Jacaranda (Engi neering) 3333
Mail Code
Phone: 818.677.6448
E

mail:
lcaretto@csun.edu
8348
Fax: 818.677.7062
College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 3
90
Fluid Mechanics
Spring 200
8
Number:
11971
Ins t ruc t or: Larry Caret t o
Apri
l 8
Home work Sol uti ons
7.
1
The
Reynolds number,
VD/
,isaveryimportantparameterinfluidmechanics.Verifythat
theReynoldsnumberisdimensionless using boththeFLTandMLTsystem forbasic
dimensions anddetermineitsvalueforethylalcoholflowingatavelocityof3m/sthrough
a2

indiamet
erpipe
.
In the FLT system, the dimensions of density, velocity, diameter and viscosity are FL

4
T
2
, LT

1
, L,
and FL

2
T. Substituting these dimensions into the definition of the Reynolds number gives:
In the MLT system, the dimensio
ns of density, velocity, diameter and viscosity are ML

3
, LT

1
, L,
and ML

1
T

1
. Substituting these dimensions into the definition of the Reynolds number gives:
Thus, the Reynolds number is dimensionless regardless of the unit syste
m used.
For ethyl alcohol the density and viscosity are 789 kg/m
3
and 0.00119 N∙s/m
2
= 0.00119 kg/m∙s,
and the reynolds number is
1.01x10
5
.
7
.
35
The flowrate over the spillway a dam is 2
7,
000 ft
3
/min. Determine the required flow ra
te for
a 1/25 scale model that is operated in accordance with Froude number similarity.
For Froude number similarity between the model (subscript m) and the actual dam (no subscript)
we must have
For a 1/25 scale model, all dimensio
ns will have this scale
. The flow rate in the model can be
written as Q
m
= V
m
A
m
, where A
m
is the area of the water flow over the spillway in the model. The
water flow in the actual dam can be written as Q = VA, where A is the actual water area over the
s
pillway. The ratio of model area to actual area will be
equal
to the length scale squared.
Apri
l 8
homewo
rk solutions
ME 390, L. S. Caretto, Spri ng 20
08
Page
2
Substituting the given value of 1/25 for the length scale gives the following result for the flowrate
ratio between the model and the actual
spillway.
.
Thus the model flow rate will be 0.00032(27,000 ft
3
/min) or
Q
m
=
8.64
ft
3
/min
.
7.38
If an airplane travels at a speed of 1120 km/h a
t
an altitude of 15 km, what is the required
speed at an altitude of 8 km to satisfy Ma
ch number similarity? Assume the air properties
correspond to those for the U.S. standard atmosphere.
For Mach number similarity, the ratio of the speed, V, to the sound speed, c, must be the same.
If we assume that air behaves as an ideal gas we can com
pute the sound speed from the
equation c
2
= kRT, where k is the heat capacity ratio. Thus, for Mach number similarity we can
write
From Table C.2 for the U.S. standard atmosphere we find the temperatures at 15 km and 8 km,
respecti
vely, are

56.50
o
C = 216.65 K and

36.94
o
C = 236.21 K. The values of k are given as a
function of temperature in Table B.4. At the two lowest temperatures,

