Opened 15 months ago
Last modified 13 months ago
#174 new requirement
Genauigkeit PT100 Schnittstelle — at Version 5
| Reported by: | (none) | Owned by: | Erik Welander |
|---|---|---|---|
| Priority: | major | Milestone: | |
| Component: | Steuerelektronik | Severity: | |
| Keywords: | Cc: | ||
| Data type: | Deadline: | ||
| Depending on: | Implements: | ||
| Memaddress: | Observerlist: | ||
| Parent: | #199 | Answer: | |
| Type: | system | Revieweraccept: | |
| Reviewerlist: | hantzsch,welander | Reviewerreject: | |
| Priority: | normal | Visibility: | |
| Writable: | Class: | functional | |
| Criticality: | must have |
Description (last modified by )
Die Analogschnittschnittstelle muss eine Genauigkeit von mindestens 0,16 % des gemessene Widerstands haben.
Herleitung:
Vor allem für Trockenkalibrierung ist die hohe Genauigkeit notwendig. Schallgeschwindigkeit soll mit einer Genauigkeit von 0,1 % bestimmt werden.
Störung (laut AGA8) bei Temperatur T = 293,15 K Stickstoff, 1 bar : dSoS/dT = 0.6 m/s/K.
Maximalfehler Schallgeschwindigkeit dSoS_max < 0,001 * 349,102 = 0,3491.
Genauigkeit Class AA PT100: 0,1 K + 0,17 % * T (in ° C) : Tsense = 20,134° C .
Maximaler Widerstand Class AA: Rmax = 100 Ohm + 7,75 Ohm = 107,75 Ohm.
Gemessener Wiederstand mit Genauigkeit r: Rmeas = Rmax * (1 + r).
Gemessene Temperatur: Tmeas = (Rmeas - 100) / 0,385 (in ° C).
Für Schallgeschwindigkeit:
( Tmeas - T) * dSoS/dT < dSoS_max
( (107,75 * (1 + r) - 100) / 0,385 - 20 ) * 0,6 < 0,3491
( (107,75 * r / 0,385 ) * 0,6 < 0,2675
r < 0,16 %
Change History (5)
comment:1 by , 15 months ago
| Description: | modified (diff) |
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| Reporter: | removed |
comment:2 by , 15 months ago
| Description: | modified (diff) |
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comment:3 by , 15 months ago
| Description: | modified (diff) |
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comment:4 by , 14 months ago
| Parent: | #104 → #199 |
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comment:5 by , 14 months ago
| Description: | modified (diff) |
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