PROCESS INSTRUMENTATION I

Transcription

1 PROCESS INSTRUMENTATION I MODULE CODE: EIPIN1B STUDY PROGRAM: UNIT 1 VUT Vaal University of Technology /10

2 EIPINI Chapter 1: Introduction to Industrial Instrumentation Page INTRODUCTION TO INDUSTRIAL INSTRUMENTATION " when you can measure what you are speaking about, and express it in numbers, you know something about it;..." Lord Kelvin ( ), Institute of Civil Engineers, London, 3rd May MEASUREMENT Measurement is defined as the determination of the existence or magnitude of a variable for monitoring and controlling purposes. 1. UNITS AND STANDARDS A measurement is done with an instrument in terms of standard units. The system of units, which is most widely used, is the SI (Systems International d'unites). The seven so called base units of the system, are the following: meter (length) kilogram (mass) second (time) Kelvin (temperature) ampere (current) candela (luminous intensity) Mole (amount of substance) Standards for these units are classified as follows: International standards International standards are defined by international agreement, representing units of measurements to the best possible accuracy allowed by measurement technology. Primary standards Primary standards are maintained at institutions in various countries. The main function is to check the accuracy of secondary standards. Secondary standards Secondary standards are employed in industry as reference for calibrating highaccuracy equipment and components. Calibration and comparison are done periodically by the involved industries against the primary standards maintained in the national standards labs. The main function of the secondary standards is to verify the accuracy of working standards. Working standards Workplace standards are used to calibrate instruments used in industrial applications and instruments used in the field, for accuracy and performance. Working standards are checked against secondary standards for accuracy.

3 EIPINI Chapter 1: Introduction to Industrial Instrumentation Page FUNCTIONAL ELEMENTS OF INSTRUMENTS Functions of instruments Instruments may be classified according to the functions they perform. Indicating function An instrument may provide the information about the value of a quantity under measurement, in the form commonly known as an indicating function. For example, the pointer and scale on a speedometer, indicates the speed of an automobile at that instant. Recording function An instrument may provide the information of the value of a quantity under measurement against time or some other variable, in the form of a written record, usually on paper. For example, an instrument may record the room temperature every second, as a graph on a strip chart. Controlling function This is one of the most important functions of an instrument, especially in the field of industrial control processes. In this case, the information provided by the instrument is used by the control system to control the original measured quantity. For example, the temperature measured in a room may be used to switch the cooling system on or off, in order to keep the room temperature within preset values Elements of instruments When examining different instruments, one soon recognizes a recurring pattern of similarity with regard to function. This leads to the concept of breaking down instruments into a number of elements according to the function of each element. Consider for example, the liquid filled pressure type thermometer in Figure 1-1. Bulb Bourdon tube Scale and pointer Tube Link and gears Figure 1-1 A temperature change results in a pressure build-up within the bulb because of the constrained thermal expansion of the filling fluid. This pressure is transmitted through the tube to a Bourdon type pressure gauge, which converts pressure to displacement. This displacement is manipulated by the linkage and gearing to give a larger pointer movement. We can now recognise the following basic functional instrument elements, using this liquid filled thermometer as an example:

4 EIPINI Chapter 1: Introduction to Industrial Instrumentation Page 1-3 Primary element: The primary element is that part of the instrument that first utilizes energy from the measured medium and produces an output depending in some way on the measured quantity. Note: For the liquid filled thermometer example, the bulb is the element in contact with the measured medium. The energy it extracts from the medium in this case is heat energy. The variable conversion from temperature to pressure is accomplished when the heat energy absorbed by the liquid in the bulb, causes an increase in pressure energy within the volume constrained liquid. Data transmission element: The data transmission element transmits data from one element to another. Note: When the elements of an instrument are physically separated, it becomes necessary to transmit the data from one to the other. It may be as simple as the tube in the liquid filled thermometer example, that transmits the pressure information from the bulb to the Bourdon tube, or as complicated as the telemetry system between a ship and the cruiser missile it has launched. Secondary element: The secondary element converts the output of the primary element, to another more suitable variable for the instrument to perform the desired function. Note: In the thermometer example, the Bourdon tube is the secondary element (or the variable conversion element, as it is often called). It responds with a movement when receiving a pressure input. Every instrument need not include a second variable conversion element, while some require several. Manipulation element: The manipulation element processes the information received from the primary or secondary element and transforms the data into a more useful form. Note: By manipulation we mean specifically a change in the numerical value of the variable according to some definite rule, while preserving the original character of the variable. In the thermometer example, a small movement of the Bourdon tube is amplified by the gears to produce a large circular movement of the pointer. A variable manipulation element does not necessarily follow a variable conversion element; it may precede it, appear elsewhere in the chain, or not appear at all. Functioning element: The functioning element is that part of the instrument that is used for indicating, recording or controlling of the measured quantity. Note: The presentation of measured information may assume many different forms. It could include the simple indication of a pointer moving over a scale or the recording of a pen moving over a chart. It may also be in the form of a digital readout or even in a form not directly detectable by human senses as in the case of a digital computer used to perform a control function according to the value of the measured variable.

5 EIPINI Chapter 1: Introduction to Industrial Instrumentation Page 1-4 In summary then, the interconnection between the various functional elements for this particular thermometer instrument, is shown in Figure 1-. It must be stressed though, that different instruments are not necessarily composed of all these elements or may not adhere to the same order of interconnection, as depicted in Figure 1-. Temperature Measured medium Bulb Tube Bourdon tube Linkage and gear Scale and pointer Primary element (Variable conversion element: temperature to pressure) Data transmission element (Pressure to pressure) Secondary element (Variable conversion element: pressure to motion) Variable manipulation element (Motion to motion) Functioning element (Data presentation: indicating function performed by moving pointer over scale) Observer Figure RANGE AND SPAN OF AN INSTRUMENT Range: The range of an instrument is the minimum and maximum values of the measured variable that the instrument is capable of measuring. Span: The span of an instrument is the arithmetic difference between the minimum and maximum range values, used to describe both the input and the output. Example: A thermometer can measure temperature between -0 ºC and 90 ºC. The range of the instrument is from -0 ºC to 90 ºC. The span of the instrument is 90 (-0) = 110 ºC.

6 EIPINI Chapter 1: Introduction to Industrial Instrumentation Page STATIC CHARACTERISTICS OF INSTRUMENTS Information about the static performance or static characteristics of an instrument, is obtained by a process called static calibration. Static calibration refers to a situation where the input is varied over some range of constant values, causing the output to vary over some range of constant values. Each reading is taken when the output has settled to a steady value. The input-output relations developed in this way comprise what is known as a static calibration. The characteristics for an instrument with ideal static calibration, is shown in Figure 1-3. Because of instrument errors, the actual static calibration of an instrument will deviate from the expected or ideal input-output relationship Output y (%) y MAX (100 %) Output span y MIN (0 %) xmin (0 %) Input span x MAX (100 %) Input x (%) Figure 1-3 (Ideal static calibration curve) As different instruments measure different variables, the input and output values may sometimes be conveniently expressed in percentage values. Of course the ideal inputoutput characteristic does not necessarily have to be a straight line but most instruments are designed to produce a linear input-output relationship. The following concepts; error of measurement, accuracy, precision, repeatability, reproducibility, resolution and sensitivity, are associated with the static characteristics of an instrument, and will subsequently be defined. Error of measurement The error of measurement is the difference between the measured value and the true value. Note: The value of the measurement error can only be evaluated when the instrument is used to measure a standard value as the true value. Accuracy The accuracy of a measurement is the closeness with which the reading approaches the true value of a variable.

7 Precision EIPINI Chapter 1: Introduction to Industrial Instrumentation Page 1-6 Precision is the closeness with which repeated measurements of the same quantity agree with each other. Students often confuse the terms precision and accuracy but a precise instrument may not be accurate. Precision simply means that if the measuring device is subjected to the same input for several times and the indicated results are tightly grouped together around some mean value (though not necessarily the true value), then the instrument is said to be of high precision. See Figure 1-4 for an interpretation of accuracy and precision. Accurate and precise Inaccurate but precise Accurate on average but imprecise Inaccurate and imprecise Figure 1-4 Two concepts related to precision, are repeatability and reproducibility. Repeatability is basically a measure of the instrument precision when the same operator in the same laboratory or the same environment, measures a constant input repeatedly, over a short time. Reproducibility is a measure of the instrument precision when a constant input is measured repeatedly, but these experiments are performed in different laboratories or locations with different ambient conditions and over a longer time span. Repeatability Repeatability is the closeness of the instrument readings when the same input is applied repeatedly under the same conditions over a short period of time. Reproducibility Reproducibility is the closeness of the instrument readings when the same input is applied under different conditions over a long period of time. Resolution Resolution is the smallest variation in the measured variable that can still be measured. For example, the resolution of an ordinary digital wristwatch is normally 1 second as it can measure the flow of time to a maximum fineness of 1 second.

