Facebook
Google
GitHub
LinkedinInstrumentation and Process Control
Instrument Calibration Tables
-
Question 1
A digital pressure transmitter has a calibrated input range of 0 to 75 PSI, and a 14-bit output (0 to 16383 “count” range). Complete the following table of values for this transmitter, assuming perfect calibration (no error).
For Counts, round down to the nearest whole value.Input Pressure (psi) Percent of Span (%) Counts (dec) Counts (hex) 0 36 62 89 95 Reveal answerInput Pressure (psi) Percent of Span (%) Counts (dec) Counts (hex) 0
0 0
000
27
36 5897
1709
46.5
62 101577
27AC
66.75
89 14580
38F4
71.25
95 15563
3CCB
-
Question 2
Suppose you wish to calibrate a thermocouple temperature transmitter to an input range of 0 to 1500 degrees F, with an output range of 4 to 20 mA. Complete the following calibration table showing the proper test temperatures and the ideal output signals at those levels:
Input Temp (℉) Percent of Span (%) Output Signal (mA) 0 25 50 75 100 Reveal answerInput Temp (℉) Percent of Span (%) Output Signal (mA) 0 0 4 375 25 8 750 50 12 1125 75 16 1500 100 20 -
Question 3
A temperature transmitter has a calibrated range of -80 to 150 degrees F and its output signal range is 4 to 20 mA. Complete the following table of values for this transmitter, assuming perfect calibration (no error).
Input Temp (℉) Percent of Span (%) Output Signal (mA) 120 -45 42 25 7.5 12.9 Reveal answerInput Temp (℉) Percent of Span (%) Output Signal (mA) 120 86.96 17.91 -45 15.22 6.435 16.6 42 10.72 -22.5 25 8 -29.69 21.88 7.5 47.94 55.63 12.9 -
Question 4
A Foxboro pneumatic square root extractor has a calibrated range of 3 to 15 PSI for both input and output. Complete the following table of values for this relay, assuming perfect calibration (no error).
Input Signal (psi) Percent of Span (%) Percent of Output (%) Output Signal (psi) 5 13 50 30 80 15 7 12 Reveal answerInput Signal (psi) Percent of Span (%) Percent of Output (%) Output Signal (psi) 5 16.67 40.82 7.899 13 83.33 91.29 13.95 9 50 70.71 11.49 6.6 30 54.77 9.573 10.68 64 80 12.6 3.27 2.25 15 4.8 4.333 11.11 33.33 7 9.75 56.25 75 12 -
Question 5
Every instrument has at least one input and at least one output. For instruments responding linearly, the correspondence between input and output is proportional:
A practical example of this is a pressure transmitter, in this case one with an input range of 0 to 1023 PSI and an output of 4-20 mA:
If you happened to measure an output current of 14.7 mA from this pressure transmitter, it would be a simple matter for you to calculate the corresponding input pressure to be 684.13 PSI.
However, students are often taken by surprise when they encounter an analog-to-digital converter (ADC) or digital-to-analog converter (DAC) and are asked to correlate input and output for such devices. What might seem a daunting task at first, though, soon reveals itself to be the same input-to-output correspondence calculations they’ve seen all along in the guise of analog sensors and other instruments.
Take for example this analog-to-digital converter, with a 10-bit output (a “count” range of 0 to 1023) and a 4-20 mA input:
Calculate the corresponding “count” output of this ADC circuit given a 6.82 mA input signal.
