Talk:Accidental release source terms

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What is this about[edit]

I'm not sure exactly what this article is about. It dives into great detail without giving, IMO, a broad enough context first. Also I don't like level 1 headlines! -- SGBailey 23:46, 15 February 2006 (UTC)[reply]

I have reduced the size of the headlines as per your suggestion. I am sorry that you don't understand what the article is about. I am sure that this Wikipedia contains hundreds of articles in technical and engineering fields that I don't understand as well.
I'm catching a plane in an hour to go to a wedding and won't return until next week. As soon as I return, I will write a lead-in paragraph and attempt to briefly explain what the article is about. Thanks for your input. -- mbeychok 15:30, 16 February 2006 (UTC)[reply]
I have now added a lead-in discussion as promised. - mbeychok 18:18, 22 February 2006 (UTC)[reply]

Explanation of reverting of EPA equation to what it was previously[edit]

User 12.4.195.228 removed the R from the equation for the U.S.EPA Method for the evaporation from a non-boiling liquid pool because presumably he looked at Equation B-7 in reference (3), from which the equation was obtained.

Equation B-7 (which indeed does not have the gas law constant R) was derived from Equation D-1 in that same reference (3). Equation D-1 does have the R, and both equations are equivalent to each other. Equation B-7 was obtained by dividing the R=82.05 into the constant in Equation D-1 to obtain the different constant used in Equation B-7 (which excludes the R). I repeat, the two equations in reference (3) are completely equivalent.

For this article, I chose to use Equation D-1 because it showed the fact that R (the gas law constant) was indeed involved. I then converted Equation D-1 so that all of the units are metric (since the D-1 used a mixture of metric and non-metric units) to arrive at the equation presented in this article.

I know the above sounds complicated ... and it requires a close reading of reference (3), but the equation as I presented it in this article was correct and the R must be retained in the equation. - mbeychok 18:13, 15 December 2006 (UTC)[reply]

More discussion of EPA equation changed by MadScientist80[edit]

Regarding your changing of the equation:

  • 0.0035 is simply 0.284 divided by R = 82.05
  • There are a great many people outside of the USA who don't know what a ft2 is ... nor do they know what a lb is. Therefore, it is necessary to use metric units for them.
  • My equation as I wrote it (using all metric units) gives exactly the same answer as the EPA's mixed up units does. Try using:
    • u = 2 m/s (in EPA equation and mine)
    • M = 90 (in EPA equation and mine)
    • A = 100 ft2 (in EPA equation) = 9.294 m2 (in mine)... since 100 ft2 = 9.294 m2
    • P = 400 mmHg (in EPA equation) = 53.33 kPa (in mine) ... since 400 mmHg = 53.33 kPa
    • T = 300 K (in EPA equation and mine)
    • R = 82.05 (in Epa equation and mine)

The EPA equation will give you 15.94 lbs/min and my equation will give you 7.23 kg/min ... which are equivalent to each other.

As I said before, if you'll give me your fax number, I will fax you a copy of the above two calculations and you can see that they give the same answer.

My email address is mbeychok@cox.net and my phone number is 949-718-1360. I will wait to hear from you on the Discussion page of the Accidental release source terms article. Wikipedia only works well when we collaborate rather than simply reverting each other. If you don't respond within 24 hours, I will be forced to revert your change again. PLEASE contact me or try the above test. - mbeychok 16:43, 18 December 2006 (UTC)[reply]

Stiver and Mackay Equation is missing wind velocity[edit]

The Stiver and Mackay equation presented here does not include a term for wind velocity, even though such a term, u, is mentioned in the list of definitions for the equation's terms that follows the equation, and is necessary to the calculation. In other references, the term k is explained as the mass transfer coefficient under prevailing wind conditions in m/s. But k is a constant here, and thus the equation is incomplete. — Preceding unsigned comment added by Rstevec (talk • contribs) 21:23, 22 January 2011 (UTC)[reply]

Rstevec, please note that the definition of the mass transfer coefficient which reads as follows:
k = mass transfer coefficient, m/s = 0.002 u
Does that not make it quite clear that k is a function of u? In view of that, I am at a loss about your comment that the Stiver and Mackay equation presented in this article lacks a term for wind velocity. mbeychok (talk) 23:56, 11 November 2011 (UTC)[reply]

External links modified[edit]

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Missing Accidental release source terms?[edit]

I have been referencing a page of Milton Beychok's original webpage and discovered that this appears to have been the basis for this Wikipedia page. However, a section of Milton's page covering liquid releases does not appear to have been transferred and this omission does not have been recognised before. I have not been able to find an explanation for this, which has been replicated in other documents based on this Wikipedia page.

The missing sections are as follows:

Liquid Discharge From A Pressurized Source Vessel: refs 1, 2 Initial instantaneous flow through the discharge opening: (1) Qi = C A [ ( 2 g d 2 H ) + ( 2 d ) ( P - PA ) ] 1/2 Final flow when the liquid level reaches the bottom of the discharge opening: (2) Qf = C A [ ( 2 d ) ( P - PA ) ] 1/2 Average flow: (3) Qavg = ( Qi + Qf ) ÷ 2 where: Q = mass flow rate, kg / s C = discharge coefficient (dimensionless, usually about 0.62) A = discharge hole area, m 2 g = local gravitational acceleration of 9.807 m / s 2 d = source liquid density, kg / m 3 P = absolute source pressure, Pa PA = absolute ambient pressure, Pa H = height of liquid above bottom of discharge opening, m When expressed in the customary USA units, the equations (1) and (2) above also contain the gravitational conversion factor gc ( 32.17 ft / s2 in USA units ) associated with the ( P - PA ) term. Since gc is 1 ( kg-m ) / ( N-s2 ) in the SI metric system of units, equations (1) and (2) above do not include it.


Liquid Discharge From A Non-Pressurized Source Vessel: refs 1, 2 Initial instantaneous flow through the discharge opening: (1) Qi = C A ( 2 g d 2 H )1/2 Final flow when the liquid level reaches the bottom of the discharge opening: (2) Qf = 0 Average flow: (3) Qavg = Qi ÷ 2 where: Q = mass flow rate, kg / s C = discharge coefficient (dimensionless, usually about 0.62) A = discharge hole area, m 2 g = local gravitational acceleration of 9.807 m / s 2 d = source liquid density, kg / m 3 H = height of liquid above bottom of discharge opening, m

Jcaiken (talk) 13:13, 23 November 2019 (UTC)[reply]

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