Pipeline Discharge and Pressure Head Calculation
Problem Statement
A pipeline connected to a reservoir discharges water to the atmosphere. The head losses between sections are given as follows:
– Loss from A to B is 1 times the velocity head.
– Loss from B to C is 1.5 times the velocity head.
– Loss from C to D is 0.5 times the velocity head.
The pipe has a diameter of 150 mm. With the datum heads at A, B, C, and D provided, calculate the pressure heads at points B and C, and determine the discharge.
Given Data
| Diameter of Pipe (d) | 150 mm = 0.15 m |
| Cross-Sectional Area (A) | A = (π/4) × (0.15)² ≈ 0.0176 m² |
| Velocity at A (Vₐ) | 0 m/s (reservoir condition) |
| Velocities in Pipe (V) | V = V_B = V_C = V_D |
| Head Loss from A to B (hₗ_AB) | hₗ_AB = V²/(2g) |
| Head Loss from B to C (hₗ_BC) | hₗ_BC = 1.5V²/(2g) |
| Head Loss from C to D (hₗ_CD) | hₗ_CD = 0.5V²/(2g) |
| Total Head Loss (hₗ_total) | hₗ_total = (1 + 1.5 + 0.5)V²/(2g) = 3V²/(2g) |
| Datum Head at A (Zₐ) | 20 m |
| Datum Head at B (Z_B) | 0 m |
| Datum Head at C (Z_C) | 15 m |
| Datum Head at D (Z_D) | 5 m |
| Pressure at A (Pₐ) | 0 (atmospheric) |
| Pressure at D (P_D) | 0 (atmospheric) |
1. Determining the Flow Velocity
Applying Bernoulli’s equation between points A and D:
Pₐ/γ + Vₐ²/(2g) + Zₐ = P_D/γ + V²/(2g) + Z_D + hₗ_total
Since Pₐ and P_D are atmospheric and Vₐ = 0, the equation simplifies to:
20 = (V²)/(2g) + 5 + 3V²/(2g)
Combining the velocity terms:
20 = 5 + (4V²)/(2g) = 5 + (2V²)/g
Rearranging:
(2V²)/g = 15
V² = (15g)/2
Using g = 9.81 m/s²:
V² ≈ (15 × 9.81)/2 ≈ 73.575 → V ≈ 8.57 m/s
2. Calculating Pressure Head at B
Apply Bernoulli’s equation between points A and B:
Pₐ/γ + Vₐ²/(2g) + Zₐ = P_B/γ + V²/(2g) + Z_B + hₗ_AB
With Pₐ = 0, Vₐ = 0, Zₐ = 20 m, Z_B = 0, and hₗ_AB = V²/(2g):
20 = P_B/γ + (V²)/(2g) + (V²)/(2g)
20 = P_B/γ + V²/g
Substituting V²/g (with V ≈ 8.57 m/s and V² ≈ 73.5):
V²/g ≈ 73.5/9.81 ≈ 7.5 m
Hence, P_B/γ = 20 − 7.5 = 12.5 m
3. Calculating Pressure Head at C
Apply Bernoulli’s equation between points B and C:
P_B/γ + V²/(2g) + Z_B = P_C/γ + V²/(2g) + Z_C + hₗ_BC
With Z_B = 0, Z_C = 15 m, and hₗ_BC = 1.5V²/(2g):
12.5 + V²/(2g) = P_C/γ + V²/(2g) + 15 + 1.5V²/(2g)
Canceling the common V²/(2g) term:
12.5 = P_C/γ + 15 + 1.5V²/(2g)
Rearranging:
P_C/γ = 12.5 − 15 − 1.5V²/(2g)
Using V²/(2g) ≈ 73.5/(19.62) ≈ 3.75 m:
P_C/γ = 12.5 − 15 − (1.5 × 3.75)
P_C/γ = −2.5 − 5.625 ≈ −8.12 m
4. Calculating the Discharge (Q)
Discharge is given by:
Q = A × V = 0.0176 m² × 8.57 m/s ≈ 0.1508 m³/s
PC/γ ≈ −8.12 m
Q ≈ 0.1508 m³/s
Physical Interpretation
This problem illustrates the energy changes occurring in a pipeline that discharges water from a reservoir to the atmosphere. Key points include:
Velocity Head and Energy Losses:
As the water flows through the pipe, its velocity remains constant in the conduit, but energy is lost in different segments of the system. The losses from A to B, B to C, and C to D are expressed as multiples of the velocity head (V²/(2g)). These losses account for friction and other dissipative effects.
Pressure Head Variations:
The pressure head at a point (P/γ) is a measure of the energy available per unit weight due to pressure. At point B, a significant pressure head (12.5 m) exists, indicating stored energy. However, by point C, the additional losses and a change in datum head reduce the available pressure head to a negative value (−8.12 m), suggesting that energy has been largely dissipated.
Elevation Head:
The different datum heads (20 m at A, 0 m at B, 15 m at C, and 5 m at D) contribute to the overall energy balance, influencing how much energy is available to overcome losses and drive the flow.
Discharge:
The computed discharge (0.1508 m³/s) represents the volumetric flow rate, determined by the product of the cross-sectional area and the velocity. This parameter is crucial for understanding how much water is transported through the pipeline.
Detailed Explanation for Students
Step 1: Understanding the System
The pipeline discharges water from a reservoir to the atmosphere. Different segments of the pipe incur head losses proportional to the velocity head. The given datum heads at various points help us track the energy changes along the flow.
Step 2: Applying Bernoulli’s Equation from A to D
By applying Bernoulli’s equation over the entire length (from A to D) and incorporating the total head loss (3V²/(2g)), we can solve for the velocity of water in the pipe. This step ensures that all energy contributions (pressure, velocity, and elevation) are balanced.
Step 3: Finding Pressure Head at Intermediate Points
With the velocity determined, Bernoulli’s equation is applied separately between A and B, and then between B and C, to determine the pressure heads at these points. The loss factors (1 for A–B and 1.5 for B–C) directly affect the energy available as pressure head.
Step 4: Calculating the Discharge
Once the velocity is known, the discharge is calculated using Q = A × V. This value indicates the rate at which water is leaving the system.
This comprehensive analysis demonstrates how energy losses in a pipeline affect both the pressure distribution and the overall flow rate. Understanding these concepts is essential for designing efficient water distribution systems.
