工程材料科学与设计(原书第2版)课后习题答案(4—8)

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 2

v o v 1v

and T1 = 35 + 273 = 308K T2 = 25 + 273 = 298K

also C v(35o C) = 2C v(25o C)

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

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.