HRR and HRRUPA
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mohamed assal
Aug 5, 2010, 8:19:38 AM8/5/10
to FDS and Smokeview Discussions
Hello members.
i have some confusion about heat release per unit of areaHHRPUA. and
heat release HRR
why the author used HRR in place of HHRPUA. he should devide by the
section, some one can tell me
see the input bellow :
----L.H. Hu etal. /J. Hazardous Materials 140 (2007) 293-298 ------
Diesel pool fire in a long (88m) channel, Max heat release rate:
0.75MW
&OBST XB= 78.61,79.39,3.61,4.39,0.00,0.10, SURF_ID='DIESEL POOL' /
&REAC ID='DIESEL'
FYI='DIESEL, C_12 H_23'
C=12.
H=23.
HEAT_OF_COMBUSTION = 45000.
SOOT_YIELD=0.1
CO_YIELD=0.015 /
&SURF ID='DIESEL POOL'
HRRPUA=750.
RAMP_Q='diesel pool'
COLOR = 'YELLOW' /
Cheers Mohamed.
i have some confusion about heat release per unit of areaHHRPUA. and
heat release HRR
why the author used HRR in place of HHRPUA. he should devide by the
section, some one can tell me
see the input bellow :
----L.H. Hu etal. /J. Hazardous Materials 140 (2007) 293-298 ------
Diesel pool fire in a long (88m) channel, Max heat release rate:
0.75MW
&OBST XB= 78.61,79.39,3.61,4.39,0.00,0.10, SURF_ID='DIESEL POOL' /
&REAC ID='DIESEL'
FYI='DIESEL, C_12 H_23'
C=12.
H=23.
HEAT_OF_COMBUSTION = 45000.
SOOT_YIELD=0.1
CO_YIELD=0.015 /
&SURF ID='DIESEL POOL'
HRRPUA=750.
RAMP_Q='diesel pool'
COLOR = 'YELLOW' /
Cheers Mohamed.
Bryan Klein
Aug 5, 2010, 10:54:34 AM8/5/10
to FDS and Smokeview Discussions
Since the author applied the 'DIESEL' surface to all sides of the
OBST, the total exposed surface area is 1.872 m^2 (assuming the bottom
is not exposed with a min_X value of 0.
With the HRRPUA of 750 the total HRR would be about 1404 kW, and based
on the comment at the top of 'Max heat release rate: 0.75 MW', I think
this was an error in the HRRPUA value. It should be closer to
HRRPUA=400.641 for this surface area.
If all the author wants is 0.75 MW released from the top surface of
the OBST, then the &OBST line needs to change from SURF_ID='DIESEL
POOL' / to something like SURF_IDS='DIESEL POOL','INERT','INERT' /
This would make the total surface area for the top of the OBST at 1.56
m^2.
Which would mean the 'DIESEL POOL' HRRPUA should then be set at
480.769 kW to get a total HRR of 0.75 MW.
It would be a nice feature to just set the desired HRR value and let
the model calculate the total area of the SURF it is applied to and
then set a HRRPUA value internally to whatever it needs to be to
create the expected results. But, until then, you are correct, the
surface area has to be calculated first and used to determine the
HRRPUA value.
Best Regards,
-Bryan Klein
On Aug 5, 7:19 am, mohamed assal <mohamedassalengineer...@gmail.com>
wrote:
OBST, the total exposed surface area is 1.872 m^2 (assuming the bottom
is not exposed with a min_X value of 0.
With the HRRPUA of 750 the total HRR would be about 1404 kW, and based
on the comment at the top of 'Max heat release rate: 0.75 MW', I think
this was an error in the HRRPUA value. It should be closer to
HRRPUA=400.641 for this surface area.
If all the author wants is 0.75 MW released from the top surface of
the OBST, then the &OBST line needs to change from SURF_ID='DIESEL
POOL' / to something like SURF_IDS='DIESEL POOL','INERT','INERT' /
This would make the total surface area for the top of the OBST at 1.56
m^2.
Which would mean the 'DIESEL POOL' HRRPUA should then be set at
480.769 kW to get a total HRR of 0.75 MW.
It would be a nice feature to just set the desired HRR value and let
the model calculate the total area of the SURF it is applied to and
then set a HRRPUA value internally to whatever it needs to be to
create the expected results. But, until then, you are correct, the
surface area has to be calculated first and used to determine the
HRRPUA value.
