Instrument test questions
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I. Fill in the blanks 1. The operating characteristics of instruments are usually divided into (static) and (dynamic) characteristics. 2. For preventing pipeline freezing, the main methods are (electric heating) and (steam heating). 3. The installation of a pressure gauge on-site generally includes tasks such as (selecting the pressure measurement point), (laying the pressure conduit), and (installing the gauge itself). 4. When measuring liquid pressure, the pressure measurement point should be located (at the bottom of the pipeline); when measuring gas pressure, it should be located (at the top of the pipeline). 5. The flap level gauge should be installed (vertically), and a (valve) should be installed between the container and the equipment to facilitate maintenance and adjustment of the gauge. 6. When the float of a float level gauge is corroded, punctured, or crushed, its indicated level will be (lower) than the actual level. 7. Float-type level gauges can be divided into external float type and internal float type. The external float type is advantageous in terms of ease of maintenance, but it is not suitable for measuring the level of liquids that are too viscous, or those that tend to crystallize or solidify easily. 8. Currently, control valves produced in China come in three types of flow characteristics: linear, equal percentage, and quick opening. 9. The handwheel mechanisms for pneumatic control valves generally include (side-mounted) handwheel mechanisms and (top-mounted) handwheel mechanisms. 10. Ball valves can be classified into (O)-type and (V)-type based on the structure of their valve cores. 11. Instruments installed directly on the pipeline should be installed (after pipeline purging but before pressure testing). When it is necessary to install them simultaneously with the pipeline, the instruments should be removed (before pipeline purging). 12. During handling and installation, dashboards, cabinets, benches, and boxes should be protected from (deformation) and (damage to the surface paint). Gas welding and cutting are strictly prohibited during installation and processing. 13. The allowable deviation in verticality for the installation of instrument boxes, insulation boxes, and protection boxes is (3 mm) ; When the height of the box is greater than 1.2 m, the allowable deviation for verticality is (4 mm) ; Allowable deviation for levelness (3mm). 14. When cables enter outdoor panels, cabinets, or boxes, it is advisable to do so from the (bottom), and (waterproof sealing) measures should be in place. 15. The wiring should not be installed above high-temperature equipment and pipes, nor below equipment and pipes containing corrosive liquids. 16. Before laying cables and wires, an visual inspection and a continuity check should be carried out; their resistance value should not be less than (5 megohms). 17. When cable trays are installed on process pipe racks, they should be located on the (side or above) of the pipes. For high-temperature pipes, it should not be above them. 18. Instrumentation engineering should carry out (circuit testing) before the system is put into operation. 19. Rotameters belong to (constant pressure drop) flowmeters. 20. The maximum measurement range when using a float-type level gauge to measure liquid level is (the length of the float). 21. Level gauges that do not come into contact with the medium include (ultrasonic), (radar), (laser) level gauges, and nuclear radiation level gauges. 22. The absolute temperature is 273.15 K, which corresponds to 0°C on the Celsius scale. 23. The higher the temperature, the greater the resistance value of materials such as platinum, nickel, and copper. 24. Measurement range: lower limit of -25 to 100°C (-25°C), upper limit (100°C), and range (125°C). 25. The bend radius of the pressure Guide tube should not be too small, as this will increase the ellipticity of the tube and lead to its damage; the bend radius of a metal tube should be no less than 6 times its outer diameter. 26. The instruments in newly installed steam pipelines should be flushed at least (2) times before startup. 27. The common modes of heat transfer are (conduction), (convection), and (radiation). 28. Instruments installed directly on equipment or pipelines shall undergo (pressure testing) along with the equipment or pipeline system after installation is completed. 29. The entry points for the wiring boxes on the instrument panel should not face upward; when this is unavoidable, (sealing) measures should be taken. 30. The installation direction of the throttle element must be such that the fluid flows from the (upstream end face) of the throttle element to its (downstream end face). 31. The sharp edge of the orifice plate or the curved side of the nozzle should face the flow direction of the side fluid. 32. The throttle element must be installed after the pipeline has been purged, and the inner diameter of the gasket used to install the throttle element should not be smaller than (the inner diameter of the pipeline). It must not protrude into the pipe wall after clamping. 33. When instrument pipelines are to be laid underground, they must pass (pressure testing) and undergo (anti-corrosion treatment) before being buried. When connecting pipes that are buried directly in the ground, (welding) must be used, and (protective sleeves) should be installed where the pipes pass through roads or emerge from the ground. 34. The (signal circuit grounding) and (shield grounding) of the instrumentation and control systems shall share the same grounding device. 35. Explosive environments include explosive (gas) environments and explosive (dust) environments. 36. The straight pipe section before the throttle orifice plate is generally required to be (10)D, while the straight pipe section after the orifice plate is generally required to be (5)D. For accurate measurement, the straight pipe section before the orifice plate should preferably be (30~50)D. 37. The compressor inlet control valve should be of the (air-shut) type. 38. The sensor of the Yokogawa EJA intelligent transmitter is of the (silicon resonator) type; it converts the parameter to be measured into the (vibration frequency of a silicon beam), and the measured differential pressure or pressure value is obtained by detecting this frequency. 39. The corrosion protection of instrument pipelines mainly relies on the metal surface (paint), which acts as a barrier against the corrosive agents in the surrounding environment. 40. The commonly used pressure calibration instruments are (piston pressure gauge) and (pressure calibration pump). 41. Classified by working principle, level measurement instruments can be divided into types such as (direct-reading), (buoyancy-type), (static pressure type), (electromagnetic type), (acoustic wave type), and (nuclear radiation type). 42. To address the problems of corrosion and blockage in pressure transfer pipelines, (flanged differential pressure transmitters) can be used. The (flange) of such transmitters is connected directly to the flange on the container, and the sealed system composed of the diaphragm box, capillary tubes, and measurement chamber is filled with (silicone oil) as the pressure transmission medium. 43. Under normal circumstances, (diaphragm-type) actuators should be used; for large diameters and high pressure differences, (piston-type) actuators are appropriate; for butterfly valves requiring high torque, (pneumatic long-stroke) actuators are suitable. 44. Elastic pressure gauges operate on the principle that the (deformation) of an elastic element is proportional to the (pressure applied) to it. 45. 1 kilogram-force = (9.8) Newtons; 1 MPa = (10×6) Pa. 46. The volume of fluid that passes through in a unit of time is called (volumetric flow rate), and it is denoted by (Q). 47. The density of a gas decreases as (temperature) increases, and increases as (pressure) increases. 48. The diaphragm chamber of the differential pressure transmitter is filled with (silicone oil), which, in addition to transmitting pressure, also provides (damping) effect, thereby ensuring a stable output from the instrument. 49. In the model number of a flanged differential pressure transmitter, an “A” indicates that (positive) offset adjustment is possible, while a “B” indicates that (negative) offset adjustment is possible. 50. When a 220VAC voltage is applied to a 10 ohm resistor, the current flowing through this resistor should theoretically be (22A). 51. An automatic control system mainly consists of four major components: (controller), (control valve), (controlled object), and (transmitter). 52. The flow rate of a fluid is in a (square root) relationship with its pressure. 53. When the pressure difference before and after the control valve is small and low leakage is required, a (single-seat) valve is generally suitable. 54. For regulating low differential pressures and high-flow gases, a (butterfly valve) can be used. 55. When both regulation and cutoff are required, an (eccentric rotary) valve can be used. 56. The five key parameters in instrument measurement are (temperature), (pressure), (level), (flow rate), and (online analysis). 57. 250 Pa = (25) mmH2O. 58. In a pipe, the fluid velocity is generally highest at the (centerline of the pipe), while it is zero at the (pipe wall). 59. A turbine flowmeter is a (velocity) type of flowmeter. 61. The process of converting alternating current into direct current is called (rectification). 60. Turbine flow transmitters should be installed (horizontally). 62. The lower the operating power of a relay, the higher its sensitivity. 