ID:--------------
Quiz: Chapter 6
Name: --------------
Time: 15 minutes
Course: 58:160, Fall 2012
The exam is closed book and closed notes.
A train travels through a tunnel as shown. The train and tunnel are circular in cross section.
Clearance is small, causing all air to be pushed from the front of the train and discharged from
the tunnel. The tunnel is 10 ft in diameter and is concrete, the train speed is 50 ft/s. assume the
concrete is very rough (  = 0 .0 5 ft ). Determine the change of pressure between the front and rear
of the train that is due to pipe friction effects. Sections 1, 4 represent the inlet and outlet of tunnel
and section 2, 3 show the rear and front of train, respectively.
is the loss coefficient for minor
loss duo to tunnel entrance/exit. The loss coefficients for tunnel entrance and exit are
and
, respectively. (specific weight of air
, kinematic viscosity of air
)
Note that:
1)
2)
3)
h l , m a jo r  f
h l , m in o r 

pa   Va
2
1
L V
2
D 2g
K
V
L
2
2g
2 g  za  pb   Vb
2
2 g  z b  hl
2
3
4
ID:--------------
Name: --------------
Quiz: Chapter 6
Course: 58:160, Fall 2012
Time: 15 minutes
ID:--------------
Quiz: Chapter 6
Name: --------------
Time: 15 minutes
Course: 58:160, Fall 2012
Solution:
With Energy equation from front of train to outlet of tunnel (from section 3 to 4) and using
continuity equations, we have
V 3  V 4  5 0 ft s
(0.5 point)
p3   V3
2
2 g  z3  p4   V4
p3   V3
2
2 g  0  0  V4
2g  0  f
2
L
p3   f
2 g  z 4  hl
2
D V4
2
L
D V4
2
2g
(2.5 points)
2g
From moody chart
 D  0 .0 5 1 0  0 .0 0 5
R e = V D    5 0   1 0   1 .5 8  1 0
4
  3 .2  1 0
6
(3 points)
f  F (  D , R e )  0 .0 3 0
So pressure at the front of train is
p3   f
L
D V2
2
2g 
 0 .0 7 6 4   0 .0 3   2 5 0 0
From inlet of tunnel to rear of train, we have
L

h l  h l , m a jo r  h l , m in o r   f

D

10  50
2
6 4 .4   2 2 .2 4 lb ft
V 1  V 2  5 0 ft s
2
and also the total loss is equal
2

K
L
 V2

 2g
(1 point)
Energy equation from outside entrance of tunnel to rear of train (from section 1 to 2) is as
p1   V1
2
2 g  z1  p 2   V 2
0  0  0  p2   0  0   K
p 2     1 .5  f
2 g  z 2  hl
2
L
D
L
V2
2
 f
D
V2
2
(2.5 points)
2g
2 g    1 .5  0 .0 3  2 5 0 0 1 0    5 0
p 2     3 4 9 .4    2 6 .6 9 lb ft
For pressure change we have
L
2
6 4 .4 
2
 p  p 3  p 2  2 2 .2 4    2 6 .6 9   4 8 .9 3 lb f t
2
(0.5 point)