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Reference Manual
00809-0100-4102, Rev AA
Section 5: Operation and Maintenance
May 2013
Operation and maintenance
Example
A range 4 differential pressure HART transmitter (2051CD4...) will be used in an application with
a static line pressure of 1200 psi (83 bar). The transmitter output is ranged with the Lower
Range Value at 500 inH20 (1, 2 bar) and the Upper Range Value at 1500 inH20 (3, 7 bar). To
correct for systematic error caused by high static line pressure, first use the following formulas
to determine the corrected values for the high trim value.
High Trim Value:
HT = (URV - (S/100 x P/1000 x LRV))
In this example:
Complete the Upper Sensor Trim procedure as described in “Sensor Trim” on page 68. In the
example above, at step 4, apply the nominal pressure value of 1500 inH
2
0. However, enter the
calculated correct upper Sensor Trim value of 1517.1 inH
2
0 with a Field Communicator.
Note
The Range Values for the upper and lower range points should be at the nominal URV and LRV. In
the example above, the values are 1500 inH
2
0 and 500 inH
2
0 respectively. Confirm the values
on the HOME screen of the Field Communicator. Modify, if needed, by following the steps in
“Set range points” on page 19.
5.4 Trim the pressure signal
5.4.1 Sensor Trim Overview
A Sensor Trim corrects the pressure offset and pressure range to match a pressure standard. The
Upper Sensor Trim corrects the pressure range and the Lower Sensor Trim (Zero Trim) corrects
the pressure offset. An accurate pressure standard is required for full calibration. A zero trim can
be performed if the process is vented, or the high and low side pressure are equal (for
differential pressure transmitters).
Zero trim is a single-point offset adjustment. It is useful for compensating for mounting position
effects and is most effective when performed with the transmitter installed in its final mounting
Where: HT = Corrected High Trim Value
URV = Upper Range Value
S = Span shift per specification (as a percent of reading)
P = Static Line Pressure in psi
URV = 1500 inH
2
O (3.74 bar)
S = -0.95%
P = 1200 psi
LT = 1500 - (-0.95%/100 x 1200 psi/1000 psi x 1500 inH
2
O)
LT = 1517.1 inH
2
0