examination (summer) hs 2018 · over acetone by exploiting their differing diffusivity. both...

1

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

Mass Transfer

Examination (Summer) HS 2018

Name:______________________________________________

Legi-Nr.:_____________________________________________

Which edition of “Diffusion” by E. L. Cussler is used? 2nd 3rd none

Exam Duration: 120 minutes

The following materials are not permitted on your table and have to be deposited in the front or the back of the examination room:

bags and jackets

exercise solutions of the mass transfer lecture (also no such handwritten notes on the summary sheet, the lecture script or in the textbook “Diffusion”)

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

Do not use red or green pen.

if any information is missing, make an assumption and justify it

2


Problem 1 (25 Points)

You are drying a polymer that was dissolved in dimethyl carbonate (C3H6O3) resulting in a

thin and rectangular polymer film with length 10 cm and width 5 cm.

a) Calculate the diffusivity of dimethyl carbonate in air at 25 °C, 1 atm.

(5 Points)

b) Assuming an average surface renewable time of 10 s, how much time is needed to

completely dry the film? Initially the film contains 10 g of C3H6O3. The lab is well

ventilated so that the C3H6O3 concentration in the air sufficiently far away from the

film stays zero at all times.

(12 Points)

If such an experiment is conducted in a flow hood (see sketch below) then:

c) Choose an appropriate mass transfer correlation and calculate the mass transfer

coefficient.

(5 Points)

d) Calculate again how long it takes to completely dry the film? Assume that the

dimethyl carbonate concentration at the interface does not change.

(3 Points)

Additional data, valid for T = 25 °C, p = 1 atm:

Vapour pressure of C3H6O3: 7570 Pa

Kinematic Viscosity of air: 0.156 cm2 s-1

Molecular Weight of C3H6O3: 90.08 g mol-1

Molecular Weight of air: 28.97 g mol-1

10 cm

Air flow

3


SOLUTION a) We use the Fuller, Schettler and Giddings equation to calculate the diffusivity.

𝐷 = 10−3𝑇1.75(1 𝑀𝑊1⁄ + 1 𝑀𝑊2)⁄

12

𝑝 [(∑ 𝑉𝑖1𝑖 )13 + (∑ 𝑉𝑖2𝑖 )

13]2

The atomic diffusion volumes are calculate with Table 5.1-4 (Cussler, 2009, 3rd edition). For C3H6O3:

∑𝑉𝑖1 = 3𝑉c + 6𝑉𝐻 + 3𝑉𝑂

= [3 ∙ 16.5 + 6 ∙ 1.98 + 3 ∙ 5.48] Å3

= 77.82 Å3 For air:

∑𝑉𝑖2 = 20.1 Å3

𝐷 = 10−32981.75(1 28.97 g mol−1⁄ + 1 90.08 g mol−1)⁄

12

1𝑎𝑡𝑚 [(77.82 Å3)13 + (20.1 Å3)

13]

2

= 0.0935 𝑐𝑚2 𝑠−1

b) We write a mass balance for the C3H6O3 evaporation using a mass transfer coefficient. The film loses mass, therefore the flow must be negative.

𝑑𝑚1

𝑑𝑡= −A k MW1 (𝑐1,𝑣𝑎𝑝 − 𝑐1∞)

We write the integral with the boundary conditions, 𝑚1(𝑡 = 0) = 𝑚10 = 10 𝑔 and 𝑚1(𝑡 = 𝑡𝑖𝑚𝑒) = 0 𝑔. We also know that 𝑐1∞ = 0 because the room is well ventilated

∫ 𝑑𝑚1

0

𝑚10

= ∫ −𝐴 𝑘 MW1

𝑡𝑖𝑚𝑒

0

𝑐1,𝑣𝑎𝑝𝑑𝑡

𝑚10 = 𝐴 𝑘 MW1𝑐1,𝑣𝑎𝑝 𝑡𝑖𝑚𝑒

Thus, we can solve for “time”. We calculate cvap with the vapour pressure and k using the surface-renewal theory.

