10 CV 35 FLUID MECHANICS NOTES UNIT-2 PRESSURE AND ITS MEASUREMENT. Dr. Nagaraj Sitaram, Professor, Civil Department, SBMJCE, Bangalore
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1 10 CV 35 FLUID MECHNICS NOTES UNIT-2 PRESSURE ND ITS MESUREMENT by UNIT-2 PRESSURE ND ITS MESUREMENT 2.0 INTRODUCTION: Fluid is a state of matter wic exibits te roerty of flow. Wen a certain mass of fluids is eld in static equilibrium by confining it witin solid boundaries (Fig.1), it exerts force along direction erendicular to te boundary in contact. Tis force is called fluid ressure (comression). Fig.1 Definition of Pressure In fluids, gases and liquids, we seak of ressure; in solids tis is normal stress. For a fluid at rest, te ressure at a given oint is te same in all directions. Differences or gradients in ressure drive a fluid flow, esecially in ducts and ies. 2.1 Definition of Pressure: Pressure is one of te basic roerties of all fluids. Pressure () is te force (F) exerted on or by te fluid on a unit of surface area (). Matematically exressed: F N = 2 m Te basic unit of ressure is Pascal (Pa). Wen a fluid exerts a force of 1 N over an area of 1m 2, te ressure equals one Pascal, i.e., 1 Pa = 1 N/m 2.Pascal is a very small unit, so tat for tyical ower lant alication, we use larger units:
2 Units: 1 kiloascal (kpa) = 10 3 Pa, and 1 megaascal (MPa) = 10 6 Pa = 10 3 kpa. 2.2 Pressure at a Point and Pascal s Law: Pascal s Princile: Pressure extends uniformly in all directions in a fluid. By considering te equilibrium of a small triangular wedge of fluid extracted from a static fluid body, one can sow (Fig.2) tat for any wedge angle θ, te ressures on te tree faces of te wedge are equal in magnitude: Fig.2 Pascal s Law Indeendent of x = y = z indeendent of θ Pressure at a oint as te same magnitude in all directions, and is called isotroic. Tis result is known as Pascal's law. 2.3 Pascal s Law: In any closed, static fluid system, a ressure cange at any one oint is transmitted undiminised trougout te system.
3 2.3.1 lication of Pascal s Law: Fig.3 lication of Pascal s Law Pressure alied to a confined fluid increases te ressure trougout by te same amount. In icture, istons are at same eigt: F F F P1 = P2 = = F Ratio 2 / 1 is called ideal mecanical advantage 2.4 Pressure Variation wit Det: Consider a small vertical cylinder of fluid in equilibrium, were ositive z is ointing vertically uward. Suose te origin z = 0 is set at te free surface of te fluid. Ten te ressure variation at a det z = - below te free surface is governed by
4 ( + ) + W = + ρg z = 0 = -ρg z d dz = -ρg or d dz = -γ Eq.(1) (as z Terefore, te ydrostatic ressure increases linearly wit det at te rate of te secific weigt γ = ρg of te fluid. Homogeneous fluid: ρ is constant By simly integrating te above equation-1: 0) d = - ρ g dz = -ρgz + C Were C is constant of integration Wen z = 0 (on te free surface), = C = 0 = (te atmoseric ressure). Hence, = -ρ gz +0 Pressure given by tis equation is called BSOLUTE PRESSURE, i.e., measured above erfect vacuum. However, for engineering uroses, it is more convenient to measure te ressure above a datum ressure at atmoseric ressure. By setting 0 = 0, = -ρ gz + 0 = -ρgz = γ = ρ g Te equation derived above sows tat wen te density is constant, te ressure in a liquid at rest increases linearly wit det from te free surface. For a given ressure intensity will be different for different liquids since, γ will be different for different liquids. P = γ
5 Hint-1: To convert ead of 1 liquid to ead of anoter liquid. S = γ γ Staandard = γ1 = γ γ = γ S 1 γ γ 21 1 S = γ = S 1 = S γ 2 γ 1 γ 1 Staandard γ S tan dard Staandard Staandard 1 = S 2 γ S tan dard 2 S 1 1 = S 2 2 Hint: 2 S x = S liquid x liquid 1x = S liquid x liquid = S liquid x liquid Pressure ead in meters of is given by te roduct of ressure ead in meters of liquid and secific gravity of te liquid. Eg: 10meters of oil of secific gravity 0.8 is equal to 10x0.8 = 8 meters of. Eg: tm ressure is 760mm of Mercury. NOTE: P = γ kn kpa 3 m m
