Transcript Drag Force

Modeling
Amy Stephens
BIEN 301
15 February 2007
Problem 5.76

A 2-ft-long model of a ship is tested in a
freshwater two tank. The measured drag
may be split into “friction” drag and “wave”
drag. The model data is below.
Tow Speed, ft/s
0.8
1.6
2.4
3.2
4.0
4.8
Friction drag, lbf
0.016
0.057
0.122
0.208
0.315
0.441
Wave drag, lbf
0.002
0.021
0.083
0.253
0.509
0.697
Sketch
Assumptions

Geometric similarity
 Kinematic similarity
 Dynamically similarity
 Scaling law is valid
 Steady flow
 Liquid
 Incompressible
Solution

The first step to solving this problem is to
calculate the Reynolds number and Froude
number for each model velocity from the
given information using the formulas below
VL
Re 

V
Fr 
gL
Note: Gravity for this
problem is 32ft/s2
The following table shows the resulting Reynolds and
Froude numbers:
V ft/s) 0.8
1.6
2.4
3.2
4.0
4.8
Re
148.5K 297K
445.5K 594K
742.5K 891K
Fr
0.1
0.3
0.5
0.2
0.4
0.6
The density and viscosity of each fluid should taken
from White Tables A.1 and A.3 (conversion necessary):
Water :   1.937 slug / ft 3 ,   2.09e  5lb * s / ft 2
Seawater :   1.986slug / ft 3 ,   2.232e  5lb * s / ft 2
From the drag force data given, the force coefficients can
be found by using the following equations:
Fdfrictio n
CFfrictio n 
V 2 L2
Fdw a ve
CFw a ve 
2 2
V L
The table below shows the resulting force coefficients:
V (ft/s) 0.8
Cdfric
Cdwave
1.6
2.4
3.2
4.0
4.8
.00322 .00287 .00273 .00262 .00254 .00247
.00040 .00106 .00186 .00318 .00410 .00390
The prototype data can now be used to find Reynolds
and Froude numbers for the prototype at the desired
velocity:
5.144e  1m / s
1 ft / s
15kn x
x
 25.3ft/s
1kn
0.3048m / s
VL 1.986 * 25.3 *150
Re p 

 3.4e8

2.232e  5
V
25.3
Frp 

 0.365
gL
32 *150
Because we assumed dynamic similarity and compressible flow,
the model and prototype Froude numbers are equal. The
calculated Froude number for the prototype can be used to
interpolate the wave force coefficient. The resulting wave force
coefficient is .00265. The wave drag force can the be calculated:
Fwave  Cf *  *V 2 L2  .00265 *1.986 * 25.32 *1502  75949lbf
The Reynolds number is out of the range of the data given, so the
friction force coefficient can be found by plotting force
coefficient versus Reynolds number and using the trend line
equation.
CFfriction  0.0181Re .1456  0.0181(3.4e8) .1456  .00104
Now we know the friction force coefficient, so
the friction force can be calculated:
Ffriction  CFfriction*  *V 2 * L2  .00104 *1.986 * 25.32 *1502  29806lbf
F  Fwave  Ffriction  75949  29806  105755lbf  105750lbf
BME Application
When working with grafts in the human body, drag force
must be considered. This is especially important if the drag
force gets large because the graft could become mobile in the
body after being knocked out of place. Modeling is also a
concept that can be used in biofluids. Before actually
implanting a graft or other material into the body, a larger
scale model should be constructed. By having a larger scale
model, force test can be performed to determine whether or
not the material is suitable for implantation into the body.
Also, chemical tests can be performed to determine corrosion
rates and physical breakdown of the material.
Questions?