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Physics 111 - Iowa State University

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Physics 111
Fluid dynamics and Static Fluid Review
Concept Checker 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 = √𝑔ℎ(B) 2gh(C) 4gh(D)ρgh(E)√2𝑔ℎ
Solution: E
(E)√2𝑔ℎTorricelli’s Theorem
Warm up problem:
Water runs through a water main of cross-sectional area 0.4 m2with 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 m2
Solution:
Av= AvV2 = .4 / .3 * 6 = 8 m/s
Practice Problem 1:
Crew members attempt to escape from a damaged submarine 100.0 m below the surface. What force must be applied to a pop-out hatch, which is 1.25 m by 0.600 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/m3
Solution:
Need to first figure out the pressure differencePressure inside:P= atmospheric pressurePressure outsideP2 = p1 +dghP1 = atmospheric pressureH= 100d= 1025 kg.m^3F=delta PAF= (125)(9.8)(100)(1.25*.600) = 7.53 * 10^5 N
Problem 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.
Solution:
Fb=Fwater– weight6.90 – 4.30 = 2.60 NW=mgM=w/g = 6.90 / 9.81 = .704 kgFb=dVgV=Fb/ dg = 2.60 / 1*10^3 * 9.81 = 2.65 * 10 ^ -4 kg/m^3D = m/v = .704 / 2.65 * 10 ^-4 m^3 = 2.65 * 10 ^3 kg/m^3
Practice Problem 3:
Through a refinery, fuel ethanol is flowing in a pipe at a velocity of 1 m/s and a pressure of 101300 Pa. The refinery needs the ethanol to be at a pressure of 2atm(202600 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!)
Solution:
1/2𝜌𝑣^2+ 𝜌𝑔ℎ+𝑃1 =1/2𝜌𝑣^2+ 𝜌𝑔ℎ2+ P2V is constant𝜌𝑔ℎ1 + 𝑃1 = 𝜌𝑔ℎ2 + 𝑃2P1-p2 / dg = delta hDelta h = -13.1 meters

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Physics 111 - Iowa State University