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```ME 363 - Fluid Mechanics
Homework #11
Due Wednesday, April 16, 2008
Fall Semester 2008
1] A student team is to design a human-powered submarine for a design competition. The overall length of the
prototype submarine is 2.24 m, and its student designers hope that it can travel fully submerged through water at
0.56 m/s. The water is freshwater (a lake) at T = 15 degrees C. The design team builds a one-eighth scale
model to test in their university’s wind tunnel. A shield surrounds the drag balance strut so that the
aerodynamic drag of the strut itself does not influence the measured drag. The air in the wind tunnel is at 25
degrees C and at atmospheric pressure. At what air speed do they need to run the wind tunnel in order to
achieve similarity?
2] This problem builds on problem 1. The students measure the aerodynamic drag on their model submarine in the
wind tunnel. They are careful to run the wind tunnel at conditions that ensure similarity with the prototype
submarine. Their measured drag force is 2.3 N. Estimate the drag force on the prototype submarine at the
conditions given in problem 1.
3] Some students want to visualize flow over a spinning baseball. Their fluids laboratory has a nice water tunnel
into which they can inject multicolored dye streaklines, so they decide to test a spinning baseball in the water
tunnel. Similarity requires that they match both the Reynolds number and the Strouhal number between their
model test and the actual baseball that moves through the air at 80 mph and spins at 300 rpm. Both the air and
the water are at 20 degrees C. At what speed should they run the water in the water tunnel, and at what rpm
should they spin their baseball?
no figure
Problem #1
Problem #2
Problem #3
Problem 1
Solution
For a scale model of a submarine being tested in air, we are to calculate the wind tunnel speed
required to achieve similarity with the prototype submarine that moves through water at a given speed.
Assumptions 1 Compressibility of the air is assumed to be negligible. 2 The wind tunnel walls are far enough
away so as to not interfere with the aerodynamic drag on the model sub. 3 The model is geometrically similar to the
prototype.
Properties
For water at T = 15oC and atmospheric pressure,  = 999.1 kg/m3 and  = 1.138  10-3 kg/ms. For
o
air at T = 25 C and atmospheric pressure,  = 1.184 kg/m3 and  = 1.849  10-5 kg/ms.
Analysis
Similarity is achieved when the Reynolds number of the model is equal to that of the prototype,
Rem 
Similarity:
pVp Lp
mVm Lm
 Rep 
m
p
(1)
We solve Eq. 1 for the unknown wind tunnel speed,

Vm  Vp  m
 p

   p  Lp 
 


   m  Lm 
 1.849  105 kg/m  s  999.1 kg/m3 
  0.560 m/s  
8  61.4 m/s

3
3  
 1.138  10 kg/m  s  1.184 kg/m 
Discussion
At this air temperature, the speed of sound is around 346 m/s. Thus the Mach number in the wind
tunnel is equal to 61.4/346 = 0.177. This is sufficiently low that the incompressible flow approximation is
reasonable.
Problem 2
Solution
We are to estimate the drag on a prototype submarine in water, based on aerodynamic drag
measurements performed in a wind tunnel.
Assumptions 1 The model is geometrically similar. 2 The wind tunnel is run at conditions which ensure similarity
between model and prototype.
Properties
For water at T = 15oC and atmospheric pressure,  = 999.1 kg/m3 and  = 1.138  10-3 kg/ms. For
o
air at T = 25 C and atmospheric pressure,  = 1.184 kg/m3 and  = 1.849  10-5 kg/ms.
Analysis
Since the Reynolds numbers have been matched, the nondimensionalized drag coefficient of the
model equals that of the prototype,
FD ,m
mVm Lm
2
2

FD ,p
pVp 2 Lp 2
(1)
We solve Eq. 1 for the unknown aerodynamic drag force on the prototype, FD,p,
2
 p  Vp   Lp 
 999.1 kg/m  s   0.560 m/s 
2
FD,p  FD,m 

2.3
N


8  10.3 N
  




 1.184 kg/m  s   61.4 m/s 
 m  Vm   Lm 
2
2
where we have used the wind tunnel speed calculated in Problem 7-36.
Discussion
Although the prototype moves at a much slower speed than the model, the density of water is much
higher than that of air, and the prototype is eight times larger than the model. When all of these factors are
combined, the drag force on the prototype is much larger than that on the model.
Problem 3
Solution
We are to calculate the speed and angular velocity (rpm) of a spinning baseball in a water channel
such that flow conditions are dynamically similar to that of the actual baseball moving and spinning in air.
Properties
For air at T = 20oC and atmospheric pressure,  = 1.204 kg/m3 and  = 1.825  10-5 kg/ms. For
o
water at T = 20 C and atmospheric pressure,  = 998.0 kg/m3 and  = 1.002  10-3 kg/ms.
Analysis
The model (in the water) and the prototype (in the air) are actually the same baseball, so their
characteristic lengths are equal, Lm = Lp. We match Reynolds number,
Rem 
pVp Lp
mVm Lm
 Rep 
m
p
(1)
and solve for the required water tunnel speed for the model tests, Vm,

Vm  Vp  m
 p

  p  Lp 
 1.002 103 kg/m  s  1.204 kg/m3 
 80.0 mph  
1  5.30 mph




5
3  
 m Lm
 1.825 10 kg/m  s  998.0 kg/m 



(2)
We also match Strouhal numbers, recognizing that n is proportional to f,
St m 
f p Lp
f m Lm
 St p 
Vm
Vp

nm Lm np Lp

Vm
Vp
(3)
from which we solve for the required spin rate in the water tunnel,
 Lp
nm  np 
 Lm
  Vm
 
  Vp

 5.30 mph 
   300 rpm 1 
  19.9 rpm

 80.0 mph 

Discussion
Because of the difference in fluid properties between air and water, the required water tunnel speed
is much lower than that in air. In addition, the spin rate is much lower, making flow visualization easier.
(4)
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