63. The ideal adjustment ratio of a control valve depends on (the valve core structure). 64. The actual adjustable ratio depends on the (valve core structure) and (piping conditions) of the control valve. 65. The signals required by actuators are divided into (continuous control voltage and control current signals) and (digital or switch signals). 66. The common methods for pipe connection are (threaded connection), (flange connection), and (welding). 67.1 mmH2O is the pressure generated by 1 millimeter of water column in pure water at a temperature of (4)°C. 68. To avoid disturbance to the liquid caused by the valve, the flow control valve should be installed behind the instrument under test. 69. The instrument device installed on a pipeline that detects a set pressure and sends an alarm signal is a (pressure switch). 70. Common cables are divided into (power cables) and (control cables). 71. To measure the current in a circuit, the ammeter must be connected in series with that circuit. To ensure that the connection of the ammeter does not affect the original state of the circuit, the (internal impedance) of the ammeter itself should be as low as possible. 72. Thermal resistance thermometers measure temperature based on the property of conductors or semiconductors (whose resistance value changes with temperature). 73. The working principle of a solenoid valve is to use the attractive force generated by the electromagnet to directly drive the moving part of the valve (the valve core) to open and close. 74. A solenoid valve is an electric actuator that uses a solenoid to drive the opening and closing of the valve. 75. The operating states of an electromagnet are either energized when current flows through it or de-energized when no current flows,; therefore, a solenoid valve can also only be in two operating states: allowing flow or blocking flow. 76. A normally open contact refers to a contact that is in an open state when the button has not been pressed. 77. A normally closed contact refers to a contact that is in a closed state when the button is not pressed. 78. Fuses are used in circuits for short-circuit or overload protection. In the International System of Units, the unit of pressure is Newton per square meter, also known as Pascal; its symbol is Pa. It is defined as the pressure resulting from a force of 1 Newton acting vertically and evenly on an area of 1 square meter. 79. The seven base units of the International System of Units are: length in meters (m), mass in kilograms (kg), time in seconds (s), electric current in amperes (A), thermodynamic temperature in kelvins (K), amount of substance in moles (moL), and luminous intensity in candela (cd). ? 80. The maximum operating voltage for the contacts of the electric contact pressure gauge is 380V for AC and 220V for DC; the maximum operating current is 1A, and the power consumption is 10W. 81. The pressure tapping methods for standard orifice plates include corner tapping and flange tapping, etc. 82. When the pressure-taking device and the temperature-sensing element are installed on the same pipe section, the pressure-taking point should be located before the temperature-sensing element in the direction of flow of the medium, with a distance of not less than 200 mm between them. 83. The types of standard throttling devices include standard orifice plates, standard nozzles, standard Venturi tubes, etc. 84. Throttling devices cannot measure the flow rate of pipes with a diameter of 50 mm or less, or 1000 mm or more. 85. A pressure switch is used to convert the measured pressure into a digital signal; its working principle is based on the lever force measurement principle. 86. Limit switches use direct contact to measure the mechanical displacement of an object in order to obtain a position signal. 87. The calibration points for instruments should be no less than (5) points. 88. When calibrating main pressures, the correction value resulting from the liquid level in the gauge tube should be taken into account. 89. The bottom of the dashboard should be above the ground level (20mm). 90. The inner diameter of the cable protection tube is generally (1.5 to 2 times) the diameter of the cable. 91. Welding may be used to make openings in (wind-pressure pipes), but the edges of the openings must be smoothed. 92. Pressure measurement points should be selected on the (straight sections) of the pipeline or flue. 93. For pressure measurement points on horizontal or inclined pipes, when the medium being measured is a gas or a gas-powder mixture, they should be installed in the (upper half) of the pipe. 94. When the medium being measured is steam or a liquid, pressure measurement points on horizontal or inclined pipes should be located in the (lower half, within an angle of 45 degrees from the horizontal). 95. To avoid eddy currents, the pressure tapping device should be installed in front of the temperature sensing element. 96. For components used to measure steam flow, pressure taps must be installed separately at the upstream and downstream sides (in the condenser). 97. Before starting the instrument, the steam and water pipelines should be flushed, generally no less than (2 times). 98. When the instrument is installed, the medium in the steam pipeline (which needs to be cooled). 99. A tightness test of the valve was conducted using a water pressure of 1.25 times the operating pressure; no leakage was observed over (5 minutes). 100. Diaphragm pressure gauges belong to (elastic pressure gauges). 101. The relationship among atmospheric pressure, gauge pressure, and absolute pressure is (absolute pressure = gauge pressure + atmospheric pressure). This post was last edited by yyong110 on 2008-7-3 16:26.]18. The double-flange level gauge shown in the figure is used to measure the level in a closed container. Given that hl = 20 cm, h2 = 20 cm, h3 = 200 cm, and h4 = 30 cm, with the density of the light medium being ρl = 0.8 g/cm³ and that of the heavy medium being ρ2 = 1.1 g/cm³, and the density of the silicone oil in the gauge’s capillary being ρ0 = 0.95 g/cm³, what are the gauge’s range and zero-point shift?
19. A float level gauge with a float length of 1 m is used to measure the level of a medium with a density of 800 kg/m³. It is calibrated using water with a density of 1000 kg/m³. What is the maximum water filling height during calibration?
20. A float level gauge is used for level measurement. The float’s length is L = 1000 mm, its outer diameter is D = 20 mm, and its weight is 376 g. The density of the medium is 0.8 g/cm³. If calibration is done using weights, what are the weights required at 0%, 25%, 75%, and 100% of the full range?
21. An electric float level transmitter is calibrated using water. The float’s length is L = 500 mm, and the density of the medium being measured is 0.85 g/cm³. Calculate the height of the float submerged in water when the output is 0%, 50%, and 100% (with the density of water taken as 1 g/cm³).
22. A pressure-type level transmitter, as shown in the figure, is used to measure the level in an open container. The density of the medium is ρ = 1200 kg/m³, with hl = 1 m and h2 = 2 m. What are the transmitter’s range and zero-point shift?
23. A double-flange differential pressure transmitter is used to measure the level in a closed container, as shown in the figure. The range of level changes is h = 1 m. In the figure, hl = 2 m, h2 = 1 m, and H = 3 m. The density of the medium being measured is ρl = 1200 kg/m³, while the density of the fluid in the transmitter’s capillary is ρ2 = 850 kg/m³. What are the transmitter’s range and zero-point shift?
24. A blowing-type level gauge is used to measure the level in an open container, as shown in the figure. Given that hl = 20 cm, the density of the medium being measured is 1000 kg/m³, and the pressure reading is 40 kPa, what is the actual liquid level height? 25. There is a proportional control valve whose maximum flow rate under standard conditions is 50 m3/h, and its minimum flow rate is 2 m3/h. If the total travel distance is 3 cm, what is the flow rate at the lcm opening? 26. What is 1/8 inch in millimeters? 27. Given that the temperature of water is 297 K (thermodynamic temperature), calculate what this temperature corresponds to in degrees Celsius. 28. The pressure gauge on the boiler drum reads X, and the barometric pressure gauge reads Y; what is the absolute pressure of the fluid inside the drum? 29. The sinusoidal current in the motor, as measured by an ammeter, is 10 A. What is the maximum value of this current? Additionally, the voltage of the AC power supply, as measured by a voltmeter, is 220 V. What is the maximum value of this voltage? 30. If the minimum resistance of the human body is 800 Ω, and it is known that a current of 50 mA passing through the body can cause respiratory paralysis, preventing the person from breaking free from the power source, what is the voltage in volts? 31. When the temperature of a mercury thermometer is 20°C, the air pressure, as measured by a mercury barometer, is 765 mmHg. Please express this pressure in Pa. 32. A winch is required to perform 50 kW•h of work within 5 hours. Ignoring other losses, what power should the motor have? 33. The pressure gauge reads P_gauge = 0.12 MPa, while the atmospheric pressure, as indicated by the barometer, is 680 mmHg. Determine the absolute pressure of the gas inside the container. 34. What is the equivalent value of a pressure of 10 mmH2O in Pa? 35. If the volume of an oxygen tank is 40 L, what is its equivalent value in m3? 1. δmax = ×100% = 1.25%
2. The range is 1°C/O.005, which equals 200°C; the lower limit is (200°C × 25%) = –50°C. Thus, the scale limits for this instrument are from –50°C to 150°C.
