A uniform tapering pipe is 20 cm diameter at one end A and 10 cm at ot...
Problem Statement
A uniform tapering pipe is 20 cm diameter at one end A and 10 cm at other end B. The pipe is 3m long, is inclined to horizontal at an angle a = tan¹ (1/4) with end A above B. If the flow velocity at section B is 0.6 m/s, determine the difference of pressure between the two sections.
Solution
Given:
- Diameter of end A = 20cm
- Diameter of end B = 10cm
- Length of pipe = 3m
- Inclination angle = tan¹ (1/4)
- Flow velocity at section B = 0.6m/s
Step 1: Calculate the inclination of the pipe
Inclination angle, a = tan¹ (1/4)
a = 14.04°
Step 2: Calculate the velocity of flow at section A
We know that the volume of fluid flowing through the pipe is constant. Hence,
Volume of fluid flowing through section A = Volume of fluid flowing through section B
π/4 x d₁² x V₁ = π/4 x d₂² x V₂
where,
- d₁ = Diameter of end A = 20cm
- d₂ = Diameter of end B = 10cm
- V₂ = Velocity of flow at section B = 0.6m/s
Substituting the values, we get
π/4 x 0.2² x V₁ = π/4 x 0.1² x 0.6
V₁ = 3.6m/s
Step 3: Calculate the pressure difference between section A and B
Using Bernoulli's equation,
P₁ + ρgh₁ + 1/2 ρV₁² = P₂ + ρgh₂ + 1/2 ρV₂²
where,
- P₁ = Pressure at section A
- P₂ = Pressure at section B
- ρ = Density of fluid
- g = Acceleration due to gravity
- h₁ = Height of section A from reference level
- h₂ = Height of section B from reference level
Assuming the reference level at section B, we have
h₁ = 3m x sin(14.04°) = 0.76m
h₂ = 0m
Also, as the fluid is incompressible, the density is constant. Hence,
P₁ + 1/2 ρV₁² + ρgh₁ = P₂ + 1/2 ρV₂² + ρgh₂
P₁ - P₂