Introduction to Fluid Statics and Manometers

Fluid Statics and Manometers Outline Introduction to Fluid Statics and Manometers • Review course introduction • Pressure independent of direction • Pressure-density-distance relationship in a static (nonmoving fluid) • Use of manometers for pressure measurements • Calculations with manometers. Larry Caretto Mechanical Engineering 390 Fluid Mechanics January 24 and 29, 2008 2 Review Review Typical Units Quantity • Dimensions and units – SI, BG and EE unit systems Density Pressure & shear stress • Fluid density, ρ, (mass/volume) and specific weight (weight/volume), γ = ρg, and specific gravity • States of matter and vapor pressure • Viscosity • Surface tension Velocity Viscosity Specific weight = ρg SI units EE units BG units kg/m3 kPa = kN/m2 lbm/ft3 slug/ft3 1 psi = 1 lbf/in2 = 144 psf = 144 lbf/ft2 m/s ft/s N·s/m2 = lbf·s/ft2 = lbf·s/ft2 = kg/m·s 32.2 lbm/ft·s slug/ft·s lbf/ft3 N/m3 Tabulated values at standard gravity 3 Review States of Matter Viscosity • Triple point: solid, liquid and vapor coexist Critical • No liquid-gas transition above Point critical point • Vapor pressure Gas for liquid-gas transition Solid Pressure Boiling line shows Pvapor = f(T) Liquid Triple Point Temperature 4 5 Newtonian Fluids have a linear variation of shearing stress with rate of shearing strain – slope is viscosity Figure 1.4 (p. 15) τ=μ ∂u ∂y 6 1 Fluid Statics and Manometers Review Surface Tension Pressure Relations • Pressure is a scalar • The force exerted by a pressure is the same in all directions • Want to see how pressure changes in a static (nonmoving) fluid • Look at balance of pressure force and fluid weight over a differential volume element, δxδyδz • Vertical force balance: γπR2h = 2πRσcosθ – Surface tension depends on fluid, temperature; wetting angle, θ, depends on fluid and surface h= 2σ cos θ γR 8 ∑ Fz = 0 = −ρ(δxδyδz )g ∂p ⎛ +⎜ p− ∂z ⎝ ∂p ⎛ −⎜ p+ ∂z ⎝ Gravity in –z direction Sum forces in each direction and divide by δxδyδz δz ⎞ ⎟δxδy 2⎠ δz ⎞ ⎟δxδy 2⎠ Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi,Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Results ∂p =0 ∂x ∂p =0 ∂y ∂p +γ=0 ∂z 9 Integrating the Result • Since ∂p/∂x =∂p/∂y = 0, p = f(z) only, and we can write ∂p/∂z = –γ as dp/dz = –γ • Multiply by dz and integrate between two points (p1, z1) and (p2, z2) p2 z2 p1 z1 ∫ dp = p2 − p1 = − ∫ γdz • To integrate γdz we have to know how γ depends on z 10 Incompressible Fluid Incompressible Fluid II • An incompressible fluid has constant density (and specific weight) • For an incompressible (constant density) fluid then p2 + γz2 = p1 + γz1 • Which pressure is z2 z1 ∫ h z1 z1 p2 + γz 2 = p1 + γz1 11 p2 z2 z2 p2 − p1 = − γdz = − γ dz = − γ ( z2 − z1 ) ∫ higher? p1 p1 = p2 + γ ( z 2 − z1 ) p1 = p2 + γh > p2 • Pressure increases with depth 12 2 Fluid Statics and Manometers Problem Solution • If the pressure at the surface of a body of water (γ = 9789 N/m3 at 20oC) is 101 kPa, what are the pressures at depths of 10 m and 100 m? • Given: p1 = 101 kPa at z1 = 0 • Find: p at z2 = –10 m and z3 = –100 m • Equation: p2 + γz2 = p1 + γz1 ⇒ Depth = 10 m p2 = p1 + γ ( z1 − z 2 ) = 101 kPa + 2 9789 N [0 m − (− 10 m)] kPa ⋅ m = 198.9 kPa 3 m 1000 N Depth = 100 m p2 = p1 + γ ( z1 − z 2 ) p3 = p1 + γ (z1 − z3 ) = 101 kPa + 9789 N kPa ⋅ m 2 [ ( ) ] 0 m − − 100 m = 1080 kPa m3 1000 N 13 ΔP = 1 Atm for Water at 20oC Pressure Head p1 = p2 + γ ( z2 + z1 ) • Fluid height z2 z1 14 p2 − p1 equivalent to h = p1 = p2 + γh > p2 γ a pressure p2 difference h • h is called pressure head p1 • For p2 – p1 = 14.696 psia = 101.325 kPa, h = 0.76 m = 29.92 in for Hg • What is h for water at this 15 Δp? Free Surface • At 20oC, γwater = 9789 N/m3 = 62.32 lbf/ft3 (p 761, text) 14.696 lb f 144 in 2 p − p1 in 2 ft 2 = = 33.96 ft h= 2 62.32 lb f γ ft 3 p − p1 h= 2 = γ 1000 N kPa ⋅ m 2 = 10.35 m 9789 N m3 101.325 kPa 16 Free Surface II • Surface of liquid open to atmosphere is called a “free surface” – Pressure, p0, is atmospheric pressure, p0 – Height, z0 = 0 – In the liquid, p + γz = p0 + γz0, where z < 0 – Depth h = z0 – z > 0 – p = p0 + γ(z0 – z) = p0 + γh • Pressure, p, at depth, h, not