Water at a pressure of 400,000Pa flows through a horizontal pipe at a speed of 1.50 m/s. The pipe narrows to 1/8 its original diameter. What is the flow speed through the narrow section?
To find the flow speed through the narrow section of the pipe, we can use the principle of conservation of mass. According to this principle, the mass flow rate of a fluid remains constant along a pipe.
The mass flow rate (m) can be calculated using the equation:
m = ρ * A * v
Where:
- m is the mass flow rate
- ρ is the density of the fluid
- A is the cross-sectional area of the pipe at a certain point
- v is the velocity of the fluid at that point
Since we are dealing with water, the density (ρ) is approximately 1000 kg/m³.
Let's assume the cross-sectional area of the pipe at the original diameter is A₁, and the cross-sectional area at the narrow section is A₂. We are given that the diameter of the pipe narrows to 1/8 of its original diameter.
Since the area of a circle is proportional to the square of its radius (A = π * r²), we can calculate the ratio of the cross-sectional areas:
(A₂ / A₁) = (r₂ / r₁)²
Given that the diameter is reduced to 1/8, the ratio of the radii will be:
(r₂ / r₁) = √(1/8)
Now, let's calculate the ratio of the cross-sectional areas:
(A₂ / A₁) = (√(1/8))² = 1/8
This means that the cross-sectional area at the narrow section is 1/8 of the original cross-sectional area.
Since the mass flow rate is constant, we can write the equation for mass flow rate at the original diameter (m₁) and at the narrow section (m₂) as follows:
m₁ = ρ * A₁ * v₁
m₂ = ρ * A₂ * v₂
Since the mass flow rate is constant, m₁ = m₂, we can equate the two expressions:
ρ * A₁ * v₁ = ρ * A₂ * v₂
Canceling ρ from both sides of the equation:
A₁ * v₁ = A₂ * v₂
Since we are interested in finding the flow speed through the narrow section (v₂), we rearrange the equation:
v₂ = (A₁ * v₁) / A₂
Given that the diameter narrows to 1/8, the ratio of the cross-sectional areas (A₂ / A₁) is 1/8. Substituting this value into the equation, we have:
v₂ = (1 * v₁) / (1/8)
Simplifying, we get:
v₂ = 8 * v₁
Therefore, the flow speed through the narrow section is 8 times the original flow speed.
In this case, the original speed is given as 1.50 m/s, so the flow speed through the narrow section will be:
v₂ = 8 * 1.50 m/s
v₂ = 12.00 m/s
Therefore, the flow speed through the narrow section of the pipe is 12.00 m/s.