Name:
College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
Solutions toMidterm Examination
1. (37 points)
The gate shown in the figure at the
right pivots about the hinge and is
held in place by the 2000 lbf
counterweight, W. Find the water
depth, h. Assume the weight of the
gate is negligible.
For the vertical part of the gate the
resultant force of the water is given by
the equation FR = hcA where hc = h/2
and A = (4 ft)h. This gives FR =
.h/2)(4 ft)h = h(2 ft).
The distance from the top of the water
to the point where this resultant force acts, yR is given by the following equation
yR 
I xc
 yc
yc A
The moment of inertia about the centroid is given by the equation I xc = ab3/12 where a is the width
of the area and b is the height. In our case, Ixc = (4 ft)h3/12. Substituting value of Ixc into the
equation for yR gives
4 ft h 3
yR 
I xc
h h h 2h
 y c  12    
h
yc A
4 ft h 2 6 2 3
2
A moment balance about the hinge is shown in the figure at the left. The distance, d, from the
hinge to the location of the resultant force is simply
d = h – yR – h = 2h/3 = h/3.
The moment balance about the hinge has only two
forces, FR and W. This balance requires that FRd =
W(3 ft).
Substituting d = h/3 and FR = h(2 ft) into this
moment balance with W = 2000 lbf and  for water =
62.4 lbf/ft3 gives


h
h 2 2 ft   2000lb f 3 ft 
3
62.4lb f
ft 3
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
2 ft  h
3
3
 6000 ft  lb f
Phone: 818.677.6448
Fax: 818.677.7062
Midterm exam solutions
ME 390, L. S. Caretto, Spring 2008
Page 2
h = 5.24 ft
2. (37 points)
Air flows through a Venturi
channel with a rectangular cross
section. The width of the
channel is 0.06 m; the heights at
point (1) and the exit are 0.04 m.
The height at the throat is 0.02
m. Compressibility and viscous
effects are negligible.
(3)
(4)
(a) What is the flow rate when
the water in the small tube
attached to the static
pressure tap at the throat is drawn up 0.10 m as shown?
Since the flow can be assumed to be inviscid and incompressible, we can use the Bernoulli
equation to relate the various state points in the flow. In particular we can relate conditions at the
throat (3) to conditions at the exit (4).
g z4  z3  
p4  p3


V42  V32
0
2
Since both locations are at the same elevation, z4 – z3 = 0. We also have p4 = 0 because this
point is a free jet (open to the atmosphere.) At the throat, the manometer equation for the water
height of 0.10 m gives p3 + (0.1 m)w = 0 or p3 = –(0.1 m)w. Finally we can use the continuity
equation, V4A4 = V3A3, to give V4(0.06 m)(0.04 m) = V3(0.06 m)(0.02 m) or V3 = 2V4. Making
these substitutions in the Bernoulli equation gives.
g  z 4  z3  
p4  p3


V42  V32
0    w 0.1m  V42  4V42
 0

 0  V4 
2

2
2 w 0.1m 
3
We are not given any data on the atmospheric pressure of the temperature of the flowing air. We
can assume that the air is at standard conditions for which we find  = 1.23 kg/m3 in the inside
front cover of the text. Using this datum and a value of 9800 N/m 3 for w (also from the tables in
the inside front cover) we can solve for the exit velocity, V4.
V4 
2 w 0.1m 

3
9800 N
0.1m  23.1m
m3

1.23kg 1N  s 2
s
3
3
m
kg  m
2
The flow rate, Q = V4A4 = (23.1 m/s)(0.06 m)(0.04 m) or
Q = 0.0554 m3/s .
(b) Determine the height, h2, at section (2) for the flow rate you found in part (a).
We can again apply Bernoulli’s equation, this time between section (2) and the exit to find the
desired height, h2. In this case Bernoulli’s equation becomes
g  z4  z2  
p4  p2

V42  V22

0
2
Midterm exam solutions
ME 390, L. S. Caretto, Spring 2008
Page 3
Since both locations are at the same elevation, z4 – z2 = 0. As before, we have p4 = 0, and, the
manometer equation for the water height h2 at station (2) is p2 + w(0.05 m) = 0 or p2 =
-(0.05 m)w. Making these substitutions and the value of V4 = 23.1 m/s found above gives.
2
 23.1m 
2

