ID:----------------Name: ------------------
Final Exam
Course: 58:160, Fall 2010
Time: 120 minutes
The exam is closed book and closed notes. Only two formula sheets are allowed.
1. A parallel-disk viscometer consists of two circular disks of radius R separated by a small gap B (with R >> B). A
fluid of constant density ρ, whose viscosity μ is to be measured, is placed in the gap between the disks. The lower
disk at z = 0 is fixed. The upper disk (at z = B) rotates with a constant angular velocity Ω. Assume that the flow is
steady and laminar. (a) Simplify the equations of continuity and show that v  v ( z, r ) . (b) Assume that the flow is
creepy (i.e. inertial term is negligible) and v ( z, r )  rf ( z ) and then simplify the equations of motion for the flow in
the parallel-disk viscometer. (c) Obtain the tangential velocity profile after writing down appropriate boundary
conditions.
2. An air flows past an object in a pipe of 2-m diameter and exits as a free jet. The velocity and pressure upstream
are uniform at V=10 m/s and pgage=50 N/m2, respectively. At the pipe exit the velocity is non-uniform as indicated.
The shear stress along the pipe wall is negligible. (a) Determine the uniform velocity at wake. (b) Determine the
force that the air puts on the object. (air=1.23 kg/m3)
3. Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes
(=0.00026m) connected in series and a pump between them. The first pipe is 20 m long and has a 6-cm diameter,
while the second pipe is 35 m long and has a 4-cm diameter. The water level in the reservoir is 30 m above the
centerline of the pipe. The pipe entrance is sharp-edged (KL=0.5), and losses associated with the connection of the
pump are negligible. Determine (a) the required pumping head and (b) the minimum pumping power to maintain the
indicated flow rate. (water=999.1 kg/m3, water=1.138×10-3 kg/m.s)
ID:----------------Name: ------------------
Final Exam
Course: 58:160, Fall 2010
Time: 120 minutes
4. During major windstorms, high vehicles such as RVs and semis may be thrown off the road and boxcars off their
tracks, especially when they are empty and in open areas. Consider a 5000-kg semi that is 8 m long, 2 m high, and 2
m wide. The distance between the bottom of the truck and the road is 0.75 m. Now the truck is exposed to winds
from its side surface. Determine the wind velocity that will tip the truck over to its side. Take the air density to be
1.1 kg/m3 and assume the weight to be uniformly distributed.
5. A laminar flow wind tunnel has a test section that is 40 cm in diameter and 60 cm in length. The air is at 20°C. At
a uniform air speed of 2.0 m/s at the test section inlet, by how much will the centerline air speed accelerate by the
end of the test section? ( vair =1.516×10-5 m2/s)
6. One end of a pond has a shoreline that resembles a half-body. A vertical porous pipe is located near the end of the
pond so that water can be pumped out. When water is pumped at the rate of 0.06 m/s through a 3-m-long pipe, what
will be the velocity at point A?
ID:----------------Name: ------------------
Final Exam
Course: 58:160, Fall 2010
Time: 120 minutes
ID:----------------Name: ------------------
Final Exam
Course: 58:160, Fall 2010
Time: 120 minutes
ID:----------------Name: ------------------
Final Exam
Course: 58:160, Fall 2010
Time: 120 minutes
ID:----------------Name: ------------------
Final Exam
Course: 58:160, Fall 2010
Time: 120 minutes
Problem 1:
(a) In steady laminar flow, the fluid is expected to travel in a circular motion. Only the tangential
component of velocity exists. The radial and axial components of velocity are zero; so, vr = 0 and vz = 0.
(1.25 point)
v
v
1 
1 v
 rvr     z  0    0  v  f ( z, r ) (0.75 point)
r r
r 
z

0
unknown
0
0.25
0.25
0.25
(b)
For a Newtonian fluid, the Navier - Stokes equation in  direction may be simplified as given below.
 direction:




 v
v
v v
v 

 vr

 vz

r
r 
z 
 t
 0;steady flow 0; creeping flow 0;continuity 0; vz 0 
0.25 
0.25
0.25
 0.25


 1   v  1  2 v  2 v
v
p
2 v

  g   
 2  2  2 r
r
 2
2

z
r
r 
 r r  r  r 
0
unknown
 0; symmetric
unknown
 0; continuity
 0; vr  0
unknown

0.25
0.25
0.25
0.25
0.25
0.25
0.25

(2.25 point)







v  rf ( z )


f
2 f f
2 f
 0     r 2     r 2 (0.75 point)
z
r
z
 r
0.25 
0.25
 0.25
0
2 f
f
 C1 
 f  C1 z  C2  v  r (C1 z  C2 ) (2 point)
2
z
z
(c)
at z  0 : v  0
at z   B : v  r
(1 point)
z  0 : v  0  r (0  C2 )  0 (0.5 point)
z   B : v  r  r (C1B  C2 )  r (0.5 point)
 C1 

; C2  0 (0.5 point)
B
v  rf ( z ) 
Problem 2:
rz
(0.5 point)
B
ID:----------------Name: ------------------
Final Exam
Course: 58:160, Fall 2010
Time: 120 minutes
(a)
min   mout
m
out
(1 point)
  A1outV1out   A2outV2out (1 point)
A1out   (1)2 / 4; V1out  ?
A2out   (2)2 / 4   (1)2 / 4; V2out 12
and
min   AinVin
(1 point)
(1 point)
Ain   (2) / 4; Vin  10
2
(
 (2)2

