Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is 1.70 105 Pa and the pipe radius is 2.50 cm. At the higher point located at y = 2.50 m, the pressure is 1.30 105 Pa and the pipe radius is 1.60 cm. (a) Find the speed of flow in the lower section.
(b) Find the speed of flow in the upper section.
(c) Find the volume flow rate through the pipe.
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To find the speed of flow in the lower section of the pipe, we can use Bernoulli's equation, which relates the pressure, density, and velocity of a fluid in steady flow.
Bernoulli's equation is given by: P₁ + (1/2)ρv₁² + ρgh₁ = P₂ + (1/2)ρv₂² + ρgh₂
Where:
P₁ and P₂ are the pressures at points 1 and 2, respectively.
ρ is the density of the fluid.
v₁ and v₂ are the velocities at points 1 and 2, respectively.
g is the acceleration due to gravity.
h₁ and h₂ are the heights of points 1 and 2, respectively.
(a) To find the speed of flow in the lower section, we need to use the given information at the lower point. We have:
P₁ = 1.70x10⁵ Pa (pressure at the lower point)
h₁ = 0 (as the reference point is set at the lower point, the height here is 0)
P₂ = ? (pressure at the higher point)
h₂ = 2.50 m (height at the higher point)
We also need the density of water, which is approximately 1000 kg/m³.
Substituting these values into Bernoulli's equation, we can solve for v₁:
P₁ + (1/2)ρv₁² + ρgh₁ = P₂ + (1/2)ρv₂² + ρgh₂
1.70x10⁵ + (1/2)(1000)v₁² + (1000)(9.81)(0) = P₂ + (1/2)(1000)v₂² + (1000)(9.81)(2.50)
1.70x10⁵ + (500)v₁² = P₂ + (500)v₂² + (24525)
Since the pipe is constricted, we can assume that the volume flow rate is constant. This implies that the velocity at both points is the same, v₁ = v₂ = v.
1.70x10⁵ + (500)v² = P₂ + (500)v² + (24525)
Rearranging the equation, we have:
1.70x10⁵ - (500)v² = P₂ - (500)v² + (24525)
Simplifying, we find:
P₂ - 1.70x10⁵ = 24525
Therefore, P₂ = 1.70x10⁵ + 24525 = 1.72x10⁵ Pa
Now, we can substitute these values back into Bernoulli's equation to solve for v₁:
1.70x10⁵ + (1/2)(1000)v₁² + (1000)(9.81)(0) = 1.72x10⁵ + (1/2)(1000)v₂² + (1000)(9.81)(2.50)
1.70x10⁵ + (500)v₁² = 1.72x10⁵ + (500)v₂² + (24525)
Since v₁ = v and substituting P₂ = 1.72x10⁵ Pa into the equation, we have:
1.70x10⁵ + (500)v² = 1.72x10⁵ + (500)v² + (24525)
Simplifying, we find:
0 = 24525
This means that the equation cannot be balanced unless there is a mistake in the given values or the calculations. Please double-check the values or equations provided or refer to the correct source of information for this problem.