CHE 410 Mass Mass Transfer ransfer Solu Soluti tion on to Prob Proble lem m Set Set 1 TA: Pengfei Zhan
08/29/2014 Problem Problem 1
A pool of contami contaminate nated d wa water ter is lined lined with a 50 cm thick containm containmen entt barrier barrier.. The contamin contaminan antt in the pit has a concentration of 2 mol/L, while the groundwater circulating around the pit flows fast enough that the contaminate contaminate concentration concentration remains 0. There is initially initially no contaminan contaminantt in the barrier material at the time of installation. The governing second order, partial differential equation for diffusion of the contaminant through the barrier is: ∂C ∂ 2 C = D 2 ∂t ∂z
(0.0.1)
a) Write all boundary and initial conditions needed to solve this equation for C(z, for C(z, t). t). b) Find the steady state solution (infinite time) for C(z) for C(z).. c) Plot the initial and steady state solutions on a C C versus z z plot. plot. Draw Draw a plausi plausible ble guess guess for an intermediate solution (at some time between t =0 =0 and infinity). Solution:
a) The boundary condition is defined by time (t (t ) and space (z (z ). ). From the information given,at t=0, before the diffusion event, the concentration within the barrier is 0mol/L. So we have: C (z, 0) = 0, 0, z > 0. 0 .
(0.0.2)
At any given time, the concentration in the pool and ground water is constant, we have: C (0, (0, t) = 2mol/L
C (50cm,t (50cm,t)) = 0mol/L
(0.0.3)
(0.0.4)
b) At steady state, the concentration is a constant at any given time t . So we have:
Solve the differential equation:
C (z, t) = C ( C (z )
(0.0.5)
dC (z ) d2 C (z ) d dC (z ) = D = D ( )=0 dt dz 2 dz dz
(0.0.6)
d dC (z ) ( )dz = dz dz 1
0dz
(0.0.7)
Direction of diffusion Barrier Pit Groundwater c=2mol/L c=0mol/L z = 0cm z = 50cm
Figure 1: Demonstration of transport process.
dC (z ) dz = C 1dz dz C (z ) = C 1 z + C 2
(0.0.8)
(0.0.9)
Apply the boundary conditions to equation (0.0.9):
C (0cm) = C 1 × 0cm + C 2 = 2mol/L C (50cm) = C 1 × 50cm + C 2 = 0mol/L
Therefore, C 1 = −0.04molL 1 cm −
−
1
and C2 = 2molL 1 .
−
(0.0.10)
The concentration profile at steady state is: C (z ) = −0.04z + 2.
(0.0.11)
* The unit of C and z is mol/L and cm respectively. c) From the boundary condition, the initial state solution of C (t =0,z ) within the barrier is plotted in Figure 2 left). At steady state, the concentration is linear, C (t → ∞ ,z ) within the barrier is plotted in Figure 2 right). At intermediate state, the concentration at any position within the barrier increases with time. So we have: ∂C (t, z ) ∂ 2 C (t, z ) > 0, thus D >0 (0.0.12) ∂t ∂z 2 2
C (z )
2mol/L
C (z )
t = 0
2mol/L
0cm
0mol/L z 0cm 50cm
z 50cm
0mol/L
steady state
Figure 2: left) initial concentration profile, right) steady state concentration profile. When second order derivative is positive, the curve concaves up instead of concaves down . An reasonable intermediate state solution should has a similar shape with the curves shown in Figure 3. C (t, z ) 2mol/L
0mol/L
intermediate states
t increases
z 50cm
0cm
Figure 3: intermediate state concentration profile. Problem 2
A balloon, initially at a volume of 0.3 cm3 and filled with an ideal gas at a pressure of 1 atm, is to be filled by flowing in helium at a rate of 1 cm3 /s and a pressure of 25 atm. The balloon will expand as helium enters to maintain a pressure inside the balloon of 1 atm. Helium may be treated as an ideal gas, and both the flow in and the temperature in the balloon are maintained at 25 C). ◦
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a) If the inlet of the balloon has an area of 0.7 mm2 , determine the velocity (in cm/s) and molar flux (in moles/cm2 s) of helium entering the balloon. b) Once filled to a volume of 1200 cm3, the balloon leaks due to the diffusion of helium through the balloon material. The flux of helium across the balloon surface can be expressed as kC, where C is the concentration of gas inside the balloon and k is a mass transfer coefficient with a value of 1x10-7 cm/s. Determine how long it takes for the balloon to shrink to a volume of 600 cm3 . You may treat the balloon as spherical. Useful expressions (where r is the sphere radius): Volume of a sphere = 4/3 π r3 Area of a sphere = 4 π r2 dr3 /dr = 3r2 c) Write the mole balance that would be used to re-examine the filling process if the diffusion of helium out of the balloon was considered during filling. Do you anticipate the filling time changing substantially due to helium diffusion? You do not need to resolve for the filling time. Solution:
a) From the problem statement, we have: F in = 1.0cm3 /s, The velocity:
Pin = 25atm,
A = 0.7mm2 ,
T = 298K (25 C ).
