Formula source of flow capacity of pressure relief deives for gas cylinders in CGA S 1.1 -2019

[yanxingqing]

Hi everyone, I am doing some pressure relief device calculations to gas cylinders according to CGA S 1.1-2019. There are two equations for flow capacity of PRD, Qa=0.154Wc for other PRD except PRV, and Qa=0.00154PWc for PRV. Qa is the flow capacity in ft3/min of free air, Wc is water capacity of the cylinder in pounds, and P is flow rating pressure in psia. I am confused about the deviation of these two equation. Why Qa is irrelevant to P for other PRD except PRV but is relevant to P for PRV? Could anyone tell me the devation of the euqations? or tell me where I can find the information on this. The information like documents, standards that require payment can also be accepted.
Thanks in advance for the help.

 

[Snickster]

Here is excerpt from CGA S 1.1-2005 for PRD:

For uninsulated cylinders for nonliquefied gas, the minimum required flow capacity of pressure relief devices,
except pressure relief valves, shall be calculated using the following formula. (For pressure relief valves, see
5.6 and 5.7.)

5.4.1 U.S. customary units

Qa=0.154Wc

Where:
Qa Flow capacity at 100 psi a test pressure in cubic feet per minute of free air
Wc Water capacity of the cylinder in pounds, but not less than 25 Ib

NOTE-The above formula expresses flow capacity requirements equal to 70% of that which will discharge through a perfect
orifice having a 0.00012 in cross-section area for each pound of water capacity of the cylinder.

Note that this equation is based on 100 psia cylinder pressure.

Here is except from CGA S 1.1-2005 for relief valve:

5.6.1 U.S. customary units
For uninsulated cylinders for nonliquefied gas, the minimum required flow capacity of pressure relief valves
shall be calculated using the following formula:

Qa=0.00154PWc

Where:
Qa Flow capacity in cubic feet per minute of free air
P Flow rating pressure in psia
Wc Water capacity of the cylinder in pounds, but not less than 12.5 Ib

Note that this equation is based on the actual pressure of the cylinder. If the cylinder pressure is 100 psia it would reduce to the same equation as for PRD. It appears that the equation for PRD only considers 100 psia cylinder pressure?

It appears the equations themselves are based on the maximum expected flow due to overfilling, fire, or thermal expansion due to heat input. I image if you convert pounds of water to cubic feet of water of cylinder volume you will get cubic feet of gas volume since volume is volume. At a given rate of heat input the pressure of the gas will increase at constant volume. To keep the pressure constant, a given flowrate of gas needs to be exhausted through the PRD/PRV.
There is no indication of exactly how they derived the required flow equation except they give a hint in the Note shown for the PRD:

NOTE-The above formula expresses flow capacity requirements equal to 70% of that which will discharge through a perfect
orifice having a 0.00012 in cross-section area for each pound of water capacity of the cylinder.

 

[Snickster]

Looking at the equation for a PRD and the Note that states it is based on flow of free air (per pound of cylinder water capacity Wc) through a perfect orifice multiplied by a factor of 0.7, the flow through a perfect orifice of 0.00012 in2 area at 100 psia and 60 F upstream with air can be found using the ideal gas equation as follows:

P(144)Q=mRT or P(144)(V)(A)=mRT

Where P is pressure in psia, Q is actual flowrate in cu. ft./sec at P&T (=Velocity ft/sec X area of orifice ft2), m is flow in lbs/sec. , R is universal gas constant, and T is temperature deg R.

For sonic velocity through the orifice with air k = 1.4 then Psonic = 0.528 Po = 0.528(100) = 52.8 psia (where Po is upstream stagnation pressure in cylinder)
And the flow is sonic since for any outlet pressure less than 52.8 psia the flow is sonic - actual outlet pressure is 14.7 psia.

And sonic velocity = SQRT(gkRT) = SQRT (gk(1545/MW)(2/(k+1))(To)) (where MW is molecular weight and To is upstream stagnation pressure in cylinder)

V = SQRT(32.2(1.4)(1545/29)(2/2.4)(520)) = 1020 ft/sec

Therefore:

52.8(144)(1020)(0.00012/144)=m(1545/29)(520)

Solve for m = 0.00028 lbs/sec This is the flow in pounds per second of air through a perfect orifice of 0.00012 in2 area, at 100 psia and 60 F upstream.

in lbs/min = 0.00028(60)= 0.0168 lbs/min

Now convert lbs/min to cu ft/min free air at 14.7 psia at 60 F using ideal gas equation:

14.7(144)Q=0.0168(1545/29)(520) Where Q is in cu ft/min

Q = 0.22 cu ft/min. This is the flow through a perfect orifice of free air at 0.0168 lbs/min.

