mass transfer (stoffaustausch) · frogs are known for cutaneous respiration (gas exchange through...

1

Prof. Dr. Sotiris E. Pratsinis Particle Technology Laboratory Sonneggstrasse 3, ML F13 http://www.ptl.ethz.ch

Mass Transfer (Stoffaustausch)

Examination August 2015

Name:______________________________________________

Legi-Nr.:_____________________________________________

Edition “Diffusion” by E. L. Cussler: none 2nd 3rd

Exam Duration: 120 minutes

The following materials are not permitted at your table and have to be deposited in front or

back of the examination room during the examination:

bags and jackets

exercises of the mass transfer lecture (also handwritten on summary sheet or textbook)

notebooks, mobile phones, devices with wireless communication ability

The following materials are permitted at your table:

1 calculator

1 copy of the book “Diffusion” (2nd or 3rd edition) by E. L. Cussler

1 printout of the lecture script

1 sheet (2 pages) summary in format DIN A4 or equivalent

Please read these points:

write your name and Legi-Nr. on each sheet of your solution

begin each problem on a new sheet

write only on the front side of each sheet

2


Problem 1 (25 points)

Frogs are known for cutaneous respiration (gas exchange through the skin via diffusion). They are able

to breath CO2 out of their bodies through their 300 µm thick skins. A group of bio-engineers wants to

investigate this effect. They approximate the system by assuming that CO2 is transported to the inner

surface of the skin. From there it diffuses to the outer surface where it is transported immediately away

due to convection. Assume that the concentration at the inner surface is constant at c = 1.2 mmol / L

for 3 s (one cycle). Assume diffusion in one dimension.

a) Are the engineers dealing with a steady or un-steady state problem? Decide in a quantitative

way.

b) Draw a detailed sketch of the problem including all relevant information.

c) How many moles of CO2 gas have gone through the frog’s inner skin surface via cutaneous

respiration in 3 s?

d) What is the percentage of moles CO2 released per cycle via cutaneous respiration compared to

the overall amount of moles CO2 set free (cutaneous + oral respiration)?

e) In absence of salt in the surrounding water the frog’s skin thickness decreases to 20 µm. How

much is the CO2 flux through the skin under these conditions if the characteristic breathing

time is 3 s?

Additional data:

- Diffusion coefficient of CO2 in the tissue: 1.92∙10-5 cm2/s

- Overall surface area for cutaneous respiration of the frog: 400 cm2

- Oral CO2 respiration rate of a frog: 0.02 g / min

- Molecular weight of CO2 = 44.01 g / mol

3



Dry ice (CO2) is placed in the bottom of a cylindrical capillary tube to cool an instrument and

evaporates to air in a room at 1 atm (Fig. 1). CO2 concentration in the room-air is 0.3 volume-% and is

kept constant by ventilation with CO2-free air. Assume the temperature in the capillary identical to that

of the dry ice.

a) Calculate the total CO2 molar flux, when the temperature of the dry ice is -86 oC.

b) By running the instrument, the temperature of dry ice and the capillary increased to -80 oC. How

much larger/smaller compared to the question (a) is the total CO2 molar flux?

c) At -80 oC how much (in percent) is the contribution of convection to the total molar flux:

i. at the surface of the dry ice and ii. at the top of the capillary?

d) To keep the CO2 concentration at 0.3 volume-%, what is the required flow rate of CO2-free air

ventilation at -80 oC?

Vapour pressure of CO2 at -86°C and -80°C is 53.3 and 90.6 kPa, respectively.

Diffusion coefficient of CO2 at -86°C and -80°C is 0.076 and 0.08 cm2/s, respectively.

Molar mass of air: 29 g/mol.

Figure 1. Schematic of the dry ice (CO2) at the bottom of a capillary tube.

Dry ice

instrument

7 cm

12mm

4



Ethylene (C2H4) is naturally produced by the tomato plant in order to ripen (=reifen) its fruits. For long

distance shipping tomatoes are even picked green (e.g. unripe) and later exposed to ethylene gas to

accelerate the ripening process. You would investigate the ethylene uptake of the tomato by exposing

it to a constant flow of air, which contains 100 ppm ethylene. The experiment is carried out in a tube

with a diameter 0.15 m. The shape of the tomato can be approximated as a sphere with diameter 7 cm.

