A 1 km long, 0.6 m diameter pipeline has a friction factor of 0.05. The height difference between the inlet and the exit of the pipe to deliver a discharge of 0.1 m3/s of water is equal to:

Question

BlackTom AI · Accepted Answer

To tackle this problem, we first restate what is being asked and identify the given data: we need the height difference (headloss) required to deliver a discharge of 0.1 m^3/s through a 1 km long pipe with D = 0.6 m and friction factor f = 0.05.

Option 1: 1.03 m
- If we plug in the numbers roughly, the headloss from Darcy–Weisbach tends to be around a few tenths of a meter for these dimensions. A value as large as 1.03 m would imply a much larger v^2/(2g) term or a much larger L/D or f, which is inconsistent with the given modest flow and long pipe. This option is unlikely.

Option 2: 0.53 m
- Compute the flow velocity: area A = πD^2/4 = π(0.6)^2/4 ≈ 0.2827 m^2. Velocity v = Q/A = 0.1 / 0.2827 ≈ 0.3536 m/s. Headloss h_f = f (L/D) (v^2/(2g)); L/D = 1000/0.6 ≈ 1666.7. v^2/(2g) = (0.3536^2)/(2×9.81) ≈ 0.125/(19.62) ≈ 0.00637. Then h_f ≈ 0.05 × 1666.7 × 0.00637 ≈ 0.530 m. This matches the computed value, making 0.53 m the correct estimate.

Option 3: 0.31 m
- This value would arise if either the velocity term or L/D factor were significantly smaller, or if a misapplication of the formula occurred (for instance using a smaller v^2/(2g) or neglecting a portion of the headloss). Given the actual calculation, 0.31 m is not consistent with the proper substitution of Q, D, f, and L.

Option 4: 0.87 m
- Achieving 0.87 m would require a larger headloss than allowed by the given parameters. With the same Q, D, f, and L, the product f(L/D)(v^2/(2g)) yields roughly 0.53 m, so 0.87 m would indicate an error in the velocity term or in the given data. This overestimates the headloss.

Option 5: 2.12 m
- A headloss of 2.12 m would imply an unusually high loss for these dimensions and flow rate, which is inconsistent with the computed value using the Darcy–Weisbach relation and the given numbers. This would correspond to either a much larger v^2/(2g) term or an erroneous application of the formula, making it clearly incorrect.

In summary, the analysis uses the Darcy–Weisbach headloss h_f = f (L/D) (v^2/(2g)) with v derived from Q and pipe area. The computed result h_f ≈ 0.53 m agrees with option 0.53 m and is consistent with the provided parameters. The other options either overestimate or underestimate the headloss and do not align with the correct substitution of Q, D, f, and L in the flow equations.

A 1 km long, 0.6 m diameter pipeline has a friction factor of 0.05. The height difference between the inlet and the exit of the pipe to deliver a discharge of 0.1 m3/s of water is equal to: 单项选择题

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