𝑐1,𝑣𝑎𝑝 =𝑝1,𝑣𝑎𝑝

𝑅 𝑇

7570 𝑃𝑎

8.315 J mol−1 K−1 298 𝐾= 3.06 𝑚𝑜𝑙 m−3 = 3.06 ∙ 10−6 𝑚𝑜𝑙 𝑐𝑚−3

𝑘 = √𝐷

𝜏= √

0.0935 𝑐𝑚2 𝑠−1

10 𝑠= 0.097 𝑐𝑚 𝑠−1

4


𝑡𝑖𝑚𝑒 =𝑚10

𝐴 𝑘 MW1𝑐1𝑣𝑎𝑝=

10 𝑔

50 𝑐𝑚2 0.097 𝑐𝑚 𝑠−1 90.08 g mol−13.06 ∙ 10−6 𝑚𝑜𝑙 𝑐𝑚−3

= 7.48 ∙ 103 𝑠 = 2.08 ℎ

c) We can use the mass transfer coefficient for a flow over a flat plate

First we need to check the flow regime:

𝑅𝑒 = 𝑣0𝐿

𝑣=

10 𝑐𝑚 𝑠−1 10 𝑐𝑚

0.156 𝑐𝑚2 𝑠−1= 641 so the flow over the plate is laminar.

𝑘 = 𝐷

𝐿0.646 (

𝐿𝑣0

𝑣)

13

(𝑣

𝐷)

13

=0.0935 𝑐𝑚2 𝑠−1

10 𝑐𝑚0.646(

10 𝑐𝑚 10 𝑐𝑚 𝑠−1

0.156 𝑐𝑚2 𝑠−1)

12

(0.156 𝑐𝑚2 𝑠−1

0.0935 𝑐𝑚2 𝑠−1)

13

= 0.195 𝑐𝑚 𝑠−1

d) We use the mass transfer coefficient from c) and the formula from b) to calculate the time.

𝑡𝑖𝑚𝑒 =𝑚10

𝐴 𝑘 MW1𝑐1,𝑣𝑎𝑝 =

10 𝑔

50 𝑐𝑚2 0.181 𝑐𝑚 𝑠−1 90.08 g mol−1 3.06 ∙ 10−6 𝑚𝑜𝑙 𝑐𝑚−3

= 3720.9 𝑠 = 1.03 ℎ𝑜𝑢𝑟𝑠

5


Problem 2 (25 points)

A heated tube 5 cm long with 0.6 cm inner diameter is used to selectively remove hydrogen over acetone by exploiting their differing diffusivity. Both acetone and hydrogen are oxidized instantly and irreversibly upon contact with the tube wall. Gases flow at a concentration of 1 ppm (25 °C, 1 atm) at a rate of 1 L/min through the tube at 300 °C in air.

a) Calculate the acetone and hydrogen diffusion coefficients at the given conditions.

(9 points)

b) What are the corresponding mass transfer coefficients for acetone and hydrogen

averaged over the entire tube length?

(6 points)

c) What is the concentration of the two gases at the exit of the filter? Clearly state the

boundary conditions and solve the differential equation.

(9 points)

d) The acetone selectivity is defined as the concentration of acetone at the tube outlet

divided by that of hydrogen. What is the selectivity for the given configuration?

(1 points)

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SOLUTION a) (9 points)

Diffusion coefficient of Hydrogen is given at 9 ℃ in Cussler (Table 5.1-1):

𝐷1(282.0 𝐾) = 0.710𝑐𝑚2

𝑠

To calculate the diffusion coefficient at the desired temperature the following formula is used:

𝐷(𝑇1) = 𝐷(𝑇2) (𝑇1𝑇2)

32

𝐷1(573.15 𝐾) = 𝐷(282.0 𝐾) ∙ (573.15 𝐾

282.0 𝐾)

32= 0.710

𝑐𝑚

𝑠∙ (573.15 𝐾

282.0 𝐾)

32= 2.06

𝑐𝑚2

𝑠

For acetone the diffusion coefficient can be calculated from the Chapman-Enskog Theory:

𝐷2 =1.86 ∙ 10−3𝑇

32 (

1𝑀𝑎𝑐𝑒

+1𝑀𝑎𝑖𝑟

)

