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.

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