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Terms

Terms.  = M/V. S.G. =  sample /  water. P = F/A. P = gh. Density Specific Gravity Pressure Gauge Pressure Absolute Pressure Pascal’s Principle. P abs = gh + P atm. Pressure applied to a confined fluid is transmitted throughout the entire fluid and acts in all directions.

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Terms

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  1. Terms  = M/V S.G. = sample/ water P = F/A P = gh Density Specific Gravity Pressure Gauge Pressure Absolute Pressure Pascal’s Principle Pabs = gh + Patm Pressure applied to a confined fluid is transmitted throughout the entire fluid and acts in all directions *In pressure problems…the force is always perpendicular to the surface area

  2. Pressure Examples Suction Cups Remove air under the cup Outside air pressure pushes against cup Difference in air pressure…”suction” Straws Reduce pressure at top of straw Greater pressure on bottom Net upward push…liquid rises Question: What is the longest straw you could use?

  3. …more Terms Archimedes Principle the buoyant force on a body immersed in a fluid is equal to the weight of the fluid displaced by the object* A F1 h FB FB = F2 – F1 Acts upward…because F2 is greater due to “h2” *FB = FgV = mFg (V = Ah) F2 Buoyant Force The buoyant force occurs because the pressure in a fluid increases with depth

  4. …more Buoyancy FB = w’ The buoyant force is equal to the weight of the liquid displaced by the object… …the buoyant force is equal to the weight of the body of fluid whose volume equals the volume of the original submerged object Objects appear to weigh less when submerged in a fluid Statue example: pg. 284

  5. Applying Archimedes Principle scale reads 13.4 kg FT = Fg (w = mg) FT = w scale reads 14.7 kg FT F’T FB m w m w F’T + FB = w w’ = m’g = F’Tscale reading based on effective weight F’T = w’ = w – FB Therefore FB = w – w’

  6. finishing up calculations… FB = FgV so (w - w’) = FB = FgV set a ratio: w = samplegV (w – w’) FgV S.G. = sample = w water (w – w’) Specific Gravity definition

  7. Fluid Flow Streamline (laminar flow) - smooth Turbulent Flow – Eddy Currents

  8. Fluid Terms Viscosity – internal friction Syrup is more viscous than water Flow Rate (f = Av) Mass Flow Rate = Δm / Δt = ( ΔV) / Δt = Av Equation of Continuity (Av)1 = (Av)2since  is constant in a fluid… **in = out (Av)1 = (Av)2** *a large cross-sectional area results in a slower fluid velocity

  9. Bernoulli’s Equation • The fluid is incompressible • The fluid’s viscosity is negligible • The flow is streamline Bernoulli’s Principle: where the velocity of a fluid is high…the pressure is low where the velocity of a fluid is low…the pressure is high *less fluid pressure results in greater fluid acceleration* P1 + ½ v12 + gy1 = P2 + ½ v22 + gy2 P + ½ v12 + gy = Constant P + gy + ½ v2

  10. Examples • A pipe of non-uniform diameter carries water. At one point in the pipe, the radius is 2 cm and the flow speed is 6 m/s. a) What is the flow rate? b) What is the flow speed at a point where the pipe constricts to a radius of 1 cm? • f = Av = πr2v = π(0.02m)2(6 m/s) = 0.0075 m3/s • v α 1/A & A α r2 so if A decreases by a factor of 4…v increases by a factor of 4…4 * (6 m/s) = 24 m/s

  11. Examples 2. If the diameter of a pipe increases from 4cm to 12 cm, what will happen to the flow speed? A = πr2 = π(1/2 d)2 = ¼ πd2…so if d increases by a factor of 3…flow speed decreases by a factor of 9.

  12. Examples 3. What does Bernoulli’s Equation tell us about a fluid at rest in a container open to the atmosphere? Because the fluid in the tank is at rest: v1 & v2 = 0 m/s P1 + gy1 = P2 + gy2 since P1 = P atm… P2 = Patm + g(y1 – y2) = Patm + gh… which is the formula for Absolute Pressure!!!

  13. Examples 4. In the figure below, a pump forces water at a constant flow rate through a pipe whose cross-sectioanl area, A, gradually decreases: at the exit point, A has decreased to 1/3 its value. If y=60cm and the flow speed of the water at point 1 is 1 m/s, what is the gauge pressure at point 1? exit P1 + ½ v12 + gy1 = P2 + ½ v22 + gy2 y1 = 0 P1 – Patm = gy2+ ½ v22 - ½ v12 P1 – Patm = gy2+ ½ (3v1)2 - ½ v12 P1 – Patm = (gy2 + 4v12) P1 – Patm = (1000 kg/m3)[(10 m/s)(0.6 m) + 4(1 m/s)2] Pg = 10000 Pa y Point 1 Pump

  14. Examples 5. The side of an above-ground pool is punctured, and water gushes out through the hole. If the total depth of the pool is 2.5 m, and the puncture is 1 m above ground level, what is the speed of the water gushing out? gy1 = ½ v22 + gy2 v2 = 2g(y1-y1) = 2gh = 30 = 5.5 m/s

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