College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
April 8 Homework Solutions
7.1
The Reynolds number, VD/, is a very important parameter in fluid mechanics. Verify that
the Reynolds number is dimensionless using both the FLT and MLT system for basic
dimensions and determine its value for ethyl alcohol flowing at a velocity of 3 m/s through
a 2-in diameter pipe.
In the FLT system, the dimensions of density, velocity, diameter and viscosity are FL -4T2, LT-1, L,
and FL-2T. Substituting these dimensions into the definition of the Reynolds number gives:
Re 
VD


Re   FL
4


T 2 LT 1 L
2
 F 11 L411( 2)T 211  F 0 L0T 0
FL T
In the MLT system, the dimensions of density, velocity, diameter and viscosity are ML-3, LT-1, L,
and ML-1T-1. Substituting these dimensions into the definition of the Reynolds number gives:
Re 
VD


Re   ML
3
LT L  M
1
1 1
11 311( 1)
L
T 11  M 0 L0T 0
ML T
Thus, the Reynolds number is dimensionless regardless of the unit system used.
For ethyl alcohol the density and viscosity are 789 kg/m 3 and 0.00119 N·s/m2 = 0.00119 kg/m·s,
and the reynolds number is
789 kg 3 m
2 in  2.54 cm 0.01 m
3
VD
s
in
cm
Re 
 m
 1.01x105 .
0.00119 kg

ms
7.35
The flowrate over the spillway a dam is 27,000 ft3/min. Determine the required flow rate 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
Frm 
Vm
gm m
 Fr 
V
g

Vm
m

with g m  g
V

For a 1/25 scale model, all dimensions will have this scale. The flow rate in the model can be
written as Qm = VmAm, where Am 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
spillway. The ratio of model area to actual area will be equal to the length scale squared.
Am   m 
 
A   
2
Vm
m

V

Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
2
Qm Vm Am
m  m 
 



   m 
Q
V A
   
  
Mail Code
8348
52
Phone: 818.677.6448
Fax: 818.677.7062
April 8 homework solutions
ME 390, L. S. Caretto, Spring 2008
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.
Q
 
. m  m 
Q   
52
 1
 
 25 
52
 0.00032
Thus the model flow rate will be 0.00032(27,000 ft3/min) or Qm = 8.64 ft3/min .
7.38
If an airplane travels at a speed of 1120 km/h at an altitude of 15 km, what is the required
speed at an altitude of 8 km to satisfy Mach 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 compute the sound speed from the
equation c2 = kRT, where k is the heat capacity ratio. Thus, for Mach number similarity we can
write
V1 V2

c1 c2

V1
k1 RT1

V2
 V2  V1
k 2 RT2
k 2T2
k1T1
From Table C.2 for the U.S. standard atmosphere we find the temperatures at 15 km and 8 km,
respectively, are -56.50oC = 216.65 K and -36.94oC = 236.21 K. The values of k are given as a
function of temperature in Table B.4. At the two lowest temperatures, -40oC and -20oC the value
of k is 1.401. We can assume that the two temperatures here, -56.50oC and -36.94oC will have
the same value of k. With this assumption, our unknown velocity, V 2, can be found as follows.
V2  V1
7.46
k 2T2
T
1120 km 236.21 K 1169 km
=
 V1 2 
h
k1T1
T1
h
216.65 K
For a certain model study involving a 1:5 scale model it is known that Froude number
similarity 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 30oC, and the model fluid is water at 70oC. 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
Frm 
Vm
gm m
Vp
 Frp 

g p p
Vm
m

with g m  g p
Vp
p
For cavitation number similarity we must have the same cavitation number in model and
prototype.
 pamb  pv m
 mVm2
2
Manipulating this equation gives

 pamb  pv  p
 pV p2
2
April 8 homework solutions
ME 390, L. S. Caretto, Spring 2008
 pamb  pv m   pamb  pv  p
 mVm2
 pV p2
pamb,m  pv,m   pamb  pv  p

Page 3
 mVm2
 pV p2
From the Froude number similarity we can replace the ratio of velocity squatred by the length
scale. This gives the following equation for the ambient pressure in the model.
pamb,m  pv ,m   pamb  pv  p
 mVm2
 
 pv,m   pamb  pv  p m m
2
 p p
 pV p
We can find the water properties needed for this equation from Table B.2. For water at 30oC the
density is 995.7 kg/m3 and vapor pressure is 4243 Pa(abs). For water at 70oC the density is
977.8 kg/m3 and vapor pressure is3.116x104 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  pv ,m   p amb  pv  p
m m
 p p
997.8 kg
1
m3
 3.116 x10 4 Pa  10.1325 x10 4 Pa  4243 Pa
995.7 kg 5
m3


pamb,m = 5.02x104 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, which 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 relationship between the model drag and the prototype drag for both
systems. Would either type of test facility be suitable for this study?
For 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.
Re m 
Vp Lp
Vm Lm
 Re p 
m
p

Vm  m L p  m 1


V p  p Lm  p 50
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 1/50th of the prototype velocity of
10 m/s. So, for the water tunnel, Vm = 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 m2/s and m =
1.46x10-5 m2/s for the model in air. This gives the following velocity scale.
1.46 x10 5 m 2
Vm  m 1
1
s


 0.261
V p  p 50 1.12 x10 6 m 2 50
s
April 8 homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 4
Multiplying the prototype velocity of 10 m/s by this scale factor gives Vm = 2.61 m/s as the
model velocity in the wind tunnel.
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 the prototype.
The relationship between the drag force on the prototype, FD,p, and the drag force on the model,
FD,m, is based on having similarity in the drag coefficient.
C D ,m 
FD,m
 V2
Am m m
2
FD, p
 C D, p 
 pV p2
Ap
 FD, p  FD,m
A p  pV p2
Am  mVm2
2
The area ratio between the model and the prototype is just the 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
FD, p  FD,m
A p  pV p2
Am  mVm2
 FD,m
L2p  p V p2
L2m  m Vm2
2
 1 
   1502  FD,m
 50 
For the water tunnel, FD,p = FD,m .
For the wind tunnel, the air density is 1.23 kg/m3 and for the prototype in water the density is 999
kg/m3. In this case the drag force relationship between the model and the prototype is
FD, p 
L2p
FD,m 2
Lm
 p V p2
 m Vm2
 999 kg  10 m  2


 1   m 3  s 
 
 4.77 FD,m
 50   1.23 kg  2.61 m 



 m 3  s 
2
For the wind tunnel, the prototype drag force is 4.77 times the model drag force .
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 prototype 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 between
the velocity of the model and that of the prototype.
Re m 
 pV p L p
 mVm Lm
 Re p 
m
p

Vm  p  m L p  p  m 20 ft 10 p  m



V p  m  p Lm  m  p 2 ft
m p
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 Rm = Rp) and the temperatures are the same for both the model and the prototype (hence
April 8 homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 5
m = p.) Finally we can substitute the values for pp = 1 atm(abs) and pm = 8 atm(abs to get a
numerical value for the velocity scale.
Vm 10 p  m 10 m p p RmTm 10 p p 101 atm




 1.25
8 atm
Vp
 m p
 p R pT p pm
pm
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 .