CENG 241 Heat and Mass Transfer
Tutorial Example 4
n
1a. Show that
J
i
0
i 1
1b. Show that DAB = DBA
2a. For the gas, the mass fraction of A, wA, is defined as wA 
A
, 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
yA 
wA
MA
wA
w
 B
MA MB
wA
M
 A
1
M
where MA, MB and M are the molar mass of A, B and the mixture respectively.
2b. Show that dy A 
dwA
2
 1 
M AM A  
M 
dw
cD dy
2c. Show that j A   DAB A and cv A  vB    AB A are equivalent.
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
c
  J  c v   c v  0
  J i  ci vi 
i i
i
i
i
1b.
i i
i i
dc A
dc
and J B   DBA B
dz
dz
dc
dc
c  c A  cB  A   B
dz
dz
J A   D AB
Using Part 1a., JA + JB = 0
 D AB
 DAB
dc A
 DBA
dz
dc A
 DBA
dz
dcB
0
dz
dc A
 0  DAB  DBA
dz
2a.1 Recalling the Ideal Gas Law
W
RT
RT  M  W
M
PV
W
RT
PAV  n A RT  y A PAV  A 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
PV  nRT  PV 
WA  WB  W or  A   B  
 M = yAMA + yBMB
 AV
2a.2
wA
MA
nA
MA


, where V is the total volume
wA
wB
n A  nB  AV   BV

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
yA 
2b.
 wA
w  dw
w  dw
dw 

 B  A  A  A  B 
M
MB  MA MA  MA MB 
dy A   A
2
 wA
wB 



M
M
B 
 A
wA + wB = 1  dwA + dwB = 0  dwA = - dwB
dy A 
2c.
1  wA dwA  wA dwA
 1 

M A M B 
M
 A

  A vA  V

dwA
 1 

M A M B 
M
 A
2
dwA
dz
j A   DAB
*
2

2
 1  dy A
  D AB 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 
 wA wB v A  vB    DAB M A M B  
,  wA  A 
 
 M  dz 
2
 cv A  vB   cDAB
M AM B
wA wB
2
 1  dy A
 
 M  dz
2
M AM B
 1  dy A
 cv A  vB   cDAB
 
M A y A M B y B  M  dz
M
M
2
 cDAB  1  dy A
 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 A  cB

= c(V - V*)
4a.
In this case, the Ideal Gas Law is modified to PV  yi  ni RT 
mi
RT ,
Mi
since they have the same pressure but occupy with different portion, yi.
PVyi M i
vi
*
RT
V 


 mi  PVyi M i
RT
 mi vi

yM v
yM
i
i i
i
i
V* 
0.2  2   10  0.4  32   2  0.4  18  12
 2.784 m/s
0.2  2  0.4  32  0.4  18
PVyi
vi
RT
V


 ni  PVyi
RT
 ni vi

yv yv
y
i i
i i
i
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  105  0.4
vO 2  V  
 2  2  97.85mol/m 2s

RT
8.314  295
c.
Molar flux of oxygen relative to the stationary axis
= cO2 vO2

PV  yO 2
1.5  105  0.4
 2  48.927mol/m 2s
vO 2 
RT
8.314  295
Edward Chan