3. Based on the basic error δm = ×100%, we get δm = ×100% = 0.75%; δml = ×100% = 1.2%. According to the accuracy class standards for common industrial instruments, a temperature measuring instrument with a measurement range of 0–800°C should be classified as grade 1 in terms of accuracy ; Thermometers with a measurement range of 600–1100°C should be classified as grade 1.5. 4. The maximum allowable error for this gauge = ×100% = 1.4%. The permissible error of the gauge is 1.5 kgf/cm2; therefore, this gauge does not meet the requirements. 10. The percentage deviation = ×100% = (205–194) / ×100% = 1.83%, which is greater than 1.0%. Hence, it does not satisfy the precision requirements for grade 1.0. 11. When the input pressure to the transmitter is P_abs, and the output current is 12 mA, then P_abs/(80-0)×16+4 = 12, which gives P_abs = 40 kPa. Since P_vacuum = P_atmospheric – P_abs = 98 – 40 = 58 kPa, the reading of the vacuum gauge is 58 kPa, with the transmitter’s output being 12 mA. 12. 445/(1000–0)×100% = 44.5% 13. According to the principle of hydrostatic pressure, P = ρgh; therefore, h = P/ρg = 49/1.25×9.8 = 4 m. 14. As can be seen from the graph, the measurement range of the level gauge is P = ρgh1 = 1.0×9.8×5 = 49 kPa. The displacement A = ρg h2 = 1.0×9.8×1 = 9.8 kPa. Thus, the calibration range of the instrument is 9.8–58.8 kPa. 15. The weight of the liquid in the tank, W1, is equal to mg, and since m = ρ1V = ρ1AH, it follows that W1 = ρ1AHg. As the density of air is ρ2, the buoyant force exerted by the air on the liquid cannot be ignored. The buoyant force W2 = ρ2Vg = ρ2AHg; therefore, the weight of the liquid in the tank is W = W1 – W2 = ρ1Ahg – ρ2Ahg = AHg(ρ1 – ρ2). 16. If the liquid level in the high-level storage tank is h, then the pressure value P is given by P = ρg(h + 5). Hence, h = p/(ρg) – 5 = 100/(1.2×9.81) – 5 = 3.5 m. 17. The output signal of the pneumatic differential pressure transmitter ranges from 0.2 to 1.0 kgf/cm2. When the liquid level is at height H, the output is 0.6 kgf/cm2; thus, H/2000×(1.0–0.2) + 0.2 = 0.6, which gives H = 1000 mm. 18. The range of the instrument is ΔP = h3(ρ2 + ρ3)g = 2×(1.1–0.8)×9.81 = 5.886 kPa. When the interface is at its lowest point, the pressures on the positive and negative sides of the instrument are respectively P+ = P1 + ρ1g(h1 + h2 + h3) + ρ2gh4 – ρ0gh = P1 + 0.8×9.81×(0.2 + 0.2 + 2) + 1.1×9.81×0.3 – 0.95×9.81h = P1 + 22.073 – 9.32h, and P– = P1 + ρ1gh1 + ρ0g(h2 + h3 + h4 – h) = P1 + 0.8×9.81×0.2 + 0.95×9.81(0.2 + 2 + 0.3 – h) = P1 + 24.868 – 9.32h. The displacement A = P+ – P– = –2.795 kPa (negative displacement). 19. To ensure that the buoyant force acting on the float when the density of the liquid medium is 800 Kg/m3 and its height is 1 m is equal to the buoyant force it experiences in water with a density of 1000 Kg/m3, water can be used to calibrate the height of the float. Thus, AL_water × ρ_water × g = AL_medium × ρ_medium × g, where L_water, ρ_water represent the height and density of water, respectively ; L denotes the height of the medium, while ρ denotes its density ; A is the cross-sectional area of the float. Therefore, L_water = ρ_medium × L_medium / ρ_water = 800 × l / 1000 = 0.8 m. 20. A float level gauge is used for measuring liquid levels; the length of the float is L = 1000 mm, its outer diameter is D = 20 mm, and its weight is 376 g. The density of the medium is 0.8 g/cm³. If calibration is carried out using weights, what weights in grams are required at 0%, 25%, 75%, and 100% of the full scale? 21. An electric float level transmitter is to be calibrated using water. It is known that the length of the float is L = 500 mm, and the density of the medium is 0.85 g/cm³. Calculate the height to which the float should be submerged when the output corresponds to 0%, 50%, and 100% of the full scale. (The density of water is taken as 1 g/cm³.) 22. The range of the transmitter is Δp = ρgh² = 1200 × 9.81 × 2 = 23.54 kPa. At the lowest liquid level, the pressure acting on the transmitter is A = ρgh₁ = 1200 × 9.81 × 1 = 11.77 kPa. In other words, the transmitter needs to shift upward by 11.77 kPa. 23. As can be determined from the given information, the range of the transmitter is ΔP = ρ1gh = 1200 × 9.81 × 1 = 11.77 kPa. When the instrument measures the lowest liquid level, the pressures in the positive and negative chambers are respectively P+ = P1 + ρ1gh2 + ρ2gh1 and P- = P1 + ρ2gH + ρ2gh1. Therefore, P+ – P- = ρ1gh2 – ρ2gH = 1200×9.81×1 – 850×9.81×3 = -13.25 kPa. Hence, the range of the transmitter is 11.77 kPa, with a negative offset of 13.25 kPa. 24. According to the principle of hydrostatic pressure, we have P = ρgh; thus, h = P/ρg = 40×10³ / (1000×9.81) = 4.08 m. Therefore, the actual liquid level height H is H = h + h1 = 4.08 + 0.2 = 4.28 m.
25. Under standard conditions, the adjustable ratio is R = Qmax/Qmin = 50 m³/h / 2 m³/h = 25. Hence, at an opening of 1 cm, the flow rate is Q = Qmin × R^(1/L) = 2 m³/h × 25^(1/3) = 5.85 m³/h.
26. 25.4 × 1/8 = 3.175 mm.
27. t = T – 273.15 = 297 – 273.15 = 23.84 (°C).
28. The value of P_table is 9.604 MPa ; Pb=101.7kPa≈0.1MPa. P_abs = P_surface + Pb = 9.604 + 0.1 = 9.704 (MPa). 29. The maximum current Im = √2 × I = √2 × 10 = 14.1 (A). The maximum voltage Um = √2 × U = √2 × 220 = 308 (V). 30. U = IR = 0.05 × 800 = 40 (V). 31. Using the conversion relationship between pressure and units, 1 mmHg = 133.3 Pa; therefore, 765 mmHg = 765 × 133.3 Pa = 101974.5 (Pa). 32. As required, the power of this motor must be sufficient to perform 50 kW•h of work within 5 hours. The formula for calculating power N is N = W/τ, where W represents the work in joules ; τ--time of work, s. Therefore, N=50/5=10(kW). 33. The formula for calculating absolute pressure from gauge pressure is P_abs = P_gauge + P_atm. Given that the gauge pressure is P_gauge = 0.12 Mpa and P_atm = 680 mmHg, then P_abs = 680×133.3 + 0.12×1000000 = 210644(Pa). 34. Since 1 mmH2O ≈ 9.81 Pa, then 10 mmH2O = 10×9.81 = 98.1(Pa). 35. Since 1 L = 10^-3 m3, then 40 L = 40×10^-3 = 0.04(m3). 1. A differential pressure gauge has a maximum differential pressure of 1600 mmH2O and a precision class of 1. What is the maximum allowable error for this gauge? If the calibration point is at 800 mmH2O, what is the allowable range of variation for the differential pressure at that point? (1 mmH2O ≈ 10 Pa). 2. There is a measuring instrument with a precision class of 0.5 and a range of 0–1000°C. Under normal conditions, its maximum absolute error is 6°C. Determine ① the maximum permissible error for this instrument ; ②Basic error ; ①allowable error ; ④Is the accuracy of the instrument satisfactory? 3. There is a pressure gauge with an accuracy class of 2.5 and a measurement range of 0–100 kPa; what is the smallest division on its scale? 4. When calibrating a pressure transmitter with a measurement range of 20–1000 kPa, only a standard gauge with a range of 0–600 kPa and an accuracy class of 0.35 is available. Can this gauge be used as a standard reference gauge?
5. A gauge indicates a gauge pressure of 150 kPa, with the local atmospheric pressure being 1000 kPa. What value will an absolute pressure gauge read in this case?
6. A pressure gauge installed 5 meters below a boiler shows a reading of 500 kPa for the pressure in the boiler’s gas phase. What is the actual pressure in the boiler’s gas phase? (γ = 10,000 N/m³)
7. An absolute pressure transmitter has a measurement range of -100–500 kPa. Calculate the pressures required to produce output values of 4 mA, 8 mA, 12 mA, 16 mA, and 20 mA respectively. What is the output when the transmitter is powered on but no pressure signal is applied? (Local atmospheric pressure = 1000 kPa)
8. The level of liquid in an open container is measured using a U-tube manometer. The density of the liquid is ρ1 = 1.2 g/cm³, and the height of the liquid column in the U-tube is 500 mm. The density of the fluid in the U-tube is ρ2 = 1.5 g/cm³. What is the height h of the liquid level in the container? (See figure below.)
9. As shown in the figure, the range of the liquid level being measured is H = 3000 mm, and the density of the liquid is ρ = 1.2 g/cm³. Calculate the pressure measurement range of the transmitter.
10. A single-flange level gauge is used to measure the level in an open container. The distances from the highest and lowest liquid levels to the gauge are hl = 3 m and h2 = 1 m, respectively. If the density of the liquid is ρ = 1 g/cm³, what are the gauge’s measurement range and zero point shift?