6 Solved Examles: Ex. 1. Calculate intensity of ressure due to a column of 0.3m of (a) (b) Mercury (c) Oil of secific gravity-0.8. Soln: (a) Given : = 0.3m of γ =? = = γ kn. 81 m 9 3 = kpa (b) Given: = 0.3m of Hg γ mercury = S.Gr. of Mercury X γ = 13.6 x 9.81 γ mercury = kn/m 3 mercury = γ mercury mercury = x 0.3 = kpa or kn/m 2 (c) Given: = 0.3 of Oil S.Gr. = 0.8 γ oil = S.Gr. of Oil X γ = 0.8 x 9.8 γ oil = kn/m 3 oil = γ oil oil = x 0.3 oil = kpa or kn/m 2 Ex.2. Intensity of ressure required at a oint is 40kPa. Find corresonding ead in (a) (b) Mercury (c) oil of secific gravity-0.9. Solution: Given Intensity of ressure at a oint 40 kpa i.e. = 40 kn/m 2 (a) Head of =? = γ 40 = = 4.077m of
7 (b) Head of mercury mercury =? γ mercury = S.Gr. of Mercury X γ = 13.6 x 9.81 γ mercury = kn/m 3 mercury = γ mercury 40 = = 0. 3m of mercury (c) Head of oil oil =? γ oil = S.Gr. of Oil X γ = 0.9 x 9.81 γ oil = kn/m 3 Ex.3 Standard atmoseric ressure is kpa Find te ressure ead in (i) Meters of (ii) mm of mercury (iii) m of oil of secific gravity 0.6. (i) Meters of = γ oil oil = γ oil 40 = = 4. 53m of oil (ii) = 9.81 x = 10.3 m of Meters of = γ mercury x mercury = (13.6x9.81) x mercury = 0.76 m of mercury (iii) = γ oil oil = (0.6 x 9.81) x = 17.21m of oil of S = 0.6 Ex.4 n oen container as to a det of 2.5m and above tis an oil of S = 0.85 for a det of 1.2m. Find te intensity of ressure at te interface of two liquids and at te bottom of te tank. (i) t te Oil - int erface (ii) t te bottom of container B B B B = γ = 10 kpa = γ = 0 il oil oil x + γ oil = ( = 10 kpa x 2. 5 = kpa γ x 9. 81) x m Oil = 0.85 x 2.5 m WTER B x
8 2.5 Tyes of Pressure: ir above te surface of liquids exerts ressure on te exosed surface of te liquid and normal to te surface. tmoseric ressure Te ressure exerted by te atmosere is called atmoseric ressure. tmoseric ressure at a lace deends on te elevation of te lace and te temerature. tmoseric ressure is measured using an instrument called Barometer and ence atmoseric ressure is also called Barometric ressure. However, for engineering uroses, it is more convenient to measure te ressure above a datum ressure at atmoseric ressure. By setting atmoere = 0, = -ρgz = ρg Unit: kpa. bar is also a unit of atmoseric ressure 1-bar = 100 kpa.= 1 kg/cm 2 bsolute ressure: bsolute ressure at a oint is te intensity of ressure at tat oint measured wit reference to absolute vacuum or absolute zero ressure. bsolute ressure at a oint is te intensity of ressure at tat oint measured wit reference to absolute vacuum or absolute zero ressure (Fig.4). bsolute ressure at a oint can never be negative since tere can be no ressure less tan absolute zero ressure. Fig.4 Definition of bsolute Pressure, Gauge Pressure and Vacuum Pressure
9 Gauge Pressure: If te intensity of ressure at a oint is measurement wit reference to atmosere ressure, ten it is called gauge ressure at tat oint. Gauge ressure at a oint may be more tan te atmoseric ressure or less tan te atmoseric ressure. ccordingly gauge ressure at te oint may be ositive or negative (Fig.4) Negative gauge ressure: It is also called vacuum ressure. From te figure, It is te ressure measured below te gauge ressure (Fig.4). bsolute ressure at a oint = tmoseric ressure ± Gauge ressure NOTE: If we measure absolute ressure at a Point below te free surface of te liquid, ten, 2 (absolute) = γ. + atm 1 = atm If gauge ressure at a oint is required, ten atmoseric ressure is taken as zero, ten, 2 (gauge) = γ. = ρg lso, te ressure is te same at all oints wit te same det from te free surface regardless of geometry, rovided tat te oints are interconnected by te same fluid. However, te trust due to ressure is erendicular to te surface on wic te ressure acts, and ence its direction deends on te geometry.