8 EIPINI Chapter 1: Introduction to Industrial Instrumentation Page 1-7 Sensitivity Sensitivity is the rate of change of the output of a system with respect to input changes. For a linear calibration curve, the sensitivity or gain K of an instrument is constant but will vary for a non-linear curve. The sensitivity at any particular input x, may be expressed as the slope of the line tangent to the calibration curve at that point. Δ y K =. Equation 1-1 Δx Example 1-1: What is the sensitivity of a linear instrument that records the following values? 0 ºC = 1.3 V and 45 ºC = 4.3 V Answer: From Equation 1-1: Δ y K = = = 1 Δx = volt per ºC 1.6 INSTRUMENT ERRORS Classification of errors No measuring instrument is entirely free from errors. We can broadly classify instrument errors into three main groups; gross errors, systematic (bias) errors and random (precision) errors. Gross errors: Gross errors are mistakes made, for instance, by the operator in gross misreading of a scale. These errors can be minimized by care and self-discipline. Systematic errors: Systematic errors affect all readings in such a way that the error of measurement has a fixed sign throughout the whole range of the instrument. These errors are usually caused by an error in the instrument, poor calibration, improper technique of the operator or loading of the instrument. Normally systematic errors are corrected by careful recalibration of the instrument. Random errors: Random errors occur because of unknown and unpredictable variations that exist in all measurement situations. This results in slightly different values obtained for each repeated measurement (scattered evenly about the mean value) of the same input. The influence of random errors on the integrity of measurements can be reduced with statistical methods and refined experimental techniques.

9 EIPINI Chapter 1: Introduction to Industrial Instrumentation Page Typical instrument errors Some errors that may be encountered while using an instrument, are errors because of non-linearity, drift, hysteresis and dead band. Non linearity Non-linearity is the maximum deviation from a straight line connecting the zero and full-scale calibration points. Note: A straight line connecting the minimum and maximum inputoutput operating points, would represent perfect linear operation of the instrument. The actual static calibration of the instrument will normally deviate from this line. Non- linearity can be expressed in a variety of ways but a widely used method is to determine the maximum deviation of the output from this line, as shown in Figure 1-5. Non-linearity is then expressed as a percentage of the maximum output value. Output y y MAX y MIN x MIN Desired linear inputoutput relationship x MAX Figure 1-5 Actual static calibration Maximum non-linearity Input x Drift Drift is the change in instrument indication over time while the input and ambient conditions are constant. Note: A typical error because of drift is a change in sensitivity. This will cause an error across the whole range of the instrument as indicated in Figure 1-6. An error because of drift is an example of a systematic error. As was mentioned before, systematic errors may normally be corrected with routine calibration of the instrument. Output y y MAX y MIN x MIN x MAX Range error Input x Figure 1-6

10 EIPINI Chapter 1: Introduction to Industrial Instrumentation Page 1-9 Hysteresis Hysteresis is the difference between the readings obtained when a given value of the measured variable is approached from below and when the same value is approached from above. Note: It is possible to find that when performing a static calibration for an instrument starting from the minimum input value to the maximum input value (also called the upscale direction), that the calibration curve obtained in this way, may differ from the static calibration obtained when the input variable is allowed to vary from the maximum value down to the minimum value (also called the downscale direction). This phenomenon, illustrated in Figure 1-7, is called hysteresis. This is usually caused by friction or backlash in the gearing of the instrument. Output y y MAX y MIN x MIN Hysteresis error for input x 0 x 0 x MAX Figure 1-7 Downscale static calibration Upscale static calibration Input x Dead band Dead band is the largest change of input to which the instrument does not respond due to friction or backlash effects Note: Dead band error is normally associated with hysteresis. Dead band operation is sometimes intentionally built into the instrument for instance in a room temperature regulator to prevent excessive on-off switching. As an example of dead band behaviour in an instrument, Figure 1-8 illustrates instrument insensitivity near zero input, typically because of friction. Output y y MAX Insensitivity near zero input y MIN x MIN Figure 1-8 x MAX Input x

11 EIPINI Chapter 1: Introduction to Industrial Instrumentation Page 1-10 Example 1-3 A displacement sensor has an input range of 0.0 to 3.0 cm. Using the calibration results given in the table, calculate: a) the input and output span. b) the maximum non-linearity as a percentage of f.s.d. (full scale deflection). c) the sensitivity of the instrument at an input of 1.0 cm. Displacement (cm) Output Voltage (mv) a) Input span = 3-0 = 3 cm and output span = 58-0 = 58 mv b) y - output voltage in mv Maximum nonlinearity = 15 mv x displacement in cm The maximum deviation from the straight line connecting the range values appears to occur when the displacement is 1.7 cm. The non-linearity at this point is approximately = 15 mv. Non-linearity expressed as percentage of full scale is (15/58) %. c) Sensitivity at x = 1 cm, is equal to the slope of the line tangent to the curve Δ y at x = 1 cm. K = = 60 4 = 9.5 mv/cm Δx 1.9 0

12 EIPINI Chapter 1: Introduction to Industrial Instrumentation Page INDUSTRIAL INSTRUMENTATION STANDARDS AND SCHEMATICS Instrument identification lettering Letter First letter A Analysis Alarm Second / third letter B Burner or combustion User s choice* C User s choice Control D User s choice E Voltage Primary element F Flow rate G User's choice Glass (sight tube) H Hand (manually initiated) High I Current Indicate J Power K Time schedule Control station L Level Light / Low M Moisture or humidity Middle N User s choice User s choice O User s choice Orifice, restriction P Pressure or vacuum Point (test connection) Q Quantity R Radiation Record or print S Speed or frequency Switch T Temperature Transmit U Multivariable Multifunction V Vibration or viscosity Valve, damper or louver W Weight or force Well X Unclassified Unclassified Y Event, state, or presence Relay or compute Z Position, dimension Driver, actuator, final control Table 1-1

13 EIPINI Chapter 1: Introduction to Industrial Instrumentation Page 1-1 * The user s choice entry in the table may be used to denote a particular meaning, and the user must describe the particular meaning(s) in the legend accompanying his drawing. The letter Y in the second position has an extended meaning of variable manipulation, and some of this instrument functions are given in table 1-. Symbol Function Σ Δ Add, subtract, multiply and divide X n ± Raise to power, square root, bias K -K Proportional reverse proportional > < > < High select, low select, high limit, low limit D or d/dt Integral, derivative Convert X to Y with X and Y selected from: X/Y P=Pressure, E=Voltage, I=Current, H=Hydraulic O=Electromagnetic or sonic, A=Analog, D=Digital Table Instrument signals and connections Primary process flow Instrument supply or connection to process Pneumatic signal Electrical signal Hydraulic signal Electromagnetic, sonic or radioactive signal Standard methods to transmit pneumatic and electrical signals The standard industrial range for pneumatic signals is 0 to 100 kpa above atmospheric pressure, which corresponds to a 0% to 100% process condition. Note that the transmitter output signal starts at 0 kpa and not 0 kpa. This 0 kpa output is called a live zero. A live zero allows control room staff to distinguish between a valid process condition of 0% (a 0 kpa reading) and a disabled transmitter or interrupted pressure line (a 0 kpa reading) providing a rough rationality check. The accepted industrial electronic standard is a 4 ma to 0 ma current signal or a 1 V to 5 V voltage signal to represent a 0 % to 100 % process condition. Again, a live zero is used to distinguish between 0% process variable (4 ma or 1 V) and an interrupted or faulted signal loop (0 ma or 0 V).

