At one point in a pipeline the water’s speed is 3.00 m/s and the gauge pressure is 5.00×104Pa. Find the gauge pressure at a second point in the line, 11.0 m lower than the first, if the pipe diameter at the second point is twice that at the first.

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Tuthornt · Accepted Answer

Step 1 The diameters of the two pipe cross-sections are d1 and d2=2d1. The area of the first is A1=πd12 and of the the second A2=πd22=π4d12=4A1 Step 2 Let&#039;s first determine the relation between the velocities based on the continuity equation v1A1=v2A2 and the fact that A2=4A1 from which we can write v2=v1A1A2 v2=v14 Step 3 Since the atmospheric pressures are the same we can write Bernoulli&#039;s equation as p1+ρgy1+12ρv12=p2+ρgy2+12ρv22 where p1 and p2 are gauge pressures. We can now insert v2=v14 to get p2=p1+ρg(y1−y2)+12ρ(v12−v1216) p2=p1+ρg(y1−y2)+1532ρv12 Step 4 At this point we can insert all the given values to get the following p2=5×104+103×9.81×11+103×1532×32 p2=162kPa

At one point in a pipeline the water’s speed is 3.00 m/s and the gauge pressure

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