Dear SAK,
Please read through the following for an understanding of the conditions.
A. NTP And STP are conventional systemized conditions uniform throughout the world under any surrounding condition of P,T etc.
B. They are independent of the conditions in the site or elsewhere.
C. They are just a way of reporting the flow required in a pre-specified already known and set conditions so that they can be uniform throughout the globe in case you have to report the values to someone else, since you would be unaware of their flow conditions.
So they can then determine the actual flow rates from either the known Std. or Normal flowrate.
1. Conditions
In chemical engineering, as per API 520-1 , 2008 (5.6.4.1) , the normal conditions are as defined below.
- NTP is typically 0 oC and 1 atm (101.325 kPa) though it varies a lot as per the convention adopted for your case, so it is better to confirm with the respective people involved.
- STP is usually 15.6 oC and 1 atm (101.325 kPa) though it varies a lot as per the convention adopted for your case, so it is better to confirm with the respective people involved
In a lot of cases the conventions are taken differently but this is as per API so if no other convention is mentioned in your case then you can consider the above API conditions for calculations.
2. Real Gas Law
You cannot assume ideal gas for your calculations since it is a real case.
So you cannot apply P1*V1/T1 = P2*V2/T2.
You must consider the compressibility of the Gas at both the respective conditions 1 and 2.
So the Real Gas Law must be applied here as follows.
P1*V1/(T1*Z1) = P2*V2/(T2*Z2)
3. Conversion
Since the Flow of Air has been obtained to be 300 NM3/hr, it means that for a condition of 0 oC and 1 atm(1.01325 bar) pressure you get a flow of air to be 300 m3/hr.
So now to get the real flow at your operating Pressure and temperature you convert this using the real gas law.
P1*V1/(T1*Z1) = P2*V2/(T2*Z2)
4. Calculation Using Compressibility factor of Air
For your case you can consider this source for Air (https://en.wikipedia...sibility_of_air)
From the source you obtain
Z1 = 0.9996 (at 273K and 1.01325 Bar which are the Normal Conditions of Flow)
Z2 = 0.9965 (at 308 K (35 oC) and 7 Bar, which is your operating condition of Flow)
So you get,
(1.01325 Bar * 300 Nm3/hr) / (0.9996 * 273 K) = (7 Bar * Q m3/hr) / (0.9965 * 308)
This gives Q = 48.84 m3/hr at 35 oC and 7 Bar pressure.
5. Conclusion
Your flow is 300 Nm3/hr at 1 bar, so it simply cannot be 300 m3/hr at 7 bar as well for a very small temp. change of 35 oC. Your supervisor seems to have misconsidered some variable so show him the procedure for calculation.
If your Control Valve was operating at Normal Conditions of 1.013 bar and 0 oC, then your supervisor is correct in saying 300 m3/hr will be the air flow rate.
But since the conditions change, the above way is the correct way to calculate it, and you obtain 48.84 m3/hr as the flow.
Regards,
Shantanu
Edited by shantanuk100, 23 November 2015 - 04:50 AM.