12

𝑝𝜎122 Ω

Where from Cussler Table 5.1-2:

𝜎12 =1

2(𝜎𝑎𝑐𝑒 + 𝜎𝑎𝑖𝑟) =

1

2(4.600 Å + 3.711 Å) = 4.156 Å

and

𝜀12𝑘𝐵= √

𝜀𝑎𝑐𝑒𝑘𝑏

∙𝜀𝑎𝑖𝑟𝑘𝑏

= √560.2 ∙ 78.6 = 209.84

𝑇𝑘𝐵𝜀12

=573.15

209.84= 2.73

Giving from Table 5.1-3:

Ω = 0.9770 +2.73 − 2.7

2.9 − 2.7∙ (0.9576 − 0.9770) = 0.9741

𝐷2 =1.86 ∙ 10−3(573.15)

32 (

158.08

+1

28.97)

12

1 𝑎𝑡𝑚 ∙ (4.156)2 ∙ 0.9741= 0.345

𝑐𝑚2

𝑠

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b) (5 Points)

The velocity in the tube can be calculated from the volumetric flow rate and the filter cross section:

𝑣0 =𝑄

𝐴=

𝑄

𝑑2

4 𝜋=(1

𝐿𝑚𝑖𝑛 ∙ 1000

𝑐𝑚3

𝐿 ∙160𝑚𝑖𝑛𝑠 )

(0.6 𝑐𝑚)2

4 𝜋= 58.95

𝑐𝑚

𝑠

So the 𝑅𝑒 is:

𝑅𝑒 = 𝑣0𝐿

𝑣= 58.95 𝑐𝑚 𝑠−1 5 𝑐𝑚

0.156 𝑐𝑚2 𝑠−1= 1889.4

The correct mass transfer correlation is “Laminar flow through circular pipe”:

𝑘𝑑

𝐷= 1.62 (

𝑑2𝑣0

𝐿𝐷)

13

For hydrogen this results in:

𝑘1 = 1.62 (𝑣0𝐷1

2

𝐿𝑑)

13

= 1.62

(

58.95

𝑐𝑚𝑠 ∙ (2.06

𝑐𝑚2

𝑠 )2

5 𝑐𝑚 ∙ 0.6 𝑐𝑚

)

13

= 7.08𝑐𝑚

𝑠

Similarly, for acetone:

𝑘2 = 1.62 (𝑣0𝐷2

2

𝐿𝑑)

13

= 1.62

(

58.95

𝑐𝑚𝑠 ∙ (0.345

𝑐𝑚2

𝑠 )2

5 𝑐𝑚 ∙ 0.6 𝑐𝑚

)

13

= 2.15𝑐𝑚

𝑠

c) (9 points)

The mass balance on a differential volume 𝐴Δ𝑧 at position z: (𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛) = (𝑓𝑙𝑜𝑤 𝑖𝑛 − 𝑓𝑙𝑜𝑤 𝑜𝑢𝑡) − (𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 𝑜𝑛 𝑓𝑖𝑙𝑡𝑒𝑟 𝑤𝑎𝑙𝑙) 0 = (𝐴𝑐𝑟𝑜𝑠𝑠𝑣

0𝑐(𝑧) − 𝐴𝑐𝑟𝑜𝑠𝑠𝑣0𝑐(𝑧 + Δ𝑧)) − 𝐴𝑚𝑎𝑛𝑡𝑙𝑒N1

The two areas are:

𝐴𝑐𝑟𝑜𝑠𝑠 =𝜋

4𝑑2

𝐴𝑚𝑎𝑛𝑡𝑙𝑒 = 𝜋𝑑Δ𝑧 And the flux is: 𝑁1 = 𝑘Δ𝑐 = 𝑘(𝑐(𝑧) − 0) = 𝑘𝑐 Combined:

0 = −𝜋

4𝑑2𝑣0Δ𝑐 − 𝜋𝑑Δ𝑧𝑘𝑐

𝑑𝑐

𝑐= −

4𝑘

𝑣0𝑑𝑑𝑧

With boundary condition:

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𝑐(𝑧 = 0) = 𝑐0 Solving the equation gives:

𝑐(𝐿) = 𝑐0 exp (−4𝑘

𝑣0𝑑𝐿)

For hydrogen at 𝑧 = 𝐿:

𝑐1(𝐿) = 1 𝑝𝑝𝑚 ∙ exp(−4 ∙7.08

𝑐𝑚𝑠

58.95𝑐𝑚𝑠 ∙ 0.6 𝑐𝑚

∙ 5 𝑐𝑚) = 0.018 𝑝𝑝𝑚 = 18 𝑝𝑝𝑏

For acetone at 𝑧 = 𝐿:

𝑐2(𝐿) = 1 𝑝𝑝𝑚 ∙ exp(−4 ∙2.15

𝑐𝑚𝑠

58.95𝑐𝑚𝑠 ∙ 0.6 𝑐𝑚

∙ 5 𝑐𝑚) = 0.296 𝑝𝑝𝑚 = 296 𝑝𝑝𝑏

d) (2 points)

The selectivity is calculated from the concentrations at the filter exit:

The selectivity is thus:

𝑆 =𝑐2(𝐿)

𝑐1(𝐿)=296 𝑝𝑝𝑏

18 𝑝𝑝𝑏= 16.44

9


Problem 3 (25 points) You want to test a new ammonia sensor and need to transport 20 L min-1 of diluted ammonia

to the sensor with minimum loss due to diffusion through the tube walls. Especially as the

laboratory ventilations generate an airstream of 1000 cm∙s-1 flowing perpendicular over the

tube. The tube is made of Teflon and has an inner diameter of 1 cm, a wall thickness of 0.1

cm, and a length of 80 cm. Assume that the ammonia is instantly removed outside of the tube

(i.e. 0 ppm).

a) Calculate the overall mass transfer coefficient for the ammonia transport from inside of a

tube to the wall, through the wall and then into the airstream. State your assumptions

clearly.

(5 Points)

b) What part represents the largest resistance to mass transfer?

(2 Points)

c) What percentage of ammonia is lost over entire length of the tube?

(14 Points)

d) What is more effective to reduce the lost ammonia, reduce the tube length or increase

the thickness of the tube wall? Reason by numbers or equations.

(4 Points)

Additional information:

Kinematic viscosity of air (0 °C): ʋ = 1.56 ∙ 10−1 𝑐𝑚2 𝑠−1

Diffusivity of ammonia in Teflon: 𝐷𝑇𝑒𝑓𝑙𝑜𝑛 = 1.64 ∙ 10−3 𝑐𝑚2 𝑠−1

Diffusivity of ammonia in air at 0 °C: 𝐷𝑎𝑖𝑟 = 0.2 𝑐𝑚2 𝑠−1

10


SOLUTION a)

Assumption: Because rinside >> l -> Membrane mass transfer coefficient

𝑘𝑤𝑎𝑙𝑙 =𝐷𝑤𝑎𝑙𝑙

𝑙=1.64∗10−3𝑐𝑚2𝑠−1

0.1 𝑐𝑚= 1.64 ∙ 10−2 𝑐𝑚/𝑠

𝑘𝑖𝑛𝑠𝑖𝑑𝑒 = 0.026 ∙𝐷𝑎𝑖𝑟𝑑𝑖𝑛𝑠𝑖𝑑𝑒

∙ (𝑑𝑖𝑛𝑠𝑖𝑑𝑒 ∙ 𝑣𝑖𝑛𝑠𝑖𝑑𝑒

ʋ)0.8

∙ (ʋ

𝐷𝑎𝑖𝑟)

13

= 0.026 ∙0.2 𝑐𝑚2𝑠−1

1 𝑐𝑚∙ (1 𝑐𝑚 ∙ 424.41 𝑐𝑚 𝑠−1

1.56 ∙ 10−1 𝑐𝑚2 𝑠−1)

0.8

∙ (1.56 ∙ 10−1 𝑐𝑚2 𝑠−1

0.2 𝑐𝑚2𝑠−1)