14 EIPINI Chapter 1: Introduction to Industrial Instrumentation Page 1-13 Process Pneumatic transmitter 100kPa output Process Electronic transmitter 0mA (5V) output 100% 100% 75% 80kPa 75% 16mA (4V) 50% 60kPa 50% 1mA (3V) 5% 40kPa 5% 8mA (V) 0% 0kPa 0% 4mA (1V) Example 1-4 A kpa output pneumatic transmitter is used to monitor the water level inside a tank. The calibrated range is 100 to 00 cm. of water above the base of the tank. Calculate the output of the transmitter when the water level is at 175 cm. above the base of the tank. Span (difference between the upper and lower limit) of the transmitter output = 100 kpa - 0 kpa = 80 kpa Fraction of measurement = ( )/(00 100) = 0.75 Output Signal = (Fraction of Measurement) (Signal Span) + Live Zero = = 80 kpa Example 1-5 An electronic transmitter with an output of 4-0 ma is calibrated for a pressure range of kpa. What pressure is represented by a 1 ma signal? Span of transmitter = 0 ma - 4 ma = 16 ma Fraction of Measurement Change = (Output Signal - Live Zero)/Signal Span = (1 4)/16 = 0.5 Actual Process Change = (Fractional Change) (Process Span) = 0.5 ( kpa) = 40 kpa Actual Process Value = Base Point + Process Change = kpa = 110 kpa. Note: One advantage of a pneumatic system is that sparks will not be produced if a transmitter malfunction occurs, making it much safer when used in an explosive environment. The biggest problem with pneumatic systems is that air is compressible. This means that a pressure transient representing a process change will only travel in the air line at sonic velocity (approximately 300 m/sec.). Long signal lines will cause substantial time delays, which is a serious drawback. Electronic signals on the other hand, travel at speeds which approach the speed of light and can therefore be transmitted over long distances without the introduction of unnecessary time delays Power supply abbreviations AS Air supply ES Electric supply GS Gas supply WS Water supply SS Steam supply HS Hydraulic supply NS Nitrogen supply

15 EIPINI Chapter 1: Introduction to Industrial Instrumentation Page Instrument symbols Instrument mounted locally (field mounted) Instrument mounted behind board (mounted behind panel in control room, not accessible to operator) Instrument mounted on board (panel mounted in control room) Instruments sharing common housing (measures two variables or single variable with two functions) Valve Valve with diaphragm actuator Valve with hand actuator Butterfly valve Orifice plate flowmeter Venturi flowmeter Rotameter flowmeter M Electric motor

16 EIPINI Chapter 1: Introduction to Industrial Instrumentation Page Schematics The key to instrument identification, is given in Figure Figure Component function (table 1-1) Component sequence number 3 Instrument function (table 1-) 4 Vendor designation 5 Panel number 6 Set point(s) 7 Application notes Example 1-6 Identify the following instruments: a) Answer: Temperature (1 st letter) recording ( nd letter) TRC controller (3 rd letter also from second column table 1-1) mounted on board b) FY Answer: Flow compute instrument mounted behind board (or rack mounted). The instrument receives a pneumatic signal and converts this signal into a pneumatic output signal representing the square root of the input signal. Exercise Identify the instrumentation blocks in the heat exchanger below Product stream to be heated PR 1a PIT 1b FIT FY FR a b c Steam supply PIC 3 TRC 4a TSH 4b M TAH 4c

17 EIPINI Chapter : Pressure Measurement Page -1. PRESSURE MEASUREMENT The purpose of this chapter is to introduce students to the definitions and units of pressure related quantities and to discuss typical methods to measure pressure..1 PRESSURE CONCEPTS AND DEFINITIONS.1.1 Pressure Pressure is defined as the force exerted over a unit area. The SI unit is newton per square meter (N/m ) or pascal (Pa). F P = Equation -1 A Weight 100 N Same force, different area different pressure Area Area 0.1 m P = 1000 Pa 0.01 m P = Pa.1. Density Density of a substance is defined as the mass of a unit volume of a substance. The SI unit is kilogram per cubic meter (kg/m 3 ). M ρ = Equation - V ρ water = 1000 kg/m 3 ρ mercury = kg/m 3 ρ transformer oil = 864 kg/m 3 ρ air = 1. kg/m Relative density (Specific gravity) Relative density of a substance is defined as the ratio of the density of the substance to the density of water. δ substance = ρ substance ρ water ρ substance = 1000 δ substance Equation -3 (Note: If the substance is a gas, the relative density is defined as the ratio of the density of the gas to the density of air at the same temperature, pressure and dryness.)

18 EIPINI Chapter : Pressure Measurement Page Absolute zero of pressure The absolute zero of pressure (or perfect vacuum), is the pressure that would exist in a chamber, if all molecules were removed from the chamber, so that no pressure forces could be exerted on the chamber walls..1.5 Absolute pressure Absolute pressure (or total pressure), is the pressure measured from absolute zero pressure..1.6 Atmospheric pressure Atmospheric pressure is the absolute pressure caused by the weight of the earth s atmosphere. (Notes: Atmospheric pressure is often called barometric pressure. Local atmospheric pressure depends on the height above sea level. Standard atmospheric pressure at sea level is kpa. or 760 mm. mercury..1.7 Gauge pressure Gauge pressure, is the difference between the absolute pressure in a medium and local atmospheric pressure, when the pressure in the medium is higher than atmospheric pressure. P gauge = P abs P atm.1.8 Vacuum pressure Vacuum pressure, is the difference between local atmospheric pressure and the absolute pressure in a medium, when the pressure in the medium is lower than atmospheric pressure. P vacuum = P atm P abs.1.9 Differential pressure Differential pressure is the difference between two pressures. Summary: A comparison of absolute pressure, atmospheric pressure, gauge pressure, vacuum pressure and differential pressure. Atmospheric pressure P atm P gauge P vacuum Differential pressure P abs Absolute pressure Absolute zero pressure (0 Pa)

19 EIPINI Chapter : Pressure Measurement Page -3. PRESSURE IN A LIQUID The cylinder in Figure -1, with a cross sectional area A meter, is filled with a liquid of density ρ kilogram/meter 3, to a height of h meter. The weight of the liquid will exert a pressure P pascal on the bottom of the container. We will now obtain an expression for P. P A h Figure -1 Volume of the liquid = V = cross-sectional area height = Ah. Mass of the liquid = m = volume density = V ρ = (Ah) ρ. Weight of the liquid = w = mg = (Ahρ) g. Pressure on the bottom of container due to weight of the liquid = w A = Ahρg/A = ρhg We conclude therefore that the pressure (pascal) caused by a liquid column h meter high and with density ρ kilogram/meter 3, is given by: P = ρhg Equation -4 where g is the gravitational acceleration. We will always use g = 9.81 m/s in pressure calculations. Note: If the absolute atmospheric pressure, exerted on the surface of the liquid, is P 0 pascal, the total pressure acting on the bottom of the container is P total = P 0 + ρhg Example -1 a) Convert a pressure of 150 cm. water, to a pressure expressed in pascal. P = ρhg = 1000 ( ) 9.81 = Pa. b) Convert a pressure of 10 kilo pascal to a pressure expressed as meter water. P = ρhg = 1000 h 9.81 h = meter Therefore 10 kpa = meter H O. c) Convert a pressure of 760 millimeter mercury to a pressure expressed in pascal. P = ρhg = ( ) 9.81 = Pa = kpa. d) Convert a pressure of 50 kpa to a pressure expressed as millimeter mercury. P = ρhg = h 9.81 h = meter Therefore 50 kpa = mm Hg.

20 EIPINI Chapter : Pressure Measurement Page -4.3 PRESSURE MEASUREMENT WITH MANOMETERS.3.1 The U tube manometer A simple U tube manometer is formed, when a glass tube, in the form of a U, is half filled with a liquid (for example mercury), as shown in Figure -. If the pressures in both legs of the manometer are the same, for instance, if both legs are open to atmospheric pressure, the manometer liquid level will lie in the same horizontal plane. This is called the zero level or zero line. P atm P atm Zero level Example - A U tube manometer is half filled with mercury. A pressure of 00 kpa is applied to the left hand leg and a pressure of 100 kpa is applied to the right hand leg. Calculate the reading h on the manometer. It is important to remember that mercury cannot be compressed by typical pressures. Therefore, if a Zero level pressure differential causes a movement of the h mercury away from the zero level, the downward movement in the one leg, will be equal to the upward movement in the other leg. Secondly, the density of air is very small in comparison with the density of the manometer liquid. The pressure contribution of the air in the tubes may therefore be neglected. Thirdly, when we compare the pressures in the two legs of the manometer, we need to remember the important theorem from hydrostatics that states: X Y The pressure at two points, in the same horizontal plane, in a liquid at rest, is the same, if a curve can be drawn from the one point to the other point, without leaving the liquid. It is now clear that we can equate the pressures in the XY plane, as this plane cuts the mercury in the same horizontal plane, and the points of intersection, may be joined via the mercury = h 9.81 h = m 00 kpa Figure kpa If the reading were taken from the zero line upwards, it would be /= m

