A venturimeter with a throat diameter of 100mm is fitted in a vertical pipeline of 200mm diameter with oil of sp.gr. 0.88 flowing upwards. The venturimeter coefficient is 0.98. The pressure gauges are fitted at tapping points, one at the throat and the other in the inlet pipe 320mm below the throat. The difference between two pressure gauge readings is 28 KN/m2. Working from Bernoulli’s equation, determine (a) the volume rate of oil through the pipe, (b) the difference in level in the two limbs of mercury if it is connected to the tapping points and connecting pipes are filled with same oil.

Venturimeter Analysis in a Vertical Pipeline

Problem Statement

A venturimeter with a throat diameter of 100 mm is fitted in a vertical pipeline of 200 mm diameter carrying oil of specific gravity 0.88 flowing upwards. The venturimeter coefficient is 0.98. Pressure gauges are installed at the tapping points: one at the throat and the other in the inlet pipe 320 mm below the throat. The difference between the two pressure gauge readings is 28 kN/m². Working from Bernoulli’s equation, determine:

(a) The volume rate of oil through the pipe
(b) The difference in level in the two limbs of mercury, if the connecting pipes are filled with oil

Given Data

Inlet Diameter (d₁)	200 mm = 0.2 m
Inlet Area (A₁)	(π/4) × (0.2)² ≈ 0.0314 m²
Throat Diameter (d₂)	100 mm = 0.1 m
Throat Area (A₂)	(π/4) × (0.1)² ≈ 0.00785 m²
Specific Gravity of Oil (S₀)	0.88
Density of Oil (ρ)	0.88 × 1000 = 880 kg/m³
Difference in Elevation (Z₂ – Z₁)	320 mm = 0.32 m
Pressure Difference (P₁ – P₂)	28 kN/m² = 28000 N/m²
Specific Gravity of Mercury (S)	13.6
Discharge Coefficient (C_d)	0.98
Acceleration due to Gravity (g)	9.81 m/s²

1. Applying Bernoulli’s Equation

Writing Bernoulli’s equation between the inlet (point 1) and the throat (point 2):

P₁/ρg + V₁²/(2g) + Z₁ = P₂/ρg + V₂²/(2g) + Z₂

Rearranging gives:

(P₁ – P₂)/(ρg) + (Z₁ – Z₂) = (V₂² – V₁²)/(2g)

Substituting the given values (noting that Z₁ – Z₂ = -0.32 m since the inlet is below the throat):

28000/(880×9.81) – 0.32 = (V₂² – V₁²)/(2×9.81)

Evaluating the left-hand side:

(28000/(8628.8)) – 0.32 ≈ 3.24 – 0.32 = 2.92 m

Hence,

(V₂² – V₁²)/(2g) = 2.92 m (a)

2. Applying the Continuity Equation

From the continuity equation:

A₁V₁ = A₂V₂ → V₂ = (A₁/A₂) V₁

With A₁ = 0.0314 m² and A₂ = 0.00785 m²:

V₂ = (0.0314/0.00785) V₁ = 4V₁ (b)

3. Solving for Velocities and Discharge

Substituting V₂ = 4V₁ into equation (a):

((4V₁)² – V₁²)/(2g) = 2.92

Simplifying:

(16V₁² – V₁²)/(2g) = (15V₁²)/(2g) = 2.92

Solving for V₁:

V₁² = (2.92×2g)/15 = (2.92×19.62)/15 ≈ 3.82 → V₁ ≈ 1.95 m/s

The discharge through the inlet is:

Q = A₁V₁ = 0.0314 × 1.95 ≈ 0.612 m³/s

Taking into account the discharge coefficient:

Q_actual = C_d × Q = 0.98 × 0.612 ≈ 0.599 m³/s

Volume Rate of Oil = 0.599 m³/s

4. Determining the Manometer Reading

The effective head from Bernoulli’s equation is:

h = (P₁ – P₂)/(ρg) + (Z₁ – Z₂) = 2.92 m

This head is related to the manometer reading (x) by:

h = x × (S/S₀ – 1)

Substituting S = 13.6 and S₀ = 0.88:

2.92 = x × (13.6/0.88 – 1) = x × (15.45 – 1) = x × 14.45

Solving for x:

x = 2.92/14.45 ≈ 0.202 m

Manometer Reading (x) ≈ 0.20 m

Conclusion

By applying Bernoulli’s equation and the continuity equation, the analysis shows that the volume rate of oil through the vertical pipeline is approximately 0.599 m³/s. Furthermore, the difference in mercury levels in the connected manometer is about 0.20 m.