Downloaded 31 times
![17
Level 1 - Pressure 1RMT Training - 05 /98
v2 = [(2 / ρ x dP) / (1- β4
)] ½
v2 = (2)½
x (1/ρ)½
x 1/ (1- β4
)½
x (dP)½
Qv2 = (πd2
/4) x (2)½
x (1/ρ)½
x 1/ (1- β4
)½
x (dP)½
Qv2 = A2 x v2
constant constant assumed
constant
velocity of approach
constant - “E”
Qv2 = k (dP/ρ)½Volumetric Flow
Qm2 = k (dP x ρ)½Mass Flow k (dP/ρ)½
x ρ
Inferring non-pressure variablesInferring non-pressure variables
FlowFlow](https://image.slidesharecdn.com/01-pressurebasic1-160304133845/75/01-pressure-basic1-17-2048.jpg)
![27
Level 1 - Pressure 1RMT Training - 05 /98
At 0% Liquid Interface (4mA)
DP = Hside - Lside
= (SG1*h1) - [(SGf*(h1-h2)) + (SG1*h2)]
Indirectly measures liquid Interface
Pbottom
Ptop
L H
Remote
Seal
Vapor
0%
100%
SG1
SG2
Dist. Betw.
Taps
(h1 - h2)
Total Liquid level must
always be above the
Top transmitter tap.
SGf
Inferring non-pressure variablesInferring non-pressure variables
InterfaceInterface
h1
h2](https://image.slidesharecdn.com/01-pressurebasic1-160304133845/75/01-pressure-basic1-27-2048.jpg)
![28
Level 1 - Pressure 1RMT Training - 05 /98
Total Liquid level must
always be above the
Top transmitter tap.
Pbottom
Ptop
L H
Remote
Seal
Vapor
0%
100%
SG1
SG2
Dist. Betw.
Taps
(h1 - h2)
At 100% Liquid Interface (20mA)
DP = Hside - Lside
= [SG2*(h1-h2) + SG1*h2)] - [(SGf*(h1-h2)) + (SG1*h2)]
Indirectly measures liquid Interface
Inferring non-pressure variablesInferring non-pressure variables
InterfaceInterface
h1
h2
SGf](https://image.slidesharecdn.com/01-pressurebasic1-160304133845/75/01-pressure-basic1-28-2048.jpg)
![46
Level 1 - Pressure 1RMT Training - 05 /98
ExerciseExercise
1. If the atmospheric pressure drop by 0.1 % and the line
pressure remains unchanged, what changes will occur in the
readings?
(A) AP reading will change.
(B) GP reading will change.
(C) Both reading will change.
(D) Both reading will not change.
[ ]
2. If a customer wants to measure vacuum, what type of transmitter
should be used?
(A) AP
(B) DP
(C) GP [ ]
Liquid flow
Line pressure = 80 psig
94.7psi 80.psi GP
Transmitter
AP
Transmitter](https://image.slidesharecdn.com/01-pressurebasic1-160304133845/75/01-pressure-basic1-46-2048.jpg)
![47
Level 1 - Pressure 1RMT Training - 05 /98
ExerciseExercise
Write down the readings in (psi) that are recorded by the transmitters
in the above application (Atmosphere = 14.7 psi).