Assume that ethylene does not diffuse through the tube walls.

a) Determine the diffusion coefficient of ethylene in air.

b) Calculate the mass transfer coefficient for ethylene absorption to the tomato from air.

c) Assume that the concentration of ethylene at the air-tomato interface is 0 ppm. What is its

ethylene absorption rate [mol/s]?

d) By measuring the tube outlet composition you detect 60 ppm of ethylene. Is this in

agreement with the concentration assumption made in (c)? If not, what is the real

interfacial concentration of ethylene?

Additional data:

Temperature, T: 20°C

Pressure, p: 1 atm

Density air, ρair: 1.293 kg/m3

Molecular weight air, Mair: 29 g/mol

Molecular weight C2H4, Meth: 28 g/mol

Dynamic viscosity of air, ηair: 1.82 ∙ 10-5 Pa s

C2H4 inlet concentration, ceth,in: 100 ppm

Gas flow through tube, Vgas: 1 L/min

Figure 2. Schematic of ripening process acceleration. Tomato are exposed to air - C2H2 stream inside

a tube.

5



A company designs a tubular reactor for the following gas phase reaction:

2𝐴 + 𝐵 → 3𝐶

The reaction is irreversible and of second order with respect to species A. Species B is in excess (im

Überfluss) in the reactor.

Process parameters:

Total gas feed into the reactor at 450 °C: 3 m3/min

Concentration of A in feed-gas: cA,0 = 15 vol%

Concentration of A in off-gas: cA,L = 0.2 vol%

Gas feed and reactor are at 450 °C and atmospheric pressure.

Reactor diameter: 0.6 m

Assume constant concentrations in radial direction.

a) Draw a detailed sketch of the problem and include the concentration profile for A and any

important information.

b) Develop the differential equation describing the concentration of species A in axial direction.

Assume steady-state.

c) Calculate the necessary reactor length to meet the requirements (0.2 vol% of A at the reactor

outlet). Assume reaction-limited mass transfer. Estimate the reaction rate constant k with the

following Arrhenius equation

k = 1.48 ∙ 105 1

𝑠∙ exp (−

6250 K

T)

d) In order to save energy costs, the customer proposes to pre-heat the feed-gas with a heat-exchanger

that is fed with the waste heat of the exhaust gas of a Brayton cycle. The new feed-gas temperature

is 395 °C. Can the reactor still be operated in the new configuration, assuming a maximum

acceptable off-gas concentration for species A of 0.3 vol% ?

6


Additional Data for all Problems:

Gas constant: R = 8.314 J/(mol·K)

1 atm=1.013 ∙105 Pa

erf 𝑥 =2

√𝜋∫ 𝑒−𝑡2

𝑑𝑡𝑥

0

Error Function

x erf(x) x erf(x) x erf(x) x erf(x) x erf(x) x erf(x)

