Hi all,
API 2000 gives the relieving rate required in terms of Nm3/hr of air.
I am a bit confused over the equations to be used for converting this Nm3/hr of air to the actual vapour flowrate. The options given are:
1. Gas Law equation PV = ZnRT. Equating the terms for Air & gas at required temp & pressure & solving for the flow rate
2. W2 = W1 x SQRT((Z1 x T1 x M2) / (M1 x Z2 x T2))
W2, Mass Flow rate of Air (Kg / hr)
W1, Mass Flow rate of Gas (Kg / hr)
Z1, Compressibility factor for Gas
T1, Temperature of Gas in °K
M1, Molecular Weight of Gas (Kg / Kmole)
M2, Molecular Weight of Air (Kg / Kmole)
Z2, Compressibility Factor for air
T2 , Temperature of air in °K
3. As mentioned in the link below
http://www.enardo.co...s_equations.htm
Ofcourse all three gives different results. It would be quite helpful if anybody can explain me which equation to be used & more importantly WHY.
Thanks in advance

Conversion Of Nm3/hr Air To Actual Vapour Flow
Started by kirangparihar, Jun 22 2006 04:08 PM
6 replies to this topic
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#1
Posted 22 June 2006  04:08 PM
#2
Posted 23 June 2006  04:21 PM
kirangparihar,
I'm not quite sure what you're asking. Please try to better explain what your problem is, as I don't think that I can address it properly without more help from you.
I believe you're asking how to convert back and forth between Nm3/hr of air and Nm3/hr of a vapor stream that could be vented from a tank. Equations 1 & 2 are essentially the same, and are probably not enough to get you where you want to go. The Enardo equations are also very basic, but aren't going to get you any closer to what you need to know. I suggest that you start with understanding the characteristics of the device you are dealing with. Let's say you're interested in a relieving scenario where you want to vent process vapors from an atmospheric tank. (This is what I think you're interested in.) So you have somehow determined the flowrate of your vapor stream that will pass through a PSV to atmosphere. Now you look at the vendor's literature and discover that their devices are rated in Nm3/hr of air. Is this the situation you are facing? To delve into this further, I need your confirmation that I'm moving in the correct direction. Is this what you seek? If not, please correct my misunderstanding of your problem.
Thanks,
Doug
I'm not quite sure what you're asking. Please try to better explain what your problem is, as I don't think that I can address it properly without more help from you.
I believe you're asking how to convert back and forth between Nm3/hr of air and Nm3/hr of a vapor stream that could be vented from a tank. Equations 1 & 2 are essentially the same, and are probably not enough to get you where you want to go. The Enardo equations are also very basic, but aren't going to get you any closer to what you need to know. I suggest that you start with understanding the characteristics of the device you are dealing with. Let's say you're interested in a relieving scenario where you want to vent process vapors from an atmospheric tank. (This is what I think you're interested in.) So you have somehow determined the flowrate of your vapor stream that will pass through a PSV to atmosphere. Now you look at the vendor's literature and discover that their devices are rated in Nm3/hr of air. Is this the situation you are facing? To delve into this further, I need your confirmation that I'm moving in the correct direction. Is this what you seek? If not, please correct my misunderstanding of your problem.
Thanks,
Doug
#3
Posted 24 June 2006  10:31 AM
Hi Doug
First of all, thanks for your reply. The problem that I am facing is simple and will I try to explain although it may seem that I should have started the topic in Students Forum indeed !!!
I have to size a Vapour Recovery Unit for the tank farm area. I have calculated the vapour loading for the VRU from API 2000 & essentially it gives the results in Nm3/hr of air (or SCFH of air).
When I started converting back this flow rate to Kg/hr (mass flow rate), I used the Enardo equations for the same wherein the term "Volumetric Flow is inversely proportional to the square root of the Molecular Weight (MW) of the vapor". This thing raised the eyebrow of the Engineers here and asked me why can't we use simple PV = nRT wherein no square root term comes into picture!!!. I had a vague answer for that.
I tried to dervive the term sqrt (M2/M1) by using the continuity equation Q = A x v wherein v (velocity) is replaced using the equation Del. P = v^2 / 2g. From here I got the term "under root of Del.P" which again can be replaced by P = nRT/V. Somewhere I got close to term sqrt (M2/M1).
Hence Proved!!!. But I got no takers for my derivation since it contains other terms too & also I am not too confident.
Also if we simply look API 520 equations We have flow rate proprotional Sqrt (Temp, & 1/M.W). Now here the problem is that the along with Molecular weight, temperature is also a term in Sqrt.
I got a lot of confusion!!!.