3. Differential Pressure Transmitter (a): [ ]
4. Gage Pressure Transmitter (b): [ ]
5. Absolute Pressure Transmitter (c): [ ]
50 psig80 psig
c a b](https://image.slidesharecdn.com/01-pressurebasic1-160304133845/75/01-pressure-basic1-47-2048.jpg)
More Related Content
01 pressure basic1
- 1. 1 Level 1 -Pressure 1RMT Training - 05 /98 Fundamental TrainingFundamental Training Level 1
- 2. 2 Level 1 -Pressure 1RMT Training - 05 /98 Topics: Slide No: • Why measure pressure? 3 • What is pressure? 4 - 5 • Pressure terminology 6 - 11 • Inferring non-pressure variables 12 - 29 • Pressure measurement technology 30 - 44 • Pressure calibrators 45 • Exercises 46 - 48 ContentsContents
- 3. 3 Level 1 -Pressure 1RMT Training - 05 /98 Why measure pressure?Why measure pressure? 4 Common Reasons4 Common Reasons Safety • prevent pressurized pipes & vessels from bursting Process Efficiency • variation of pressure below or above a set-point will result in scrap rather than useable product in some manufacturing process Cost Saving • preventing unnecessary expense of creating more pressure or vacuum than is required saves money Inferred Measurement of Other Variables • rate of flow through a pipe • level of fluid in a tank • density of fluid • how two or more liquids in a tank interface
- 4. 4 Level 1 -Pressure 1RMT Training - 05 /98 What is pressure?What is pressure? The Same Weight, Different PressureThe Same Weight, Different Pressure 1 sq ins 100 sq ins 1 sq ins100 sq ins Weight = 100lb Pressure = Pressure =1lb/in² 100 lb/in²
- 5. 5 Level 1 -Pressure 1RMT Training - 05 /98 What is pressure?What is pressure? Liquid & Gas PressuresLiquid & Gas Pressures LIQUIDS The pressure exerted by a liquid is influenced by 3 main factors. 1. The height of the liquid. 2. The density of the liquid. 3. The pressure on the surface of the liquid. GASES The pressure exerted by a gas is influenced by 2 main factors. 1. Volume of the gas container. 2. Temperature of the gas Note. Gases are compressible whereas liquids are not
- 6. 6 Level 1 -Pressure 1RMT Training - 05 /98 I/P PT PIC • Pressure Loop Issues: – May be a Fast Process » Liquid » Small Volume – May Require Fast Equipment Pressure terminologyPressure terminology Pressure Control LoopPressure Control Loop
- 7. 7 Level 1 -Pressure 1RMT Training - 05 /98 Pressure terminologyPressure terminology Engineering UnitsEngineering Units Pressure is defined as FORCE applied over a unit AREA. P = F/A Examples of pressure units: Units of force per unit area Pascals Pa N / m2 (Newtons / square metre) psi lbs/in2 (Pounds / square inch) Bar Bar = 100,000 Pa Units referenced to columns of liquids ins. water gauge in H2O mm water gauge mm H2O ins. mercury in Hg mm mercury mm Hg Atmosphere atm Pressure applied by a 1 inch column of mercury with a density of 13.5951 g/cm³. Pressure exerted by the earth’s atmosphere at sea level (approximately 14.6959psi) Pressure applied by a 1 inch column of water at 20°C.
- 8. 8 Level 1 -Pressure 1RMT Training - 05 /98 Gage(psig) - Level of pressure relative to atmospheric – Positive or negative in magnitude Atmospheric Pressure Approx. 14.7 psia Absolute Gage Compound Range Barometric Range Pressure Total Vacuum (Zero Absolute) Absolute(psia) - based from zero absolute pressure - no mass Typical atm reference: 14.73 psia Compound Range (psig) - Gage reading vacuum as negative value Differential(psid) - difference in pressure between two points Pressure terminologyPressure terminology Reference PressureReference Pressure
- 9. 9 Level 1 -Pressure 1RMT Training - 05 /98 Absolute Zero Total Vacuum Atm. Pressure 14.7 psia 5 psig ? Psia19.7 5 psi vacuum ? Psia ? Psig-5 9.7 Assume: Patm = 14.7psia; 28 inches H2O per psi 1000 in H2O = ___________ psi35.71 Pressure terminologyPressure terminology QuizQuiz
- 10. 10 Level 1 -Pressure 1RMT Training - 05 /98 Pressure terminologyPressure terminology Measurable PressuresMeasurable Pressures The four most common types of measurable pressures used in the process control industries are: 1. Head Pressure or Hydrostatic Pressure.Head Pressure or Hydrostatic Pressure. Pressure exerted by a column of liquid in a tank open to atmosphere, HEAD PRESSURE = HEIGHT x DENSITY 2. Static Pressure, Line Pressure, or Working pressureStatic Pressure, Line Pressure, or Working pressure Pressure exerted in a closed system 3. Vapor PressureVapor Pressure The temperature at which a liquid boils, or turns into a vapor varies depending on the pressure. The higher the pressure, the higher the boiling point. 4. VacuumVacuum Absolute pressure below atmospheric pressure ( a compound range gage transmitter will read a negative pressure)
- 11. 11 Level 1 -Pressure 1RMT Training - 05 /98 Pressure terminologyPressure terminology Measurable PressureMeasurable Pressure Typical Vapor Pressure Curve Pressure(log) Temperature liquid gas Higher Altitute Lower Altitute (Sea Level) T1 T2 Vapor pressure increases with temperature. • Liquid boils when its vapor pressure equals atmospheric pressure.
