material science for chemical engineers

1. MATERIAL SCIENCE
COMPETITION DAY
Eng. Kareem. H. Mokhtar

2. Find the diameter of an 18 m long steel wire
that will stretch no more than .009 m when a
load of 3800 N is hung on the end of the
wire.TheYoung’s Modulus for steel wire is
200x109 N/m2 .
5 Points

3. ■ Y=
𝑆𝑡𝑟𝑒𝑠𝑠
𝑠𝑡𝑟𝑎𝑖𝑛
=
𝐹
𝐴
delta 𝐿
𝐿𝑜
=
3800∗18
𝐴∗0.009
■ A= 3.8 x 10 -5 m2 = (Pi/4) D2
Answer

4. Human bone has aYoung’s Modulus of 14.5
x108 N/m2. Under compression, it can
withstand a stress of about 160 x 106
N/m2(stress) before breaking. Assuming a
human adult thigh bone (femur) is about 0.5
m long, how much change in length the bone
withstand before breaking.
5 Points

5. Answer
■ Y=
𝑆𝑡𝑟𝑒𝑠𝑠
𝑠𝑡𝑟𝑎𝑖𝑛
=
𝐹
𝐴
𝐿
𝐿𝑜
=
160∗106
𝑑𝑒𝑙𝑡𝑎 𝐿
0.5
■ Delta L = 0.055 m

6. As shown in the binary phase diagram,
an alloy with a composition C0 = 48% is
cooled down from liquid to form solid. At
the temperature T, Cs =26%, CL = 72%,
the fraction
of the solid phase is
7 Points

7. Answer
■ Solid fraction =
72−48
72−26
= 0.52

8. A cylindrical rod has a diameter and length of
10mm is to be deformed using a tensile load of
27500N. It must NOT experience either plastic
deformation or length reduction of more than
0.02 mm, which one of these material will you
choose
10 Points

9. Answer
■ Stress=
𝐹
𝐴
=
27500
𝑝𝑖
4
0.01 2
= 3.5 x 108 = 350 Mpa
■ First criteria: do not enter plastic region  only steel and titanium alloy
■ Second criteria:
■ For steel :Y=207x109 =
𝑆𝑡𝑟𝑒𝑠𝑠
𝑠𝑡𝑟𝑎𝑖𝑛
=
3.5 𝑥 108
𝑑𝑒𝑙𝑡𝑎 𝐿
0.01
 delta L = 0.0169 mm
■ ForTi:Y=107x109 =
𝑆𝑡𝑟𝑒𝑠𝑠
𝑠𝑡𝑟𝑎𝑖𝑛
=
3.5 𝑥 108
𝑑𝑒𝑙𝑡𝑎 𝐿
0.01
 delta L = 0.0327 mm

10. A cylindrical alloy is stressed in compression,
if its original and final diameters are 20 and
20.025 mm respectively and its original and
final length are 7.5cm and 7 cm.
compressibility of this alloy is 9x 10-10 m2/N,
find the depth at which the alloy is
submerged
7 Points

11. Answer
■ B=
1
𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦
= 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑥 𝑔 𝑥 ℎ
𝑑𝑒𝑙𝑡𝑎 𝑉
𝑉𝑜
=
1000𝑥 9,8𝑥 ℎ
1.06x 10−6/ 2.36x 10−5
 h= 5041.5 m
■ Vo= 2.36x 10-5 m3
■ Vf= 2.25x 10-5 m3
■ deltaV= 1.06x 10-6 m3

12. ■ Using the Cu-Ni phase diagram
■ Compute the amount of each phase for
Cu-Ni alloy that has 40% Cu at:
– 1250 C
– 1325 C
– 1450 C
■ Compute the amount of Cu in each phase
at:
– 1250 C
– 1320 C
– 1450 C
7 Points
5 Points

13. Answer
■ At 1250 --> all the phase is solid
■ At 1450  all liquid
■ At 1325  solid =
60−56
65−56
= 0.44
– Liq = 0.66
• B) at 1250  amount of Cu in solid = 40%
• At 1325  amount of Cu in solid = 35%
• Amount of Cu in solid = 44%

14. If we have Cu-20% Ni, Cu, and Ni and we
want to produce 100Kg of alloy that has
40% Cu, find 2 different ways to produce
the alloy and pick the best one knowing
that
1 Kg of Ni = 25$
1 Kg of Cu= 20$
1 Kg of Cu-20%Ni= 23$
8 Points

15. Answer
■ To produce 100Kg of alloyCu-60% Ni
■ First method (using Cu and Ni)
■ We need 40Kg Cu and 60Kg Ni
■ Second method (using Ni- and Cu-20%Ni)
– Total amount of Ni = amount of Ni in (Cu-20%Ni) + amount of Ni
■ 60 = 0.2 x + (100-x)
■ X = 50KG
■ So we need 50Kg of alloy Cu-20% Ni and 50 kg of pure Ni
■ To pick one calculate the cost in each case and choose the lower cost

16. Winners
Mohamed Farid AhmedTarek
Ahmed Mohamed Ali Asmaa Ashraf
KareemTharwat

17. Mid term Final notes

18. Next Section Rules
■ Send your presentation and report before Saturday 28/10 midnight
■ Late submissions will lose 25% of the final mark
■ Be on time
■ The presentation is Simi-formal
■ Your report must not be copied (plagiarism will not be accepted)
■ You should use at least one book as a reference
■ You should attend others presentation, any funny comments will be corrected by
severe punishment