College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
March 4 Homework Solutions
4.56
Water flows in a branching pipe shown in
figure P4.56 at the right, with uniform velocity
at each inlet and outlet. The fixed control
volume coincides with the system at time t –
20 s. Make a sketch to indicate (a) the
boundary of the system at time t = 20.2s, (b)
the fluid that left the control volume during
that 0.2 s interval, and (c) the fluid that
entered the control volume during that
interval.
The fluid travels a distance Vt during the time
periodt = 0.2 s. At point 1, this distance is V1t = (2 m/s)(0.2 s) = 0.4 m. At point 2, this
distance is V2t = (1 m/s)(0.2 s) = 0.2 m. At point 3, this distance is V3t = (1.5 m/s)(0.2 s) = 0.3
m. Using these dimensions we can prepare the sketch of the system and the control volume
shown below.
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
March 4 homework solutions
3.61
ME 390, L. S. Caretto, Spring 2007
Page 2
The wind blows across a field with an
approximate velocity profile shown in
Figure P4.61 shown at the right. Use
equation 4.16 with the parameter b equal to
the velocity to determine the momentum
flow rate across the vertical surface A-B
which is of unit depth into the paper.
Equation 4.16 gives the following expression
for the flow of a quantity b
B   bV  ndA
A
In this case, b is actually a vector quantity, V = Vi, where i is the unit vector in the x direction.
From the graph we see that V = (1.5/s)y between 0 and 10 feet and V = 15 ft/s between 10 and
20 ft. We can thus write our integral as follows noting that n = i, and b is the vector quantity, V
  V(Vi  i)dA  ViVdA  i V 2 d wy 
B   bV  ndA  B



A
A
A
A
Here we have used the fact that i▪I = 1 and the differential area, dA = wdy where w is the width of
one ft into the page. Substituting the relationship between V and y given above gives the
following result for the integral.
2
20 ft
10 ft  1.5  2

 15 ft 
B  i  V 2 d wy   iw  V 2 dy  iw     y 2 dy   
 dy 
s 
 0  s 

A
y
10 ft 
10
ft
2
 1.5 2 y 3

 1.5  2 10 ft 3  15 ft  2


ft 3 
 15 ft  20 ft 
 

 iw  

 iw  

 y
 10 ft   iw  3000 2 
s  10 dr 
s 
3
s 
 s  3
s 







0


Using the standard density of air  = 0.00238 slugs/ft3 and the given unit width, w = 1 ft, gives the
final answer as
2

0.00238 slugs 
ft 3 
ft 3  1 lb f  s


B  iw 3000 2  i1 ft 
3000 2
 7.14i lb f


s 
ft 3
s  slug  ft


5.7
Water flows along the centerline of a 50-mmdiameter pipe with an average velocity of 10 m/s
and out radially between two large circular disks as
shown in Figure P 5.7 at the right. The disks are
parallel and spaced 10 mm apart. Determine the
average velocity of the water at a radius of 300 mm
in the space between the disks.
This is a basic continuity equation problem with
constant density so that Q = V1A1 = V2A2. The desired
velocity is found as follows
A1 10 m
 25 mm

 1.04 m/s .
A2
s 2 300 mm10 mm
2
V2  V1
March 4 homework solutions
5.11
ME 390, L. S. Caretto, Spring 2007
Page 3
At cruise conditions air flows into a jet engine at a steady rate of 65 lb m/s. Fuel enters the
engine at a steady rate of 0.60 lbm/s. The average velocity of the exhaust gases is 1500 ft/s
relative to the engine. If the engine exhaust effective cross section is 3.5 ft2, estimate the
density of the exhaust gases in lb m/ft3.
This is a problem of mass conservation and use of the continuity equation. The exhaust mass is
he sum of the air mass flow rate and the fuel mass flow rate: 65 lbm/s + 0.60 lbm/s = 56.60 lbm/s.
This mass flow rate of exhaust must be equal to the density-velocity-area product, VA, for the
exhaust. We can thus solve for the unknown density.
65.60 lbm
m
s


1500
ft
VA
3.5 ft 2
s

5.12
 = 0.0125 lbm/ft3

Air at standard atmospheric conditions is drawn into a compressor at the steady rate of 30
m3/min. The compressor pressure ratio, pexit/pinlete, is 10 to 1. Through the compressor
p/n remains constant with n = 1.4. If the average velocity in the compressor discharge
pipe is not to exceed 30 m/s calculate the minimum discharge pipe diameter required.
We can apply the continuity equation to this problem with 1 as the inlet and 2 as the outlet.
1V1 A1  1Q1   2V2 A2   2V2
D22
4
Solving this equation for D2 in terms of Q1 gives.
.
D2 
4 1Q1
2V2
Since p/n = constant, p1/1n = p2/2n. Making this substitution and substituting numerical data
gives.
.
5.23
D2 
p
4 1Q1
  1
2V2
 p2



1n
4Q1
1
  
V2
 10 
1 1.4
30 m 3
4
s
30 m

s
D2 = 0.064 m .
The Hoover Dam backs up the Colorado River and creates Lake Mead, which is
approximately 115 miles long and has a surface area of approximately 225 square miles. If
during flood conditions the Colorado River flows into the lake at a rate of 45,000 cfs and
the outflow from the dam is 8000 cfs, how many feet per 24-hour day will the lake level
rise?
This is a transient continuity problem. If we assume that the density and area are constant we can
write the equation as follows
dm d V  d Ah
dh


 A
 m in  m out   Qin  Qout 
dt
dt
dt
dt
If we assume that the flow rates are constant so that the derivative becomes a finite difference we
have the following result.
March 4 homework solutions
A
h
  Qin  Qout 
t
ME 390, L. S. Caretto, Spring 2007
Page 4
 45000 ft 3 8000 ft 3 
3600 s

24 h 

s
s
h
Q  Qout t 

 h  in

2
A
225 mi2  5280 ft 
 mi 
h = 0.0510 ft