- 12. 12 Level 1 -Pressure 1RMT Training - 05 /98 Flow Restriction in Line cause a differential Pressure Line Pressure QV= K DP Orifice Plate Inferring non-pressure variablesInferring non-pressure variables FlowFlow
- 13. 13 Level 1 -Pressure 1RMT Training - 05 /98 Theoritical equations come from 3 sources: Continuity Equation • Flow into pipe equals flow out of pipe and is the same at all pipe cross sections (Conservation of Mass) Bernoulli’s Equation • (Conservation of Energy for fluid in a pipe) Experimentally Determined Correction Factors • Discharge Coefficient • Gas Expansion Factor Qm= K DP Inferring non-pressure variablesInferring non-pressure variables FlowFlow
- 14. 14 Level 1 -Pressure 1RMT Training - 05 /98 The volume flowing into a pipe equals the volume flowing out of pipe, assuming constant density A1V1 A2V2Flow Flow v1 = A2/A1 x v2 v1 = d2 /D2 x v2 πd2 /4 x πD2 /4 Continuity Equation A1v1 = A2v2 A = area of pipe cross section v = velocity d/D = β v1 = β2 x v2 Inferring non-pressure variablesInferring non-pressure variables FlowFlow
- 15. 15 Level 1 -Pressure 1RMT Training - 05 /98 Bernoulli’s Equation cancel - off for level pipe v1 v2 P1 P2 D d Three energies: Kinetic (1/2ρv2 ) Potential (ρgh) Static Pressure (P) Flow The total energy before the restriction in the pipe must equal the total energy after the restriction. Inferring non-pressure variablesInferring non-pressure variables FlowFlow
- 16. 16 Level 1 -Pressure 1RMT Training - 05 /98 P 1 P 2 ..1 2 ρ v 2 2 ..1 2 ρ v 1 2 common P 1 ..1 2 ρ v 1 2 ..ρ g h 1 P 2 ..1 2 ρ v 2 2 ..ρ g h 2 Before restriction After restriction dP = ½ ρ (v2 2 - v1 2 ) 2 / ρ x dP = v2 2 - v1 2 V1 2 = (β2 x V2)2 2 / ρ x dP = v2 2 - β4 x v2 2 2 / ρ x dP = (1- β4 ) v2 2 common subject v2 2 = (2 / ρ x dP) / (1- β4 ) Re-arranged Inferring non-pressure variablesInferring non-pressure variables FlowFlow
- 17. 17 Level 1 -Pressure 1RMT Training - 05 /98 v2 = [(2 / ρ x dP) / (1- β4 )] ½ v2 = (2)½ x (1/ρ)½ x 1/ (1- β4 )½ x (dP)½ Qv2 = (πd2 /4) x (2)½ x (1/ρ)½ x 1/ (1- β4 )½ x (dP)½ Qv2 = A2 x v2 constant constant assumed constant velocity of approach constant - “E” Qv2 = k (dP/ρ)½Volumetric Flow Qm2 = k (dP x ρ)½Mass Flow k (dP/ρ)½ x ρ Inferring non-pressure variablesInferring non-pressure variables FlowFlow
- 18. 18 Level 1 -Pressure 1RMT Training - 05 /98 (i) What would be the differential at 10m³/s? Quiz: If an orifice plate creates a differential of 50 kPa at 30m³/s DP2 = 5.6kPa (ii) What would be the flow rate at 30kPa differential? 30/Qv2 = √50/ √30 Qv = K √DP Qv1 √DP1 --- = ---- Qv2 √DP2 30/10 = √50/ √DP2 Qv2 = 23.26m³/s Qv = K √DP Qv1 √DP1 --- = ---- Qv2 √DP2 Inferring non-pressure variablesInferring non-pressure variables FlowFlow
- 19. 19 Level 1 -Pressure 1RMT Training - 05 /98 H P P P P D Liquid Hydrostatic Pressure - The liquid will rise to the same level in each vessel regardless of its diameter & shape. Which shape gives higher pressure at the bottom of the vessel? Unit Area (eg. per cm2 ) Similar height of column will have same mass acting on the same unit area SAME PRESSURE Inferring non-pressure variablesInferring non-pressure variables LevelLevel
- 20. 20 Level 1 -Pressure 1RMT Training - 05 /98 The hydrostatic pressure exerted by the column of liquid depends on the S.G. (or density) of the liquid and its vertical height. Density of liquid = D Average cross-section area of vessel = A Vertical height of liquid = H Volume of liquid, V = Total weight of liquid, M = = Pressure at the bottom of liquid = weight of liquid cross-section area = = H x A D x V A x H D x H With reference to inches or mm WATER S.G x H D x (D x A x H) / A Inferring non-pressure variablesInferring non-pressure variables LevelLevel
- 21. 21 Level 1 -Pressure 1RMT Training - 05 /98 P = r x g x height x area / area mass x g r x volume Density = mass/volume = r P= force / area g = gravitational acceleration height x area Phead = r x g x h Pascal Inferring non-pressure variablesInferring non-pressure variables LevelLevel