10 Solved Examle: Convert te following absolute ressure to gauge ressure: (a) 120kPa (b) 3kPa (c) 15m of H 2 O (d) 800mm of Hg. Solution: (a) abs = atm + gauge gauge = abs atm = = 18.7 kpa (b) gauge = = kpa gauge = 98.3 kpa (vacuum) (c) abs = atm + gauge (d) 15 = gauge gauge = 4.7m of abs = atm + gauge 800 =760 + gauge gauge = 40 mm of mercury 2.6 Vour Pressure: Vaor ressure is defined as te ressure at wic a liquid will boil (vaorize) and is in equilibrium wit its own vaor. Vaor ressure rises as temerature rises. For examle, suose you are caming on a ig mountain (say 3,000 m in altitude); te atmoseric ressure at tis elevation is about 70 kpa and te boiling temerature is around 90 C. Tis as consequences for cooking. For examle, eggs ave to be cooked longer at elevation to become ard-boiled since tey cook at a lower temerature. ressure cooker as te oosite effect. Namely, te tigt lid on a ressure cooker causes te ressure to increase above te normal atmoseric value. Tis causes to boil at a temerature even greater tan 100 C; eggs can be cooked a lot faster in a ressure cooker! Liquid Fig.5
11 Vaor ressure is imortant to fluid flows because, in general, ressure in a flow decreases as velocity increases. Tis can lead to cavitation, wic is generally destructive and undesirable. In articular, at ig seeds te local ressure of a liquid sometimes dros below te vaor ressure of te liquid. In suc a case, cavitation occurs. In oter words, a "cavity" or bubble of vaor aears because te liquid vaorizes or boils at te location were te ressure dis below te local vaor ressure. Cavitation is not desirable for several reasons. First, it causes noise (as te cavitation bubbles collase wen tey migrate into regions of iger ressure). Second, it can lead to inefficiencies and reduction of eat transfer in ums and turbines (turbo macines). Finally, te collase of tese cavitation bubbles causes itting and corrosion of blades and oter surfaces nearby. Te left figure below sows a cavitating roeller in a tunnel, and te rigt figure sows cavitation damage on a blade. 2.7 Measurement of Pressure: Measurement of ressure Barometer Simle manometer Piezometer column Bourdon gage Pressure transducer Barometer: barometer is a device for measuring atmoseric ressure. simle barometer consists of a tube more tan 760 mm long inserted in an oen container of mercury wit a closed and evacuated end at te to and oen tube end at te bottom and wit mercury extending from te container u into te tube. Strictly, te sace above te liquid cannot be a true vacuum. It contains mercury vaor at its saturated vaor ressure, but tis is extremely small at room temeratures (e.g Pa at 20 o C).
12 Te atmoseric ressure is calculated from te relation P atm = ρg were ρ is te density of fluid in te barometer. at o = γ mercury x y + vaor = atm Wit negligible vaor = 0 atm = γ mercury x y Simle Manometer: Simle monometers are used to measure intensity of ressure at a oint. Tey are connected to te oint at wic te intensity of ressure is required. Suc a oint is called gauge oint Tyes of Simle Manometers a) Piezometers X Common tyes of simle manometers are a) Piezometers b) U-tube manometers c) Single tube manometers d) Inclined tube manometers Piezometer consists of a glass tube inserted in te wall of te vessel or ie at te level of oint at wic te intensity of ressure is to be measured. Te oter end of te iezometer is exosed to air. Te eigt of te liquid in te iezometer gives te ressure ead from wic te intensity of ressure can be calculated. 12mm. Pie X rrangement for te measurement negative or vacuum or section ressure To minimize caillary rise effects te diameters of te tube is ket more tan X Pie
13 Merits Simle in construction Economical Demerits Not suitable for ig ressure intensity. Pressure of gases cannot be measured. (b) U-tube Manometers: X Manometer reading Manometric liquid Pie Tank X X Pie U-tube manometers consists of a glass tube bent in U-Sae, one end of wic is connected to gauge oint and te oter end is exosed to atmosere. U-tube consists of a liquid of secific of gravity oter tan tat of fluid wose ressure intensity is to be measured and is called monometric liquid.
