Homework Problem Set 2 Solutions (1)

homework problem set 2 solution s 3.5.

An Al film film was deposited deposited at a rate of ~1 m/min m/min in vacuum at 25 ºC, ºC, and it i t was estim es timated ated that the the oxygen -3 content of the film was 10 . What was the partial pressure of oxygen in the system? For this problem, you could use equation 3-24 on p. 116 directly if you get the units right. I find, however, that I need to back up a bit to make the units work out more clearly. So, I used equation 2-8 on p. 61 for the gas impingement flux, and the expression given for evaporant vapor impingement rate on p. 116.

N A P g a

)

(

1

2 M g RT

=

2

PM a

=

 N A d ˙

(

)

1

 d 2 M g RT ˙

M a

2

-3

and we know that the ratio of the impingement rates is 10 . This gives the following expression for the oxygen oxygen gas gas pressure pressure P: P: ˙

P

P

(

)

0.001 d 2 M g RT =

2

M a

(

0.001 2.7g / cm =

1

3

)(1.667



10

6



)[

]

cm / s 2 (0.032kg / mole)(8.314 J / mole  K )298K 26.98g / mole

1 2

10 

4

cm 2

m2

and assuming my units work out, and I didn’t make any calculation errors, P

3.6.

=

2.1  10

4

Pa

In order to deposit films of the alloy Y Ba2Cu3, the metals metals Y , Ba, Ba, and Cu Cu are evaporated evaporated from three point sources. The latter are situated at the corners of an equilateral triangle whose side is 20 cm. Directly above the centroid of the source array, and parallel to it, lies a small substrate; the deposition system geometry is effectively a tetrahedron, each side being 20 cm long. -3 a. If the the Y source is heated to 1740 1740 K to produce produce a vapor vapor pressure of 10 10 torr, to what temperature must the Cu source be heated to maintain film stoichiometry? b. Rather ather than than a point source, a surface source is used us ed to to evaporate evaporate Cu. How must must the the Cu source source temperature be changed to ensure deposit stoichiometry? c. If the the source configuration in part (a) is employed, what minimum minimum O2 partial partial pressure press ure is required required to deposit stoichiometr stoichiometric ic superconducting superconducting oxide Y Ba2Cu3O7 films by a reactive evaporation process? The The atom atomic ic weigh eightts are are Y = 89, Cu =63.5 3.5, Ba Ba = 137, and and O =16. NOTE NOTE:: If you find a better way way to work t his pro blem, please please let let us know!

a. For this this problem, we we know the desired desired ratio ratio of Cu to Y based on the the compound compound stoichiometry, and we are concerned with the ratio of the fluxes of two atom types in the vapor stream above the melt. Thus, it is a problem that deals with preferential vaporization, and we can apply Equation 3-11 to solve for a desired vapor pressure of the Cu. It appears that, given the tetrahedral configuration, we can ignore all

THIN FILMS MATERIALS and the DESIGN OF EXPERIMENTS

HOMEWORK 2 SOLUTIONS, PAGE 2 OF 12

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geometrical effects since the point sources are the same distance from the substrate and oriented in the same manner. From the given stoichiometry, we know that we need a ratio of fluxes of Cu atoms to Y atoms of 3:1. Let’s assume that we can also arrange for a 1:1 molar ratio of Y to Cu in point sources, i.e., XCu = X Y . Let’s also assume that the activity coefficients of Y and Cu are equal, i.e., Cu =  Y (this may or may not be the case, but finding these values in the literature could be difficult). Cu Y

(

=

(

 Y X Y PY (0) M Cu

12

) 12 )

 Cu X CuPCu (0) M Y

=

12

( ) 12 P (0)( M ) PCu (0) M Y Y

3 =

Cu

1

Rearranging and filling in some values, and solving for the vapor pressure of Cu gives: 12

PCu (0)

=

3PY (0)( M Cu ) 12

( M Y )

(

)

12

3 10 torr (63.5 g / mole) =

3



12

(89g / mole)

=

3



2.5  10 torr

From Figure 3-1, a temperature of about 1670 K is required to attain this vapor pressure for C u. b. Surface source implies that, at the same source temperature, the flux of Cu atoms will be greater that that observed from a point source. Thus, the temperature of the source should be decreased to attain the proper stoichiometry. c. It seems like there should be a way to solve this problem by treating O2 as an impurity and using Table 3-2, but I can’t quite get it to work out that way. So, I’m going to base my solution on the desired ratio of the gas impingement fluxes and use Equations 2-8 and 2-9. 