influenced by size or shape of container • Pressure, p, at depth, h, not influenced by size or shape of container 17 18 3 Fluid Statics and Manometers Reference Pressure Gage Pressure • Free surface equation: p = p0 + γ(z0 – z) • Apply this to two different pressures • For taking pressure differences, we can use any reference pressure • Many pressure measurement methods measure the difference between actual and atmospheric pressure • We can used this measured pressure difference, called gage pressure, directly in Δp calculations • pabsolute = pgage + patmosphere – p1 = p0 + γ(z0 – z1) – p2 = p0 + γ(z0 – z2) • Find p2 – p1 from these equations – Result: p2 – p1 = p0 + γ(z0 – z2) – [p0 + γ(z0 – z1)] = γ(z1 – z2) independent of p0 or z0 • Reference pressure cancels in taking pressure differences 19 Gage and Absolute Pressure 20 Barometric Pressure • Mercury barometer used to measure atmospheric pressure – Top is evacuated and fills with mercury vapor – Patm = γh + pvapor – pvapor = 0.000023 psia = 0.1586 Pa at 68oF (20oC) – h = 760 mm = 29.92 in for standard atmosphere Figure 2.7, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. 21 Figure 2.8, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. 22 What a Barometer Measures Gage/Absolute Notation • It actually measures the local pressure • A barometer in a undersea submersible, pressurized to 4 times atmospheric pressure would measure this level • Weather barometer readings are corrected to mean sea level • Standard atmosphere: 760 mm Hg, 760 torr, 29.921 in Hg, 101.325 kPa, 14.696 psia, 2116.2 psfa, 1013.25 mbar • For pressure differences a specification of gage or absolute is not required • Traditional notation is psig (or psfg) and psia (or psfg) for gage and absolute pressure, respectively • Can also use kPa(abs) or kPa(gage) • Munson uses psi or kPa for gage pressures and psi(abs) or kPa(abs) for absolute pressures 23 24 4 Fluid Statics and Manometers Variable Density Variable Density II • Problem: integrate dp/dz = –γz when density (and hence γ) is not constant • Simple solution: for gases γ is small so that p does not change much with z • Result: the pressure change of 0.1181 kPa is only 0.12% of patm = 101.325 kPa • Simple solution: for gases with small elevation changes we can assume that the specific weight is constant! • This is not valid for changes of several kilometers as in the atmosphere – E. g. air at atmospheric pressure and T = 20oC has γ = 11.81 N/m3 – If γ = 11.81 N/m3 were constant an elevation change of 10 m gives Δp = (11.81 N/m3)(10 m) = 118.1 N/m2 = 0.1181 kPa – Standard atmosphere defined and used for aerospace designs 25 26 U. S. Standard Atmosphere (See appendix C for more data) Constant specific weight is a good assumption for gases except for large elevation changes Figure E2.2, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Piezometer Figure 2.6, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. 27 Open means p = patm Problem • A passage between a container (such as the pipe at A) and the atmosphere is called a piezometer tube • For a piezometer tube the pressure in the fluid, pA = p1 = patm + γ1h1 • patm(gage) = 0 Figure 2.9, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. 28 • Find the pressure at point A if the fluid is water at 20oC, h1 = 0.2 m, and patm = 101 kPa • From table B.2, γ1 = γwater = 9.789 kN/m3 at 20oC • pA = p1 = patm + γ1h1 29 Figure 2.9, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. 30 5 Fluid Statics and Manometers Solution Simple U-Tube Manometer • h1 = 0.2 m, patm = 101 kPa, and γ1 =γwater = 9.789 kN/m3 • pA = patm + γ1h1 PA = 101 kPa + 9.789 kN m 3 • Manometers measure pressure by measuring height differences (0.2 m ) kPa ⋅ m • Point A is fluid (γ = γ1) in a pipe 2 1 kN • pA = 103 kPa (absolute) • pA = 1.96 kPa (gage) • h1 and h2 are measured • Gage fluid has γ = γ2 • What is pressure at A? 31 Figure 2.9, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Simple U-Tube Manometer II • Right side: p3 = patm + γ2h2 • Left side: p2 = pA + γ1h1 32 Figure 2.10, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Simple U-Tube Manometer III • Result for absolute pressure: pA = patm + γ2h2 – γ1h1 patm • p3 = p2 gives patm + γ2h2 = pA + γ1h1 p2 = p3 = p0 – γ2h0 • Conclusion: pA = patm + γ2h2 – γ1h1 • Result for gage pressure: pA = γ2h2 – γ1h1 h0 p 33 0 Figure 2.10, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. • Given: Known specific weights and measured heights shown above • Find: pA – pB Equation: pα + γzα = pβ + γzβ 35 FigureE2.5, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. 