21 EIPINI Chapter : Pressure Measurement Page Using a U tube manometer to measure differential pressure, gauge pressure and absolute pressure Differential pressure: To measure the difference between two unknown pressures, the one pressure is applied to one leg and the other to the second leg, as shown in Figure -3 (a). The reading h is directly proportional to the pressure difference P 1 -P. Comparing pressures in the XY plane: P 1 = P + ρhg X P 1 h P Zero level Y P 1 -P = ρhg Figure -3 (a) Gauge pressure. The arrangement to measure gauge pressure, is shown in Figure -3 (b). A pressure P 1, larger than atmospheric pressure, is applied to one leg, and atmospheric pressure to the other. The reading h will be indicative of the pressure difference P 1 P atm or the gauge pressure. P 1 h P atm Zero level Comparing pressures in the XY plane: P 1 = P atm + ρhg P 1 P atm = ρhg X Y P gauge = ρhg Figure -3 (b) Absolute pressure. In order to measure absolute pressure, it is necessary to compare the unknown pressure with zero pascal, as shown in Figure -3 (c). For that purpose, all the air must be removed from one leg, to form a perfect vacuum. That leg is then sealed. The two mercury levels will take on their zero line position, only if zero pascal is applied to the open leg. Equating pressures in the XY plane: P abs = 0 + ρhg P abs = ρhg X P abs h Vacuum (0 Pa) Zero level Y Figure -3 (c)

22 Example -3 A u-tube manometer is filled with two liquids, one liquid with a relative density of 1 and the other with a relative density of Calculate the pressure difference, P1 P, applied across the manometer. Comparing pressures on the XY line: P =P P = P P 1 P = = kpa. EIPINI Chapter : Pressure Measurement Page -6 Example -4 δ=13.6 You are requested to design a scale plate for a U-tube manometer that uses zeal oil, with relative density of 0.88, as manometer liquid. You are told that the maximum differential pressure to be measured, will be 10 kpa. From the zero line upward, the following values must be marked off on the scale plate:.5 kpa, 5 kpa, 7.5 kpa and 10 kpa. Calculate the distances between the markings, and sketch the designed plate. P 1 P = ρhg, so for P 1 -P =10 kpa: =880 h 9.81 h = 1158 mm. 579mm 434.5mm 89.5mm mm Zero line 10 kpa 7.5 kpa 5 kpa.5 kpa 0 kpa Distance from zero line to 10 kpa marking=579 mm. Intervals= mm. Example -5 The distance from the zero level to the top of a 100 kpa 00 kpa mercury manometer is 1 meter, when both tubes P x are open to an atmospheric pressure of 100 kpa. 1-(h/) 1 m The right hand tube is now sealed off and a pressure of 00 kpa is applied to the left hand tube. Calculate the manometer reading h. h When 00 kpa is applied to the left hand tube, the pressure in the sealed tube, will rise to a new, higher than 100 kpa, pressure which we will call P x. If the cross sectional area of the X Y tube is A, we may use Boyle s law to obtain an expression for P x. The volume of the air in the right hand tube is 1 A when open to 100 kpa and sealed, with 00 kpa applied to the left hand tube, it is (1-h/) A. Using Boyle s law, P 1 V 1 =P V : [1 A] = P x [(1-h/) A] P x = /[1-h/]. (1) Comparing pressures on the XY line: = P x h () (1) in (): = /[1-h/] h = 100/[1-h/] h 00 = 100/[1-h/] h 00 (1-h/) = h (1-h/) h= h 66.71h 66.71h 33.4h = 0 {ax +bx+c=0 x=[-b± (b -4ac)]/a} h = [33.4± ( )]/ = 0.5 m or h=3m (unacceptable) 1 m P 1 P X δ=1 Y 0.5 m h/

23 EIPINI Chapter : Pressure Measurement Page The well type manometer (cistern type manometer) The well type manometer, is essentially a manometer with one of the limbs (the well or reservoir) having a large cross sectional area of A 1, and the second limb, a glass tube, with much smaller cross sectional area, of A. Cross sectional Area of tube = A Low P High P 1 h Zero level Cross sectional Area of well = A 1 X d A 1 A Manometer liquid density = ρ Y Figure -4 (a) Figure -4 (b) When the two limbs are open, as shown in Figure -4 (a), the manometer liquid meniscuses will fall on the zero line. If a pressure differential, P 1 - P (P 1 > P ), is applied to the instrument, in Figure -4 (b), the rise and fall of the manometer liquid in the two limbs will be different (h > d). The level h, in the glass tube, to which the manometer liquid rises above the zero line, can be measured, while the fall in the liquid level d, in the well, can not be observed, and as such, will be eliminated from our equations below. Comparing the pressures in the two limbs, on level XY, in figure -4 (b): P 1 = P + ρ(h+d)g (1) Also, the volume of manometer liquid, leaving the well, is equal to the volume of manometer liquid, entering the tube: A 1 d = A h d = () in (1): P 1 = P + ρ A h.. () A 1 A h + h A 1 g A P 1 P = ρhg 1 + Equation -5 A 1

24 EIPINI Chapter : Pressure Measurement Page The inclined limb manometer The inclined manometer is a variation of the well type in that the tube is not vertical, but leaning to one side. Referring to Figure -5, the movement L, of the manometer liquid along the tube, is amplified with respect to its vertical height h. This facilitates the detection of small changes in applied differential pressure. P Low P 1 High L h Zero level α X d Y Cross sectional Area of well = A 1 Cross sectional Area of tube = A Figure -5 Deriving the relationship between the applied pressure differential P 1 -P, and the manometer reading L, is very similar to that of the well type. The only difference is that the tube and horizontal does not form an angle of 90, but an angle α. Comparing the pressures on level XY, in figure -5: P 1 = P + ρ(h+d)g (1) Equating rise and fall of manometer liquid: A 1 d = A L A d = L.. () A 1 And from figure -5: h = Lsinα. (3) A () and (3) in (1): P 1 = P + ρ Lsinα + L g A 1 A P 1 P = ρlg sin α + Equation -6 A 1

25 EIPINI Chapter : Pressure Measurement Page -9 Example -6 An inclined limb manometer is used for the measurement of pressure. The inclined limb forms an angle of 30 degrees with the horizontal plane. The relative density of the manometer fluid is 0.8. The internal diameter of the well is 3 cm and the internal diameter of the inclined limb is 1 mm. a) Calculate the maximum applied pressure (in pascal), for a maximum scale reading (L) of 100 cm on the scale attached to the inclined limb. b) The range of the above inclined manometer must be extended so that the maximum pressure that can be applied to the manometer is increased by 1000 pascal, by using a different manometer fluid, without changing the construction of the manometer. Calculate the relative density of the manometer fluid that is required. π(3 10 ) π( ) = = A /A 1 = / = 0.16 {or A /A 1 =(D /D 1 ) =(1/30) =0.16} From equation -6: P 1 P = ρlg(sinα+a /A 1 ) = [sin ] = 7848 ( ) = = 5180 Pa. b) (P 1 P ) new = = 6180 Pa. P 1 P = ρlg(sinα+a /A 1 ) 6180 = ρ ρ = 6180/6.475 = kg/m 3 δ new = πd a) A 1 = 1 4 Example -7 = πd and A = = 4 The reading h on a well type mercury manometer, is 73 cm when measuring a pressure of 100 kpa. a) Calculate the ratio of well diameter to the diameter of the tube. b) Determine the change in level that the well mercury experiences. a) From equation -5: A P 1 P = ρhg 1 + A = ( ) 9.81 [1 + (A /A 1 )] 1 + (A /A 1 ) = /[13600 ( ) 9.81] = 1.07 (A /A 1 ) = 0.07 (D /D 1 ) = 0.07 D /D 1 = D 1 /D = (ratio of well to tube diameter) A b) d = h A1 d = = mm.