Best Regards,
-Bryan Klein
On Aug 5, 7:19 am, mohamed assal <mohamedassalengineer...@gmail.com>
wrote:
mohamed assal
Aug 5, 2010, 12:10:26 PM8/5/10
to FDS and Smokeview Discussions
Thanks Bryan
for your clear answer;
thank you so Much
Cheers mohamed
for your clear answer;
thank you so Much
Cheers mohamed
mohamed assal
Aug 5, 2010, 12:40:56 PM8/5/10
to FDS and Smokeview Discussions
Bryan u did a mistakes in your calculation,
the total exposed surface area is not 1.872 m2
because the area of square = L x L
that's give : 2(0.78x0.78)+0.078x4= 1.5288 m2
the HRRPUA must be 490.58 KW.
and the HRR is applicated to just the top surface S=0.6084 m2
the 'DIESEL POOL' HRRPUA should then be set at
1232.74 kW to get a total HRR of 0.75 MW.
i think i'ts clear now for you and for the author of this input
Thank you so much.
Cheers Mohamed
the total exposed surface area is not 1.872 m2
because the area of square = L x L
that's give : 2(0.78x0.78)+0.078x4= 1.5288 m2
the HRRPUA must be 490.58 KW.
and the HRR is applicated to just the top surface S=0.6084 m2
the 'DIESEL POOL' HRRPUA should then be set at
i think i'ts clear now for you and for the author of this input
Thank you so much.
Cheers Mohamed
Bryan Klein
Aug 5, 2010, 5:03:55 PM8/5/10
to FDS and Smokeview Discussions
Looks like we both had some problems with calculations and
assumptions. :)
I multiplied the top surface length by 2 instead of the other length,
and you multiplied the top surface area by 2 which would include the
bottom surface. (Unless it is supposed to be exposed and burning?)
So here is what I calculate (with improved math):
&OBST XB= 78.61,79.39,3.61,4.39,0.00,0.10
79.39 - 78.61 = 0.78 (X)
4.39 - 3.61 = 0.78 (Y)
0.10 - 0.00 = 0.1 (Z)
Top Surface
0.78 * 0.78 = 0.6084
Side Surfaces
0.78 * 0.10 = 0.078
Total Area of All Side Surfaces
0.078 * 4 = 0.312
Total Exposed Surface Area
0.6084 + 0.312 = 0.9204 m^2
or
(0.78 * 0.78) + (0.078 * 4) = 0.9204
(Bottom Surface is not included assuming 0.00 is bottom of domain and
would not be exposed.)
Required HRRPUA value for total HRR of 0.75 MW with sides burning.
750 kW / 0.9204 m^2 = 814.8631 kW/m^2
If this is a pan/pool fire then probably only the top surface should
be burning with an HRRPUA of
750 kW / 0.6084 m^2 = 1232.74 kW/m^2
Sorry for the initial miscalculation, good catch. I hope my math is
better this time.
It's the simple things that sometimes pass by unnoticed.
-Bryan
On Aug 5, 11:40 am, mohamed assal <mohamedassalengineer...@gmail.com>
assumptions. :)
I multiplied the top surface length by 2 instead of the other length,
and you multiplied the top surface area by 2 which would include the
bottom surface. (Unless it is supposed to be exposed and burning?)
So here is what I calculate (with improved math):
&OBST XB= 78.61,79.39,3.61,4.39,0.00,0.10
79.39 - 78.61 = 0.78 (X)
4.39 - 3.61 = 0.78 (Y)
0.10 - 0.00 = 0.1 (Z)
Top Surface
0.78 * 0.78 = 0.6084
Side Surfaces
0.78 * 0.10 = 0.078
Total Area of All Side Surfaces
0.078 * 4 = 0.312
Total Exposed Surface Area
0.6084 + 0.312 = 0.9204 m^2
or
(0.78 * 0.78) + (0.078 * 4) = 0.9204
(Bottom Surface is not included assuming 0.00 is bottom of domain and
would not be exposed.)
Required HRRPUA value for total HRR of 0.75 MW with sides burning.
750 kW / 0.9204 m^2 = 814.8631 kW/m^2
If this is a pan/pool fire then probably only the top surface should
be burning with an HRRPUA of
750 kW / 0.6084 m^2 = 1232.74 kW/m^2
Sorry for the initial miscalculation, good catch. I hope my math is
better this time.
It's the simple things that sometimes pass by unnoticed.
