# Question #1 (10 marks)

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```					     CHEE 412 – Transport Phenomena in Chemical Engineering
Test #2 – March 8, 2007
Note:
 Closed book test, the use of notes and textbooks is not permitted
 This test consists of two (2) questions (Question #1 – 10 marks and Question #2 – 10
marks)
 Time for this test is 1 hour

Question #1 (10 marks)
An electrical wire with radius ro = 0.50 mm is made of copper which has an electrical
conductivity, kE = 5.1  107 ohm-1 m-1, and thermal conductivity, kC = 380 W/mK. It is
insulated to an outer radius r1 = 1.50 mm with a plastic coating which has a thermal conductivity,
kP = 0.350 W/mK.

l i
In s u a t o n
W ire

T0               A ir
T1          T   2

r0
r0
r1
r1
The ambient air is at T2 = 38C and the heat transfer coefficient from the outer insulated
surface to the surrounding air is 8.5 W/m2K. Determine the maximum current in amperes that
can flow at steady state in the wire without any portion of the plastic insulation getting heated
above its maximum allowable temperature of 93C.
Notes:
The heat generated per unit volume by electrical dissipation in the wire is
J2
ψG       ,
kE
Where J is the current density (amp/m2) and kE is the electrical conductivity (ohm-1 m-1). The
resulting units for G are (amp2  ohm/m3) or (W/m3). The current through the wire, I (amp), is
then given by I  J  ro2 .
Question #2 (10 marks)

The diagram below illustrates the gas-phase diffusion in the neighbourhood of a catalytic
surface. Component A diffuses through a stagnant film of thickness  to the catalytic surface
where it is instantaneously converted to B by the reaction,

A 3B

The product B diffuses away from the catalytic surface, back through the stagant film.
(a) Determine the rate at which A enters the gas film if this is a steady-state process.
(b) Evaluate the concentration profile yA = f(z).

z= 0
A
                                          B
z=
General balance relationships:
           
Without generation:        A  -          0
            

With generation:             A      G
 V 
  CP T, for energy and heat transfer

Conserved properties:             CA , for mass transfer
     U, for momentum transfer

  k  CP , thermaldiffusivity for energy and heat transfer

Transport diffusivities:           DA , molar/massdiffusivity for mass transfer
     , kinematic viscosityfor momentum transfer


              A()             V()
Cartesian coordinates x, y or z      A, a constant      (A)(x)
Cylindrical coordinates   r              2rL             r2L
Spherical coordinates     r              4r2           (4/3)r3

The linear first order differential equation,
d  dy 
 dx   m y ,
2

dx  
with general solution, y = f(x), 0  x  L
has the following solutions:

(i)    With boundary conditions, y(0) = yo and y(xx = L) = 0,
y
 e- m x
yO
(ii)   With boundary conditions, y(0) = yO and y(x = L) = 0,
y     emx          e-mx
          
yO 1 - e2mL 1 - e- 2mL
(iii) With boundary conditions, y(0) = yO and y(x = L) = 0
y      emx          e-mx
          
yO 1  e2mL 1  e- 2mL
Bulk motion in mass transfer:

The x-direction molar flux of a species A is given by the relationship,
NA            d yA      N      N 
 - D CT        yA  A  B 
A             dx        A      A 
P
where NB/A is the molar flux of species B (reference to species other than A) and CT       .
RT
CHEE 412 – Test #2 Solutions

Question #1
Heat transfer through the insulation:

