A 2.5m long pipeline tapers uniformly from 10cm diameter to 20cm diameter at its upper end. The pipe centerline slopes upwards at an angle of 300 to the horizontal and the flow direction is from smaller to bigger cross-section. If the pressure at lower and upper ends of the pipe are 2bar and 2.4bar respectively, determine the flow rate and the pressure at the mid-length of the pipeline. Assume no energy losses.

1. Setting up the Problem – Areas and Elevations

Cross-sectional area at the lower end (section 1):
A₁ = π × (d₁/2)² = π × (0.1/2)² = π × 0.0025 = 0.00785 m²

Cross-sectional area at the upper end (section 2):
A₂ = π × (d₂/2)² = π × (0.2/2)² = π × 0.01 = 0.0314 m²

Taking datum at section 1 (lower end):
Z₁ = 0 m
Z₂ = 2.5 × sin(30°) = 2.5 × 0.5 = 1.25 m
Elevation at mid-length (section 3):
Z₃ = 1.25 × sin(30°) = 1.25 × 0.5 = 0.625 m

2. Applying the Continuity Equation

Using the continuity equation to relate the velocities at both ends:

Substituting the cross-sectional areas:

Solving for V₁ in terms of V₂:

So the velocity at the lower end (V₁) is 4 times the velocity at the upper end (V₂).

3. Applying Bernoulli’s Equation to Calculate Flow Rate

Using Bernoulli’s equation between section 1 (lower end) and section 2 (upper end), assuming no energy losses:

Substituting known values and V₁ = 4V₂:

Simplifying:

Rearranging to solve for V₂:

Therefore:

Calculate the flow rate:
Q = A₁ × V₁ = 0.00785 × 10.56 = 0.083 m³/s

4. Determining Pressure at Mid-Length (Section 3)

Diameter at mid-length:
d₃ = (d₁ + d₂)/2 = (0.1 + 0.2)/2 = 0.15 m

Cross-sectional area at mid-length:
A₃ = π × (d₃/2)² = π × (0.15/2)² = π × 0.00563 = 0.01767 m²

Velocity at mid-length:
Using continuity equation: Q = A₃ × V₃
V₃ = Q/A₃ = 0.083/0.01767 = 4.7 m/s

Applying Bernoulli’s equation between section 1 and section 3:

Substituting known values:

Solving for P₃: