Re: [foxboro] Steam FLow Calc
- From: "Loyd Greer" <LGREER@xxxxxxxx>
- To: <foxboro@xxxxxxxxxxxxx>
- Date: Wed, 20 Apr 2005 10:32:04 -0400
Brian,
I don't know if this will help or not but here is a copy of a block that I got
from Foxboro. It Compensates for temp and press.
NAME = TOWER_9:STM_CALC1
TYPE = CALC
DESCRP =
PERIOD = 2
PHASE = 0
LOOPID =
RI01 = :FI_6911.PNT
HSCI1 = 50000
LSCI1 = 0.0
DELTI1 = .10
EI1 = LBS/HR
RI02 = UTILITY:PI_6927.PNT
HSCI2 = 250.0
LSCI2 = 0.0
DELTI2 = .25
EI2 = PSI
RI03 = :TI_6925.PNT
HSCI3 = 500.
LSCI3 = 0.0
DELTI3 = .5
EI3 = DEG F
RI04 = 0
HSCI4 = 250.0
LSCI4 = 0.0
DELTI4 = .50
EI4 = LBS/HR
RI05 = 0.0
HSCI5 = 250.0
LSCI5 = 0.0
DELTI5 = .50
EI5 = %
RI06 = 0.0
HSCI6 = 100.0
LSCI6 = 0.0
DELTI6 = 1.0
EI6 = %
RI07 = 0.0
HSCI7 = 100.0
LSCI7 = 0.0
DELTI7 = 1.0
EI7 = %
RI08 = 0.0
HSCI8 = 100.0
LSCI8 = 0.0
DELTI8 = 1.0
EI8 = %
BI01 = 0
BI02 = 0
BI03 = 0
BI04 = 0
BI05 = 0
BI06 = 0
BI07 = 0
BI08 = 0
BI09 = 0
BI10 = 0
BI11 = 0
BI12 = 0
BI13 = 0
BI14 = 0
BI15 = 0
BI16 = 0
II01 = 0
II02 = 0
LI01 = 0
LI02 = 0
HSCO1 = 50000
LSCO1 = 0.0
EO1 = LBS/HR
HSCO2 = 50000
LSCO2 = 0.0
EO2 = LBS.HR
HSCO3 = 100.0
LSCO3 = 0.0
EO3 = %
HSCO4 = 100.0
LSCO4 = 0.0
EO4 = %
MA = 1
INITMA = 1
TIMINI = 0
M01 = 14.7
M02 = 264.7
M03 = 460
M04 = 870
M05 = 0.0
M06 = 0.0
M07 = 0.0
M08 = 0.0
M09 = 0.0
M10 = 0.0
M11 = 0.0
M12 = 0.0
M13 = 0.0
M14 = 0.0
M15 = 0.0
M16 = 0.0
M17 = 0.0
M18 = 0.0
M19 = 0.0
M20 = 0.0
M21 = 0.0
M22 = 0.0
M23 = 0.0
M24 = 0.0
STEP01 = IN RI03
STEP02 = LAC M03
STEP03 = ADD
STEP04 = STM M05
STEP05 = CST
STEP06 = IN RI02
STEP07 = LAC M01
STEP08 = ADD
STEP09 = LAC M02
STEP10 = DIV
STEP11 = LAC M04
STEP12 = MUL
STEP13 = LAC M05
STEP14 = DIV
STEP15 = SQRT
STEP16 = IN RI01
STEP17 = MUL
STEP18 = OUT RO01
STEP19 = END
STEP20 =
STEP21 =
STEP22 =
STEP23 =
STEP24 =
STEP25 =
STEP26 =
STEP27 =
STEP28 =
STEP29 =
STEP30 =
STEP31 =
STEP32 =
STEP33 =
STEP34 =
STEP35 =
STEP36 =
STEP37 =
STEP38 =
STEP39 =
STEP40 =
STEP41 =
STEP42 =
STEP43 =
STEP44 =
STEP45 =
STEP46 =
STEP47 =
STEP48 =
STEP49 =
STEP50 =
END
Loyd Greer
Great Lakes Chemical
>>> blong@xxxxxxx 04/20/05 09:28AM >>>
yes, I should have mentioned I need compensation. Using CHAR blocks is a
good thought and I may try that. It will most likely be better than what
I'm doing now. Not sure Foxboro offers, Alex?
Thanks,Brian
-----Original Message-----
From: foxboro-bounce@xxxxxxxxxxxxx
[mailto:foxboro-bounce@xxxxxxxxxxxxx]On Behalf Of Gregory A Hurwitt
Sent: Wednesday, April 20, 2005 8:18 AM
To: foxboro@xxxxxxxxxxxxx
Subject: Re: [foxboro] Steam FLow Calc
Are you doing mass flow compensation? If so, are you doing it with the
standard compensation term: sqrt ((P/P0)*(T0/T))? This compensation
equation is based on the ideal gas law, and steam does not make a good
ideal gas.
For steam mass flow compensation we have on occasion done a "two-way"
characterization where we set up two CHARC blocks to give separate
compensation factors for pressure and temperature over a range of interest.
It's cumbersome, and it's only accurate over a limited range, but it's
better than ideal gas compensation for steam.
Does Invensys still offer a physical properties package for I/A? Does it
include compensation factors for steam? This might be a good solution.
Greg Hurwitt
BASF Freeport
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| | |"BrianLong" <blong@xxxxxxx> |
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| | | |
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| Subject: [foxboro] Steam FLow Calc
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We are using a CALC block to calculate steam flow. The calculation is not
as accurate as I'd like. Does anyone have an accurate "standard" way to
calculate steam flow in K# / HR.
Thanks,
Brian Long
Arkansas Kraft
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_______________________________________________________________________
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Systems (formerly The Foxboro Company). Use the info you obtain here at
your own risks. Read http://www.thecassandraproject.org/disclaimer.html
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