Instrumentation and Control Questions – Formulas & Equations
In this article, you will find the important instrumentation and control questions related to the formulas and equations used in industrial measurement.
Instrumentation and Control Questions
By the end of this article, you will have a better understanding of how these formulas and equations are used in real-world applications.
What is the formula for converting a temperature in degrees Celsius to degrees Fahrenheit?
A. °F = (°C * 9/5) + 32
B. °F = (°C – 32) * 5/9
C. °F = °C + 32
D. °F = °C * 5/9
Answer: A
Explanation: To convert a temperature in degrees Celsius to degrees Fahrenheit, use the formula °F = (°C * 9/5) + 32, where °C is the temperature in degrees Celsius.
What is the formula for converting pressure in pounds per square inch (PSI) to kilopascals (kPa)?
A. kPa = PSI * 6.895
B. kPa = PSI / 6.895
C. kPa = PSI * 0.145
D. kPa = PSI / 0.145
Answer: A
Explanation: To convert pressure in pounds per square inch (PSI) to kilopascals (kPa), use the formula kPa = PSI * 6.895, where PSI is the pressure in pounds per square inch.
What is the formula for converting the output signal of a level transmitter in mA to a level measurement in meters (m)?
A. Level (m) = (mA output – mA min) / (mA max – mA min) * (max level – min level) + min level
B. Level (m) = (mA output – mA min) * [(max level – min level) / (mA max – mA min)] + min level
C. Level (m) = (mA output – mA min) / (mA max – mA min) * max level
D. Level (m) = (mA output – mA min) * max level / (mA max – mA min)
Answer: B
Explanation: The formula for converting the output signal of a level transmitter in mA to a level measurement in meters (m) is
Level (m) = (mA output – mA min) * [(max level – min level) / (mA max – mA min)] + min level
Where mA output is the current output signal from the transmitter, mA max and mA min are the maximum and minimum current output signals of the transmitter, and max level and min level are the maximum and minimum levels of the process being measured.
A flow transmitter has a range of 0-100 liters per minute (LPM) and a 4-20 mA output signal. What is the flow rate in LPM if the transmitter is outputting a signal of 12 mA?
A. 40 LPM
B. 50 LPM
C. 80 LPM
D. 100 LPM
Answer: B
Explanation:
In a 4-20 mA current loop standard, 4 mA corresponds to the minimum process value (0 LPM in this case), and 20 mA corresponds to the maximum process value (100 LPM in this case).
The flow rate for any given mA signal within this range can be calculated as follows:
Determine the span of the mA range and the flow range:
mA span = 20 mA – 4 mA = 16 mA
Flow span = 100 LPM – 0 LPM = 100 LPM
Calculate the ratio of the difference between the actual mA signal and the minimum mA signal to the total mA span:
Ratio = (12 mA – 4 mA) / 16 mA = 0.5
Multiply this ratio by the flow span:
Flow rate = 0.5 * 100 LPM = 50 LPM
Therefore, if the transmitter is outputting a signal of 12 mA, the flow rate is 50 liters per minute.
A level transmitter has a range of 0-15 meters and a 4-20 mA output signal. If the transmitter is outputting a signal of 17.5 mA, what is the level measurement in meters?
A. 6.36 meters
B. 12.6 meters
C. 10.56 meters
D. 11.58 meters
Answer: B
Explanation:
In a 4-20 mA current loop standard, 4 mA corresponds to the minimum process value (0 meters in this case), and 20 mA corresponds to the maximum process value (15 meters in this case).
The level measurement for any given mA signal within this range can be calculated as follows:
Determine the span of the mA range and the level range:
mA span = 20 mA – 4 mA = 16 mA
Level span = 15 meters – 0 meters = 15 meters
Calculate the ratio of the difference between the actual mA signal and the minimum mA signal to the total mA span:
Ratio = (17.5 mA – 4 mA) / 16 mA = 0.84375
Multiply this ratio by the level span:
Level measurement = 0.84375 * 15 meters = 12.65625 meters
Therefore, if the transmitter is outputting a signal of 17.5 mA, the level measurement is approximately 12.65625 meters.
A pressure transmitter has a range of 0-56 kPa and a 4-20 mA output signal. If the transmitter is outputting a signal of 8.5 mA, what is the pressure measurement in kPa?
A. 15.75 kPa
B. 40.58 kPa
C. 20.02 kPa
D. 25.63 kPa
Answer: A
Explanation:
The pressure measurement for any given mA signal within this range can be calculated as follows:
Determine the span of the mA range and the pressure range:
mA span = 20 mA – 4 mA = 16 mA
Pressure span = 56 kPa – 0 kPa = 56 kPa
Calculate the ratio of the difference between the actual mA signal and the minimum mA signal to the total mA span:
Ratio = (8.5 mA – 4 mA) / 16 mA = 0.28125
Multiply this ratio by the pressure span:
Pressure measurement = 0.28125 * 56 kPa = 15.75 kPa
Therefore, if the transmitter is outputting a signal of 8.5 mA, the pressure measurement is 15.75 kPa.