Reveal answerCount = 180 (rounding low) or 181 (rounding high)
Notes:First, converting the given 6.82 mA input value into a per-unit figure:
$${6.82 \hbox{ mA} - 4 \hbox{ mA} \over 20 \hbox{ mA} - 4 \hbox{ mA}} = 0.17625 \hbox{ per unit}$$
We know that this is a 10-bit converter, therefore its count range (expressed in decimal) must be 0 to 1023 counts. Thus, we may multiply the 0.17625 per-unit figure by the maximum count value of 1023 to see how many counts this ADC will output at 6.82 mA:
$$(0.17625)(1023 \hbox{ counts}) = 180.30375 \hbox{ counts}$$
Since we don’t know anything about how this particular ADC digitizes the signal, we cannot predict whether the output will be rounded up, rounded down, or perhaps even oscillate between the upper and lower rounded values:
Output “count” value = 180 or 181
-
Question 6
The ADC0804 is an example of an integrated circuit analog-to-digital converter (ADC), converting an analog input voltage signal into an 8-bit binary output:
When operated from a 5.0 volt DC power supply in its simplest mode, the ADC0804 converts any DC input voltage between 0.0 volts and 5.0 volts into an 8-bit number at the command of a clock pulse. A 0.0 volt input yields a binary output (or “count”) of {\tt 00000000}, of course, while a 5.0 volt input yields a count of {\tt 11111111}.
Complete this table of numbers, relating various DC input voltages with count values (expressed in binary, hex, and decimal) for an ADC0804 having an input range of 0.0 to 5.0 volts DC:
DC Input Voltage Binary Count Hex Count Decimal Count 0.0 V 0000 0000 0011 0011 51 2.2 V 70 B3 179 1100 1100 CC 5.0 V 1111 1111 Reveal answerPartial answer:
DC Input Voltage Binary Count Hex Count Decimal Count 0.0 V 0000 0000 1.0 V 0011 0011 51 2.2 V 0111 0000 70 112 3.51 V 1011 0011 B3 179 4.0 V 1100 1100 CC 204 5.0 V 1111 1111 FF Notes:The input voltage range of this ADC is 0.0 to 5.0 volts DC, and the output “count” range is 0 to 255 (because it outputs an 8-bit unsigned binary number which has this counting range). Therefore, the relationship between the input voltage and the output count value (in decimal) is a simple proportionality:
$${V_{in} \over 5} = {\hbox{Count} \over 255}$$
Note that all count values are shown rounded {\it down} to the nearest integer value:
DC Input Voltage Binary Count Hex Count Decimal Count 0.0 V 0000 0000 00 0 1.0 V 0011 0011 33 51 2.2 V 0111 0000 70 112 3.51 V 1011 0011 B3 179 4.0 V 1100 1100 CC 204 5.0 V 1111 1111 FF 255 -
Question 7
An analog-to-digital converter has a 12-bit binary output and an analog input voltage range of 0.0 to + 5.0 volts. Calculate:
{\bullet} The digital output (in hexadecimal) at 0.0 volts analog input =
{\bullet} The digital output (in hexadecimal) at 5.0 volts analog input =
{\bullet} The digital output (in hexadecimal) at 2.1 volts analog input =
{\bullet} The analog input corresponding to a digital output of 2D0 =
{\bullet} The analog input corresponding to a digital output of F14 =
Reveal answer{\bullet} The digital output (in hexadecimal) at 0.0 volts analog input = 000
{\bullet} The digital output (in hexadecimal) at 5.0 volts analog input = FFF
{\bullet} The digital output (in hexadecimal) at 2.1 volts analog input = 6B7
{\bullet} The analog input corresponding to a digital output of 2D0 = 0.88 V
{\bullet} The analog input corresponding to a digital output of F14 = 4.71 V
Notes:12-bit conversion yields a max value of 4095.
First, find the ratio of actual analog input / max analog input (e.g., 2.1/5). Multiply this value by 4095 to determine the actual decimal value of the output (in this case, 1719).
Convert this decimal value to hexadecimal by dividing the final actual decimal value by 256, then dividing that remainder by 16, then using the final remainder value.
