Answer:
The velocity in the water tunnel is 128.717m/s
Explanation:
To find the result we need to Apply dynamic symiliraty, that is,
[tex]\frac{V_m I_m}{\upsilon_m}= \frac{VI}{\upsilon}[/tex]
Where [tex]\upsilon_i[/tex] is the kinematic viscosity, V the velocity and I the lenght.
To find the kinematic viscosity it is necessary search in the table with this properties (I attached it)
To this kind of fluids, the properties to 15.6°c (Seawater) and 20°c (Water)are:
[tex]\upsilon_m= 1.17*10^{-6}m^2/s[/tex] (Seawater)
[tex]\upsilon_m = 1.004*10^{-6}m^2/s[/tex] (Water)
Replacing,
[tex]\frac{V_m I_m}{\upsilon_m}= \frac{VI}{\upsilon}[/tex]
Solving to V_m
[tex]V_m = \frac{I}{I_m}\frac{\upsilon_m}{\upsilon}V[/tex]
[tex]V_m= \frac{5}{1} * \frac {1.004*10^{-6}}{1.17*10^{-6}} 30[/tex]
[tex]V_m = 128.717m/s[/tex]
The velocity in the water tunnel is 128.717m/s
