In a network with two parallel pipes D1 = 0.15 m and D2 = 0.20 m, same length and roughness, to have equal head loss; If total Q = 0.6 m^3/s, compute Q1 and Q2.

• A. Q1 = 0.197 m^3/s; Q2 = 0.403 m^3/s
• B. Q1 = 0.300 m^3/s; Q2 = 0.300 m^3/s
• C. Q1 = 0.100 m^3/s; Q2 = 0.500 m^3/s
• D. Q1 = 0.450 m^3/s; Q2 = 0.150 m^3/s

Answer: (A)

When two parallel pipes have the same length and roughness, they experience the same head loss. For the Darcy–Weisbach loss, with identical L and roughness, the head loss is proportional to V^2/D. Since V = Q/A and A ∝ D^2, this leads to h_f ∝ Q^2 / D^5.

Equal head losses in the two pipes give the relation Q1^2 / D1^5 = Q2^2 / D2^5. Taking the square root gives the flow ratio Q1/Q2 = (D1/D2)^{5/2}. With D1 = 0.15 m and D2 = 0.20 m, D1/D2 = 0.75, so (0.75)^{2.5} ≈ 0.487. Thus Q1 ≈ 0.487 Q2 and the total flow 0.60 m^3/s satisfies Q1 + Q2 = 0.60, giving (0.487 + 1) Q2 ≈ 0.60, so Q2 ≈ 0.403 m^3/s and Q1 ≈ 0.197 m^3/s.

Therefore, Q1 ≈ 0.197 m^3/s and Q2 ≈ 0.403 m^3/s.