- 22. 22 Level 1 -Pressure 1RMT Training - 05 /98 Inferring non-pressure variablesInferring non-pressure variables LevelLevel XMTR HL Ullage or Vapor S.G Phead Phead = S.G x Height 0% 100% Height DP Transmitter at the bottom of the tank measures HEAD. HEAD = pressure at the bottom of a column of liquid with known relative density (S.G) Height = Phead / S.G Cancelled off since both L & H sides of transmitter experience it.
- 23. 23 Level 1 -Pressure 1RMT Training - 05 /98 Quiz: Open Tank What is the level if Pmax = 120 inH2O, s.g.= 1.2? XMTR HL ?Height = Phead / S.G Height = 120 / 1.2 Height = 100 inches Inferring non-pressure variablesInferring non-pressure variables LevelLevel
- 24. 24 Level 1 -Pressure 1RMT Training - 05 /98 Quiz: Closed Tank Dry leg: no fluid in low side impulse piping, or leg Ph = 105 psi Pl = 100 psi What is level if s.g. = 1.0? Ptop= Ullage XMTR HL dP = 5 psi = 5 x 28 inH2O Height = 140 / 1.0 Height = 140 inches Phead Inferring non-pressure variablesInferring non-pressure variables LevelLevel
- 25. 25 Level 1 -Pressure 1RMT Training - 05 /98 Pbottom = Ptop = Pbottom - Ptop = Hence, S.G = Ptop Phead(top) Pbottom Ptop Phead(bottom) h1 h2 Liquid level must be above the Top transmitter tap. H H S.G X h2 S.G X h1 S.G (h2 - h1) diff. Pressure / dist. betw. taps Inferring non-pressure variablesInferring non-pressure variables DensityDensity
- 26. 26 Level 1 -Pressure 1RMT Training - 05 /98 Ullage Pbottom Ptop 50” H H Quiz: Determined the S.G of the process fluid if Ptop = 20 psi Pbottom = 22 psi Distance between taps = 50 inches Assuming 1 psi = 28”H2O S.Gprocess = DP / dist. betw. Taps = 56 / 50 = 1.12 DP = (22 -20) = 2 psi = 56”H2O Inferring non-pressure variablesInferring non-pressure variables DensityDensity
- 27. 27 Level 1 -Pressure 1RMT Training - 05 /98 At 0% Liquid Interface (4mA) DP = Hside - Lside = (SG1*h1) - [(SGf*(h1-h2)) + (SG1*h2)] Indirectly measures liquid Interface Pbottom Ptop L H Remote Seal Vapor 0% 100% SG1 SG2 Dist. Betw. Taps (h1 - h2) Total Liquid level must always be above the Top transmitter tap. SGf Inferring non-pressure variablesInferring non-pressure variables InterfaceInterface h1 h2
- 28. 28 Level 1 -Pressure 1RMT Training - 05 /98 Total Liquid level must always be above the Top transmitter tap. Pbottom Ptop L H Remote Seal Vapor 0% 100% SG1 SG2 Dist. Betw. Taps (h1 - h2) At 100% Liquid Interface (20mA) DP = Hside - Lside = [SG2*(h1-h2) + SG1*h2)] - [(SGf*(h1-h2)) + (SG1*h2)] Indirectly measures liquid Interface Inferring non-pressure variablesInferring non-pressure variables InterfaceInterface h1 h2 SGf
- 29. 29 Level 1 -Pressure 1RMT Training - 05 /98 Application Example: • Transmitter calibrated from 120”H2Oto 132”H2O • Determine % of interface of Liquid A with respect to Liquid B Vapor 0% 100% SG1= 1.0 SG2= 1.1 Pbottom Ptop L H Remote Seal 10 ft Liquid A Liquid B 123 inH2O If transmitter reads 123 inH2O % interface = (3/12) * 100% = 25% Inferring non-pressure variablesInferring non-pressure variables InterfaceInterface
- 30. 30 Level 1 -Pressure 1RMT Training - 05 /98 Barometer Used to measure Barometric Pressure Reference is 0 psia, due to low vapor pressure of Hg. General operating principle: Phead Patm Barometric Pressure = Atmospheric Pressure 29.9 inHgWhat is the barometric Pressure? • Tube completely filled with mercury & Invert into the container filled with mercury. • The mercury level in the tube will drop until it reaches an equilibrium. • This equilibrium height is a measure of atmospheric pressure. Phead = Patm Pressure measurement technologyPressure measurement technology Pressure GaugesPressure Gauges