N A P

=

(2 MRT )

=

1



3.513  10

P

22

( MT )

2

2

atoms or molecules/cm s

1 2

when P is in torr and T is in K. I think we can ignore most of the information provided in the problem and simply determine the flux of one atom type (Y, Ba, or Cu) and base the necessary O2 flux on that number. Since we have all the information on Y , let’s go with that:

Y

=

3.513 10

(10

22

3

torr

)

((89g / mole)(1740K ))

16

1

= 8.927 10

2

atoms/cm s

2

We need an O:Y ratio of 7:1, or an O2:Y ratio of 3.5:1. Assuming the oxygen gas is input at room temperature (298 K), we have: O

PO

3.9

=

2

=

2

3.5Y

8.7 10

=

4

17

3.1243 10

2

molecules/cm s = 3.513  10

22

PO2

((32 g / mole)(298K ))

1

2

torr

A tungsten evaporation source is rated at 1000 W and operates at 120 V. If the filament heater wire is 20 cm long and 0.75 mm in diameter, estimate the temperature (T) the source will reach when powered. Compare your answer with an alternative estimate of the temperature assuming all of the input electrical power dissipated is thermally radiated from the filament surface. For the first part of the problem, simply plug all your known values into Equation 3-25 and solve for T.



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P

[T

=

V

2

I R

=

]

T (0)

2

V =

R

2

[

]

 (0) T T (0)

=

L Ac

2

V Ac

n =

 T

P (0) L

=

T (0)1.2

T (0)1.2

V Ac P (0) L

P (0) L

=

293K 1.2

293K 1.2

2



2

2

3

8

2

(0.75 10 m) 4 (1000W )(5.5 10 m)(0.20m) (120V )

V Ac

=



(0.75 10 m) 4 (1000W )(5.5 10 m)(0.20m) (120V )

2

2

T

n

3

8

=

58,700K

So, obviously this number is nonsense, since tungsten melts around 3700 K. Can you find an error in my calculation? For the second part of this problem, we simply need to plug numbers into Equation 3-26 and research a value for the emissivity of tungsten. Using a quick Google search, I found a tungsten emissivity value of about 0.4 at higher temperatures. P

=

T =

 A

s

(T

4

 T (0)4 )

+

T (0)

P 4

A 

s

4

= 4

1000W

(

0.4 5.6  10

8



2

4

W m K

)(

(0.75  10



3



)

m)(0.20 m)

+

4

293



3100K

This number is a bit more reasonable!

3.10.

Given: The specific energy for e-beam evaporation in a 100mm diameter water-cooled hearth for Zr,EZr, is 61.5 kW·hr/kg. The thermal conductivity of Zr, Zr, is 30 W/K·m (= 0.3 W/K·cm). The average charge thickness,l, is 1 cm. Task:

Compare EZr with that predicted in the book.

Solution: We need to calculate the power density terms on page 124 of the text. These terms give the power losses associated with evaporating 1018 atoms/cm2·s in W/cm2. First convert EZr into PZr:

61.5  kW  hr



=

(

33.5 W

cm

25 Kg  1018 atoms  10 3 W 3600 sec 1.515  10  )    2 Kg Zr  hr kW cm  sec  atom  Zr

(

2

)

(

)

(

)

Then look up some constants: o

Heat of sublimation, Hs,Zr, @ 298K: 600.8 kJ /mol (http://www.speclab.com/elements/zirconium.htm ) =3.75e24 eV/mol =6.2276 eV/atom

o

Temperature of Source, Ts, from Table 3.3: 1987°C = 2260K

o

Reference Temperature, T0, is room temperature: 293K

)



