Homework Problem Set 2 Solutions (1)

7/28/2019 Homework Problem Set 2 Solutions (1)

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homework prob lem se t 2 so lu t ion s

3.5. An Al film was deposited at a rate of ~1 m/min in vacuum at 25 C, and it was estimated that the oxygen

content of the film was 10-3. 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 thegas impingement flux, and the expression given for evaporant vapor impingement rate on p. 116.

g

a

=

NAP

2MgRT( )

1

2

NAd

Ma

= PMa

d2MgRT( )12

and we know that the ratio of the impingement rates is 10-3. This gives the following expression for the

oxygen gas pressure P:

P =0.001d2MgRT( )

12

Ma

P =0.001 2.7g /cm

3( ) 1.667106cm /s( ) 2 0.032kg /mole( ) 8.314J/mole K( )298K[ ]1

2

26.98g /mole

104cm2

m2

and assuming my units work out, and I didnt make any calculation errors,

P = 2.1104Pa

3.6. In order to deposit films of the alloy YBa2Cu3, the metals Y, Ba, and Cu are evaporated from three pointsources. The latter are situated at the corners of an equilateral triangle whose side is 20 cm. Directly abovethe centroid of the source array, and parallel to it, lies a small substrate; the deposition system geometry iseffectively a tetrahedron, each side being 20 cm long.a. If the Y source is heated to 1740 K to produce a vapor pressure of 10

-3torr, to what temperature must

the Cu source be heated to maintain film stoichiometry?b. Rather than a point source, a surface source is used to evaporate Cu. How must the Cu source

temperature be changed to ensure deposit stoichiometry?c. If the source configuration in part (a) is employed, what minimum O2 partial pressure is required to

deposit stoichiometric superconducting oxide YBa2Cu3O7 films by a reactive evaporation process?The atomic weights are Y = 89, Cu = 63.5, Ba = 137, and O = 16.

NOTE: If you find a better way to work this problem, please let us know!

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


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THIN FILMS MATERIALS and the DESIGN OF EXPERIMENTS HOMEWORK 2 SOLUTIONS, PAGE 2 OF 12

ENGR 3899

geometrical effects since the point sources are the same distance from the substrate and oriented inthe same manner. From the given stoichiometry, we know that we need a ratio of fluxes of Cu atoms toY atoms of 3:1. Lets assume that we can also arrange for a 1:1 molar ratio of Y to Cu in point sources,

i.e., XCu = XY. Lets 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

=

CuX

CuPCu

(0) MY( )1 2

YX

YPY

(0) MCu( )

1 2=

PCu

(0) MY( )1 2

PY

(0) MCu( )

1 2=

3

1

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

PCu(0) =3PY(0) MCu( )

1 2

MY( )1 2

=

3 103torr( ) 63.5g /mole( )1 2

89g /mole( )1 2

= 2.5103torr

From Figure 3-1, a temperature of about 1670 K is required to attain this vapor pressure for Cu.

b. Surface source implies that, at the same source temperature, the flux of Cu atoms will be greater thatthat observed from a point source. Thus, the temperature of the source should be decreased to attainthe proper stoichiometry.

c. It seems like there should be a way to solve this problem by treating O2 as an impurity and using Table3-2, but I cant quite get it to work out that way. So, Im going to base my solution on the desired ratioof the gas impingement fluxes and use Equations 2-8 and 2-9.

=N

AP

2MRT( )1

2

= 3.5131022 P

MT( )1

2

atoms or molecules/cm2s

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 ofone atom type (Y, Ba, or Cu) and base the necessary O2 flux on that number. Since we have all theinformation on Y, lets go with that:

Y = 3.5131022

103torr( )

(89g /mole)(1740K)( )1

2

= 8.9271016

atoms/cm2s

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 roomtemperature (298 K), we have:

O2= 3.5Y = 3.124310

17molecules/cm

2s = 3.51310

22PO2

(32g /mole)(298K)( )1

2

PO2= 8.710

4

torr

3.9 A tungsten evaporation source is rated at 1000 W and operates at 120 V. If the filament heater wire is 20cm 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 electricalpower 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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ENGR 3899

P = I2R =

V2

R=

V2

(0) T T(0)[ ]n

L Ac

T T(0)[ ]n

=

V2A

c

P(0)

L T = T(0)

V2A

c

P(0)

L

1.2 = 293K

120V( )2

40.75 10

3m( )

2

1000W

( ) 5.510

8m

( ) 0.20m

( )

1.2

T = T(0)V

2A

c

P(0)L1.2 = 293K

120V( )2

40.75 10

3m( )

2

1000W( ) 5.5 108m( ) 0.20m( )1.2

= 58,700K

So, obviously this number is nonsense, since tungsten melts around 3700 K. Can you find an error in mycalculation?

For the second part of this problem, we simply need to plug numbers into Equation 3-26 and research avalue for the emissivity of tungsten. Using a quick Google search, I found a tungsten emissivity value ofabout 0.4 at higher temperatures.

