Steady Flow in a Constricted Hose: Calculating Volume Flow Rate
Calculating Volume Flow Rate of Olive Oil in a Constricted Hose
Given Data:
- Density of olive oil, ρ = 875 kg/m³
- Diameter of hose at inlet, d₁ = 3.20 cm = 0.032 m
- Diameter of hose at outlet, d₂ = 1.25 cm = 0.0125 m
- Change in pressure between inlet and outlet, ΔP = -4900 Pa (negative as the flow is from higher pressure to lower pressure)
Assumptions:
- Steady, ideal flow
- Incompressible fluid
Calculate the area of the hose at the inlet and outlet:
Area at inlet, A₁: π(d₁/2)² = π(0.032/2)² = 8.0425 × 10⁻⁴ m²
Area at outlet, A₂: π(d₂/2)² = π(0.0125/2)² = 1.5708 × 10⁻⁴ m²
Apply Bernoulli's equation to relate pressure, velocity, and density at the inlet and outlet:
Pressure at inlet + (1/2)ρv₁² = Pressure at outlet + (1/2)ρv₂²
Rearrange Bernoulli's equation to solve for velocity at the inlet:
Velocity at inlet, v₁ = √[(2(P₁ - P₂)) / ρ]
Substitute given values to find velocity at the inlet:
Velocity at inlet, v₁ = √[(2(-4900)) / (875)] = 3.90 m/s
Calculate the volume flow rate at the inlet:
Volume Flow Rate, Q = v₁A₁ = (3.90)(8.0425 × 10⁻⁴) = 0.00310 m³/s
Therefore, the volume flow rate of olive oil in the constricted hose is 0.00310 m³/s based on the given data.