Water is flowing at a rate of 3 m/sec in a horizontal pipe under a pressure of 200,000 N/m². If the pipe narrows to half its original diameter
-What is the new speed of flow? -What is the new pressure?
-How does the flow rate through the narrow section compare with the flow
rate through the wider section?
To solve this problem, we can use the principle of continuity equation and the Bernoulli's equation.
1. New Speed of Flow:
According to the principle of continuity, the mass flow rate remains constant in an incompressible fluid. Therefore, we can write:
A1 * V1 = A2 * V2
where A1 and V1 represent the cross-sectional area and velocity of flow at the wider section of the pipe, and A2 and V2 represent the corresponding values at the narrower section.
Since the area is related to the diameter squared (A = π * D²/4), when the diameter is halved, the area becomes one-fourth of its original value. Therefore, we have:
A2 = 1/4 * A1
Let's assume the original diameter is D1 and the new diameter is D2. Since D2 is half of D1, we have:
D2 = 1/2 * D1
Substituting these values into the continuity equation, we get:
(π * D1²/4) * 3 m/sec = (π * (D1/2)²/4) * V2
(π * D1²) * 3 = (π * (D1/2)²) * V2
Canceling out π and 4 on both sides:
3 * D1² = (D1/2)² * V2
Simplifying and solving for V2 gives:
V2 = 3 * 4 / 1²
V2 = 12 m/sec
Therefore, the new speed of flow is 12 m/sec.
2. New Pressure:
According to Bernoulli's equation, the pressure changes as the speed of flow changes in a pipe. The equation is given by:
P1 + (1/2) * ρ * V1² = P2 + (1/2) * ρ * V2²
where P1 and V1 represent the initial pressure and velocity of flow, and P2 and V2 represent the corresponding values at the narrower section.
Since the density (ρ) and velocity (V) remain constant, the equation simplifies to:
P1 = P2 + (1/2) * ρ * (V2² - V1²)
Since the density of water is constant, we can substitute the given values into the equation:
P1 = 200,000 N/m²
V1 = 3 m/sec
V2 = 12 m/sec
P2 = 200,000 + (1/2) * 1000 * (12² - 3²)
P2 = 200,000 + 0.5 * 1000 * (144 - 9)
P2 = 200,000 + 0.5 * 1000 * 135
P2 = 200,000 + 0.5 * 135,000
P2 = 200,000 + 67,500
P2 = 267,500 N/m²
Therefore, the new pressure is 267,500 N/m².
3. Flow Rate Comparison:
The flow rate through a pipe is given by:
Q = A * V
where Q represents the flow rate and A and V represent the cross-sectional area and velocity of flow, respectively.
Since we know the velocity at the wider section is 3 m/sec and at the narrower section is 12 m/sec, and the area at the narrower section is one-fourth of the wider section, we can compare the flow rates as follows:
Q2 / Q1 = (A2 * V2) / (A1 * V1)
Since A2 = 1/4 * A1:
Q2 / Q1 = [(1/4) * A1 * V2] / (A1 * V1)
Q2 / Q1 = (1/4) * (V2 / V1)
Q2 / Q1 = (1/4) * (12/3)
Q2 / Q1 = 1
Therefore, the flow rate through the narrow section is the same as the flow rate through the wider section.
To find the new speed of flow, we can use the principle of continuity. According to this principle, the product of the cross-sectional area of the pipe and the speed of flow remains constant.
Let's assume the original diameter of the pipe is D, and the new diameter is D/2.
1. Cross-sectional area of the original pipe (A1) can be calculated using the formula for the area of a circle: A1 = π(D/2)² = (π/4)D².
2. Cross-sectional area of the narrow section (A2) can be calculated similarly: A2 = π[(D/2)/2]² = (π/16)D².
Now, using the principle of continuity, we can set up the equation:
A1 * v1 = A2 * v2,
where v1 is the original speed of flow and v2 is the new speed of flow.
Applying our previously calculated values:
(π/4)D² * 3 = (π/16)D² * v2.
Simplifying the equation:
(π/4) * 3 = (π/16) * v2,
3/4 = (1/16) * v2,
v2 = 3/4 * 16,
v2 = 12 m/sec.
Therefore, the new speed of flow is 12 m/sec.
To find the new pressure, we can use Bernoulli's equation, which relates the pressure and speed of flow in a fluid.
1. Assuming no loss of height, we can cancel out the potential energy term in Bernoulli's equation.
2. The equation can then be simplified to:
P1 + (1/2) * ρ * (v1)² = P2 + (1/2) * ρ * (v2)²,
where P1 is the original pressure, P2 is the new pressure, v1 is the original speed of flow, v2 is the new speed of flow, and ρ is the density of water.
Substituting the given values:
200,000 + (1/2) * ρ * (3)² = P2 + (1/2) * ρ * (12)².
Simplifying the equation:
200,000 + (9/2) * ρ = P2 + 72 * ρ,
P2 = 200,000 + (9/2) * ρ - 72 * ρ.
As we don't have a specific value for the density of water, we cannot calculate the new pressure accurately without that information. However, we can say that the new pressure will be different from the original pressure due to the change in speed of flow.
The flow rate through the narrow section will be greater than the flow rate through the wider section. This is because the flow rate is directly proportional to the cross-sectional area of the pipe, and the new cross-sectional area of the narrow section is smaller than the original cross-sectional area. As the speed of flow through the narrow section increases, the flow rate will also increase.