1719/256 = 6 remainder 183
183/16 = 11 (B in hex) remainder 7
Answer = 6B7 -
Question 8
An analog-to-digital converter has a 16-bit binary output and an analog input voltage range of 0.0 to + 10.0 volts. Calculate:
{\bullet} The digital output (in hexadecimal) at 0.0 volts analog input =
{\bullet} The digital output (in hexadecimal) at 10.0 volts analog input =
{\bullet} The digital output (in hexadecimal) at 3.4 volts analog input =
{\bullet} The analog input corresponding to a digital output of 3D6A =
{\bullet} The analog input corresponding to a digital output of C005 =
Reveal answer{\bullet} The digital output (in hexadecimal) at 0.0 volts analog input = 0000
{\bullet} The digital output (in hexadecimal) at 10.0 volts analog input = FFFF
{\bullet} The digital output (in hexadecimal) at 3.4 volts analog input = 5709
{\bullet} The analog input corresponding to a digital output of 3D6A = 2.40 V
{\bullet} The analog input corresponding to a digital output of C005 = 7.5 V
Notes:16-bit conversion yields a max value of 65,535.
First, find the ratio of actual analog input / max analog input (e.g., 3.4/10). Multiply this value by 65535 to determine the actual decimal value of the output (in this case, 22281).
Convert this decimal value to hexadecimal by dividing the final actual decimal value by 4096, then dividing that remainder by 256, then dividing that remainder by 16, then using the final remainder value.
22281/4096 = 5 remainder 1801
1801/256 = 7 remainder 9
9/16 = 0 remainder 9
Answer = 5709
This answer is still hex, although it contains no characters A-F.
-
Question 9
An analog-to-digital converter (ADC) has a calibrated input range of 0 to 10 volts, and a 12-bit output (0 to 4095 “count” range). Complete the following table of values for this converter, assuming perfect calibration (no error):
Input Voltage (V) Percent of Span (%) Decimal Count Hex Count 0 25 50 75 100 Reveal answerInput Voltage (V) Percent of Span (%) Decimal Count Hex Count 0 0 0 0 2.5 25 1023 3FF 5.0 50 2047 7FF 7.5 75 3071 BFF 10 100 4095 FFF -
Question 10
An essential component of any digital control system is an analog-to-digital converter, or ADC. This is necessary to convert the analog process variable measurement into a digital number for the control algorithm to process. Another essential component is a digital-to-analog converter, or DAC, which does the exact opposite.
In a system using 4-20 mA analog currents to relay instrument signals, there is an ADC located at the process variable input of the controller, and a DAC located at the output of the controller:
In a Fieldbus system, the communication is all digital, which places the ADC at the transmitter and the DAC at the transducer:
In either case, we need to “scale” the binary count of the ADC and DAC to their respective analog variable values.
Consider a flow control system, with a flow transmitter ranging from 0 to 200 GPM and a pneumatic control valve operating on a pressure range of 3 to 15 PSI (instrument air). Assuming the ADC has a resolution of 16 bits (a digital conversion range of 0000 to FFFF) and the DAC has a resolution of 14 bits (a digital conversion range of 0000 to 3FFF), complete the calibration tables for the ADC signal and for the DAC signal.
Calibration table for process variable signal (ADC)
Measurement Digital Output (Hex) 0 GPM 0000 100 GPM 200 GPM FFFF Calibration table for process variable signal (DAC)
Digital Input (Hex) Measurement 0000 0 PSI 7.5 PSI 3FFF 15 PSI Reveal answerADC calibration table:
Measurement Digital Output (Hex) 0 GPM 0000 100 GPM 7FFF 200 GPM FFFF DAC calibration table:
Digital Input (Hex) Measurement 0000 0 PSI 1FFF 7.5 PSI 3FFF 15 PSI -
Question 11
An analog-to-digital converter (ADC) has a calibrated input range of 0 to 10 volts, and a 16-bit output (0 to 65535 “count” range). Complete the following table of values for this converter, assuming perfect calibration (no error):
Input Voltage (V) Percent of Span (%) Decimal Count Hex Count 0 25 50 75 100 Reveal answerInput Voltage (V) Percent of Span (%) Decimal Count Hex Count 0 0 0 0000 2.5 25 16383 3FFF 5.0 50 32767 7FFF 7.5 75 49151 BFFF 10 100 65535 FFFF -
Question 12
Suppose you wish to calibrate an RTD temperature transmitter to an input range of 50 to 200 degrees F, with an output range of 4 to 20 mA. Complete the following calibration table showing the proper test temperatures and the ideal output signals at those levels:
Input Temp (℉) Percent of Span (%) Output Signal (mA) 0 25 50 75 100 Reveal answerInput Temp (℉) Percent of Span (%) Output Signal (mA) 50 0 4 87.5 25 8 125 50 12 162.5 75 16 200 100 20 -
Question 13
A digital pressure transmitter has a calibrated input range of 50 to 200 PSI, and a 10-bit output (0 to 1023 “count” range). Complete the following table of values for this transmitter, assuming perfect calibration (no error):
Input Pressure (PSI) Percent of Span (%) Decimal Counts Hex Counts 7 22 39 56 78 Reveal answerInput Pressure (PSI) Percent of Span (%) Decimal Counts Hex Counts 60.5 7 71 047 83 22 225 0E1 108.5 39 398 18E 134 56 572 23C 167 78 797 31D -
Question 14
A digital level transmitter has a calibrated input range of 20 to 170 inches of liquid level, and a 10-bit output (0 to 1023 “count” range). Complete the following table of values for this transmitter, assuming perfect calibration (no error):
Input Level (in) Percent of Span (%) Decimal Counts Hex Counts 11 28 55 73 92 Reveal answerInput Level (in) Percent of Span (%) Decimal Counts Hex Counts 36.5 11 112 070 62.0 28 286 11E 102.5 55 562 232 129.5 73 746 2EA 158.0 92 941 3AD -
Question 15
An analog-to-digital converter (ADC) has a calibrated input range of 0 to 5 volts, and a 12-bit output. Complete the following table of values for this converter, assuming perfect calibration (no error):
Input Voltage (V) Percent of Span (%) Decimal Counts Hex Counts 1.6 3022 40 A2F Reveal answerInput Voltage (V) Percent of Span (%) Decimal Counts Hex Counts 1.6 32 1310 51E 3.69 73.8 3022 BDE 2.0 40 1638 666 3.18 63.66 2607 A2F Notes:The fact that this is a 12-bit ADC tells us all we need to know to calculate its maximum count value:
$$\hbox{Count}_{max} = 2^{12} - 1 = 4096 - 1 = 4095$$
Therefore, the count range is 0 to 4095 counts, matching the input voltage range of 0 to 5 volts, proportional. Relating input voltage to count value is as simple as solving for this proportionality. Each count is worth ${1 \over 4095}$ of 5 volts, or 1.221 mV.
Performing calculations for the first table row, we take the given input voltage of 1.6 volts and solve for the percentage of range as well as the proportional count value:
$${1.6 \hbox{ V} \over 5 \hbox{ V}} = 32
$$N = 1310.4 \hbox{ counts}$$
Since there is no such thing as a non-integer count value, we must round down in the case of A-D converters (to 1310 counts). The last step is simply converting the decimal value of 1310 into hexadecimal.
-
Question 16
Suppose you wish to calibrate a pneumatic pressure transmitter to an input range of -10 to +50 inches of mercury, with an output range of 3 to 15 PSI. Complete the following calibration table showing the test pressures to use and the allowable low/high output signals for a calibrated tolerance of +/- 0.5%
Input Pressure (“ Hg) Percent of Span (%) Nominal Output (PSI) Min Output (PSI) Max Output (PSI) 0 25 50 75 100 Reveal answerInput Pressure (“ Hg) Percent of Span (%) Nominal Output (PSI) Min Output (PSI) Max Output (PSI) -10 0 3 2.985 3.015 5 25 6 5.97 6.03 20 50 9 8.955 9.045 35 75 12 11.94 12.06 50 100 15 14.925 15.075 -
Question 17
A tachogenerator is used to measure the rotary speed of a machine. Its calibrated range is 0 to 1500 RPM (revolutions per minute) and its corresponding signal output is 0 to 10 volts DC. Given these range values, calculate the output voltages for the following input shaft speeds, and then describe how you were able to correlate the different speeds to output voltage values:
Shaft Speed (RPM) Output Voltage (V DC) 100 350 500 750 890 975 1230 1410 1500 Reveal answerShaft Speed (RPM) Output Voltage (V DC) 100 0.67 350 2.3 500 3.3 750 5.0 890 5.9 975 6.5 1230 8.2 1410 9.4 1500 10.0 Notes:This sort of problem may be solved using simple proportions. We know that the voltage output by a tachogenerator will be directly proportional to shaft speed. In other words, doubling the shaft speed doubles the voltage. This makes the math very simple.
All we have to do is determine what percentage of full speed (1500 RPM) our given speed is, and then multiply the full-speed voltage (10 volts) by that percentage.
We may express the relationship between shaft speed ($S$) and voltage ($V$) as a proportionality:
$$S \propto V$$
We may express this relationship a little more precisely by inserting a constant of proportionality ($k$) and changing the proportional symbol to an equality:
$$S = kV$$
Now, all we need to do is solve for $k$ using a pair of simultaneous values for $S$ and $V$. Since we were told the output of the tachogenerator is 10 volts at 1500 RPM:
$$1500 = k 10$$
$$k = 150$$
Knowing that $k$ = 150 for this tachogenerator (with $S$ in units of RPM and $V$ in units of volts) allows us to solve for any speed given any voltage by way of simple multiplication:
$$S = 150 V$$
If the tachogenerator drives a voltmeter, we could even post this $k$ value on the face of the meter so that an operator could multiply the voltage measurement by 150 to obtain shaft speed in RPM.
Another way to think about this sort of problem is to use a technique we call “unity fractions”. Since we know tachogenerator speed and voltage to be proportional to one another, and we know they share the same zero point (0 RPM = 0 volts), we may form a fraction out of the equality 1500 RPM = 10 volts and use it in such a way as to cancel the unit of RPM and replace it with volts.
We know that any fraction where the numerator and denominator are the same quantity ($5 \over 5$, for example) has a value of 1, or unity. We also know that multiplying any value by 1 does not change that value (for example, $570 \times 1 = 570$). So, if we form a fraction out of two physically equal quantities—in this case, the fraction 10 volts divided by 1500 RPM, since those two quantities are physically equal to one another in the context of the tachogenerator—this “unity fraction” has a physical value of 1, and we may safely multiply it by any other value without changing that other value’s quantity. The point of doing this is to cancel out the unwanted unit and replace it with another unit.
To see how this works, let’s use this technique to calculate tachogenerator voltage at 1230 RPM:
$$\left({1230 \hbox{ RPM} \over 1} \right) \left( {10 \hbox{ volts} \over 1500 \hbox{ RPM}} \right) = 8.2 \hbox{ volts}$$
The unit of “RPM” cancels from the top of the left-hand fraction and the bottom of the right-hand fraction, leaving the unit of “volts” as the only one remaining. Since the unity fraction of $10 \hbox{ volts} \over 1500 \hbox{ RPM}$ has a physical value of 1, multiplying it by the fraction $1230 \hbox{ RPM} \over 1$ does not alter that value. We only cancel out the unit of RPM, to tell us how many volts the tachogenerator will output at that speed.
We can use this same “unity fraction” to translate from volts to RPM, too! Imagine a situation where the tachogenerator was spinning fast enough to output a voltage of 7.1 volts. How fast is this? We can use this same unit-cancellation technique to find out:
$$\left({7.1 \hbox{ volts} \over 1} \right) \left( 1500 \hbox{ RPM} \over {10 \hbox{ volts}} \right) = 1065 \hbox{ RPM}$$
Since the unity fraction of $10 \hbox{ volts} \over 1500 \hbox{ RPM}$ has a physical value of 1, we may flip it upside-down without changing its value ($5 \over 5$ is still 1, whether you flip the fraction upside-down or not!). In this case, we needed to flip the unity fraction upside-down in order to make the unit of “volts” cancel out and leave us with the unit of “RPM” as the only one left standing.
While the technique of “unity fractions” may seem over-complicated, especially for a simple proportion problem like this one, it is very powerful.
-
Question 18
Suppose you wish to calibrate an electronic pressure transmitter to an input range of 0 to 50 PSI, with an output range of 4 to 20 mA. Complete the following calibration table showing the proper test pressures and the ideal output signal levels at those pressures:
Input Pressure (PSI) Percent of Span (%) Output Signal (mA) 0 25 50 75 100 Reveal answerInput Pressure (PSI) Percent of Span (%) Output Signal (mA) 0 0 4 12.5 25 8 25 50 12 37.5 75 16 50 100 20 -
Question 19
Suppose you wish to calibrate an electronic pressure transmitter to an input range of 0 to 200 PSI, with an output range of 4 to 20 mA. Complete the following calibration table showing the proper test pressures and the ideal output signal levels at those pressures:
Input Pressure (PSI) Percent of Span (%) Output Signal (mA) 0 25 50 75 100 Reveal answerInput Pressure (PSI) Percent of Span (%) Output Signal (mA) 0 0 4 50 25 8 100 50 12 150 75 16 200 100 20 -
Question 20
In instrumentation parlance, a transducer is a calibrated device used to convert one standardized signal into another standardized signal. One very common form of transducer is an I/P transducer, which converts an electric current signal into a pneumatic pressure signal:
The symbols shown above are standard for process and instrumentation diagrams (P\&ID’s), where an electric cable is shown as a dashed line, a pneumatic pipe or tube shown as a line with double hash-marks periodically drawn through it, and the instrument is a circle with letters (in this case, “Y” represents some signal relay, computing element, transducer, or converter).
The most popular range for electric current signals is 4 to 20 mA DC. The most common range for pneumatic (air pressure) signals is 3 to 15 PSI. Therefore, the most common type of I/P transducer has an input range of 4-20 mA and an output range of 3-15 PSI. Both of these ranges are there to represent some measured or manipulated quantity in an instrument system. That is, 0
Complete all the missing data in the following calibration table for this I/P transducer, and then describe how you were able to correlate the different percentages of range with specific current and pressure signal values:
Input Current (mA) Percent of Span (%) Output Pressure (PSI) 4 0 3 10 20 25 30 40 12 50 9 60 70 75 80 90 20 100 15 Reveal answerInput Current (mA) Percent of Span (%) Output Pressure (PSI) 4 0 3 5.6 10 4.2 7.2 20 5.4 8 25 6 8.8 30 6.6 10.4 40 7.8 12 50 9 13.6 60 10.2 15.2 70 11.4 16 75 12 16.8 80 12.6 18.4 90 13.8 20 100 15 -
Question 21
Suppose you wish to calibrate a current-to-pressure (“I/P”) transducer to an output range of 3 to 15 PSI, with an input range of 4 to 20 mA. Complete the following calibration table showing the proper test pressures and the ideal input signal levels at those pressures:
Input Current (mA) Percent of Span (%) Output Pressure (PSI) 35 80 95 Reveal answerInput Current (mA) Percent of Span (%) Output Pressure (PSI) 9.6 35 7.2 16.8 80 12.6 19.2 95 14.4 -
Question 22
An electronic level transmitter has a calibrated range of 0 to 2 feet, and its output signal range is 4 to 20 mA. Complete the following table of values for this transmitter, assuming perfect calibration (no error).
Input Level (ft) Percent of Span (%) Output Signal (mA) 1.6 7.1 40 Reveal answerInput Level (ft) Percent of Span (%) Output Signal (mA) 1.6 80 16.8 0.3875 19.375 7.1 0.8 40 10.4 -
Question 23
A pneumatic level transmitter has a calibrated range of 0 to 5 feet, and its output signal range is 3 to 15 PSI. Complete the following table of values for this transmitter, assuming perfect calibration (no error).
Input Level (ft) Percent of Span (%) Output Signal (PSI) 3.2 4 50 2.4 11.3 18 Reveal answerInput Level (ft) Percent of Span (%) Output Signal (PSI) 3.2 64 10.68 0.4167 8.33 4 2.5 50 9 2.4 48 8.76 3.458 69.17 11.3 0.9 18 5.16 -
Question 24
Suppose you wish to calibrate an electronic pressure transmitter to an input range of 0 to 50 inches of water, with an output range of 4 to 20 mA. Complete the following calibration table showing the proper test pressures and the ideal output signal levels at those pressures:
Input Pressure (“ H2O) Percent of Span (%) Output Signal (mA) 5 33 61 Reveal answerInput Pressure (“ H2O) Percent of Span (%) Output Signal (mA) 2.5 5 4.8 16.5 33 9.28 30.5 61 13.76 -
Question 25
An electronic pressure transmitter has a calibrated range of 0 to 200 inches of mercury, and its output signal range is 4 to 20 mA. Complete the following table of values for this transmitter, assuming perfect calibration (no error):
Input Pressure (“ Hg) Percent of Span (%) Output Signal (mA) 24 19 11.7 Reveal answerInput Pressure (“ Hg) Percent of Span (%) Output Signal (mA) 24 12 5.92 38 19 7.04 96.25 48.13 11.7 -
Question 26
A pneumatic differential pressure transmitter has a calibrated range of -100 to +100 inches of water column (” H2O), and its output signal range is 3 to 15 PSI. Complete the following table of values for this transmitter, assuming perfect calibration (no error).
Input Pressure (“ H2O) Percent of Span (%) Output Pressure (PSI) 0 -30 8 13 65 10 Reveal answerInput Pressure (“ H2O) Percent of Span (%) Output Pressure (PSI) 0 50 9 -30 35 7.2 -16.67 41.67 8 66.67 83.33 13 30 65 10.8 -80 10 4.2 -
Question 27
A turbine flowmeter measuring cooling water for a large power generator uses an electronic circuit to convert its pickup coil pulses into a 4-20 mA analog current signal. The “K factor” for the turbine element is 99 pulses per gallon, and the 4-20 mA analog output ranges from 0 to 500 GPM flow. Complete the following table of values for this transmitter, assuming perfect calibration (no error).
Flow (GPM) Signal Frequency (Hz) Percent of Span (%) Output Signal (mA) 250 412 305 780 63 49 10 16 Reveal answerFlow (GPM) Signal Frequency (Hz) Percent of Span (%) Output Signal (mA) 250 412.5 50 12 412 679.8 82.4 17.18 184.8 305 36.79 9.915 472.7 780 94.55 19.13 315 519.8 63 14.08 245 404.3 49 11.84 187.5 309.4 37.5 10 375 618.8 75 16 Notes:The key to calculating the signal frequency is to convert the gallons per minute to gallons per second (divide by 60).
Then, multiply the number of gallons in each second by 99, which is the number of pulses provided for every gallon.
