Presentation on theme: "Terms Density Specific Gravity Pressure Gauge Pressure"— Presentation transcript:
1 Terms Density Specific Gravity Pressure Gauge Pressure Absolute PressurePascal’s Principle = M/VS.G. = sample/ waterP = F/AP = ghPabs = gh + PatmPressure 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 Straws Reduce pressure at top of straw Remove air under the cupOutside air pressure pushes against cupDifference in air pressure…”suction”StrawsReduce pressure at top of strawGreater pressure on bottomNet upward push…liquid risesQuestion: What is the longest straw you could use?
3 …more Terms Archimedes Principle Buoyant Force the buoyant force on a body immersed in a fluid is equal to the weight of the fluid displaced by the object*AF1hFBFB = F2 – F1Acts upward…because F2 is greater due to “h2”*FB = FgV = mFg (V = Ah)F2Buoyant ForceThe buoyant force occurs because the pressure in a fluid increases with depth
4 Objects appear to weigh less when submerged in a fluid …more BuoyancyFB = 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 objectObjects appear to weigh less when submerged in a fluidStatue example: pg. 284
5 Applying Archimedes Principle scale reads 13.4 kgscale reads 14.7 kgFTF’TFBmwmFT = Fg (w = mg)FT = wwF’T + FB = ww’ = m’g = F’T scale reading based on effective weightF’T = w’ = w – FBTherefore FB = w – w’
6 finishing up calculations… FB = FgVso (w - w’) = FB = FgVset a ratio: w = samplegV(w – w’) FgVS.G. = sample = wwater (w – w’)Specific Gravity definition
8 *a large cross-sectional area results in a slower fluid velocity Fluid TermsViscosity – internal frictionSyrup is more viscous than waterFlow Rate (f = Av)Mass Flow Rate = Δm / Δt = ( ΔV) / Δt = AvEquation of Continuity(Av)1 = (Av)2 since is constant in a fluid…**in = out (Av)1 = (Av)2***a large cross-sectional area results in a slower fluid velocity
9 *less fluid pressure results in greater fluid acceleration* Bernoulli’s EquationThe fluid is incompressibleThe fluid’s viscosity is negligibleThe flow is streamlineBernoulli’s Principle:where the velocity of a fluid is high…the pressure is lowwhere 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 + gy2P + ½ v12 + gy = ConstantP + gy + ½ v2
10 ExamplesA 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) = m3/sv 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 Examples2. 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 Examples3. 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/sP1 + gy1 = P2 + gy2since P1 = P atm…P2 = Patm + g(y1 – y2) = Patm + gh…which is the formula for Absolute Pressure!!!
13 Examples4. 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?exitP1 + ½ v12 + gy1 = P2 + ½ v22 + gy2y1 = 0P1 – Patm = gy2 + ½ v22 - ½ v12P1 – Patm = gy2 + ½ (3v1)2 - ½ v12P1 – Patm = (gy2 + 4v12)P1 – Patm = (1000 kg/m3)[(10 m/s)(0.6 m) + 4(1 m/s)2]Pg = PayPoint 1Pump
14 Examples5. 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 + gy2v2 = 2g(y1-y1) = 2gh = 30 = 5.5 m/s