In simple words, for converting Nm3/hr of Air to Actual m3/hr (or Kg/hr), which of the 3 equations should I use
Thanks in advance
Kiran Parihar
First of all, thanks for your reply. The problem that I am facing is simple and will I try to explain although it may seem that I should have started the topic in Students Forum indeed !!!
I have to size a Vapour Recovery Unit for the tank farm area. I have calculated the vapour loading for the VRU from API 2000 & essentially it gives the results in Nm3/hr of air (or SCFH of air).
When I started converting back this flow rate to Kg/hr (mass flow rate), I used the Enardo equations for the same wherein the term "Volumetric Flow is inversely proportional to the square root of the Molecular Weight (MW) of the vapor". This thing raised the eyebrow of the Engineers here and asked me why can't we use simple PV = nRT wherein no square root term comes into picture!!!. I had a vague answer for that.
I tried to dervive the term sqrt (M2/M1) by using the continuity equation Q = A x v wherein v (velocity) is replaced using the equation Del. P = v^2 / 2g. From here I got the term "under root of Del.P" which again can be replaced by P = nRT/V. Somewhere I got close to term sqrt (M2/M1).
Hence Proved!!!. But I got no takers for my derivation since it contains other terms too & also I am not too confident.
Also if we simply look API 520 equations We have flow rate proprotional Sqrt (Temp, & 1/M.W). Now here the problem is that the along with Molecular weight, temperature is also a term in Sqrt.
I got a lot of confusion!!!.
In simple words, for converting Nm3/hr of Air to Actual m3/hr (or Kg/hr), which of the 3 equations should I use
Thanks in advance
Kiran Parihar
#4
Posted 26 June 2006  02:06 AM
Kiran Parihar:
You are making some simple conversions much harder than they are by looking at too many confusing references. So let's forget all the equations you've looked at and start from fundamentals!!
Converting Nm3/h to actual m3/hr:
First of all, you must find out what reference temperature (Tr) and reference pressure (Pr) were used for defining the Nm3/h that you obtained. A Nm3 is usually (but not always) defined as being at 273.15 °K and 101.325 kPa (kiloPascals). Since other reference conditions may have been used, you should find out exactly what reference temperature and pressure was used to define your Nm3/h ... I want to stress that most strongly.
Now, using the universal gas law twice:
(1) ... (Pa)(Va) = (Za)(n)(R')(Ta)
(2) ... (Pr)(Vr) = (Zr)(n)(R')(Tr)
Dividing equation (1) by equation (2), we get:
Va = (Vr)(Za/Zr)(Ta/Tr)(Pr/Pa)
where:
Va = actual m3/hr
Vr = Nm3/h
Za = compressibility factor at Ta and Pa, dimensionless
Zr = compressibility factor at Tr and Pr, dimensionless
Ta = the absolute temperature at which you want the actual m3/h, in °K
Pa = the absolute pressure at which you want the actual m3/h, in kPa (i.e., kiloPascals)
Tr = the absolute reference temperature that applies to your Nm3/h, in °K
Pr = the absolute reference pressure that applies to your Nm3/h, in kPa (i.e., kiloPascals)
Unless your actual condition pressure (Pa) is quite high or you need some very high accuracy, in most cases you can ignore using the compressibility factors.
Converting either Nm3/h or actual m3/h to kg/h:
Again, start with the universal gas law:
(3) ... (P)(V) = (Z)(n)(R')(T) ..... and since n = w/M, we get:
(4) ... w = (M)(P)(V) / [(Z)(R')(T)]
where:
w = kg/h
M = molecular weight of the gas
T = absolute temperature, in °K
P = absolute pressure, in kPa (i.e., kiloPascals)
R' = the universal gas law constant of 8.3145
Z = the compressibility, dimensionless
Equation (4) could be used to get kg/h based on using either the Z, P and T of your actual m3/h or the reference Z, P and T of your Nm3/h. Either way should give the same answer.
I am assuming that up to this point, everything is based on air (M = 28.96). So finally, to convert to your actual gas:
kg/h of gas = (kg/h of air)(M of gas) / (M of air)
To clear up confusion about square root term:
If your Nm3/h of air was obtained from a flow meter of some type that works by measuring a ΔP across an orifice or some other device and you want to recalibrate the meter to use a different set of reference temperature and pressure, then the square root term comes into play. But, in this case, it seems to me that you are not trying to recalibrate a meter. You are simply tring to convert the Nm3/h of air (that you got from a meter or some other way) to kg/h of your actual gas. In that case, the above equations are all you need!
You are making some simple conversions much harder than they are by looking at too many confusing references. So let's forget all the equations you've looked at and start from fundamentals!!