P = AsT

4T(0)

4( )

T =P

As

+ T(0)4

4 =1000W

0.4 5.6108

W m2

K4( ) (0.75 103 m)(0.20m)( )

+ 2934

4 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, is61.5 kWhr/kg. The thermal conductivity of Zr, Zr, is 30 W/Km (= 0.3 W/Kcm). 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 givethe power losses associated with evaporating 1018 atoms/cm2s in W/cm2.

First convert EZr into PZr:

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: 1987C = 2260K

o Reference Temperature, T0, is room temperature: 293K

61.5 kW hrKgZr

10

18 atomscm

2 sec( ) 3600 sec hr( ) 1.515 1025 Kg

atom

Zr

103 WkW

)( )

= 33.5 Wcm

2( )


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ENGR 3899

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

PZr=PZr,s+PZr,k+

PZr,r+

PZr,c =0.16Hs,Zr +2.07e-5Ts + 5.67e-12 (Ts

4-T0

4) +Zr(Ts-T0)/l:

1.00 + 0.04 +67.28 + 590 = 658 W/cm2

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 dont like this answer.

3.14 Given: Two small area sources 100 cm apart and 50 cm below a planar substrate. The source plane isparallel 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) =15PB(0) at 1500K.

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: a) , let =lA/h and = lB/h, so that

Assume MA =MB, so that d0A = 10d0B. For the point is question, dA/dB = 60/40, and lA+lB = 100 orlA/50+lB/50 =2 =+. Substitute d0A/d0B = 10, (2- )

2=

2and dA/dB =1.5 and do some algebra:

, the positive root of this is =0.772, so that

lB=38.6 cm and lA=61.4 cm

b) P=P0exp(-H/RT); R=1.987 cal/mol-K. PA/PB =10 at 1300K and 15 at 1500K, so

a)

&

b)

Divide a) by b) and simplify to get d = (He,B- He,A) = -2.09x10-5

cal/mol

100 cm

h=50 cm

lA lB

22

0

1

1

+

=

h

ld

d

( ) ( )( )( )22

0

22

022

,022

,0

1

1

1&

1

+

+=

+=

+=

B

A

B

AB

B

A

A

d

d

d

ddd

dd

01551015215102

=++

1.2583/

0

0

1.2583/

0

1.2583/

0 exp

exp

exp10 d

B

A

H

B

H

A

B

A

P

P

P

P

P

P===

5.2980/

0

0

5.2980/

0

5.2980/

0 expexp

exp15 d

B

A

H

B

H

A

B

A

P

P

P

P

P

P===


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ENGR 3899

3.16 Given: A source initially at a distance, h, from a substrate. During deposition the source recedes adistance h.

Task: Derive an expression for the fractional change in film thickness, d/d0, at any point along the filmsubstrate, l, as a function ofh/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:

Multiply the second term in the denominator by a judicious choice of 1 (h2/h

2) to get:

22

0

1

1

++

=

hh

ld

d

22

0

1

1

1

+

+

=

hh

hld

d


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ENGR 3899

Montgomery Chapter 3Experiments with a Single Factor: The Analysis of Variance

Solutions

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

Mixing

Technique Tensile Strength (lb/in2)

1 3129 3000 2865 28902 3200 3300 2975 31503 2800 2900 2985 30504 2600 2700 2600 2765

(a) Test the hypothesis that mixing techniques affect the strength of the cement. Use = 0.05.

Design Expert Output

Response: Tensile Strengthin lb /in^2ANOVA for Selected Factor ial Model

Analysi s o f vari ance tab le [Part ial sum of squares]Sum of Mean F

Source Squares DF Square Value Prob > FModel 4.897E+005 3 1.632E+005 12.73 0.0005 significant

A 4.897E+005 3 1.632E+005 12.73 0.0005Residual 1.539E+005 12 12825.69Lack of Fit 0.000 0Pure Error 1.539E+005 12 12825.69Cor Total 6.436E+005 15

The Model F-value of 12.73 implies the model is significant. There is onlya 0.05% chance that a "Model F-Value" this large could occur due to noise.

Treatment Means (Adjusted, If Necessary)

Estimated StandardMean Error

1-1 2971.00 56.632-2 3156.25 56.633-3 2933.75 56.634-4 2666.25 56.63

Mean Standard t for H0

Treatment Difference DF Error Coeff=0 Prob > |t|1 vs 2 -185.25 1 80.08 -2.31 0.03921 vs 3 37.25 1 80.08 0.47 0.65011 vs 4 304.75 1 80.08 3.81 0.00252 vs 3 222.50 1 80.08 2.78 0.01672 vs 4 490.00 1 80.08 6.12


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ENGR 3899

2700 2800 2900 3000 3100

Tensile Strength

Scaled t Distributi on

(3) (2)(1)(4)

Based on examination of the plot, we would conclude that 1 and 3 are the same; that 4 differs from 1 and 3 ,

that 2 differs from 1 and 3 , and that 2 and 4 are different.