11. A double-flange transmitter is used to measure the level in a closed container. The zero point and measurement range of the transmitter have been calibrated. However, due to maintenance needs, the transmitter’s installation position is lowered by a distance of h2 = 0.5 m, as shown in the figure. Given that h = 5 m, hl = 1 m, the density of the fluid in the container is ρ1 = 1.0 g/cm³, and the density of the oil filling the transmitter’s capillary tube is ρ2 = 0.8 g/cm³, how will the zero point and measurement range of the transmitter change after it is lowered? 12. It is known that the length of the float for a measuring level is L = 600 mm. The density of the heavy medium to be measured is ρ2 = 1.25 g/cm3, while the density of the light medium is ρl = 0.8 g/cm3. When using water as a reference for calibration, how high should the water level be at the zero point and within the measurement range? (The density of water is taken as 1.0 g/cm3.) 13. Determine the transfer function of the system shown in the diagram, and calculate its dynamic parameters. 14. In a ratio control system, the range of transmitter G1 is 0–8000 m3/h, while the range of transmitter G2 is 0–10000 m3/h. After taking the square root of the flow rate, a pneumatic ratio controller or a pneumatic multiplier is used. If G2/G1 = K = 1.2 is to be maintained, what values should be set for the ratio coefficients in the controller and the multiplier? 15. If the flow coefficient C of the control valve is 50, and the pressure difference across the valve is 16×100 kPa with a fluid density of 0.81 g/cm3, what is the maximum flow rate that can pass through? 16. A temporary oil tank has a volume of 8.1 m3. An oil filter with a flow rate of 150 L/min is used to fill this tank. How long will it take to fill the tank? 17. There is a pressure gauge with a range of 0–25 MPa and an accuracy class of 1. During verification, at 20 MPa, the reading was 19.85 when rising and 20.12 when falling; calculate the variation of this gauge. Is this form valid? 18. A pressure gauge is installed 10 meters vertically above the bottom of a water tank, and its reading is 1.9. What is the pressure at the bottom of the tank? (Find the gauge pressure at the bottom.) 19. To thread an M24×2 screw, the required thread depth is 40 mm; determine the drilling depth needed. 20. Use the formula to calculate the diameter of the round rod before threading it with the following thread sleeve. (1)M24 ; (2)M30。 21. Use a Φ6 round steel bar to bend into a ring with an outer diameter of 200 mm; determine the cutting length of the steel bar. 22. To manufacture a circular tube with an outer diameter of 100 mm and a height of 50 mm, formed by bending and welding 5 mm thick steel plates, calculate its blanking dimensions. 1. The maximum allowable error of the instrument is 1600×1% = 16 mmH2O. At a calibration point of 800 mmH2O, the allowable range of differential pressure is (800 ± 16) mmH2O; in other words, the differential pressure can vary within the range of 784–816 mmH2O. 2. ① The maximum permissible error δ = ×100% = 0.6% ; ②Basic error = 0.6% ; ③Allowable error = ±0.5% ; ④The accuracy of the instrument is not satisfactory, as its basic error is 0.6%, which is greater than the allowable error of ±0.5% for this instrument. 3. The maximum absolute error of this instrument is Δmax = 2.5% × (100 – 0) = 2.5 kPa. Since the division values on the instrument’s scale should not be smaller than the absolute error corresponding to its allowable error, the scale can have at most (100 – 0) / 2.5 = 40 divisions. 4. The output pressure range of the pressure transmitter is 20–1000 kPa, with a maximum allowable absolute error of ±80×1% = ±0.8 kPa. When using a standard pressure gauge of O~600 kPa, its maximum allowable error is ±600×0.35% = ±2.1 kPa. It can be seen that the error of the calibration table is greater than that of the instrument being calibrated; therefore, it cannot be used as a standard output table. In calibration work, the allowable error for a standard table should generally be one-third of the allowable error for the instrument being calibrated. 5. P_abs = P_table + P_large = 150 + 100 = 250 kPa. 6. When the instrument is installed 5 meters below the boiler, this introduces an error equivalent to a water column height of 5 meters; thus, the actual pressure is P = P_side – P_influence = 500 – Hγ = 500 – 5×10000/1000 = 450 kPa. 7. The measurement range of the instrument is from -100 to 500 kPa; hence, its full scale range is 500–600 kPa. Assuming that when the input pressure is P_abs, the transmitter outputs the corresponding current value, then: (P_abs/600)×16 + 4 = 4, so P_abs1 = 0 kPa; (P_abs2/600)×16 + 4 = 8, so P_abs2 = 0 kPa; (P_abs3/600)×16 + 4 = 12, so P_abs3 = 0 kPa; (P_abs4/600)×16 + 4 = 16, so P_abs4 = 0 kPa; (P_abs5/600)×16 + 4 = 20, so P_abs5 = 0 kPa. Even in the absence of an input pressure, the transmitter still experiences a pressure of 1 atmosphere, so the output current is J = (100/600)×16 + 4 = 6.67 mA. 8. The pressure generated by the U-tube is P′ = ρ2gh′. The pressure generated when the liquid level in the container is at height h is P = ρ1gh. When P′ = P, then ρ2gh′ = ρ1gh, and thus h = ρ2h′/ρ1 = 1.5×500/1.2 = 625 mm. 9. When there are minor bubbles escaping from the lower end of the air duct, the air pressure inside the duct is almost equal to the static pressure generated by the liquid level, i.e., P = ρgH = 1.2×9.8×3 = 35.28 kPa. Therefore, the pressure measurement range of the transmitter is from 0 to 35.28 kPa. 10. As can be seen from the diagram, the effective measurement range of the instrument is h1 – h2 = 2 m. Therefore, the range of the instrument Δp is given by ΔP = (h1 – h2)ρg = (3 – 1) × 1 × 9.8 = 19.62 kPa. When the liquid level is at its lowest, the pressures acting on the positive and negative chambers of the instrument are respectively P+ = ρg h2 = 1 × 9.81 × 1 = 9.81 kPa, and P- = 0. Thus, the displacement of the gauge is P+ – P- = 9.81 – 0 = 9.81 kPa (positive displacement).
11. Before the installation position of the instrument was lowered, its range was ΔP1 = ρ2gh = 1 × 9.81 × 5 = 49.05 kPa. The displacement A1 was calculated as A1 = ρ2gh1 – ρ2g(h1 + h) = –ρ2gh = –0.8 × 9.81 × 5 = -39.24 kPa. Hence, the adjusted range of the instrument is -39.24 to 9.81 kPa. After the instrument is installed at a lower position, its range becomes ΔP2=ρ1gh=l×9.81×5=49.05 kPa. The displacement A2 is given by A2=ρ2g (h1+ h2)–ρ2g (h1+ h2+h)=–ρ2gh=–0.8×9.81×5=–39.24 kPa. The calibration range of the instrument is -39.24 to 9.81 KPa; in other words, when the transmitter is moved downward, its zero point and range remain unchanged. At the lowest level, the instrument reads zero, as the float is completely submerged in the light medium. Therefore, for water calibration, it should be AL_waterρ_water = AL_ρ1 → L_water = L_ρ1/ρ_water = 600×0.8/1 = 480 mm. At the highest level, the instrument reads full scale, as the float is completely submerged in the heavy medium. Thus, for water calibration, it should be AL_waterρ_water = AL_ρ2 → L'_water = L_ρ2/ρ_water = 600×1.25/1 = 750 mm. Clearly, this height exceeds the length of the float; therefore, a correction must be made during calibration. Based on the calculations above, the range of variation in water level is ΔL_water = 750 – 480 = 270 mm. During calibration, the zero point needs to be adjusted to 600 – 270 = 330 mm; thus, the water level at the full scale point will be 600 mm. After calibration, the zero point must be adjusted back to its original position before the device can be used again. 13. The system consists of a proportional plus integral plus first-order inertia element, and its transfer function can be expressed as G(s) = K/s(Ts+1). With a corner frequency of ω = 10, it follows that Tl = 1/10 = 0.1. Since 20lg|G(jω)| = 6, we have 20lgK – 20lgω = 6; therefore, K = 20. Hence, the transfer function is G(s) = 20/s(0.1s+1). 14. ① When a pneumatic ratio controller is used, since the flow rate passes through a differential pressure transmitter and then through a square root converter, the formula for determining the set value is Kˊ=K×G1max/G2max=1.2×8000/10000=0.96. Therefore, the ratio value in the ratio controller should be set to 0.96. ②When using a pneumatic multiplier, the formula for determining the set value is PB=80K×G1max/G2max + 20 = 80×1.2×8000/10000 + 20 = 96.8 kPa; therefore, the set value for the pneumatic multiplier should be 96.8 kPa. 15. Volumetric flow rate Q = C√ΔP/ρ = 50√16/0.81 = 222 m3/h; Mass flow rate Q = C√ΔPρ = 50√16×0.81 = 180 t/h.
16. 8.1 m3 = 8.1×1000 = 8100 L; 8100/150 = 54 minutes.
17. Variation = |20.12 – 19.85| = 0.27 MPa. The allowable error for this gauge is ±25×1% = 0.25 MPa. Since the variation is greater than the allowable error, this gauge is not acceptable.
18. P_bottom = P_gauge + P_liquid_column. Since P_liquid_column = 10 mH2O ≈ 0.1 MPa, then P_bottom = 1.9 + 0.1 = 2.0 MPa.
19. Drilling depth = Required drilling depth + 0.7d (where d is the outer diameter of the thread) = 40 + 0.7×24 = 56.8 mm.
20. (1) D ≈ d – 0.13t (where d is the outer diameter of the thread and t is the pitch) = 24 – 0.13×3 = 23.61 mm. (2) D ≈ d – 0.13t = 30 – 0.13×3.5 = 29.54 mm.
