How many Flops per iteration in Fluent solver

tuananhtran56 · September 13, 2017, 05:27

Hi all,
I'm newbie on the Fluent. I have a question that how many Flops need for each iteration in Fluent solver. Assuming that my mesh about 14M cells.
Thank you.

spx · October 5, 2017, 10:04

Hi, I don’t know the answer to your question-but with only cell count there is no way of knowing.

Many other parameters and settings will affect your required computational power (and time).
Such factors may include:
- pressure-velocity coupling (for eg pressure based solver)
- node count
- transient v steady
- BCs
- Turbulence model
- multi/single phase
Etc. and many more

As it really is a case by case problem, you’ll have to run it to find out.

LuckyTran · October 8, 2017, 15:41

FLOPS is also dependent on your hardware. Even with all other settings the same, no one can tell you what the FLOPS throughput for you would be.

September 13, 2017, 05:27	How many Flops per iteration in Fluent solver	#1
tuananhtran56 New Member tuananh Join Date: Sep 2015 Posts: 4 Rep Power: 10	Hi all, I'm newbie on the Fluent. I have a question that how many Flops need for each iteration in Fluent solver. Assuming that my mesh about 14M cells. Thank you.

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October 5, 2017, 10:04		#2
spx New Member spx Join Date: Sep 2017 Posts: 1 Rep Power: 0	Hi, I don’t know the answer to your question-but with only cell count there is no way of knowing. Many other parameters and settings will affect your required computational power (and time). Such factors may include: - pressure-velocity coupling (for eg pressure based solver) - node count - transient v steady - BCs - Turbulence model - multi/single phase Etc. and many more As it really is a case by case problem, you’ll have to run it to find out.

October 8, 2017, 15:41		#3
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,673 Rep Power: 65	FLOPS is also dependent on your hardware. Even with all other settings the same, no one can tell you what the FLOPS throughput for you would be.