26 .3.5 Liquids used in manometers EIPINI Chapter : Pressure Measurement Page Transformer oil Relative density: Applications: Useful when measuring small pressure differences. Suitable for pressure measurement in ammonia gas installations. Advantages: Low density for measuring small pressure differences. Unaffected by ammonia. Can be easily seen. Does not readily evaporate. Disadvantages: Tends to cling to inside of tubes. Density of transformer oil varies Aniline Relative density: 1.05 Applications: Suitable for pressure measurement in low pressure gas or air installations, with the exception of ammonia and chlorine. Advantages: Low density for measuring small pressure differences. Evaporates slowly. Does not mix with water. Can be easily seen. Disadvantages: Attacks paint. Poisonous, penetrates the skin and causes blood poisoning. Aniline darkens on contact with air Dibutylphathalate Relative density: Applications: Suitable for pressure measurement in ammonia gas installations. Advantages: Does not mix with water. Disadvantages: Carbon Tetrachloride Relative density: Applications: Useful when measuring higher pressure differences. Suitable for measuring pressure in chlorine gas installations. Advantages: Not attacked by chlorine. Disadvantages: Not easily seen. Readily evaporates Tetrabromoethane Relative density:.964 Applications: Useful when measuring higher pressure differences. Suitable for pressure measurement in ammonia gas installations. Advantages: Evaporates slowly. High density. Disadvantages: Bromoform Relative density:.9 Applications: Useful where pressure measurement demands manometer liquid with density between water and mercury. Advantages: Density that falls between water and mercury. Disadvantages: Density uncertain. Poisonous. Freezes easily. Subject to attack. Attacks rubber Mercury Relative density: 13.6 Applications: Pressure measurements in compressed gas, and in water and steam applications. Advantages: High density. Can be easily seen. Mercury does not: i) evaporate, ii) mix with other liquids, iii) wet sides of tubes. Disadvantages: Expensive. Mobility and density are affected by contamination.

27 EIPINI Chapter : Pressure Measurement Page ELASTIC PRESSURE SENSORS.4.1 The C type bourdon tube gauge Bourdon tube pressure gauges are usually used where relatively large static pressures are to be measured. A typical bourdon tube pressure gauge is shown in Figure -6. The Bourdon tube pressure gauge consists of a C-shaped tube with one end sealed. The sealed end is connected by a mechanical link to a pointer on the dial of the gauge. The other end of the tube is fixed and open to the pressure being measured. The inside of the Bourdon tube experiences the measured pressure, while the outside of the tube is exposed to atmospheric pressure. Therefore, the tube responds to changes in P measured P atm. Increasing this pressure will tend to straighten out the tube and move the pointer to a higher scale position Pointer and scale Hairspring Bourdon tube Adjustable link Range adjust Pinion gear Pivot point Sector gear Tube cross section Pressure connection Figure -6

28 EIPINI Chapter : Pressure Measurement Page Bellows pressure sensor The bellows element is basically a flexible metallic cylinder with a ripple profile, which can expand when a pressure differential exists between the interior pressure of the bellows and the pressure surrounding the bellows. In Figure -7(a), a bellows pressure sensor is used to measure a differential pressure P 1 P. Bellows element P 1 High pressure P Low pressure Moving end Spring Pressure indication Figure -7 (a) Differential pressure The bellows element may also be used to measure gauge pressure if P is equal to atmospheric pressure, as depicted in Figure -7 (b). Absolute pressure may be measured, see Figure -7 (c), if all air is removed from the bellows enclosure, so that the pressure in the bellows, acts against a vacuum (0 Pa). Atmospheric pressure Vacuum (0 Pa) P 1 P 1 Pressure indication Figure -7 (b) Gauge pressure Pressure indication Figure -7 (c) Absolute pressure

29 EIPINI Chapter : Pressure Measurement Page Diaphragm pressure sensors Diaphragms are round flexible disks, formed from thin metallic sheets with concentric corrugations. Two diaphragms may be used together to form a diaphragm capsule. Figure -8 (a) shows the structure of a single diaphragm while Figure -8 (b) and Figure -8 (c), indicate the design of convex and nested diaphragm capsules, respectively. Figure -8 (a) (single) Figure -8 (b) (Convex) Figure -8 (c) (Nested) Figure -9 (a), shows a diaphragm used to measure a pressure difference, P 1 - P, while in Figure -8 (b), the same function is fulfilled with a diaphragm capsule. Diaphragm P Capsule P P 1 P 1 Pressure indication Figure -9 (a) (Diaphragm) Pressure indication Figure -9 (b) (Capsule) Capsules are sometimes filled with silicone oil and a solid plate mounted in the center of the capsule to protect against over-pressure. Pressure is then applied to both side of the diaphragm (Figure -10) and it will deflect towards the lower pressure. Most pneumatic differential pressure transmitters (discussed in section.6) are built around the pressure capsule concept. Pressure indication P 1 (High pressure) Backup plate Force bar Seal and pivot Silicone oil P (Low pressure) Capsule Figure -10

30 EIPINI Chapter : Pressure Measurement Page FORCE-BALANCE GAUGE CALIBRATOR This instrument is also known as the piston type gauge or the dead weight tester. Its main purpose is to calibrate other pressure gauges. The deadweight tester consists of a pumping piston that screws into the oil filled reservoir, a primary piston that carries the dead weight, and the gauge under test (Figure -11). The primary piston (of cross sectional area A), is loaded with the amount of weight (W) that corresponds to the desired calibration pressure (P = W/A). When the screw is rotated, the pumping piston pressurizes the whole system by pressing more oil into the reservoir cylinder, until the dead weight lifts off its support. The gauge under test is also exposed to the oil pressure that at this stage is equal to the calibration pressure. Mass pieces Gauge under test Oil Platform Primary piston Secondary (pumped) piston Screw Figure -11 Example -8 A dead weight tester has a primary piston with a diameter of 1.5 cm. The mass of the platform and primary piston together, is 300 gram. Calculate the mass m, of the mass pieces, that must be placed on the platform to check a gauge at 150 kpa. Pressure = Weight of masspieces + weight of platform and primary piston Area of primary piston m ( ) 9.81 = ( ) π 4 ( ) ( ) = 9.81m m = m = 3.57 m =.403 kg. The total mass of the mass pieces to be placed on the platform is therefore.403 kg.

31 EIPINI Chapter : Pressure Measurement Page The pneumatic differential pressure transmitter (DP cell) The purpose of this instrument is to measure a differential pressure P high P low, and convert the measured value into a standard output pressure that varies between 0 kpa and 100 kpa. The measured value may then be transmitted as a pressure variable, to a station some distance away. A simplified schematic of a pneumatic pressure transmitter is given in Figure -1. Restriction Regulated air supply Pilot relay Flapper Nozzle Cross flexure Feedback bellows Output pressure P 0 Zero adjustment (0 kpa) Liquid filled diaphragm capsule A L 1 L B Pivot point (range wheel adjust) Range bar Force bar Pivot and seal Capsule flexure Low pressure (P ) Figure -1 High pressure (P 1 ) The operation of the differential transmitter is governed by the flapper and nozzle feedback mechanism, which keeps the range bar, pivoted by the range wheel, in balance. The upper part of the range bar and force bar is connected by a flexible plate. When the input pressure differential, P 1 -P, increases, the force bar will pivot in a clockwise direction, and that will in turn cause the range bar to pivot clockwise. The flapper will therefore move towards the nozzle and airflow from the supply, through the nozzle, will consequently be reduced (blocked by flapper). This will result in a lower pressure drop across the restriction in the supply line and thus a higher pressure will be presented to the feedback bellows via the pilot relay that serves as a pneumatic buffer amplifier between the nozzle and feedback bellows (for clarity, a direct connection via the pilot relay, is shown in Figure. -1).

32 EIPINI Chapter : Pressure Measurement Page -16 The feedback bellows will now push the range bar in an anti-clockwise direction, thereby restoring balance of both the range bar and the force bar, but at a higher output pressure value of P 0, indicative of an increased value of P 1 -P. Similarly, when P 1 -P decreases, the flapper will be pushed away from the nozzle, thereby increasing the airflow through the nozzle resulting in a higher pressure drop across the restriction and a lower pressure transmitted to the feedback bellows. Balance will thus be restored, but at a lower value of P 0. The zero adjustment represents a pressure of 0 kpa in opposition to P 0, so that when the pressure differential, P 1 P, is zero, the output must still be 0 kpa. To simplify the discussion, let us assume that the effective clockwise moment at point A is (P 1 - P )L 1 while the anti-clockwise moment at point B is (P 0 0)L. Equating these moments around the range wheel: (P 1 P )L 1 = (P 0-0)L L 1 P 0 = (P 1 P ) + 0 kilopascal....(1) L The ratio L 1 /L is adjusted during calibration, by changing the position of the range wheel, to ensure that P 0 equals 100 kpa when (P 1 -P ) reaches it s maximum value. Setting this ratio equal to m, we can rewrite equation (1) as: P 0 = m (P 1 P ) + 0 kilopascal Equation -7 In Equation -7, the variables P 0, P 1 and P, must be expressed in kilopascal. A graphical representation of Equation -7 is given in Figure -13 Output P 0 [kpa] 100 P 0 = m (P 1 P ) + 0 where m = 80/(P 1 -P ) MAX 80 0 (P 1 -P ) MAX Figure Input (P 1 -P ) (P 1 -P ) MAX [kpa] Example -9 A differential pressure transmitter is correctly calibrated for a process variable that varies from 0 kpa to 170 kpa. Determine the output of the DP transmitter when the process variable reaches 90 kpa.