34 Figure 2.10, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. p A = p1 + γ1h1 p1 = p2 = p3 p3 = p4 + γ 2 h2 p5 = p4 p B = p5 + γ1 (h1 + h2 ) 36 FigureE2.5, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. 6 Fluid Statics and Manometers FigureE2.5, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Combine first set of equations p A = p1 + γ1h1 p1 = p2 = p3 p3 = p4 + γ 2 h2 p A = p1 + γ1h1 = p3 + γ1h1 = p4 + γ 2 h2 + γ1h1 Combine result from previous page with second set of equations p A = p5 + γ 2 h2 + γ1h1 = p B − γ1 (h1 − h2 ) + γ 2 h2 + γ1h1 p A − p B = (γ 2 − γ1 )h2 37 FigureE2.5, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Incline Manometer Problem p A = p4 + γ 2 h2 + γ1h1 p5 = p4 p B = p5 + γ1 (h1 + h2 ) 38 Incline Manometer Problem II p1 = p A + γ1h1 • Incline used to increase accuracy for small pressure differences • Want to find pB – pA when we know γ1, γ2, γ3, h1, l2, h3, and θ p2 = p B + γ 3h3 p1 = p2 + γ 2 ( z2 − z1 ) = p2 + γ 2 l 2 sin θ ⇒ p B = p2 − γ 3h3 p B = p2 − γ 3h3 = p1 − γ 2 l 2 sin θ − γ 3h3 p B = p A + γ1h1 − γ 2 l 2 sin θ − γ 3h3 p A − pB = γ 3h3 + γ 2 l 2 sin θ − γ1h1 Equation: pα + γzα = pβ + γzβ 39 Figure 2.9, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Solving Manometer Problems • Basic equation: pressures at two depths open in same fluid: p2 = p3 + γ(z3 – z2) = p3 + γh 3 • “Open” means p = patm – patm = 101.325 kPa = 14.696 psia – For gage pressure, patm = 0 • Same pressures at same level on two sides of a manometer – p 2 = p3 41 40 Solving Manometer Problems II • Write equations for (1) pressures at two open depths in same fluid and (2) equal 3 pressures at same level (with same fluid) at all branches in manometer. • Eliminate intermediate pressures from equations to get desired ΔP • Watch units for length, psi or psf, N or kN • For gases γΔz ≈ 0 42 7 Fluid Statics and Manometers Problem Problem Continued • Given: air at 0.50 psig, pA = 2 psig, and γoil = 54.0 lbf/ft3 and other data shown on diagram. Find: z and h p A − pair = γ oil z z = 4 ft • Given: air at 0.50 psig, pA = 2 psig, and γoil = 54.0 lbf/ft3 and other data shown on diagram. Find: h popen = p A + γ oil (2 ft ) = 0 psig p A − pair γ oil 2 lb f 0.5 lb f − 2 in 2 = in 54.0 lb f ft 2 3 ft 144 in 2 z= popen + γ mano h γ mano = SGmano γ H 2O ⎛ 62.4 lb f = 3.05⎜⎜ 3 ⎝ ft Figure P2.27, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. 43 Problem Concluded • popen = 0 psig, γmano = 190.3 h=? p A + γ oil (2 ft ) = popen + γ mano h p A + γ oil (2 ft ) − popen h = 2.08 ft h= γ mano 0 2 lb f 144 in 2 54.0 lb f (2 ft ) + ft 2 ft 3 in 2 h= 190.3 lb f ft 3 45 Problem 2.38 Part 2 PoceanP left Pshell = Pshell + γ Hg (0.735 m) = Pright = Pocean + water Pshell = γHghbaro hbaro = 765 mm Hg 44 A hemispherical shell on the ocean floor has an internal barometric pressure of 765 mm Hg. A mercury manometer measures the differential pressure between the sea outside and the shell interior as shown in the diagram. What is the pressure at the ocean surface? 46 Problem 2.38 Part 3 γ sea (10 m + 0.36 m) Equal pressure line Figure P2.27, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Problem 2.38 lbf/ft3, Figure P2.27, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. ⎞ 190.3 lb f ⎟= ⎟ ft 3 ⎠ Use specific weight data from Table 1-6 (ignore difference in temperature) γHg = 133 kN/m3 γsea water = 10.1 kN/m3 47 • Rearrange equations from previous slide and substitute given data Pleft = Pshell + γ Hg (0.735 m) = Pright = Pocean + γ sea (10.36 m) water Pocean = Pshell + γ Hg (0.735 m) − γ sea (10.36 m) water 101.745 kN ⎛ 133 kN ⎞ Pshell = γ Hg hbaro = ⎜ ⎟(0.765 m) = m2 ⎝ m3 ⎠ 101.745 kN 133 kN 10.1 kN Pocean = + (0.735 m) − (10.36 m) m3 m3 m3 94.9 kN Pocean = = 94.9 kPa 48 m2 8