The Note for the equation for PRD states that the ideal flow is multiplied by a factor of 0.7 to get the actual flow considering losses through the orifice:

Qa = 0.7 (0.22) = 0.154 cu ft/min which is the constant in the equation for PRD (in other words this is cu ft/min of free air per pound of water of cylinder volume)

This proves that Qa shown in the equation for PRD is equal to 0.154 times the ideal flowrate of air (of 0.22 cu ft/min) through a 0.00012 in2 orifice at 100 psia and 60 F upstream considering the flow is sonic through the orifice, since if you plug into the equation Wc = 1 pound of water then Qa = 0.154 cu ft/min.

Qa = 0.154Wc

Now how did they come up with this flowrate required based on all the different gasses that could be in the cylinder and all of the relieving conditions such as overfilling, fire and heating from external sources other than fire that is the question. I think they are basically looking at the equivalent volume that each pound of water occupies = 1/62.4 = 0.016 cu ft = volume of gas, and somehow calculating for some rate of heat input or overfill per volume of gas, what the maximum worse case relief rate would need to be for the worst case gas contained in the cylinder in terms of cu ft/min air so that the cylinder is not overpressurized.

 

[yanxingqing]

Looking at the equation for a PRD and the Note that states it is based on flow of free air (per pound of cylinder water capacity Wc) through a perfect orifice multiplied by a factor of 0.7, the flow through a perfect orifice of 0.00012 in2 area at 100 psia and 60 F upstream with air can be found using the ideal gas equation as follows:

P(144)Q=mRT or P(144)(V)(A)=mRT

Where P is pressure in psia, Q is actual flowrate in cu. ft./sec at P&T (=Velocity ft/sec X area of orifice ft2), m is flow in lbs/sec. , R is universal gas constant, and T is temperature deg R.

For sonic velocity through the orifice with air k = 1.4 then Psonic = 0.528 Po = 0.528(100) = 52.8 psia (where Po is upstream stagnation pressure in cylinder)
And the flow is sonic since for any outlet pressure less than 52.8 psia the flow is sonic - actual outlet pressure is 14.7 psia.

And sonic velocity = SQRT(gkRT) = SQRT (gk(1545/MW)(2/(k+1))(To)) (where MW is molecular weight and To is upstream stagnation pressure in cylinder)

V = SQRT(32.2(1.4)(1545/29)(2/2.4)(520)) = 1020 ft/sec

Therefore:

52.8(144)(1020)(0.00012/144)=m(1545/29)(520)

Solve for m = 0.00028 lbs/sec This is the flow in pounds per second of air through a perfect orifice of 0.00012 in2 area, at 100 psia and 60 F upstream.

in lbs/min = 0.00028(60)= 0.0168 lbs/min

Now convert lbs/min to cu ft/min free air at 14.7 psia at 60 F using ideal gas equation:

14.7(144)Q=0.0168(1545/29)(520) Where Q is in cu ft/min

Q = 0.22 cu ft/min. This is the flow through a perfect orifice of free air at 0.0168 lbs/min.

The Note for the equation for PRD states that the ideal flow is multiplied by a factor of 0.7 to get the actual flow considering losses through the orifice:

Qa = 0.7 (0.22) = 0.154 cu ft/min which is the constant in the equation for PRD (in other words this is cu ft/min of free air per pound of water of cylinder volume)

This proves that Qa shown in the equation for PRD is equal to 0.154 times the ideal flowrate of air (of 0.22 cu ft/min) through a 0.00012 in2 orifice at 100 psia and 60 F upstream considering the flow is sonic through the orifice, since if you plug into the equation Wc = 1 pound of water then Qa = 0.154 cu ft/min.

Qa = 0.154Wc

Now how did they come up with this flowrate required based on all the different gasses that could be in the cylinder and all of the relieving conditions such as overfilling, fire and heating from external sources other than fire that is the question. I think they are basically looking at the equivalent volume that each pound of water occupies = 1/62.4 = 0.016 cu ft = volume of gas, and somehow calculating for some rate of heat input or overfill per volume of gas, what the maximum worse case relief rate would need to be for the worst case gas contained in the cylinder in terms of cu ft/min air so that the cylinder is not overpressurized.

Thanks for your reply, Snickster. I really appreciate it. You are right, the first formula expresses that for 1 lb cylinder, the flow capacity is 0.154 cu ft/min, which equals to 70% of free air through a perfect orifice having a 0.00012 sq inch cross-section area at 100 psi and 60 F. I am still confused about the following questions.
(1) Why does they use 0.00012 sq inch cross-section for 1 lb cylinder as the reference value in overpressure situation? Are there any tests or analyses about this? I mean, why do they think 0.00012 sq inch area for 1 lb cylinder is enough in all ovrepressure situation?
(2) The euqations are different for PRV and other PRD. There is P in euqation for PRV, and isn't in equation for other PRD, why? In my opinion, the required flow capacity of cylinder in overpressure situation is only related to the reason of overpressure, and is not related to the type of PRD. Can 'Qa=0.154Wc' be used in any flow rating pressure?
(3) Are there any interpretation to CGA S 1.1? these two equations are so hard to be understood.
Thanks again for your time and reply again.