0.00 0.00000 0.52 0.53790 1.04 0.85865 1.56 0.97263 2.08 0.99673 2.60 0.99976

0.02 0.02256 0.54 0.55494 1.06 0.86614 1.58 0.97455 2.10 0.99702 2.62 0.99979

0.04 0.04511 0.56 0.57162 1.08 0.87333 1.60 0.97635 2.12 0.99728 2.64 0.99981

0.06 0.06762 0.58 0.58792 1.10 0.88021 1.62 0.97804 2.14 0.99753 2.66 0.99983

0.08 0.09008 0.60 0.60386 1.12 0.88679 1.64 0.97962 2.16 0.99775 2.68 0.99985

0.10 0.11246 0.62 0.61941 1.14 0.89308 1.66 0.98110 2.18 0.99795 2.70 0.99987

0.12 0.13476 0.64 0.63459 1.16 0.89910 1.68 0.98249 2.20 0.99814 2.72 0.99988

0.14 0.15695 0.66 0.64938 1.18 0.90484 1.70 0.98379 2.22 0.99831 2.74 0.99989

0.16 0.17901 0.68 0.66378 1.20 0.91031 1.72 0.98500 2.24 0.99846 2.76 0.99991

0.18 0.20094 0.70 0.67780 1.22 0.91553 1.74 0.98613 2.26 0.99861 2.78 0.99992

0.20 0.22270 0.72 0.69143 1.24 0.92051 1.76 0.98719 2.28 0.99874 2.80 0.99992

0.22 0.24430 0.74 0.70468 1.26 0.92524 1.78 0.98817 2.30 0.99886 2.82 0.99993

0.24 0.26570 0.76 0.71754 1.28 0.92973 1.80 0.98909 2.32 0.99897 2.84 0.99994

0.26 0.28690 0.78 0.73001 1.30 0.93401 1.82 0.98994 2.34 0.99906 2.86 0.99995

0.28 0.30788 0.80 0.74210 1.32 0.93807 1.84 0.99074 2.36 0.99915 2.88 0.99995

0.30 0.32863 0.82 0.75381 1.34 0.94191 1.86 0.99147 2.38 0.99924 2.90 0.99996

0.32 0.34913 0.84 0.76514 1.36 0.94556 1.88 0.99216 2.40 0.99931 2.92 0.99996

0.34 0.36936 0.86 0.77610 1.38 0.94902 1.90 0.99279 2.42 0.99938 2.94 0.99997

0.36 0.38933 0.88 0.78669 1.40 0.95229 1.92 0.99338 2.44 0.99944 2.96 0.99997

0.38 0.40901 0.90 0.79691 1.42 0.95538 1.94 0.99392 2.46 0.99950 2.98 0.99997

0.40 0.42839 0.92 0.80677 1.44 0.95830 1.96 0.99443 2.48 0.99955 3.00 0.99998

0.42 0.44747 0.94 0.81627 1.46 0.96105 1.98 0.99489 2.50 0.99959 3.02 0.99998

0.44 0.46623 0.96 0.82542 1.48 0.96365 2.00 0.99532 2.52 0.99963 3.04 0.99998

0.46 0.48466 0.98 0.83423 1.50 0.96611 2.02 0.99572 2.54 0.99967 3.06 0.99998

0.48 0.50275 1.00 0.84270 1.52 0.96841 2.04 0.99609 2.56 0.99971 3.08 0.99999

0.50 0.52050 1.02 0.85084 1.54 0.97059 2.06 0.99642 2.58 0.99974 3.10 0.99999

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.00 0.50 1.00 1.50 2.00 2.50 3.00

x

erf

x

2

0

2x

terf x e dt

7


Solution 1

a) via Fourier number:

1

𝐹𝑜=

𝑙2

𝐷 ∙ 𝑡>

(0.03 𝑐𝑚) 2

1.92 ∙ 10−5 𝑐𝑚2

𝑠 ∙ 3 𝑠 = 𝟏𝟓. 𝟔𝟑 ≫ 𝟏

This problem therefore has to be treated with time dependence, by applying equations

derived for the semi-infinite slab problem.

b)

b) Equation from Cussler (Semi-infinite slab; diffusion across the interface x = 0):

𝑗(𝑡) = (𝑐0 − 𝑐∞)√𝐷

𝜋 ∙ 𝑡

Integration over time t and with c∞ = 0 gives:

∆𝑗 = ∫ 𝑗(𝑡)𝑑𝑡𝑡𝑚

0

= ∫ (𝑐0√𝐷

𝜋 ∙ 𝑡 ) 𝑑𝑡

𝑡𝑚

0

= 2 𝑐0√𝐷 𝑡𝑚

𝜋=

= 2 ∙ 1.2 ∙ 10−6 𝑚𝑜𝑙

𝑐𝑚3 ∙

√1.92 ∙ 10−5 𝑐𝑚2

𝑠 ∙ 3 𝑠

𝜋= 1.03 ∙ 10−8

𝑚𝑜𝑙

𝑐𝑚2

∆𝐽 = ∆𝑗 ∙ 𝐴 = 1.33 ∙ 10−8 𝑚𝑜𝑙

𝑐𝑚2 ∙ 400 𝑐𝑚2 = 𝟒. 𝟏𝟏 ∙ 𝟏𝟎−𝟔 𝒎𝒐𝒍

8


d) given: MWCO2, oral CO2 respiration rate Noral = 0.02 g / min

wanted: percentage x

𝑛𝑜𝑟𝑎𝑙 = 𝑁𝑜𝑟𝑎𝑙

60/3 ∙ 𝑀𝑊𝐶𝑂2

= 0.02

𝑔𝑚𝑖𝑛

20 ∙ 44.01 𝑔

𝑚𝑜𝑙

= 2.27 ∙ 10−5 𝑚𝑜𝑙

𝑥 = ∆𝐽

∆𝐽 + 𝑛𝑜𝑟𝑎𝑙=

4.11 ∙ 10−6 𝑚𝑜𝑙

4.11 ∙ 10−6 𝑚𝑜𝑙 2.27 ∙ 10−5 𝑚𝑜𝑙= 0.1532 = 𝟏𝟓. 𝟑𝟐 %

e)

1

𝐹𝑜=

𝑙2

𝐷 ∙ 𝑡>

(0.002 𝑐𝑚) 2

1.92 ∙ 10−5 𝑐𝑚2

𝑠 ∙ 3 𝑠 = 0.07 ≪ 1

This problem therefore may be be treated without time dependence, by applying equations

derived for the thin film problem.

𝑗 = 𝐷

𝑙 ∙ (𝑐1 − 𝑐2) =

1.92 ∙ 10−5 𝑐𝑚2

𝑠0.002 𝑐𝑚

∙ (1.2 ∙ 10−6 𝑚𝑜𝑙

𝑐𝑚3− 0) = 𝟏. 𝟏𝟓 ∙ 𝟏𝟎−𝟖

𝒎𝒐𝒍

𝒄𝒎𝟐

9


Solution 2

(a) The total flux of CO2 in the capillary at the surface of dry ice is given by (Cussler’s book,

3rd ed., Equation 3.3-11):

11

10

1ln

1

LyDcn

l y

1[1] (1)

where

51.013 1065.2

8.314 187

totp Pac

JRTK

mol K

mol/m3 (2)

The CO2 concentration at the top of the capillary is 0.3%, so the mole fraction at 7l L cm

is:

1 1( ) 0.003Ly l L y (3)

At the surface of dry ice, ( 0l cm ), the mole fraction of CO2 is:

1 11 1 10 1

53.3kPa( ) 0.527

101.3kPa

Ideal gas lawsat satsat

tot

c py l L y y

c p (4)

Therefore:

25

3

1

0.076 6.52 101 0.003

ln7 1 0.527

cm mol

s cmncm

7

25.28 10

mol

cm s

(5)

(b) At -80 oC, the mole fraction of CO2 at the surface of dry ice is

1 11 10 1

90.6kPa( 0) 0.896

101.3kPa

Ideal gas lawsat satsat

tot

c py l y y

c p (6)

where

51.013 1063.1

8.314 193

totp Pac

JRTK

mol K

mol/m3 (7)

Thus,

25

3

2

0.081 6.31 101 0.003

ln7 1 0.896

cm mol

s cmncm

6

21.65 10

mol

cm s

(8)

Therefore:

6

22

712

1.65 10

3.13

5.28 10

moln cm s

moln

cm s

times (9)

(c)The total molar flux within the capillary is given by Eq. (1):

0.25 [2]

0.25 [3] 0.25 [4]

0.25 [5]

1 [6]

1 [7] 0.5 [8] For the soltuion

3 [11] Fort he solution

0.25 [9]

0.25 [10]

0.5 [12]

0.5 [14] For the soltuion

0.5 [13]

0.5 [13]

2 [14]

2 [15] For the solution

10


11

10

1ln

1

LyDcn

l y

(10)

The diffusive flux within the capillary depends on z as shown by (Cussler’s book, 3rd ed.,

Equation 3.3-12):

1 11 10

10 10

1 11 ln

1 1

z

lL Ly yDc

j yl y y

(11)

