Estimate the head loss across a 90-degree bend and a throttling globe valve, given velocity head V^2/(2g).

• A. Bend h_L ≈ 0.1–0.2 [V^2/(2g)]; valve h_L ≈ 0.5 [V^2/(2g)].
• B. Bend h_L ≈ 0.9–1.5 [V^2/(2g)]; valve h_L ≈ 2–10 [V^2/(2g)].
• C. Bend h_L ≈ 5 [V^2/(2g)]; valve h_L ≈ 20 [V^2/(2g)].
• D. Bend h_L ≈ [V^2/(2g)]; valve h_L ≈ [V^2/(2g)].

Answer: (B)

In piping, head loss from fittings and devices is written as h_L = K × (V^2/(2g), where V^2/(2g) is the velocity head. For a 90-degree bend, the flow must change direction and often separates and becomes turbulent, so the loss coefficient K is of order unity. In practice, bend losses fall roughly in the 0.9 to 1.5 range of the velocity head, depending on bend radius and surface finish. A throttling globe valve, on the other hand, constricts the flow and creates substantial energy dissipation due to flow contraction and turbulence through the narrow passage; its loss coefficient can vary widely with opening but commonly lies in the 2 to 10 range of the velocity head, increasing as the valve closes. Therefore, the estimates Bend h_L ≈ 0.9–1.5 [V^2/(2g)] and Valve h_L ≈ 2–10 [V^2/(2g)] reflect typical behavior, matching the best understanding of how these components affect head loss.