Fluid StaticsThe word “statics” is derived from Greek word “statikos”= motionless For a fluid at rest or moving in such a manner that there is no relative motion between particles there are no shearing forces present: Rigid body approximation STATIC FLUID APPLICATION Vessel thickness design, Measurement of pressure, Separation of fluids with different density, Hydraulic jack Design of ship Principle of Fluid Static • P = P 0 + µgd F = µ g V • P = F/A 05 Density is defined as the ratio of the mass of fluid to its volume. It is denoted by the Greek symbol, µ. µ = V m 3 kgm -3 If the density is constant (most liquids), the flow is incompressible. If the density varies significantly (eg some gas flows), the flow is compressible. (Although gases are easy to compress, the flow may be treated as incompressible if there are no large pressure fluctuations) µ water = 998 kgm -3 µ air =1.2kgm -3 kg m Mass per unit volume (e.g., @ 20 o C, 1 atm) Water = 1000 kg/m 3 = 62.3 lbm/ft 3 Mercury = 13,500 kg/m 3 Air = 1.22 kg/m 3 Density • Densities of gasses increase with pressure • Densities of liquids are nearly constant (incompressible) for constant temperature • Specific volume = 1/density 950 960 970 980 990 1000 0 50 100 Temperature (C) D e n s i t y ( k g / m 3 ) Specific Weight • Weight per unit volume (e.g., @ 20 o C, 1 atm) ¸ water = (998 kg/m 3 )(9.807 m 2 /s) = 9790 N/m 3 [= 62.4 lbf/ft 3 ] ¸ air = (1.205 kg/m 3 )(9.807 m 2 /s) = 11.8 N/m 3 [= 0.0752 lbf/ft 3 ] ] / [ ] / [ 3 3 ft lbf or m N g µ ¸ = API gravity is an alternative method of comparing the densities of different petroleum substances. The API system of gravity measurement has units called 'Degrees API' (ºAPI). The device used for the measurement of API and specific gravity is the 'HYDROMETER'. Specific gravity = {Brix/(258.6-[Brix/258.2]*227.1)}+1 Brix and specific gravity refer to the amount of sugar in a water solution. Commonly used in beer and wine making, the amount of sugar in the unfermented wort (for beer) or must (for wine) determine the level of alcohol in the finished product. API gravity is calculated from the Specific Gravity as follows: - °API = (141.5 ÷ SG) - 131.5 Example 2: An oil has an API gravity of 42.0°. Calculate its S.G. S.G. = 141.5 ÷ (131.5 + 42) = 141.5 ÷ 173.5 = 0.816 SG From the above formulae, it is found that pure water (S.G. = 1.000) has an API gravity of 10°. As fluid density decreases, the API gravity increases. Specific Gravity • Ratio of fluid density to density water at specified T dan P (e.g., @ 20 o C, 1 atm) 3 / 9790 m kg SG liquid water liquid liquid µ µ µ = = Water SG water = 1 Mercury SG Hg = 13.6 Air SG air = 1 3 / 205 . 1 m kg SG gas air gas gas µ µ µ = = TEKANAN Gaya per satuan luas, dimana gaya tegak lurus luasan. p= A m 2 Nm -2 (Pa) N F p atmosfir = 1.013X10 5 Nm -2 Pa (Pascal) 1 psi = 6895 Pa 0 = F Hydrostatic Pressure F = gaya dari atas + gaya dari bawah + gaya gravitasi = 0 0 = A A A ÷ A A ÷ A A z y x g y x P y x P b a µ g z P P b a µ ÷ = A ÷ Tekanan atas Pb Tekanan bawah Pa Z a Z b Densitas=µ Az g dz dP µ ÷ = Incompressible fluid Liquids are incompressible i.e. their density is assumed to be constant: P gh P o + µ = By using gage pressures we can simply write: gh P µ = ) z z ( g P P 1 2 1 2 ÷ µ ÷ = ÷ P o is the pressure at the free surface (P o =P atm ) When we have a liquid with a free surface the pressure P at any depth below the free surface is: Units for Pressure Unit Definition or Relationship 1 pascal (Pa) 1 kg m -1 s -2 1 bar 1 x 10 5 Pa 1 atmosphere (atm) 101,325 Pa 1 torr 1 / 760 atm 760 mm Hg 1 atm 14.696 pounds per sq. in. (psi) 1 atm Tekanan pada permukaan air danau adalah 105 kPa. Hitung tekanan pada kedalaman 35.0 m dibawah permukaan air. ( )( )( ) atm 3.4 kPa 343 m 35 m/s 8 . 