33 EIPINI Chapter : Pressure Measurement Page -17 From Equation -7, the output of the transmitter is given by: P 0 = m (P 1 P ) + 0 When (P 1 -P ) = 170 kpa, the output is P 0 = = m m = P 0 = (P 1 P ) + 0 kilopascal If (P 1 -P ) = 90 kpa.: P 0 = = 6.35 kpa. The Pilot Relay In Figure -1, the pressure developed by the nozzle, is enhanced by a pilot relay. Theoretically, without a pilot relay, as shown in Figure -14, the restriction, flapper, nozzle and feedback bellows mechanism, would still function properly and respond to the force applied to the flapper, with an output pressure related to the force. The practical problem however, is that an increase in output pressure, must be accompanied by an increase in air flow through the very narrow restriction, while a decrease in pressure, must be accompanied by air bleeding away through the small nozzle opening. The response of the output pressure to changes in flapper movement, will inevitably be slow. Air supply Restriction Output Flapper and nozzle F Pivot Figure -14 Feedback bellows The pilot relay alleviates this problem by allowing the nozzle pressure to operate a small diaphragm which in turn controls the output pressure of the pilot relay in such a way, that it will follow the nozzle pressure, but this time, the output pressure is derived directly from the more powerful air supply. In Figure -15, the arrangement of the flapper and nozzle, assisted by a pilot relay, is shown. Air supply Spring Supply valve (ball) Valve stem Vent Restriction Output Exhaust valve (cone) Diaphragm Flapper and nozzle F Pivot Figure -15 Feedback bellows When the force F moves the flapper towards the nozzle, the airflow through the nozzle will be reduced, thereby causing a smaller pressure drop across the restriction, so that more of the supply pressure will arrive at the diaphragm

34 EIPINI Chapter : Pressure Measurement Page -18 chamber of the pilot relay, pushing the valve stem to the left. Moving the valve stem to the left, will have a dual effect. Firstly the supply valve will allow more of the air supply to reach the output (increasing the output pressure), and secondly, the exhaust valve will close a bit more, making it more difficult for the newly established higher output pressure, to relax itself through the vent. Balance will again be restored by the higher pressure in the feedback bellows, that opposes the disturbing force. Similarly, when the external force pulls the flapper away from the nozzle, the air flow through the nozzle will increase. The increased air flow will cause more of the available supply pressure to fall across the restriction, making less pressure available on the nozzle side of the restriction. The diaphragm will slacken, as it is now exposed to a lower pressure and the valve stem will move to the right. The supply valve will begin to close, thereby restricting the flow of air from the supply to the output (thereby decreasing the output pressure) and at the same time, the exhaust valve will open more, thus providing a wider escape route for the original high output pressure, facilitating in this way with the rapid change in output pressure from a higher value to a lower value. As always, the feedback bellows, now receiving a lower pressure, will oppose the external force and bring the flapper back into balance. The flapper/nozzle, pilot relay arrangement, is an important pneumatic mechanism and is also used in other instruments, such as the pneumatic control valve discussed in Chapter 6, in addition to the differential pressure transmitter..7 Strain gauges Many pressure instruments such as an electronic differential pressure transmitter, may need to develop a standard electrical signal of 4 to 0 ma or 1 to 5 V, instead of the standard 0 to 100 kpa pressure signal. The strain gauge is one of the devices used to convert a pressure or force into an electrical signal. The majority of strain gauges are foil types, shown in Figure -16. They consist of a pattern of resistive foil which is mounted on a backing material and operate on the principle that as the foil is subjected to stress, the resistance of the foil changes in a defined way. Alignment marks Backing material Grid Figure -16 Solder tabs

35 EIPINI Chapter : Pressure Measurement Page -19 For a metal wire, the electrical resistance is given by: R = ρ a l, where R is the resistance of the wire (Ω), ρ is the metal s resistivity (Ω-m), l the length of the wire (m) and a the cross sectional area of the wire (m ). The resistance will increase with increasing length of the wire or as the cross sectional area decreases. When force is applied, as indicated in Figure -17, the overall length of the wire tends to increase while the cross-sectional area decreases. This increase in resistance is proportional to the force that produced the change in length and area. The gauge factor (GF) of the strain gauge is defined as: GF = ΔR/R, Δl/l where ΔR is the change in resistance, corresponding to a change in length, Δl. Wire without tension Force Cross sectional area decreases Length increases Figure -17 Force Wire under tension (stress) The fractional change in length Δl/l is called the strain ε, so that the gauge factor may be expressed as: GF = ΔR/R, Equation -8 ε where ε = Δ l l. Equation -9 The value of GF for a metallic strain gauge is. The strain gauge pattern can be bonded to the surface of a pressure capsule or embedded inside the capsule. The change in the process pressure will cause a resistive change in the strain gauge, which can be used to produce a 4 to 0 ma or 1 to 5 V signal. To facilitate with converting a change in resistance to a corresponding voltage change, a Wheatstone bridge, shown in Figure -18, is used. The Wheatstone bridge is excited with a stabilised DC supply and the bridge can be zeroed at the null point of measurement. As stress is applied to the bonded strain gauge, a resistive change takes place and unbalances the Wheatstone bridge. This results in a signal output, related to the stress value.

36 EIPINI Chapter : Pressure Measurement Page -0 Gauge in tension (R + ΔR) F E R 1 R V 0 + R 3 R 4 Strain gauge Figure -18 Using the voltage division rule, the output voltage of the bridge is easily obtained as: V 0 = R R + R 4 R R R 3 E. From this equation it is apparent that when = (which implies R R R R R that R 1 R 4 = R R 3 or R 4 R = R ), the voltage output V0 will be zero. Under these 3 1 conditions, the bridge is said to be balanced. Any change in resistance in any arm of the bridge will now result in a nonzero output voltage. Therefore, if we replace R 4 in Figure -16 with an active strain gauge, any change in the strain gauge resistance will unbalance the bridge and produce a nonzero output voltage, related to the stress. Let us assume that when the bridge is in balance, the nominal values of the bridge arms are R 1 = R, R = R, R 3 = R and R 4 = R. Now if R 4 is put under tension (stress), R 4 will change its value to R + ΔR and the bridge output will become: V 0 = V 0 = V 0 = (R + ΔR) R + (R + ΔR) R R + R E = (R + ΔR) (R + ΔR) (R + ΔR) Δ 4R R E + ΔR From Equation -8: ΔR = (GF)Rε. E R 4 + R R + + Δ Δ R R 1 E Using this expression for ΔR in the expression for V 0 above: (GF)Rε R(GF) ε V 0 = E = E 4R + (GF)Rε 4R[1+ (GF) ε/] R 3 +

37 EIPINI Chapter : Pressure Measurement Page -1 V 0 = (GF) ε (GF) ε E Equation -10 Equation -10 is the bridge equation for one strain gauge in the bridge or what is known as a quarter bridge. Other structures are possible, such as one active and one dummy strain gauge or two active strain gauges (half-bridge) or four active strain gauges (full bridge). The bridge output voltage is typically very small and additional electronic circuitry is needed to amplify the signal and condition it for a 4 to 0 ma or a 1 to 5 V signal. Example -10 A strain gauge, imbedded in a silicone filled pressure capsule, is used to measure a differential pressure P 1 P. The strain gauge is connected to a quarter Wheatstone bridge arrangement shown in the figure below. Each of the resistors in the three fixed arms, has a resistance of 10 Ω. The strain gauge has a nominal resistance of 10 Ω and the bridge is therefore in balance if the capsule experiences no stress. The gauge factor of the strain gauge is two (GF = ). The pressure cell is put under stress by applying a differential pressure P 1 P = 100 kpa which results in a strain of ε = in the strain gauge. Calculate the amplifier gain required to produce an output of 1 volt from the Wheatstone output voltage V 0, when P 1 - P = 100 kpa. 10 Ω 10 Ω 10 V V Ω Strain gauge P Pressure capsule P 1 A Output From Equation -10, (GF) ε V 0 = E = = V (= 4.88 mv) (GF) ε The amplifier gain A must therefore be 1/( ) = 40.19

