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Question 1 - Crystal structures of titanium

This assignment (the first of two) is due 5.00 pm Monday , September 10 . You may hand it in at the lecture or hand it in directly to my office before September 10.  This assignment is worth 17% of your final mark.  You must sign and submit the attached declaration sheet. Please print this question sheet and write your answers directly onto it. When appropriate, give details about how you obtained your answer.

Q1.   The metal titanium, which is useful in making alloys to resist high temperatures and extremely corrosive environments, adopts two crystal structures – close-packed hexagonal (α-phase) at room temperature and body-centred cubic (β-phase) above 882 °C. Under favourable circumstances, the α phase can dissolve up to 32 atomic % (nearly 15 wt %) of oxygen whereas the β phase can dissolve very little oxygen at the transition temperature (882 °C) and a little more at higher temperatures.

Oxygen atoms have a radius of approximately 0.7 x 10-10 m. a) Given that the “distance of closest approach” between two titanium atoms is approximately 2.92 x 10-10 m, calculate the radii of the octahedral and tetrahedral interstices in both the hexagonal and cubic forms of titanium.

  1. a) to explain why the solubility of oxygen is different in the two structures of titanium?
  2. c) OPTIONAL BONUS question – Can you explain how a metal that is capable of dissolving so much oxygen below 882 °C can simultaneously have excellent oxidation resistance?

Q2. a) In a 10 litre container, 1 mole of CO2 gas and 3 moles of N2 gas are mixed at 25 °C. Assuming they are ideal gasses, what is the partial pressure of CO2 in the container? (3 marks).

b) An engineer (trained at a different university) wants to use a brass (Copper-Zinc alloy) component at a temperature of 400 °C. She remembers something about zinc having low melting and boiling points from her studies and asks you if there will be any problems. Calculate the equilibrium vapour pressure of Zn over brass at 400 °C given that the boiling point of zinc at 1 atm. pressure is 910 °C and the latent heat of sublimation ?Hf = 130 kJ/mol. What advice would you give the engineer?

c) Ethylene glycol (C2H4(OH)2) dissolved in water provides the standard ‘anti-freeze’ coolant for watercooled engines. How many grams of ethylene glycol would need to be dissolved in 5 kg of pure water in order to depress the freezing point of water by 18 °C? (The molal freezing point depression constant for water Kf  = 1.86 K mol-1 kg and the relevant atomic masses are: C = 12g, H = 1g and O = 16g.) Note: ethylene glycol is an organic compound and does not break up or dissociate when it dissolves in water i.e. i = 1. Hint: first calculate the molality (m) of the ethylene glycol solution. (4 marks).

Q3. The fuel economy of vehicles is an important consideration as we approach the end of our fossil fuel reserves and experience growing environmental concerns about emissions. a) The molecule: iso-octane C8H18 is reasonable “representative molecule” for petrol. Liquid iso-octane can burn to produce carbon monoxide CO and water according to the following unbalanced reaction:         

  1. b) CO can oxidize further to carbon dioxide CO2 according to the reaction:

Sum the above two chemical equations in an appropriate way in order to determine the amount of heat evolved when one mole of C8H18 burns completely to form CO2. (3 marks).

c) A small-medium car in Australia currently typically uses about 9 litres of fuel per 100 km in a city environment. Given that the density of iso-octane C8H18 is 705.5 kg m-3 at 25°C and the atomic masses of carbon and hydrogen are C = 12g and H = 1g, (i) how many moles of C8H18  are there in these 9 litres of C8H18? (ii) what is the maximum amount of energy released by burning them? (iii) If the work done in moving the car around the city is on average 300 kJ/km, how efficiently is the fuel used (in % of the chemical energy liberated)?  (4 marks).

Q4. (10 marks, 2 per section). The zinc-tin (Zn-Sn) phase diagram is important in the canned goods industry as alloys from within it are used to coat the inside of food containers and food processing equipment. Answer the following questions about the Zn-Sn phase diagram shown overleaf. Express all compositions as Zn-x wt% Sn, e.g. the eutectic composition (which is explicitly shown in this particular diagram) is written as Zn-91.2 wt% Sn.  