-Bryan
On Aug 5, 11:40 am, mohamed assal <mohamedassalengineer...@gmail.com>
mohamed assal
Aug 6, 2010, 4:19:47 AM8/6/10
to fds...@googlegroups.com
Bryan i'ts right,
and really i'ts pan / diesel pool fire,
thank you for the explanations.
we should take HRRPUA =1232.74 kw/m2
Cheers
Mohamed
and really i'ts pan / diesel pool fire,
thank you for the explanations.
we should take HRRPUA =1232.74 kw/m2
Cheers
Mohamed
2010/8/6, Bryan Klein <kl...@thunderheadeng.com>:
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Chris
Aug 6, 2010, 10:12:14 AM8/6/10
to FDS and Smokeview Discussions
Is this value for HRRPUA not too high for Diesel? I have values for
Heptane/Toluol-Mix Pool fires going up to 720 kW/m2.
Cheers
Chris
On 6 Aug., 10:19, mohamed assal <mohamedassalengineer...@gmail.com>
> Mobile : +213790541241- Zitierten Text ausblenden -
>
> - Zitierten Text anzeigen -
Heptane/Toluol-Mix Pool fires going up to 720 kW/m2.
Cheers
Chris
On 6 Aug., 10:19, mohamed assal <mohamedassalengineer...@gmail.com>
>
> - Zitierten Text anzeigen -
Bryan Klein
Aug 6, 2010, 11:33:54 AM8/6/10
to FDS and Smokeview Discussions
NUREG 1805 has values for different liquid fuels.
http://www.nrc.gov/reading-rm/doc-collections/nuregs/staff/sr1805/
Chapter 3 - Estimating Burning Characteristics of Liquid Pool Fire,
Heat Release Rate, Burning Duration, and Flame Height
http://www.nrc.gov/reading-rm/doc-collections/nuregs/staff/sr1805/ch3-6.pdf
Page 3-6 presents numbers for Diesel Oil.
Mass Burning Rate - 0.044 (kg/m^2-sec)
Effective Heat of Combustion - 44,400 (kJ/kg)
Density - 918 (kg/m^3)
0.044 * 44,400 = 1953.6 kW/m^2 (HRRPUA)
The vent discussed earlier of 0.78*0.78 m (0.6084 m^2) should result
in an HRR of 1188.57 kW.
0.044 * 44400 * 0.6084 * (1 - (e^((-100) * 0.88))) = 1188.57024
(This assumes a Diameter based on the Area given, with an empirical
constant of 100 based on the * note in the table on page 3-6 for
conservative estimation.)
To get the HRR down to 750 kW total output, you would have to have the
top surface of the vent at 0.3839 m^2 which would be approx. 0.6196 m
length on the sides of a square instead of 0.78 m.
These numbers could be debated, or tweaked to fit whatever you are
trying to do. But, I think that NUREG 1805 is a good source for
reference values and methods of estimation.
Out of curiosity, the values for Gasoline would give you an HRRPUA of
2403.5 kW/m^2.
I hope I didn't mess up the math this time. :)
-Bryan
http://www.nrc.gov/reading-rm/doc-collections/nuregs/staff/sr1805/
Chapter 3 - Estimating Burning Characteristics of Liquid Pool Fire,
Heat Release Rate, Burning Duration, and Flame Height
http://www.nrc.gov/reading-rm/doc-collections/nuregs/staff/sr1805/ch3-6.pdf
Page 3-6 presents numbers for Diesel Oil.
Mass Burning Rate - 0.044 (kg/m^2-sec)
Effective Heat of Combustion - 44,400 (kJ/kg)
Density - 918 (kg/m^3)
0.044 * 44,400 = 1953.6 kW/m^2 (HRRPUA)
The vent discussed earlier of 0.78*0.78 m (0.6084 m^2) should result
in an HRR of 1188.57 kW.
0.044 * 44400 * 0.6084 * (1 - (e^((-100) * 0.88))) = 1188.57024
(This assumes a Diameter based on the Area given, with an empirical
constant of 100 based on the * note in the table on page 3-6 for
conservative estimation.)
To get the HRR down to 750 kW total output, you would have to have the
top surface of the vent at 0.3839 m^2 which would be approx. 0.6196 m
length on the sides of a square instead of 0.78 m.
These numbers could be debated, or tweaked to fit whatever you are
trying to do. But, I think that NUREG 1805 is a good source for
reference values and methods of estimation.
Out of curiosity, the values for Gasoline would give you an HRRPUA of
2403.5 kW/m^2.
I hope I didn't mess up the math this time. :)
-Bryan
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