A 20cmx10cm venturimeter is mounted in a vertical pipeline carrying oil of sp.gr. 0.8 flowing upwards. The throat section is 20cm above the entrance section of the venturimeter. The differential U-tube manometer shows a gauge deflection of 25cm. Calculate the discharge of the oil and the pressure difference between the entrance and the throat section. Take Cd =0.96

Venturimeter Discharge and Pressure Difference Calculation

Venturimeter Discharge and Pressure Difference Calculation

Problem Statement

A 20 cm x 10 cm venturimeter is mounted in a vertical pipeline carrying oil of specific gravity 0.8 flowing upward. The throat section is located 20 cm above the inlet of the venturimeter. A differential U-tube manometer indicates a gauge deflection of 25 cm. With a discharge coefficient (Cd) of 0.96, calculate:

  • The discharge of oil (Q)
  • The pressure difference between the inlet and the throat (P₁ – P₂)

Given Data

Inlet Diameter (d₁) 20 cm = 0.2 m
Inlet Area (A₁) (π/4) × (0.2)² ≈ 0.0314 m²
Throat Diameter (d₂) 10 cm = 0.1 m
Throat Area (A₂) (π/4) × (0.1)² ≈ 0.00785 m²
Specific Gravity of Oil (S₀) 0.8
Density of Oil (ρ) 0.8 × 1000 = 800 kg/m³
Elevation Difference (Z₂ – Z₁) 20 cm = 0.2 m
Manometer Deflection (x) 25 cm = 0.25 m
Specific Gravity of Mercury (S) 13.6
Discharge Coefficient (Cd) 0.96
Acceleration due to Gravity (g) 9.81 m/s²

1. Determining the Head (h)

The head is calculated using the manometer reading:

h = x × (S/S₀ – 1)

Substituting the values:
h = 0.25 m × (13.6/0.8 – 1) = 0.25 × (17 – 1) = 0.25 × 16 = 4 m

2. Calculating the Discharge (Q)

The discharge is given by:

Q = Cd × √(2gh) × (A₁ × A₂) / √(A₁² – A₂²)

Substituting the values:
Q = 0.96 × √(2 × 9.81 × 4) × (0.0314 × 0.00785) / √(0.0314² – 0.00785²)

Evaluating the above expression gives approximately Q ≈ 0.0689 m³/s.

Discharge (Q) = 0.0689 m³/s

3. Calculating the Pressure Difference (P₁ – P₂)

The head can also be expressed as:

h = (P₁ – P₂)/(ρg) + (Z₁ – Z₂)

Rearranging for (P₁ – P₂):
(P₁ – P₂) = ρg × (h – (Z₁ – Z₂))

Note: Since the throat is 20 cm above the inlet, Z₁ – Z₂ = -0.2 m (or considering the magnitude, 0.2 m added to h). Thus, the effective head for pressure difference becomes: h + 0.2 = 4 + 0.2 = 4.2 m.

Now, substituting the values:
(P₁ – P₂) = 800 kg/m³ × 9.81 m/s² × 4.2 m ≈ 32962 N/m²

Pressure Difference (P₁ – P₂) ≈ 32962 N/m²

Conclusion

The venturimeter analysis shows that the discharge of oil is approximately 0.0689 m³/s, and the pressure difference between the inlet and the throat sections is about 32962 N/m². The calculation uses the manometer head corrected for the difference in specific gravities, the geometric areas of the inlet and throat, and the discharge coefficient.

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