Calculate theoretical friction loss for a 45 mm hose line 75 meters long operating at 375 l/min.

Calculate theoretical friction loss for a 45 mm hose line 75 meters long operating at 375 l/min.

• A. 300 kPa
• B. 259.45 kPa
• C. 200 kPa
• D. 350 kPa

Answer: (B)

Friction loss in a hose line comes from the pressure drop created by flowing water along the length of the hose, which can be estimated with the Darcy-Weisbach equation: h_f = f (L/D) (V^2 / (2g)). Start by finding the flow velocity from the given flow and diameter. The hose is 45 mm (0.045 m) in diameter and the flow is 375 L/min, which is 0.375 m^3/min = 0.00625 m^3/s. The cross-sectional area is A = πD^2/4 ≈ π(0.045)^2/4 ≈ 0.001589 m^2, so the velocity V = Q/A ≈ 0.00625 / 0.001589 ≈ 3.93 m/s. With water, the Reynolds number is very high at this velocity and diameter, indicating turbulent flow. A typical friction factor for smooth bore hose at this condition is about f ≈ 0.02. Now compute the headloss: L/D = 75 / 0.045 ≈ 1666.7, V^2/(2g) = 3.93^2 / (2×9.81) ≈ 15.45 / 19.62 ≈ 0.787. So h_f ≈ 0.02 × 1666.7 × 0.787 ≈ 26.3 meters of water. Convert to pressure: p = ρ g h_f ≈ 1000 × 9.81 × 26.3 ≈ 257–258 kPa. This matches the given value, about 259 kPa.

Friction loss in a hose line comes from the pressure drop created by flowing water along the length of the hose, which can be estimated with the Darcy-Weisbach equation: h_f = f (L/D) (V^2 / (2g)). Start by finding the flow velocity from the given flow and diameter. The hose is 45 mm (0.045 m) in diameter and the flow is 375 L/min, which is 0.375 m^3/min = 0.00625 m^3/s. The cross-sectional area is A = πD^2/4 ≈ π(0.045)^2/4 ≈ 0.001589 m^2, so the velocity V = Q/A ≈ 0.00625 / 0.001589 ≈ 3.93 m/s.

With water, the Reynolds number is very high at this velocity and diameter, indicating turbulent flow. A typical friction factor for smooth bore hose at this condition is about f ≈ 0.02. Now compute the headloss: L/D = 75 / 0.045 ≈ 1666.7, V^2/(2g) = 3.93^2 / (2×9.81) ≈ 15.45 / 19.62 ≈ 0.787. So h_f ≈ 0.02 × 1666.7 × 0.787 ≈ 26.3 meters of water.

Convert to pressure: p = ρ g h_f ≈ 1000 × 9.81 × 26.3 ≈ 257–258 kPa. This matches the given value, about 259 kPa.