Water flows through a 4.0-cm-diameler horizontal pipe ata speed uf 1.3 m/s. The pipe then narrows down to a diameter of 2.0 cm. Ignoring viscosity, what is the pressure difference between the wide and narrow sections of the pipe
figure the flow in the smaller pipe using the law of continuinty
Speed1*area1=Speed2*area2
then, having the two speeds, use bernouli's equation to find pressure.
4.2
To find the pressure difference between the wide and narrow sections of the pipe, we can use the principle of continuity equation:
A₁v₁ = A₂v₂
where A₁ and A₂ are the cross-sectional areas of the wide and narrow sections of the pipe, and v₁ and v₂ are the velocities of the water flow through those sections, respectively.
Given:
Diameter of wide section (D₁) = 4.0 cm
Diameter of narrow section (D₂) = 2.0 cm
Velocity of water flow (v₁) = 1.3 m/s
First, we need to calculate the areas of the wide and narrow sections:
Area of wide section (A₁) = (π/4) * D₁²
= (π/4) * (4.0 cm)²
= π * (2.0 cm)²
= 4π cm²
Area of narrow section (A₂) = (π/4) * D₂²
= (π/4) * (2.0 cm)²
= π cm²
Now, we can substitute the given values into the continuity equation:
A₁v₁ = A₂v₂
(4π cm²)(1.3 m/s) = (π cm²)(v₂)
Simplifying the equation:
v₂ = (4π cm²)(1.3 m/s) / π cm²
v₂ = 5.2 m/s
Now, we can calculate the pressure difference between the wide and narrow sections of the pipe using Bernoulli's equation:
ΔP = 1/2 ρ (v₂² - v₁²)
where ΔP is the pressure difference, and ρ is the density of water.
Assuming the density of water (ρ) is 1000 kg/m³, we can calculate the pressure difference:
ΔP = 1/2 (1000 kg/m³) (5.2 m/s)² - (1.3 m/s)²
= 1/2 (1000 kg/m³) (27.04 m²/s²) - (1.69 m²/s²)
= 13,520 N/m² - 1,156 N/m²
= 12,364 N/m²
Therefore, the pressure difference between the wide and narrow sections of the pipe is 12,364 N/m².
To find the pressure difference between the wide and narrow sections of the pipe, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid.
Bernoulli's equation is given by:
P1 + 1/2 * ρ * v1² + ρ * g * h1 = P2 + 1/2 * ρ * v2² + ρ * g * h2
Where:
P1 and P2 are the pressures at points 1 and 2, respectively.
v1 and v2 are the velocities at points 1 and 2, respectively.
ρ is the density of the fluid.
g is the acceleration due to gravity.
h1 and h2 are the heights at points 1 and 2, respectively.
In this case, we can assume that the height difference between the wide and narrow sections of the pipe is negligible, and there is no change in the height of the fluid. Therefore, h1 and h2 can be ignored.
The equation can be simplified to:
P1 + 1/2 * ρ * v1² = P2 + 1/2 * ρ * v2²
Since we are interested in the pressure difference, we can rearrange the equation to solve for ΔP (pressure difference):
ΔP = P2 - P1 = 1/2 * ρ * (v1² - v2²)
Now let's plug in the given values:
d1 = 4.0 cm = 0.04 m (diameter of the wide section)
d2 = 2.0 cm = 0.02 m (diameter of the narrow section)
v1 = 1.3 m/s (velocity in the wide section)
To find v2, we can use the continuity equation, which states that the flow rate is constant at different sections of an incompressible fluid. The flow rate (Q) is given by:
Q = A1 * v1 = A2 * v2
Where:
A1 and A2 are the cross-sectional areas at points 1 and 2, respectively.
We can use the equation for the area of a circle:
A = π * r²
Where:
r is the radius of the circle.
Let's find the values for A1 and A2:
A1 = π * (d1/2)² = π * (0.04/2)²
A2 = π * (d2/2)² = π * (0.02/2)²
Now we can calculate v2 using the continuity equation:
A1 * v1 = A2 * v2
v2 = (A1 * v1) / A2
Finally, we can calculate the pressure difference (ΔP):
ΔP = 1/2 * ρ * (v1² - v2²)
To complete the calculation, we need the density of water (ρ), which is approximately 1000 kg/m³.
Now you have all the necessary information to calculate the pressure difference between the wide and narrow sections of the pipe.