Title HPGE 25 Solutions.pdf ✅

25 Hydraulics: Pipe Flow (miscellaneous) Solutions SITUATION 1. A 20-mm thick, non-rigid steel pipe with a diameter of 450 mm discharges freely under a head of 200 m. Rapid closure of the valve stops a discharge of 0.32 m3 s . Consider ESteel = 200 GPa and EB (water) = 2100 MPa. ▣ 1. Determine the composite modulus of elasticity. [SOLUTION] 1 E = 1 Ef + D Ept 1 E = 1 2100 MPa + 450 mm (200000 MPa)(20 mm) E = 1698.69 MPa ▣ 2. What water hammer pressure would develop in the walls of the pipe near the valve? [SOLUTION] ΔP = ρcV Solve for the celerity: c = √ E ρ c = √ 1698.69 × 106 Pa 1000 kg m3 c = 1303.3 m s For the velocity: V = Q A V = 0.32 m3 s π 4 (0.45 m) 2 V = 2.012 m s
ΔP = (1000 kg m3 ) (1303.3 m s ) (2.012 m s ) ΔP = 2622.36 kPa ▣ 3. What is the maximum stress in the pipe? [SOLUTION] Compute for the total pressure P = Pstatic + ΔP P = (9.81 kN m3 ) (200 m) + 2622.36 kPa P = 4584.36 kPa Compute for the hoop stress: σ = PD 2t σ = (4584.36 kPa)(0.45 m) 2(0.02 m) σ = 51.57 MPa SITUATION 2. A 3.2 km rigid pipe with a diameter of 600 mm discharges oil (Sg = 0.8, Eb=1800 MPa) at a rate of 0.5 m3 s . ▣ 4. Determine the celerity of the pressure wave. [SOLUTION] c = √ E ρ c = √ 1800 × 106 Pa 0.8 (1000 kg m3 ) c = 1500 m s ▣ 5. What water hammer pressure would develop in the walls of the pipe if a valve at the outlet were closed in 4 seconds? [SOLUTION] Compute for the critical time (roundtrip time)