I was trying to calculate emissions and the old emission calculation uses 387 scf/lbmol for off-gas to a boiler? How do I ensure that it is 387 scf/lbmol because all I found was conversions of 359 scf/lbmol or 379 scf/lbmol at STP. Where did they get 387 scf/lbmol from?
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Volumetric Flow Rate 1
- Thread starter newengr
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Montemayor
Chemical
newengr:
If you are truly a new Chemical Engineer out in industry, welcome to the real world.
Lesson #1: Always note the basis for your calculations and statements. 379 scf/lbmol is the "standard" volume that a lb mol of gas occupies. What inept engineers fail to state is that it is based on a certain base temperature and pressure (70 oF & 14.696 psia, I believe; I'm not at my home at present). The important point here is to learn that you can't just throw constants or "standard" numbers around without stating what the basis is (or are). Calculations that don't specifically state their basis are specifically worthless.
Lesson #2: Learn to demand and not be satisfied with anything less than specifics regarding any engineering calculation and decision you will be involved with. I realize that the era(s) of mentor engineering that I was raised under are now gone. This is a stronger reason for demanding what is due to you: The whole truth, not just scraps of truth. If you are to develop into a top-notch, experienced engineer you must have access to accurate, detailed, and fully explained calculations and decisions - in both mathematical and graphical modes.
The 387 figure is probably some sloppy, inept engineer's "standard" for the lbmol volume at a higher temperature base (or lower pressure) - which is perfectly OK, but lacks the identity of the base chosen; the problems start when this sloppiness causes errors, mistakes, lost time and effort in other engineers that inherit this type of inept calculations or communications. As an engineer you will be effective and successful to the degree that you are able to communicate your needs, orders, instructions, and wishes to your fellow workers - most of which are relying on you as their leader or in charge of the direction(s) to take.
It is not sufficient or enough to simply generate the absolutely correct mathematical solution to a problem. It must be communicated accurately, concisely, and efficiently in order to allow others around and below you to use the information correctly and in the precise manner that it was intended to be administered. Details are important! They make up the whole.
I hope other engineers that read this and who believe that they are real "hot shots" simply because they can multiply, divide and compute faster and more accurately than others, come to realize that they are (inspite of their mechanical and mental talents) ineffective and totally useless if they can't communicate this intelligence accurately and concisely to others. It's great to dominate and know your stuff; but if you can't communicate and explain it to others, you are a dead loss from a practical engineering and investment point of view. There are a lot of professional, registered engineers out there right now that couldn't explain to others how to build an effective outhouse if they had to - not because they don't know how, but because they don't know how to communicate.
Art Montemayor
Spring, TX
If you are truly a new Chemical Engineer out in industry, welcome to the real world.
Lesson #1: Always note the basis for your calculations and statements. 379 scf/lbmol is the "standard" volume that a lb mol of gas occupies. What inept engineers fail to state is that it is based on a certain base temperature and pressure (70 oF & 14.696 psia, I believe; I'm not at my home at present). The important point here is to learn that you can't just throw constants or "standard" numbers around without stating what the basis is (or are). Calculations that don't specifically state their basis are specifically worthless.
Lesson #2: Learn to demand and not be satisfied with anything less than specifics regarding any engineering calculation and decision you will be involved with. I realize that the era(s) of mentor engineering that I was raised under are now gone. This is a stronger reason for demanding what is due to you: The whole truth, not just scraps of truth. If you are to develop into a top-notch, experienced engineer you must have access to accurate, detailed, and fully explained calculations and decisions - in both mathematical and graphical modes.
The 387 figure is probably some sloppy, inept engineer's "standard" for the lbmol volume at a higher temperature base (or lower pressure) - which is perfectly OK, but lacks the identity of the base chosen; the problems start when this sloppiness causes errors, mistakes, lost time and effort in other engineers that inherit this type of inept calculations or communications. As an engineer you will be effective and successful to the degree that you are able to communicate your needs, orders, instructions, and wishes to your fellow workers - most of which are relying on you as their leader or in charge of the direction(s) to take.
It is not sufficient or enough to simply generate the absolutely correct mathematical solution to a problem. It must be communicated accurately, concisely, and efficiently in order to allow others around and below you to use the information correctly and in the precise manner that it was intended to be administered. Details are important! They make up the whole.
I hope other engineers that read this and who believe that they are real "hot shots" simply because they can multiply, divide and compute faster and more accurately than others, come to realize that they are (inspite of their mechanical and mental talents) ineffective and totally useless if they can't communicate this intelligence accurately and concisely to others. It's great to dominate and know your stuff; but if you can't communicate and explain it to others, you are a dead loss from a practical engineering and investment point of view. There are a lot of professional, registered engineers out there right now that couldn't explain to others how to build an effective outhouse if they had to - not because they don't know how, but because they don't know how to communicate.
