P HYSICS 111 Fluid dynamics and Static Fluid Review
C ONCEPT C HECKER 1:. An open bottle is filled with a liquid which is flowing out trough a spigot located at the distance h below the surface of the liquid. What is the velocity of the liquid leaving the bottle? What theorem is this? (A) v = √ ℎ (B) 2gh (C) 4gh (D) ρgh (E)√2 ℎ
S OLUTION : E (E)√2 ℎ Torricelli’s Theorem
W ARM UP PROBLEM : Water runs through a water main of cross-sectional area 0.4 m 2 with a velocity of 6 m/s. Calculate the velocity of the water in the pipe when the pipe tapers down to a cross-sectional area of 0.3 m 2
S OLUTION : Av= Av V2 =.4 /.3 * 6 = 8 m/s
P RACTICE P ROBLEM 1: Crew members attempt to escape from a damaged submarine m below the surface. What force must be applied to a pop-out hatch, which is 1.25 m by m, to push it out at that depth? Assume that the air inside the submarine is at atmospheric pressure and that the density of the ocean water is 1025 kg/m 3
S OLUTION : Need to first figure out the pressure difference Pressure inside: P= atmospheric pressure Pressure outside P2 = p1 + dgh P1 = atmospheric pressure H= 100 d= 1025 kg.m^3 F=delta PA F= (125)(9.8)(100)(1.25*.600) = 7.53 * 10^5 N
P ROBLEM 2: A paperweight, when weighed in air, has a weight of w = 6.90 N. When completely immersed in water, however, it has a weight of w6 *u1", = 4.30 N. Find the density of the paperweight.
S OLUTION : Fb = Fwater – weight 6.90 – 4.30 = 2.60 N W=mg M=w/g = 6.90 / 9.81 =.704 kg Fb = dVg V= Fb/ dg = 2.60 / 1*10^3 * 9.81 = 2.65 * 10 ^ -4 kg/m^3 D = m/v =.704 / 2.65 * 10 ^-4 m^3 = 2.65 * 10 ^3 kg/m^3
P RACTICE P ROBLEM 3: Through a refinery, fuel ethanol is flowing in a pipe at a velocity of 1 m/s and a pressure of Pa. The refinery needs the ethanol to be at a pressure of 2 atm ( Pa) on a lower level. How far must the pipe drop in height in order to achieve this pressure? Assume the velocity does not change. (Hint: Use the Bernoulli equation. The density of ethanol is 789 kg/m3 and gravity g is 9.8 m/s2. Pay attention to units!)
S OLUTION : 1/ 2 ^2 + ℎ + 1 = 1/ 2 ^2 + ℎ 2 + P2 V is constant ℎ = ℎ P1-p2 / dg = delta h Delta h = meters