 (1) 2
4
4
 V1out  4 m / s
)(12) 
 (1) 2
4
V1out 
 (2) 2
4
(10)
(0.5 point)
(b)
 F  m
x
V
out out x
 minVin x
(1.5 point)
LHS:
F
x
  FRx  pin _ gage Ain (I)
(2 point)
RHS:
m
V  minVin x   A1outV12out   A2outV22out   AinVin2 (1.5 point)
out out
  FRx  pin _ gage Ain   A1outV12out   A2 outV22out   AinVin2
  FRx  (50)(
 (2)2
4
 FRx  110.71 N
)  (1.23)(
 (1)2
4
)(4) 2  (1.23)(
(0.5 point)
Problem 3:
p1


V12
p V2
 z1  2  2  z2  hl  hpump (1 point)
2g
 2g
V1  0 (0.5 point)
 (2) 2
4

 (1) 2
4
)(12) 2  (1.23)(
 (2) 2
4
)(10) 2
ID:----------------Name: ------------------
Final Exam
Course: 58:160, Fall 2010
Time: 120 minutes
V2  Vpipe #2 (0.5 point)
p1  p2  patm (0.5 point)
 hpump 
2
Vpipe
#2
2g
 hl  z1
Pipe 1:
Vpipe 
Re 

Q
0.018

 6.366 m / s (0.5 point)
A  (0.06)2 / 4
Vpipe D (999.1)(6.366)(0.06)

 335300

1.138 103
Turbulent (0.25 point)
0.00026
 0.0043 (0.25 point)
D
0.06
 f  F (Re,  / D)  0.02941 Moody Chart (0.5 point)

hl1  ( f
L
V2
20
6.3662
  KL )
 ((0.02941)
 0.5)
 21.3 m (1.5 point)
D
2g
0.06
2(9.81)
Pipe 2:
Vpipe 
Re 

Q
0.018

 14.32 m / s (0.5 point)
A  (0.04)2 / 4
Vpipe D (999.1)(14.32)(0.04)

 502900

1.138 103
Turbulent (0.25 point)
0.00026
 0.0065 (0.25 point)
D
0.04
 f  F (Re,  / D)  0.03309 Moody Chart (0.5 point)

hl 2  f
L V2
35 14.322
 (0.03309)
 302.6 m (1 point)
D 2g
0.04 2(9.81)
Total head loss:
hl  hl1  hl 2  21.3  302.6  323.9 m (0. 5 point)
 hpump
2
Vpipe
#2
14.322

 hl  z1 
 323.9  30  304.4 m (0.5 point)
2g
2(9.81)
Also:
Ppump   ghpumpQ  (999.1)(9.81)(304.4)(0.018)  53.7 kW (1 point)
Problem 4:
From table:
L/D=2/2 => CD=2.2 (1 point)
When the truck is first tipped, the wheels on the wind-loaded side of the truck will be off the ground, and
thus all the reaction forces from the ground will act on wheels on the other side. Taking the moment about
ID:----------------Name: ------------------
Final Exam
Course: 58:160, Fall 2010
Time: 120 minutes
an axis passing through these wheels and setting it equal to zero gives the required drag force and
consequently the wind speed:
M  0
 M drag  M weight  0
(3 point)
M drag  FD  armdrag (1 point)
FD  1/ 2  AV 2CD  1/ 2(1.1)(8)(2)V 2 (2.2)
 19.36V 2
(1 point)
armdrag  2 / 2  0.75  1.75 m (1 point)
M weight  mg  armweight (1 point)
armweight  2 / 2  1 (1 point)
 19.36V 2 (1.75)  49050(1)  0 (1 point)
 V  38.05 m / s
Problem 5:
The Reynolds number at the end of the test section is:
Re x 
Vx


(2)(0.6)
 7.92 104 (1 point)
(1.516 105 )
In fact, Rex is lower than the critical Reynolds number, Rex,critical ≈ 1 × 105, for a smooth flat plate with a
clean free stream. Since the walls are smooth and the flow is clean, we may assume that the boundary layer
on the wall remains laminar throughout the length of the test section. We estimate the displacement
thickness at the end of the test section, (1 point)
* 
1.72 x
1.72(0.6)

 3.67 103 m  3.67 mm (3 point)
4
Re x
7.92 10
The effective radius at the end of the test section is reduced by δ*. We apply conservation of mass to
calculate the air speed at the end of the test section,
Vend Aend  Vbeginning Abeginning  Vend
 R2
 Vbeginning
 ( R   * )2
(3 point)
Therefore:
Vend  (2.0)
(0.2)2
 2.08 m / s (1 point)
(0.2  3.67 103 )2
Thus the air speed increases by approximately 4% through the test section, due to the effect of
displacement thickness. (1 point)
Problem 6:
ID:----------------Name: ------------------
Final Exam
Course: 58:160, Fall 2010
For half body:
  Ur sin  
m
 (2 point)
2
So that:

 U sin  (2 point)
r
1 
m
(2 point)
vr 
 U cos  
r 
2 r
v  
At point A:
v  U sin(0)  0 (0.5 point)
vr  U cos(0) 
m
(0.5 point)
2 (15)
Therefore:
vTotal  vr 2  v 2  vr  U cos(0) 
m
(0.5 point)
2 (15)
Since flow rate is 0.06 m3/s in a 3-m long pipe, the source strength is 0.06/3. (1 point)
Then:
b
m
m
0.06 / 3
 5 U 

 6.37 104 m / s (1 point)
2 U
2 b 2 (5)
Therefore:
m
2 (15)
0.06 / 3
 6.37 104 cos(0) 
(0.5 point)
2 (15)
vTotal  U cos(0) 
 8.49 104 m / s
Time: 120 minutes