Astudent team is to design a human-powered submarine for a design competition. the overall length of the prototype submarine is 4.85 m, and its student designers hope that it can travel fully submerged through water at 0.440 m/s. the water is freshwater (a lake) at t = 15 °c. the design team builds a one-fifth 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°c and at one standard atmosphere pressure. 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 5.75 n. "estimate the drag force" on the prototype submarine at the given conditions.

drag force FDp = 26.93μN

Explanation: to begin, let us define the parameters given

velocity of the submarine Vp = 0.440 m/s

For water at T = 15⁰C;

S = 999.1 kg/m³  and  μ= 1.138*10⁻³kg/ms

For air at T = 25⁰C;

S = 1.184 kg/m³  and  μ= 1.849*10⁻⁵kg/ms

To begin, recall that Reynolds number;

Rе = рVL/μ = SmVmLm/μm = SpVpDp/μp

     Vm = Vp (μm/μp)(Sp/Sm)(Lp/Lm)

       Vm = 0.440 [1.849*10⁻⁵/ 1.138*10⁻³] × [999.1/1.184] × Lp/(Lp/s)

        Vm = 30.163 m/s

having gotten  the velocity as 30.163m/s, we can solve for the Drag coefficient;

Drag coefficient (CD) = 2FD/SL²V²

where the similarity between model and prototype is;

(CD)p = (CD)m

comparing we have;

2FDp/SpLp²Vp² = 2FDm/SmLm²Vm²

FDp = FDm (Sp/Sm) (Vp/Vm)² (Lp/Lm)²

  FDp= 6 * (999.1/1.184) (0.44/30.163) (Lp/Lp/5)²

FDp = 26.93μN  

answer:

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