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

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 25.0 cm and a torpedo model aligned along the long axis of the pipe.The model has a 5.00 cm diameter and is to be tested with water flowing past it at 2.50 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?

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Answer #1

(a) cross-section area A = (π/4)(25cm)² = 491 cm²

reduced area Ar = (π/4)(25² - 5²)cm² = 471 cm²

The flow rate must be the same at both points: Q = v * Ar = v' * A

2.5m/s * 471cm² = v' * 491cm²

v' = 2.4 m/s

(b) Bernoulli: Δp = ½ρ(v² - v'²) = ½ * 1000kg/m³ * (2.5² - 2.4²)m²/s²

Δp = 24 500 Pa

Kindly rate :)

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