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Module 2: Hydrostatics Module 2: Hydrostatics . Hydrostatic pressure and devices: 2 lectures . Forces on surfaces: 2.5 lectures . Buoyancy, Archimedes, stability: 1.5 lectures Mech 280: Frigaard Lectures 1-2: Hydrostatic pressure . Should be able to: . Use common pressure terminology . Derive the general form for the pressure distribution in static fluid . Calculate the pressure within a constant density fluids . Calculate forces in a hydraulic press . Analyze manometers and barometers . Calculate pressure distribution in varying density fluid . Calculate pressure in fluids in rigid body motion in non-inertial frames of reference Mech 280: Frigaard Pressure . Pressure is defined as a normal force exerted by a fluid per unit area . SI Unit of pressure is N/m2, called a pascal (Pa). Since the unit Pa is too small for many pressures encountered in engineering practice, kilopascal (1 kPa = 103 Pa) and mega-pascal (1 MPa = 106 Pa) are commonly used . Other units include bar, atm, kgf/cm2, lbf/in2=psi . 1 psi = 6.695 x 103 Pa . 1 atm = 101.325 kPa = 14.696 psi . 1 bar = 100 kPa (close to atmospheric pressure) Mech 280: Frigaard Absolute, gage, and vacuum pressures . Actual pressure at a give point is called the absolute pressure . Most pressure-measuring devices are calibrated to read zero in the atmosphere. Pressure above atmospheric is called gage pressure: Pgage=Pabs - Patm . Pressure below atmospheric pressure is called vacuum pressure: Pvac=Patm - Pabs. Mech 280: Frigaard Pressure at a Point . Pressure at any point in a fluid is the same in all directions . Pressure has a magnitude, but not a specific direction, and thus it is a scalar quantity . Later we will talk about the stress tensor: . Used to define the stress vector (traction) across any surface . Pressure is (minus) the mean normal stress σ xx σ xy σ xz − p 0 0 τ xx τ xy τ xz σ = σ σ σ = − + τ τ τ = − δ +τ ij yx yy yz 0 p 0 yx yy yz p ij ij σ σ σ − τ τ τ zx zy zz 0 0 p zx zy zz . 1st index = direction of unit normal, 2nd index = direction of force . In a fluid at rest the shear stresses are zero . Pressure is the part of the stress that remains Mech 280: Frigaard Trapezoidal paradox: fluids & solids? . Is the pressure at the bottom larger h in A or B? . If material is solid? . If material is liquid? W W . Paradox was resolved by Blaise Pascal . Shown is a Pascal vase from Yale University Mech 280: Frigaard Pascal - 1623–1662 . Blaise Pascal was a French mathematician, physicist, and religious philosopher. He was a child prodigy who was educated by his father. Pascal's earliest work was in the natural and applied sciences where he made important contributions to the construction of mechanical calculators, the study of fluids, and clarified the concepts of pressure and vacuum by generalizing the work of Evangelista Torricelli. Pascal also wrote powerfully in defense of the scientific method. Inventions include the hydraulic press (using hydraulic pressure to multiply force) and syringe. Following a mystical experience in late 1654, he abandoned his scientific work and devoted himself to philosophy and theology. His two most famous works date from this period: the Lettres provinciales and the Pensées. However, he had suffered from ill-health throughout his life and his new interests were ended by his early death two months after his 39th birthday. Mech 280: Frigaard Variation of Pressure with Depth . In the presence of a gravitational field, pressure increases with depth because more fluid rests on deeper layers. To obtain a relation for the variation of pressure with depth, consider rectangular element . Force balance in z-direction gives ∑ Fzz= ma = 0 P21∆− x Px ∆−ρ gxz ∆∆=0 . Dividing by ∆x and rearranging gives ∆P = P21 − P =ργ gz ∆=s ∆ z . We can also write this in differential form, dividing by ∆z: ∂P = ρg = γ or more generally, ∇P = ρge = γ e ∂z s g s g Mech 280: Frigaard Scuba Diving and Hydrostatic Pressure Mech 280: Frigaard Example 1: 1 a) Compute the pressure on a scuba diver at 100 ft. b) What is the danger of the diver 100 ft ascending too quickly? 