Sir
I want to know Standard temperature & pressure (STP) and Normal Temperature & pressure(NTP) in a gas sytem and their conversion factor. How can I covert 1000 m3/hr of gas ( with P,T ,MW given ) to sm3/hr and Nm3/hr.
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Standard And Normal T P
Started by Guest_Guest_*, Jan 24 2005 01:02 AM
5 replies to this topic
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#1
Guest_Guest_*
Posted 24 January 2005 - 01:02 AM
#2
Guest_Guest_*
Posted 24 January 2005 - 08:49 AM
It depends - standard and normal are neither as 'standard' nor as 'normal' as the name implies. Different people/companies do sometimes specify them differently.
The definitions you will commonly see are as follows:
Normal conditions = 1 atmosphere pressure, temperature of 0 degC.
Standard conditions = 1 atmosphere pressure, temperature of 15 degC.
If you want to convert a vapour volumetric flowrate to Nm3/hr or Sm3 per hour, the easiest way is to use:
(P1 * V1)/T1 = (P2 * V2)/T2 .....................................................................(EQN 1)
Where P = pressure, V = volumetric flowrate, T = temperature, with 1 and 2 referring to initial and final conditions.
Another more convaluted way of doing it with gases is to remember that 1 kg-mole of an ideal gas occupies 22.4 m3. You can use this with a known molar flowrate, plus knowing the temperature, pressure and moleculare weight, with the ideal gas equation to get your volumetric flow.
The conversion I've given above as equation 1 only works if the fluid you are considering is vapour at both conditions. If you were looking at an air separation plant for example, there is a chance that your streams could be liquid or 2-phase under process conditions. In this case you would need to do the following:
1) use m' = v' * (rho) ------------ m' and v' are mass and volumetric flows respectively, (rho) is the density
2) use N' = m' / (MW) -------- N' and m' are mass and molar flows respectively, (MW) is the molecular weight
3) use p * v'' = N' * R * T ------- p and T are pressure and temperature under the specified conditions (normal or standard, you choose), v'' is the standard/normal volumetric flow (dependoing on which you chose), N' is your molar flowrate.
I use these equations with SI units, exceptions being where the conversion to another unit cancels out (e.g. using pressure in psi in eqn 1), or where a common factor has been introduced to both sides of the equation (such as hr^-1, in all equations).
Note that temperatures should be always be in absolute units, such as Kelvin.
The definitions you will commonly see are as follows:
Normal conditions = 1 atmosphere pressure, temperature of 0 degC.
Standard conditions = 1 atmosphere pressure, temperature of 15 degC.
If you want to convert a vapour volumetric flowrate to Nm3/hr or Sm3 per hour, the easiest way is to use:
(P1 * V1)/T1 = (P2 * V2)/T2 .....................................................................(EQN 1)
Where P = pressure, V = volumetric flowrate, T = temperature, with 1 and 2 referring to initial and final conditions.
Another more convaluted way of doing it with gases is to remember that 1 kg-mole of an ideal gas occupies 22.4 m3. You can use this with a known molar flowrate, plus knowing the temperature, pressure and moleculare weight, with the ideal gas equation to get your volumetric flow.
The conversion I've given above as equation 1 only works if the fluid you are considering is vapour at both conditions. If you were looking at an air separation plant for example, there is a chance that your streams could be liquid or 2-phase under process conditions. In this case you would need to do the following:
1) use m' = v' * (rho) ------------ m' and v' are mass and volumetric flows respectively, (rho) is the density
2) use N' = m' / (MW) -------- N' and m' are mass and molar flows respectively, (MW) is the molecular weight
3) use p * v'' = N' * R * T ------- p and T are pressure and temperature under the specified conditions (normal or standard, you choose), v'' is the standard/normal volumetric flow (dependoing on which you chose), N' is your molar flowrate.
I use these equations with SI units, exceptions being where the conversion to another unit cancels out (e.g. using pressure in psi in eqn 1), or where a common factor has been introduced to both sides of the equation (such as hr^-1, in all equations).
Note that temperatures should be always be in absolute units, such as Kelvin.
#3
Posted 24 January 2005 - 03:01 PM
As a supplement to the response you got from Dave:
(1) There are many, many different definitions of standard conditions and the metric normal conditions. In the USA, standard conditions are often defined as 14.696 psia and 60 °F. The American Gas Association (AGA) uses that but also sometimes uses 14.73 psia and 60 °F. The USA Environmental Protection Agency (EPA) uses 14.696 psia and 68 °F (20 °C).
