Respuesta :
Answer:
Explanation:
Given
scale i.e. [tex]L_r=1:15[/tex]
Using Reynolds number similarity
[tex](Re)_m=(Re)_p[/tex]
[tex](\frac{Vl}{\nu })_m=(\frac{Vl}{\nu })_p[/tex]
Properties of air
[tex]\nu _{air}=1.57\times 10^{-4} ft/s[/tex]
Properties of sea water
[tex]\nu _{sea}=1.26\times 10^{-5} ft/s[/tex]
[tex]\left ( \frac{V_ml_M}{\nu _m}\right )=\left ( \frac{V_pl_p}{\nu _p}\right )[/tex]
[tex]V_p=V_m\left ( \frac{l_m}{l_p}\right )\left ( \frac{\nu _p}{\nu _m}\right )[/tex]
[tex]V_p=180\times \frac{1}{15}\times \frac{1.26\times 10^{-5}}{1.57\times 10^{-4}}[/tex]
[tex]V_p=0.963\ ft/s[/tex]
