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Help with unit conversion

n8cole

Chemical
Jun 24, 2024
20
I am trying to make a pressure relief valve calculation in Excel for determining the minimum required area of discharge to get a desired mass flow rate. I am using the ACHIE relief valve sizing article and the example in it to validate my calculations. However, when I convert the metric units to imperial units, the formulas to not calculate the correct discharge area. Based on the example problem in the article, my discharge area should be 0.167 sq.ft, but my Excel sheet is calculating ~0.64 sq.ft. I could really use a fresh pair of eyes on this one. below is a table of the orginial metric values, my conversion factors, and the new imperial value. As well as a link to the ACHIE pressure relief valve sizing article, see pg 75 (pg 8 of the pdf) for the example problem.

I suspect my error is either with the C value or maybe my pressure conversion.

Article: https://www.aiche.org/sites/default/files/cep/20131068_r.pdf

Table:
VariablesMetric ValuesMetric UnitsConversion FactorsImperial ValuesImperial Units
W50kg/s2.205110.23lb/s
P_MAWP8barg2088.5416,708.32lbf/ft^2
Built-up Backpressure10%unfired pressure vessels10%%
P_backpressure0.7barg2088.541,461.98lbf/ft^2
T473K391.73deg F
P_set7barg2088.5414,619.78lbf/ft^2
MW100kg/kmol1100lb/lbmol
Y1.31.3
z11
P_max8.8barg2088.5418,379.15lbf/ft^2
P_atm1.013bar2088.542,115.69lbf/ft^2
P_19.813bara2088.5420,494.84lbf/ft^2
Pb% (figure 7)17.5%%17.5%%
Kb (figure 7)11
Kd0.9750.975
C7.32E-03sqrt(kg kmol-K)/N-s0.166461.70E-02sqrt(lb-lbmol-F)/lbf-s
gc1unitless1
A1.55E-02m^210.75840.64323ft^2
Rg8314Pa-m^3/kmol-K0.185871545.347sqrt(lb-lbmol-F)/lbf-s
 
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Your gc is wrong.

gc is functionally unitless. However, gc in imperial units is 32.2 (lbm*ft)/(lbf*s^2). gc in metric is 1 kg*m/N*s^2.

F = m*a, right? No, F = m*a/gc.

In metric, this means 9.81N = 1 kg * 9.81 m/s2

However, in Imperial units, 1lbm * 32.2 ft/s^2 = 32.2 lbm*ft/s^2. For whatever historical reason, it was decided 1lbm should numerically equal 1 lbf. So gc was set at 32.2 to make 1lbm exert a force of 1 lbf.

Since gc in metric is equal to 1, it is often ignored and is even left out of formulae. Be aware that when looking at equations that involve acceleration due to gravity, g, that gc should also be there.
 
I have not worked through the article in detail but it looks questionable to me and I would advise you to compare it with other sources.

I agree with Latexman and TiCl4 that gc is wrong. And it is wrongly described in the article as being the gravitational constant, which it is not (although it has the same numerical value). gc is used because the US customary units are not internally consistent. If you measure the force in lbf then you should measure the mass in slugs. If you cheat and use lbm then you need gc as a conversion factor.

Also it looks strange to me that pressures are in bar, but the gas constant is given in Pascal.

I would be very suspicious of any results from these formulas.
 
I have not worked through the article in detail but it looks questionable to me and I would advise you to compare it with other sources.

I agree with Latexman and TiCl4 that gc is wrong. And it is wrongly described in the article as being the gravitational constant, which it is not (although it has the same numerical value). gc is used because the US customary units are not internally consistent. If you measure the force in lbf then you should measure the mass in slugs. If you cheat and use lbm then you need gc as a conversion factor.

Also it looks strange to me that pressures are in bar, but the gas constant is given in Pascal.

I would be very suspicious of any results from these formulas.
To “work around” this problem, I solved for the area by converting from imperial to metric, using the article’s constants and formulas, and then converting back to imperial units. I compared the answer my sheet gives me with other sources I have, and the results were close. At most, it’s off by -3 sq. in. I definitely need to tweak it some more.
 
Your gc is wrong.

gc is functionally unitless. However, gc in imperial units is 32.2 (lbm*ft)/(lbf*s^2). gc in metric is 1 kg*m/N*s^2.

F = m*a, right? No, F = m*a/gc.

In metric, this means 9.81N = 1 kg * 9.81 m/s2

However, in Imperial units, 1lbm * 32.2 ft/s^2 = 32.2 lbm*ft/s^2. For whatever historical reason, it was decided 1lbm should numerically equal 1 lbf. So gc was set at 32.2 to make 1lbm exert a force of 1 lbf.

Since gc in metric is equal to 1, it is often ignored and is even left out of formulae. Be aware that when looking at equations that involve acceleration due to gravity, g, that gc should also be there.
Yeah, I knew my problem was with my gc, but correcting it would cause my C variable to be wrong. I ended up converting imperial units to metric units, solving using the method in the article, then converting my final answer to imperial units. There is still issue with this method, but my answer matches that in the article.

Also, I like to believe lbf was set to equal 32.174 lbm-ft/s^2, so weight scales made more sense. Of course, I haven't investigated this theory at all, and with a little extra thought it doesn't really make much sense. At least Fahrenheit makes more sense than Celsius.
 
Hi,
Consider this resource to support your work .You will find input about 2 different units system.
Good luck
Pierre
 

Attachments

  • safety valves calculation CROSBY.pdf
    1.5 MB · Views: 7
Hi,
For a small investment, consider buying from Katmar software the best in the class units convertor, Uconeer.
Pierre
 

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