  V2
2
2
p  p2 V4  V2
0    w 0.05m   s 
g z4  z2   4

 0

0

2

2
Solving this equation for V2 and substituting data for  and w as before gives
9800 N
2
2
0.05m  1332m 2
2
3


23
.
1
m
2

0
.
05
m
23
.
1
m




m
w
V22  







1.23kg 1N  s 2

s2
 s 
 s 
m3 kg  m
Taking the square root gives V2 = 36.5 m/s. Finally, we can apply the continuity equation to find
the height, h2
V4 A4  V2 A2
 V4 w0.04m   V2 wh2
 h2  0.04m 
V4
23.1m s
 0.04m 
V2
36.5m s 2
h2 = 0.02534 m
(c) Determine the pressure at section (1) required to produce this flow.
This time we apply the Bernoulli equation between points (1) and (4).
g z4  z1  
p4  p1


V42  V12
0
2
Since both locations are at the same elevation, z4 – z1 = 0. As before, we have p4 = 0, Since the
flow areas at both points (1) and (4) are the same, the two velocities are the same. These
substitutions in the Bernoulli equation give
g z4  z1  
p4  p1

V42  V12
0  p1

 0
0  0
2

p1 = 0
3. (37 points)
Water flows through the 20o
reducing bend shown in the figure
at the right at a rate of 0.025 m3/s.
The flow is frictionless, gravitational
effects are negligible, and the
pressure at section (1) is 150 kPa.
Determine the x and y components
of the force required to hold the
bend in place.
We can use the general momentum
balance in both the x and y directions
to get two equations for the two
Midterm exam solutions
ME 390, L. S. Caretto, Spring 2008
Page 4
required force components. The general equation is shown below.
mVk cv
t

Noutlets
Ninlets
o1
i 1
 oVo AoVk ,o   iVi AiVk ,i   Fk
Here we have one inlet and one outlet and we assume steady flow. This gives the following
balance equations in the two directions. Note that the pressure force is in the +x direction at
station 1 with no y component and we have to resolve the components of the pressure force at
station 2; the x component is –p2A2cos 20o, because it acts in the minus-x direction; the y
component is p2A2sin20o, because it acts in the +y direction. Fx and Fy denote the x and y
components of force required to hold the bend in place.
2V2 A2Vx, 2  1V1 A1Vx,1  Fx  p1 A1  p2 A2 cos 20o
2V2 A2V y ,2  1V1 A1V y ,1  Fy  p2 A2 sin 20o
For this system, Vx,1 = V1, Vy,1 = 0, Vx,2 = V2cos20o, and Vy,2 = –V2sin20o. The negative sign for
Vy,2 is because it is in the minus-y direction.
The continuity equation tells us that the mass flow rate is the same at both inlet and outlet. In
addition, the water has constant density so the volume flow rate is constant. We can compute the
mass flow rate and the velocities at inlet and outlet from continuity relations.
0.025m3
Q
Q
3.18m
s
V1 



2
2
A1 D1
s
 0.1m
4
4
0.025m3
Q
Q
12.7m
s
V2 



2
2
A2 D2
s
 0.5m 
4
4
Using the density of water as 998 kg/m 3 we find the mass flow rate as
m  Q 
998kg 0.025m3 24.95 kg

s
s
m3
m  2V2 A2  1V1 A1
We can substitute this mass flow rate and the following velocity components into our momentum
equations: Vx,1 = V1, Vy,1 = 0, Vx,2 = V2cos20o, and Vy,2 = –V2sin20o. This gives:


m V2 cos 20o  V1  Fx  p1 A1  p2 A2 cos 20o


m  V2 sin 20o  Fy  p2 A2 sin 20o
We are given P1 = 150 kPa, but we do not know the value of P2. Since the flow is frictionless and
we have already assumed that it has constant density, we can use Bernoulli’s equation to relate
points (1) and (2) and solve for P2.
g z2  z1  
p2  p1


V22  V12
0 
2
p2  p1  
V22  V12
 g z2  z1 
2
We are told that gravitational effects are negligible so we can ignore the gz term. This gives
2
2
V22  V12
998kg 1  12.7m   3.18m   kPa  m  s 2
p2  p1  
 150kPa 
 74.5kPa
 
 

2
m3 2  s   s   1000kg
We now have the required information to compute the resultant forces.
Midterm exam solutions
ME 390, L. S. Caretto, Spring 2008


Fx  p2 A2 cos 20o  p1 A1  m V2 cos 20o  V1 
150kPa
Page 5
1000 N 
0.05m 2
74
.
5
kPa
cos 20o 
2
4
kPa  m 
0.1m 2  24.95 kg  12.7 m
3.18 m  1N  s 2
cos 20o 



4
s
s  kg  m
 s

Fx = –882 N
Fy   p2 A2 sin 20o  m V2 sin 20o 
2
 24.95 kg  12.7 m
1000 N 
0.05m 2
o
o  1N  s

74
.
5
kPa
sin
20

sin
20




4
s
kPa  m2 
 s
 kg  m

Fy = –156 N