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Emissivity, Zr, at Ts, http://www.osti.gov/energycitations/product.biblio.jsp?osti_id=5313392 I assumed linear decay in the temperature range cited: 0.455 4

4

P Zr=PZr,s+P Zr,k+P Zr,r+P Zr,c =0.16 Hs,Zr +2.07e-5·Ts + 5.67e-12· ·(Ts -T0 ) + Zr(Ts-T0)/l: 2

1.00 + 0.04 +67.28 + 590 = 658 W/cm Conductance losses dominate, as stated in the text. Theoretical value is 20x the measured value. The only way to bring these values in line with one another is to use a ridiculously low Ts. I don’t like this answer. 3.14

Given: Two small area sources 100 cm apart and 50 cm below a planar substrate. The source plane is parallel to the substrate plane. The sources are operated at 1300K and the vapor pressure of element A (in source A) is 10x that of element B in Source B. P A(0) =15P B(0) at 1500K. l A

l B

h=50 cm 100 cm

Tasks: a) Find the distance along the substrate where the film is 60 at% A – 40at% B. b) Find the difference in the heats of vaporization of A & B, d = ( He,B- He,A). Solutions:

d

a)

dA =

d0, A

d0

1

=

  l  2  1+      h    

(1+  )

2 2

& dB =

d0,B

, let =lA/h and = lB/h, so that

2

 2 2

(1+  )

dA dB

=

( ) (1+  )

d0A 1+  d0B

2 2

2 2

Assume MA =MB, so that d0A = 10d0B. For the point is question, dA/dB = 60/40, and lA+lB = 100 or 2 2 lA/50+lB/50 =2 = +. Substitute d0A/d0B = 10, (2- ) =  and dA/dB =1.5 and do some algebra: 10 

15 

2

+ 2 15 + 10  5 15 = 0 , the positive root of this is =0.772, so that lB=38.6 cm and lA=61.4 cm

b) P=P 0exp(- H/RT); R=1.987 cal/mol-K. P A/PB =10 at 1300K and 15 at 1500K, so a)

PA PB

=

10

=

 H

/ 2583.1

 H

/ 2583.1

P0A exp

P0B exp

=

P0A P0B

expd / 2583.1

& b)

 H

PA PB

=

15

=

P0A exp

 H

P0B exp

/ 2980.5 / 2980.5

=

P0A P0B

expd / 2980.5

-5

Divide a) by b) and simplify to get d = ( He,B- He,A) = -2.09x10 cal/mol



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3.16

Given: A source initially at a distance, h, from a substrate. During deposition the source recedes a distance h. Task: Derive an expression for the fractional change in film thickness, d/d0, at any point along the film substrate, l, as a function of h/h. Solution: After time t the source has receded h so that the source to substrate distance is now h+ h. The fractional thickness equation (3-22) becomes: d d0

=

1

  l  2  1+     h + h    

2

2

2

Multiply the second term in the denominator by a judicious choice of 1 (h /h ) to get: d d0

=

1 2   l     h   1+  h     1+ h    

2



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Montgo mery Chapter 3 Experim ents with a Single Factor: The Analysi s of Variance

Solutions 3-1 The tensile strength of portland cement is being studied. Four different mixing techniques can be used economically. The following data have been collected:

Mixing Technique

2

Tensile Strength (lb/in ) 3129 3000 3200 3300 2800 2900 2600 2700

1 2 3 4

2865 2975 2985 2600

2890 3150 3050 2765

(a) Test the hypothesis that mixing techniques affect the strength of the cement. Use Design Expert Output Response: Tensile Strengthin lb /in^2 ANOVA for Sel ect ed Fac to r ial Mod el An aly si s o f vari anc e tab le [ Part ial su m o f s qu ares ] Sum of Mean Source Squares DF Square Model 4.897E+005 3 1.632E+005 A

4.897E+005

3

Residual

1.539E+005

12

Lack of Fit

0.000

Pure Error

1.539E+005

12

Cor Total

6.436E+005

15

1.632E+005

F Value 12.73

Prob > F 0.0005

12.73

0.0005

= 0.05.