A temperature transmitter using a 3-wire RTD has a resistance of 105.8 ohms at 0°C and 148.5 ohms at 100°C. What is the temperature measurement in °C if the transmitter is currently measuring a resistance of 110.2 ohms?
A. 10.3°C
B. 25.3°C
C. 50.8°C
D. 75.5°C
Answer: A
Explanation:
The temperature-resistance relationship of an RTD is typically approximated as linear over small temperature ranges. The resistance value of an RTD increases with temperature, and the relationship is typically defined at two points: 0°C and another temperature, in this case, 100°C.
First, calculate the resistance change per degree Celsius:
Resistance change per degree = (Resistance at 100°C – Resistance at 0°C) / (100°C – 0°C)
Resistance change per degree = (148.5 Ohms – 105.8 Ohms) / 100
Resistance change per degree = 0.427 Ohms/°C
Then, to find the temperature corresponding to a resistance of 110.2 Ohms, you subtract the 0°C resistance from 110.2 Ohms and divide by the resistance change per degree:
Temperature = (Measured Resistance – Resistance at 0°C) / Resistance change per degree
Temperature = (110.2 Ohms – 105.8 Ohms) / 0.427 Ohms/°C
Temperature = ~10.3°C
So, if the transmitter is currently measuring a resistance of 110.2 Ohms, the temperature measurement is approximately 10.3°C.
A pressure transmitter has a range of 0-350 psi and a 4-20 mA output signal. What is the output signal in mA if the pressure measurement is 100 psi?
A. 10.47 mA
B. 12.25 mA
C. 8.57 mA
D. 16.89 mA
Answer: C
Explanation:
The mA output for any given pressure within this range can be calculated as follows:
Determine the span of the mA and pressure ranges:
mA span = 20 mA – 4 mA = 16 mA
Pressure span = 350 psi – 0 psi = 350 psi
Calculate the ratio of the difference between the desired pressure and the minimum pressure to the pressure span:
Ratio = (100 psi – 0 psi) / 350 psi = 0.2857
Multiply this ratio by the mA span and add the minimum current (4 mA):
mA output = (0.2857 * 16 mA) + 4 mA = 8.5714 mA
So, the mA output at a pressure of 100 psi is approximately 8.57 mA.
A level transmitter using a displacement-type sensor has a range of 0-50 inches and a 4-20 mA output signal. If the transmitter is currently measuring a displacement of 30 inches, what is the output signal in mA?
A. 8.6 mA
B. 12.6 mA
C. 13.6 mA
D. 20.6 mA
Answer: C
Explanation:
In the scenario of a 4-20 mA current loop system, the minimum value of the process (0 inches for this level transmitter) is represented by 4 mA, while the maximum value (50 inches in this case) is denoted by 20 mA.
To find the corresponding mA output for a specific measurement (like 30 inches), we follow these steps:
Calculate the range for both mA and the measurement:
The difference in mA = 20 mA – 4 mA = 16 mA
The difference in level measurement = 50 inches – 0 inches = 50 inches
Identify the proportion of the actual measurement to the maximum measurement:
The proportion = (30 inches – 0 inches) / 50 inches = 0.6
Apply this proportion to the mA range, then add the minimum current of 4 mA:
Current output = (0.6 * 16 mA) + 4 mA = 13.6 mA
Thus, for a level displacement of 30 inches, the level transmitter would provide a current output of 13.6 mA.
A temperature transmitter using a thermocouple has an output voltage of 10.5 mV. If the reference junction is at 25°C and the hot junction is at 125°C, what is the type of thermocouple being used?
A. J-type
B. K-type
C. T-type
D. E-type
Answer: B
Explanation: Using the standard thermocouple reference tables, we can determine the type of thermocouple based on the output voltage and the temperature difference between the hot and reference junctions. For a voltage of 10.5 mV and a temperature difference of 100°C, we can see that this corresponds to a K-type thermocouple.
A flow transmitter using a magnetic flow meter has a range of 0-200 GPM and a 4-20 mA output signal. If the transmitter is currently measuring a flow rate of 45 GPM, what is the output signal in mA?
A. 8.5 mA
B. 12.4 mA
C. 7.6 mA
D. 20.3 mA
Answer: C
Explanation:
The process of figuring out the output signal in mA for a specific flow rate, in a system that uses a 4-20 mA current loop protocol, involves a few steps. For this particular magnetic flow transmitter, it’s set up so that 4 mA corresponds to 0 GPM and 20 mA is equivalent to 200 GPM.