◦
F in 1.0cm3 /s 100mm2 v= = = 142.9cm/s. 0.7mm2 cm2 A
(0.0.13)
(0.0.14)
The molar flux: J = vC in = v
P in 142.9cm/s × 25atm = = 0.146mol · cm 2 s. 3 1 1 RT 82.06cm mol K atm × 298K −
−
−
(0.0.15)
b) Since the balloon is filled, the inlet term in molar balance is zero. The outlet flux of helium is expressed by kC . The molar balance is: dCV = −J diffusionA = −kCA. (0.0.16) dt We know the pressure and temperature remain constant inside the balloon, from the ideal gas law, C =P /RT , the concentration C stays constant. Thus, from (0.0.16), we have: dV = −kA. dt
(0.0.17)
Both V and A are dependent on the radius of the sphere r . With V = 4/3 π r3 and A = 4 π r 2 . Plug these expressions into equation (0.0.17): d 4 3 ( πr ) = −k4πr 2 . dt 3
(0.0.18)
Further simplify the equation: d 4 3 4 dr 3 4 dr 3 dr 4 dr dr ( πr ) = π = π = π3r2 = 4πr 2 = −k4πr 2 dt 3 3 dt 3 dr dt 3 dt dt 4
(0.0.19)
dr = −k. (0.0.20) dt The initial volume V0 =1200cm3 , the final volume Vτ =600cm3 . So we can calculate the initial and final diameter of the sphere (r 0 and r τ ): r0 = (
3V 0 1/3 ) = 6.59cm, 4π
rτ = (
3V τ 1/3 ) = 5.23cm. 4π
(0.0.21)
From (0.0.20), we have:
r
r0
τ
τ
dr = −
kdt
=⇒
r|rr = −kt|τ 0 τ
0
=⇒
τ =
0
r0 − rτ . k
(0.0.22)
The time it takes for the balloon to shrink to a volume of 600cm3 is: τ =
r0 − rτ = 1.36 × 107 = 157.4 days. k
(0.0.23)
c) If we consider the helium gas flowing in, the mole balance becomes: dCV = F in C in − J diffusionA = F in C in − kCA. dt
(0.0.24)
From ideal gas law C=P/RT, we have: P dV P in P = F in − kA. RT dt RT RT
(0.0.25)
From(0.0.19),we have: dV/dt =4πr 2 dr/dt . So we can simplify (0.0.25) and obtain: dr F in P in 25 = − k = ( − 1 × 10 7 ) cm/s. 2 2 dt 4πr P 4πr −
(0.0.26)
outlet
inlet
The change of radius comes from both inlet helium gas and outlet diffusion. However, the inlet contribution is much larger than the outlet. So the filling time won’t change substantially due to helium diffusion . Another way to understand this is from class example, it takes less than 1 second to fill the balloon. From problem 2b, the time it takes to allow half of the helium gas diffuse out of the balloon is 157.4 days. So diffusion is a much slower process.
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