Pressure Calculation in Tapered Pipe with Energy Considerations
Problem Statement
The water is flowing through a taper pipe of length 50m having diameters 40cm at the upper end and 20cm at the lower end, at the rate of 60 lps. The pipe has a slope of 1 in 40. Find the pressure at the lower end if the pressure at the higher level is 24.525 N/cm² under two conditions:
- Assuming no loss of energy
- With a loss of head of 0.2m
Given Data
| Pipe length | 50 m |
| Diameter at upper end (d₂) | 40 cm = 0.4 m |
| Diameter at lower end (d₁) | 20 cm = 0.2 m |
| Flow rate (Q) | 60 lps = 0.06 m³/s |
| Slope | 1 in 40 |
| Pressure at upper end (P₂) | 24.525 N/cm² = 245250 N/m² |
| Density of water (ρ) | 1000 kg/m³ |
| Acceleration due to gravity (g) | 9.81 m/s² |
1. Clarification of Pipe Orientation
From the problem statement:
- Section 2 is at the upper end with diameter 40 cm (larger diameter)
- Section 1 is at the lower end with diameter 20 cm (smaller diameter)
2. Key Principles and Equations
We’ll use the Bernoulli equation to solve this problem, which states that the total energy at two points in a fluid flow is constant (assuming no energy losses).
P₁/ρg + V₁²/(2g) + Z₁ = P₂/ρg + V₂²/(2g) + Z₂ + hₗ
Where:
P = Pressure (N/m²)
ρ = Density of fluid (kg/m³)
g = Acceleration due to gravity (9.81 m/s²)
V = Velocity of fluid (m/s)
Z = Elevation from datum (m)
hₗ = Head loss (m)
3. Calculating Cross-sectional Areas
Cross-sectional area at lower end (section 1):
A₁ = π × (d₁/2)² = π × (0.2/2)² = π × 0.01 = 0.0314 m²
Cross-sectional area at upper end (section 2):
A₂ = π × (d₂/2)² = π × (0.4/2)² = π × 0.04 = 0.1256 m²
4. Calculating Velocities
Using the continuity equation Q = A × V:
Velocity at lower end (V₁):
V₁ = Q/A₁ = 0.06/0.0314 = 1.91 m/s
Velocity at upper end (V₂):
V₂ = Q/A₂ = 0.06/0.1256 = 0.48 m/s
5. Calculating Elevation Difference
Given the slope of the pipe is 1 in 40:
Slope = tan θ = 1/40
θ = tan⁻¹(1/40) = 1.43°
If we set our datum at the lower end of the pipe (Z₁ = 0), then:
Z₂ = 50 × sin(1.43°) = 50 × 0.025 = 1.25 m
6. Case (a): Assuming No Loss of Energy
Applying Bernoulli’s equation with hₗ = 0:
P₁/ρg + V₁²/(2g) + Z₁ = P₂/ρg + V₂²/(2g) + Z₂
Rearranging to solve for P₁:
P₁/ρg = P₂/ρg + V₂²/(2g) + Z₂ – V₁²/(2g) – Z₁
Substituting values:
P₁/ρg = 245250/(1000×9.81) + (0.48)²/(2×9.81) + 1.25 – (1.91)²/(2×9.81) – 0
P₁/ρg = 25 + 0.012 + 1.25 – 0.186
P₁/ρg = 26.076
Therefore:
P₁ = 26.076 × 1000 × 9.81 = 255806 N/m²
7. Case (b): With a Loss of Head of 0.2m
Applying Bernoulli’s equation with hₗ = 0.2 m:
P₁/ρg + V₁²/(2g) + Z₁ = P₂/ρg + V₂²/(2g) + Z₂ + hₗ
Rearranging to solve for P₁:
P₁/ρg = P₂/ρg + V₂²/(2g) + Z₂ – V₁²/(2g) – Z₁ – hₗ
Substituting values:
P₁/ρg = 245250/(1000×9.81) + (0.48)²/(2×9.81) + 1.25 – (1.91)²/(2×9.81) – 0 – 0.2
P₁/ρg = 25 + 0.012 + 1.25 – 0.186 – 0.2
P₁/ρg = 25.876
Therefore:
P₁ = 25.876 × 1000 × 9.81 = 253846 N/m²
8. Visualization of Results
No Energy Loss
25.58 N/cm²
With 0.2m Head Loss
25.38 N/cm²
9. Physical Interpretation
The analysis of this tapered pipe flow reveals several important hydraulic principles:
- Energy Conservation: In the ideal case with no losses, the sum of pressure energy, kinetic energy, and potential energy remains constant throughout the flow system.
- Pressure Increase: The pressure at the lower end (25.58 N/cm² with no losses) is higher than at the upper end (24.525 N/cm²) despite the elevation decrease. This is because the increase in velocity at the constriction (from 0.48 m/s to 1.91 m/s) results in a decrease in kinetic energy that must be balanced by an increase in pressure energy.
- Effect of Losses: When energy losses (0.2m head loss) are considered, the pressure at the lower end decreases (to 25.38 N/cm²) compared to the ideal case. This reduction (0.2 N/cm²) represents the energy dissipated due to factors like friction, turbulence, and flow separation.
- Conversion Between Energy Forms: The problem demonstrates how potential energy (due to elevation), kinetic energy (due to velocity), and pressure energy are interconverted in fluid flow systems while maintaining the principle of energy conservation.
- Practical Implications: In real-world applications, accounting for energy losses is crucial for accurate pressure predictions, especially in systems with significant constrictions or expansions where flow separation and turbulence can lead to substantial energy dissipation.
10. Conclusion
We have successfully calculated the pressure at the lower end of the tapered pipe under two different conditions:
- Case (a) – No energy loss: P₁ = 255806 N/m² = 25.58 N/cm²
- Case (b) – With 0.2m head loss: P₁ = 253846 N/m² = 25.38 N/cm²
The difference between these two results (0.2 N/cm²) directly corresponds to the 0.2m head loss in the system, demonstrating how energy losses in a fluid system manifest as reduced pressure. This analysis highlights the importance of considering energy losses in hydraulic systems for accurate pressure predictions, especially in pipes with varying cross-sections where flow separation and turbulence can lead to significant energy dissipation.