21. The cutting length is calculated based on the diameter of the center of the finished ring. L = (200 – 6) × 3.14 = 194 × 3.14 = 609.16 mm. 22. The material required is calculated based on the center of the tube wall after bending and welding. L = (100 – 5) × 3.14 = 298.3 mm. 1. The pressure of the medium to be measured is 1.0 MPa, and the ambient temperature around the instrument is ta = 55°C. A Bourdon tube pressure gauge is used for measurement, with a required measurement accuracy of δa = 1%. Determine the measurement range Pd and accuracy class δd of this pressure gauge. (The temperature coefficient β is 0.0001.) 2. A single-flange level gauge is used to measure the liquid level in an open container. The gauge has been calibrated, but due to maintenance requirements, its installation position was lowered by a distance ΔH. Calculate how the gauge’s indication will change as a result, and what should be done to address this issue. 3. Explain through derivation how a float level gauge measures the liquid interface, and what are the conditions under which it functions. 4. In industry, the hydraulic pressure at the bottom of storage tanks is often used to calculate the mass of the liquid inside. As shown in the diagram, a pipe is inserted into the tank, and compressed air is sent into this pipe. When bubbles emerge from the liquid surface, the valve is closed to stop the flow of air. The pressure at point B is measured to be 0.16 MPa. If the cross-sectional area of the tank is 500 m² and the distance from the tank bottom to the side walls is 8 m, what is the mass of the liquid in the tank? 5. A float level transmitter is used to measure the water level in a boiler’s drum. Under normal operating conditions, the pressure of saturated steam is 10.00 MPa (absolute pressure), with a density of saturated water of ρW = 688.4 kg/m³ and a density of saturated steam of ρS = 55.6 kg/m³. The length of the float is L = 800 mm. This float transmitter was calibrated using water with a density of ρl = 1000 kg/m³, ignoring the effect of saturated steam (i.e., assuming ρS = 0). Calculate the water level indicated by the gauge when the actual water level in the drum drops to 0% under normal operating conditions. 6. Simplify the block diagram shown in the figure and write down the closed-loop transfer function. 7. Simplify the block diagram of the transfer function shown in the figure and write down the transfer function. 8. Determine the transfer function for the logarithmic amplitude frequency response shown in the figure, and calculate the dynamic parameters. 9. In a certain ratio control system, the range of the primary flow meter transmitter is F1 = 0–1600 m3/h, while the range of the secondary flow meter transmitter is F2 = 0–2000 m3/h ; A differential pressure signal is generated by the orifice plate, and a square-root function is available; the process requirement is F2/F1 = 3. Determine what coefficient the ratio controller should have. 10. In a ratio control system composed of pneumatic unit instruments, the range of the pneumatic ratio controller is 0–4, while the signal range is 0.2–1.0 bar (1 bar = 10^5 Pa). During operation, it was measured that the output signal of the ratio controller was 0.6 bar. What is the ratio coefficient at this time? 11. In a ratio control system using electric type III instruments, the range of the primary flow rate is 0–400 Nm³/h, and that of the secondary flow rate is 0–600 Nm³/h. During operation, the instrument operator measured that the output current of the ratio controller was 12 mA. Assuming the use of a square root function, what is the ratio of the primary to secondary flow rates? (The range of the ratio controller is 0–4.) 12. There are two pneumatic diaphragm control valves, each with an adjustable ratio of R1 = R2 = 50. The maximum flow rate for the first valve is Q1max = 100 m³/h, while that for the second valve is Q2max = 10 m³/h. What is the adjustable ratio when using split-range control? 13. What is the weight of a steel pipe with a diameter of Φ273×10 in kilograms per meter? (Given that the density of steel is ρ = 7800 kg/m³) 1. According to the regulations, if the value to be measured does not exceed 2/3 of the instrument’s range Pd, then 2/3Pd ≥ Pa, which implies Pd ≥ 3/2 Pa = 3/2 × 1.0 = 1.5 MPa. Therefore, the measurement range of the pressure gauge should be set at 0–1.6 MPa. Temperature-induced error: ΔPF = Pdβ(ta – 25) = 1.5 × 0.0001 × (55 – 25) = 0.0045 MPa. Allowable error for the pressure gauge: δd = Required measurement accuracy – Temperature-induced error = 1 × 1% – 0.0045 = 0.0055 MPa. The accuracy of the gauge is (0.0055 / 1.6 – 0) × 100% = 0.34%. Therefore, a pressure gauge with a measurement range of 0–1.6 MPa and an accuracy class of 0.25 should be selected. 2. Assume that the lowest liquid level in the medium is at the center of the flange. When the liquid level is low, the pressure acting on the positive-pressure side of the level gauge is Hρg (where ρ is the density of the liquid filling the capillary of the level gauge). During calibration, this Hρg value must be adjusted upward. However, if after the migration the gauge moves downward by a distance ΔH, then at the lowest liquid level the pressure on the positive-pressure side of the gauge becomes Hρg + ΔHρg; in other words, an additional ΔHρg is present in the positive-pressure chamber. As a result, the output of the level gauge increases. To ensure accurate measurements in this case, it is necessary to migrate ΔHρg in the positive direction as well. 3. Let the weight of the floating ball be G, and its volume be V. The volumes of the floating ball in the two media are V1 and V2 respectively, with the densities of the two media being ρ1 and ρ2. According to the measurement principle of the float level gauge, the weight G of the float equals the sum of the buoyant forces acting on it in the two media at the two interfaces; that is, G = V1ρ1 + V2ρ2. Therefore, V2 = (G – V1ρ1) / ρ2 = (G – Vρ1) / (ρ1 – ρ2). It can be seen from this formula that if ρ1, ρ2, V, and G remain constant, then V2 also remains constant. When the interface changes, the float will change in sync with the interface, thereby causing the level gauge to indicate the actual liquid level in the container. This is possible provided that the weight G of the float, its volume V, and the specific gravities ρ1 and ρ2 of the two media do not change. 4. Since the density of compressed air is higher than that at normal pressure, the pressure resulting from the weight of the air must be taken into account; that is, P2 = P1 + Δρ2gh. Here, Δρ represents the increase in air density due to compression. According to the gas law, Pa/P2 = (Pa + P1) / (P2 + ΔPa), which leads to Δρ2 = P1ρ2/Pa. Here, Pa denotes atmospheric pressure ; ρ2 is the air density. Substituting equation (2) into equation (1) gives P2 = P1 + P1ρ2gh/Pa = P1(1+ρ2gh/Pa). From the tables, ρ2 = 1.2 kg/m3 and Pa = 1.01×10^5 Pa. Therefore, P2 = P1(1+1.2×9.81×8/1.01×10^5) Pa, which implies that P1 = 0.9991 P2. Since P1 = mg/A, it follows that m = P1A/g. Thus, the mass of the liquid in the tank is m = 0.9991×0.16×10^6×500/9.81 = 8147.6 tons. 5. For verification, when the water level is considered to be 100%, the buoyant force F100 = LsρWg. Since ρs is assumed to be 0, the buoyant force at a water level of 0% is F0 = 0. In fact, when the water level is 0%, the buoyant force generated is F0actual = LSρsg. Therefore, when the actual water level is 0%, the value indicated by the gauge for the water level is h = LSρsg / LSρWg = ρs/ρW = 55.6/688.4 = 8.1%. 6. The original diagram becomes the following; the transfer function G(S) = C/R = G1G2/(1 + G2H1 + G1G2). 7. The original diagram is simplified to this form: the transfer function G = C/R = G2/(1 – G1)/(1 – G1 + G1G2). The symbols in the first column remain unchanged, indicating that the system is stable. 8. The system consists of a first-order inertial element plus a proportional element; with a transition frequency of ω=10, the inertial element is 1/(0.1s+1). Therefore, 201gK = 20, and K = 10; thus the transfer function of this system is G(s) = 10/(0.1s + 1) • 9. K’ = K × F1max/F2max = 3 × 1600/2000 = 2.4. 10. ① (0.6 – 0.2)/(1.0 – 0.2) = 0.5; ② K’ = 4 × 0.5 = 2. 11. ① (12 – 4)/(20 – 4) = 0.5; ② 0.5 × 4 = 2.0; ③ 2.0 × 400/600 = 1.3. 12. The minimum flow rate for the first valve is Qlmin = 100/50 = 2 m3/h, while the minimum flow rate for the second valve is Q2min = 10/50 = 0.2 m3/h. Hence, the overall adjustable ratio between the two valves is R = Qlmax/Q2min = 100/0.2 = 500. 