water is flowing at 3 m/s in a horizontal pipe under a pressure of 200 kpa. the pipe, which narrows to half its original diameter, is laid out on a level ground? a) what is the speed of flow in the narrow section? b) what is the pressure in the narrow section? c)how do the volume flow rates in the two sections?
To answer these questions, we can apply the principles of continuity equation and Bernoulli's equation.
a) To find the speed of flow in the narrow section, we can use the continuity equation, which states that the product of cross-sectional area and velocity is constant for an incompressible fluid.
The equation can be written as:
A1 * v1 = A2 * v2
Where:
A1 is the cross-sectional area of the wider section
v1 is the speed of flow in the wider section
A2 is the cross-sectional area of the narrow section (half of A1)
v2 is the speed of flow in the narrow section (what we need to find)
From the given information, we know that A2 = 1/2 * A1 and v1 = 3 m/s.
Plugging these values into the continuity equation:
A1 * 3 = (1/2 * A1) * v2
v2 = 6 m/s
Therefore, the speed of flow in the narrow section is 6 m/s.
b) To find the pressure in the narrow section, we can use Bernoulli's equation, which relates the pressure, density, speed, and height at two different points in a fluid flow.
The equation can be written as:
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2
Where:
P1 is the pressure in the wider section (200 kPa)
P2 is the pressure in the narrow section (what we need to find)
ρ is the density of water (assumed constant)
v1 is the speed of flow in the wider section (3 m/s)
v2 is the speed of flow in the narrow section (6 m/s)
g is the acceleration due to gravity (assumed constant)
h1 and h2 are the heights of the wider and narrow sections (both on level ground, so h1 and h2 are the same)
Since both sections are at the same height, the terms involving height cancel out.
Plugging in the known values:
200 kPa + 1/2 * ρ * (3 m/s)^2 = P2 + 1/2 * ρ * (6 m/s)^2
Simplifying, we find:
200 kPa + 4.5 * ρ = P2 + 18 * ρ
Since ρ is constant on both sides and cancels out, we get:
200 kPa + 4.5 = P2 + 18
P2 = 200 kPa + 4.5 - 18
P2 = 186.5 kPa
Therefore, the pressure in the narrow section is 186.5 kPa.
c) The volume flow rate in a pipe is given by the equation:
Q = A * v
Where:
Q is the volume flow rate
A is the cross-sectional area
v is the velocity
The volume flow rate is directly proportional to the product of cross-sectional area and velocity.
Since the cross-sectional area of the narrow section is half of the wider section (A2 = 1/2 * A1), and the speed of flow in the narrow section is twice the wider section (v2 = 2 * v1), the volume flow rate in the narrow section would be:
Q2 = (1/2 * A1) * (2 * v1)
Q2 = A1 * v1
Therefore, the volume flow rate remains the same in both sections.
To answer these questions, we can use the principles of continuity equation and Bernoulli's equation for the flow of an ideal fluid. Let's break down each question and explain how to find the answers.
a) What is the speed of flow in the narrow section?
To find the speed of flow in the narrow section, we can use the principle of continuity. The continuity equation states that the volume flow rate is constant in an incompressible fluid within a closed system.
The equation for continuity is:
A₁v₁ = A₂v₂
where A₁ and A₂ are the cross-sectional areas of the pipe in the original and narrow sections, respectively, and v₁ and v₂ are the speeds of flow in those sections.
Given that the pipe narrows to half its original diameter, we know that the cross-sectional area in the narrow section (A₂) is one-fourth of the original cross-sectional area (A₁). Since the area is proportional to the square of the diameter, A₂ = (1/4)A₁.
From the given information, the speed of flow in the original section is 3 m/s (v₁).
Using the continuity equation, we can solve for the speed of flow in the narrow section (v₂):
A₁v₁ = A₂v₂
v₂ = (A₁/A₂) * v₁
v₂ = (A₁ / (1/4)A₁) * 3
v₂ = 4 * 3
v₂ = 12 m/s
Therefore, the speed of flow in the narrow section is 12 m/s.
b) What is the pressure in the narrow section?
To find the pressure in the narrow section, we can use Bernoulli's equation. Bernoulli's equation relates the pressure, speed, and height of a fluid in a streamline.
The equation for Bernoulli's equation in this case is:
P₁ + (1/2)ρv₁² + ρgh₁ = P₂ + (1/2)ρv₂² + ρgh₂
In this equation, P₁ and P₂ are the pressures in the original and narrow sections, respectively, ρ is the density of the fluid, v₁ and v₂ are the speeds of flow in those sections, and h₁ and h₂ are the heights of the fluid in the respective sections (level ground, so h₁ = h₂).
From the given information, the pressure in the original section is 200 kPa (P₁), and we have already found the speed of flow in the narrow section as 12 m/s (v₂).
Using the Bernoulli's equation, we can solve for the pressure in the narrow section (P₂):
P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂²
200 kPa + (1/2)ρ(3 m/s)² = P₂ + (1/2)ρ(12 m/s)²
(1/2)ρ(9 m²/s²) = (1/2)ρ(144 m²/s²)
0.5ρ(9 m²/s²) = 0.5ρ(144 m²/s²)
The density (ρ) cancels out, implying that the pressure difference is independent of density. So, we can say that the pressure difference is the same in both sections.
Now, we only need to calculate the pressure difference between the sections:
200 kPa = P₂
Therefore, the pressure in the narrow section is also 200 kPa.
c) How do the volume flow rates in the two sections compare?
According to the principle of continuity, the volume flow rate is constant in an incompressible fluid within a closed system.
The volume flow rate (Q) is determined by the cross-sectional area of a pipe multiplied by the speed of flow: Q = Av.
Since the pipe narrows to half its original diameter, the cross-sectional area in the narrow section (A₂) is one-fourth of the original cross-sectional area (A₁).
Applying the formula, we get:
Q₁ = A₁v₁
Q₂ = A₂v₂
Given that A₂ = (1/4)A₁, and if we assume that the density (ρ) is the same for both sections, we can conclude that the volume flow rate in the narrow section (Q₂) is one-fourth (1/4) of the volume flow rate in the original section (Q₁).
In summary:
a) The speed of flow in the narrow section is 12 m/s.
b) The pressure in the narrow section is 200 kPa.
c) The volume flow rate in the narrow section is one-fourth of the volume flow rate in the original section.
a = 12 m/s
b = 132.5 kpa