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The non conformity error depends on the ratio between diameter and
length of the vessel: for ratio ≥5 to 1 the error is ≤0.25%. The
polynomium found with mathematical method gives an error of ±0.15%.
c) Cilindrical vessel with elliptical or pseudoelliptical ends (see Fig. 1c).
Transmitter measuring the whole vessel heigth.
The same polynomium can be used also for the cylindrical vessel with
elliptical or pseudoellipticall ends. To obtain the volume contained in the
vessel can be used the following empyrical formula:
2
V = Out • (d/1.12838)
• (L + 2/3 m)
where m is the length of the minor ellipse axis (see Fig.1c)
The non conformity error depends on the ratio between the diameter
and the length of the vessel: for ratio ≥5 to 1 the error is ≤0.25%. The
polynomium found with mathematical method gives an error of ±0.15%.
2.2 SPHERICAL TANK
Spherical tank (see Fig.1d). Transmitter measuring the whole vessel
height.
The following polynomium gives the volume of the spherical section in
relation to the heigth h of the liquid in the tank.
Out = 3 h
2
- 2 h
3
This formula is geometrical and then his conformity is perfect.
Being both the input h and the output Out normalized, i.e. in the range
0 to 1 (or 0% to 100%), the sphere diameter D corresponding to a
volume equal to 1 (100%) will be "normalized" by a "K" factor of the
following value:
K = 2 •
3
The volume of the liquid contained in the tank, at heigth = h will be
V = Out • (D/1.2407)
where D = sphere diameter .
2.3 CYLINDRICAL VESSEL AND SPHERICAL TANK WITH
PARTIAL LEVEL MEASUREMENT
Cases a) to d) but with partial level measurement (Fig. 2a)
In these cases two methods can be used:
1) Plot the changes in volume in relation to the level changes and,
using a mathematical method, find the relevant polynomium.
2) Use the polynomium coefficients for cases a) to d) and calibrate the
transmitter range to cover the full diameter of the vessel or tank: the
changes in volume for the h changes between h
correct. Of course the transmitter will transmit, when the level is
≤h
, the volume corresponding to h
0
All transmitted volumes are % of the total volume of the vessel.
If it is required the partial volume starting from h
solved for h
with negative sign: for example for h
0
A
= - 0.02 + 0.297 • 0.2 + 2.83 • 0.2
0
The polynomium coefficients for the example will be:
A
0
Out = - 0.14179 + 0.297 h + 2.83 h
Note : The accuracy of all above numerical values can not be guaranteed.
General notes for level measurement
The level transmitter calibration is effected by the transmitter installation conditions, i.e. if the reference connection is
empty (dry leg) or liquid filled (wet leg). In the first case (dry leg) the calibration in affected by the specific gravity of the
measured liquid and the atmosphere above the liquid at process condition, whereas in the second case (wet leg), it is
affected by the specific gravity of the liquid in the connecting pipe(s).
- 36 -
. . . ADDENDUM FOR SELECTABLE OUTPUT FUNCTIONS
√ 3/ (4 π) = 1.2407
3
and h
0
: the same apply for level ≥h
0
(i.e. the volume at h
0
= 20%
0
2
- 4.255 • 0.2
A
A
A
1
2
2
- 4.255 h
will be
max
.
max
= 0) then the A
0
0
3
4
+ 3.5525 • 0.2
A
A
3
4
5
3
4
+ 3.5525 h
-1.421 h
Fig. 1c
Fig. 1d
Fig. 2a
coefficient should be equal to the polynomium
5
-1.421 • 0.2
= - 0.14179
5
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