4-20mA Conversions
Generally it is possible to generate a “custom” linear equation in the form of y = mx + b for any linear responding instrument relating input directly to output, a more general approach may be used to relate input to output values by translating all values into (and out of) per unit quantities. A “per unit” quantity is simply a ratio between a given quantity and its maximum value. A half-full glass of water could thus be described as having a fullness of 0.5 per unit. The concept of percent (“per one hundred”) is very similar, the only difference between per unit and percent being the base value of comparison: half-full glass of water has a fullness of 0.5 per unit (i.e. 1/2 of the glass’s full capacity), which is the same thing as 50 percent (i.e. 50 on a scale of 100, with 100 representing complete fullness).
Let’s now apply this concept to a realistic 4-20 mA signal application. Suppose you were given a liquid level transmitter with an input measurement range of 15 to 85 inches and an output range of 4 to 20 milliamps, respectively, and you desired to know how many milliamps this transmitter should output at a measured liquid level of 32 inches. Both the measured level and the milliamp signal may be expressed in terms of per unit ratios, as shown by the following graphs:
So long as we choose to express process variable and analog signal values as a per unit ratios ranging from 0 to 1, we see how m (the slope of the line) is simply equal to the span of the process variable or analog signal range, and b is simply equal to the lower-range value (LRV) of the process variable or analog signal range. The advantage of thinking in terms of “per unit” is the ability to quickly and easily write linear equations for any given range.
In fact, this is so easy that we don’t even have to use a calculator to compute m in most cases, and we never have to calculate b because the LRV is explicitly given to us. The instrument’s input equation is y = 70x+15 because the span of the 15 to 85 inch range is 70, and the LRV is 15. The instrument’s output equation is y = 16x+4 because the span of the 4-to-20 milliamp range is 16, and the LRV is 4.
If we manipulate each of the y = mx + b equations to solve for x (per unit of span), we may express the relationship between the input and output of any linear instrument as a pair of fractions with the per unit value serving as the proportional link between input and output:
The question remains, how do we apply these equations to our example problem: calculating the milliamp value corresponding to a liquid level of 32 inches for this instrument? The answer to this question is that we must perform a two-step calculation: first, convert 32 inches into a per unit ratio, then convert that per unit ratio into a milliamp value.
First, the conversion of inches into a per unit ratio, knowing that 32 is the value of y and we need to solve for x:
32 = 70x + 15
32 − 15 = 70x
x = 0.2429 per unit (i.e. 24.29%)
Next, converting this per unit ratio into a corresponding milliamp value, knowing that y will now be the current signal value using m and b constants appropriate for the 4-20 milliamp range:
y = 16x + 4
y = 16(0.2429) + 4
y = 3.886 + 4
y = 7.886 mA
y = 16(0.2429) + 4
y = 3.886 + 4
y = 7.886 mA
Instead of deriving a single custom y = mx+b equation directly relating input (inches) to output (milliamps) for every instrument we encounter, we may use two simple and generic linear equations to do the calculation in two steps with “per unit” being the intermediate result. Expressed in general form, our linear equation is:
y = mx + b
Value = (Span)(Per unit) + LRV
Value = (URV − LRV)(Per unit) + LRV
Thus, to find the per unit ratio we simply take the value given to us, subtract the LRV of its range, and divide by the span of its range. To find the corresponding value we take this per unit ratio, multiply by the span of the other range, and then add the LRV of the other range.
Example 1:
Given a pressure transmitter with a measurement range of 150 to 400 PSI and a signal range of 4 to 20 milliamps, calculate the applied pressure corresponding to a signal of 10.6 milliamps.
Also Read : Formula for Linear % to Square root % conversion
Solution:
Take 10.6 milliamps and subtract the LRV (4 milliamps), then divide by the span (16 milliamps) to arrive at 41.25% (0.4125 per unit). Take this number and multiply by the span of the pressure range (400 PSI − 150 PSI, or 250 PSI) and lastly add the LRV of the pressure range (150 PSI) to arrive at a final answer of 253.125 PSI.
