This preview shows pages 1–2. Sign up to view the full content.
EMSE 201 – Introduction to Materials Science & Engineering
Fall 2010
Case Western Reserve University
1 of 4
Case School of Engineering
Solution to Homework #9
Explain
how answers were obtained. Give appropriate
units
for all numerical answers.
1)
(5 points)
The constant A in Equation 17.2 is 12
π
4
R/5
θ
D
3
, where R is the gas constant and
θ
D
is the Debye
temperature (K). Estimate
θ
D
for aluminum, given that the specific heat is 4.60 J/kgK at 15 K. (C&R problem
17.3)
Solution
(from the publisher’s solution manual; used with permission)
Determine the magnitude of
A
, as
A
=
C
v
T
3
=
c
v
A
Al
T
3
=
(4.60
J
/
mol

K
)(1
kg
/1000
g
)(26.98
g
/
mol
)
(15
K
)
3
= 3.68
×
10
5
J/molK
4
(2 pts)
As stipulated in the problem statement
A
=
12
π
4
R
5
θ
D
3
Or, solving for
D
D
=
12
4
R
5
A
⎛
⎝
⎜
⎞
⎠
⎟
1/3
=
(12)(
)
4
(8.31
J
/
mol

K
)
(5)(3.68
×
10
−
5
J
/
mol

K
4
)
⎡
⎣
⎢
⎤
⎦
⎥
1/3
=
375
K
2)
(6 points)
To what temperature must a cylindrical rod of tungsten 15.025 mm in diameter and a plate of 1025
steel having a circular hole 15.000 mm in diameter have to be heated for the rod to just fit into the hole?
Assume that the initial temperature is 25°C.
(
C&R problem 17.7)
Solution
(modified from the publisher’s solution manual; used with permission)
Solution of this problems requires the use of Equation 17.3a, which is applied to the
diameters of both the rod and hole. That is
d
f
−
d
0
d
0
=
α
l
(
T
f
−
T
0
)
Solving this expression for
d
f
yields
d
f
=
d
0
1
+
α
l
(
T
f
−
T
0
)
⎡
⎣
⎢
⎤
⎦
⎥
Now all we need do is to establish expressions for
d
f
(steel) and
d
f
(W), set them equal to one
another, and solve for
T
f
.
According to Table 17.1,
α
l
(steel) = 12.0
×
10
6
(
°
C)
1
and
α
l
(W) =
4.5
×
10
6
(
°
C)
1
.
Thus
d
f
(steel) =
d
f
(W)
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '09

Click to edit the document details