14 Manometric liquids Manometric liquids sould neiter mix nor ave any cemical reaction wit te fluid wose ressure intensity is to be measured. It sould not undergo any termal variation. Manometric liquid sould ave very low vaour ressure. Manometric liquid sould ave ressure sensitivity deending uon te magnitude. Of ressure to be measured and accuracy requirement. Gauge equations are written for te system to solve for unknown quantities. To write te gauge equation for manometers Stes: 1. Convert all given ressure to meters of and assume unknown ressure in meters of s. 2. Starting from one end move towards te oter keeing te following oints in mind. ny orizontal movement inside te same liquid will not cause cange in ressure. Vertically downward movement causes increase in ressure and uward motion cause decrease in ressure. Convert all vertical columns of liquids to meters of by multilying tem by corresonding secify gravity. Take atmoseric ressure as zero (gauge ressure comutation). 3. Solve for te unknown quantity and convert it into te required unit.
15 Solved Problem: 1. Determine te ressure at for te U- tube manometer sown in fig. lso calculate te absolute ressure at in kpa. X 750mm 500mm Water Hg (S = 13.6) X 750mm 500mm Water Hg (S = 13.6) Let be te ressure ead at in meters of x13.6 = 0 = 6.05m of = γ = 9.81x 6.05 = kpa ( gauge ressure) abs = atm + gauge = abc = kpa
16 2. For te arrangement sown in figure, determine gauge and absolute ressure at te oint M. 250mm X M 750 mm Oil (S = 0.8) Mercury (13.6) Let M be te ressure ead at te oint M in m of, M x x 13.6 = 0 M = 4 m of = γ = kpa abs = abs kpa 3. If te ressure at t is 10 kpa (Vacuum) wat is te value of x? 200mm x x Mercury Oil (S = 1.2) = 10 kpa (Vacuum) = - 10 kpa
17 γ 10 = = m of 9.81 = m of x1.2 + x (13.6) = 0 x = m 4. Te tank in te accomanying figure consists of oil of S = Determine te kn ressure gauge reading in 2. m ir 25 cm 3.75 m S = 0.75 Mercury Let te ressure gauge reading be m of 3.75 x x 13.6 = 0 = m of = γ = kpa = kpa (Vacuum)
18 5. closed tank is 8m ig. It is filled wit Glycerine u to a det of 3.5m and linseed oil to anoter 2.5m. Te remaining sace is filled wit air under a ressure of 150 kpa. If a ressure gauge is fixed at te bottom of te tank wat will be its reading. lso calculate absolute ressure. Take relative density of Glycerine and Linseed oil as 1.25 and 0.93 resectively. ir 150 kpa M X Linseed oil 2 m 2.5 m N Glycerin S = m P H M M = 150 kpa 150 = 9.81 = m of Let N be te ressure gauge reading in m of. N -3.5 x x 0.93 =15.29 N = m of = 9.81 x = kpa (gauge) abs = kpa 6. vertical ie line attaced wit a gauge and a manometer contains oil and Mercury as sown in figure. Te manometer is oened to atmosere. Wat is te gauge reading at? ssume no flow in te ie. -3 x x x 13.6 = 0 = m of = γ x = 9.81 x 21.99
19 S = m 37.5 cm S = 13.6 = kpa (gauge) abs = abs = kpa DIFFERENTIL MNOMETERS Differential manometers are used to measure ressure difference between any two oints. Common varieties of differential manometers are: (a) Two iezometers. (b) Inverted U-tube manometer. (c) U-tube differential manometers. (d) Micromanometers. (a) Two Pizometers I x x B B Te arrangement consists of two izometers at te two oints between wic te ressure difference is required. Te liquid will rise in bot te iezometers. Te difference in elevation of liquid levels can be recorded and te ressure difference can be calculated. It as all te merits and demerits of iezometer.
