Function Code 7 – Square Root
GENERAL DESCRIPTION
S1
This function computes the square root of the input signal in
engineering units. The output equals a factor (k) times the
square root of the input. The equation that represents this
function is:
(7)
N
√
Y = S2 〈 S1〉
where:
<S1> = Input value.
S2
= Gain value (k) in engineering units.
Y
= Output value (Y = 0 if <S1> ≤ 0).
UTILIZATION
CBC01
IMCOM03
IMMFP02
INPPT01
NMFC01
CLC01
IMCOM04
IMMFP03
INSEM01
NMFC02
CLC02
IMLMM02
IMMPC01
NAMM02
SLC01
CLC03
IMMFC03
IMQRC01
NCOM02
SLC21
CLC04
IMMFC04
INIIT02
NGCM01
CSC01
IMMFC05
INIPT01
NGCM02
IMAMM03
IMMFP01
INIPT02
NLMM01
OUTPUTS
Blk
Type
Description
N
R
Output value equals square root of input value multiplied by the
gain value (k)
SPECIFICATIONS
Spec
Tune
Default
Type
Range
S1
N
6
I
Note 1
S2
Y
1.000
R
Full
Description
Block address of input
Gain value (k) in engineering units (EU)
NOTE:
1. Maximum values are: 254 for COM modules and CLC01/02/03/04
1023 for SLC01/21 and IMLMM02
2046 for NMFC01/02, IMMFC05, CBC01 and IMMPC01
9998 for IMMFC03/04 and IMMFP01/02/03
I-E96-200B
15 FEB 1995
Square Root
7-1
Function Code 7
®
APPLICATIONS
Specification S2 is the gain (k) applied to the value 〈 S1〉 and
can be any real number. It is used to scale an input signal to a
meaningful or easy to work with output signal. Figure 7-1
shows an example of how function code 7 can be used. In the
example, a flow rate of zero to 50,000 pounds per hour is
being measured by a differential pressure transducer whose
output range is zero to 200 inches of water. The flow is a function of the square root of the differential pressure multiplied
by some constant (k). The equation for this example is:
Flow = k diff. pressure
If it is known that the flow is 50,000 pounds per hour at a
transmitter output indicating 200 inches of water differential
pressure, the required constant (k) can be calculated as follows:
50,000 pounds per hour = k 200
50,000 pounds per hour = k (14.142)
50, 000
---------------- = k
14, 142
k = 3,535.534
0-200
in. H 2O
S1
(27)
201
AI
S1
(7)
300
√
0
TO
50,000
lb/hr
S2 = 3535.534
50,000
45,000
40,000
35,000
30,000
lb/hr
25,000
20,000
15,000
10,000
5,000
0
0
20
40
60
80
100
120
140
160
180
200
in. H 2O
TP45115B
Figure 7-1. Converting a Pressure Signal
to a Flow Rate Using Function Code 7
Square Root
7-2
15 FEB 1995
I-E96-200B
Function Code 7
Many nonlinear inputs need to be converted to linear outputs. Figure 7-2 illustrates converting a nonlinear pressure
signal to a linear flow signal using function code 7.
∆ PRESSURE
TRANSMITTER
NONLINEAR
PRESSURE S1
(27)
201
AI
S1
(7)
300
√
LINEAR FLOW
S2 = 10
PRESSURE
100
%
90
90
80
80
70
70
60
60
50
%
50
40
40
30
30
20
20
10
10
0
0
10
20
30
40
50
%
60
FLOW
100
70
80
90
100
0
0
10
20
30
40
50
60
70
%
80
90
100
TP22010B
Figure 7-2. Converting a Nonlinear Pressure Input to a Linear Flow Output
I-E96-200B
15 FEB 1995
Square Root
7-3
®