9 kg/m 1000 2 3 atm atm = = = = ÷ = A + = dg P P P dg P P µ µ Kerapatan air segar hydrostatic • The pressure in a homogeneous, incompressible fluid at rest depends on the depth of the fluid relative to some reference plane, and it is not influenced by the size or shape of the tank or container Fluid is the same in all containers Pressure is the same at the bottom of all containers h Tekanan di permukaan planet Venus adalah 95 atm. How far below the surface of the ocean on Earth do you need to be to experience the same pressure? ( )( ) m 950 N/m 10 9.5 m/s 8 . 9 kg/m 1025 N/m 10 9.5 atm 94 atm 1 atm 95 2 6 2 3 2 6 atm = × = × = = + = + = d d dg dg dg P P µ µ µ Density of sea water KEDALAMAN OIL RESERVOIR What the depth below the surface of oil reservoir that produce oil with relative density 0.8 and wellhead pressure of 120 kN/m2? µ water = 1000 kg/m3, and p atmosphere = 101kN/m2. Vertical plane surfaces F Vertical rectangular wall (wall width =W) H h Here the pressure varies linearly with depth: P=µgh P The lock gate of a canal is rectangular, 20 m wide and 10 m high. One side is exposed to the atmosphere and the other side to the water. What is the net force on the lock gate? Vertical plane surfaces • For an infinitesimal area dA the normal force due to the pressure is dF = p dA • Find resultant force acting on a finite surface by integration ( ) Wh d h g } = µ } = dA P F For vertical rectangular wall: F = ½ µ g W H 2 dh h gW } = µ Storage Tanks They are used in a variety of industries like Petroleum refining Chemical Power Food & beverage Pharmaceutical MAIN COMPONENTS OF PRESSURE VESSEL Following are the main components of pressure Vessels in general • Shell • Head • Nozzle • Support SHELL It is the primary component that contains the pressure. Pressure vessel shells in the form of different plates are welded together to form a structure that has a common rotational axis. Shells are either cylindrical, spherical or conical in shape. SHELL Horizontal drums have cylindrical shells and are constructed in a wide range of diameter and length. The shell sections of a tall tower may be constructed of different materials, thickness and diameters due to process and phase change of process fluid. Shell of a spherical pressure vessel is spherical as well. HEAD • All the pressure vessels must be closed at the ends by heads (or another shell section). • Heads are typically curved rather than flat. • The reason is that curved configurations are stronger and allow the heads to be thinner, lighter and less expensive than flat heads. • Heads can also be used inside a vessel and are known as intermediate heads. • These intermediate heads are separate sections of the pressure vessels to permit different design conditions. NOZZLE • A nozzle is a cylindrical component that penetrates into the shell or head of pressure vessel. • They are used for the following applications. • Attach piping for flow into or out of the vessel. • Attach instrument connection (level gauges, Thermowells, pressure gauges). • Provide access to the vessel interior at MANWAY. • Provide for direct attachment of other equipment items (e.g. heat exchangers). SUPPORT • Support is used to bear all the load of pressure vessel, earthquake and wind loads. • There are different types of supports which are used depending upon the size and orientation of the pressure vessel. • It is considered to be the non-pressurized part of the vessel. structure must be designed to resist deformation and collapse under all the conditions of loading. The loads to which a process vessel will be subject Major loads 1. Design pressure: including any significant static head of liquid. 