significant

12825.69

0 12825.69

The Model F-value of 12.73 implies the model is significant. There is only a 0.05% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adj usted, If Necessary) Estimated Standard Mean Error 1-1 2971.00 56.63 2-2 3156.25 56.63 3-3 2933.75 56.63 4-4 2666.25 56.63 Mean Treatment Difference 1 vs 2 -185.25 1 vs 3 37.25 1 vs 4 304.75 2 vs 3 222.50 2 vs 4 490.00 3 vs 4 267.50

DF 1 1 1 1 1 1

Standard Error 80.08 80.08 80.08 80.08 80.08 80.08

t for H0 Coeff=0 -2.31 0.47 3.81 2.78 6.12 3.34

Prob > |t| 0.0392 0.6501 0.0025 0.0167 <0.0001 0.0059

The F-value is 12.73 with a corresponding P-value of .0005. Mixing technique has an effect. (b) Construct a graphical display as described in Section 3-5.3 to compare the mean tensile strengths for the four

mixing techniques. What are your conclusions?

Syi .

=

MSE n

12825.7 =

4

=

56.625



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Scaled t Distributi on

(4)

(3)

2700

2800

(1)

2900

(2)

3000

3100

Tensile Strength

Based on examination of the plot, we would conclude that that

μ 2

differs from

μ 1

and

μ 3 ,

and that

μ 2 and μ 4

μ 1

and

μ 3

are the same; that μ 4 differs from

μ 1 and μ 3 ,

are different.

(c) Use the Fisher LSD method with a =0.05 to make comparisons between pairs of means.

LSD

=



LSD

=

LSD

=

2MSE

t 2

,N  a

t0.025,16 4

n 2(12825.7 ) 4

2.179 6412.85

Treatment 2 vs. Treatment 4 =3156.250 Treatment 2 vs. Treatment 3 =3156.250 Treatment 2 vs. Treatment 1 =3156.250 Treatment 1 vs. Treatment 4 =2971.000 Treatment 1 vs. Treatment 3 =2971.000 Treatment 3 vs. Treatment 4 =2933.750 -

=

174.495

2666.250 =490.000 > 174.495 2933.750 =222.500 > 174.495 2971.000 =185.250 > 174.495 2666.250 =304.750 > 174.495 2933.750 = 37.250 <174.495 2666.250 =267.500 > 174.495

The Fisher LSD method is also presented in the Design-Expert computer output above. The results agree with the graphical method for this experiment. (d) Construct a normal probability plot of the residuals. What conclusion would you draw about the validity of the

normality assumption? There is nothing unusual about the normal probability plot of residuals.



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Normal plot of residuals

99 95 90 80 70 50 30 20 10 5 1

-181.25

-96.4375

-11.625

73.1875

158

Residual

(e) Plot the residuals versus the predicted tensile strength. Comment on the plot.

There is nothing unusual about this plot. Residuals vs. Predicted 158

73.1875

-11.625

2 -96.4375

-181.25

2 66 6. 25

2 78 8. 75

2 91 1. 25

3 03 3. 75

3 15 6. 25

P redicted

(f)

Prepare a scatter plot of the results to aid the interpretation of the results of this experiment.

Design-Expert automatically generates the scatter plot. The plot below also shows the sample average for each treatment and the 95 percent confidence interval on the treatment mean.



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One Factor Plot 3300

3119.75

2939.51

2759.26

2 2579.01 1

2

3

4

Technique

3-6 A pharmaceutical manufacturer wants to investigate the bioactivity of a new drug. A completely randomized single-factor experiment was conducted with three dosage levels, and the following results were obtained.

Dosage 20g 30g 40g

24 37 42

Observations 28 37 44 31 47 52

(a) Is there evidence to indicate that dosage level affects bioactivity? Use

30 35 38 =0.05.