-
Question 28
Suppose you wish to calibrate a turbine flowmeter to an input range of 0 to 600 gallons per minute. The “K factor” for this turbine is 20 pulses per gallon, making the input frequency range 0 to 200 Hz for this flow range. The output signal range is 4 to 20 mA. Complete the following calibration table showing the proper test frequencies and the ideal output signals at those levels:
Flow (GPM) Signal Frequency (Hz) Percent of Span (%) Output Signal (mA) 97 25 60 400 Reveal answerFlow (GPM) Signal Frequency (Hz) Percent of Span (%) Output Signal (mA) 97 32.33 16.17 6.587 150 50 25 8 360 120 60 13.6 400 133.3 66.6 14.67 -
Question 29
Suppose you wish to calibrate a turbine flowmeter to an input range of 0 to 300 gallons per minute. The “K factor” for this turbine is 80 pulses per gallon, making the input frequency range 0 to 400 Hz for this flow range. The output signal range is 4 to 20 mA. Complete the following calibration table showing the proper test frequencies and the ideal output signals at those levels:
Flow (GPM) Signal Frequency (Hz) Percent of Span (%) Output Signal (mA) 17 130 45 300 Reveal answerFlow (GPM) Signal Frequency (Hz) Percent of Span (%) Output Signal (mA) 51 68 17 6.72 130 173.33 43.33 10.933 135 180 45 11.2 300 400 100 20 -
Question 30
Suppose you wish to calibrate a turbine flowmeter to an input range of 0 to 800 gallons per minute. The “K factor” for this turbine is 30 pulses per gallon, making the input frequency range 0 to 400 Hz for this flow range. The output signal range is 4 to 20 mA. Complete the following calibration table showing the proper test frequencies and the ideal output signals at those levels:
Flow (GPM) Signal Frequency (Hz) Percent of Span (%) Output Signal (mA) 55 25 63 700 Reveal answerFlow (GPM) Signal Frequency (Hz) Percent of Span (%) Output Signal (mA) 55 27.5 6.875 5.1 200 100 25 8 504 252 63 14.08 700 350 87.5 18 -
Question 31
An electronic pressure transmitter has a calibrated range of -10 to 60 PSI, and its output signal range is 4 to 20 mA. Complete the following table of values for this transmitter, assuming perfect calibration (no error).
Input Pressure (PSI) Percent of Span (%) Output Signal (mA) 0 -5 12 13.6 40 22 Reveal answerInput Pressure (PSI) Percent of Span (%) Output Signal (mA) 0 14.29 6.286 -5 7.143 5.143 25 50 12 32 60 13.6 18 40 10.4 5.4 22 7.52 -
Question 32
A pneumatic level transmitter has a calibrated range of 150 inches to 300 inches of liquid level, with an output range of 3 to 15 PSI. Complete the following table of values for this transmitter, assuming perfect calibration (no error):
Input Level (in) Percent of Span (%) Output Pressure (PSI) 4 7.2 189 Reveal answerInput Level (in) Percent of Span (%) Output Pressure (PSI) 156 4 3.48 202.5 35 7.2 189 26 6.12 -
Question 33
Suppose you wish to calibrate a pneumatic level transmitter to an input range of 0 to 400 inches, with an output range of 3 to 15 PSI. Complete the following calibration table showing the proper test levels and the ideal output signals at those levels:
Input Level (in) Percent of Span (%) Output Pressure (PSI) 0 25 50 75 100 Reveal answerInput Level (in) Percent of Span (%) Output Pressure (PSI) 0 0 3 100 25 6 200 50 9 300 75 12 400 100 15 -
Question 34
An ultrasonic level transmitter has a calibrated range of 40 to 75 inches and its output signal range is 4 to 20 mA. Complete the following table of values for this transmitter, assuming perfect calibration (no error).
Input Level (in) Percent of Span (%) Output Signal (mA) 47 6 75 60 15.1 34 Reveal answerInput Level (in) Percent of Span (%) Output Signal (mA) 47 20 7.2 44.38 12.5 6 66.25 75 16 60 57.14 13.14 64.28 69.38 15.1 51.9 34 9.44