- 31. 31 Level 1 -Pressure 1RMT Training - 05 /98 dP = H (SGfill fluid - SGprocess fluid) – Reference side can be: • Sealed (AP reference) • Open to atmosphere(GP reference) • Connected to reference pressure(DP reference) – Typically used for low pressures, non process control Manometers U-tube with one side reference, one side measured pressure H How to check for dP ? Pressure measurement technologyPressure measurement technology Pressure GaugesPressure Gauges
- 32. 32 Level 1 -Pressure 1RMT Training - 05 /98 Mechanical The mechanical element techniques convert applied pressure into displacement. The displacement may be converted into electrical signal with help of Linear Variable Displacement Transformer (LVDT). Pressure measurement technologyPressure measurement technology Pressure GaugesPressure Gauges
- 33. 33 Level 1 -Pressure 1RMT Training - 05 /98 Output to Actuator (or Relay) Constant flowrate maintained (Compressed air) Nozzle Flapper Bourdon Tube Process Pressure Pressure measurement technologyPressure measurement technology Pneumatic Pressure CellsPneumatic Pressure Cells Pneumatic Controller Relay’s modulated output is the controller output which is usually a pneumatic signal that adjusts the final control element (Control valve)
- 34. 34 Level 1 -Pressure 1RMT Training - 05 /98 Disadvantages – Reconfiguration costly – Losses occur over long piping runs – Performance levels are not comparable to electronic instrumentation Pressure Transmitter Produce a linear output proportional to input pressure Zero Scale: Full Scale: 3 psig 15 psig Pressure measurement technologyPressure measurement technology Pneumatic Pressure CellsPneumatic Pressure Cells
- 35. 35 Level 1 -Pressure 1RMT Training - 05 /98 – Made up of 2 main elements: • Transducer - Electronic sensor module that registers process variable and outputs a corresponding usable electrical signal eg. resistance, millivolts, capacitance, etc. • Electronics - Convert transducer output to a standard output signal eg. 4 - 20 mA, 1 - 5 V dc, digital signal, etc. Pressure measurement technologyPressure measurement technology Electronic Pressure TransmittersElectronic Pressure Transmitters
- 36. 36 Level 1 -Pressure 1RMT Training - 05 /98 Transmitter Signal from sensor module (Transducer) Signal To Controller Process Variable (Standard signals) Sensing Diaphragm (Line / Static Pressure) Example of Application Transmitter configured to operate from: 0 to 50 psi Electronic Output: 4 to 20 mA This mean 0% reading (0 psi) represents 4 mA and 100% reading (50 psi) represents 20 mA. What will be the output current at 25 psi reading? 4 + (25/50)*16 = 12 mA Pressure measurement technologyPressure measurement technology Electronic Pressure TransmittersElectronic Pressure Transmitters
- 37. 37 Level 1 -Pressure 1RMT Training - 05 /98 Characterized by the type of sensing element: – Variable capacitance – Variable Resistance (Wheatstone bridge) • Strain gauge » Thin -film strain gauge » Diffused, strain gauge – Variable inductance – Variable reluctance – Vibrating wire – Piezoelectric Pressure measurement technologyPressure measurement technology Electronic Pressure Sensor ModulesElectronic Pressure Sensor Modules
- 38. 38 Level 1 -Pressure 1RMT Training - 05 /98 Variable Capacitance • Process pressure transmitted thru isolating diaphragm • Distortion of sensing diaphragm proportional to the differential pressure • Position of sensing diaphragm detected by capacitor plates • Differential capacitance translated to 4-20mA or 10-50mA output dc signal. Pressure measurement technologyPressure measurement technology Electronic Pressure Sensor ModulesElectronic Pressure Sensor Modules
- 39. 39 Level 1 -Pressure 1RMT Training - 05 /98 Variable Resistance / Piezo-Resistive Thin Film Strain Gauge Diffused Strain Gauge • Process pressure transmitted thru isolating diaphragm • Very small distortion in sensing diaphragm • Applies strain to a wheatstone bridge circuit • Change in resistance translated to 4-20mA or 1-5V dc signal • GP XMTRs - ref. side of sensor exposed to atm. Pressure • AP XMTRs - sealed vacuum reference. Pressure measurement technologyPressure measurement technology Electronic Pressure Sensor ModulesElectronic Pressure Sensor Modules