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

4951748564121792

4

7128252

2

4160250

2

...LSD

).(tLSD

n

MStLSD

,.

aN,

E

==

=

=

Treatment 2 vs. Treatment 4 =3156.250 - 2666.250 =490.000 > 174.495Treatment 2 vs. Treatment 3 =3156.250 - 2933.750 =222.500 > 174.495

Treatment 2 vs. Treatment 1 =3156.250 - 2971.000 =185.250 > 174.495Treatment 1 vs. Treatment 4 =2971.000 - 2666.250 =304.750 > 174.495Treatment 1 vs. Treatment 3 =2971.000 - 2933.750 = 37.250 174.495

The Fisher LSD method is also presented in the Design-Expert computer output above. The results agree with thegraphical method for this experiment.

(d) Construct a normal probability plot of the residuals. What conclusion would you draw about the validity of thenormality assumption?

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


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ENGR 3899

Residual

Normal plot of residuals

-181.25 -96.4375 -11.625 73.1875 158

1

5

10

20

30

50

7080

90

95

99

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

There is nothing unusual about this plot.

22

Predicted

Residuals vs. Predicted

-181.25

-96.4375

-11.625

73.1875

158

2666.25 2788.75 2911.25 3033.75 3156.25

(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 eachtreatment and the 95 percent confidence interval on the treatment mean.


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ENGR 3899

Technique

One Factor Plot

1 2 3 4

2579.01

2759.26

2939.51

3119.75

3300

22

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

Dosage Observations

20g 24 28 37 30

30g 37 44 31 3540g 42 47 52 38

(a) Is there evidence to indicate that dosage level affects bioactivity? Use =0.05.

Minitab Output

One-way ANOVA: A ctivity versus Dosage

Analysis of Variance for ActivitySource DF SS MS F PDosage 2 450.7 225.3 7.04 0.014Error 9 288.3 32.0

Total 11 738.9

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 =0.0500Individual error rate =0.0209

Critical value =3.95

Intervals for (column level mean) - (row level mean)

20g 30g

30g -18.1774.177

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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ENGR 3899

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

Residual

Percent

86420-2-4-6-8

99

95

90

80

70

60

50

40

30

20

10

5

1

Normal Probability Plot of the Residuals(response is Activity)

Fitted Value

Residual

464442403836343230

8

6

4

2

0

-2

-4

-6

-8

Residuals Versus the Fitted Values

(response is Activity)

3-10 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:

Temperature Density

100 21.8 21.9 21.7 21.6 21.7125 21.7 21.4 21.5 21.4150 21.9 21.8 21.8 21.6 21.5175 21.9 21.7 21.8 21.4

(a) Does the firing temperature affect the density of the bricks? Use = 0.05.


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ENGR 3899

No, firing temperature does not affect the density of the bricks. Refer to the Design-Expert output below.

Design Expert Output

Response: DensityANOVA for Selected Factor ial Model

Analysi s o f vari ance tab le [Part ial sum of squares]Sum of Mean F

Source Squares DF Square Value Prob > FModel 0.16 3 0.052 2.02 0.1569 not significantA 0.16 3 0.052 2.02 0.1569Residual 0.36 14 0.026Lack of Fit 0.000 0Pure Error 0.36 14 0.026Cor Total 0.52 17

The "Model F-value" of 2.02 implies the model is not significant relative to the noise. There is a15.69 % chance that a "Model F-value" this large could occur due to noise.

Treatment Means (Adjusted, If Necessary)Estimated Standard

Mean Error1-100 21.74 0.0722-125 21.50 0.0803-150 21.72 0.072

4-175 21.70 0.080

Mean Standard t for H0

Treatment Difference DF Error Coeff=0 Prob > |t|1 vs 2 0.24 1 0.11 2.23 0.04251 vs 3 0.020 1 0.10 0.20 0.84651 vs 4 0.040 1 0.11 0.37 0.71562 vs 3 -0.22 1 0.11 -2.05 0.06012 vs 4 -0.20 1 0.11 -1.76 0.09963 vs 4 0.020 1 0.11 0.19 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 withFishers LSD method to decide which mean is difference.

(c) Analyze the residuals from this experiment. Are the analysis of variance assumptions satisfied? There isnothing unusual about the residual plots.

Residual

Normal plot of residuals

-0.3 -0.175 -0.05 0.075 0.2

1

5

10

2030

50

70

80

90

95

99

22

22

22

Predicted

Residuals vs. Predicted

-0.3

-0.175

-0.05

0.075

0.2

21.50 21.56 21.62 21.68 21.74


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ENGR 3899

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

21.2 21.3 21.4 21.5 21.6 21.7 21.8

Mean Density

Scaled t Distribution

(125) (175,150,100)

Homework Problem Set 2 Solutions (1)

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