13

= 2.67 𝑐𝑚/𝑠

𝑘𝑜𝑢𝑡𝑠𝑖𝑑𝑒 = 0.80 ∙𝐷𝑎𝑖𝑟𝑑𝑜𝑢𝑡𝑠𝑖𝑑𝑒

∙ (𝑑𝑜𝑢𝑡𝑠𝑖𝑑𝑒 ∙ 𝑣𝑜𝑢𝑡𝑠𝑖𝑑𝑒

ʋ)0.47

∙ (ʋ

𝐷𝑎𝑖𝑟)

13

= 0.80 ∙0.2 𝑐𝑚2𝑠−1

1.2 𝑐𝑚∙ (1.2 𝑐𝑚 ∙ 1000 𝑐𝑚 𝑠−1

1.56 ∙ 10−1 𝑐𝑚2 𝑠−1)

0.47

∙ (1.56 ∙ 10−1 𝑐𝑚2 𝑠−1

0.2 𝑐𝑚2𝑠−1)

13

= 8.23 𝑐𝑚/𝑠

𝐾𝑜𝑣𝑒𝑟𝑎𝑙𝑙 =1

1𝑘𝑖𝑛𝑠𝑖𝑑𝑒

+1

𝑘𝑤𝑎𝑙𝑙+

1𝑘𝑜𝑢𝑡𝑠𝑖𝑑𝑒

=1

10.0164 𝑐𝑚/𝑠

+1

2.67 𝑐𝑚/𝑠+

18.23 𝑐𝑚/𝑠

= 1.63 ∙ 10−2 𝑐𝑚/𝑠

b)

Mass transfer is restricted by the Diffusion through the wall as

𝑘𝑤𝑎𝑙𝑙

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𝑐(𝑧) = 𝑐𝑜𝑢𝑡𝑠𝑖𝑑𝑒 + 𝐴 ∙ 𝑒−

2𝐾𝑜𝑣𝑒𝑟𝑎𝑙𝑙𝑣𝑖𝑛𝑠𝑖𝑑𝑒∙𝑟𝑖𝑛𝑠𝑖𝑑𝑒

∙𝑧

With BC.

𝑐(𝑧 = 0) = 𝑐𝑖𝑛𝑙𝑒𝑡

𝐴 = 𝑐𝑖𝑛𝑙𝑒𝑡 − 𝑐𝑜𝑢𝑡𝑠𝑖𝑑𝑒

𝑐(𝑧) = 𝑐𝑜𝑢𝑡𝑠𝑖𝑑𝑒 + (𝑐𝑖𝑛𝑙𝑒𝑡 − 𝑐𝑜𝑢𝑡𝑠𝑖𝑑𝑒) ∙ 𝑒−


∙𝑧

With

𝑐𝑜𝑢𝑡𝑠𝑖𝑑𝑒 = 0

1 −𝑐(𝑧 = 𝐿)

𝑐(𝑧 = 0)= 1 −

𝑐𝑖𝑛𝑙𝑒𝑡 ∙ 𝑒−


∙𝐿

𝑐𝑖𝑛𝑙𝑒𝑡= 1 − 𝑒

−2𝐾𝑜𝑣𝑒𝑟𝑎𝑙𝑙

𝑣𝑖𝑛𝑠𝑖𝑑𝑒∙𝑟𝑖𝑛𝑠𝑖𝑑𝑒∙𝐿

= 1 − 𝑒−

2∙1.63∙10−2 𝑐𝑚𝑠

4.24∙102 𝑐𝑚𝑠∙0.5 𝑐𝑚

∙80 𝑐𝑚

= 0.012 = 1.2%

Thus ammonia can be efficiently transferred without significant loss.

d) It does not matter, in the case where the mass transfer is dominated by diffusion through the wall reducing the length L or increasing the wall thickness l is equivalent.

𝐾𝑜𝑣𝑒𝑟𝑎𝑙𝑙 ≈ 𝑘𝑤𝑎𝑙𝑙 =𝐷𝑤𝑎𝑙𝑙𝑙

1 − 𝑒−

2∙𝐷𝑤𝑎𝑙𝑙𝑙

∙𝐿

𝑣𝑖𝑛𝑠𝑖𝑑𝑒∙𝑟𝑖𝑛𝑠𝑖𝑑𝑒 ≈ 1 − 𝑒−𝐴∙𝐿

𝑙 with A being a constant.

