written 5.0 years ago by |
Also determine the ratio of drag (Resistance) between the model and its prototype . Take the value of kinematic viscosities for sea water and air as 0.012 stokes and 0.016 stokes respectively. The density for sea water and air is given as 1030 kg/m$^{3}$ and 1.24/m$^{3}$ respectively.
Solution:
For prototype
Speed , $V_{p}=10 m/s$
Fluid sea water
Kinematic viscosity , $Vp=0.012 \ stokes$
=$0.012 cm^{2}$/s
=$0.012 \times 10^{-4} m^{2}/s$
density, $\rho_p=1030kg/m^{3}$
for model:-
fluid = air
Kinematic viscosity $V_{m}=0.016 \ stokes$
=$0.016 cm^{2}/s$
=$0.016\times 10^{-4}m^{2}/s$
Density $\rho_m=1.24kg/m^{3}$
Also $\frac{length \ of \ prototype}{length \ of \ model}$=$\frac{L_p}{L_m} = 30.0$
Let velocity of air in model = $V_{m}$
For dynamic similarity between model and sub marine the viscous resistance is to be over come and hence for fully submerged sub marine , the Reynolds number for model and prototype should be same
$\frac{\rho p . V_p . D_p}{\mu_p}$=$\frac{\rho_m V_m D_m}{\mu_m}$
$\frac{V_p D_p}{(\mu / \rho)_p}$=$\frac{V_m D_m}{(\mu / \rho)_m}$
$\frac{V_p D_p}{V_p}=\frac{V_m D_m}{V_m}$
$V_m=\frac{V_m}{V_p}\times \frac{D_p}{D_m}\times V_p$
=$\frac{0.016 \times 10^{-4}}{0.012\times 10^{-4}} \times 30 \times 10m/s$
=$\frac{0.016}{0.012}\times 30\times 10$
=400m/s
Ratio of drag force(resistance)
Drag force= mass $\times$ Acceleration
=$\rho L^{3}\times\frac{V}{t}$
=$\rho L^{2}.\frac{L}{t}\times V$
=$\rho L^{2}V^{2}$ -----------since [$\frac{L}{t}=V$]
$\frac{F_p}{F_m}= \frac{\rho_p L_p^2 V_p^2}{\rho_m L_m^2 V_m^2}$
= $\frac{\rho_p}{\rho_m} \times (\frac{L_p}{L_m})^2 \times (\frac{V_p}{V_m})^2 $
=$\frac{1030}{1.24}\times 30^{2}\times(\frac{10}{400})^{2}$
=467.22