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[1] Water flowing in a pipe is determined to be moving at the velocities given in the diagram below. The higher level is 3 meters above the lower one and the pressure in the lower portion is measured to be 200 kPa. Determine the pressure inside the upper pipe Treat the water as an ideal fluid obeying Bernoullis equation. Consider the path connecting poin in the lower pipe with point 2 in the upper pipe a streamline 200 kPa 2.0 m/s ater Bernoullis equation is Using p-200 kPa-2.00 × 105 Pa. p 1000 kg/m*, y-y,-3.0m, v,-2.0m/s, and v,-3.0m/s, P2- 2.00 x 105 Pa (10 kg/m)[(2m/s)2 - (3m/s)] +(10 kg/m(9.81m/s)(-3m) - 1.68 x 105 Pa If the upper pipe has an open end, we can also determine the velocity of the water just outside the opening Pin = 168 kPa, v.-3.0 m/s Pour = Parm-10 1 kPa 168kPa + ½ ( 10, kg/m)(3m/s)-101 kPa ÷ ½ (10 kg/m) w (168kPa-101 kPa)/1% (10 kg/m)] + 9 (m/s)-vout2-143 (m/s) Vout [143] m/s = 12 m/s2] Suppose you have an incompressible liquid flowing through the horizontal pipe described below. Determine the velocity and pressure in the narrow section. Remember you can use the continuity relation relating cross-sectional areas and velocities as well as Bernoullis Law. Al-3. 14 cm2 v,-0.4 m/s P,-150 kPa A,- .3 14 cm2 V,-,, P,-? A1 A2

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