13. First, determine the volume of metal per meter of steel pipe, and then multiply it by the density. The metal volume of the tube = Cross-sectional area of the tube × Length. The cross-sectional area of the tube = Area of the outer diameter – Area of the inner diameter. The area of the outer diameter = {D/2}²π = 0.2372/4 × π = 58.50 × 10⁻³ (m²). The area of the inner diameter = {(D – 2δ)/2}²π = (0.237 – 2 × 0.01)²/4 × π = 50.25 × 10⁻³ (m²). Therefore, the cross-sectional area of the tube = 58.5 × 10⁻³ – 50.25 × 10⁻³ = 8.25 × 10⁻³ (m²). The metal volume of the tube = 8.25 × 10⁻³ × 3 = 8.25 × 10⁻³ (m³). The weight per meter of tube is: Volume × Density, that is, 8.25 × 10⁻³ × 7800 = 64.35 (kg). A bridge with an accuracy of 0.5 grade has negative readings on its lower scale, representing 25% of the full range; the allowable absolute error for this instrument is 1°C. Determine the upper and lower limits of this instrument’s measurement range. The range is 1℃/0.005 = 200℃; the lower limit is -(200℃×25%) = -50℃. The scale of this instrument spans from -50℃ to 150℃. 2. A vacuum pressure gauge has a range of –100 to 500 kPa. During calibration, the maximum error occurs at 200 kPa; the readings on the calibration gauge are 194 kPa at the upper limit and 205 kPa at the lower limit. Does this gauge meet the accuracy requirement of class 1.0? The percentage decrease = ×100% = (205–194) / ×100% = 1.83% > 1.0%, so it does not meet the precision requirement of level 1.0. 3. An 1151 absolute pressure transmitter with a measurement range of 0–80 kPa (absolute pressure). For calibration, a standard vacuum gauge with a scale range of 100–0 kPa is used. If the local atmospheric pressure is 98 kPa, what will be the reading indicated by the vacuum gauge when the transmitter outputs 12 mA? When the input pressure to the transmitter is P_abs, and the output current is 12 mA, then P_abs/(80-0) × 16 + 4 = 12, which gives P_abs = 40 KPa. Since P_vacuum = P_atmospheric – P_abs, we have 98 – 40 = 58 KPa. Therefore, the reading of the vacuum gauge is 58 KPa, with the transmitter’s output remaining at 12 mA. 4. Calculate the calibrated range and drift amount of the pressure-type level gauge as shown in the figure below. Given that h1 = 5 m, h2 = 1 m, and ρ = 1.0 g/cm3, it can be determined from the figure that the measurement range of the level gauge is: P = ρgh1 = 1.0 × 9.8 × 5 = 49 KPa; the drift amount A = ρgh2 = 1.0 × 9.8 × 1 = 9.8 KPa. Therefore, the calibrated range of the instrument is 9.8–58.8 KPa. 5. A single-flange level gauge is used to measure the liquid level in an open container. The distances from the highest and lowest liquid levels to the gauge’s installation point are respectively h1 = 3 m and h2 = 1 m, as shown in the figure below. If the density of the liquid being measured is ρ = 1 g/cm3, what are the gauge’s range and zero-point adjustment? As can be seen from the diagram, the effective measurement range of the gauge is h1–h2 = 2 m. Therefore, the gauge’s pressure range ΔP = (h1 – h2)pg = (3 – 1) × 1 × 9.81 = 19.62 KPa. When the liquid level is at its lowest, the pressures in the gauge’s positive and negative chambers are respectively P+ = pg h2 = 1.0 × 9.81 × 1 = 9.81 KPa, and P– = 0. Thus, the displacement of the level gauge is P+ – P– = 9.81 – 0 = 9.81 KPa (positive displacement). 6. A dual-flange transmitter is used to measure the liquid level in a closed container; the zero point and range of the level gauge have been calibrated. However, due to maintenance needs, the installation position of the gauge was moved downward by a distance of h2 = 0.5 m, as shown in the diagram below. Given that h=5m, h1=1m, the density of the medium inside the container is ρ1=1.0g/cm3, and the density of the oil filling the transmitter’s capillary is ρ2=0.8g/cm3, determine how the zero point and range of the transmitter will change before and after it moves downward Before the instrument was installed at a lower position, its measurement range was △P1=p1gh=1.0×9.81×5=49.05 KPa. The displacement amount A1 was calculated as A1=p2gh1 – p2g(h1+h) = -p2gh=-0.8×9.81×5=-39.24 KPa. Therefore, the adjusted measurement range of the instrument is -39.24 to 9.81 KPa. After the instrument is installed at a lower position, its measurement range ΔP2 becomes ΔP2 = p1gh = 1.0 × 9.81 × 5 = 49.05 KPa. The displacement amount A2 is equal to A2 = p2g(h1 + h2) – p2g(h1 + h2 + h) = –p2gh = –0.8 × 9.81 × 5 = –39.24 KPa. Thus, the calibrated range of the instrument is from –39.24 KPa to 9.81 KPa; in other words, after the transmitter is moved to a lower position, its zero point and measurement range remain unchanged. 7. Simplify the block diagram shown below and write the closed-loop transfer function. The transfer function is G(S) = C/R = G1G2/(1 + G2H1 + G1G2). 8. If the flow coefficient C of the control valve is 50, and the pressure difference across the valve is 16×100 kPa with a fluid density of 0.81 g/cm3, what is the maximum flow rate that can pass through? Volumetric flow rate, mass flow rate. 9. There are two pneumatic diaphragm sleeve control valves, with an adjustable ratio of R1=R2=50. The maximum flow rate for the first control valve is Qmax=100 m3/h, while that for the second control valve is Q2max=10 m3/h. What is the adjustable ratio when using split-range control? Q1min=100/500=20m3/h ; Q2min=100/500=20m3/h ; R=Q1max/Q2min=100/0.2=500 10. The control system for a certain tower is shown in the figure below; please draw the block diagram of this control system. The block diagram of its control system is as follows: 1. Draw the symbols for AND gates, OR gates, and RS flip-flops, and list the truth tables showing how their outputs change with the inputs (the inputs are IN1 and IN2, while the output is Q). IN1 IN2 Q IN1 IN2 Q R=IN1 S=IN2 Q 0 0 0 0 0 0 0 0 Stay at 0 1 0 0 1 1 0 1 1 1 0 0 1 0 1 1 0 0 1 1 1 1 1 1 1 1 Uncertain 2. Draw a 3-to-2 logic diagram with an unlock switch. in1 OR AND bypass1 in2 OR AND OR bypass2 in3 OR AND bypass3 3. Draw a schematic diagram of the structure of a two-chamber equilibrium container. Solution: Q = S * U = R2 × 5 × 3600 = d2 × 5 × 3600 = (159 – 5 × 2)2 × 5 × 3600 = 314 m3/h. 10. The measurement range of the differential pressure transmitter used in conjunction with the throttling device is 0–40 kPa; the scale of the secondary meter ranges from 0–30 t/h. The value obtained from the secondary meter is taken to the square root, with a 2% reduction for low signal levels. ①If the secondary gauge indicates 10 t/h, what is the input differential pressure of the transmitter? What is the output current of the transmitter? ② If the input differential pressure is 0.01 kPa, what will be displayed on the secondary gauge? Solution: ① Using the formula, △P_real = 2 × △P_measured = 2 × 40 = 4.44 KPa; I = (2 / 16) + 4 = 0.125 + 4 = 5.78 mA. ② Using the same formula, Q_real/Q_measured = SQRT(0.01/40) < 2%. Since the cutoff point for low signals is 2%, the secondary gauge will display 0. 1. There is a measuring instrument with an accuracy class of 0.5 and a range of 0–1000℃. When calibration is carried out under normal conditions, the maximum absolute error is 6°C; determine the 1st and maximum reference errors of this instrument ; 2. Basic error ; 3. Allowable error ; 4. Whether the accuracy of the instrument is satisfactory. Solution: 1. Maximum allowable error δ = ×100% = 0.6%. 2. Basic error = 0.6%. 3. Permissible error = ±0.5%. 4. The instrument’s accuracy does not meet the requirements. Because the basic error of this gauge is 0.6%, which is greater than the allowable error of ±0.5% specified for the instrument. 2. There is a pneumatic differential pressure transmitter with a measurement range of 25,000 Pa; the corresponding maximum flow rate is 50 t/h. The process requirement is that an alarm be triggered at a flow rate of 40 t/h. Question: (1) What should the alarm value be set at without a square root calculator? (2) When a square root calculator is used, what should the alarm value be set at? Solution: 1. Without a square root function, the differential pressure corresponding to a flow rate of 40 t/h is ΔP1 = 25000 × (40/50)² = 16000 Pa. The output voltage corresponding to this flow rate is P_out1 = (16000/25000) × 80 + 20 = 71.2 KPa. Therefore, the alarm value S is 71.2 KPa. 2. With a square root function, since ΔQ = K × ΔP, the differential pressure corresponding to a flow rate of 40 t/h is ΔP2 = 25000 × (40/50) = 20000 Pa. The output voltage corresponding to this flow rate is P_out2 = (20000/25000) × 80 + 20 = 84 KPa. Hence, the alarm value S is 84 KPa. 3. The mass flow rate of the superheated steam is 100 t/h, and the inner diameter of the pipe used is D = 200 mm. If the density of the steam is ρ=38 kg/m3, what is the average flow velocity of the steam in the pipe? Solution: The relationship between mass flow rate M and volume flow rate Q is given by: Q = M/ρ = 100×1000/38 = 2631 m3/h = 0.731 m3/s. The cross-sectional area of the pipe, S, is S = (π/4)D2 = (π/4)×(0.2)2 = 0.0314 m2. Since Q = S×u, the average flow velocity of the steam, u, is u = Q/S = 0.731/0.0314 = 23.3 m/s. 4. The original measurement range of the 1151GP pressure transmitter was 0–100 KPa; now its zero point has shifted by 100%. Then 1. What becomes the measurement range of the instrument? 