In the metric nations, normal conditions are often defined as 1 atmosphere absolute pressure and 0 °C. However, the International Union of Pure and Applied Chemistry (IUPAC) defines the normal conditions as being 1 bar absolute pressure and 0 °C.
(2) The most important thing to remember is to ALWAYS determine what temperature and pressure basis is being used for the particular value of standard or normal conditions with which you are concerned ... ALWAYS!
(3) As pointed out by Dave, his equation (1) will enable you to convert from one pressure and temperature basis to another. As he also pointed out, be sure to use absolute temperatures (either °K or °R) and absolute pressures in his equation (1). In some cases, where the two pressure and temperature conditions differ by large amounts, you may have to use compressibility factors as follows:
(P1 * V1 * Z1)/T1 = (P2 * V2 * Z2)/T2
(1) There are many, many different definitions of standard conditions and the metric normal conditions. In the USA, standard conditions are often defined as 14.696 psia and 60 °F. The American Gas Association (AGA) uses that but also sometimes uses 14.73 psia and 60 °F. The USA Environmental Protection Agency (EPA) uses 14.696 psia and 68 °F (20 °C).
In the metric nations, normal conditions are often defined as 1 atmosphere absolute pressure and 0 °C. However, the International Union of Pure and Applied Chemistry (IUPAC) defines the normal conditions as being 1 bar absolute pressure and 0 °C.
(2) The most important thing to remember is to ALWAYS determine what temperature and pressure basis is being used for the particular value of standard or normal conditions with which you are concerned ... ALWAYS!
(3) As pointed out by Dave, his equation (1) will enable you to convert from one pressure and temperature basis to another. As he also pointed out, be sure to use absolute temperatures (either °K or °R) and absolute pressures in his equation (1). In some cases, where the two pressure and temperature conditions differ by large amounts, you may have to use compressibility factors as follows:
(P1 * V1 * Z1)/T1 = (P2 * V2 * Z2)/T2
#4
Posted 27 January 2005 - 12:18 PM
Just one addition. The equations Dave gives are correct. He correctly points out that they work only if no condensation occurs.
It is, howevever customary in flue gas treatment business (flue gas cleaning) to indicate gas volume flowrate in Nm3/hr (1 atm abs., 0 degree C), even if there is condensation.
A typical case would be a flue gas cleaning plant, handling 100000 m3/hr containing, say, 14% volume water (wet basis).
If we assume that the gas is at 0,98 atm and 200 degree C we would say that the normal flowrate is
100000*(273)/(273+200)*(0,98/1)=56560 Nm3/hr EVEN with water condensing. It is obvious that because at 0 degree most of the water would condense, the real flow rate we would have at zero degree C is some 14% less. My point is that Normal flowrate is just a convention.and Milton is very right to stress the importance of giving the reference T and P.
It is, howevever customary in flue gas treatment business (flue gas cleaning) to indicate gas volume flowrate in Nm3/hr (1 atm abs., 0 degree C), even if there is condensation.
A typical case would be a flue gas cleaning plant, handling 100000 m3/hr containing, say, 14% volume water (wet basis).
If we assume that the gas is at 0,98 atm and 200 degree C we would say that the normal flowrate is
100000*(273)/(273+200)*(0,98/1)=56560 Nm3/hr EVEN with water condensing. It is obvious that because at 0 degree most of the water would condense, the real flow rate we would have at zero degree C is some 14% less. My point is that Normal flowrate is just a convention.and Milton is very right to stress the importance of giving the reference T and P.
#5
Posted 22 July 2008 - 09:21 AM
QUOTE (mbeychok @ Jan 24 2005, 09:01 PM) <{POST_SNAPBACK}>
As a supplement to the response you got from Dave:
In the metric nations, normal conditions are often defined as 1 atmosphere absolute pressure and 0 °F. However, the International Union of Pure and Applied Chemistry (IUPAC) defines the normal conditions as being 1 bar absolute pressure and 0 °C.
In the metric nations, normal conditions are often defined as 1 atmosphere absolute pressure and 0 °F. However, the International Union of Pure and Applied Chemistry (IUPAC) defines the normal conditions as being 1 bar absolute pressure and 0 °C.
Milton,
Is that "0 °F" there, right after the phrase "metric nations", a typo?
I suspect it should be 0°C.
Cheers,
- Steve.
#6
Posted 27 July 2008 - 06:59 AM
I wonder no body has pointed one more parameter i.e. RH
standard condition defines some relative humidity so if you are dealing with air you have got to account RH to your volumetric flow rate.
standard condition defines some relative humidity so if you are dealing with air you have got to account RH to your volumetric flow rate.
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