38 EIPINI Chapter 3: Flow Measurement Page FLOW MEASUREMENT The purpose of this chapter is to introduce students to the definitions and units of flow related quantities and to discuss typical methods to measure volumetric flow and flowrate. 3.1 VOLUMETRIC FLOW AND FLOWRATE Volumetric flow Volumetric flow is the total volume of a liquid or gas passing a given point over a certain period of time, and is measured in cubic meter (m 3 ). Note: An example of volumetric flow measurement is municipal water meters that measures the total volume of water used by the customer over a month period. Another example is measuring the total volume of petrol at a gas station, when filling up a car s tank. Flowrate Flowrate is the volume of a liquid or gas passing a given point per unit time, and is measured in cubic meter per second (m 3 /s). Note: The flowrate (q) may also be expressed as the product of the velocity (v) of the flow and the cross sectional area (A) of the pipe through which the flow occurs: q = Av Equation 3-1 v A Total volume transported in 1 second = q = Av 3. VISCOSITY v Distance cylinder travels in 1 second Viscosity is a measure of a fluid's resistance to flow and is measured in poiseuille (PI). Note: Not all liquids are the same. Some are thin and flow easily. Others are thick and sticky. Honey or syrup will pour more slowly than water. A liquid's resistance to flow is called its viscosity. Imagine two layers of a liquid at a distance y from each other and with layer area A, as shown in Figure 3-1. If we assume that the bottom plate is the layer of stationary liquid molecules, clinging to the wall of the pipeline, then the force F that we must apply to move the top plate at a constant velocity v relative to the bottom plate, will be indicative of the fluid s flow resistance.

39 EIPINI Chapter 3: Flow Measurement Page 3- F v y v Figure 3-1 The quantity A F, is called the shear stress in the fluid and the ratio y v is called the velocity gradient (or shear rate). For typical liquids (Newtonian liquids), the shear stress is proportional to the velocity gradient and the constant of proportionality is called the viscosity η of the liquid. F/A η = v/y Equation 3- The SI units for viscosity are the poiseuille (PI) or pascal-second (Pa-s) or newtonsecond per square meter (N-s/m ). Another common (cgs) unit used to express viscosity is the poise (1 poise = 0.1 PI). Some examples of viscosity of liquids (at 0 ºC): η air = PI η water = PI η mercury = PI η oil = 1 PI (typical) η honey = 100 PI η peanut butter = 500 PI Notes: i) Pressure has very little effect on viscosity. ii) Viscosity is not related to density. iii) The viscosity of liquids decrease while the viscosity of gasses increase with increasing temperature. iv) The viscosity of a liquid divided by the density of the liquid is called the kinematic viscosity of the liquid. 3.3 STREAMLINED FLOW AND TURBULENT FLOW Streamlined flow In a streamlined flow (also called laminar flow), all the particles in the liquid, flow in the same direction and parallel to the walls of the pipe, and the streamlines are smooth (Fig 3-a). Turbulent flow In a turbulent flow, the particles in the stream, flow axially as well as radially, and the streamlines are in a chaotic pattern of ever changing swirls and eddies (Fig 3-b). Figure 3- (a) Laminar flow Figure 3- (b) Turbulent flow

40 The Reynolds number The Reynolds number for a flowstream is given by: EIPINI Chapter 3: Flow Measurement Page 3-3 R e = Dvρ η Equation 3-3 where D is the pipe diameter (m), v is the flow speed of the fluid (m/s), ρ is the density of the fluid (kg/m 3 ) and η is the viscosity of the fluid (PI). At low Reynolds numbers (generally below R e = 000) the flow is streamlined while at high Reynolds numbers (above R e =3000) the flow becomes fully turbulent. Flow straighteners (straightening vanes) When flow is measured and the flow is not streamlined, errors may arise in the readings obtained. This problem can be prevented by installing flow straighteners or straightening vanes, inside the pipe as shown in Figure 3-3. Flow Flow Figure 3-3 Example 3-1 The average velocity of water at room temperature in a tube of radius 0.1 m is 0. ms -1. Is the flow laminar or turbulent? R e = ( )/0.001 = > 3000 hence turbulent. 3.4 POISEUILLE S LAW The flowrate of a streamlined liquid in a horizontal pipe is given by: πr 4 q = (p 1 p ) 8ηL Equation 3-4 where q is the liquid s flowrate (m 3 /s), R is the radius of the pipe (m), η is the viscosity of the fluid (PI), L is the length of the pipe (m) and p 1 p is the pressure differential across the pipe (Pa). The variables used in Equation 3-4, are illustrated in Figure 3-4, and the typical parabolic velocity profile associated with laminar flow in a pipe, is also shown. R p 1 q p L Figure 3-4

41 EIPINI Chapter 3: Flow Measurement Page 3-4 Example 3-: Calculate the flowrate of water in a pipe with diameter of 0.15 m and length 100 m that discharges into air (p =100 kpa) while the pump at the other end maintains a π(0.075) 4 pressure of 10 kpa. Ans: q = ( ) = m 3 /s. 8 (0.001) ENERGY OF A LIQUID IN MOTION Pressure energy Pressure energy is the energy which a liquid has by virtue of its internal pressure. A body of liquid with volume V meter 3 under pressure p newton/meter, possesses pressure energy equal to pv joule. Pressure energy per unit volume (when V equals 1 cubic meter), equals p joule. Kinetic energy Kinetic energy is the energy a liquid has by virtue of its motion. A body of liquid with mass m kilogram, moving at velocity v meter/second, possesses kinetic energy equal to ½mv joule. If the density of the liquid is ρ kilogram/meter 3, then the kinetic energy per unit volume (m = ρv; if V = 1, then m = ρ), equals ½ρv joule. Potential energy Potential energy is the energy that a liquid has by virtue of its height above a given plane. A body of liquid with mass m kilogram and a height h meter above a reference plane, possesses potential energy equal to mgh, where g is the gravitational acceleration constant. If the density of the liquid is ρ kilogram/meter 3, then the potential energy per unit volume, equals ρgh joule. 3.6 BERNOULLI S LAW If an incompressible fluid is in a streamlined flow with no friction, the sum of the pressure energy, the kinetic energy and potential energy per unit volume, is constant at every point in the flow. p + 1 ρv + ρgh = constant. Equation 3-5 (a) Or alternatively, at point 1 and in the stream: p 1 1 ρv 1 ρv + ρgh 1 = p + + ρgh 1 +. Equation 3-5 (b)

42 Static pressure EIPINI Chapter 3: Flow Measurement Page 3-5 Static pressure is the pressure that would be measured by a pressure gauge moving with the flow. Dynamic pressure Dynamic pressure is the pressure exerted by a flow because of the flow velocity. Stagnation pressure Stagnation pressure is the sum of the static and dynamic pressure in a flow. Note: Referring to equation 3-5a, the energy per unit volume, p, is the static pressure and the kinetic energy per unit volume ½ρv, is the dynamic pressure (energy per unit volume may be associate with pressure as [Joule/meter 3 ] is equivalent to [Pascal] and interested students are encouraged to verify this dimensional equivalence). The static and dynamic pressures taken together, is called the stagnation pressure (or impact pressure), which is the pressure realized when a flowing fluid is brought to rest. p stag = p + ½ρv p stag = p stat + p dyn.. Equation THE PITOT TUBE One of the earliest flow meters, operation based on Equation 3-6, is the Pitot tube. Stagnation (impact) Static pressure p stat pressure p stag The L shaped tube, has the pitot opening directly facing the oncoming flow stream, as shown in Figure 3-5. At the Pitot opening, the stream is brought to rest, and the impact or stagnation pressure is measured. The static pressure is measured v Impact hole Figure 3-5 at right angles to the flow direction. The flow velocity and therefore the flowrate q = Av, may be determined from the difference between the stagnation pressure and the static pressure: v = (p p ) stag stat ρ. Equation 3-7 A disadvantage of the Pitot tube is that it measures the flowrate only at one point. An annubar flowmeter overcomes this problem by positioning several Pitot tubes across the pipe diameter, providing a better average. Note: The Pitot tube measures flowrate, by making direct use of Equation 3-6, p dyn = p stag p stat ½ρv = p stag p stat v = [(p stag p stat )/ρ].

