Oil with sp gr 0.75 is flowing through a 15cm diameter pipe under a pressure of 105KN/m2. If the total energy relative to a datum plane 2.5m below the center of the pipe is 18m, determine the flow rate of oil.

Oil Flow Rate Calculation

Problem Statement

Oil with specific gravity 0.75 is flowing through a 15 cm diameter pipe under a pressure of 105 kN/m². If the total energy relative to a datum plane 2.5 m below the center of the pipe is 18 m, determine the flow rate of oil.

Given Data

  • Specific gravity of oil (sg) = 0.75
  • Pipe diameter (d) = 15 cm = 0.15 m
  • Pressure (P) = 105 kN/m² = 105 kPa
  • Datum head (z) = 2.5 m (below pipe center)
  • Total energy (E) = 18 m

1. Calculate Cross-Sectional Area of Pipe

The cross-sectional area of the pipe is given by:

A = π × (d/2)² = π × (0.15/2)² = π × 0.075² = π × 0.005625

A = 0.01767 m²

2. Apply Bernoulli’s Equation

According to Bernoulli’s equation, the total energy at any point in a fluid flow is:

E = z + P/(ρg) + V²/(2g)

Where:
E = Total energy (head) in meters
z = Elevation head (datum head) in meters
P/(ρg) = Pressure head in meters
V²/(2g) = Velocity head in meters

For our problem:
E = 18 m
z = 2.5 m
P = 105 kPa
ρ = 0.75 × 1000 kg/m³ = 750 kg/m³ (since specific gravity is 0.75)
g = 9.81 m/s²

Rearranging the Bernoulli equation to find velocity (V):

V²/(2g) = E – z – P/(ρg)
V²/(2g) = 18 – 2.5 – 105000/(750×9.81)
V²/(2g) = 18 – 2.5 – 14.27
V²/(2g) = 1.23
V² = 1.23 × 2g = 1.23 × 2 × 9.81 = 24.13
V = √24.13 = 4.91 m/s

3. Calculate Flow Rate

The volumetric flow rate (Q) is the product of cross-sectional area and fluid velocity:

Q = A × V = 0.01767 m² × 4.91 m/s = 0.0867 m³/s

The flow rate of oil is 0.0867 m³/s

Verification

We can verify our solution by checking if the calculated velocity satisfies the Bernoulli equation:

E = z + P/(ρg) + V²/(2g)
E = 2.5 + 105000/(750×9.81) + 4.91²/(2×9.81)
E = 2.5 + 14.27 + 1.23
E = 18 m

This confirms our calculation is correct.

Additional Notes

  • The flow rate can also be expressed as 86.7 liters per second (0.0867 × 1000)
  • This calculation assumes steady, incompressible flow with negligible frictional losses
  • The specific gravity of 0.75 means the oil is lighter than water, which has implications for energy calculations
  • For engineering applications, this flow rate would need to be considered when selecting pumps and designing the piping system

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