Question 1 - Crystal structures of titanium

This assignment (the first of two) is due 5.00 pm Monday , September 10 . You may hand it in at the lecture or hand it in directly to my office before September 10.  This assignment is worth 17% of your final mark.  You must sign and submit the attached declaration sheet.  Please print this question sheet and write your answers directly onto it. When appropriate, give details about how you obtained your answer.  

Q1.   The metal titanium, which is useful in making alloys to resist high temperatures and extremely corrosive environments, adopts two crystal structures – close-packed hexagonal (α-phase) at room temperature and body-centred cubic (β-phase) above 882 °C. Under favourable circumstances, the α phase can dissolve up to 32 atomic % (nearly 15 wt %) of oxygen whereas the β phase can dissolve very little oxygen at the transition temperature (882 °C) and a little more at higher temperatures. Oxygen atoms have a radius of approximately 0.7 x 10-10 m. a) Given that the “distance of closest approach” between two titanium atoms is approximately 2.92 x 10-10 m, calculate the radii of the octahedral and tetrahedral interstices in both the hexagonal and cubic forms of titanium. (5 marks).

Sol.

As given in question,

The distance of closest approach = 2.92 x 10-10 m

For BCC structure,

Since, for BCC structure = radius =1.46x Ans

For HCP

radius = 1.1921 x 10-10 m Ans.

  1. a) to explain why the solubility of oxygen is different in the two structures of titanium? (3 marks).

Answer

The solubility of oxygen in with titanium in α phase is explained by large number of unfilled shells and the donor-acceptor properties oxygen atoms and titanium, this case explanation is also viable for other analogue of titanium such as zirconium and hafnium of α -phase situation. Due to this reason, there is wide range of interstitial solid solutions is being made. It also helps in transformation of ordering of structure and some lower oxide-suboxide is also created due to solubility of oxygen in α-phase and

  1. b) Use your answers to part Answer: When heating of titanium goes above 882oC, the intermolecular spaces are getting wider as compared to α -titanium, in this condition, there is little more spaces for oxygen occurs and results in little more solubility. But it also decreases the ductility of titanium because of slip is converted from waviness to plane slip. This causes less ductile material as compared to pure titanium.
  2. c) OPTIONAL BONUS question – Can you explain how a metal that is capable of dissolving so much oxygen below 882 °C can simultaneously have excellent oxidation resistance?   (2 marks).

Answer – This is due to reason that, titanium has passive oxide films on their surface. This property is like stainless steel, which completely depends upon oxide films for its corrosion resistance property.

Q2. a) In a 10 litre container, 1 mole of CO2 gas and 3 moles of N2 gas are mixed at 25 °C. Assuming they are ideal gasses, what is the partial pressure of CO2 in the container? (3 marks).

Answer- As given in question,

CO2 = 1 mole

Question 2 - Partial pressure of CO2 in a gas mixture

N2 = 3 moles,

V = 10 lt

T = 273oC + 25oC = 298oK

R = 8.206 x 10-2 L atm /(K.mol)

For ideal gas law we know that

PV = RT (For one mole of mixture)

P = 8.206 x 10-2 X 298/ 10 = 2.445388

Now, P(CO2) = 1 x 2.445388 = 2.445388

P(N2) = 3 * 2.445388 = 7.336164

Total pressure = P(CO2)+ P(N2) = 2.445388 + 7.336164 = 9.781552 atm Ans

b) An engineer (trained at a different university) wants to use a brass (Copper-Zinc alloy) component at a temperature of 400 °C. She remembers something about zinc having low melting and boiling points from her studies and asks you if there will be any problems. Calculate the equilibrium vapour pressure of Zn over brass at 400 °C given that the boiling point of zinc at 1 atm. pressure is 910 °C and the latent heat of sublimation ?Hf = 13 kJ/mol. What advice would you give the engineer? (3 marks).

Answer- As given in question,

T1 = 400+273 = 673oK, T2 = 910+273 = 1183o K, ?Hf = 130000, P = 1 mol K =8.314 J/mol

From Clausius Clapeyron equation,

Putting the value as per given above

P = 0.3673 atm Ans

In this condition, we can suggest that, brass component can be made at closed environment.

