Manometer Reading in a Sudden Expansion of a Pipe
Problem Statement
A 15 cm diameter pipe is suddenly expanded to a 25 cm diameter pipe. The head loss at the sudden expansion from section 1 to 2 is given by:
hₗ = (V₁ – V₂)²/(2g)
For a discharge of 45 lps, calculate the manometer reading (h). The manometer connects between the two sections with water on one side and mercury on the other.
Given Data
| Diameter at Section 1 (d₁) | 15 cm = 0.15 m |
| Area at Section 1 (A₁) | A₁ = (π/4) × (0.15)² ≈ 0.0176 m² |
| Diameter at Section 2 (d₂) | 25 cm = 0.25 m |
| Area at Section 2 (A₂) | A₂ = (π/4) × (0.25)² ≈ 0.049 m² |
| Discharge (Q) | 45 lps = 0.045 m³/s |
| Datum Head at Section 1 (Z₁) | 0.5 m |
| Datum Head at Section 2 (Z₂) | 0 m |
| Acceleration due to Gravity (g) | 9.81 m/s² |
| Specific Weight of Water (γ_water) | ≈ 9810 N/m³ |
| Specific Weight of Mercury (γ_Hg) | ≈ 13.6 × 9810 N/m³ |
1. Calculating Velocities in Each Section
The velocities at sections 1 and 2 are given by:
V₁ = Q/A₁ = 0.045/0.0176 ≈ 2.55 m/s
V₂ = Q/A₂ = 0.045/0.049 ≈ 0.92 m/s
2. Determining the Head Loss (hₗ)
The head loss due to the sudden expansion is given by:
hₗ = (V₁ – V₂)²/(2g)
Substituting the values:
hₗ = (2.55 – 0.92)²/(2×9.81) = (1.63)²/19.62 ≈ 2.66/19.62 ≈ 0.14 m
3. Applying Bernoulli’s Equation
For sections 1 and 2, Bernoulli’s equation (including head loss) is:
P₁/γ + V₁²/(2g) + Z₁ = P₂/γ + V₂²/(2g) + Z₂ + hₗ
With Z₁ = 0.5 m, Z₂ = 0, and V₁ and V₂ as calculated:
P₁/γ + (2.55²)/(2×9.81) + 0.5 = P₂/γ + (0.92²)/(2×9.81) + 0 + 0.14
Evaluating the velocity head terms:
(2.55²)/(2×9.81) ≈ 6.50/19.62 ≈ 0.33 m
(0.92²)/(2×9.81) ≈ 0.85/19.62 ≈ 0.043 m
Rearranging, the difference in pressure heads becomes:
P₁/γ – P₂/γ = [ (0.043 + 0.14) – (0.33 + 0.5) ] = 0.183 – 0.83 ≈ -0.647 m
Multiplying by γ (9810 N/m³):
P₁ – P₂ ≈ -0.647 × 9810 ≈ -6359.6 Pa
4. Determining the Manometer Reading (h)
The manometer connects the two sections, with water on one side and mercury on the other. Its equation is:
P₁ + γ_water (y) + γ_Hg (h) = P₂ + γ_water (h)
Here, y is the vertical distance from the connection point at section 1 to the manometer, given as 0.5 m.
Rearranging and substituting the known pressure difference (P₁ – P₂ = -6359.6 Pa):
-6359.6 + 9810(0.5 – h) + (13.6 × 9810) h = 0
Solving this equation gives:
h ≈ 0.0117 m
Physical Interpretation
This problem explores the energy changes when water flows through a sudden expansion in a pipe. The key points are:
Sudden Expansion and Velocity Change:
When the pipe expands from 15 cm to 25 cm, the velocity drops from 2.55 m/s to 0.92 m/s. The difference in kinetic energy is responsible for the head loss.
Head Loss:
The head loss, calculated using hₗ = (V₁ – V₂)²/(2g), quantifies the energy dissipated due to turbulence and flow separation at the expansion. In this case, about 0.14 m of head is lost.
Pressure Difference:
The Bernoulli equation reveals a pressure difference between the two sections (P₁ – P₂ ≈ -6359.6 Pa). This difference drives the manometer reading.
Manometer Reading:
The manometer, with water on one side and mercury on the other, converts the pressure difference into a height difference. The small reading (≈0.0117 m) reflects the high density of mercury, which produces a significant pressure head per unit height.
Detailed Explanation for Students
Step 1: Determining Flow Velocities
Calculate the cross-sectional areas of the pipe before and after expansion. Then, use the discharge (Q) to find the velocities in each section (V₁ and V₂).
Step 2: Computing Head Loss
Use the formula for head loss at a sudden expansion:
hₗ = (V₁ – V₂)²/(2g)
This loss represents the energy dissipated due to the abrupt change in pipe diameter.
Step 3: Applying Bernoulli’s Equation
Bernoulli’s equation is applied between sections 1 and 2. With the known datum heads and calculated velocity heads, you determine the pressure difference (P₁ – P₂) that results from the energy losses.
Step 4: Using the Manometer Equation
The manometer equation relates the pressure difference to the height difference in a fluid column, accounting for the different densities of water and mercury. Here, a small height difference (h) in mercury corresponds to a large pressure difference.
This step-by-step approach links theoretical concepts with practical measurements, helping you understand how energy losses in fluid flow manifest as observable manometer readings.
