Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is P1 = 1.85 104 Pa, and the pipe diameter is 5.0 cm. At another point y = 0.30 m higher, the pressure is P2 = 1.05 104 Pa and the pipe diameter is 2.50 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.
To solve this problem, we can apply Bernoulli's equation, which states that the sum of the pressure, kinetic energy, and potential energy per unit volume remains constant along a streamline in steady, ideal flow.
First, let's define our variables:
P1 = pressure at the lower point = 1.85 × 10^4 Pa
P2 = pressure at the higher point = 1.05 × 10^4 Pa
D1 = diameter at the lower point = 5.0 cm = 0.05 m
D2 = diameter at the higher point = 2.50 cm = 0.025 m
y = vertical displacement between the two points = 0.30 m
v1 = speed of flow at the lower point (what we're trying to find)
v2 = speed of flow at the higher point (what we're trying to find)
A1 = cross-sectional area at the lower point = (π/4) × D1^2
A2 = cross-sectional area at the higher point = (π/4) × D2^2
ρ = density of water = 1000 kg/m^3 (assume it's constant)
Now let's find the answers to the questions:
(a) Find the speed of flow in the lower section:
Using Bernoulli's equation, we can write:
P1 + (1/2)ρv1^2 + ρgy1 = P2 + (1/2)ρv2^2 + ρgy2
Since the pipe is constricted and the flow is ideal (no energy losses due to friction), we can neglect the pressure difference between the two points and assume ρgy1 ≈ ρgy2. Also, since the flow is steady, the kinetic energy per unit volume is the same at both points, so (1/2)ρv1^2 ≈ (1/2)ρv2^2.
Therefore, the equation simplifies to:
P1 ≈ P2 + (1/2)ρv2^2
Solving for v1, we have:
v1^2 ≈ 2(P1 - P2)/ρ
v1 ≈ √(2(P1 - P2)/ρ)
Substituting the given values:
v1 ≈ √(2(1.85 × 10^4 - 1.05 × 10^4)/(1000))
v1 ≈ √(2(0.8 × 10^4)/(1000))
v1 ≈ √(0.16 × 10^4)
v1 ≈ 4 × 10^2 m/s
v1 ≈ 400 m/s
Therefore, the speed of flow in the lower section is approximately 400 m/s.
(b) Find the speed of flow in the upper section:
Using the same reasoning as in part (a), the equation becomes:
P1 + (1/2)ρv1^2 + ρgy1 = P2 + (1/2)ρv2^2 + ρgy2
Since the pressure difference is the same as before and ρgy1 ≈ ρgy2, we can simplify the equation to:
(1/2)ρv1^2 ≈ (1/2)ρv2^2
This means that the kinetic energy per unit volume is the same at both points, so:
v1^2 ≈ v2^2
Taking the square root of both sides, we have:
v1 ≈ v2
Therefore, the speed of flow in the upper section is approximately the same as the speed of flow in the lower section, which is 400 m/s.
(c) Find the volume flow rate through the pipe:
The volume flow rate (Q) is given by:
Q = A1v1 = A2v2
Substituting the given values:
Q = (π/4)(0.05^2)(400) = (π/4)(0.025^2)(400)
Q = 0.001964 m^3/s
Therefore, the volume flow rate through the pipe is approximately 0.001964 m^3/s.
To solve this problem, we can apply Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a steady ideal flow situation.
Bernoulli's equation states:
P + 0.5ρv^2 + ρgh = constant,
where P is the pressure, ρ is the density of the fluid, v is the velocity of the flow, g is the acceleration due to gravity, and h is the height of the fluid.
Let's solve part (a) first:
(a) Find the speed of flow in the lower section.
Given:
P1 = 1.85 x 10^4 Pa (pressure at the lower point)
D1 = 5.0 cm = 0.05 m (diameter at the lower point)
D2 = 2.50 cm = 0.025 m (diameter at the higher point)
y = 0.30 m (height difference)
To find the speed of flow in the lower section, we can use the continuity equation:
A1v1 = A2v2,
where A is the cross-sectional area of the pipe and v is the velocity.
The area of the pipe is given by A = πr^2, where r is the radius.
Let's start by calculating the radius at both points:
r1 = D1/2 = 0.05/2 = 0.025 m,
r2 = D2/2 = 0.025/2 = 0.0125 m.
The cross-sectional areas are given by:
A1 = πr1^2,
A2 = πr2^2.
Now, let's find the speed of flow at the lower section, v1:
Using Bernoulli's equation, we can write:
P1 + 0.5ρv1^2 + ρgh1 = P2 + 0.5ρv2^2 + ρgh2.
Since the fluid is flowing at the same height, h1 = h2 = 0, the equation simplifies to:
P1 + 0.5ρv1^2 = P2 + 0.5ρv2^2.
Rearranging and substituting the values, we get:
0.5ρv1^2 = P2 - P1.
Plugging in the pressure values, we have:
0.5ρv1^2 = 1.05 x 10^4 Pa - 1.85 x 10^4 Pa.
Substituting the density of water (ρ = 1000 kg/m^3), we get:
0.5 x 1000 kg/m^3 x v1^2 = -0.8 x 10^4 Pa.
Simplifying, we find:
v1^2 = -0.8 x 10^4 Pa x 2 / (1000 kg/m^3).
Taking the square root of both sides, we obtain:
v1 = sqrt((-0.8 x 10^4 Pa x 2) / (1000 kg/m^3)).
Calculating this expression will give us the speed of flow in the lower section.