  1. c) Ethylene glycol (C2H4(OH)2) dissolved in water provides the standard ‘anti-freeze’ coolant for water-cooled engines. How many grams of ethylene glycol would need to be dissolved in 5 kg of pure water in order to depress the freezing point of water by 18 °C? (The molal freezing point depression constant for water Kf  = 1.86 K mol-1 kg and the relevant atomic masses are: C = 12g, H = 1g and O = 16g.) Note: ethylene glycol is an organic compound and does not break up or dissociate when it dissolves in water i.e. i = 1. Hint: first calculate the molality (m) of the ethylene glycol solution. (4 marks).

Answer: The molar mass of glycol (C2H4(OH)2) = {(12 x 2) + (1x6) +(16x2)} = 62 g/mol

Now Solvent (b) = 5 kg = 5000 gm

Now molality concentration (m) = …………….(i)

Freezing point depression

Putting the value from equation (i) and (ii)

18 = 1.86x

Therefore, 3000 g of ethylene glycol needed to dissolve 5 kg of pure water

Q3. The fuel economy of vehicles is an important consideration as we approach the end of our fossil fuel reserves and experience growing environmental concerns about emissions. a) The molecule: iso-octane C8H18 is reasonable “representative molecule” for petrol. Liquid iso-octane can burn to produce carbon monoxide CO and water according to the following unbalanced reaction:         

C8H18 (l) +  O2(g) →  CO (g)  +   H2O (l)

First, balance this equation. Next, calculate how much heat in kJ is evolved in the burning of one mole of C8H18 to CO? The enthalpies of formation are: ?Hf° (C8H18 (l)) = -208.5 kJ mol-1, ?Hf° (CO (g)) =   110.6 kJ mol-1, ?Hf° (H2O (l)) = -284.9 kJ mol-1. (3 marks).

 Answer-

The balanced equation can be given as

Question 3 - Burning of iso-octane and heat evolved

As given in question

From equation adding the above equation,

 The for one mole Ans

  1. b) CO can oxidize further to carbon dioxide CO2 according to the reaction:

                  CO(g) + ½ O2 (g)  → CO2 (g)    ?H = -285 kJ mol-1.  

Sum the above two chemical equations in an appropriate way in order to determine the amount of heat evolved when one mole of C8H18 burns completely to form CO2. (3 marks).

Answer – As per previous equation,

CO(g) + ½ O2 (g) → CO2 (g)    ?H = -285 kJ mol-1.  

c) A small-medium car in Australia currently typically uses about 9 litres of fuel per 100 km in a city environment. Given that the density of iso-octane C8H18 is 705.5 kg m-3 at 25°C and the atomic masses of carbon and hydrogen are C = 12g and H = 1g, (i) how many moles of C8H18  are there in these 9 litres of C8H18? (ii) what is the maximum amount of energy released by burning them? (iii) If the work done in moving the car around the city is on average 300 kJ/km, how efficiently is the fuel used (in % of the chemical energy liberated)?  (4 marks).

Answer- As given in question,

Density of C8H18 = 705 kg-m3

(i) I mole of C8H18 = 12x8 + 1 x 18 = 114 gm.

We know that mass = volume x density = 0.702 x 1000 = 702 gm

Since 1 litre of C8H18 contains 702 gm

Then 9 litres, = 9 *x 702 = 6318 gm

Number of moles in 6318 gm of C8H18 = 6318/114 = 55.42 moles

(ii) Energy evolved by burning 55.42 moles of C8H18 = -3240.4 x 55.42 = 179582.968 KJ Ans.

(iii) Movement of car = 100/9 km/litre

Distance travelled by consuming chemical energy = 179582.968/300 = 598.61 km

Efficiency of car w.r.t chemical energy = 100 * 100/598.61 = 16.705% Ans.

Q4. (10 marks, 2 per section). The zinc-tin (Zn-Sn) phase diagram is important in the canned goods industry as alloys from within it are used to coat the inside of food containers and food processing equipment. Answer the following questions about the Zn-Sn phase diagram shown overleaf. Express all compositions as Zn-x wt.% Sn, e.g. the eutectic composition (which is explicitly shown in this particular diagram) is written as Zn-91.2 wt% Sn.  