2. Maximum weight of the vessel and contents, under operating conditions. 3. Maximum weight of the vessel and contents under the hydraulic test conditions. 4. Wind loads. 5. Earthquake (seismic) loads. 6. Loads supported by, or reacting on, the vessel. As a general guide the wall thickness of any vessel should not be less than the values given below; the values include a corrosion allowance of 2 mm: Vessel diameter (m) Minimum thickness (mm) • The cross-sectional area is pi times the diameter squared divided by 4. • The cross-sectional area of tank A is: • The volume V is A x H: • The weight of the water W A is: • Therefore the pressure is: • This is the pressure in pounds per square feet, one more step is required to get the pressure in pounds per square inch or psi. There is 12 inches to a foot therefore there is 12x12 = 144 inches to a square foot. • The pressure p at the bottom of tank A in psi is: PRESSURE MEASUREMENT TECHNIQUES PRESSURE MEASUREMENT Many techniques have been developed for the measurement of pressure and vacuum. Instruments used to measure pressure are called pressure gauges or vacuum gauges. A manometer could also be referring to a pressure measuring instrument, usually limited to measuring pressures near to atmospheric. The term manometer is often used to refer specifically to liquid column hydrostatic instruments. A vacuum gauge is used to measure the pressure in a vacuum ABSOLUTE, GAUGE AND DIFFERENTIAL PRESSURES Absolute pressure is zero-referenced against a perfect vacuum, so it is equal to gauge pressure plus atmospheric pressure. Gauge pressure is zero-referenced against ambient air pressure, so it is equal to absolute pressure minus atmospheric pressure. Negative signs are usually omitted. Differential pressure is the difference in pressure between two points. ABSOLUTE AND GAGE PRESSURE • ABSOLUTE PRESSURE: The pressure of a fluid is expressed relative to that of vacuum (=0) • GAGE PRESSURE: Pressure expressed as the difference between the pressure of the fluid and that of the surrounding atmosphere gage atm abs P P P + = atmospheric pressures, deep vacuum pressures must be absolute Atmospheric pressure is typically about 101.325 kPa or 100 kPa or 30 inHg at sea level, but is variable with altitude and weather. a vacuum of 26 inHg gauge is equivalent to an absolute pressure of 30 inHg (typical atmospheric pressure) − 26 inHg = 4 inHg. Tire pressure and blood pressure are gauge pressures by convention Measurement of Pressure Manometers are devices in which one or more columns of a liquid are used to determine the pressure difference between two points. –U-tube manometer –Inclined-tube manometer A manometer is a U-shaped tube that is partially filled with liquid. Both ends of the tube are open to the atmosphere. The difference in fluid height in a liquid column manometer is proportional to the pressure difference. THE U-TUBE MANOMETER. Liquid column Measurement of Pressure Differences m a m b b m m b a gR Z g P P R Z g P P µ µ µ + + = + + = ) ( ) ( 3 2 Apply the basic equation of static fluids to both legs of manometer, realizing that P 2 =P 3 . ) ( b a m b a gR P P µ µ ÷ = ÷ Cylinder of gas A container of gas is connected to one end of the U-tube C B’ B A d gd P gd P P P P gd P P P B C B C B B µ µ µ = = ÷ = ÷ + = = gauge atm ' B' B P P = atm c P P = Point A is the original location of the top of the fluid before the gas cylinder is connected. Using a u-tube manometer to measure gauge pressure of fluid density 700 kg/m3, and the manometric fluid is mercury, with a relative density of 