Let’s break down how to get the output signal for a flow rate of 45 GPM:
First, we have to calculate the total range of both the mA signal and the flow rate:
The total range for the mA signal is 20 mA – 4 mA, which equals 16 mA.
The total range for the flow rate is 200 GPM – 0 GPM, which equals 200 GPM.
Next, we compute what fraction the actual flow rate is of the total flow rate:
This ratio is (45 GPM – 0 GPM) / 200 GPM, which equals 0.225.
The final step is to apply this ratio to the total mA range and add the minimum mA value (4 mA):
The resulting mA output is (0.225 * 16 mA) + 4 mA, which equals 7.6 mA.
So, with a flow rate of 45 GPM, the magnetic flow transmitter would produce an output signal of approximately 7.6 mA.
A temperature transmitter using a thermocouple has an output voltage of 10 mV. If the reference junction is at 25°C and the hot junction is at an unknown temperature, what is the temperature of the hot junction if the type of thermocouple being used is a K-type?
A. 255.5°C
B. 300.5°C
C. 350.5°C
D. 209.5°C
Answer: D
Explanation:
To solve this problem, we need to understand the voltage-to-temperature conversion characteristics for a type K thermocouple.
The Seebeck coefficient (voltage-to-temperature conversion factor) for a type K thermocouple is roughly 54.2 µV/°C. To find the temperature differential, we convert the output voltage to µV and divide by the Seebeck coefficient:
10 mV = 10,000 µV
Temperature differential = 10,000 µV / 54.2 µV/°C ≈ 184.5°C
This gives the temperature difference between the hot junction and the reference junction.
The temperature of the hot junction is then the sum of the reference junction temperature and the temperature differential:
Hot junction temperature = 25°C + 184.5°C = 209.5°C
So, the hot junction is approximately at 209.5°C. Please note that this is a rough estimation, and the actual value could vary slightly due to factors such as non-linearity of the thermocouple and other sources of error.
A level transmitter using a differential pressure sensor has a range of 0-40 inches of water column (inWC) and a 4-20 mA output signal. If the transmitter is currently measuring a differential pressure of 28.5 inWC, what is the output signal in mA?
A. 15.4 mA
B. 12.4 mA
C. 10.4 mA
D. 19.4 mA
Answer: A
Explanation:
To calculate the corresponding current output of a differential pressure sensor, you’d apply the process in the 4-20mA current loop standard. This standard sets 4mA as the lower limit (0 inches of water column, or inWC, in this case), and 20mA as the upper limit (40 inWC for this sensor).
Here are the steps to find the current output:
Compute the spans of the current and level:
Current span = 20mA – 4mA = 16mA
Level span = 40 inWC – 0 inWC = 40 inWC
Determine the ratio of the current differential pressure to the level span:
Ratio = (28.5 inWC – 0 inWC) / 40 inWC = 0.7125
Multiply this ratio by the current span and add the minimum current (4mA):
Current output = (0.7125 * 16mA) + 4mA = 15.4mA
In conclusion, if the level transmitter is measuring a differential pressure of 28.5 inches of water column, the output signal would be approximately 15.4 mA.
A flow transmitter using an orifice plate has a range of 0-1000 GPM and a 4-20 mA output signal. If the transmitter is currently measuring a differential pressure of 50 psi across the orifice plate, what is the output signal in mA?
A. 9.4 mA
B. 15.8 mA
C. 4.8 mA
D. 8.6 mA
Answer: C
Explanation:
The exact relationship between the differential pressure across an orifice plate and the corresponding flow rate can be complex, depending on factors such as fluid properties, orifice diameter, pipeline diameter, and the shape of the orifice. Generally, the flow rate is proportional to the square root of the differential pressure.
Without the detailed specifications of the orifice plate and the flow characteristics, it would be difficult to give an exact mA output corresponding to a 50 psi differential pressure.
However, if we assume a linear relationship between flow and pressure (which is a simplification and not typically accurate in real-world conditions), we can estimate as follows:
Determine the span for the mA output:
mA span = 20 mA – 4 mA = 16 mA
Assuming a linear relationship between pressure and flow rate, and considering the maximum pressure as 1000 psi (as inferred from the flow rate), calculate the ratio of the current differential pressure to the total span:
Ratio = 50 psi / 1000 psi = 0.05
Multiply this ratio by the span of the mA output and add the minimum mA output (4 mA):
mA output = (0.05 * 16 mA) + 4 mA = 4.8 mA
So, under these assumptions, a differential pressure of 50 psi might correspond to an output signal of approximately 4.8 mA. Please note, this is a simplification, and actual output may vary. To get the exact output, more details about the orifice plate and fluid properties are required.