Begin by labelling all parts of the phase diagram with Greek characters α, β etc as done in Chapter 2 of the Text.

Answer:

The phase diagram of given Zn-SN is labelled above, since there is not significant α and β, phase, therefore only α+L and β+L and α + β phase is available. To identify the composition at different point we have assigned points from B1 to B9, their composition is as given below.

Points

Composition

B1

Zn 10 wt.% Sn

B2

Zn 20 wt.% Sn

B3

Zn 30 wt.% Sn

B4

Zn 40 wt.% Sn

B5

Zn 50 wt.% Sn

B6

Zn 60 wt.% Sn

B7

Zn 70 wt.% Sn

B8

Zn 80 wt.% Sn

B9

Zn 91.2 wt.% Sn (Given)

  1. a) An industrially important composition is Zn-48.5 wt% Sn. What is the chemical composition (in wt%) of the very first solid to appear in a liquid alloy of this overall composition that is cooled slowly from 400°C?  

Question 4 - Zinc-Tin Phase Diagram

Answer: AT temp TL 397oC and at Zn 12.4 wt.% Sn

  1. b) At what temperature does this solid begin to form in the liquid?

Answer: .AT temperature TE = 198.5 oC

  1. c) For the same composition alloy:  

(ii) What are the compositions (in wt %) of the two phases that are present at 250°C?

Answer: AT 250o C, the phases are (α+L) with Zn 85 wt.% Sn and (liquid) from Zn 85 wt.% Sn to Zn 99.9 wt.% Sn.

(iii) What are the relative fractions of the two phases that are present at 250°C?

Answer: At 250o C, the phases are (α+L) with f(α+L) is 0.85   and fl = 0.15

  1. d) For the same alloy:

 (iii) What are the compositions (in wt %) of the two phases present at 180°C?

Answer: Only one phase present i.e. i.e. Solid (α + β) phase, starting from Zn 0 wt.% Sn to Zn 100 wt.% Sn

(iv) What are the relative fractions/amounts of the two phases present at 180°C?

Answer: Since only one phases is present at 180oC and its fraction is 0 to 1.

  1. e) What is the lowest temperature a liquid Zn-Sn alloy can exist at and what is the composition of that liquid?

Answer: The lowest temperature is eutectic temperature which is TE = 198.5 oC. The composition is Zn-91.2 wt.% Sn

  1. f) OPTIONAL BONUS question – for this alloy, draw the microstructure just above and just below the eutectic temperature. Calculate the fraction of free α phase, beta phase and eutectic alpha phase in this alloy at room temperature. (5 marks).

 Fraction of Free alpha phase if given by

And free beta phase =

 Alpha in eutectic mixture = Total α – primary α = 0.904-0.088 = 0.90312

Q5. The concentration of a reactant labelled ‘A’ in a new polymerization reaction (i.e. to form a new polymer material) is observed every 14 s and the results are given in the table below

  1. a) Plot the data appropriately on the two graph templates given and use the graphs to decide whether this is a first order or a second order reaction. (5 marks).

Answer: After plotting the graph with the given data, which is as follows,

T

A

lnA

1/lnA

0

4

1.386294

0.721348

14

3.3333

1.203963

0.83059

28

2.857143

1.049822

0.952542

42

2.5

0.916291

1.091357

56

2.2222

0.798498

1.252352

70

2

0.693147

1.442695

84

1.818182

0.597837

1.672696

98

1.666667

0.510826

1.957614

112

1.538462

0.430783

2.321353

126

1.428571

0.356675

2.803676

140

1.3333

0.287657

3.476362

154

1.25

0.223144

4.48142

168

1.176471

0.162519

6.153116

182

1.111111

0.105361

9.491222

196

1.052632

0.051294

19.49557

210

1

0

#DIV/0!

The second graph which is showing time Vs 1/ln[A], is not showing any significant result, but from first graph, the order of the reaction is zero order reaction.

  1. b) What is the rate constant? (5 marks).

The rate constant provides the relation between rate of the reaction and concentration of the reactant. The rate constant is proportionality factor in the mathematical formulation of rate law.

