Question 13 of 18advanced🔧 ApplyNumerical5 marks

Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0×104.0 \times 10⁻³ kg s⁻¹, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3×101.3 \times 10³ kg m⁻³ and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].

Correct Answer

Given: length L = 1.5 m, radius r = 1.0 cm = 0.01 m, mass flow rate = 4.0×104.0 \times 10⁻³ kg s⁻¹, density ρ = 1.3×101.3 \times 10³ kg m⁻³, viscosity η = 0.83 Pa s. Volume flow rate Q = mass flow rate / ρ = (4.0×104.0 \times 10⁻³) / (1.3×101.3 \times 10³) = 3.08×103.08 \times 10⁻⁶ m³ s⁻¹. For laminar flow through a horizontal tube, Poiseuille's law gives Q = (π r⁴ ΔP) / (8 η L). Rearranging, ΔP = (8 η L Q) / (π r⁴). Substituting values: ΔP = (8×0.838 \times 0.83 × 1.5×3.081.5 \times 3.08 × 10⁻⁶) / (3.14 × (0.01)⁴). Compute r⁴ = 1×101 \times 10⁻⁸ m⁴. Thus ΔP ≈\approx 975 Pa. To verify laminar flow, calculate Reynolds number Re = (ρ v D) / η, where average velocity v = Q / (π r²) ≈\approx 0.0098 m s⁻¹, diameter D = 0.02 m. Re = (1.3×101.3 \times 10³ × 0.0098×0.020.0098 \times 0.02) / 0.83 ≈\approx 0.31, which is far below the critical value of 2000, confirming laminar flow. Hence, the pressure difference is about 975 Pa.

Exercise: EXERCISES | Q: 9.13 | (Page 21)
For More Understanding

Explanation

The solution uses Poiseuille's law for laminar flow through a cylindrical tube, which relates volume flow rate to pressure difference, viscosity, length, and radius. The mass flow rate is converted to volume flow rate using density. The Reynolds number is calculated to verify laminar flow assumption.

Solution Steps

  1. Step 1: Convert radius to meters: r = 1.0 cm = 0.01 m.

  2. Step 2: Calculate volume flow rate Q = (mass flow rate) / density = (4.0×104.0 \times 10⁻³ kg s⁻¹) / (1.3×101.3 \times 10³ kg m⁻³) = 3.08×103.08 \times 10⁻⁶ m³ s⁻¹.

  3. Step 3: Apply Poiseuille's formula: Q = (π r⁴ ΔP) / (8 η L). Rearranged: ΔP = (8 η L Q) / (π r⁴).

  4. Step 4: Substitute values: η = 0.83 Pa s, L = 1.5 m, Q = 3.08×103.08 \times 10⁻⁶ m³ s⁻¹, r⁴ = (0.01)⁴ = 1×101 \times 10⁻⁸ m⁴, π = 3.14.

  5. Step 5: Compute ΔP = (8×0.838 \times 0.83 × 1.5×3.081.5 \times 3.08 × 10⁻⁶) / (3.14×13.14 \times 1 × 10⁻⁸) ≈\approx 975 Pa.

  6. Step 6: Check laminar flow: average velocity v = Q / (π r²) = 3.08×103.08 \times 10⁻⁶ / (3.14 × (0.01)²) ≈\approx 0.0098 m s⁻¹. Reynolds number Re = (ρ v D) / η, where D = 2r = 0.02 m. Re = (1.3×101.3 \times 10³ × 0.0098×0.020.0098 \times 0.02) / 0.83 ≈\approx 0.31 < 2000, so flow is laminar.

PreviousNext