1PRESSURE & HYDROSTATICS Pressure = F / A • Pressure is typically measured with respect to a reference level • Reference could be atmosphere or vaccum. • The reference level = atmospheric pressure (95-105kPa) • Pressure measured with respect to atmosphere = Gage pressure • Pressure measured with respect to perfect vaccum = Absolute pressure The three related by the equation – Pabs = Pgage + Patm See Figure 3.12 Points – • Perfect vacuum = lowest possible pressure • Absolute pressure will always be positive • Gage pressure above atmospheric is +ve • Gage pressure below atmospheric will be –ve • Units = Pa (abs) or Pa (gage) – always mention what the reference is! • Range of atmospheric pressure = 95 kPa to 105 kPa3Assumed as 101 kPa in the text. Or 14.7 psi(a) Problem 3.1 Given Pressure (gage) = 155 kPa Atmospheric Pressure = 98 kPa What is the Absolute Pressure????? Absolute pressure = Pressure and Elevation: Pressure is directly related to elevation and changes with elevation - example – swimming pool or diving deep into a water body The greater the depth/elevation of fluid – the greater the pressure Elevation in water – measured with respect to a reference; positive upwards; negative downwards See Figure 3.24 It is advisable to choose the lowest point for a reference level so that you don’t have to deal with negative elevation values! ****Change in pressure = Δp = γ h Where γ is the specific weight of the fluid and h is the elevation5******Key Points: • Equation is only valid for homogenous liquids at rest • Points on the same horizontal level have the same pressure • Pressure varies linearly with change in elevation/depth • Change in pressure is proportional to the specific weight of the fluid Above equation does not apply to gases because the specific weight of gas changes with elevation – However the change in gas pressure with elevation is small! An increase in elevation of 300 m changes gas pressure by only 3.4kPa Problem 3.5 Determine the pressure at a depth of 5m for water. Δp = γ h Specific wt of water = 9.81 kN/m3 Therefore, Δp = 9.81 x 5 = 49.05 kN/m2 = 49.05 k Pa6Problem 3.7 (SI units) Compute the gage pressures at pts. A, B, C, D, E and F. Specific gravity of Oil = 0.9 Always start with the point of know pressure – Pt A in this case.7 Answers: Pa = 0 (gage) – atmospheric pressure γ oil = 0.9 x 9.81 = 8.83 kN/m2 Pb = 3.0 x 8.83 Pb = 26.5 kN/m2 = 26.5 kPa Pc = 6.0 x 8.83 Pc = 53 kN/m2 Pd = Pb (same level) Pd = 26.5 kN/m2 Pe = Pa – same level as A Pe = 0 kN/m2 Pf = 0 – 1.5*8.83 Pf = -13.2 kN/m28Key observations from example above – • Pressure increases with depth • Pressure changes linearly • Points at the same elevation have the same pressure • Pressure decreases at higher elevations93.6 Pascal’s paradox Pressure depends only on the elevation and the type of the fluid; NOT on the size of the fluid container See Figure 3.7 All containers have the same pressure at the bottom!! -referred to as the Pascal’s Paradox10The role of elevation also comes into play with respect to pressure in a water distribution system – See Figure 3.8 The Supply point should be higher than the receiving points to allow pressure to be maintained by gravity.113.7 Manometers Instrument to measure pressure. • Simplest kind – U tube manometer • One end – open to the atmosphere • Other end – connected to the fluid whose pressure is to be measured • Contains liquid (gage fluid)– whose deflection indicates the pressure • Gage liquid – should not mix with the other liquid12Procedure for measurement: • Start from the point which is exposed to atmosphere, and move towards the point at which the pressure is desired. • If you move down in the fluid, pressure increases; and vice-versa13Problem 3.8 (SI units) Compute the pressure at A?14γm = 9.81 x 13.54 = 132.8 kN/m3 P1 = 0 P2 = 0.25 x 132.8 = 33.2 kN/m2 P3 = P2 = 33.2 kN/m2 P4 = P3 – 0.4 x 9.81 = P3 - 3.92 = 29.28 kN/m2 Answer = Pa = 29.28 kN/m2 = 29.28kPa15Problem 3.9 (US Units) Determine the difference in pressure between points A and B Specific wt of water = 62.4 lb/ft3.16 P1 = Pa + 33.75 x γo P1 = P2 P3 = P1 – 29.5 x γw P4 = P3 – 4.25 x γo Pb = P4 = Pa + 33.75 x γo – 29.5 x γw – 4.25 x γo Or Pb – Pa = 33.75 x γo – 29.5 x γw – 4.25 x γo = 29.5 γo – 29.5 x γw = 29.5 (γo – γw) γo = 0.86 x 62.4 = 53.7 lb/ft3 Pb-Pa = 29.5 in x (53.7 – 62.4) lb/ft3 = 29.5 in x (-8.7 lb/ft3) x (1 ft3/ 1728 in3) Answer : Pb – Pa = -0.15 lb/in217Other types of manometers: Well-type manometer Inclined well-type manometer18Barometers • Device for measuring atmospheric pressure. • Filled with mercury • Mercury filled tube is inverted in mercury bath. Mercury column drops a little – filled with mercury vapor at 0.17 Pa. The height of the mercury provides the atmospheric pressure 0 + γmh = Patm Patm = γmh19• Mercury depth decreases 1.0 inch every 1000 ft of increase in altitude. (pressure will decrease as you go up in the atmosphere). • Specific wt of mercury changes with temp! So adjustments with temp have to be made!20Pressure gages and transducers Gage - Pressure sensed mechanically. Pressure transducer – pressure measured at one point, displayed at another – pressure sensed mechanically and converted into an electrical signal.2122Assignment # 2 • 3.48M • 3.54M • 3.62M • 3.63E • 3.65M • 3.66M Some other types of problems –23Q. A pressure gage at 19.0 ft from bottom of tank reads = 13.19 psi. Another at 14 ft, reads = 15.12 psi Compute – specific wt, density, and specific gravity of fluid in tank We have two known pressure points and the distance between them! Δp = γ Δh Apply the equation (15.12 – 13.19) * 144 = γ * (19.0-14.0) Therefore γ = 55.6 lb/ft3 Remember γ = ρg Therefore, ρ = 55.6/ 32.2 = 1.73 slug/ft3 SG = γf/ γw = 55.6/62.4 = 0.891. Q.24 A reservoir of CCl4 has mass of 500 kg and a volume of 0.315 m3. Find the weight, density, specific weight and specific gravity. m = 500 kg g = 9.81 m/s2 W = mg = 500*9.81 = 4905 N = 4.905 kN Density = ρ = m/V = 500/0.315 = 1587 kg/m3 Specific wt = γ = W/V = 4.905/0.315 = 15.57 kN/m3 SG = 15.57/9.81 = 1.5925Q. Fluid is Oil = SG = 0.85