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 25 cm and a torpedo model aligned along the long axis of the pipe. The model has a 5.9 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

Area 1 = pi r1^2

r1 = Diameter1/2 = 0.25/2 = 0.125 m

A1 = 3.14 * 0.125* 0.125 = 0.0490625 m^2

A2 = /4*(D^2 - d^2) = 3.14/4*(0.25^2 - 0.059^2) = 0.0463 m^2

a) now use the equation of continuity as A1*V1 = A2*V2

where A is area and V is velocity

so

0.04906 * V1 = 0.0463* 3.3

V1 = 3.11m/s (answer part A)
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b) use bernoullis theroem as P1 + 1/2*V1^2 = P2 + 1/2**V2^2

where P is pressure, V is velocity , rho is density (of water = 1000 kg/m^3)

(P1 - P2) = 1/2**(V2^2 - V1^2)

(P1 - P2) = 1/2*1000*(3.3^2 - 3.11^2)

(P1-P2) = 609.5 Pa (answer part B)

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