13.6. What is the gauge pressure if: a). h1 = 0.4m and h2 = 0.9m? b) h1 stayed the same but h2 = -0.1m? THE U-TUBE MANOMETER. Inclined Manometer • To measure small pressure differences need to magnify R m some way. o µ µ sin ) ( 1 b a b a gR P P ÷ = ÷ Measurement of Pressure The atmospheric pressure can be measured with a barometer. – For mercury barometers atmospheric pressure (101.33kPa) corresponds to h=760 mmHg (= 29.2 in) – If water is used h = 10.33 m H 2 O (= 34 ft) vapor atm p gh p + µ = The Bourdon pressure gauge uses the principle that a flattened tube tends to change to be straightened or larger circular cross-section when pressurized. Although this change in cross-section may be hardly noticeable, and thus involving moderate stresses within the elastic range of easily workable materials, the strain of the material of the tube is magnified by forming the tube into a C shape or even a helix, such that the entire tube tends to straighten out or uncoil, elastically, as it is pressurized The Bourdon pressure gauge Bourdon tubes measure gauge pressure, relative to ambient atmospheric pressure Stationary parts: A: Receiver block. This joins the inlet pipe to the fixed end of the Bourdon tube (1) and secures the chassis plate (B). The two holes receive screws that secure the case. B: Chassis plate. The face card is attached to this. It contains bearing holes for the axles. C: Secondary chassis plate. It supports the outer ends of the axles. D: Posts to join and space the two chassis plates. Moving Parts: Stationary end of Bourdon tube. This communicates with the inlet pipe through the receiver block. Moving end of Bourdon tube. This end is sealed. Pivot and pivot pin. Link joining pivot pin to lever (5) with pins to allow joint rotation. Lever. This is an extension of the sector gear (7). Sector gear axle pin. Sector gear. Indicator needle axle. This has a spur gear that engages the sector gear (7) and extends through the face to drive the indicator needle. Due to the short distance between the lever arm link boss and the pivot pin and the difference between the effective radius of the sector gear and that of the spur gear, any motion of the Bourdon tube is greatly amplified. A small motion of the tube results in a large motion of the indicator needle. Hair spring to preload the gear train to eliminate gear lash and hysteresis. Schematic drawing of a simple mercury barometer with vertical mercury column and reservoir at base Barometers Measuring Pressure Barometers A barometer is used to measure the pressure of the atmosphere. The simplest type of barometer consists of a column of fluid. p 1 = 0 vacuum h p 2 = p a p 2 - p 1 = µgh p a = µgh examples water: h = p a /µg =10 5 /(10 3 *9.8) ~10m mercury: h = p a /µg =10 5 /(13.4*10 3 *9.8) ~800mm Atmospheric pressure is equivalent to a column of mercury 76.0 cm tall. gd P µ = Temperature variation with altitude for the U.S. standard atmosphere COMPRESSIBLE FLUID Gases are compressible i.e. their density varies with temperature and pressure µ =P M /RT – For small elevation changes (as in engineering applications, tanks, pipes etc) we can neglect the effect of elevation on pressure ( ¸ ( ¸ ÷ ÷ = = = o o RT z z g p p T f or ) ( exp : const T 0 0 g dz dp µ ÷ = CONSTANT Temperature µ = = = = RT p V M M RT nRT pV Linear Temperature Gradient ) ( 0 0 z z T T ÷ ÷ = o } ÷ ÷ ÷ = } z z p p z z T dz R g p dp 0 0 ) ( 0 0 o R g T z z T p z p o o ( ¸ ( ¸ ÷ ÷ = 0 0 0 0 ) ( ) ( Atmospheric Equations • Assume linear R g T z z T p z p o o ( ¸ ( ¸ ÷ ÷ = 0 0 0 0 ) ( ) ( • Assume constant 0 0 ) ( 0 ) ( RT z z g e p z p ÷ ÷ = Temperature variation with altitude for the U.S. standard