In order to calculate the contribution of convection to the total flux, we can use the general

expression for the total molar flux from Table 3.2-1 (Cussler’s book, 3rd ed.):

*111 vcjn

*

1 1

1 1

1c v j

n n (12)

.10&11Eq

*

1 1 110

1 1 10

11 1 1

1

z

lLc v j y

yn n y

(13)

Therefore, the contribution of convection at -80°C at the two positions is:

i. at 0l :

1010

1

*1 )1(1 yyn

vc

*

1

1

0.896 89.6%c v

n

at l L (=7 cm):

1

*

1 110

1 10

*

11

1

11 1

1

0.003 0.3%

L

L

c v yy

n y

c vy

n

(d)The mole flow rate of CO2 by the evaporation is

2

2 26 6

2 2

1.21.65 10 1.87 10

4CO

cm mol molN An

cm s s

(14)

Thus, to keep the CO2 concentration,

2

2

2

2

0.003

0.003

CO

Air CO

CO

Air CO

N

N N

NN N

(15)

66

4

1.87 101.87 10

0.003

6.2 10 0.833 / min

mol smol s

mol s L

1.5 [16]

1.5 [17]

1 [18]

0.5 [19]

0.25 [25]

0.5 [21]

0.5 [22]

0.25 [24]

0.5 [23]

1 [23] 0.5 [26]

0.5 [27]

2.5 [27]

11


Solution 3 (a)

The Chapman-Enskog equation can be found in Cussler Ch. 5, equation 5.1-1:

𝐷 = 1.86 ∙ 10−3 ∙𝑇3/2∙(𝑀1

−1+𝑀2−1)

1/2

𝑝∙𝜎122 ∙Ω

0.5 [1] (1)

D is the diffusion coefficient [cm2/s], T the temperature [K], M the molecular weights of species 1 and

2 [g/mol], p the pressure [atm], σ12 the averaged collision diameter of species 1 and 2 [Angstrom] and

the Lenard-Jones interaction potential [-]. The collision diameters of ethylene and air can be found

in Cussler Tab. 5.1-2:

σeth = 4.163 A 0.25 [2]

σair = 3.711 A 0.25 [3]

With these values we can compute the average collision diameter σ12:

𝜎12 = 0.5 ∙ (𝜎𝑒𝑡ℎ + 𝜎𝑎𝑖𝑟) = 0.5 ∙ (4.163 𝐴 + 3.711 𝐴) = 3.937 𝐴 (2) 0.25 [4] 0.5 [5]

With the tables in Cussler (Tab. 5.1-2 and 5.1-3) we can determine the interaction potential. First we

read out the values form Tab. 5.1-2: 𝜀𝑒𝑡ℎ

𝑘𝐵= 224.7 𝐾 0.25 [6] (3)

𝜀𝑎𝑖𝑟

𝑘𝐵= 78.6 𝐾 0.25 [7] (4)

With which we determine:

𝑇∙𝑘𝐵

𝜀12= 𝑇 ∙ √

𝑘𝐵

𝜀𝑒𝑡ℎ∙

𝑘𝐵

𝜀𝑎𝑖𝑟= (20 + 273)𝐾 ∙ √

1

224.7 𝐾∙

1

78.6 𝐾= 2.2047 (5)

0.5 [8] 0.5 [9]

By Interpolating with (5) in Tab. 5.1-3 one gets :

Ω =(2.2047−2.1)

(2.3−2.1)∙ (1.026 − 1.057) + 1.057 = 1.0408 (6)

0.5 [10] 0.5 [11]

The diffusion coefficient can be computed from equation (1):

𝐷𝑒𝑡ℎ = 1.86 ∙ 10−3 ∙(20+273𝐾)

32∙((28

𝑔

𝑚𝑜𝑙)

−1+(29

𝑔

𝑚𝑜𝑙)

−1)

12

1 𝑎𝑡𝑚∙(3.937𝐴)2∙1.0408= 𝟎. 𝟏𝟓𝟑𝟐 𝒄𝒎𝟐/𝒔 2 [12] (7)

(b)

The mass transfer correlation can be found in Cussler Tab. 8.3-2 and 8.3-3. This system described by a

fluid-solid interface 0.25 [13], so that Tab. 8.3-3 is of relevance. The “forced convection around a

solid sphere” 0.25 [14] describes the tomato in the tube around which the ethylene containing air

flows.