12


Problem 4 (25 points)

Titanium nitride nanoparticles (TiN, diameter = 30 nm) strongly absorb light and can be used

for solar energy harvesting. However, they may oxidize to TiO2 by reaction with oxygen in air

that deteriorates their performance.

a) Make a detailed sketch of the oxidation process, include all diffusion and reaction steps

of educts and products.

(5 Points)

b) At room temperature (25 °C), oxidation is very slow. Calculate the time needed for the

pure TiN core to shrink to 10 nm when TiN particles are exposed to pure oxygen.

(10 Points)

c) In solar thermophotovoltaic applications, operating temperature is often around 1000

°C leading to very fast reactions. Determine now the time required for such TiN cores

to be reduced to 10 nm in diameter. Would the particles be suitable for this

application? Also, give two options how the lifetime of these particles could be

increased.

(10 Points)

The reaction is first order for all species.

p= 1 bar

atmosphere: air

DO2 in TiO2=4*10-12 cm2/s (@ 1000 °C)

DN2 in TiO2=8*10-12 cm2/s

Densities: TiN = 5.2 g/cm3, TiO2 = 4.2 g/cm3

Molecular weight: MWTiN=62 g/mol, MWTiO2=79 g/mol

10 * 13610 exp( )*

reaction

m L kJ

mol s RT

13


SOLUTION a) (5 points)

b) This is a “Shrinking-Core” problem, as described in the Cussler on page 466, Case C. The rate

limiting step is the reaction of TiN to TiO2.

We define species 1 = O2, and species 2 = TiN, species 3 = TiO2, species 4 = N2.

Mass balance:

3 2

2 2

4( ) 43

dR c R r

dt (1)

The reaction rate r2 can be written as:

2 1 2reactionr c c (2)

Combining equations (1) and (2) and after some rearranging:

1reaction

dRc

dt (3)

Integrating eq (3) gives:

0 1( ) reactionR t R c t (4)

Rearrange for t

0

1

final

reaction

R Rt

c

(5)

The concentration c1 can be calculated using the ideal gas law: 5

1

100.0403

8.314 (273 25)

n p Pa molc

JV RT LK

molK

(6)

14


And the reaction rate constant using the Arrhenius function

10 14

136 '000* *

10 *exp( ) 1.45*10* *

8.314 *(273 25)*

reaction

J

m L m LmolJmol s mol s

Kmol K

(7)

Plugging in all the numbers in (5) gives:

9

0 7

141

(15 5) *10

1.71*10 4755*

0.0403 1.45*10*

final

reaction

mnmR R nmt s hours

mol m Lc

L mol s

(8)

c) As the temperature is increased, the reaction occurs very fast. Thus, now the diffusion of

O2 through the “ash” (in this case TiO2) to the unreacted surface of the TiN core is the rate

limiting step (Cussler p. 466, Case E).

The mass transfer coefficient through the TiO2 layer, which changes in thickness over time,

can be written as:

2 2

0

O inTiODk

R R

(9)

This gives the following mass balance:

3 2 12

0

4( ) 43

DcdR c R

dt R R

(10)

10

2

( )DcdR

R Rdt c

(11)

Integration leads to:

0.510

2

2( )

DcR R t

c (12)

The concentration c2 can be calculated with the density and molecular weight of TiN:

3

2 3 3

5.4

0.087 87097

62

TiN

TiN

g

mol molcmcgMW cm m

mol

(13)

Rearranging (12) for the time, and plugging all values:

15


9 22 3

0 e 2

2 212 41

2 3

(10 *10 ) *87097( R )

270.152

2*4*10 *10 *0.0403 *1000

nd

m molnm

R c nm mt scm m mol LDc

s cm L m

Not suitable for this application. Options to increase lifetime:

Change initial radius, i.e. start with larger particles

Introduce a diffusion barrier, such as a coating.

Limit the oxygen concentration in the environment, i.e. work under inert atmosphere

Lower the operational temperature

16


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

examination (summer) hs 2018 · over acetone by exploiting their differing diffusivity. both...

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