From the above equation, y = -0.0063x + 1.2081, the term -0.0063 is the rate constant for given data

Q6. (Difficult, 10 marks) A (low carbon steel) camshaft needs to be case-hardened by the process of ‘carburization’ of the region near the surface. In order to do this, the component is embedded in carbon at a high temperature. This gives a constant surface concentration Cs of carbon of 4.6 x 10-29 atoms m-3. The carbon is allowed to diffuse into the steel for a certain time t (120 minutes) at a temperature of 975°C. The parameters that describe the diffusion coefficient (D) of carbon in Fe are the following: Do = 2.7 x 10-7 m2 sec-1 and Q = 75 kJ mol-1. Assume there is no carbon already in the component.  First, using the appropriate equation in Chapter 3, calculate the diffusion coefficient of carbon in Fe at 975°C. The ideal gas constant R is: 8.314 Jmol-1K-1 and °K = °C +273. Make sure that you properly convert any units.  

Answer: As given in question,

Do = 2.7 x 10-7 m2/sec, Q = 75 kJ/mol, t = 975+273 = 1248 KR = 8.314 J/mol/K

The diffusion constant can be calculated by given formula

Putting the appropriate value

m2/sec Ans

Next, calculate the average depth of penetration d (the average distance an average atom moves in the diffusion time t of 120 minutes), see the equation in Chapter 3 of the Text.

Answer The average depth of penetration is given as

Finally, calculate the detailed concentration depth profile C(x) of the diffusion of carbon into Fe after the diffusion time of 120 minutes. To calculate C(x), use the relevant solution to Fick’s Second Law (also called the Diffusion Equation) for the case of diffusion from a constant surface concentration Cs. This solution is:                       

                   C(x,t) = Cs [1-erf (x/(2(Dt)1/2))]

where erf is known as the error function, x is the distance into the material from the surface and t is the diffusion time. Note that the time t is fixed in this case. It’s best to present this depth profile by calculating the ratio C(x)/Cs and plotting this ratio as a function of distance x. Unless you are a bit lucky and your calculator happens to have the erf function built in, it’s probably best to use MATLAB or EXCEL (which have the function erf immediately available) for this calculation. For a suitable range for the depth x, you should choose 0 < x < 4d or thereabouts.  

Answer: As per given equation

                   C(x,t) = Cs [1-erf (x/(2(Dt)1/2))]

Cs = 4.6 x 10-29 atoms/m3, C0 = 0, x = 0 to 4*0.06213, t = 5000, 6000, 7000, 8000 sec

Now putting value in equation

Now I must make a table for different value of x and Cx which is as follows,

x

X/2*(dt)^0.5

Cx

0

0

4.6E-29

0.04

0.455251925

2.39E-29

0.08

0.910503851

9.1E-30

0.12

1.365755776

2.46E-30

0.16

1.821007701

4.61E-31

0.2

2.276259627

5.91E-32

0.24

2.731511552

5.15E-33

0.28

3.186763478

3.03E-34

0.32

3.642015403

1.19E-35

0.36

4.097267328

3.15E-37

The graph plotted against Cx Vs t is as follows.

As per Cs profile for depth x the suitable Cs value is 1.4 x 10-29 atoms/m3

Q7.  a) Label the following schematic stress-strain diagram with the yield strength (σy), ultimate tensile strength (UTS), strain to fracture (εf) and Young’s modulus (E). (2 marks).

Answer:

  1. b) From the graph, determine the numerical values of the quantities in part (a). In calculating Young’s modulus, you need to know the strain at the yield point which is 0.0012 (difficult to read from the graph). (4 marks).

Answer:- Young’s Modules =   Ans

  1. c) Force is applied to a cylinder of this material until it reaches its yield stress and held at that level. The cylinder was initially 100 mm long and 5 mm in diameter. What is its new length with the force applied and what happens when the force is relaxed? (2 marks).

Answer :  E = 200000 MPa, L = 100 mm,

Area of cross section =

Now, Stress = 0.0012 x 200000 = 240 MPa

We know that F = Stress x  area =  1.223 x 1010 Newton.

The change in length =

Then new length will be = 100+0.12 = 100.012 mm Ans

  1. d) A strip of this material is stretched by 25% in tension and the stress is then relaxed. What is the new yield strength of the material? (2 marks).