2 PV= PV If you hold your breath on ascent, your lung 11 2 2 V P3.95 atm volume would increase by a factor of 4, which 12= = ≈ 4 would result in embolism and/or death. V21 P1 atm Mech 280: Frigaard Example 2: The glass flask is filled with water as shown. The column has inner diameter 3mm and the bulb of the flask has burst pressure of 3kPa. Is it safe to add 1ml to the flask? Column 8cm 10cm Bulb Mech 280: Frigaard Variation of Pressure with Depth . Pressure in a fluid at rest is independent of the shape of the container . See the Trapezoid paradox . Pressure is the same at all points on a horizontal plane in a given fluid. Consider force balance in x-direction Mech 280: Frigaard Pascal’s Law . Pressure applied to a confined fluid increases the pressure throughout by the same amount. If the pistons are at same height: FF12 FA 22 PP12=→=→= AA12 FA 11 . Pascal observed: a small force could generate a large force . Ratio A2/A1 is called ideal mechanical advantage Example 3: Hydraulic lift What is the height h of oil required in the narrow tube to lift the load in the 3 large tube? (ρoil = 900 kg/m ) Mech 280: Frigaard Fluids in the news: Mech 280: Frigaard Review: hydrostatic pressure basics . From last lecture: . General equation of hydrostatics: ∇P = ρgeg = γseg • Usually we align z with eg: ∂P = ρg = γ ∂z s . Constant density (and gravity): ∆P = P2 − P1 = ρg∆z = γs∆z z2 . Variable density: ∆P = ∫ ρg dz z1 Mech 280: Frigaard The Manometer . An elevation change of ∆z in constant density fluid at rest corresponds to ∆P/ρg . A device to measure pressure based on this is called a manometer . A manometer consists of a U-tube containing one or more fluids, e.g. mercury, water, alcohol, oil . Heavy fluids such as mercury are used if large PP12= pressure differences are = + ρ anticipated P2 Patm gh . Pressure range limited by height and fluid density Mech 280: Frigaard Measuring Pressure Drops . Manometers are well suited to measure pressure drops across valves, pipes, heat exchangers, etc. Relation for pressure drop P1-P2 is obtained by starting at point 1 and adding or subtracting ρgh terms until we reach point 2. If fluid in pipe is a gas, then ρ2>>ρ1 and P1-P2= ρgh Mech 280: Frigaard Multi-fluid Manometer . For multi-fluid systems . Pressure change across a fluid column of height h is ∆P = ρgh . Pressure increases downward, and decreases upward. Multi-fluid manometer . Two points at the same elevation in a continuous fluid are at same pressure. Pressure can be determined by adding and subtracting ρgh terms. P2++ρρ 11 gh 2 gh 2 + ρ 33 gh = P 1 Mech 280: Frigaard Example 1: Calculate the air pressure in the pressurized tank, if h1 = 0.18 m, h2 = 0.2m and h3 = 0.25m. The density of the mercury, water and oil are 13,600 kg/m3,1000 kg/m3 and 850 kg/m3, respectively. Mech 280: Frigaard The Barometer . Atmospheric pressure is measured by a device called a barometer; thus, atmospheric pressure is often referred to as the barometric pressure. PC can be taken to be zero since there is only Hg vapor above point C, and it is very low relative to Patm. Change in atmospheric pressure due to elevation has many effects: Cooking, nose bleeds, engine performance, aircraft performance. PC +=ρ gh Patm Patm = ρ gh Mech 280: Frigaard Static pressure in the atmosphere? . Throughout atmosphere density varies significantly. How does the pressure vary? . Assume Ideal gas: d P P 2 d P P g 2 d z = −ρ g = − g or ∫ = ln 2 = − ∫ d z RT 1 P P1 R 1 T . For an isothermal atmosphere with T = To? g(z2 − z1) P2 = P1 exp − RTo . How realistic is constant temperature? . How realistic is constant gravitational acceleration? Mech 280: Frigaard Example 2: Up to an altitude of approximately 36,000ft (l l,000 m), the mean atmospheric temperature decreases nearly linearly and can be represented by: T = To - B z, where B is the lapse rate. The following values are assumed to apply for air, from 2 2 sea level to 36,000 ft: To = 288.16˚K (15 ˚C); B = 6.5˚K/km; R = 287 m /(s ˚K) Calculate atmospheric pressure as a function of z. g /(RB) Bz g P = Pa 1 − where = 5.26(forair) To RB Mech 280: Frigaard Example 3: Determine the pressure and density as a function of depth for a fluid in which P=Cρn Mech 280: Frigaard Accelerating fluids in rigid-body motion . In rigid body motion there is no shearing in the fluid . Pressure is again only component of the stress . There are 2 special cases where a body of fluid can undergo rigid-body motion: . linear acceleration, and . rotation of a cylindrical container. Newton's 2nd law of motion can be used to derive an equation of motion for a fluid that acts as a rigid body ∇P = ρgeg − ρa . In Cartesian coordinates: ∂∂∂PPP =−=−=−+ρρρa,, a( ga) ∂∂∂xyzxy x Mech 280: Frigaard Example 4: Linear Acceleration The truck, length 12m is 1/4 full with milk. It accelerates at constant 1m/s2. How much higher is the level of milk at the back of the tanker than at the front? ax≠==0, aa yz 0 ∂P ∂∂ PP =ρρag, = 0, = − ∂xx ∂∂ yz Total differential of P dP=−−ρρ ax dx gdz P21−=− Pρρ ax ( x 21 − x) − gz( 21 − z) Pressure difference between 2 points a ∆=zz − z =−x ( xx −) ss2 s 1g 21 Find the rise by selecting 2 points on free surface P2 = P1 Rotation in a Cylindrical Container The container is rotating about the z-axis with constant angular velocity ω.
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