Converting Nm3/h to actual m3/hr:
First of all, you must find out what reference temperature (Tr) and reference pressure (Pr) were used for defining the Nm3/h that you obtained. A Nm3 is usually (but not always) defined as being at 273.15 °K and 101.325 kPa (kiloPascals). Since other reference conditions may have been used, you should find out exactly what reference temperature and pressure was used to define your Nm3/h ... I want to stress that most strongly.
Now, using the universal gas law twice:
(1) ... (Pa)(Va) = (Za)(n)(R')(Ta)
(2) ... (Pr)(Vr) = (Zr)(n)(R')(Tr)
Dividing equation (1) by equation (2), we get:
Va = (Vr)(Za/Zr)(Ta/Tr)(Pr/Pa)
where:
Va = actual m3/hr
Vr = Nm3/h
Za = compressibility factor at Ta and Pa, dimensionless
Zr = compressibility factor at Tr and Pr, dimensionless
Ta = the absolute temperature at which you want the actual m3/h, in °K
Pa = the absolute pressure at which you want the actual m3/h, in kPa (i.e., kiloPascals)
Tr = the absolute reference temperature that applies to your Nm3/h, in °K
Pr = the absolute reference pressure that applies to your Nm3/h, in kPa (i.e., kiloPascals)
Unless your actual condition pressure (Pa) is quite high or you need some very high accuracy, in most cases you can ignore using the compressibility factors.
Converting either Nm3/h or actual m3/h to kg/h:
Again, start with the universal gas law:
(3) ... (P)(V) = (Z)(n)(R')(T) ..... and since n = w/M, we get:
(4) ... w = (M)(P)(V) / [(Z)(R')(T)]
where:
w = kg/h
M = molecular weight of the gas
T = absolute temperature, in °K
P = absolute pressure, in kPa (i.e., kiloPascals)
R' = the universal gas law constant of 8.3145
Z = the compressibility, dimensionless
Equation (4) could be used to get kg/h based on using either the Z, P and T of your actual m3/h or the reference Z, P and T of your Nm3/h. Either way should give the same answer.
I am assuming that up to this point, everything is based on air (M = 28.96). So finally, to convert to your actual gas:
kg/h of gas = (kg/h of air)(M of gas) / (M of air)
To clear up confusion about square root term:
If your Nm3/h of air was obtained from a flow meter of some type that works by measuring a ΔP across an orifice or some other device and you want to recalibrate the meter to use a different set of reference temperature and pressure, then the square root term comes into play. But, in this case, it seems to me that you are not trying to recalibrate a meter. You are simply tring to convert the Nm3/h of air (that you got from a meter or some other way) to kg/h of your actual gas. In that case, the above equations are all you need!
#5
Posted 26 June 2006  12:55 PM
Kiran:
There's not much I can add to Milton's excellent and comprehensive reply to your question. In fact, your question alerted me to my own ignorance of the principles involved in venting devices. Though I'm finding this question a bit difficult to express, I am curious if a conservation vent, PSV, etc. would be considered to be mass or a volumetric devices. I know that for gases conservation vent and PSV's and rupture disks, etc. tend to be rated in Nm3 or SCF per unit of time. If they are volumetric devices, then if the molecular weight changed, they would still be capable of passing the same Nm3 or SCF as before (constant volume = constant moles). If they are mass flow devices, then if the molecular weight were to increase, then the Nm3 or SCF rating would have to decrease to keep the kg/hr or lb/hr constant (but moles vary). Could anyone explain how these devices operate?
Thanks,
Doug
There's not much I can add to Milton's excellent and comprehensive reply to your question. In fact, your question alerted me to my own ignorance of the principles involved in venting devices. Though I'm finding this question a bit difficult to express, I am curious if a conservation vent, PSV, etc. would be considered to be mass or a volumetric devices. I know that for gases conservation vent and PSV's and rupture disks, etc. tend to be rated in Nm3 or SCF per unit of time. If they are volumetric devices, then if the molecular weight changed, they would still be capable of passing the same Nm3 or SCF as before (constant volume = constant moles). If they are mass flow devices, then if the molecular weight were to increase, then the Nm3 or SCF rating would have to decrease to keep the kg/hr or lb/hr constant (but moles vary). Could anyone explain how these devices operate?
Thanks,
Doug
#6
Posted 03 January 2008  12:58 PM
Dear Kirangparihar,
regarding your query on the 22 of Jun 2006 4:08 p.m.
