Two pipes of equal and constant diameter leave a water pumping station and dump water out of an open end that is open to the atmosphere (see the following figure). The water enters at a pressure of two atmospheres and a speed of (v1 = 1.0 m/s). One pipe drops a height of 10 m. What is the velocity of the water as the water leaves each pipe?
Given: Inlet pressure P1=2 atm=2.026×105 Pa Inlet velocity v1=1.0 m/s Outlet pressure P2=P3=1 atm=1.013×105 Pa Density of water ρ=1000 kg/m3 Acceleration due to gravity g=9.8 m/s2 1. For the horizontal pipe: The problem states the pipes have a constant diameter. According to the equation of continuity: A1v1=A2v2 Since the diameter is constant, the cross-sectional area A1=A2. Therefore: v2=v1=1.0 m/s 2. For the pipe with a 10 m drop: Applying Bernoulli's equation between the inlet (point 1) and the outlet (point 3): P1+21ρv12+ρgh1=P3+21ρv32+ρgh3 Taking the outlet level as the reference (h3=0), then h1=10 m. 2.026×105+21(1000)(1.0)2+(1000)(9.8)(10)=1.013×105+21(1000)v32+0 202600+500+98000=101300+500v32 301100=101300+500v32 199800=500v32 v32=500199800 v32=399.6 v3=399.6≈19.99 m/s Final velocities: Horizontal pipe: 1.0 m/s Dropping pipe: 20.0 m/s