Question

Water flows through a water hose at a rate of Q1 = 740 cm3/s, the diameter of the hose is d1 = 2.83 cm. A nozzle is attached to the water hose. The water leaves the nozzle at a velocity of v2 = 12.4 m/s.(a) Express the cross-sectional area of the hose, A1, in terms of d1.

(b) Calculate the numerical value of A1 in cm2.
(c) Express the speed of the water inside the hose, v1, in terms of V1 and A1.
(d) Calculate the numerical value of v1 in m/s.
(e) Express the cross-sectional area of the nozzle, A2, in terms of v1, v2 and A1.
(f) Calculate the numerical value of A2 in cm2

Answer 1

Given Information:

Inlet flow rate = V₁ = 740 cm3/s

Inlet diameter = d₁ = 2.83 cm

Exit velocity = v₂ = 12.4 m/s

Answer:

(a) A₁ = πd₁²/4

(b) A₁ = 6.29 cm²

(c) v₁ = V₁/A₁

(d) v₁ = 117.65 m/s

(e) A₂ = A₁(v₁/v₂)

(f) A₂ = 59.68 cm²

Step-by-step explanation:

(a) Express the cross-sectional area of the hose, A₁ in terms of d₁

We know that area is given by

A₁ = πr₁²

we know that diameter is related to radius as

r₁ = d₁/2

A₁ = π(d₁/2)²

A₁ = πd₁²/4

(b) Calculate the numerical value of A₁ in cm²

The numerical value of inlet area A₁ of water hose is

A₁ = π(2.83)²/4

A₁ = 6.29 cm²

(c) Express the speed of the water inside the hose, v₁ in terms of V₁ and A₁

The volumetric flow rate is area times velocity

V₁ = A₁v₁

which becomes

v₁ = V₁/A₁

(d) Calculate the numerical value of v₁ in m/s

The numerical value of inlet velocity v₁ of water hose is

v₁ = 740/6.29

v₁ = 117.65 m/s

(e) Express the cross-sectional area of the nozzle, A₂ in terms of v₁, v₂ and A₁

As we already know V₁ = A₁v₁ and V₂ = A₂v₂ therefore,

A₁v₁ = A₂v₂

A₂ = A₁(v₁/v₂)

(f) Calculate the numerical value of A₂ in cm²

The numerical value of exit area A₂ of water hose is

A₂ = 6.29*(117.65/12.4)

A₂ = 59.68 cm²