Answer:
[tex]v_2\\[/tex] =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
[tex]A_1v_1=A_2v_2[/tex]
Given Data:
velocity of liquid in 12 in diameter pipe=1.65 ft/s=[tex]v_1\\[/tex]
[tex]D_1\\[/tex] is pipe diameter of 12 in
[tex]D_2[/tex] is the diameter of jet= 3 in
Required:
[tex]v_2\\[/tex] velocity in a 3-in-diameter jet exiting from a nozzle attached to the pipe=?
Solution:
From the equation of continuity
[tex]A_1v_1=A_2v_2[/tex]
[tex]v_2=\frac{A_1v_1}{A_2}\\[/tex]
It will become:
[tex]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[/tex]
[tex]v_2\\[/tex] =velocity in a 3-in-diameter jet exiting from a nozzle attached to the pipe=26.4 ft/s
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