Example 2:
Given a temperature transmitter with a measurement range of −88 degrees to +145 degrees and a signal range of 4 to 20 milliamps, calculate the proper signal output at an applied temperature of +41 degrees.
Solution:
Take 41 degrees and subtract the LRV (-88 degrees) which is the same as adding 88 to 41, then divide by the span (145 degrees − (−88) degrees, or 233 degrees) to arrive at 55.36% (0.5536 per unit). Take this number and multiply by the span of the current signal range (16 milliamps) and lastly add the LRV of the current signal range (4 milliamps) to arrive at a final answer of 12.86 milliamps.
Example 3:
Given a pH transmitter with a measurement range of 3 pH to 11 pH and a signal range of 4 to 20 milliamps, calculate the proper signal output at 9.32 pH.
Solution:
Take 9.32 pH and subtract the LRV (3 pH), then divide by the span (11 pH − 3 pH, or 8 pH) to arrive at 79% (0.79 per unit). Take this number and multiply by the span of the current signal range (16 milliamps) and lastly add the LRV of the current signal range (4 milliamps) to arrive at a final answer of 16.64 milliamps.
What is Zero Suppression ?
In some cases, the level transmitter has to be mounted X meters below the base of an open tank as shown in below figure.
The liquid in the high pressure impulse line exerts a constant pressure (P = S . X) on the high pressure side.
That is, the pressure on the high pressure side of the DP Transmitter is always higher than the actual pressure exerted by the liquid column in the tank by (SG . X) – so the reading will be in error high.
This constant pressure would cause an output signal that is higher than 4 mA when the tank is empty and above 20 mA when it is full.
When the liquid level is at H meters, pressure on the high pressure side of the transmitter will be:
Phigh = S·H + S.X + Patm
Plow= Patm
ΔP = Phigh – Plow = S . H + S . X
The transmitter has to be negatively biased by a value of S.X so that the output of the transmitter is proportional to the tank level (S . H) only.
The above procedure is called Zero Suppression and it can be done during calibration of the transmitter.
What is Zero Elevation ?
When a wet leg installation is used,the low pressure side of the level transmitter will always experience a higher hydrostatic pressure than the high pressure side.
This is due to the fact that the height of the wet leg (X) is always just greater than the maximum height of the liquid column (H) inside the tank.
When the liquid level is at H meters, we have:
Phigh =Pgas + S . H
Plow =Pgas + S . X
ΔP = Phigh – Plow = S . H – S . X = -S(X . H)
The differential pressure ΔP sensed by the transmitter is always a negative value (i.e. the low pressure side has a higher pressure than high pressure side).
To properly calibrate the transmitter, a positive bias (S . X) is needed to elevate the transmitter output.
This positive biasing technique is called zero elevation.
Span = (x) (GL)
HW at minimum level = (z) (GS) + (y) (GL)
HW at maximum level = (z) (GS) + (x + y) (GL)
Where
GL = Specific gravity of tank liquid
GS = Specific gravity of seal liquid
HW = Equivalent head of water
x, y, and z as shown in above figure
Example:
Open tank with x = 80 inches
y = 5 inches
z = 10 inches
GL = 0.8
GS = 0.9
Span = (80)(0.8) = 64 inches
HW at minimum level = (10)(0.9) + (5)(0.8) = 13 inches
HW at maximum level = (10)(0.9) + (5 + 80)(0.8) = 77inches
Calibrated Range = 13 to 77 inches head of water
Zero elevation calculation Closed Tank With Wet Leg
Span = (x)(GL) Wet Leg
HW at minimum level = (y)(GL) – (d)(GS)
HW at maximum level = (x + y)(GL) – (d)(GS)
Where
GL = Specific gravity of tank liquid
GS = Specific gravity of seal liquid
HW = Equivalent head of water
Example:
Closed tank with x = 70 inches y = 20 inches,
and d = 100 inches
GL = 0.8 Seal Liquid
GS = 0.9
Span = (70)(0.8) = 56 inches
HW at minimum level = (20)(0.8) – (100)(0.9) = -74 inches<
HW at maximum level = (70 + 20)(0.8) – (100)(0.9) = -18 inches
Calibrated Range = -74 to -18 inches head of water
(Minus signs indicate that the higher pressure is applied to the low pressure side of the transmitter.)