Answer: Generally yield strength is derived from graphical solution; In general 0.2% offset is taken for elastic limit, i.e. As per above said, if the material is stretched by 25%, then its yield strength. Then their new lengths will 24.8% bigger that previous length. Its yield point is also shifted to some higher end.

  1. e) OPTIONAL BONUS question – if Poisson’s ratio for the material is 0.33, what is the diameter of the stressed rod in part (c)? (2 marks).

Answer: It is clear that when there is longitudinal strain on length of cylinder, its diameter will reduce i.e. it will be under lateral strain,

From the formulae, Poisson ratio µ = Lateral strain/longitudinal strain

Putting the value of longitudinal and poison ration,

Lateral strain = 0.0012 x 0.33 = 0.00396

The new diameter will be = 5 – 0.00396*5 = 5-0.0198 = 4.9802 mm Ans

 Q8.  a) Write a brief summary of each of the strengthening mechanisms that can operate in metals. Use dot points and try to stay within the space provided. (5 marks).

Answer: There are number of strengthening mechanism is applied, but the main operation is listed below.

  • Grain Size reduction- It is well known that grain boundaries are the barrier for slip in any material. If barrier is more disoriented, the strength of the barrier is increased. In the same way if grain side is reduced the barrier will also get strength.  
  • Solid solution- A solid solution with little impurity can distorts the atom and stress is generated, On the other hand stress can produce heavy barrier to dislocation movement
  • Strain Hardening (Cold work)-If any material passes through yield strength then it comes under the cold work. This process is generally happened at room temperature and commonly the cross sectional area is being deformed. In this process, the material entangles with each other internally, and makes difficult to dislocate. This process makes overall material stronger.
  • Annealing- Annealing is well known process in metal working, in which material is heated to above the recrystallization temperature and then cooled down. The main purpose behind this process is to improve cold work property by increasing ductility and retaining most of the hardness. This also help to reduce the dislocation significantly
  • Precipitation hardening.-When supersaturated solid solution is processed in several steps, the fine particle is precipitated from coarse particle. In this process strength and hardness can be increased by precipitating fine particle
  1. b) Consider the grain size strengthening effect in high-strength steel with a = 350 MPa and b=20 MPa/mm-1/2what yield strength will an alloy with a mean grain diameter of 20x10-6m have? (3 marks).

 Answer-From Hall petch relation we know that,

(Putting the value in given equation)

 = 350MPa + 4472MPa = 4822.136 Ans.

  1. c) What will the yield strength be if the manufacturing process is changed so that the mean grain diameter of the steel produced is now 5x10-6m have? (2 marks).

 Answer- Again putting the Value in above equation

 = 350 + 8944.272 = 9294.272 Ans.

References

Baz?ant, Z, Scaling of structural strength. in , Amsterdam, Elsevier Butterworth-Heinemann, 2005.

Case, J, H Chilver, & C Ross, Strength of materials and structures. in , London, Arnold, 1999.

Dutta, S, & A Lele, Metallurgical thermodynamics kinetics and numericals. in , New Delhi, S. Chand, 2012.

Ghosh, A, Textbook of materials and metallurgical thermodynamics. in , New Delhi, Prentice-Hall of India, 2003.

Hartog, J, Strength of Materials. in , Dover Publications, 2012.

Hopkins, D, Problems in Metallurgical Thermodynamics and Kinetics. in, Elsevier Science, 1977.

Peters, A, & J Warn, Concise chemical thermodynamics. in , Boca Raton, Taylor & Francis, 2010.

Steinmann, P, & G Maugin, Mechanics of Material Forces. in , Dordrecht, Springer-Verlag New York Inc., 2006.

Atkins, P, & J De Paula, Physical chemistry. in , New York, W.H. Freeman and Co., 2010.

Barrow, G, Physical chemistry. in , New Delhi, McGraw-Hill Publishing, 2008.

Moelwyn-Hughes, E, Physical chemistry. in , Cambridge, Cambridge University Press, 2015.

Morris, J, A biologist's physical chemistry. in , London, E. Arnold, 1987.

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