Pressure Calculation for Oil Flow in a Pipe
Problem Statement
Oil flows from a tank through 140 m of 15 cm diameter pipe and then discharges into the air. Given that the head loss from point 1 to point 2 is 0.55 m of oil, determine the pressure needed at point 1 to cause 0.02 m³/s of oil to flow.
Given Data
| Pipe Diameter (d₂) | 15 cm = 0.15 m |
| C/S Area at section 2 (A₂) | A₂ = (π/4)×(0.15)² ≈ 0.0176 m² |
| Discharge (Q) | 0.02 m³/s |
| Velocity at point 1 (V₁) | 0 m/s (at the tank) |
| Datum Head at point 1 (Z₁) | 25 m |
| Datum Head at point 2 (Z₂) | 30 m |
| Head Loss (hₗ) | 0.55 m of oil |
| Pressure at point 2 (P₂) | 0 (atmospheric) |
| Oil Specific Gravity | 0.8 (oil density = 0.8×1000 = 800 kg/m³) |
| Acceleration due to Gravity (g) | 9.81 m/s² |
1. Calculating Velocity at Point 2 (V₂)
Since Q = A₂ × V₂, then:
V₂ = Q / A₂ = 0.02 / 0.0176 ≈ 1.14 m/s
2. Applying Bernoulli’s Equation
Bernoulli’s equation between points 1 and 2 including head loss is:
P₁/γ + V₁²/(2g) + Z₁ = P₂/γ + V₂²/(2g) + Z₂ + hₗ
Given that V₁ = 0 and P₂ is atmospheric (0 gauge pressure), this simplifies to:
P₁/γ + Z₁ = V₂²/(2g) + Z₂ + hₗ
3. Substituting Known Values
Substitute:
– Z₁ = 25 m
– Z₂ = 30 m
– hₗ = 0.55 m
– V₂ ≈ 1.14 m/s
– g = 9.81 m/s²
– γ (specific weight of oil) = 0.8 × 1000 × 9.81 ≈ 7848 N/m³
The equation becomes:
P₁/7848 + 25 = (1.14²)/(2×9.81) + 30 + 0.55
Calculate the velocity head at point 2:
(1.14²)/(2×9.81) = (1.2996)/(19.62) ≈ 0.0662 m
So:
P₁/7848 + 25 = 0.0662 + 30 + 0.55 = 30.6162 m
4. Solving for Pressure at Point 1 (P₁)
Rearranging:
P₁/7848 = 30.6162 − 25 = 5.6162 m
Therefore, P₁ = 5.6162 × 7848 ≈ 43037 Pa
Converting to kPa:
P₁ ≈ 43.04 kPa
Physical Interpretation
This problem demonstrates how energy differences drive fluid flow from a pressurized tank through a pipe. Here are the key aspects:
Elevation Head: The tank’s datum head (25 m) and the discharge point’s datum head (30 m) indicate that, in addition to providing pressure, the elevation difference influences the energy balance.
Head Loss (hₗ): Energy losses due to friction and other dissipative effects in the 140 m pipe amount to 0.55 m of oil head. This loss must be overcome by the pressure at point 1.
Velocity Head: The kinetic energy of the oil is represented by the velocity head at the pipe exit. In this problem, even though the flow rate is moderate (0.02 m³/s), the small cross-sectional area produces a measurable velocity (≈1.14 m/s).
By balancing these energy components using Bernoulli’s equation, we determine the necessary pressure at point 1. This pressure must overcome both the potential energy difference and the head loss to achieve the desired flow rate.
Detailed Explanation for Students
Step 1: Identify Energy Terms
Recognize that the energy in a flowing fluid is composed of elevation head, velocity head, and pressure head. In this case, the water discharges into the air, so atmospheric pressure applies at the discharge point.
Step 2: Use Bernoulli’s Equation with Head Loss
Write Bernoulli’s equation between the tank (point 1) and the pipe exit (point 2), including the head loss term. Since the reservoir and discharge are both at atmospheric pressure, the pressure terms simplify.
Step 3: Compute Velocity and Head Loss Contributions
Determine the velocity at the pipe exit using the known discharge and cross-sectional area. Calculate the velocity head and subtract the head loss from the available elevation head difference.
Step 4: Solve for the Unknown Pressure
Rearranging the equation provides the pressure head required at point 1, which is then converted into pressure (in Pascals or kPa) using the specific weight of oil.
This methodical approach shows how to account for energy losses and elevation differences to find the pressure required to maintain a specific flow rate.