An RTD temperature sensor has a resistance of 110 Ω at 0°C and a resistance of 121 Ω at 100°C. What is the temperature of the sensor if its resistance is measured to be 115 Ω?
A. 27.7°C
B. 45.5°C
C. 60.9°C
D. 72.2°C
Answer: B
Explanation:
To determine the temperature of an RTD temperature sensor when its resistance is measured to be 115 Ω, we can use the concept of linear interpolation between the two known data points: resistance at 0°C and resistance at 100°C.
Given:
Resistance at 0°C = 110 Ω
Resistance at 100°C = 121 Ω
Measured Resistance = 115 Ω
We can set up a proportion to find the unknown temperature:
(115 Ω – 110 Ω) / (121 Ω – 110 Ω) = (Unknown Temperature – 0°C) / (100°C – 0°C)
Simplifying the equation:
5 Ω / 11 Ω = Unknown Temperature / 100°C
Cross-multiplying:
11 Ω * Unknown Temperature = 5 Ω * 100°C
Dividing both sides by 11 Ω:
Unknown Temperature = (5 Ω * 100°C) / 11 Ω
Calculating:
Unknown Temperature ≈ 45.45°C
Therefore, when the resistance of the RTD temperature sensor is measured to be 115 Ω, the corresponding temperature is approximately 45.45°C.
A level transmitter using a hydrostatic pressure sensor has a range of 0-1000 inches of water column (inWC) and a 4-20 mA output signal. If the specific gravity of the liquid being measured is 0.8 and the transmitter is currently measuring a hydrostatic pressure of 800 inWC, what is the output signal in mA?
A. 10.42 mA
B. 14.24 mA
C. 16.84 mA
D. 18.45 mA
Answer: B
Explanation:
To calculate the output signal in mA for a level transmitter using a hydrostatic pressure sensor, we need to consider the relationship between the hydrostatic pressure, specific gravity, and the mA output.
The hydrostatic pressure can be calculated using the equation:
Hydrostatic pressure (in inches of water column) = Height of liquid (in inches) * Specific gravity
Let’s calculate the hydrostatic pressure using the given specific gravity of 0.8 and a height of liquid of 800 inches:
Hydrostatic pressure = 800 in * 0.8 = 640 inWC
Now, we can determine the mA output based on the range of the transmitter and the hydrostatic pressure:
Determine the span of the mA output and the hydrostatic pressure:
mA span = 20 mA – 4 mA = 16 mA
Pressure span = 1000 inWC – 0 inWC = 1000 inWC
Calculate the ratio of the hydrostatic pressure to the pressure span:
Ratio = 640 inWC / 1000 inWC = 0.64
Multiply this ratio by the mA span and add the minimum mA output:
mA output = (0.64 * 16 mA) + 4 mA = 14.24 mA
Therefore, with a hydrostatic pressure of 800 inWC and a specific gravity of 0.8, the output signal of the level transmitter would be approximately 14.24 mA.
A flow transmitter using a differential pressure sensor has a range of 0-500 inches of water column (inWC). If the transmitter is currently measuring a differential pressure of 200 inWC, what is the output signal in mA? Also, calculate it in percentage?
A. 20%
B. 40%
C. 60%
D. 56%
Answer: B
Explanation:
To determine the output signal in mA for a flow transmitter measuring a differential pressure of 200 inWC within a range of 0-500 inWC, we need to consider the linear relationship between the differential pressure and the output signal in mA.
The output signal can be calculated using the following formula:
mA output = (Differential Pressure / Full Scale Range) * (Maximum mA – Minimum mA) + Minimum mA
Substituting the given values:
mA output = (200 inWC / 500 inWC) * (20 mA – 4 mA) + 4 mA
mA output = 0.4 * 16 mA + 4 mA
mA output = 6.4 mA + 4 mA
mA output = 10.4 mA
Therefore, with a differential pressure of 200 inWC, the output signal of the flow transmitter is approximately 10.4 mA.
To calculate the output signal as a percentage of full scale (%FS), we can use the formula:
%FS = (Output Signal – Minimum mA) / (Maximum mA – Minimum mA) * 100
Substituting the values:
%FS = (10.4 mA – 4 mA) / (20 mA – 4 mA) * 100
%FS = 6.4 mA / 16 mA * 100
%FS = 40%
Therefore, the output signal of the flow transmitter, expressed as a percentage of full scale (%FS), is 40%.
A level transmitter using a capacitance sensor has a range of 0-10 feet and a 4-20 mA output signal. If the transmitter is currently measuring a level of 6 feet, what is the output signal in mA?