Art Montemayor
Spring, TX
I would like to second (third?) what Art has said about numbers being useless if the basis from which they come is unknown. But beware - even when you specify exactly what you mean, your readers may not read it that way.
In the early part of my career I worked in a plant at an altitude of 5300 feet, where the atmospheric pressure was only 12.3 PSI. We were aware of the potential problems that this could cause (from bitter experience) and we always highlighted it in any enquiries, but suppliers invariably ignored that, and designed for 14.7 PSI. You would be surprised at the subtle and far reaching effects this has. (Horror stories censored for sensitive readers.)
Just for the record the numbers that newengr quoted come from the following conditions:
379 scf/lbmol is what I would call "standard" conditions - these are 60 F and 14.7 PSI
387 scf/lbmole is another "standard" - this time 70 F and 14.7 PSI
Beware - some people call 68 F and 14.7 PSI the "standard" conditions. This would be 385 scf/lbmol.
359 scf/lbmole is what is usually termed "Normal" conditions - 0 C and 760 mmHg. This is relatively rarely used with scf, but is the most common condition used with metric or SI figures. You will often see in European communications the units Nm3/h. These are Normal metre cubed per hour.
Whenever I give a gas rate in volumetric terms I always give it in mass terms as well. Kilograms and pounds remain the same where ever they are - even on the moon or in deep space - and a competent engineer will always be able to convert the mass flows into the volumetric flows in the conditions he/she needs.
Katmar
In the early part of my career I worked in a plant at an altitude of 5300 feet, where the atmospheric pressure was only 12.3 PSI. We were aware of the potential problems that this could cause (from bitter experience) and we always highlighted it in any enquiries, but suppliers invariably ignored that, and designed for 14.7 PSI. You would be surprised at the subtle and far reaching effects this has. (Horror stories censored for sensitive readers.)
Just for the record the numbers that newengr quoted come from the following conditions:
379 scf/lbmol is what I would call "standard" conditions - these are 60 F and 14.7 PSI
387 scf/lbmole is another "standard" - this time 70 F and 14.7 PSI
Beware - some people call 68 F and 14.7 PSI the "standard" conditions. This would be 385 scf/lbmol.
359 scf/lbmole is what is usually termed "Normal" conditions - 0 C and 760 mmHg. This is relatively rarely used with scf, but is the most common condition used with metric or SI figures. You will often see in European communications the units Nm3/h. These are Normal metre cubed per hour.
Whenever I give a gas rate in volumetric terms I always give it in mass terms as well. Kilograms and pounds remain the same where ever they are - even on the moon or in deep space - and a competent engineer will always be able to convert the mass flows into the volumetric flows in the conditions he/she needs.
Katmar
Not too much to add, yet one shouldn't forget these volumes correspond to ideal gases. I.e., they refer to gases that comply with PV=nRT, which, BTW, is an exact relationship for real gases in the limit as P=>0 provided possible dissociation of molecules is excluded.
The ideal gas is a "model" fluid because simple equations, as this one, may describe it. Equations as this one are applicable as very good approximations for actual gases.
For n=1, V=RT/P. When selecting as standard conditions P=1 atm, T=273 K (=492oR), with the gas constant R=0.0821 L.atm/(g-mol.K), one obtains:
This is the famous 22.4 L/g-mol, or 22.4 m3/kg-mol at standard conditions, mentioned in General Chemistry books, which converted to british units would be
Using the same law one can find the molar volume at other conditions following:
Assume pressure equals always 1 atm, the volume at other (absolute) temperatures would be:
60oF=520oR; 70oF=530oR
V60= 359(520/492)~ 379 ft3/lb-mol
V70= 359(530/492)~ 387 ft3/lb-mol
as Katmar as explained.![[pipe]](/data/assets/smilies/pipe.gif "[pipe] [pipe]")
The ideal gas is a "model" fluid because simple equations, as this one, may describe it. Equations as this one are applicable as very good approximations for actual gases.
For n=1, V=RT/P. When selecting as standard conditions P=1 atm, T=273 K (=492oR), with the gas constant R=0.0821 L.atm/(g-mol.K), one obtains:
V=(0.0821)(273.1)/(1) = 22.4 L/g-mol
This is the famous 22.4 L/g-mol, or 22.4 m3/kg-mol at standard conditions, mentioned in General Chemistry books, which converted to british units would be
(22.4)(454 g/lb)/(28.3 L/ft3) = 359 ft3/lb-mol
Using the same law one can find the molar volume at other conditions following:
(PV/T)2=(PV/T)1
Assume pressure equals always 1 atm, the volume at other (absolute) temperatures would be:
V2=V1(T2/T1)
60oF=520oR; 70oF=530oR
V60= 359(520/492)~ 379 ft3/lb-mol
V70= 359(530/492)~ 387 ft3/lb-mol
as Katmar as explained.
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