Question

(a) What is the pressure drop (in N/m2) due to the Bernoulli effect as water goes into a 3.20 cm diameter nozzle from a 9.10 cm diameter fire hose while carrying a flow of 42.0 L/s?

(b) To what maximum height (in m) above the nozzle can this water rise? (The actual height will be significantly smaller due to air resistance.)

Answer 1

Ans: We know the Bernoulli equation

P₁ + 1/2( V₁²) + gh₁ = P₂ + 1/2( V2²) + gh₂ -------------------------------- (1)

Where,

P_1, P₂ be the pressure at the fire hose and nozzle

is the density of fluid

V₁ , V₂ be the velocities of the fluid at fire hose and nozzle

when two points are at the same height (h₁ = h₂ )

P₁ + 1/2( V₁²) = P₂ + 1/2( V2²) -------------------------------- (2)

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Diameter of the fire hose (d₁) = 9.10 cm = 0.0910 m.

Radius of the fire hose (r₁) = 0.0455 m.

Diameter of the nozzle (d2) = 3.20 cm = 0.0320 m.

Radius of the fire nozzle (r₂ ) = 0.016 m.

Density of fluid () = 1000 kg/m³.

Flow (Q) = 40 L/s

Q=A₁V₁

V₁ = Q/ A₁ = 40 x 10^-3/ r₁²

= 40 x 10^-3 / 3.14 x (0.0455)²

= 6.153 m/s

  Q=A2V2

V2 = Q/ A2 = 40 x 10^-3/ r₁²

= 40 x 10^-3 / 3.14 x (0.016)²

= 57.6 m/s

Therefor eq. (2) becomes

P₁ + 1/2( V₁²) = P₂ + 1/2( V2²)

P₂ -   P₁ = 1/2( V2²) - 1/2( V₁²)

  P = (1/ 2) (V2² - V₁²)

= (1/2) x 1000 ( 57.6² - 6.15²)

= 1.639 x 10⁶ N/m²

therefor the pressure drop is 1.639 x 10⁶ N/m²

We have to calculate the max height of the fluid therefor eq. becomes

Patm + 1/2( V₁²) + gh₁ = Patm + 1/2( V2²) + gh₂ ----------- (3)

Here, h₁ is the height of the fluid that leaving

and h₂ is the max height that fluid reaching

Eq (3) becomes

  

Patm + 1/2( V₁²) + 0 = Patm + 0 + gh₂

h₂ = {(1/2)V₁² } / g

h₂ = {0.5 x (57.6)² }/ 9.8

h₂ = 169.2 m

Therefor the max height that fluid raise to the above nozzle is 169.2 m

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