ModelingModeling

Amy Stephens

BIEN 301

15 February 2007

Problem 5.76Problem 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

SketchSketch

AssumptionsAssumptions

Geometric similarityKinematic similarityDynamically similarityScaling law is validSteady flowLiquidIncompressible

SolutionSolution

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

gL

VF

VL

r

eR

Note: Gravity for this problem is 32ft/s2

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 0.2 0.3 0.4 0.5 0.6

The following table shows the resulting Reynolds and Froude numbers:

The density and viscosity of each fluid should taken from White Tables A.1 and A.3 (conversion necessary):

23

23

/*5232.2 ,/986.1 :Seawater

/*509.2 ,/937.1 :Water

ftslbeftslug

ftslbeftslug

22

22

LV

FC

LV

FC

wave

wave

friction

friction

dF

dF

From the drag force data given, the force coefficients can be found by using the following equations:

The table below shows the resulting force coefficients:

V (ft/s) 0.8 1.6 2.4 3.2 4.0 4.8

Cdfric .00322 .00287 .00273 .00262 .00254 .00247

Cdwave .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:

365.0150*32

3.25

84.35232.2

150*3.25*986.1R

25.3ft/s /3048.0

/1 x

1

/1144.5 x 15

e

gL

VF

ee

VLsm

sft

kn

smekn

rp

p

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:

lbfLVCF fwave 75949150*3.25*986.1*00265.** 2222

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.

00104.)84.3(0181.0Re0181.0 1456.1456. eC frictionF

Now we know the friction force coefficient, so the friction force can be calculated:

lbflbfFFF

lbfLVCF

frictionwave

Ffriction friction

1057501057552980675949

29806150*3.25*986.1*00104.*** 2222

BME ApplicationBME 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?Questions?