I know that my reply comes late but I saw the discussion only now. My response is:

Your objective I believe is to calculate the equivalent flow of your substance that will give you the same relief valve size as the flow of air calculated by API2000. First of all, if I were you I would send the equivalent flow of air to the valve vendor and the supplier then will size the valve accordingly. If you insist converting the mass flow of air to the equivalent mass flow of the substance that gives you the same orifice size then I would go with your approach number 2.
Note equation 4A and 4B in API2000. For the same orifice area we should have:
A = Wair*SQRT(Mair*Tair*Zair) = Wreal*SQRT(Mreal*Treal*Zreal).
Wair in Nm3/hr
Wreal in Nm3/hr
If you convert the Nm3/hr to kg/hr you will get
So W_air*SQRT(Mreal*Tair*Zair) = W_real*SQRT(Mair*Treal*Zreal)
which is your option 2.
W_air in kg/hr
W_real in kg/hr
Now API2000 assumes that the air is at 15degC. You can prove that by deriving equations (1B) and (1A) in the same standard. The compressibility of air is 1 and we assume here that the isentropic exponent in equations (4A) and (4B) is the same for air and for the vapour you are looking at.
I have crossed checked the result with:
1. Vendor data (Shand & Jurs)
2. Equation (1A) and (1B) of the same standard that basically converts the vapour release to air flow at 15degC using the same formula.
mbeychock to my understanding assumes that the same molar flow of air and gas will give the same relief valve orifice size, hence he is using the equation:
kg/hgas = (kg/hr air) *(M of gas) / (M of air).
This is not correct according to API200 eq. (4A) and (4B) that dictate that the orifice area is proportional to the molar flow and to the square root of temperature and molecular weight....

For all the above I followed standards rather than judgement and intuition.
Hope it helps for any more info please communicate via emial I can send you the derivations and the vendor data.
regarding your query on the 22 of Jun 2006 4:08 p.m.
I know that my reply comes late but I saw the discussion only now. My response is:

Your objective I believe is to calculate the equivalent flow of your substance that will give you the same relief valve size as the flow of air calculated by API2000. First of all, if I were you I would send the equivalent flow of air to the valve vendor and the supplier then will size the valve accordingly. If you insist converting the mass flow of air to the equivalent mass flow of the substance that gives you the same orifice size then I would go with your approach number 2.
Note equation 4A and 4B in API2000. For the same orifice area we should have:
A = Wair*SQRT(Mair*Tair*Zair) = Wreal*SQRT(Mreal*Treal*Zreal).
Wair in Nm3/hr
Wreal in Nm3/hr
If you convert the Nm3/hr to kg/hr you will get
So W_air*SQRT(Mreal*Tair*Zair) = W_real*SQRT(Mair*Treal*Zreal)
which is your option 2.
W_air in kg/hr
W_real in kg/hr
Now API2000 assumes that the air is at 15degC. You can prove that by deriving equations (1B) and (1A) in the same standard. The compressibility of air is 1 and we assume here that the isentropic exponent in equations (4A) and (4B) is the same for air and for the vapour you are looking at.
I have crossed checked the result with:
1. Vendor data (Shand & Jurs)
2. Equation (1A) and (1B) of the same standard that basically converts the vapour release to air flow at 15degC using the same formula.
mbeychock to my understanding assumes that the same molar flow of air and gas will give the same relief valve orifice size, hence he is using the equation:
kg/hgas = (kg/hr air) *(M of gas) / (M of air).
This is not correct according to API200 eq. (4A) and (4B) that dictate that the orifice area is proportional to the molar flow and to the square root of temperature and molecular weight....

For all the above I followed standards rather than judgement and intuition.
Hope it helps for any more info please communicate via emial I can send you the derivations and the vendor data.
#7
Posted 05 January 2008  06:46 AM
Also late and maybe adding to the confusion. A large chemical company says the following in their guidelines for pressure relief:
An actual volumetric flow rate of gas or vapour at tank conditions can be converted to the equivalent air flow rate at reference conditions by multiplying the actual volumetric flow rate by: SQRT(rho_A/rho_R)
Where:
rho_A: Vapour density at tank conditions
rho_R: Air density at reference conditions
Using this method gives slightly different values compared to the Enardo method but can give massive differences compared to the gas law method (depending on molecular weight of the gas).
An actual volumetric flow rate of gas or vapour at tank conditions can be converted to the equivalent air flow rate at reference conditions by multiplying the actual volumetric flow rate by: SQRT(rho_A/rho_R)
Where:
rho_A: Vapour density at tank conditions
rho_R: Air density at reference conditions
Using this method gives slightly different values compared to the Enardo method but can give massive differences compared to the gas law method (depending on molecular weight of the gas).
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