Luuk21
Chemical
- Oct 21, 2023
- 24
Hi,
I have a question regarding the calculation of the required capacity for a Low-Pressure storage tank, using API-2000.
I'm using the 5th edition of API-2000, we don't have the newer ones available.
The case is that 100 m3/h liquid ethanol is pumped into an ethanol storage tank, displacing 100 m3 of ethanol vapor. The set pressure of the valve is 1 barg / 15psig (still within limits API-2000), temperature 25 degrees, vapor density 3 kg/m3.
If I understand correctly, the required relief capacity for this particular case is:
202 Nm3/h - based on Table 1B - Normal Venting Requirements, API-2000 5t edition.
However, I'm getting told that you should convert the relieved gas to Normal Cubic Meter Air, which is 1m3 of Air at 0 degrees (273.15K). This because the vendor spec of the valve is in this unit.
According to our tool, this would lead to:
Relieving flow @normal cond. FV,N = FV,R · (T2 · P1 · Z2) / (T1 · P2 · Z1)
FV,N = 200 * (273.15 * 2 * 1) / (298.15 * 1.013 * 1) = 200 * 1.83 = 362 [Nm3/h] Ethanol
Density @ normal cond. rhoN = Mw*1.013/273.15/0.083144
rhoN = 46.068 * 1.013 / 273.15 / 0.083144 = 2.05 [kg/m3] Ethanol density at zero deg C & Patm
Equivalent diagram flow for vapor relief FV,dia = FV,N · sqrt[(rhoN·T1·1.013) / (rhoAir·273.15·P1)]
FV,dia = 362 * sqrt[(2.05 * 298.15 * 1.013) / (1.2 * 273.15 * 2)] = 352 [m3/h] equivalent flow Normal Cubic Meter Air
My question is;
Is this correct? Do you have to compensate for the Normal conditions or can you just apply Table 1B? My feeling is that you can simply apply Table 1B, since the distinction is made between low boiling / flashpoint liquids and high boiling / flashpoint liquids.
Then, liquid relief. Yes, I know that breather valves are not designed for liquid relief, but our tool does indicate there is a relation to determine the capacity:
FV,dia,L = FV,L / sqrt(rhoair / rhoL)
In case of a 30 m3/h overfilling this leads to;
FV,dia,L = 30 / SQRT(1.2/875) = 810 Nm3/h
I could not find any reference on this conversion. It makes sense since the correction in conversion to Normal Cubic Meters are is done Square-root. However, liquid behaves different than gas. Is this approach conservative?
See attachment for clarification
I have a question regarding the calculation of the required capacity for a Low-Pressure storage tank, using API-2000.
I'm using the 5th edition of API-2000, we don't have the newer ones available.
The case is that 100 m3/h liquid ethanol is pumped into an ethanol storage tank, displacing 100 m3 of ethanol vapor. The set pressure of the valve is 1 barg / 15psig (still within limits API-2000), temperature 25 degrees, vapor density 3 kg/m3.
If I understand correctly, the required relief capacity for this particular case is:
202 Nm3/h - based on Table 1B - Normal Venting Requirements, API-2000 5t edition.
However, I'm getting told that you should convert the relieved gas to Normal Cubic Meter Air, which is 1m3 of Air at 0 degrees (273.15K). This because the vendor spec of the valve is in this unit.
According to our tool, this would lead to:
Relieving flow @normal cond. FV,N = FV,R · (T2 · P1 · Z2) / (T1 · P2 · Z1)
FV,N = 200 * (273.15 * 2 * 1) / (298.15 * 1.013 * 1) = 200 * 1.83 = 362 [Nm3/h] Ethanol
Density @ normal cond. rhoN = Mw*1.013/273.15/0.083144
rhoN = 46.068 * 1.013 / 273.15 / 0.083144 = 2.05 [kg/m3] Ethanol density at zero deg C & Patm
Equivalent diagram flow for vapor relief FV,dia = FV,N · sqrt[(rhoN·T1·1.013) / (rhoAir·273.15·P1)]
FV,dia = 362 * sqrt[(2.05 * 298.15 * 1.013) / (1.2 * 273.15 * 2)] = 352 [m3/h] equivalent flow Normal Cubic Meter Air
My question is;
Is this correct? Do you have to compensate for the Normal conditions or can you just apply Table 1B? My feeling is that you can simply apply Table 1B, since the distinction is made between low boiling / flashpoint liquids and high boiling / flashpoint liquids.
Then, liquid relief. Yes, I know that breather valves are not designed for liquid relief, but our tool does indicate there is a relation to determine the capacity:
FV,dia,L = FV,L / sqrt(rhoair / rhoL)
In case of a 30 m3/h overfilling this leads to;
FV,dia,L = 30 / SQRT(1.2/875) = 810 Nm3/h
I could not find any reference on this conversion. It makes sense since the correction in conversion to Normal Cubic Meters are is done Square-root. However, liquid behaves different than gas. Is this approach conservative?
See attachment for clarification