𝑘∙𝑑

𝐷= 2 + 0.6 ∙ (

𝑑∙v0

𝜈)

1/2∙ (

𝜈

𝐷)

1/3 0.5 [15] (8)

k is the mass transfer coefficient [cm/s], d the diameter of the tomato [cm], D the diffusion coefficient

of ethylene in air [cm2/s], v0 the flow velocity of the gas [cm/s] and ν the kinematic viscosity of the gas

[cm2/s].

v0 is computed by dividing the volume flow, Vgas, by the cross section area of the tube:

12


v0 = 𝑉𝑔𝑎𝑠

𝐴𝑡𝑢𝑏𝑒=

𝑉𝑔𝑎𝑠

𝜋∙(0.5∙𝑑𝑡𝑢𝑏𝑒)2 = 1∙1000

60𝑐𝑚3/𝑠

𝜋∙(0.5∙15𝑐𝑚)2 =16.667 𝑐𝑚3/𝑠

176.714 𝑐𝑚2 = 0.094 𝑐𝑚/𝑠 (9)

0.25 [16] 0.5 [17]

The kinematic viscosity is determined from the dynamic viscosity and the density of air:

𝜈𝑎𝑖𝑟 =𝜂𝑎𝑖𝑟

𝜌𝑎𝑖𝑟=

1.82∙10−5𝑃𝑎 𝑠

1.293 𝑘𝑔/𝑚3 = 0.1408 𝑐𝑚2/𝑠 (10)

0.25 [18] 0.5 [19]

The mass transfer coefficient can be determined with (8)-(10):

𝑘𝑒𝑡ℎ = 2 ∙𝐷𝑒𝑡ℎ

𝑑𝑡𝑜𝑚+ 0.6 ∙ (

𝑑𝑡𝑜𝑚∙v0

𝜈𝑎𝑖𝑟)

1/2∙ (

𝜈𝑎𝑖𝑟

𝐷𝑎𝑖𝑟)

1/3∙

𝐷𝑒𝑡ℎ

𝑑𝑡𝑜𝑚 0.75 [20] (11)

𝑘𝑒𝑡ℎ = 2 ∙0.1532

𝑐𝑚2

𝑠

15 𝑐𝑚+ 0.6 ∙ (

15 𝑐𝑚∙0.094cm

s

0.1408𝑐𝑚2

𝑠

)

1

2

∙ (0.1408

𝑐𝑚2

𝑠

0.1532𝑐𝑚2

𝑠

)

1

3

∙0.1532

𝑐𝑚2

𝑠

15 𝑐𝑚= 𝟎. 𝟎𝟕𝟏𝟒 𝒄𝒎/𝒔 (12)

2 [21]

(c)

The mass transfer of ethylene is described with the following equation:

𝑛𝑒𝑡ℎ = 𝐴𝑡𝑜𝑚 ∙ 𝑘𝑒𝑡ℎ ∙ (𝑐𝑒𝑡ℎ,𝑖𝑛 − 𝑐𝑒𝑡ℎ,𝑖𝑛𝑡) 0.5 [22] (13)

Here, neth is the mass transfer rate [mol/s], Atom is the surface area of the tomato [cm 2], keth the mass

transfer coefficient [cm/s], ceth,bulk the bulk gas concentration of ethylene [mol/cm3] and ceth,int the

interfacial gas concentration of ethylene [mol/cm3].