40
o
C and

20
o
C the value
of k is 1.401. We can assume that the two temperatures here,

56.50
o
C a
nd

36.94
o
C will have
the same value of k. With this assumption, our unknown velocity, V
2
, can be found as follows.
=
7.46
For a certain model study involving a 1:5 scale model it is known that Froude number
s
imilarity must be maintained. The possibility of cavitation is also to be investigated, and
it is assumed that the cavitation number must be the same for the model and prototype.
The prototype fluid is water at 30
o
C, and the model fluid is water at 70
o
C.
If the prototype
operates at an ambient pressure of 101 kPa(abs), what is the required ambient pressure
for the model system.
For Froude number similarity between the model (subscript m) and the
prototype (
subscript
p
) we
must have
For cavitation number similarity we must have the same cavitation number in model and
prototype.
Manipulating this equation gives
Apri
l 8
homewo
rk solutions
ME 390, L. S. Caretto, Spri ng 20
08
Page
3
From the Froude number similarity we can replace the ratio of velocity squat
red by the length
scale. This gives the following equation for the ambient pressure in the model.
We can find
the
water properties
needed for this equation from
Table B.2. For water at 30
o
C the
density is 995.7 kg/m
3
and vapor pre
ssure is 4243 Pa(abs). For water at 70
o
C the density is
977.8 kg/m
3
and vapor pressure is3.116x10
4
Pa(abs).
Substituting these
water properties and the length sacle of 1:5 into the equation for the ambient
pressure in the model gives
p
amb,m
=
5.02x10
4
Pa(abs)
7.55
The drag on a small completely submerged solid body having a characteristic length of 2.5
mm and moving with a velocity of 10 m/s through water is to be determined with the aid of
a model. The length scale is to be 50, w
hich indicates that the model is to be
larger
than
the prototype. Investigate the possibility of using either an unpressurized wind tunnel or a
water tunnel for this study. Determine the required velocity in both the wind and water
tunnels, and the relat
ionship between the model drag and the prototype drag for both
systems. Would either type of test facility be suitable for this study?
Fo
r completely submerged flow we must have Reynolds number similarity between
the model (m)
and prototype (p). With the
stated scale factor, the Reynolds number scaling gives the following
relationship between the velocity of the model and that of the prototype.
If we use the same temperature for both the prototype and the model in the water tunnel,
the
kinematic viscosity will be the same in both cases and the velocity scale will be the same as the
length scale: 1/50. This means that the model velocity will be 1/50
th
of the prototype velocity of
10 m/s. So, for the water tunnel,
V
m
=
0.200
m/s
.
If
we use a wind tunnel, the velocity scale will depend on the viscosity ratio. Using the standard
data for kinematic viscosity we have for the prototype in water,
p
= 1.12x10

6
m
2
/s and
m
=
1.46x10

5
m
2
/s for the model in air. This gives the following velocity scale.
Apri
l 8
homewo
rk solutions
ME 390, L. S. Caretto, Spri ng 20
08
Page
4
Multiplying the prototype velocity of 10 m/s by this scale factor gives
V
m
=
2.61
m/s
as the
model velocity in the wind tunn
el.
Both of these velocities seem reasonable and we could use either as the test medium. The
computation of the drag force for the prototype from drag force measurements on the model
requires that the loss coefficient be the same for both the model and th
e prototype.
The relationship between the drag force on the prototype, F
D,p
, and the drag force on the model,
F
D,m
, is based on having similarity in the drag coefficient.
The area ratio between the model and the prototype is just th
e square of the length scale. For
the water tunnel, the density ratio will be one and the velocity ratio was found above to be 1/50.
For the water tunnel then
For the
water
tunnel,
F
D,p
=
F
D,m
.
For the wind tunnel,
the air densit
y is 1.23
kg
/m
3
and for the prototype in water the density is 999
kg/m
3
. In this case the drag force relationship between the model and the prototype is
For the
wind
tunnel,
the
prototype
drag
force
is
4.77
times
the
model
drag
for
ce
.
7.57
If the unpressurized wind tunnel of Problem 7.56 were replaced with a tunnel in which the
air can be pressurized isothermally to 8 atm(abs), what range of air velocities would be
required to maintain Reynolds number similarity for the same prototy
pe velocities given in
Problem 7.56
–
20 mph and 90 mph. For the pressurized tunnel the maximum
characteristic model length that can be accommodated is 2 ft, whereas the maximum
characteristic prototype length remains at 20 ft.
This is another completely
submerged flow for which we must have Reynolds number similarity
between the model (m) and prototype (p). With the stated maximum lengths of 2 ft for the model
and 20 ft for the prototype, the Reynolds number scaling gives the following relationship betwe
en
the velocity of the model and that of the prototype.
We can find the density from the ideal gas law,
= p/RT
. We can substitute this equation for
both densities in the velocity

scale equation and cancel terms since the both use the same fluid
(hence R
m
= R
p
) and the temperatures are the same for both the model and the prototype (hence
Apri
l 8
homewo
rk solutions
ME 390, L. S. Caretto, Spri ng 20
08
Page
5
m
=
p
.) Finally
we can substitute the values for p
p
= 1 atm(abs) and p
m
= 8 atm(abs to get a
numerical value for the velocity scale.
Thus, the model velocities corresponding to the desired prototype velocities of 20 mph and 90
mph are simply 1.25
times these velocities or
25
mph
and
112.5
mph
.
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