A. 10.4 mA
B. 12.8 mA
C. 18.6 mA
D. 13.6 mA
Answer: D
Explanation:
To determine the output signal in mA for a level transmitter measuring a level of 6 feet within a range of 0-10 feet using a capacitance sensor, we can use the following formula:
mA output = [((Level – Minimum Level) / (Maximum Level – Minimum Level)) * (Maximum mA – Minimum mA)] + Minimum mA
Substituting the given values:
Level = 6 feet
Minimum Level = 0 feet
Maximum Level = 10 feet
Minimum mA = 4 mA
Maximum mA = 20 mA
mA output = ((6 feet – 0 feet) / (10 feet – 0 feet)) * (20 mA – 4 mA) + 4 mA
mA output = (6 feet / 10 feet) * 16 mA + 4 mA
mA output = 0.6 * 16 mA + 4 mA
mA output = 9.6 mA + 4 mA
mA output = 13.6 mA
Therefore, with a level of 6 feet, the output signal of the level transmitter is approximately 13.6 mA.
A temperature transmitter using a thermocouple sensor has a range of 0-500°C and a 4-20 mA output signal. If the transmitter is currently measuring a temperature of 320°C, what is the output signal in percent (%FS)?
A. 33%
B. 50%
C. 77%
D. 64%
Answer: D
Explanation:
To determine the output signal in mA for a temperature transmitter measuring a temperature of 320°C within a range of 0-500°C using a thermocouple sensor, we can use the following formula:
mA output = ((Temperature – Minimum Temperature) / (Maximum Temperature – Minimum Temperature)) * (Maximum mA – Minimum mA) + Minimum mA
Substituting the given values:
Temperature = 320°C
Minimum Temperature = 0°C
Maximum Temperature = 500°C
Minimum mA = 4 mA
Maximum mA = 20 mA
mA output = ((320°C – 0°C) / (500°C – 0°C)) * (20 mA – 4 mA) + 4 mA
mA output = (320°C / 500°C) * 16 mA + 4 mA
mA output = 0.64 * 16 mA + 4 mA
mA output = 10.24 mA + 4 mA
mA output = 14.24 mA
Therefore, with a temperature of 320°C, the output signal of the temperature transmitter is approximately 14.24 mA.
To calculate the output signal as a percentage of full scale (%FS), we can use the formula:
%FS = ((mA output – Minimum mA) / (Maximum mA – Minimum mA)) * 100
Substituting the values:
%FS = ((14.24 mA – 4 mA) / (20 mA – 4 mA)) * 100
%FS = (10.24 mA / 16 mA) * 100
%FS = 64%
Therefore, the output signal of the temperature transmitter, expressed as a percentage of full scale (%FS), is 64%.
A pressure transmitter has a range of 0-110 psi and a 4-20 mA output signal. If the transmitter is currently measuring a pressure of 70.5 psi, what is the output signal in milliamps (mA)?
A. 12.8 mA
B. 16.0 mA
C. 14.3mA
D. 20.0 mA
Answer: C
Explanation:
To determine the output signal in milliamps (mA) for a pressure transmitter measuring a pressure of 70.5 psi within a range of 0-110 psi using a 4-20 mA output signal, we can use the following formula:
mA output = ((Pressure – Minimum Pressure) / (Maximum Pressure – Minimum Pressure)) * (Maximum mA – Minimum mA) + Minimum mA
Substituting the given values:
Pressure = 70.5 psi
Minimum Pressure = 0 psi
Maximum Pressure = 110 psi
Minimum mA = 4 mA
Maximum mA = 20 mA
mA output = ((70.5 psi – 0 psi) / (110 psi – 0 psi)) * (20 mA – 4 mA) + 4 mA
mA output = (70.5 psi / 110 psi) * 16 mA + 4 mA
mA output = 0.6418 * 16 mA + 4 mA
mA output = 10.3088 mA + 4 mA
mA output = 14.3088 mA
Therefore, with a pressure of 70.5 psi, the output signal of the pressure transmitter is approximately 14.3088 mA.
A flow transmitter using a magnetic flowmeter has a range of 0-1000 GPM and a 4-20 mA output signal. If the transmitter is currently measuring a flow rate of 950 GPM, what is the output signal in percent (%FS)?
A. 98%
B. 91%
C. 99%
D. 95%
Answer: D
Explanation:
To determine the output signal in mA for a flow transmitter measuring a flow rate of 950 GPM within a range of 0-1000 GPM using a magnetic flowmeter with a 4-20 mA output signal, we can use the following formula:
mA output = ((Flow Rate – Minimum Flow Rate) / (Maximum Flow Rate – Minimum Flow Rate)) * (Maximum mA – Minimum mA) + Minimum mA
Substituting the given values:
Flow Rate = 950 GPM
Minimum Flow Rate = 0 GPM
Maximum Flow Rate = 1000 GPM
Minimum mA = 4 mA
Maximum mA = 20 mA
mA output = ((950 GPM – 0 GPM) / (1000 GPM – 0 GPM)) * (20 mA – 4 mA) + 4 mA
mA output = (950 GPM / 1000 GPM) * 16 mA + 4 mA
mA output = 0.95 * 16 mA + 4 mA
mA output = 15.2 mA + 4 mA
mA output = 19.2 mA
Therefore, with a flow rate of 950 GPM, the output signal of the flow transmitter is approximately 19.2 mA.
