Solution for assignment 1

Solution for Assignment 1

1. Which property of a material is used for Casting it into a desired shape

(a) Strength
(b) Fluidity
(c) Ductility
(d) Formability

2. The line graph that represents change of phase of matter from liquid to solid is

(a) Cooling curve
(b) Melting curve
(c) Solid curve
(d) Fusion line

3. For any transformation to take place from one phase of a material to another, free energy
change

(a) Should be positive
(b) Should be negative
(c) Should be zero
(d) Has no role

4. Which of the following statement is not true for heterogeneous nucleation?

(a) Foreign particles assessed in nucleation.
(b) Foreign particles alter the liquid solid interface energy.
(c) Amount of supercooling required to start nucleation is reduced.
(d) Probability of nucleus at all the sites in the domain is same

5. Which of the following statement is true regarding growth process?

(a) It occurs in the direction of heat transfer
(b) It occurs in the direction opposite to that of heat transfer
(c) It is followed by nucleation process
(d) It is not required in case of metals as they solidify at constant temperature

6. Constitutional supercooling is experienced in case of

(a) pure metals
(b) alloys with large freezing range
(c) alloys with eutectic composition
(d) pure metals having large melting temperature

7. Centreline feeding resistance (CFR) means ratio of time of solidification

(a) Of whole casting to that at mould walls
(b) At mould Centre line of casting to that at mould walls of casting
(c) At mould Centre line of casting to total solidification time of casting
(d) At mould wall to that at mould centerline of casting

8. If CFR value is high, it means that

(a) Material is difficult to be fed
(b) Material can be fed easily
(c) Material is of eutectic composition
(d) Material is a pure metal

9. Width of the freeze wave lines i.e. Duration of time between start and finish of solidification
lines, in case of sand mould as compared to metal mould will be

(a) More
(b) Les
(c) Equal
(d) Cannot be said

10. Pure aluminium melts at 667oC. At 690oC, free energy of molten aluminium with respect to
free energy of aluminium in solid state is

(a) Larger
(b) Smaller
(c) Equal
(d) Can't be said

11. Match the items in two columns provided below:

(A) Degree of (i) Should be negative for transformation to take

undercooling place

(B) Free energy change (ii) Portion of the cooling curve when metal is

(C) Rate of nucleation getting solidified by losing latent heat of fusion.

(D) Thermal arrest (iii) Decrease in temperature below

melting/equilibrium temperature

(iv) Determine the rate of formation of new self-

organized structure

(a) A-i, B-ii, C-iii, D-iv

(b) A-iii, B-i, C-iv, D-ii

(c) A-i, B-iv, C-iii, D-ii

(d) A-i, B-iii, C-ii, D-iv

12. With height of cylinder as two times its diameter, the ratio of solidification time for a sphere
to that of cylinder (of the same diameter) will be

(a) 36/25
(b) 25/36
(c) 64/25
(d) 25/64
Solution:

V/A for sphere = 34π(D2 )3 = D
4π(D2 )2 6

V/A for cylinder =2((π4π4DD22))×+(22πDD)2 = D
5

Solidification time for sphere / Solidification time for cylinder = (V/A)2sphere / (V/A)2cylinder

= 25/36 (Ans)

13. The cylindrical mould have diameter-to-height ratio as 1.0. Considering total time of
solidification as 2.0 min and mould constant as 1956000 sec/m2, the dimensions of the
mould will be

(a) D = H = 7.4 cm
(b) D = H = 4.7 cm
(c) D = H = 6.7 cm
(d) D = H = 13.4 cm
Solution:
D/H =1, Time of solidification = Ts =2 min

Cm = 1956000 sec/m2 = (1956000)/ (60 × 1002) =3.26 min/ cm2

V = πD2H /4 = π D3/ 4, though D/H =1

A = (2 ×πD2/4) + (πDH)

Ts = Cm (V/A)2 = 3.26 × (D2/36) = 2

D = 4.7 = H (Ans)

14. The ratio of surface energy term to volume energy term for critical radius during
solidification process is

(a) -3/2
(b) -2/3
(c) -5/6
(d) -7/12
Solution:

Surface energy term = 4 π r2 γ

Volume energy term = (4/3) π r3 Δg

Critical radius = (-2 γ)/Δg = r*

Ratio of surface energy term to volume energy term = [4 π (r*)2 γ] / [(4/3) π (r*)3 Δg]

= -3/2 (Ans)

15. Enthalpy of fusion of ice is 6.02 kJmol-1. The energy of ice water interface is 0.076 Jm-1.
Molar volume of ice can be taken as 19 cm3. The critical radius of nucleation of ice from
water at -40oC will be

(a) 33 nm
(b) 41 nm
(c) 33 Ao
(d) 41 Ao
Solution:

Δ h = (6.02 × 103) / (19 × 10-6) J m-3

γ = 0.076 J m-1

Δ T = -40 °C = -40 K

Tm = 273 K

Critical radius = (-2 γ) / (Δg) where Δg = (Δh × ΔT) / (Tm)

= 33 A0 (Ans)