2. What is the range of the instrument? 3. At what pressure values will the instrument output 4, 12, and 20 mA? Solution: The original measurement range of the instrument was 0–100 KPa; now the zero point has shifted upward by 100%. Therefore: 1. The measurement range has become 100–200 KPa. 2. The range of the gauge is 200–100 = 100 KPa. 3. When the input is 100 KPa, the instrument’s output is 4 mA. When the input is 150 KPa, the instrument's output is 12 mA. When the input is 200 KPa, the instrument's output is 20 mA. 5. Convert the following process flow diagram into a system block diagram. Answer: 6. There is a differential pressure gauge used to measure the liquid level in a tank. Given that H = 5 m, h0 = 1.2 m, and H0 = 7.2 m, determine: 1. The amount of migration, and indicate whether it is positive or negative migration ; 2. What is the range of this transmitter? (Assume the liquid level density ρ = 0.96 g/cm3) Solution: When the liquid level is at zero, P0 = ρg(h0 – H0) = -56.448 kPa, indicating negative migration. At full liquid level, P = ρg(H + h0 – H0) = -9.408 kPa. The range of values is: -56.448 kPa to -9.408 kPa. 7. For a certain throttling device steam flow meter, the originally designed steam pressure is 2.94 MPa and the temperature is 400°C. In actual use, the steam pressure at the measured point is 2.84 MPa (with constant temperature). When the flow meter indicates 102 t/h, what is the actual flow rate in tons per hour? (Vapor density table: at 2.94 MPa and 400°C, ρ = 9.8668 kg/m3) ; At 2.84 MPa and 400°C, ρ = 9.5238 kg/m3). Solution: Actual M = Design M × SQRT(Actual ρ / Design ρ) = 102 t/h × SQRT(9.5238 / 9.8668) = 102 t/h × 0.9825 = 100.2 t/h. Therefore, the actual flow rate is 100.2 t/h. 8. A single-tube mercury manometer is currently being used to measure the pressure inside a certain air tank. If the local gravitational acceleration is the standard value (i.e., g=980.665 cm/s2) and the local ambient temperature is 30°C, with a gauge pressure reading of 420 mmHg on the pressure gauge, what is the absolute pressure inside the tank in mmHg, and what is its equivalent in Pa? (The atmospheric pressure at standard gravity is 760 mmHg; the densities of mercury at 0°C and 30°C are 13.5951 and 13.52 g/cm3, respectively.) Solution: The absolute pressure inside the tank is P_abs = P_atmospheric + P_gauge = 760 + 420 × (13.52/13.5951) = 760 + 417.68 = 1177.68 mmHg. 1177.68 mmHg = 1177.68 × 13.5951 × 9.80665 = 157.011 KPa. The absolute pressure inside the tank is 1177.68 mmHg, which is approximately equal to 157.011 KPa. 9. Flow rate is measured using a differential pressure transmitter; the differential pressure is 25 Kpa, the range of the secondary meter is 0–200 T/h, and the output current of the differential pressure transmitter is 4–20 mA. Determine the corresponding differential pressure values and current values when the flow rate is 80 T/h and 100 T/h Solution: According to the differential pressure flow formula: F2 = K × Δp. Δp_x/Δp = (Fmax/F)². Therefore, Δp_x = Δp × (Fmax/F)² = 2500 × (80/200)² = 4000 Pa. I = (Fmax/F) × 16 + 4 = 6.56 mA. Δp_x = 2500 × (100/200)² = 6250 Pa. I = (100/200)² × 16 + 4 = 8 mA. 10. It is given that in a 24VDC power supply circuit, there are 2 12V, 1.5W indicator lights and 1 12V, 1.5W indicator light in series. What is the actual power of each light bulb? Solution: (1) The resistance of each bulb is as follows: For 12V, 1.5W: 12×12/1.5 = 96 ohms; for 12V, 1W: 12×12/1 = 144 ohms. (2) The resistance of the series circuit is: 96×2 + 144 = 336 ohms. (3) The current in the series circuit is: 24/336 = 0.0714 A. (4) The power of each indicator light is as follows: For 12V, 1.5W: 0.07142×96 = 0.49 W; for 12V, 1W: 0.07142×144 = 0.724 W. Therefore, after connecting them in series, the power of each 12V, 1.5W indicator light is 0.49 W ; The power of the 12V, 1 W indicator light is 0.74 W. 11. Complete the timing diagram based on the following PLC ladder diagram: Ladder diagram: Timing diagram: 10S Solution: 1) If the internal resistance of an electrical appliance is 24 ohms, the supply voltage is 60V, and the internal resistance of the power supply is 1 ohm, what is the current flowing through the electrical appliance? Solution: I=U/R = 60/(24+1) = 2.4A. 2) The original measurement range of the 1151GP pressure transmitter was 0–100 KPa; now the zero point has shifted by 100%. Then 1. What becomes the measurement range of the instrument? 2. What is the range of the instrument? 3. At what pressure values will the instrument output 4, 12, and 20 mA? Solution: The original measurement range of the instrument was 0–100 KPa; now the zero point has shifted upward by 100%. Therefore: 1. The measurement range has become 100–200 KPa. 2. The range of the gauge is 200–100 = 100 KPa. 3. When the input is 100 KPa, the instrument’s output is 4 mA. When the input is 150 KPa, the instrument's output is 12 mA. When the input is 200 KPa, the instrument's output is 20 mA. 3) There are two temperature measuring instruments, with measurement ranges of 0–800°C and 600–1100°C respectively. It is known that their maximum absolute error is ±6°C for each. Determine their accuracy grades respectively. Solution: Using the basic error formula δm = Δmax/(A_upper – A_lower) × 100%, we get δm1 = 6/(800 – 0) × 100% = 0.75%, and δm2 = 6/(1100 – 600) × 100% = 1.2%. According to the accuracy class standards for common industrial instruments, temperature measuring instruments with a measurement range of 0–800°C should have an accuracy class of 1, while those with a measurement range of 600–1100°C should have an accuracy class of 1.5. 4) There is a pneumatic differential pressure transmitter with a measurement range of 25,000 Pa; the corresponding maximum flow rate is 50 t/h. The process requirement is that an alarm be triggered at a flow rate of 40 t/h. Question: (1) What should the alarm value be set at without a square root calculator? (2) When a square root calculator is used, what should the alarm value be set at? Solution: 1. Without a square root function, the differential pressure corresponding to a flow rate of 40 t/h is ΔP1 = 25000 × (40/50)² = 16000 Pa. The output corresponding to this flow rate is P_out1 = (16000/25000) × 80 + 20 = 71.2 KPa. Therefore, the alarm value S is 71.2 KPa. 2. With a square root function, since ΔQ = K × ΔP, the differential pressure corresponding to a flow rate of 40 t/h is ΔP2 = 25000 × (40/50) = 20000 Pa. The output corresponding to this flow rate is P_out2 = (20000/25000) × 80 + 20 = 84 KPa. Hence, the alarm value S is 84 KPa. 5) When a certain throttling device was designed, the density of the fluid was 520 Kg/m³; however, in actual use, the density is 480 Kg/m³. If, during design, the differential pressure transmitter outputs 100 KPa corresponding to a flow rate of 50 t/h, what will be the corresponding flow rate in actual use? Solution: Using the basic formula based on mass flow rate, and since all conditions other than the fluid density remain unchanged, with the designed fluid density ρ_design = 520 Kg/m³ and the actual fluid density ρ_actual = 480 Kg/m³, and the designed flow rate when the transmitter outputs 100 KPa being M_design = 50 t/h, then the actual flow rate M_actual is… 6) While calibrating an 1151AP absolute pressure transmitter, it was found that as long as power was supplied, the instrument would output 12 mA. Is this phenomenon normal? (If the measurement range of the gauge is 50–150 KPa gauge pressure, and the atmospheric pressure at that time is 100 KPa) Solution: An absolute pressure transmitter measures absolute pressure; its measurement range is 50–150 KPa. In other words, when an input of 50 KPa gauge pressure is applied, the transmitter outputs 4 mA ; When an absolute pressure of 150 KPa is applied, the instrument outputs 20 mA ; At atmospheric pressure, that is, when an input pressure of 100 Kpa is applied, the instrument’s output is ×(100-50)+4=12mA. The instrument’s output is currently exactly 12mA, so it is functioning normally. 