To calculate the output signal as a percentage of full scale (%FS), we can use the formula:
%FS = ((mA output – Minimum mA) / (Maximum mA – Minimum mA)) * 100
Substituting the values:
%FS = ((19.2 mA – 4 mA) / (20 mA – 4 mA)) * 100
%FS = (15.2 mA / 16 mA) * 100
%FS = 95%
Therefore, the output signal of the flow transmitter, expressed as a percentage of full scale (%FS), is 95%.
What is the formula for calculating the level (L) in a cylindrical tank with a diameter (D) of 2 meters, a height (H) of 4 meters, and a volume (V) of 25 cubic meters?
A. L = V / (π * (D/2)^2)
B. L = V / (π * D^2 * H / 4)
C. L = V / (π * D * H^2 / 4)
D. L = V / (π * D^2 / 4)
Answer: A
Explanation: The level (L) in a cylindrical tank with a diameter (D) of 2 meters, a height (H) of 4 meters, and a volume (V) of 25 cubic meters can be calculated using the formula L = V / (π * (D/2)^2), where D is the diameter in meters, H is the height in meters, and V is the volume in cubic meters.
What is the formula for converting a flow rate reading in liters per minute (LPM) to gallons per minute (GPM)?
A. GPM = LPM / 3.785
B. GPM = LPM * 0.264
C. GPM = LPM / 0.264
D. GPM = LPM * 3.785
Answer: B
Explanation: To convert a flow rate reading in liters per minute (LPM) to gallons per minute (GPM), use the formula GPM = LPM * 0.264.
What is the formula for calculating the resistance of a thermistor?
A. R = V/I
B. R = (V^2)/P
C. R = (V/I)e^(B/T)
D. R = V/P
Answer: C
Explanation: The resistance of a thermistor can be calculated using the Steinhart-Hart equation: R = R0e^(B/T), where R0 is the resistance at a known temperature, B is a constant, T is the temperature in Kelvin, and e is the natural logarithm base.
What is the formula for calculating the pressure (P) in a system with a force (F) of 200 N and an area (A) of 0.5 m²?
A. P = F/A
B. P = A/F
C. P = F* A
D. P = A/ F
Answer: A
Explanation: In a system with a force (F) of 200 N and an area (A) of 0.5 m², the pressure (P) can be calculated using the formula P = F/A, where F is the force in newtons and A is the area in square meters.
What is the formula for calculating the volumetric flow rate (Q) in a system with a velocity (v) of 3 m/s and a cross-sectional area (A) of 2 m²?
A. Q = vA
B. Q = A/v
C. Q = v/A
D. Q = Av
Answer: D
Explanation: In a system with a velocity (v) of 3 m/s and a cross-sectional area (A) of 2 m², the volumetric flow rate (Q) can be calculated using the formula Q = A*v, where A is the cross-sectional area in square meters.
What is the formula for calculating the level (L) in a cylindrical tank with a diameter (D) of 2 meters and a measured height (h) of 3 meters?
A. L = h
B. L = D + h
C. L = (h/D)2
D. L = (h/D)1
Answer: D
Explanation: In a cylindrical tank with a diameter (D) of 2 meters and a measured height (h) of 3 meters, the level (L) can be calculated using the formula L = (h/D)*1.
What is the formula for calculating the output voltage (Vout) of a 4-20 mA pressure transmitter with a range of 0-100 psi and a supply voltage (Vs) of 24 VDC?
A. Vout = Vs(I – 4)/16
B. Vout = Vs(I – 20)/(-16)
C. Vout = Vs*(I – 4)/(16100)
D. Vout = Vs(I – 20)/(16100)
Answer: C
Explanation: In a 4-20 mA pressure transmitter with a range of 0-100 psi and a supply voltage (Vs) of 24 VDC, the output voltage (Vout) can be calculated using the formula Vout = Vs(I – 4)/(16*100), where I is the current in mA.
What is the formula for converting a flow rate in gallons per minute (GPM) to liters per minute (LPM)?
A. LPM = GPM * 3.785
B. LPM = GPM / 3.785
C. LPM = GPM * 0.264
D. LPM = GPM + 3.785
Answer: A
Explanation: To convert a flow rate in gallons per minute (GPM) to liters per minute (LPM), use the formula LPM = GPM * 3.785, where GPM is the flow rate in gallons per minute.