A     0

k d  C P T 
  r; A   Ar   2  r L;   - P
dT
-k
 CP    dr           dr
                dT                     d  dT 
2  r L  k P dr   0
d
Or             r    0
dr                                          dr  dr 
    
dT                    dT C1
r      C1 Or              
dr                   dr     r
T  C1 ln r   C2
At r = r0, T = T0:         T0  C1 ln r0   C2
At r = r1, T = T1:         T1  C1 ln r1   C2
r 
Subtracting:                T0 - T1  C1 ln  0 
r 
 1
T -T
 C1  0 1
ln r0 r1 
q  dT
The heat flux is given by the relationship:    - k P
A  dr
dT
With A = 2  r L: q  - 2  r L k
dr
dT C1 1 T0 - T1 
      
dr      r    r ln r0 r1 
1 T0 - T1 
q  - 2 r L kP
r ln r0 r1 
T0 - T1 
 q  2 L kP
ln r1 r0 
Convection from the surface of the insulation to the air is given by:
q  2  r1 L h T1 - T2 
Write these heat transfer rates in terms of resistances:
q ln r1 r0 
 T0 - T1 
2 L kP

 T1 - T2 
q
2  r1 L h
Add these equations:
q ln r1 r0          q
               T0 - T2
2 L kP          2  r1 L h
T0 - T2
     q
ln r1 r0      1

2  L k P 2  r1 L h
The heat generation per unit volume by electrical dissipation is
J2                  I
ψG            where J 
kE                  r02
But we can also write:
q
ψG 
 r02 L
q          I2                      I2 L
                            Or q 

 r02 L  r02 2 k E                 r02 k E
Equating these two expressions for the heat transfer rate:
I2 L                T0 - T2

 r0 k E
2
ln r1 r0       1

2  L k P 2  r1 L h
1                                             1
                   2                                                   2
 T - T   r 2 k                                  2 T - T  k 
I        0    2 0   E
                Or         I         0     2    E
  r0
 ln r1 r0     1                                  ln r1 r0       1 
 2 k  2 r h                                     k            
r1 h 
         P       1                                       P             
1
                                           2
 2  273  93 - 273  38  5.1 10 7 
I                                                0.5  10
-3

      ln 1.5 0.5           1             


          0.350       1.5  10 -3  8.5    


I = 13.0 amps
Question #2
A 3B
The general balance with no generation is:

A     0 ;  = z and A() = A(z) = A, a constant.

NA               dy        N       N 
      - D A CT A  y A  A  B                with N A  - 3 N B
A                dz        A       A 
N               dy        N                       dy         N
 A  - D A C T A  y A  A 1  3  - D A C T A - 2 y A A
A                dz        A                       dz        A
N        D C dy A
 A - A T
A      1  2 y A dz
Substitute this expression in the general balance,
d   D A C T dy A                     d  1         dy A 
A  -                 0     Or                       0
dx   1  2 y A dz 
                                 dx 1  2 y A dz 
1     dy A                                   dy A
 C1             Or                       C1 dz
1  2 y A dz                                   1  2 yA

ln 1  2y A   C1z  C 2 - Integrated differential equation
1
2
ln 1  2y o   C1 0  C 2  C 2
1
At z = 0, assume yA = yo:
2
ln 1  0  C1  C 2
1
At z = , yA = 0 (instantaneous reaction):
2
C
 C 2  ln 1  2y o                                                   ln 1  2y o 
1                                                    1
And         C1  - 2  -
2                                                  2
Substitute C1 and C2 into the integrated differential equation,
ln 1  2 y A   -      ln 1  2 y o   ln 1  2 y o 
1                        z                   1
2                       2                   2
1  2 yA                                            z
ln 1  2 y A  - ln 1  2 y o   ln                - ln 1  2 y o   ln 1  2 y o  
z
 1 2 y          
         o 
1  2 yA                 z                                               z
 1  2 y o            Or      1  2 y A  1  2 y o 1  
1  2 yo
1                z  
       yA  1  2 y o   - 1 - Concentration profile – part (b)
1-
2                   
NA     D A C T dy A                  dy A
-                                     C1 1  2 y A  (from above)
A     1  2 y A  dz                 dz
NA
ln 1  2y o 
1
          - D A C T C1 and C1  -
A                                     2
N       D C
 A  A T ln 1  2 y o            Flux of species A – part (a)
A         2

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