43 EIPINI Chapter 3: Flow Measurement Page THE FLOW EQUATION In this section, we will derive the flow equation for a horizontal flow stream. Given that the flow stream is horizontal, gravitational forces and potential energy of the flow, will be neglected. A flow restriction will be formed by allowing a section of the pipe to become narrower than the rest of the pipe. We will then show that the flowrate q, may be determined from the square root of the pressure difference across the restriction. Figure 3-6 h Flowdirection A 1 p 1 v 1 p A ρ ρ v h 1 Flowrate = q X Y Referring to Figure 3-6, consider a unit volume in the flow stream with mass ρ (given that the density of the stream is ρ). Let us now follow this unit volume in the flow stream, as it passes the point X (flow area A 1 ), travelling with velocity v 1 and under pressure p 1, and then later passing the point Y (with smaller flow area A ) at a higher velocity v and under the influence of a smaller pressure p. According to Bernoulli s theorem for a steady stream, the total energy content (pressure energy plus kinetic energy) of the unit volume should stay constant. 1 ρv p = ρv + p.. Equation 3-8 (a) 1 And flow continuity demands that the flow rate q must be the same at X and Y: q = A 1 v 1 = A v, therefore v 1 = A A v. Equation 3-8 (b) 1 Using Equation 3-8 (b), v 1 can be eliminated from Equation 3-8 (a), which will allow us to obtain an expression for v and to determine the flowrate from q = A v. 1 1 ρ[ (A /A )v ] + p = ρv + p ρ(a p ρv p 1 1 / A 1 ) v + 1 = + ρv 1 (A A / 1 ) (p p = (p 1 p ) = ) v 1 ρ 1- (A /A 1 )

44 EIPINI Chapter 3: Flow Measurement Page 3-7 v = (p 1 p ) ρ 1 (A /A ) Equation 3-8 (c) The flow rate q is given by: q = A v Equation 3-8 (d) From Equations 3-8 (c) and 3-8 (d) it then follows that, q = (p 1 p ) A ) Equation 3-8 (e) ρ 1 (A /A 1 Equation 3-8 (e), expresses essentially what we wanted to show, namely, the flow rate q varies with the square root of the pressure difference across the restriction. Equation 3-8 (e) may be simplified further, if we choose to specifically measure the pressure difference p 1 p by allowing the liquid in the stream to rise up in the two vertical tubes, as shown in Figure 3-6, and then take the reading h. p 1 = P atm + ρ(h 1 + h)g and p = P atm + ρh 1 g. p 1 p = ρhg.... Equation 3-8 (f) Using p 1 p from Equation 3-8 (f) in Equation 3-8 (e): q = ρhg A = ρ (A /A 1 ) 1 A hg 1 (A /A 1 ). Equation 3-8 (g) We now define a calibration constant: k = A g/[1 (A A 1 ) / ].... Equation 3-8 (h) From Equations 3-8 (f) and 3-8 (g), we may conclude that, q = k h Equation 3-9 Note: Although we assumed a horizontal flow stream, it can be shown that Equation 3-9 is equally valid for inclined flow streams or even vertical flow streams. Please remember that h in Eq. 3-9 still implies the pressure difference p 1 p (pascal).

45 EIPINI Chapter 3: Flow Measurement Page 3-8 Example 3-3 A flow rate meter, uses a restriction in the flow stream, to measure the flow rate of a liquid in a horizontal pipe. The pressure difference across the restriction is determined by allowing the liquid into two vertical tubes installed on top of the pipe and on both sides of the restriction. When the flow rate is 0.1 cubic meter per second, the level difference of the liquid in the tubes is 0.3 meter. Calculate the flow rate when the level difference of the liquid in the two tubes is 0.6 meter. Answer: Using Equation 3-9: q 1 = k h 0.1 = k 0.3 k = 0.1/ 0.3 = q = k h = = m 3 /s Example 3-4 A cylindrical object with volume V f = m 3, density ρ f = 000 kg/m 3 and cross sectional area v v A f = m p, is suspended in the centre of a vertical A tapered tube by water with density ρ l = 1000 kg/m 3, rushing upwards with a velocity v 1 meter/sec. driven forward by pressure p 1 pascal when the tube s crosssectional area is A 1 = m A 1 p 1 v 1 and speeding up to a q velocity v meter/sec. when it reaches the restricted crosssectional area A = m p p, around the object, where 1 q 0.005m the pressure has diminished to p pascal. Calculate the 0.004m 0.001m flow rate of the water. Answer: There are two forces operating on the object, a gravitational force F w pulling it downwards and a drag force F d caused by the water stream, pulling it upwards. F w = Weight of the object Weight loss of the object in the water = ρ f V f g ρ l V f g (Weight loss, according to Archimedes s law, equals weight of water displaced by object) = 000 ( ) ( ) 9.81 = N F d = Force caused by pressure p 1 (up) Force caused by pressure p (down) = p 1 Cross-sectional area of object p Cross-sectional area of object = p 1 A f p A f = p 1 ( ) p ( ) = (p 1 p ) newton For this object to be stationary and stay suspended at one position in the water, the water drag force acting on the object, must equal the gravitational force. F d = F w (p 1 -p ) = p 1 p = 45.3 Pa From Equation 3-8 (e): q = A {(p 1 -p )/ρ l [1-(A /A 1 ) ]} = ( ) {[ 45.3]/[1000 (1 ( / ) )]} = ( ) {(490.6)/[1000 (1-0.04)]}=( ) (490.6/960) = = m 3 /sec. F d F w

46 EIPINI Chapter 3: Flow Measurement Page 3-9 Example 3-5 A flow rate meter, uses a restriction in the flow stream, to measure the flow rate of a liquid in a horizontal pipe. When the flow rate of the stream reaches its maximum value (100%), the differential pressure meter also registers its maximum reading (100%). Assuming that the pressure meter will show a zero reading when the flow rate is zero, calculate the flow rate when the pressure meter indicates 0%, 40%, 60% and 80% of its full scale reading. Draw a graph of percentage flow rate versus percentage differential pressure. It is given that the flow rate q is 100% when the differential pressure h is 100%. Using Equation 3-9, q = k h, 100 = k 100 k = 10 For h = 0%, q = 10 0 = 44.7%. For h = 40%, q = = 63.5% For h = 60%, q = = 77.46%. For h = 80%, q = = 89.44% Flowrate q (percent of full scale) q = 10 h or h = 0.001q (y = ax ) kpa 0 36 kpa kpa 68 kpa 84 kpa 100 kpa Differential pressure h (percent of full scale) If we did use a differential pressure transmitter to measure the pressure difference, then 0% input pressure would correspond to 0 kpa, 0% input pressure to 36 kpa, 40% input pressure to 5 kpa, 60% input pressure to 68 kpa, 80% input pressure to 84 kpa and 100% input pressure to 100 kpa. These values are also shown on the differential pressure axis of the graph. Example 3-6: Use Equation 3-8 (c) to derive the simplified Bernoulli s equation, p 1 p 4v, that medical doctors use when they examine a patient s blood circulation. Assume δ blood = 1, p 1 p is measured in mm. Hg and (A /A 1 ) << 1. v = {[(p 1 -p ) pascal ]/[ρ blood (1-(A /A 1 ) ]} [(p 1 -p ) pascal /1000] because A << A 1 and ρ blood 1000 kg/m 3. Also (p 1 -p ) pascal = [13600 (p 1 -p ) mmhg 9.81]/1000 v = [ 13600(p 1 -p ) mmhg 9.81/1000 ] = 0.7(p 1 -p ) p 1 -p 4v

47 EIPINI Chapter 3: Flow Measurement Page 3-10 Allowing a pressure differential to develop across a restriction in a flow stream, is a popular method to measure the flowrate in a liquid as well as a gaseous flow. Two important meters that utilise this method are the venturi tube and the orifice plate. 3.9 THE VENTURI TUBE Operation and construction The venturi tube, illustrated in Figure 3-7, has a converging conical inlet, a cylindrical throat, and a diverging conical outlet cone. It has no projections into the fluid, no sharp corners, and no sudden changes in contour. The inlet section decreases the area of the fluid stream causing the velocity to increase and the pressure to decrease. The low pressure is measured in the centre of the cylindrical throat since the pressure will be at its lowest value, and neither the pressure nor the velocity is changing. The outlet or recovery cone allows for the recovery of pressure such that total pressure loss is only 10% to 5%. The high pressure is measured upstream of the entrance cone. High pressure tap (upstream tap) Low pressure tap (downstream tap) d Flow D Inlet cone (19º 3º) d Throat Outlet cone (5º 15º) D/ d/ Figure Advantages and disadvantages of the venturi tube Advantages Pressure loss is small Operation is simple and reliable Disadvantages Highly expensive Occupies considerable space.