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FLUID MECHANICS - CASE STUDY SOLUTION

An Attack Submarine

 

The submarine model is built with a scale of 1:10. That is, Lp/Lm = 10

Since submarines usually cruise well below the water surface, the Froude number is not important in this study. In addition, it is assumed that the compressibility effects are negligible, hence the Mach number has no role in this study as well. In order for dynamic similitude, both the Reynolds number and dimensionless drag should be identical for the model and prototype.

The Reynolds number for the prototype is calculated as

     Rep = ρpVpLpp = (1,030)(4)(10Lm)/(1.2)(10-3)

           = 34.3×106 Lm

From the properties table of water, the viscosity and density of fresh water at a temperature of 40oC are 0.7×10-3 N-s/m2 and 992 kg/m3, respectively.

Equate the Reynolds number for the prototype and model and rearrange terms to yield an expression for the model's velocity

     Rem = Rep
      
ρmVmLmm = 34.3×106 Lm
     992 Vm / 0.7×10-3= 34.3×106
     Vm = 24.2 m/s

Now, equating the dimensionless drag force for the prototype and model yields

     

In this problem, it was assumed that both the Re and dimensionless drag force relationship where known. They could have been derived using the Buckingham Pi Theorem.

     
   
 
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