Chan