Models of torpedoes are sometimes tested in a horizontal pipe of flowing water,

Question

Models of torpedoes are sometimes tested in a horizontal pipe of flowing water, much as a wind tunnel is used to test model airplanes. Consider a circular pipe of internal diameter 28 cm and a torpedo model aligned along the long axis of the pipe. The model has a 4.9 cm diameter and is to be tested with water flowing past it at 3.3 m/s. (a) With what speed must the water flow in the part of the pipe that is unconstricted by the model? (b) What will the pressure difference be between the constricted and unconstricted parts of the pipe?

Explanation / Answer

a)
By flow continuity, the water volume flow through the unconstricted part must be equal to the water volume flow through the constricted part:

Let q_un and q_con are the water volume flow per unit time in the unconstrict and constrict parts respectively:

By continuity:
q_un = q_con

Thus:
A_un*v_un = A_con*v_con
where A is the cross section area of water flow and v is the velocity of the water flow.

In this case:
A_un = pi*(27/2)^2
A_con = pi*[(27/2)^2 - (5.2/2)^2]
v_con = 3.1
v_un can be calculated

Note that although v is in m/s and A is in cm^2, there is no need to  standardize the unit of spatial dimensions. This is because the conversion factor on both sides of the equation will be cancelled out.


b)
By Bernallius (or some name with similar spelling...) equation:
P1 + pgh1 + pv1^2/2 = P2 + pgh2 + pv2^2/2
where p is the density of water in SI unit (kg/m^3), which u can easily find it online.

In this case, the pipe is horizontal so h1 = h2. From (a) we have v1 and v2. The pressure difference P1 - P2 can be easily calculated.

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