Answer: 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