Water flows through a water hose at a rate of Q1 = 620 cm3/s, the diameter of the hose is d1 = 2.27 cm. A nozzle is attached to the water hose. The water leaves the nozzle at a velocity of v2 = 12.8 m/s.

a. Enter an expression for the cross-sectional area of the hose, A1, in terms of its diameter, d1. sig.gif?tid=6C65-32-87-4C-B016-17686

b. Calculate the numerical value of A1, in square centimeters.

c. sig.gif?tid=6C65-32-87-4C-B016-17686Enter an expression for the speed of the water in the hose, v1, in terms of the volume flow rate Q1 and cross-sectional area A1. sig.gif?tid=6C65-32-87-4C-B016-17686

d. Calculate the speed of the water in the hose, v1 in meters per second

e. sig.gif?tid=6C65-32-87-4C-B016-17686Enter an expression for the cross-sectional area of the nozzle, A2, in terms of v1, v2 and A1. sig.gif?tid=6C65-32-87-4C-B016-17686

f. Calculate the cross-sectional area of the nozzle, A2 in square centimeters.

Question

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Respuesta :

moya1316 moya1316 · Answer 1 · 2019-12-05T00:34:03+01:00

Answer:

a) A₁ = π d₁² / 4 , b) A₁ = 4.05 cm² , c) v₁ = Q / A₁ , d) v₁ = 153 m / s , e) A₂ = A₁ v₁ / v₂, f) A₂= 48.4 cm²

Explanation:

This is a fluid mechanics exercise, let's use the continuity equation

Q = A₁ v₁ = A₂ v₂

Where Q is the flow, A are the areas and v the speeds

a) the area of the hose (A₁) that has a circular section is

A₁ = π r₁²

Since the radius is half the diameter

A₁ = π (d₁ / 2)²

A₁ = π d₁² / 4

b) let's calculate

A₁ = π 2.27²/4

A₁ = 4,047 cm²

A₁ = 4.05 cm²

c) Let's use the left part of the initial equation

Q = A₁ v₁

v₁ = Q / A₁

d) let's calculate the value

v₁ = 620 / 4.05

v₁ = 153 m / s

e) We use the right part of the equation

A₁ v₁ = A₂ v₂

A₂ = A₁ v₁ / v₂

f) Calculate

A₂ = 4.05 153/12.8

A₂= 48.4 cm²