- 40. 40 Level 1 -Pressure 1RMT Training - 05 /98 • Piezoelectric crystal is a natural or a synthetic crystal that produces a voltage when pressure is applied to it. • Voltage produce by crystal increases with increases in pressure and vice-versa. • The produced small voltage is then amplified to a standard control signal. Piezoelectric Amplifier & electronics Control Signal Piezoelectric Crystal Diaphragm Process Pressure Pressure measurement technologyPressure measurement technology Electronic Pressure Sensor ModulesElectronic Pressure Sensor Modules
- 41. 41 Level 1 -Pressure 1RMT Training - 05 /98 • Inductance is the opposition to a change in current flow • Alternating current pass through the coil • Elastic element connected to core • Applied pressure deflects elastic element • Position of core changes relative to coil resulting in change in inductance • Resistor connected in series with inductor to measure change in voltage. Variable Inductance Pressure measurement technologyPressure measurement technology Electronic Pressure Sensor ModulesElectronic Pressure Sensor Modules
- 42. 42 Level 1 -Pressure 1RMT Training - 05 /98 • Reluctance is a property of magnetic circuit • A moving magnetic element located between two coils • Coil turn electromagnet when excited by AC source • Position of element with respect to the coils determines differential magnetic reluctance • Thus differential inductance within the coils • A bridge is used to measure changes in a circuit Variable Reluctance Pressure measurement technologyPressure measurement technology Electronic Pressure Sensor ModulesElectronic Pressure Sensor Modules
- 43. 43 Level 1 -Pressure 1RMT Training - 05 /98 • Wire located in magnetic field vibrate when current pass through it • Wire movement within field induces current into it • Induced voltage amplified as output signal • Vibration frequency depends on wire tension • Elastic element connected to wire. • Frequency of wire vibration become a function of measured pressure • Direct digital output signal Vibrating Wire Pressure measurement technologyPressure measurement technology Electronic Pressure Sensor ModulesElectronic Pressure Sensor Modules
- 44. 44 Level 1 -Pressure 1RMT Training - 05 /98 – Sensor (transducer) module is part of the transmitter. – Sensor will become active only when the transmitter is powered. (Attenuation) – Output Electronics in the transmitter translates the userable electrical signal from the sensor into a standard output signal. Output Electronics Sensor Module Output Electronics Sensor Module Diaphragm Seal Pressure measurement technologyPressure measurement technology Electronic Pressure Sensor ModulesElectronic Pressure Sensor Modules
- 45. 45 Level 1 -Pressure 1RMT Training - 05 /98 ISO Require calibration device to be 4 times more accurate than the accuracy of the instrument being calibrated. If the reference accuracy of a 3051C transmitter is 0.075% of span, – What should the accuracy of the C/V pressure source be? – the equipment for calibrating the pressure source? If the diameter of the ball on a dead weight tester is 0.75 inches. The weight of a plate is 723g. – What is the pressure required to freely float that plate on the dead weight tester (g/cm2 )? Pressure calibratorsPressure calibrators ISO RequirementISO Requirement
- 46. 46 Level 1 -Pressure 1RMT Training - 05 /98 ExerciseExercise 1. If the atmospheric pressure drop by 0.1 % and the line pressure remains unchanged, what changes will occur in the readings? (A) AP reading will change. (B) GP reading will change. (C) Both reading will change. (D) Both reading will not change. [ ] 2. If a customer wants to measure vacuum, what type of transmitter should be used? (A) AP (B) DP (C) GP [ ] Liquid flow Line pressure = 80 psig 94.7psi 80.psi GP Transmitter AP Transmitter
