Fluid Mechanics: Solving Problems with Pressure and Volume

1. Strategy: To find W, we need mass, which we can get from ρ and volume of the cylinder. x ρ P left = P right because pressure should be equal at any depth Po + ρoil g L = Po + ρmetalg x Subst ρmetal = m/vol = m / Ax ρoil g L = (m / Ax) g x A ρoil g L = m So W = mg = m A ρoil g L or choice a.

2. P + ½ ρv2 +ρgh = P + ½ ρv2 +ρgh 200,000 + ½ x1000 x 4^2 + 1000x10x0 = 150,000 + ½ 1000 v^2 + 1000x10x5 50,000 + 8000 = 500 V^2 + 50,000 58,000 – 50,000 = 500 V^2 8000/500 = V^2 16 = V^2 v = 4 m/s, so the speed doesn’t change in this case. The drop in pressure P is balanced by an increase in ρgh only. Av must = Av: since the pipe doesn’t get thinner or thicker, v must be the same. P = 200,000 Pa P = 150,000 Pa What would happen to v in and v out if the height we raised above 5 m? Both would decrease by the same amount How high would you have to raise it to stop flow? 20 m where pgh = 200, 000 Pa. H= 5 meters V = 4 m/s