20 (b) Inverted U-tube manometers: S M y 1 X x y 2 S 1 x Inverted U-tube manometer is used to measure small difference in ressure between any two oints. It consists of an inverted U-tube connecting te two oints between wic te ressure difference is required. In between tere will be a ligter sensitive manometric liquid. Pressure difference between te two oints can be calculated by writing te gauge equations for te system. Let and B be te r ead at and B in meters of (Y 1 S 1 ) + (x S M ) + (y 2 S 2 ) = B. B = S 1 y 1 S M x S 2 y 2, B = γ ( B ) (c) U-tube Differential manometers x S 1 y 1 x y 2 x B S 2 S M differential U-tube manometer is used to measure ressure difference between any two oints. It consists of a U-tube containing eavier manometric liquid, te two limbs of wic are connected to te gauge oints between wic te ressure difference
21 is required. U-tube differential manometers can also be used for gases. By writing te gauge equation for te system ressure difference can be determined. Let and B be te ressure ead of and B in meters of + S 1 Y 1 + x S M Y 2 S 2 = B B = Y 2 S 2 Y 1 S 1 x S M Solved Problems: (1) n inverted U-tube manometer is sown in figure. Determine te ressure difference between and B in N/M 2. Let and B be te ressure eads at and B in meters of. 30 cm S = cm 120 cm Water x B Water x (190 x 10-2 ) + (0.3 x 0.9) + (0.4) 0.9 = B B = 1.23 meters of B = γ ( B ) = 9.81 x 1.23 B = kpa B = x 10 3 N/m 2
22 2. In te arrangements sown in figure. Determine te o. 2N /cm 2 IR IR 25 cm of Mercury (Vacuum) = x 13.6 = m of 1.5 m 4 m Water KERO S = 0.8 2N/cm 2 = 2 x N/m 2 = 20 kpa = = meters of S = ( ) 0.8 = 3.4 = 3.6 m 3. In te figure given, te air ressure in te left tank is 230 mm of Mercury (Vacuum). Determine te elevation of gauge liquid in te rigt limb at. If te liquid in te rigt
23 tank is. 21 kpa m ir x B C (5 m) Oil m S = 0.8 Water (2 m) m y S = 1.6 c Pc = γ B = 230mm of Hg = 0.23 x 13.6 B = m of x y x 1.6 (y + 2) = x y x 1.6 y 2 = 2.14 y = m Elevation of = Elevation of = m = 2.14mof c
24 4. Comute te ressure different between M and N for te system sown in figure. M 0.2 m S = m S = 1.15 N Let M and N be te ressure eads at M and N in m of. m + y x x (0.3 y + 0.2) 1.15 = n m y x y x 1.15 = n m = n n m = 0.391meters of n m = γ ( N m ) = 9.81 x n m = kpa
25 5. Petrol of secify gravity 0.8 flows u troug a vertical ie. and B are te two oints in te ie, B being 0.3 m iger tan. Connection are led from and B to a U tube containing Mercury. If te ressure difference between and B is 18 kpa, find te reading of manometer. R x 0.3 m S = 0.8 x (y) x Hg B = 18kPa B = P P B γ B =1. 835m of + y x 0.8 x 13.6 (0.3 + y x) 0.8 = B B = 0.8y x y 0.8 x B =12.8 x = 12.8x x = m
26 6. cylindrical tank contains to a eigt of 50mm. Inside is a small oen cylindrical tank containing kerosene at a eigt secify gravity 0.8. Te following ressures are known from indicated gauges. B = 13.8 kpa (gauge) C = kpa (gauge) Determine te gauge ressure and eigt. ssume tat kerosene is revented from moving to te to of te tank. ir Water Kerosene 50 mm S = 0.8 B C C = kpa C = m of B = 13.8 kpa B = meters of = = meters of = x 9.81 = kpa B x 0.8 (0.05 ) = = = = = 0.02 m
27 7. Find te ressure different between and B if d 1 = 300mm, d 2 = 150mm,d 3 = 460mm, d 4 = 200mm and d m C Water B Water d m d m d m Let and B be te ressure ead at and B in m of ( Sin 45) 13.6 = B - B = 7.88m of B = (7.88 ) (9.81) B = kpa 8. Wat is te ressure in te fig given below? Take secific gravity of oil as 0.8. ir Oil S = m Water 4.6 m 0.3 m Hg + (3 x 0.8) + ( ) (13.6) = 0 = 2.24 m of oil = 9.81 x 2.24 = kpa
28 9. Find d in te system sown in fig. If = 2.7 kpa Oil S = 0.6 ir 0.05 m 2.0 mm = 2.7 kpa 10 mm S = 1.4 d 300 mm Hg = γ = = m of + (0.05 x 0.6) + ( )0.6 + (0.01x13.6) (0.03 x13.6) d x1.4) = d = 0 d = m or d = 49.4 mm
29 10. Determine te absolute ressure at for te system sown in fig. ir S = Water Oil x S = (0.25 x 0.8) + (0.15 x 0.7) + (0.3 x 0.8)-(0.6) = 0 = m of = x 9.81 = kpa abs = abs = kpa SINGLE COLUMN MNOMETER: Single column manometer is used to measure small ressure intensities. S y B B 1 B B 1 C 1 C 1 U tube 2 (rea=a) C 1 Sm single column manometer consists of a sallow reservoir aving large cross sectional area wen comared to cross sectional area of U tube connected to it. For any cange in ressure, cange in te level of manometeric liquid in te reservoir is small ( ) and cange in level of manometric liquid in te U- tube is large.