What is the formula for calculating the level (L) in a rectangular tank with a length (L) of 5 meters, a width (W) of 3 meters, and a measured height (h) of 2 meters?
A. L = h
B. L = W + h
C. L = L + h
D. L = (h/L)5(W)
Answer: D
Explanation: In a rectangular tank with a length (L) of 5 meters, a width (W) of 3 meters, and a measured height (h) of 2 meters, the level (L) can be calculated using the formula L = (h/L)5(W).
What is the formula for calculating the volumetric flow rate (Q) in a system with a velocity (v) of 5 m/s and a pipe diameter (D) of 0.5 meters?
A. Q = (πD^2/4)v
B. Q = v/D
C. Q = D/v
D. Q = (D^2/4)v
Answer: A
Explanation: In a system with a velocity (v) of 5 m/s and a pipe diameter (D) of 0.5 meters, the volumetric flow rate (Q) can be calculated using the formula Q = (πD^2/4)*v, where D is the pipe diameter in meters.
What is the formula for calculating the percentage (%) of a measured value (M) in relation to a range (R) of 0-100 units?
A. % = (M / R) * 100
B. % = (R / M) * 100
C. % = (M / R)
D. % = (R / M)
Answer: A
Explanation: To calculate the percentage (%) of a measured value (M) in relation to a range (R) of 0-100 units, use the formula % = (M / R) * 100, where M is the measured value and R is the range.
What is the formula for converting a flow rate reading in cubic meters per hour (m3/h) to gallons per minute (GPM)?
A. GPM = m3/h * 4.402
B. GPM = m3/h / 0.227
C. GPM = m3/h * 0.227
D. GPM = m3/h / 4.402
Answer: B
Explanation: To convert a flow rate reading in cubic meters per hour (m3/h) to gallons per minute (GPM), use the formula GPM = m3/h / 0.227.
What is the formula for converting the output signal of a flow transmitter in mA to a flow rate in liters per minute (L/min)?
A. Flow rate (L/min) = (mA output – mA min) / (mA max – mA min) * (max flow rate – min flow rate) + min flow rate
B. Flow Rate (L/min) = ((mA output – mAmin) / (mAmax – mAmin)) * (Flowmax – Flowmin) + Flowmin
C. Flow rate (L/min) = (mA output – mA min) / (mA max – mA min) * max flow rate
D. Flow rate (L/min) = (mA output – mA min) * max flow rate / (mA max – mA min)
Answer: B
Explanation:
To convert the output signal of a flow transmitter in mA to a flow rate in liters per minute (L/min), you typically need the calibration parameters of the specific flow transmitter, which include the flow range and the calibration curve equation.
Assuming a linear calibration curve, the formula can be expressed as:
Flow Rate (L/min) = ((mA output – mAmin) / (mAmax – mAmin)) * (Flowmax – Flowmin) + Flowmin
Where:
- mA output is the current output of the flow transmitter in mA
- mAmin and mAmax are the minimum and maximum mA values of the transmitter’s range
- Flowmin and Flowmax are the corresponding minimum and maximum flow rates of the transmitter’s range
However, it’s important to note that this formula assumes a linear relationship between the mA output and the flow rate, which may not always be accurate.
Different flow transmitters may have different calibration curves and non-linear characteristics. Therefore, it is crucial to refer to the manufacturer’s documentation or consult the specific calibration curve provided for accurate conversion from mA to flow rate.
What is the formula for calculating the pressure (P) in a system with a force (F) of 1000 newtons and an area (A) of 2 square meters?
A. P = F / A
B. P = A / F
C. P = F * A
D. P = A * F
Answer: A
Explanation: In a system with a force (F) of 1000 newtons and an area (A) of 2 square meters, the pressure (P) can be calculated using the formula P = F / A, where F is the force in newtons and A is the area in square meters.
What is the formula for converting a flow rate in gallons per minute (GPM) to cubic meters per hour (m3/hr)?
A. m3/hr = GPM / 13.21
B. m3/hr = GPM * 0.227
C. m3/hr = GPM / 158.5
D. m3/hr = GPM * 0.063
Answer: A
Explanation: To convert a flow rate in gallons per minute (GPM) to cubic meters per hour (m3/hr), use the formula m3/hr = GPM / 13.21, where GPM is the flow rate in gallons per minute.
What is the formula for calculating the frequency of a vibrating wire sensor?
A. f = (1/(2L))sqrt(T/μ)
B. f = (1/L)*sqrt(T/μ)
C. f = (2/L)sqrt(T/μ)
D. f = (1/(2L))sqrt(μ/T)
Answer: A
Explanation: The frequency of a vibrating wire sensor can be calculated using the formula f = (1/(2L))*sqrt(T/μ), where L is the length of the wire, T is the tension in the wire, and μ is the mass per unit length of the wire.
