A venturimeter is used for measurement of discharge of water in a horizontal pipeline. The ratio of the upstream pipe diameter and throat is 2:1 and upstream diameter is 300mm. Mercury manometer connected at the pipe and throat shows the reading of 0.24m and the loss of head through the meter is 1/8 of the throat velocity head. Calculate the discharge in the pipe using the continuity and energy equations.

Venturimeter Discharge Measurement in a Horizontal Pipeline

Problem Statement

A venturimeter is used for the measurement of discharge of water in a horizontal pipeline. The ratio of the upstream pipe diameter to the throat diameter is 2:1, with the upstream diameter being 300 mm. A mercury manometer connected at the pipe and the throat shows a reading of 0.24 m, and the head loss through the meter is 1/8 of the throat velocity head. Using the continuity and energy equations, calculate the discharge in the pipe.

Given Data

Upstream (Pipe) Diameter (d₁)	300 mm = 0.3 m
Throat Diameter (d₂)	d₁/2 = 0.15 m
Cross-sectional Area (A₁)	(π/4) × (0.3)² ≈ 0.0707 m²
Cross-sectional Area (A₂)	(π/4) × (0.15)² ≈ 0.01767 m²
Manometer Reading (x)	0.24 m
Specific Gravity of Mercury (S)	13.6
Specific Gravity of Water (S₀)	1
Head Loss (h_L)	1/8 of throat velocity head = (1/8)(V₂²/(2g))
Acceleration due to Gravity (g)	9.81 m/s²

1. Determining the Effective Head (h)

The head corresponding to the mercury manometer reading is given by:

h = x × (S/S₀ – 1)

Substituting the given values:

h = 0.24 × (13.6/1 – 1) = 0.24 × (13.6 – 1) = 0.24 × 12.6 ≈ 3.02 m

2. Applying Bernoulli’s Equation

For a horizontal pipeline, Bernoulli’s equation between the inlet (point 1) and the throat (point 2) with an added head loss (h_L) is:

P₁/ρg + V₁²/(2g) = P₂/ρg + V₂²/(2g) + h_L

Rearranging for the difference in kinetic energy:

(P₁ – P₂)/(ρg) = (V₂² – V₁²)/(2g) + h_L

Since the effective head from the manometer is:

h = (P₁ – P₂)/(ρg)

we have:

h = (V₂² – V₁²)/(2g) + (1/8)(V₂²/(2g))

Combining the terms:

h = [V₂²/(2g)] [1 – (V₁²/V₂²) + 1/8]

Noting that from the continuity equation (next step) we will express V₂ in terms of V₁.

3. Applying the Continuity Equation

The continuity equation for incompressible flow is:

A₁V₁ = A₂V₂ → V₂ = (A₁/A₂) V₁

With A₁ ≈ 0.0707 m² and A₂ ≈ 0.01767 m²:

V₂ = (0.0707/0.01767) V₁ ≈ 4V₁

4. Solving for Velocity and Discharge

Substituting V₂ = 4V₁ into the energy equation:

h = [(V₂² – V₁²)/(2g)] + [1/8 × V₂²/(2g)]

Rewriting with V₂ = 4V₁:

h = [(16V₁² – V₁²)/(2g)] + [1/8 × (16V₁²)/(2g)]

Simplify the terms:

h = (15V₁²/(2g)) + (2V₁²/(2g)) = (17V₁²)/(2g)

Equate to the effective head:

(17V₁²)/(2g) = 3.02

Solving for V₁:

V₁² = (3.02 × 2g) / 17 ≈ (3.02 × 19.62)/17 ≈ 3.56 → V₁ ≈ 1.86 m/s

The discharge through the pipe is then:

Q = A₁V₁ = 0.0707 × 1.86 ≈ 0.131 m³/s

Discharge (Q) ≈ 0.131 m³/s

Conclusion

By applying the continuity and energy equations (with the inclusion of a head loss equal to 1/8 of the throat velocity head), the discharge in the horizontal pipeline is determined to be approximately 0.131 m³/s.