7) There is a differential pressure transmitter with a measurement range of 0–10000 Pa. The manual for this instrument specifies that a negative drift of 100% is possible (the maximum drift amount being -100%). What is the maximum drift amount for this instrument? Solution: Negative migration refers to shifting the zero point in the negative direction without changing the range; the percentage of this shift relative to the range is the percentage of migration. Migration amount = 10000 × 100% = 10000 Pa; therefore, a transmitter operating in the 0–10000 Pa range can be migrated to the -10000–0 Pa range. 8) When the detection system of the nickel-chromium/nickel-silicon thermocouple is in operation, the temperature at its cold end is t0 = 30°C. The instrument measures the thermoelectromotive force E(t, t0) = 39.17 mV. Determine the actual temperature of the medium being measured. Solution: Using the table, we find that E(30º, 0º) = 39.17 mV. Moreover, E(t, 0º) = E(t, 30º) + E(30º, 0º) = 39.17 + 1.2 = 40.37 mV. From the table, the corresponding value for 977°C is also obtained. 9) When using a differential pressure transmitter to measure flow rate, the differential pressure is 25 KPa; the range of the secondary meter is 0–200 T/h, and the output current of the differential pressure transmitter is 4–20 mA. What are the corresponding differential pressure values and current values when the flow rate is 80 T/h and 100 T/h? Solution: According to the differential pressure flow formula: F2 = K × Δp. Δp/x/Δp = (Fmax/F)², so Δp = Δp × (Fmax/F)² = 2500 × (80/200)² = 4000 Pa. I = (Fmax/F) × 16 + 4 = 6.56 mA. Δp = 2500 × (100/200)² = 6250; I = (100/200)² × 16 + 4 = 8 mA. 10) When using a differential pressure transmitter to measure flow, the differential pressure is 25 KPa. The range of the secondary meter is 0–200 T/h, and the counter advances 1000 counts per hour at full scale. The output current of the differential pressure transmitter is 4–20 mA. Determine the number of counts per hour when the flow rate is 80 T/h and 100 T/h, as well as the current value represented by each count. Solution: Let the number of counts be X. 1) For a flow rate of 80 T/h: 80/200 = X/1000 → X = 80 × 1000/200 = 400 counts/hour. 2) For a flow rate of 100 T/h: 80/200 = X/1000 → X = 80 × 1000/200 = 500 counts/hour. 3) The flow rate represented by each count is 200 T/h / 1000 counts/h = 0.2 T per count. 11) If the control valve is fully open, the pressure difference before and after the valve is 400 KPa, and the flow rate of clean water is 100 m³ per hour. What is the flow coefficient C of the valve, and what is its CV value? Solution: Given that Q = 100 m3/h, ΔP = 400 KPa, r = 1 gf/cm3, Cv = 1.17, and C = 1.17 × 50 = 58.5. 12) Use a glass U-tube to measure the vacuum level at PA in container B. The installation diagram is as follows. From the diagram, it can be seen that the vacuum level at PA is 380 mmHg, while the external atmospheric pressure is 760 mmHg. What is the absolute pressure at PA? Solution: P_abs = P_surface + P_atmosphere. P_surface = -380 mmHg = -50.67 KPa; P_atmosphere = 760 mmHg = 101.33 KPa. Therefore, P_abs = -50.67 KPa + 101.33 KPa = 50.66 KPa. The absolute pressure at PA is 50.66 KPa. 13) For a throttling device used as a steam flow meter, the designed steam pressure is 2.94 MPa and the temperature is 400°C. In actual use, the steam pressure at the measured point is 2.84 MPa (with constant temperature). When the flow meter indicates 102 t/h, what is the actual flow rate? Water vapor density table: at 2.94 MPa and 400°C, ρ = 9.8668 kg/m3 ; At 2.84 MPa and 400°C, ρ = 9.5238 kg/m3; the actual flow rate is 100.2 t/h. 14) A metal wire with a length of 200 M and a cross-sectional area of 0.20 cm2 has a resistance of 10 ohms. Determine its mechanical properties Solution: According to Ohm’s law, R = ρt/s. Since ρ = Rs/t = 10 × 0.00002/200 = 0.000001 (Ωm), the electrical conductivity of the metal is 1.0×10-6 ohm-meters. 15) Calculate the total resistance within a junction box containing three devices connected in parallel. Solution: Let the internal resistance of each instrument box be r, and the total internal resistance of the parallel-connected instrument boxes be R. According to the rules for resistors in parallel, 1/R = 1/r1 + 1/r2 + 1/r3. Therefore, 1/R = 3/r, and thus R = r/3. 16) Given that the power supply for the instruments is 24 V and the output power is 32 W, determine the output current. Solution: P=UI; I=P/U. Therefore, I=32 W/24 V=1.33 A. 17) When using a WZG-type thermistor for temperature measurement, its resistance value is 71.02 ohms. If R0=53 ohms and A=4.25×10^-31 per degree, what is the temperature in degrees Celsius? Solution: Rt = R0(1 + At). t = (Rt – R0) / (R0A). t = (71.02 – 53) / 53 × 4.25 × 10^-3. t = 80°C. 18) A platinum-rhodium-platinum thermocouple is used to measure the temperature inside a furnace; in the absence of any other compensation methods, the electromotive force measured is 10.638 mV, and the temperature at the free end of the thermocouple is 50°C. Determine the actual temperature. Solution: The value corresponding to 50°C as per the table is 0.299 mV. Therefore, E(t, 0°C) = 10.638 + 0.299 = 10.938 mV. Using the table, the corresponding temperature is t = 1118°C. 19) When using a DDZ-Ⅲ type differential pressure transmitter to measure flow rate, with a flow range of 0–16 m3/h, what is the output signal when the flow rate is 12 m3/h? Solution: The differential pressure is proportional to the square of the flow rate, and the output current of the transmitter is proportional to this differential pressure; therefore, the 4-20mA current output by the transmitter is proportional to the square of the flow rate. That is: (I-4)/(20-4) = Q22/Q12 = 122/162; therefore, I = 4 + 122 = 13 mA. Answer: The output signal is 13 mA. 20) It is known that a 24VDC power supply unit is used to power several 12V indicator lights in series. If the indicator lights are to function properly, how many such indicator lights can be connected in series? If each indicator light has a resistance of 5 ohms, what is the current flowing through each one? Solution: In series, the voltage is divided as 24/12=2; therefore, 2 lamps can be connected in series. The current is I=U/R=24/(5×2)=2.4 A. The current flowing through each indicator lamp is 2.4 A. 21) It is given that a container has a cylindrical shape, with a base area of 4.0 square meters. When it is completely filled with water, its height is 35 meters. What is the pressure at the base area? Solution: F = ρghS = 1000 × 35 × 4 = 140,000 Kg. Answer: The pressure on the base area is 140,000 Kg. 22) A pressure gauge with a grade of 1.5 and a scale range of 0–100 KPa was tested; it was found that the error at 50 KPa was the greatest, at 1.4 KPa, while the errors at other points on the scale were all less than 1.4 KPa. Is this pressure gauge qualified? Solution: The maximum reading error of this gauge = ×100% = 1.4% of P_, so the gauge exhibits positive drift. 3. When the liquid level h is 2.5 m, the output of the instrument is given by I = (h – h2) / (h1 – h2) × (20 – 4) + 4 = (2.5 – 1) / (3 – 1) × 16 + 4 = 16 mA. Figure 39 shows the use of a differential pressure transmitter to measure the liquid level in a closed container. It is known that h2 = 200 cm, h3 = 140 cm, and the density of the liquid being measured is 0.85 g/cm3. Find the transmitter’s range and offset. Solution: ΔP = h2ρg = 200×0.85×980.7×(100/1000) = 16671.9 Pa. At the lowest liquid level, P+ = h3ρg = 140×0.85×980.7×(100/1000) = 11670.3 Pa; P_ = 0. Since P+ > P_, the gauge shows a positive drift. The migration amount is 11670.3 Pa, and the range is 16671.9 Pa. 40) As shown in the figure, the boiler bubble level is measured using a two-chamber equilibrium vessel. Given that P1 = 3.82 MPa, ρ_vap = 19.7 Kg/m3, ρ_liquid = 800.4 Kg/m3, ρ_cold = 915.8 Kg/m3, h1 = 0.6 m, and h2 = 1.5 m, determine the range and zero shift of the differential pressure transmitter. Solution: The range of the differential pressure level gauge is ΔP = h1(ρ_liquid – ρ_vapor)g = 0.6×(800.4 – 19.7)×9.807 = 4593.8 Pa. When the liquid level is at its lowest, the forces acting on the positive and negative chambers of the gauge are as follows: P+ = P1 + h1ρ_vapor×g + h2ρ_water×g = 3.82×10^6 + 0.6×19.6×9.807 + h2ρ_water×g = 3820115.3 + h2ρ_water×g; P_- = ρ1 + h1ρ_cold×g + h2ρ_water×g = 3.82×10^6 + 0.6×915.8×9.807 + h2ρ_water×g = 3825388.8 + h2ρ_water×g. Thus, the drift amount of the gauge is P = P+ – P_- = -5273.5 Pa. Since P+ < P_, there is negative drift. The measurement range of the gauge is from -5273.5 to (-5273.5 + 4593.8) = 679.7 Pa