Velocity and Flow Rate from a Horizontal Pipe
Problem Statement
A 20 cm diameter horizontal pipe is attached to a reservoir as shown. If the total head loss between the water surface in the reservoir and the water jet at the end of the pipe is 1.8 m, determine the velocity and flow rate of water being discharged from the pipe.
Given Data
| d₂ | 20 cm = 0.2 m |
| C/S Area at section 2 (A₂) | A₂ = (π/4)×(0.2)² ≈ 0.0314 m² |
| Velocity at point 1 (V₁) | 0 m/s (at the reservoir) |
| Datum Head at point 1 (Z₁) | 4.5 m |
| Datum Head at point 2 (Z₂) | 0 m |
| Head Loss (hₗ) | 1.8 m |
| Pressure at point 1 (P₁) | 0 (atmospheric) |
| Pressure at point 2 (P₂) | 0 (atmospheric) |
1. Applying Bernoulli’s Equation
For points 1 (reservoir surface) and 2 (pipe exit), Bernoulli’s equation including head loss is:
P₁/γ + V₁²/(2g) + Z₁ = P₂/γ + V₂²/(2g) + Z₂ + hₗ
With atmospheric conditions at both points, P₁ and P₂ are zero.
2. Substituting Known Values
Since V₁ = 0, the equation simplifies to:
0 + 0 + 4.5 = 0 + V₂²/(2g) + 0 + 1.8
Rearranging:
V₂²/(2g) = 4.5 − 1.8 = 2.7 m
3. Solving for Velocity at Point 2 (V₂)
Multiply both sides by 2g (with g = 9.81 m/s²):
V₂² = 2.7 × 2 × 9.81 = 2.7 × 19.62 ≈ 52.97
Therefore, V₂ ≈ √52.97 ≈ 7.28 m/s
4. Calculating Flow Rate (Q₂)
Flow rate is given by:
Q₂ = A₂ × V₂ ≈ 0.0314 m² × 7.28 m/s ≈ 0.228 m³/s
Q₂ ≈ 0.228 m³/s
Physical Interpretation
This problem uses Bernoulli’s equation to relate the energy states between the water surface in a reservoir and the discharge jet from a horizontal pipe. Notice the following:
Elevation Head (Z): The water surface is at 4.5 m while the pipe exit is at 0 m. This elevation difference provides the potential energy for the flow.
Head Loss (hₗ): Energy is lost due to friction and other factors within the system, here amounting to 1.8 m. This loss reduces the energy available for converting to kinetic energy.
Velocity Head (V²/(2g)): The remaining energy is converted into kinetic energy at the pipe exit. The resulting velocity head is calculated from the net energy after subtracting head loss.
The computed velocity (7.28 m/s) and flow rate (0.228 m³/s) reflect the conversion of potential energy (from elevation) minus the losses into the kinetic energy of the jet.
Detailed Explanation for Students
Step 1: Setting Up the Problem
We start by identifying the key energy components between the reservoir (point 1) and the pipe exit (point 2): elevation head, velocity head, and head loss.
Step 2: Using Bernoulli’s Equation
Bernoulli’s equation (with head loss) expresses conservation of energy. Since both the reservoir surface and the pipe exit are exposed to atmospheric pressure, the pressure terms cancel out.
Step 3: Accounting for Energy Losses
The head loss of 1.8 m represents the energy lost due to friction and turbulence. This loss is subtracted from the available elevation head.
Step 4: Converting the Remaining Energy to Kinetic Energy
The net energy (elevation head minus head loss) is then converted into kinetic energy at the pipe exit. This is expressed by the term V²/(2g), which allows us to solve for the exit velocity.
Step 5: Calculating Flow Rate
Once the velocity is known, the flow rate is determined using the cross-sectional area of the pipe (Q = A × V). This provides the volume of water discharged per second.
This systematic approach shows how energy differences drive fluid motion and how losses affect the overall flow.
