A Venturimeter is to be fitted in a horizontal pipe of 0.15m diameter to measure a flow of water which may be anything up to 240m3/hour. The pressure head at the inlet for this flow is 18m above atmospheric and the pressure head at the throat must not be lower than 7m below atmospheric. Between the inlet and the throat there is an estimated frictional loss of 10% of the difference in pressure head between these points. Calculate the minimum allowable diameter for the throat.

Venturimeter Analysis

Venturimeter Analysis for Flow Measurement

Problem Statement

A Venturimeter is to be fitted in a horizontal pipe of 0.15 m diameter to measure a flow of water which may be anything up to 240 m3/hour. The pressure head at the inlet for this flow is 18 m above atmospheric, and the pressure head at the throat must not be lower than 7 m below atmospheric. Between the inlet and the throat, there is an estimated frictional loss of 10% of the difference in pressure head between these points. Calculate the minimum allowable diameter for the throat.

Given Data

Diameter of pipe (d1) 0.15 m
Cross-sectional area (A1) π/4 × (0.15)2 = 0.01767 m2
Discharge (Q) 240 m3/hour = 240/3600 = 0.0667 m3/s
Velocity at inlet (V1) 3.77 m/s
Pressure head at inlet (P1/ρg) 18 m
Pressure head at throat (P2/ρg) -7 m
Frictional loss (hL) 0.1 × (18 – (-7)) = 2.5 m
Acceleration due to Gravity (g) 9.81 m/s2

1. Calculate Inlet Area and Velocity

The cross-sectional area of the pipe is:

A1 = π/4 × (0.15 m)2 = 0.01767 m2

The discharge in SI units is:

Q = 240 m3/hour = 0.0667 m3/s

Therefore, the velocity at the inlet is:

V1 = Q / A1 = 0.0667 / 0.01767 ≈ 3.77 m/s

2. Determine the Frictional Loss

With pressure heads:

P1/ρg = 18 m, P2/ρg = -7 m

The frictional head loss is:

hL = 0.1 × (18 – (-7)) = 0.1 × 25 = 2.5 m

3. Apply Bernoulli’s Equation

For a horizontal pipe (Z1 = Z2), Bernoulli’s equation between the inlet (section 1) and the throat (section 2) is:

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

Rearranging:

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

Substituting known values:

18 – (-7) = (V22 – (3.77)2)/(2 × 9.81) + 2.5

Simplifying:

25 = (V22 – 14.22)/(19.62) + 2.5

Solving for V22:

(V22 – 14.22) = (25 – 2.5) × 19.62 = 22.5 × 19.62 = 441.45
V22 = 441.45 + 14.22 = 455.67
V2 ≈ √455.67 ≈ 21.34 m/s

4. Calculate the Minimum Throat Diameter

The cross-sectional area at the throat is:

A2 = Q / V2 = 0.0667 / 21.34 ≈ 0.003126 m2

Equating the area to the circular cross-section:

A2 = π/4 × d22

Solving for d2:

d2 = √(4A2/π) = √(4 × 0.003126/π) ≈ 0.063 m

Thus, the minimum allowable throat diameter is approximately 63 mm.

Throat Diameter (d2) ≈ 0.063 m (63 mm)

Conclusion

For a maximum flow of 240 m3/hour with a pressure head of 18 m at the inlet and ensuring that the throat pressure head does not drop below -7 m (after accounting for a frictional loss of 2.5 m), the analysis shows that the minimum allowable throat diameter is approximately 63 mm.

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