工程材料科学与设计原书第2版课后习题答案4—8章复习过程

此文档仅供学习和交流

Solutions to Chapter 4

1. FIND: What material has a property that is hugely affected by a small impurity level?

SOLUTION: Electrical conductivity spans a wide range. Incorporation of a few parts per million impurities can change electrical conductivity orders of magnitude. Small cracks in brittle materials decrease their tensile strength by orders of magnitude. Small additions of impurity can change the color of gems.

COMMENTS: These are but a few examples.

2. COMPUTE: The temperature at which the vacancy concentration is one half that of

25o C.

GIVEN: C 2 = C C 25v C 35v

o o EQUATION:⎪⎪⎭

⎫ ⎝⎛RT Q - = C fv v exp where C v = vacancy concentration

Q fv = activation energy for vacancy information

R = gas constant 8.314 J/mole-K

T = absolute temperature

In the present problem C)25(C = C C);

35(C = C o v 2v o v 1v

and T 1 = 35 + 273 = 308K

T 2 = 25 + 273 = 298K

also C v(35o C) = 2C v(25o C) 此文档仅供学习和交流

此文档仅供学习和交流 Thus,

Solving for Q fv we get Q fv = 52893.5 J/mole.

Using this value of Q fv , the C v (25o C) can be calculated

The problem requires us to calculate the temperature at which the vacancy concentration is ½ C v (25o C).

½ C v (25o C) = 2.675 x 10-10

Thus

for solving T, we get: T = 288.63K or 15.63o C.

3. COMPUTE: C)80( C 3 = (T) C o v v

GIVEN: C) 80( C 4

1 = C) 25( C o v o v EQUATION:⎪⎭⎫ ⎝⎛298.R Q - C) 25( C Sv o v exp

此文档仅供学习和交流

Dividing (1) by (2) we get:

Solving for Q, we get: Q = 22033.56 J/mole

= exp(-7.511)

= 5.46 x 10-4

The problem requires computing a temperature at which C v = 3C v (80o C).

3C v (80o C) = 3 x 5.46 x 10-4 = 1.63 x 10-3

⎪⎭⎫ ⎝⎛T x 8.3122033.56- = 10 x 1.633-ex p

solving for T, we get: T = 413.05K or 140.05o C

4.

5.

FIND: Are Al and Zn completely soluble in solid solution? If Al-Zn system obeys all the Hume-Rothery rules. Then it is expected to show complete solubility.

(i)

The atomic radii of Al and Zn are 0.143nm and 0.133 nm respectively. The difference in their radii is 7.5% which is less than 15%.

(ii)

The electronegativities of Al and Zn are 1.61 an 1.65 respectively which are also very similar.

(iii)

The most common valence of Al is +3 and +2 for Zn.

(iv)

Al has an FCC structure where Zn has a HCP structure.

It appears that Al-Zn system obeys 3 out of 4 Hume-Rothery rules. In this case they are not expected to be completely soluble.