atmosphere Compressible Isentropic v p C C P constant P = = = ¸ µ µ ¸ ¸ 1 1 y P P T T 1 1 1 ÷ | | . | \ | = ¸ ( ¸ ( ¸ | | . | \ | A | | . | \ | ÷ ÷ = ( ¸ ( ¸ | | . | \ | A | | . | \ | ÷ ÷ = ÷ 1 1 2 1 1 1 2 1 1 1 1 RT z gM T T RT z gM P P ¸ ¸ ¸ ¸ ¸ ¸ Application: bottom hole conditions in gas wells Separation of fluid with different densiti How it works… GRAVITY DECANTER A A A A b B Z Z Z Balance c Hydrostati µ µ µ 2 1 = + A B A B T A A Z Z Z µ µ µ µ ÷ | . | \ | ÷ = 1 2 1 When ρ B ≈ρ A interface location is very sensitive to height of heavy liquid overflow leg. This leg is often has adjustable height to give the best separation. DECANTER • It is proposed to use a gravity decanter to separate a light petroleum oil (density 50.0 lbm/ft3) from water (density 62.3 lbm/ft3). Its desire to maintain a total depth of 30 in. in the vessel and to have exactly equal depth of oil and water. What should be the height , expressed in inch of the water discharge leg above the bottom of the vessel. Centrifugal decanters When the density difference between two immiscible liquids is small gravitation forces may be too weak to separate them in a reasonable time. In this case we can use centrifugal forces to amplify the forces exerted on the liquids. Centrifugal separations are important in many food industries such a breweries, vegetable oil processing, fruit juice processing. They are also used to separate emulsions into their components. Hydrostatic Equilibrium in a Centrifugal Field 2 2 60 2 | . | \ | = = N mr mr F c t e mg F g = 2 60 2 | . | \ | = N g r F F g c t Typically N≈1000 and r≈1m. Fc/Fg≈110. Neglect g. Hydrostatic Equilibrium in Centrifugal Field dm r dF r at dr element on Force 2 e = dr rb dm tµ 2 = dr r b dF 2 2 2 e tµ = ( ) 2 1 2 2 2 1 2 2 2 2 r r P P dr r rb dF dP ÷ = ÷ = = µ e µ e t Continuous Centrifugal Decanter ? Why P P P P B i A i ÷ = ÷ A B B A B A i r r r µ µ µ µ ÷ | . | \ | ÷ = 1 2 2 Continuous Centrifugal Decanter Consequences: •ρ A ρ B within 3% r i unstable •r B constant r A increased r i shifted toward bowl wall •In commercial units r A and r B are usually adjustable Example Consider a 90˚ elbow in a 2 in. pipe. A pipe tap is drilled through the wall of the elbow on the inside curve, and another though the outer wall directly across from the first. The radius of curvature of the inside bend is 2 in. and that of the outside of the bend is 4 in. The pipe is carrying water, and a manometer containing an immiscible oil with S.G. of 0.90 is connected across the two taps. If the reading of the manometer is 7 in., what is the average velocity of the water in the pipe? Design of hydraulic jack Direction of fluid pressure on boundaries Furnace duct Pipe or tube Heat exchanger Dam Pressure is a Normal Force (acts perpendicular to surfaces) It is also called a Surface Force In a fluid confined by solid boundaries, pressure acts perpendicular to the boundary – it is a normal force. Pascal’s Principle the Principle of transmission of fluid-pressure "pressure exerted anywhere in a confined incompressible fluid is transmitted equally in all directions throughout the fluid such that the pressure ratio (initial difference) remains the same The 1 pound load on the 1 square inch area causes an increase in pressure on the fluid in the system. This pressure is distributed equally throughout and acts on every square inch of the 10 square inch area of the large piston. As a result, the larger piston lifts up a 10 pound weight. The larger the cross-section area of the second piston, the larger the mechanical advantage, and the more weight it lifts. when there is an increase in pressure at any point in a confined fluid, there is an equal increase at every other point in the container. Gaya force F 1 bekerja pada piston A 1 . 