ME 320 Fall 2014
Homework #2 (Due Monday, September 15, 2014) Revised September 9, 2014
1. In a rectangular brick (as is used to build houses) having vertices at (0, 0, 0) m, (0, 0, 0.20) m,
(0, 0.10, 0) m, (0, 0.10, 0.20) m, (0.07, 0, 0) m, (0.07, 0, 0.2

ME 320 Fall 2016
Homework #1 Solutions
1. The key issue here is to identify the initial and final states.
Initial state: Mass M ice of ice at 0C, 20 ounces of Gatorade at 115F.
Final state:
Mass M ice M w,liq of ice at 0C, 20 ounces of Gatorade at 4C, mas

ME 320 Fall 2014
Homework #3 Solutions
1.
dTCu
d 2TCu
d
=
k
k
0 x 0.2 mm when kCu is constant.
Cu
Cu
dx
dx
dx 2
d 2TC
d dTCu
=
k
k
0 =
0.2 mm x 2.2 mm when kC is constant.
C
C
dx
dx
dx 2
=
0
The general solutions of the differential equations are T

ME 320 Fall 2014
Homework #4 Solutions
1. The (nonhomogeneous) ODE governing the problem is
dT 2 hP
(T T ) = 0
dx 2 kAc
dT ( L)
The boundary conditions are T (0) = Tb and k
=
h [T ( L) T ]
dx
kAc
dT 2
hP(T T ) = 0
dx 2
There are many ways to solve the O

ME 320 Fall 2014
Homework #6 Solutions
1. a)
1
2
3
4
5
x1
x2
x3
x4
x5
x=0
x = 3 cm
x = 6 cm
x = 9 cm
x = 12 cm
T = 20 C
T = 40 C
where the red bar indicates the extent of the heated region. The differential equation is
d 2T
5 W m d 2T 0.2 W
d 2T 4 K
0= k

ME 320 Fall 2014
Homework #2 Solutions
1. The heat transfer rate into the block is
Q = q nin dA
A
where q is the heat flux vector, nin is the inward unit normal, and A is the surface area of the brick.
The volume is a rectangular parallelepiped, so that t

ME 320 Fall 2014
Homework #5 Solutions
1. a) There is no heat flux on either constant-z boundary, and the temperature distribution is at steady
state. Hence, there will be no variation T with z. For constant conductivity, the partial
differential equation