Question

If the velocity of a liquid is 1.65 ft/s in a 12-in-diameter pipe, what is the velocity in a 3-in-diameter jet exiting from a nozzle attached to the pipe?

Answer 1

Answer:

$v_2$ =velocity in a 3-in-diameter jet exiting from a nozzle attached to the pipe=26.4 ft/s

Explanation:

From the equation of continuity

$A_1v_1=A_2v_2$

Given Data:

velocity of liquid in 12 in diameter pipe=1.65 ft/s=$v_1$

$D_1$ is pipe diameter  of 12 in

$D_2$ is the diameter of jet= 3 in

Required:

$v_2$ velocity in a 3-in-diameter jet exiting from a nozzle attached to the pipe=?

Solution:

From the equation of continuity

$A_1v_1=A_2v_2$

$v_2=\frac{A_1v_1}{A_2}$

It will become:

$v_2=v_1\frac{\pi D_1^2/4}{\pi D_2^2/4}\\v_2=v_1\frac{ D_1^2}{ D_2^2}\\v_2=1.65*\frac{ 12^2}{ 3^2}\\v_2=26.4 ft/s$

$v_2$ =velocity in a 3-in-diameter jet exiting from a nozzle attached to the pipe=26.4 ft/s

Answer 2

Answer:

v₂ = 26.4 ft/s

The velocity in a 3-in-diameter jet exiting from a nozzle attached to the pipe is 26.4 ft/s

Explanation:

Since the same volume of liquid pass through the two pipes per unit time.

V₁ = V₂ ......1

And,

V = Av ......2

Where, V = volumetric flow rate,

A is cross sectional area of pipe

v is the speed of fluid.

Substituting equation 2 to 1

A₁v₁= A₂v₂ ........3

The cross sectional area of a pipe can be defined mathematically as;

A = πd²/4 .......4

Substituting into equation 3

πd₁²v₁/4 = πd₂²v₂/4

divide both sides by π/4

d₁²v₁ = d₂²v₂

making v₂ the subject of formula

v₂ = d₁²v₁/d₂²..........5

Given:

d₁ = 12 inches

v₁ = 1.65 ft/s

d₂ = 3 inches

Substituting the values;

v₂ = (1.65 × 12²)/3²

v₂ = 26.4 ft/s

The velocity in a 3-in-diameter jet exiting from a nozzle attached to the pipe is 26.4 ft/s