1 1 2 2 2 2 1 1 A A A F A F 2 point at 1 point at F F P P | | . | \ | = = A = A F 2 1 500 N 10 5000 N 100 50,000 N 1 2 A A F 1 = 500 N A hydraulic press has an input cylinder 1 inch in diameter and an output cylinder 6 inches in diameter. Assuming 100% efficiency, find the force exerted by the output piston when a force of 10 pounds is applied to the input piston. If the input piston is moved through 4 inches, how far is the output piston moved? Exercises: a. 360 pounds b. 1/9 inch The input and output pistons of a hydraulic jack are respectively 1 cm and 4 cm in diameter. A lever with a mechanical advantage of 6 is used to apply force to the input piston. How much mass can the jack lift if a force of 180 N is applied to the lever and efficiency is 80%? Exercises: 1410.6 kg Design of Ship The buoyant force When an object is placed in a fluid, the fluid exerts an upward force we call the buoyant force. The buoyant force comes from the pressure exerted on the object by the fluid. Because the pressure increases as the depth increases, the pressure on the bottom of an object is always larger than the force on the top - hence the net upward force. F 1 F 2 h 1 h 2 H Buoyancy The net force due to pressure in the vertical direction is: F B = F 2 - F 1 = (P bottom - P top ) (AxAy) The pressure difference is: P bottom – P top = µ g (h 2 -h 1 ) = µ g H Combining: F B = µ g H (AxAy) Thus the buoyant force is: F B = µ g V Buoyant Force (F B ) weight of fluid displaced ARCHIMEDES’ PRINCIPLE –F B = µ fluid V displaced g –F g = mg = µ object V object g –object sinks if µ object > µ fluid –object floats if µ object < µ fluid – object floats if µ fluid = µ object Think this Which weighs more: 1. A large bathtub filled to the brim with water. 2. A large bathtub filled to the brim with water with a battle-ship floating in it. 3. They will weigh the same. Tub of water Tub of water + ship Overflowed water Weight of ship = Buoyant force = Weight of displaced water 16 1. Go up, causing the water to spill out of the glass. 2. Go down. 3. Stay the same. COBA PIKIRKAN B = µ W g V displaced W = µ ice g V ice ÷ µ W g V Must be same! ice-cube Sebuah balok es mengambang diatas segelas air, sampai permukaan air rata pada pinggiran. Ketika es meleleh maka air di dalam akan : ARCHIMEDES EXAMPLE A cube of plastic 4.0 cm on a side with density = 0.8 g/cm 3 is floating in the water. When a 9 gram coin is placed on the block, how much sinks below water surface? h koin ARCHIMEDES EXAMPLE mg F b Mg E F = m a F b – Mg – mg = 0 µ g V disp = (M+m) g V disp = (M+m) / µ h A = (M+m) / µ h = (M + m)/ (µ A) = (51.2+9)/(1 x 4 x 4) = 3.76 cm M = ρ plastic V cube = 0.8x4x4x4 = 51.2 g h koin ( ) ( ) ( )( ) m 5 . 1 m 10 * m 20 kg/m 1000 kg 10 0 . 3 0 3 5 = × = = = = = = = = ÷ = ¿ A m d m Ad m V g m g V g m w F w F F w b b w b w w b w w w B B µ µ µ µ Wadah segi 4 berdasar datar diisi dengan coal, massanya 3.0×10 5 kg. Panjang 20 m dan lebar 10m, mengambang diair. Berapa kedalaman wadah masuk ke dalam air. w F B Sepotong logam dilepaskan dibawah air water. Volume metal 50.0 cm 3 dan SG 5.0. Hitung percepatan initialnya, saat v=0 tidak ada gaya drag. w F B w F B ma w F F B = ÷ = ¿ Vg F B water µ = F B adalah berat fluida yang dipindahkan oleh benda | | . | \ | ÷ = ÷ = 1 ρ ρ ρ ρ obj ect obj ect water obj ect obj ect water V V g g V Vg a g m F m w m F a B B ÷ = ÷ = 0 . 5 gravity specific water obj ect = = µ µ µ water = 1000 kg/m 3 (at 4 °C). 2 obj ect obj ect water m/s 8 . 7 1 0 . 5 1 1 . . 1 1 ρ ρ ÷ = | . | \ | ÷ = | . | \ | ÷ = | | . | \ | ÷ = g G S g V V g a