30 To derive exression for ressure ead at : BB and CC are te levels of manometric liquid in te reservoir and U-tube before connecting te oint to te manometer, writing gauge equation for te system we ave, + y x S 1 x S m = 0 Sy = S m 1 Let te oint be connected to te manometer. B 1 B 1 and C 1 C 1 are te levels of manometeric liquid. Volume of liquid between BBB 1 B 1 = Volume of liquid between CCC 1 C 1 = a 2 = a 2 Let be te ressure ead at in m of. + (y + ) S ( ) Sm = 0 = ( ) Sm (y + ) S = Sm + 1 Sm + 2 Sm ys S = (Sm S) + 2 Sm = a 2 (Sm S) + 2 Sm It is enoug if we take one reading to get 2 If a is made very small (by increasing ) ten te I term on te RHS will be negligible. Ten = 2 Sm
31 INCLINED TUBE SINGLE COLUMN MNOMETER: x 2 C C y B B B B C C 1 θ 8 m Inclined tube SCM is used to measure small intensity ressure. It consists of a large reservoir to wic an inclined U tube is connected as sown in fig. For small canges in ressure te reading 2 in te inclined tube is more tan tat of SCM. Knowing te inclination of te tube te ressure intensity at te gauge oint can be determined. = a 2 sinθ ( Sm S) + 2 sinθ. Sm If a is very small ten = ( 2 = Sinθ) S m MECHNICL GUGES: Pressure gauges are te devices used to measure ressure at a oint. Tey are used to measure ig intensity ressures were accuracy requirement is less. Pressure gauges are searate for ositive ressure measurement and negative ressure measurement. Negative ressure gauges are called Vacuum gauges. Mecanical gauge consists of an elastic element wic deflects under te action of alied ressure and tis deflection will move a ointer on a graduated dial leading to te measurement of ressure. Most oular ressure gauge used is Bordon ressure gauge.
32 BSIC PRINCIPLE: Elastic Element (Posor Bronze) Link Sector Pinion Graduated Dial Togauge Point Te arrangement consists of a ressure resonsive element made u of osor bronze or secial steel aving ellitical cross section. Te element is curved into a circular arc, one end of te tube is closed and free to move and te oter end is connected to gauge oint. Te canges in ressure cause cange in section leading to te movement. Te movement is transferred to a needle using sector inion mecanism. Te needle moves over a graduated dial. Bourdon gage: Is a device used for measuring gauge ressures te ressure element is a ollow curved metallic tube closed at one end te oter end is connected to te ressure to be measured. Wen te internal ressure is increased te tube tends to straigten ulling on a linkage to wic is attaced a ointer and causing te ointer to move. Wen te tube is connected te ointer sows zero. Te bourdon tube, sketced in figure. It can be used for te measurement of liquid and gas ressures u to 100s of MPa.
33 2.7.4 Electronic Pressure Measuring Devices: Electronic Pressure transducers convert ressure into an electrical outut. Tese devices consist of a sensing element, transduction element and signal conditioning device to convert ressure readings to digital values on dislay anel. Sensing Elements: Te main tyes of sensing elements are Bourdon tubes, Diaragms, Casules, and Bellows.
34 Pressure Transducers: transducer is a device tat turns a mecanical signal into an electrical signal or an electrical signal into a mecanical resonse (e.g., Bourdon gage transfers ressure to dislacement). Tere are a number of ways to accomlis tis kind of conversion Strain gage Caacitance Variable reluctance Otical Normally Electronic Pressure transducers are costly comared to conventional mecanical gauges and need to be calibrated at National laboratories before ut in to use.
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