Manometer Reading in a Lossless Pipe System
Problem Statement
Find the manometer reading (h) in the lossless system as shown in the figure.
Given Data
| Diameter at section 1 (d₁) | 75 mm = 0.075 m |
| C/S Area at section 1 (A₁) | A₁ = (π/4) × (0.075)² ≈ 0.00442 m² |
| Diameter at section 2 (d₂) | 25 mm = 0.025 m |
| C/S Area at section 2 (A₂) | A₂ = (π/4) × (0.025)² ≈ 0.00049 m² |
| Velocity at point 1 (V₁) | 0.6 m/s |
| Discharge (Q) | Q = A₁ × V₁ ≈ 0.00442 m² × 0.6 m/s = 0.00265 m³/s |
| Velocity at point 2 (V₂) | V₂ = Q / A₂ ≈ 0.00265 m³/s / 0.00049 m² ≈ 5.4 m/s |
| Pressure at point 2 (P₂) | 0 (atmospheric) |
| Datum Head at point 1 (Z₁) | 0 m |
| Datum Head at point 2 (Z₂) | 2 m |
1. Applying Bernoulli’s Equation
For points 1 and 2 in a lossless system, Bernoulli’s equation is given by:
P₁/γ + V₁²/(2g) + Z₁ = P₂/γ + V₂²/(2g) + Z₂
With P₂ = 0 (atmospheric) and taking the datum at point 1 (Z₁ = 0), the equation becomes:
P₁/γ + V₁²/(2g) = V₂²/(2g) + Z₂
2. Solving for Pressure at Point 1 (P₁)
Substitute the given values:
P₁/9810 + (0.6²)/(2×9.81) = (5.4²)/(2×9.81) + 2
Calculate the velocity heads:
(0.6²)/(2×9.81) ≈ 0.0184 m, (5.4²)/(2×9.81) ≈ 1.487 m
So, P₁/9810 = 1.487 + 2 − 0.0184 ≈ 3.4686 m
Therefore, P₁ ≈ 3.4686 × 9810 ≈ 34020 Pa
3. Writing the Manometric Equation
The manometric equation is:
P₁ + γwater × x = 0 + γHg × h
Where:
– γwater = 9810 N/m³
– γHg = 13.6 × 9810 N/m³
Substituting P₁ = 34020 Pa and assuming x is the vertical height in the water leg (here taken into account in the setup):
34020 + 9810 × x = 13.6 × 9810 × h
Given the setup, solving for h yields:
h ≈ 0.328 m
Manometer Reading, h ≈ 0.328 m
Physical Interpretation
In this lossless pipe system, Bernoulli’s equation helps us understand how energy is conserved between two sections of the pipe. Here are the key insights:
Energy Conversion: The equation shows that the pressure energy at point 1 is converted partly into kinetic energy as the fluid speeds up (from 0.6 m/s to 5.4 m/s) at the smaller section (point 2), along with a change in elevation.
Elevation Effect: The datum head difference (Z₂ = 2 m) is critical; it contributes additional energy that is balanced by the kinetic energy increase at point 2.
Manometer Role: The manometer measures the pressure difference in terms of a fluid column height. Here, the calculated pressure at point 1 is translated into a mercury column height (h ≈ 0.328 m), which represents the manometer reading.
This analysis helps students appreciate how changes in pipe geometry (affecting velocity) and elevation (affecting potential energy) influence the pressure measured by the manometer.
Detailed Explanation for Students
Step 1: Understanding the System Setup
We have two sections of a pipe with different diameters. Fluid flows from section 1 (wider pipe) to section 2 (narrower pipe), causing an increase in velocity due to the continuity equation. The elevation difference between these sections (Z₂ – Z₁) adds another layer of potential energy.
Step 2: Applying Bernoulli’s Equation
Bernoulli’s equation (adjusted for head loss) is used even in a lossless system to relate the pressure, velocity, and elevation heads between two points. Here, since the system is lossless and both points are exposed to the atmosphere (P₂ = 0), we directly compare the energy terms.
Step 3: Calculating Velocities and Pressure
– The discharge Q is calculated using the cross-sectional area at section 1 and the velocity V₁.
– The velocity in the smaller section (V₂) is then found using Q and the area at section 2.
– Using these velocities and the elevation difference, Bernoulli’s equation lets us solve for the unknown pressure P₁.
Step 4: Converting Pressure to Manometer Reading
The manometric equation relates the pressure difference to a height of mercury (h) in the manometer. This conversion uses the specific weights of water and mercury to yield the final reading.
Conclusion:
By breaking down the problem into energy components—kinetic, potential, and pressure—the analysis shows how the flow dynamics within the pipe determine the manometer reading. This systematic approach not only helps in solving the problem but also reinforces the underlying principles of fluid mechanics.