Atom is computed from its known diameter:

𝐴𝑡𝑜𝑚 = 4 ∙ 𝜋 ∙ (0.5 ∙ 𝑑𝑡𝑜𝑚)2 = 4 ∙ 𝜋 ∙ (0.5 ∙ 7 𝑐𝑚)2 = 153.983 𝑐𝑚2 (14) 0.25 [23] 0.5 [24]

The bulk concentration of ethylene, ceth,bulk, is assumed to be equivalent to the inlet concentration,

ceth,in. The unit ppm must be converted to mol/cm3 via the ideal gas law:

𝑐𝑒𝑡ℎ,𝑖𝑛 =100 𝑝𝑝𝑚

106 ∙𝑛

𝑉=

100 𝑝𝑝𝑚

106 ∙𝑝

𝑅∙𝑇=

100 𝑝𝑝𝑚

106 ∙1∙105𝑃𝑎

8.314𝐽

𝑚𝑜𝑙 𝐾∙(20+273)𝐾

∙ (10−6𝑚3/𝑐𝑚3) 0.5 [25]

𝑐𝑒𝑡ℎ,𝑖𝑛 = 4.11 ∙ 10−9 𝑚𝑜𝑙/𝑐𝑚3 0.75 [25] (15)

Therefore, neth can now be computed with (13)-(15):

𝑛𝑒𝑡ℎ = 153.983 𝑐𝑚2 ∙ 0.0714 𝑐𝑚/𝑠 ∙ (4.11 ∙ 10−9𝑚𝑜𝑙/𝑐𝑚3 − 0 𝑚𝑜𝑙/𝑐𝑚3)

𝑛𝑒𝑡ℎ = 𝟒. 𝟓𝟏𝟑 ∙ 𝟏𝟎−𝟖𝒎𝒐𝒍/𝒔 2 [26] (16)

(d)

A simple mass balance around the tube enclosure allows us to determine the amount of absorbed

ethylene:

[consumed] = [in] – [out] 0.25 [27]

Or in other words:

𝑛𝑒𝑡ℎ = 𝑛𝑒𝑡ℎ,𝑖𝑛 − 𝑛𝑒𝑡ℎ,𝑜𝑢𝑡 0.25 [28] (17)

The inlet flow concentration of ethylene:

𝑛𝑒𝑡ℎ,𝑖𝑛 = 𝑉𝑔𝑎𝑠 ∙ 𝑐𝑒𝑡ℎ,𝑖𝑛 =1∙1000

60𝑐𝑚3/𝑠 ∙ 4.11 ∙ 10−9𝑚𝑜𝑙/𝑐𝑚3 = 6.842 ∙ 10−8 𝑚𝑜𝑙/𝑠 (18)

0.25 [29] 0.5 [30]

13


When assuming ceth,int = 0 ppm, neth,out is therefore:

𝑛𝑒𝑡ℎ,𝑜𝑢𝑡 = 𝑛𝑒𝑡ℎ,𝑖𝑛 − 𝑛𝑒𝑡ℎ = (6.842 − 4.513) ∙ 10−8𝑚𝑜𝑙/𝑠 = 2.328 ∙ 10−8𝑚𝑜𝑙/𝑠 (19)

0.5 [31]

With the reverse computation form (18) we get:

𝑐𝑒𝑡ℎ,𝑜𝑢𝑡 =𝑛𝑒𝑡ℎ,𝑜𝑢𝑡

𝑉𝑔𝑎𝑠=

2.328∙10−8 𝑚𝑜𝑙/𝑠1∙1000

60𝑐𝑚3/𝑠

= 1.397 ∙ 10−9 𝑚𝑜𝑙/𝑐𝑚3 0.5 [32] (20)

Which is equivalent to 34 ppm 1 [33] when using the inverse of (15)

Since we detect a greater concentration (60 ppm), the assumed interfacial surface concentration is

incorrect 0.25 [34]. With (17) we determine the actual absorption 0.25 [35] of ethylene:

𝑛𝑒𝑡ℎ = 6.842 ∙ 10−8𝑚𝑜𝑙/𝑠 −60 𝑝𝑝𝑚

106 ∙1∙105𝑃𝑎

8.314𝐽

𝑚𝑜𝑙 𝐾∙(20+273)𝐾

∙ (10−6𝑚3/𝑐𝑚3) ∙1∙1000

60𝑐𝑚3/𝑠

𝑛𝑒𝑡ℎ = 6.842 ∙ 10−8𝑚𝑜𝑙/𝑠 − 4.105 ∙ 10−8𝑚𝑜𝑙/𝑠

𝑛𝑒𝑡ℎ = 𝟐. 𝟕𝟑𝟕 ∙ 𝟏𝟎−𝟖 𝒎𝒐𝒍/𝒔 2 [36] (21)