Minitab Output One-way ANOVA: A ctivity versus Dosage Analysis of Variance for Activity Source DF SS MS F Dosage 2 450.7 225.3 7.04 Error 9 288.3 32.0 Total 11 738.9

P 0.014

There appears to be a different in the dosages. (b) If it is appropriate to do so, make comparisons between the pairs of means. What conclusions can you draw? Because there appears to be a difference in the dosages, the comparison of means is appropriate. Minitab Output Tukey's pairwise comparisons Family error rate =0.0500 Individual error rate =0.0209 Critical value =3.95 Intervals for (column level mean) - (row level mean) 20g

30g

30g

-18.177 4.177

40g

-26.177 -19.177 -3.823 3.177

The Tukey comparison shows a difference in the means between the 20g and the 40g dosages. (c) Analyze the residuals from this experiment and comment on the model adequacy.



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There is nothing too unusual about the residual plots shown below.

Normal Probability Pl ot of the Residuals (response is Activity) 99

95 90 80

t 70 n e c 60 r e 50 P

40 30 20 10 5

1

-8

-6

-4

-2

0 Residual

2

4

6

8

42

44

46

Residuals Versus the Fitted Values

(response is Activity) 8 6 4 l a 2 u d i s e 0 R

-2 -4 -6 -8 30

32

34

36

38

40

Fitted Value

An experiment was run to determine whether four specific firing temperatures affect the density of a certain type of brick. The experiment led to the following data: 3-10

Temperature 100 125 150 175

21.8 21.7 21.9 21.9

21.9 21.4 21.8 21.7

Density 21.7 21.5 21.8 21.8

(a) Does the firing temperature affect the density of the bricks? Use

21.6 21.4 21.6 21.4 = 0.05.

21.7 21.5



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No, firing temperature does not affect the density of the bricks. Refer to the Design-E xpert output below. Design Expert Output Response: Density ANOVA fo r Selec ted Fact or ial Mod el An aly si s o f vari anc e tab le [ Part ial su m o f s qu ares ] Sum of Mean Source Squares DF Square Model 0.16 3 0.052 A

0.16

3

0.052

Residual

0.36

14

0.026

Lack of Fit

0.000

0

Pure Error

0.36

14

Cor Total

0.52

17

F Value 2.02

Prob > F 0.1569

2.02

0.1569

not significant

0.026

The "Model F-value" of 2.02 implies the model is not significant relative to the noise. There is a 15.69 % chance that a "Model F-value" this large could occur due to noise. Treatment Means (Adj usted, If Necessary) Estimated Standard Mean Error 1-100 21.74 0.072 2-125 21.50 0.080 3-150 21.72 0.072 4-175 21.70 0.080 Mean Treatment Difference 1 vs 2 0.24 1 vs 3 0.020 1 vs 4 0.040 2 vs 3 -0.22 2 vs 4 -0.20 3 vs 4 0.020

DF 1 1 1 1 1 1

Standard Error 0.11 0.10 0.11 0.11 0.11 0.11

t for H0 Coeff=0 2.23 0.20 0.37 -2.05 -1.76 0.19

Prob > |t| 0.0425 0.8465 0.7156 0.0601 0.0996 0.8552

(b) Is it appropriate to compare the means using the Fisher LSD method in this experiment?

The analysis of variance tells us that there is no difference in the treatments. There is no need to proceed with Fisher’s LSD method to decide which mean is difference. (c) Analyze the residuals from this experiment. Are the analysis of variance assumptions satisfied? There is nothing unusual about the residual plots. Normal plot of residuals

Residuals vs. P redicted 0.2

99 95

2

0.075

90 80 70 50

2

-0.05

30 20

2

10

-0.175

5 1 -0.3

-0.3

-0.175

-0.05

Residual

0.075

0.2

21.50

21.56

21.62

P redicted

21.68

21.74



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(d) Construct a graphical display of the treatments as described in Section 3-5.3. Does this graph adequately summarize the results of the analysis of variance in part (b). Yes. Scaled t Distribution

(125)

21.2

21.3

21.4

(175,150,100)

21.5

21.6

Mean Density

21.7

21.8

Homework Problem Set 2 Solutions (1)

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