Consider an oil flow of 13 m3/h in a long, straight and smooth pipe with diameter of 6 cm. What is the pressure loss per meter? Oil absolute viscosity is 0,1 kg/ms and density is 900 kg/m3. Friction factor for laminar flow can be calculated fromf=64Re f = \frac{64}{Re} .Round off the answer to an integer in Pa/m, but enter the answer without the unit.

Question

BlackTom AI · Accepted Answer

We start by restating the given data and identifying the required quantity. The task asks for the pressure loss per meter (ΔP/L) for oil flowing in a long, straight, smooth pipe with D = 0.06 m, Q = 13 m3/h, oil viscosity μ = 0.1 kg/(m·s), and density ρ = 900 kg/m3. The laminar friction factor is f = 64/Re, with Re = ρVD/μ.

Step 1: Convert the flow rate to a usable form and compute the flow velocity.
- Q = 13 m3/h = 13/3600 = 0.003611111... m3/s.
- Cross-sectional area A = πD^2/4 = π(0.06)^2/4 = π(0.0036)/4 ≈ 0.00282743 m2.
- Velocity V = Q/A ≈ 0.003611111 / 0.00282743 ≈ 1.2777 m/s.

Step 2: Calculate Reynolds number to determine the flow regime.
- Re = ρVD/μ = 900 × 1.2777 × 0.06 / 0.1 ≈ 689–690 (approximately 690).
- Since Re < 2100, the flow is laminar, which justifies using f = 64/Re.

Step 3: Compute the friction factor for laminar flow.
- f = 64/Re ≈ 64 / 690 ≈ 0.0928 (more precisely about 0.0927–0.0928).

Step 4: Use the Darcy–Weisbach equation for pressure drop per unit length in laminar flow.
- ΔP/L = f × (ρ V^2) / (2 D).
- Dynamic pressure term: ρ V^2 / 2 ≈ 900 × (1.2777)^2 / 2 ≈ 900 × 1.632 / 2 ≈ 900 × 0.816 ≈ 734.4 Pa.
- Divide by D: (ρ V^2) / (2 D) ≈ 734.4 / 0.06 ≈ 12,240 Pa/m.
- Multiply by f: ΔP/L ≈ 0.0927 × 12,240 ≈ 1,135 Pa/m.

Step 5: Round to the requested format.
- The calculation gives ΔP/L ≈ 1134.9 Pa/m, which rounds to 1135 Pa/m when expressed as an integer without units.

Therefore, the pressure loss per meter is approximately 1135 (Pa/m). If you compare this with the rounded intermediate values, you’ll see the consistency across slightly different rounding steps.

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