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

part a

cross-section area , Ar' = (/4)(25cm)² = 490.87 cm²
reduced area , Ar = (/4)(25² - 5.9²)cm² = 463.534 cm²

using equation of continuity

The flow rate must be the same at both points: Q = v * Ar = v' * Ar'
3.3m/s * 463.534 cm² = v' * 490.87 cm²
v' = 3.11622 m/s

part b

Bernoulli:theorem

p = ½(v² - v'²) = ½ * 1000kg/m³ * (3.3² - 3.11622²)m²/s²
p = 589.586 kg/m-s2 =589.586 n/m2 = 589.586 Pa

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.