- 47. 47 Level 1 -Pressure 1RMT Training - 05 /98 ExerciseExercise Write down the readings in (psi) that are recorded by the transmitters in the above application (Atmosphere = 14.7 psi). 3. Differential Pressure Transmitter (a): [ ] 4. Gage Pressure Transmitter (b): [ ] 5. Absolute Pressure Transmitter (c): [ ] 50 psig80 psig c a b
- 48. 48 Level 1 -Pressure 1RMT Training - 05 /98 ExerciseExercise 6. What is the differential pressure (P1 - P2) in kPa being applied to the manometer in the the above application ? S.G of Process Fluid @ Temp + Pressure = 1.0 P2P1 S.G. = 13.6 200mm (Note 1 mm H2O = 9.8 Pa)
Editor's Notes
- #6 Liquids. Provided the density and surface pressure remain the same, then the pressure measured at a depth of 5 feet will be the same in a 5000 gallon tank as in a 50 gallon tank. If the liquid density doubles, then the pressure measured at a depth of 5 feet will also double. If the the pressure on the surface of the liquid increases, then the pressure measured at a depth of 5 feet would increase by the same amount. Gases Boyle’s law states that the pressure of a gas varies proportionally to the volume it occupies provided the temperature is held constant. That is, if you transfer the gas from a 5000ft³ tank at a pressure of 1000psia to a 1000ft³ tank, if you maintain the same temperature (e.g.300°C) the pressure would increase to 5000psia Charles’s law states that the pressure of a gas varies proportionally to the absolute temperature provided the volume is held constant. For instance, if in the above example we reduce the temperature to 14°C then the pressure would drop to 2500psia
- #7 Identify the instruments in the diagram. Considerations: Speed - Pressure loops may respond to changes in loads and changes in control equipment actions either slowly or quickly. For example: A large volume gas system (such as a long transmission pipeline or a large gas storage vessel) may respond slowly to changes in loads and control equipment actions. A small volume liquid system will respond to changes very quickly to changes in loads and control equipment actions. Some applications, for example compressor bypass or surge control, by their very nature require very fast response of the control equipment. Equipment Requirements - Fast-acting processes will require instruments and control valves that can react quickly to changes in the process variable.
- #8 1 Torr = 1 mm Hg 12 inH2O = 1 ft H2O 1 psi = 27.7296 inH2O 1 inHg = 25.4 mm Hg 1 bar = 14.5038 psi 1 kg/cm2 = 980.665 mbars AtmospheresAtmos.1 PascalsPa N / m2101325 Pa psilbs/in214.696 psia BarBar1.01325 Bar abs ins. water gaugein H2O408.07 in. H2O at 15°C mm water gaugemm H2O10365 mm H2O at 15°C ins. mercuryin Hg29.921 in Hg at 0°C mm mercurymm Hg760mm Hg at 0°C
- #11 HEAD PRESSURE. If the height is measured in inches, and the Density is replaced by the Specific Gravity, then the Head Pressure will be in inches of water
- #13 Infer Flow - flow is calculated (indirect measurement) You have a line pressure of 1000PSI. As the flow comes to the orificeplate the pressure will increase because of the blockage. As the process makes it through the blockage the pressure will drop and then gradually work its way back to 1000PSI. This pressure drop is inverse to the flow. In other words if the flow increases the pressure on the low side will go lower. The simple equation to convert the pressure drop in on the bottom of the slide. The “K” constant is made of many variables but is represented by one number in non-compensated flow application. These variables do change as flow changes and that is what the 3095 addresses. The square root of of DP is a characteristic of a head producing flow device, the relationship between the flow velocity and DP it creates. There are different places to measure this pressure drop. This diagram is flange taps. (k) - reflects the characteristics of the restrictor and flow conditions.
