College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
Solutions to Exercise Five –
Control Volumes
1
(1)
Two rivers merge to form a larger river as
shown in Figure P5-13 at the right. At a
location downstream from the junction
(before the two streams completely
merge) The nonuniform profile is as
shown and the depth is 6 ft. Determine
the value of V. (Problem and figure
copied from Munson et al., Fluid
Mechanics text.)
(2)
(3)
(2)
(1)
This problem basically involves the continuity
equation with a non-uniform velocity at one
part. We can apply the continuity equation to
the three streams (numbering 1 and 2 and
the two initial streams and 3 as the merged stream to obtain the following results, after assuming
that the density is constant.
.
1  m
2  m
3
m
 1V1 A1   2V2 A2   3V3 A3
 V1 A1  V2 A2  V3 A3
The average velocity at point 3 is the integration of the velocity profile across the L = 100 ft width
of the river (since the depth is constant).
V3 
1
1
V ( A)dA 

A
DL
xL
 V ( x)d ( Dx ) 
x 0
xL
1
1
V ( x)dx 

L x 0
100 ft
x  70 ft
 Vdx 
x 0
1
100 ft
x 100 ft
 0.8Vdx
x  70 ft
The integration is simply the integration of a constant so the results become
x  70 ft
x 100 ft

1
1
x  70 ft
x 100 ft
x0Vdx  100 ft x70 0ft.8Vdx  100 ft Vxx0  0.8Vxx70 ft
V
70 ft  0  0.8V 100 ft  70 ft   0.94V

100 ft
1
V3 
100 ft

Note that we could have used our intuition to write this equation for V3 by saying that 70% of the
river has a velocity V and the other 30% has a velocity 0.8V so that the average velocity is 0.7V +
0.3(0.8V) = 0.94V. The integration used above is a general approach that will apply to any
velocity profile.
We can also compute the areas of the three streams from the given width and depths: A 1 =
(80 ft)(5 ft) = 400 ft2; A2 = (50 ft)(3 ft) = 150 ft2; A3 = (100 ft)(6 ft) = 600 ft2. Substituting these
areas, the two given velocities and the equation that V3 = 0.94V into the continuity equation for
this problem gives
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Exercise five solutions
ME 390, L. S. Caretto, Spring 2008
. V1 A1  V2 A2  V3 A3









Page 2

4 ft
3 ft
400 ft 2 
150 ft 2  0.94V 600 ft 2
s
s

Solving this equation for V gives

4 ft
3 ft
400 ft 2 
150 ft 2
s
s
 3.63 ft/s
.V 
0.94V 600 ft 2
2

It takes you one minute to fill you’re your car’s fuel tank with 8.8 gallons of gasoline. What
is the approximate average velocity of he gasoline leaving the 0.60-inch diameter nozzle at
this pump. (Problem 5.27 from Munson et al., Fluid Mechanics text.)
This is an unsteady problem in which there is only one inlet stream. Consequently the continuity
 in   inVin Ain  dmcv dt . If we assume that the mass flow
equation for this process becomes m
rate is constant (ignoring initial start-up and final shut-down transient flows) we can integrate this
 in t   inVin Ain t  mcv . We are given the volume of
equation to obtain the result that m
gasoline added, not the mass, but we know that mcv = Vcv, where Vcv is the volume.
Substituting this result into the m equation gives
 inVin Ain t  Vcv
Vin 
3
 Vin 
Vcv
4 Vcv

Ain t  Din2 t
8.8 gal  0.13368 ft 3 1 min 144 in 2  9.99 ft
4 Vcv
4

s
 Din2 t  0.6 in 2 1 min  1 gal
60 s
ft 2
An evaporative cooling
tower (see figure P5.10 at
the right) is used to cool
water from 110oF to 80oF.
Water enters the tower at
a rate of 250,000 lbm/hr.
Dry air (no water vapor)
flows into the tower at a
rate of 151,000 lbm/hr. If
the rate of wet air flow
out of the tower is
165,900 lbm/hr, determine
the rate of water
evaporation in lbm/hr and
the rate of cooled water
flow in lbm/hr. (Problem
5.10 from Munso n et al.,
Fluid Mechanics text.
Figure from solutions
manual.)
In this problem we have to consider not only the continuity equation for conservation of total
mass, but also the coservation of each species (air and water) in the absence of any chemical
Exercise five solutions
ME 390, L. S. Caretto, Spring 2008
Page 3
raction. There are only two locations with air flow so we know that the mass flow rate of dry air
leaving the tower at section 2 must be the same as the mass flow entering at section 3: 151,000
lbm/hr. Since the total flow rate of dry air plus water at section 2 is 156,900 lb m/hr, the difference
between this total flow and the flow of dry air is 156,900 lbm/hr – 151,000 lbm/hr = 5,900
lbm/hr.must be the flow of water that evaporated and leaves the tower . The flow of
cooled water at section 4 is simply the original flow of 250,000 lbm/hr minus the evaporated water
flow rate of 5,900 lbm/hr. This gives a cooled water flow rate of 244,100 lbm/hr .