What is the formula for calculating the pressure of a fluid in a closed vessel?
A. P = F/A
B. P = ρgh
C. P = (mg)/(Acosθ)
D. P = (F*A)/V
Answer: B
Explanation: The pressure of a fluid in a closed vessel can be calculated using the formula P = ρgh, where ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth of the fluid.
What is the formula for calculating the flow rate of a fluid through an orifice plate?
A. Q = KCdA(2gh)^0.5
B. Q = Av
C. Q = CdA(2gh)^0.5
D. Q = (2gh)^0.5
Answer: C
Explanation: The flow rate of a fluid through an orifice plate can be calculated using the formula Q = CdA(2gh)^0.5, where Cd is the discharge coefficient, A is the area of the orifice, g is the acceleration due to gravity, and h is the difference in pressure between the upstream and downstream sides of the orifice.
What is the formula for converting the resistance of a 100-ohm copper RTD to a temperature reading in Celsius (°C)?
A. T = (Rt – R0) / (0.00428 * R0)
B. T = (Rt / R0 – 1) / 0.00428
C. T = (Rt – R0) / (0.00428 * Rt)
D. T = (Rt / R0 – 1) / (0.00428 * R0)
Answer: A
Explanation: To convert the resistance of a 100-ohm copper RTD to a temperature reading in Celsius (°C), use the formula T = (Rt – R0) / (0.00428 * R0), where Rt is the measured resistance and R0 is the resistance at 0°C.
What is the formula for calculating the temperature coefficient of resistance (TCR) for a 100-ohm copper RTD?
A. TCR = (R100 – R0) / (100 * R0 * T)
B. TCR = (R100 – R0) / (100 * R0 * (T – 273))
C. TCR = (R100 / R0 – 1) / (100 * T)
D. TCR = (R100 / R0 – 1) / (100 * (T – 273))
Answer: A
Explanation: The formula for calculating the temperature coefficient of resistance (TCR) for a 100-ohm copper RTD is TCR = (R100 – R0) / (100 * R0 * T), where R100 is the resistance at 100°C, R0 is the resistance at 0°C, and T is the temperature in Celsius (°C).
What is the formula for calculating the temperature difference between two RTDs with different resistance values?
A. ΔT = (R1 – R2) / (R0 * α)
B. ΔT = (R1 / R2 – 1) / (R0 * α)
C. ΔT = (R1 – R2) / (R0 * (α1 + α2) / 2)
D. ΔT = (R1 / R2 – 1) / (R0 * (α1 + α2) / 2)
Answer: C
Explanation: The formula for calculating the temperature difference between two RTDs with different resistance values is ΔT = (R1 – R2) / (R0 * (α1 + α2) / 2), where R1 and R2 are the resistance values of the two RTDs at a specific temperature, R0 is the reference resistance at 0°C, and α1 and α2 are the temperature coefficients of resistance for the two RTDs.
What is the formula for calculating the resistance of a thermistor at a specific temperature?
A. R = R0 * exp(B * (1/T – 1/T0))
B. R = R0 * exp(B * (T – T0))
C. R = R0 * (1 + B * (T – T0))
D. R = R0 * (1 + B * (1/T – 1/T0))
Answer: A
Explanation: The formula for calculating the resistance of a thermistor at a specific temperature is R = R0 * exp(B * (1/T – 1/T0)), where R is the resistance at the specific temperature T, R0 is the resistance at the reference temperature T0, and B is a constant that depends on the type.
What is the formula for converting the resistance of a 100-ohm platinum RTD to a temperature reading in Celsius (°C)?
A. T = (Rt – R0) / (0.00385 * R0)
B. T = (Rt / R0 – 1) / 0.00385
C. T = (Rt – R0) / (0.00385 * Rt)
D. T = (Rt / R0 – 1) / (0.00385 * R0)
Answer: A
Explanation: To convert the resistance of a 100-ohm platinum RTD to a temperature reading in Celsius (°C), use the formula T = (Rt – R0) / (0.00385 * R0), where Rt is the measured resistance and R0 is the resistance at 0°C.
What is the formula for calculating the temperature coefficient of resistance (TCR) for a 100-ohm platinum RTD?
A. TCR = (R100 – R0) / (100 * R0 * T)
B. TCR = (R100 – R0) / (100 * R0 * (T – 273))
C. TCR = (R100 / R0 – 1) / (100 * T)
D. TCR = (R100 / R0 – 1) / (100 * (T – 273))
Answer: A
Explanation: The formula for calculating the temperature coefficient of resistance (TCR) for a 100-ohm platinum RTD is TCR = (R100 – R0) / (100 * R0 * T), where R100 is the resistance at 100°C, R0 is the resistance at 0°C, and T is the temperature in Celsius (°C).
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Thank you. Good Share.