- #14 <number>
- #15 <number>
- #16 <number>
- #17 <number> As stated in the continuity equation, if the area is reduced the velocity must increase. If the velocity (kinetic energy) increases, the static pressure must decrease to keep total energy equal. So, with the potential energy equal before and after, (horizontal pipe), the difference in pressure is inversely proportional to the difference in velocity.
- #18 <number> As stated in the continuity equation, if the area is reduced the velocity must increase. If the velocity (kinetic energy) increases, the static pressure must decrease to keep total energy equal. So, with the potential energy equal before and after, (horizontal pipe), the difference in pressure is inversely proportional to the difference in velocity.
- #19 <number> (i) Qv = K DP Qv1 DP1 -----=--------- Qv2 DP2 30/10= 50/ DP2 DP2=5.6kPa (ii) 30/Qv2= 50/ 30 Qv2=23.26m³/s
- #20 <number>
- #21 <number>
- #23 <number>
- #26 To measure Relative Density (Specific Gravity) Can use one differential transmitter with two remote seals at the two points.
- #28 Interface is the separation boundary between two immersible liquid. SG1- Lower specific gravity SG2- higher specific gravity Both the SGs must be predetermined and should remain constant for accurate measurement. Can use one differential transmitter with two remote seals at the two points.
- #34 It is a FORCE BALANCE pressure transmitter (Closed loop feedback device). In a force balance unit, pressure displaces the sensing element (diaphragm). The amount of displacement is detected and the element is returned to a null or zero displacement position by a restoring force pneumatically. Thus a pneumatic pressure will be maintained exactly proportional to differential pressure and is used as a standard output signal (3 to 15 psi).
- #35 It is a FORCE BALANCE pressure transmitter (Closed loop feedback device). In a force balance unit, pressure displaces the sensing element (diaphragm). The amount of displacement is detected and the element is returned to a null or zero displacement position by a restoring force pneumatically. Thus a pneumatic pressure will be maintained exactly proportional to differential pressure and is used as a standard output signal (3 to 15 psi).
- #37 We normally use a transmitter to measure the process signals. The transmitter changes the output of the primary element into a control signal that can be utilized by the process controller; typically a current, voltage, HART compliant, or digital signal. The primary element senses a change in the value, quantify, or quality of the process variable and provides an output in the form of a change in current, voltage, motion, force, pressure, etc. The sensor output is generally a very low level signal that must be amplified, conditioned, or converted before it is useful in a process control system. Other devices are used to change signals from one type to another. These units are often called transducers or converters; and they are used to match or provide the necessary signal type to be compatible with another device. We will discuss transmitters for the “BIG FOUR”, those being pressure, temperature, level, and flow. Some other common measurements are oxygen, pH, conductivity, position, weight, and composition.
- #39 In the DP Sensor, process pressure is transmitted through the isolating diaphragm and fill fluid to the sensing diaphragm in the center of the capacitance cell. Capacitor plates on both sides of the sensing diaphragm detect its position. The differential capacitance between the sensing diaphragm and the capacitor plates is proportional to process pressure.
- #40 The AP Sensor is fabricated utilizing a processing method called Chemical Vapor Decomposition (CVD). Resistive sensors have 4 resistors in the diaphragm, connected as a wheatstone bridge.The resistors are formed by vapour deposition in a thin film on top of a silicon substrate.(Thin-film strain gauge) OR diffusion / embedding inside the silicon. (Piezoresistive strain gauges)- solid state device. Hydraulically (sealed fluid) connected to the high pressure side of the transmitter. Process pressure is transmitted through the fill fluid to the sensing element, creating a very small deflection of the silicon substrate. The resulting strain on the substrate changes the bridge resistance in proportion to the the pressure applied. (Bridge unbalanced) GP Sensor is fabricated in the same technique as AP sensor, however the reference side of the silicon substrate is vented to atmosphere instead of concealed in the vacuum.
- #46 0.001875% 0.00469% 2.55 g/cm2
- #47 1.B 2C
- #48 3.80 - 50 = 30 psi 4.50 psi 5.80 + 14.7 = 94.7 psi
- #49 DP = 200 (13.6 -1.0) = 2520 mm H2O = 2520 * 9.8/1000 kPa = 24.7 kPa