Zero Suppression and Zero Elevation Calculations
What is Zero Suppression ?
In some cases, the level transmitter has to be mounted X meters below the base of an open tank as shown in below figure.
The liquid in the high pressure impulse line exerts a constant pressure (P = S . X) on the high pressure side.
That is, the pressure on the high pressure side of the DP Transmitter is always higher than the actual pressure exerted by the liquid column in the tank by (SG . X) – so the reading will be in error high.
This constant pressure would cause an output signal that is higher than 4 mA when the tank is empty and above 20 mA when it is full.
When the liquid level is at H meters, pressure on the high pressure side of the transmitter will be:
Phigh = S·H + S.X + Patm
Plow= Patm
ΔP = Phigh – Plow = S . H + S . X
The transmitter has to be negatively biased by a value of S.X so that the output of the transmitter is proportional to the tank level (S . H) only.
The above procedure is called Zero Suppression and it can be done during calibration of the transmitter.
What is Zero Elevation ?
When a wet leg installation is used,the low pressure side of the level transmitter will always experience a higher hydrostatic pressure than the high pressure side.
This is due to the fact that the height of the wet leg (X) is always just greater than the maximum height of the liquid column (H) inside the tank.
When the liquid level is at H meters, we have:
Phigh =Pgas + S . H
Plow =Pgas + S . X
ΔP = Phigh – Plow = S . H – S . X = -S(X . H)
The differential pressure ΔP sensed by the transmitter is always a negative value (i.e. the low pressure side has a higher pressure than high pressure side).
To properly calibrate the transmitter, a positive bias (S . X) is needed to elevate the transmitter output.
This positive biasing technique is called zero elevation.
Zero Suppression and Zero Elevation Example Calculation
Zero suppression calculation
Span = (x) (GL)
HW at minimum level = (z) (GS) + (y) (GL)
HW at maximum level = (z) (GS) + (x + y) (GL)
Where
GL = Specific gravity of tank liquid
GS = Specific gravity of seal liquid
HW = Equivalent head of water
x, y, and z as shown in above figure
Example:
Open tank with x = 80 inches
y = 5 inches
z = 10 inches
GL = 0.8
GS = 0.9
Span = (80)(0.8) = 64 inches
HW at minimum level = (10)(0.9) + (5)(0.8) = 13 inches
HW at maximum level = (10)(0.9) + (5 + 80)(0.8) = 77inches
Calibrated Range = 13 to 77 inches head of water
Zero elevation calculation Closed Tank With Wet Leg
Span = (x)(GL) Wet Leg
HW at minimum level = (y)(GL) – (d)(GS)
HW at maximum level = (x + y)(GL) – (d)(GS)
Where
GL = Specific gravity of tank liquid
GS = Specific gravity of seal liquid
HW = Equivalent head of water
Example:
Closed tank with x = 70 inches y = 20 inches,
and d = 100 inches
GL = 0.8 Seal Liquid
GS = 0.9
Span = (70)(0.8) = 56 inches
HW at minimum level = (20)(0.8) – (100)(0.9) = -74 inches<
HW at maximum level = (70 + 20)(0.8) – (100)(0.9) = -18 inches
Calibrated Range = -74 to -18 inches head of water
(Minus signs indicate that the higher pressure is applied to the low pressure side of the transmitter.)