With (13) and the result from (21), ceth,int can be determined:

𝑐𝑒𝑡ℎ,𝑖𝑛𝑡 = (𝑐𝑒𝑡ℎ,𝑖𝑛 −𝑛𝑒𝑡ℎ

𝐴𝑡𝑜𝑚∙𝑘𝑒𝑡ℎ) = 4.105 ∙ 10−9 𝑚𝑜𝑙

𝑐𝑚3 −2.737∙10−8𝑚𝑜𝑙

𝑠

153.94 𝑐𝑚2∙0.0714𝑐𝑚

𝑠

0.25 [36]

𝑐𝑒𝑡ℎ,𝑖𝑛𝑡 = 𝟏. 𝟔𝟏𝟔 ∙ 𝟏𝟎−𝟗𝒎𝒐𝒍/𝒄𝒎𝟑 1 [37] (22)

This is equivalent to 39.4 ppm 2 [38]

14


Solution 4:

a)

b)

Mass Balance for species A:

dVcA

dt= A(nA,in − nA,out + rA∆x) 2[5-6] mass balance with hom. reaction term

dcA

dt=

(jA(x)−jA(x+∆x))+v0(cA(x)−cA(x+∆x))

∆x+ rA 1[7]

dcA

dt= −

djA

dx− v0 dcA

dx+ rA 1[8]

With Fick’s law:

dcA

dt=

d2cA

dx2 − v0 dcA

dx+ rA 1[9]

The reaction is irreversible and of second order with respect to species A:

𝑟𝐴 = −2𝑘 𝑐𝐴2 2[10-11]

With this the axial concentration of species A is described by

dcA

dt= D

d2cA

dx2 − v0 dcA

dx− 2k 𝑐𝐴

2

With the assumption of steady-state

0 = Dd2cA

dx2 − v0 dcA

dx− 2k 𝑐𝐴

2 1[12]

c)

Estimation of reaction rate constant at 450 °C:

k = 1.48 ∙ 105 1

𝑠∙ exp (−

6250 K

T) = 1.48 ∙ 105

1

𝑠∙ exp (−

6250 K

(273+450)K) = 26.06

1

𝑠 1[13]

15


with the assumption that the process is reaction limited

0 = −v0 dcA

dt− 2k 𝑐𝐴

2 2[14-15]

∫dcA

𝑐𝐴2

𝑐𝐴,𝐿

𝑐𝐴,0= − ∫

2k

v0

𝐿

0 dz 1[16]

(1

𝑐𝐴,𝐿−

1

𝑐𝐴,0) =

2k

v0 ∙ L 1[17]

Calculation of the volume average velocity

v0 =

𝐴=

𝜋𝑑2

4

=3

𝑚3

𝑚𝑖𝑛

3.14∙(0.6𝑚)2

4

= 0.177 𝑚

𝑠 2[18-19]

Finally, the reactor length should be

𝐿 =v0

2∙k(

1

𝑐𝐴,𝐿−

1

𝑐𝐴,0) =

0.177 𝑚

𝑠

2∙26.061

𝑠

∙ (1

0.002−

1

0.15) = 1.68 𝑚 2[20-21]

d)

Estimation of reaction rate constant at 395 °C:

k = 1.48 ∙ 105 1

𝑠∙ exp (−

6250 K

T) = 1.48 ∙ 105

1

𝑠∙ exp (−

6250 K

(273+395)K) = 12.79

1

𝑠 2[22-23]

Calculation of off-gas concentration of species A

𝑐𝐴,𝐿 = (2𝑘𝐿

𝑣0 +1

𝑐𝐴,0)

−1

= (2∙12.79

1

𝑠∙1.68𝑚

0.177 𝑚

𝑠

+1

0.15)

−1

= 0.004 = 0.4 𝑣𝑜𝑙% 2[24-25]

The outlet concentration of species A would be above the acceptable upper limit of 0.3 vol%!

mass transfer (stoffaustausch) · frogs are known for cutaneous respiration (gas exchange through...

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