Square Root
Function Code 7
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 for this function is:
(7 )
N
S1
Y = S2
á S1ñ
where:
<S1> = Input value.
S2
= Gain value (k) in engineering units.
= Output value (Y = 0 if <S1> £ 0).
Y
NOTES:
1. When function code 7 is utilized as a shaping algorithm for function code
222 (analog in/channel), its tunable specifications are not adaptable.
2. When function code 7 is used as a shaping algorithm, it can not at the
same time also be used as a logic function because the block output will not
respond to the specification S1 input. Function code 7 should not be referenced by function blocks other than function code 222 utilizing it as a shaping
algorithm.
3. Multiple instances of function code 222 function blocks may utilize the
same function code 7 function block as a shaping algorithm. The function code
7 shaping algorithm function block is not required to be in the same segment as
the function code 222 blocks.
Outputs
Blk
Type
N
R
Description
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
NOTES:
1. Maximum values are:
WBPEEUI210504B0
Description
Block address of input
Gain value (k) in engineering units (EU)
9,998 for the BRC-100, IMMFP11/12
31,998 for the HAC
7-1
Applications
Specification S2 is the gain (k) applied to the value á S1ñ 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
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.
7-2
WBPEEUI210504B0
A I/I
(1 21 )
2 01
0-20 0
in. H O
2
S1
(7)
3 00
Ö
0
TO
5 0,00 0
lb/hr
S 2 = 35 35 .53 4
5 0,0 00
4 5,0 00
4 0,0 00
3 5,0 00
3 0,0 00
lb /h r
2 5,0 00
2 0,0 00
1 5,0 00
1 0,0 00
5,0 00
0
0
20
40
80
60
100
12 0
1 40
160
1 80
20 0
in . H O
2
T 02 026 A
Figure 7-1. Converting a Pressure Signal to a Flow Rate
, P R ES S U R E
TR A N S M ITTE R
A I/I
N O N LIN E A R
P R ES S U RE
(121 )
201
S1
(7)
3 00
Ö
L IN E AR F LO W
S 2 = 10
PR ESSU RE
%
100
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 6 0
%
70
80
9 0 100
0
F LOW
0
10
20
30
40
5 0 60
70
80
90 100
%
T 0202 7A
Figure 7-2. Converting a Nonlinear Pressure Input to a Linear Flow Output
WBPEEUI210504B0
7-3
WBPEEUI210504B0