MTF Test Stand 4 Relief Sizing

Roger Rabehl

11/6/00

**Introduction**

A primary system of relief devices and a secondary system are required
by paragraph 4.1.1 of CGA S-1.3-1995 \[2\]. Paragraph 1.7 of this code
says, \"CGA believes that reclosing pressure relief devices on a
container shall be able to handle all operational emergency conditions
except fire. . . . The operational emergency conditions referred to here
shall include but not be limited to loss of vacuum, runaway fill. . .
.\" Nevertheless, for loss of insulating vacuum to air, the venting rate
from a liquid helium vessel is so large that a rupture disk is the only
practical solution, and that is the approach taken here. The rupture
disk (the secondary system) is sized for fire or loss of vacuum to air,
and the primary relief should handle any other \"operational emergency\"
condition, including quenching.

With the primary relief sized per paragraph 5.2 and retested every 5
years, a secondary relief consisting of a rupture disk may be set to a
maximum of 150% of the MAWP (paragraph 4.1.2.3(2)).

The MAWP of Stand 4 = 85 psig = 100 psia. Therefore, the primary relief
must be set at 85 psig or less, and the rupture disk at 1.5 x 85 = 127
psig or less. The two relief valves are set at 50 psig and 55 psig, and
the rupture disk is set at 100 psig.

# Primary relief sizing

The primary relief size neglecting unusual events outside of the realm
of CGA considerations like quenching is estimated per paragraph 5.2.2
\[2\]:

$$Q_{a} = \frac{\text{590} - T}{4(\text{1660} - T)}\text{FG}_{i}\text{UA}$$

This sizes the relief for loss of vacuum to helium. The parameters in
this equation and the result are evaluated in the attached spreadsheet.
The primary relief size must be at least 224 SCFM air.

# Fire relief sizing

The fire relief size is estimated per paragraph 5.3.3 \[2\]:

$$Q_{a} = \text{FG}_{i}\text{UA}^{0\text{.}\text{82}}$$

The parameters and result are also shown in the attached spreadsheet.
The fire relief size must be at least 1169 SCFM air.

# Loss of Insulating Vacuum Relief

Loss of insulating vacuum to air is also evaluated on the attached
spreadsheet. A good reference for air condensation on a helium vessel is
a paper by Lehmann and Zahn \[4\]. For a superinsulated vessel, the
result is 0.6 W/cm^2^ of vessel surface area, or about 0.03 g/s/cm^2^ of
vessel area.

The surface area was estimated using the known dimensions of the feed
can helium vessel, cold masses, and external heat exchanger pipes. For
the cold mass, additional heat transfer resistance was added to account
for the conduction from the cold mass skin to the cold mass cooling
holes. The cold mass skin is not in direct contact with a large volume
of liquid helium. Aside from the end domes, the bulk of the helium is in
the cold mass cooling holes. There are then two heat transfer
resistances in series: condensation on the cold mass skin, and
conduction from the cold mass skin to the cooling holes. For the
condensation of air:

$$R^{\text{''}} = \frac{\Delta T}{q^{\text{''}}} = \frac{\text{300 K}}{0\text{.}\text{6 W/cm}^{2}} = \text{ 500 }\frac{\text{cm}^{2} - K}{W}$$

The conduction resistance from the cold mass skin to the cooling holes
was approximated as a cylindrical geometry with its outer radius at the
outer radius of the cold mass and its inner radius at the centers of the
cooling holes, as shown in Figure 1.

![](media/image1.wmf)

**[Figure 1]{.underline}** Geometry used for cold mass conduction
resistance.

$$R^{\text{''}} = \frac{R_{i}\text{ ln}\left( \frac{R_{o}}{R_{i}} \right)}{k} = \frac{\left( \text{20}\text{.}\text{8 cm} \right)\text{ ln}\left( \frac{\text{20}\text{.}8\text{cm}}{\text{16}\text{ cm}} \right)}{0\text{.}\text{03 W/cm-K}} = \text{ 182 }\frac{\text{cm}^{2} - K}{W}$$

The conduction resistance increases the total heat transfer resistance
by about 1/3. This is accounted for in the relief sizing calculations by
using ¾ of the cold mass skin surface area with the 0.6 W/cm^2^ heat
flux.

The flow rate is estimated to be 4090 g/s. The result is that a 1.47
inch diameter orifice is required, equivalent to a 2691 SCFM air
capacity for 137 psia at the rupture disk. This calculation includes a 6
psi pressure drop along the relief piping.

This calculation assumes a 9 K temperature in the piping leading to the
rupture disk. The flow rate is 4.09 kg/s with a temperature rise from
7.9 K (the temperature at which the specific heat input at 142 psia is a
minimum) to 9 K. This corresponds to a required heat input of 40.2 kW
from the relief piping. Using 3.8 W/cm^2^ as the heat input from bare
relief piping \[4\], a 40.2 kW heat input corresponds to 12.4 ft of 3
IPS pipe, which would have a total surface area of 1.06 m^2^. The relief
piping as constructed has approximately 10.5 ft of 3 IPS piping (0.89
m^2^ of surface area), 6.5 ft of 3 IPS flexhose (1.3 m^2^, assuming the
corrugated flexhose has a surface area of 2.5 times that of straight
pipe), and 2 ft of 2 IPS piping (0.1 m^2^ of surface area). The
estimated total installed surface area is 2.3 m^2^.

The addition of foam pipe insulation will greatly reduce the heat
transfer to the venting helium. The heat transfer resistance from
atmosphere to bare relief piping is given by:

$$R^{\text{''}} = \frac{\Delta T}{q^{\text{''}}} = \frac{\text{295 K}}{3\text{.}\text{8 W/cm}^{2}} = \text{ 77}\text{.}\text{6 }\frac{\text{cm}^{2} - K}{W}$$

A typical thermal conductivity for pipe insulation is k = 0.03 W/m-K =
0.0003 W/cm-K. For a thin, 0.25 in (0.64 cm) layer of pipe insulation:

$$R^{\text{''}} = \frac{t}{k} = \frac{0\text{.}\text{64 cm}}{0\text{.}\text{0003 W/cm-K}} = \text{ 2133 }\frac{\text{cm}^{2} - K}{W}$$

The use of even a thin layer of pipe insulation greatly increases the
heat transfer resistance. The heat flux is reduced from 3.8 W/cm^2^ to
0.13 W/cm^2^. The length of 3 IPS now required to raise the temperature
of the venting helium to 9 K is very long, and the 9 K temperature used
in the relief calculations is therefore conservatively high. Pipe
insulation will be placed on the system relief piping.

# Quench Relief

The primary relief must also be sized for magnet quenches. The rate at
which a powered Q2a&b magnet will deposit energy into the liquid helium
can be estimated using data from quench testing of dipole prototypes at
CERN \[3\]. A 13 kA quench of a dipole prototype produces an initial
spike in pressure and helium discharge flow. After approximately 2
seconds, the helium discharge flow is steady at 3 kg/s at a magnet
pressure of 1 atm. Using the heat of vaporization at 1 atm, a 3 kg/s
discharge flow corresponds to a heat rate of 62 kW. This heat rate is
then scaled with the coil area in contact with the liquid helium. The
dipole prototype is 10 m long with two 50 mm diameter bore tubes,
yielding 3.1 m^2^ of coil/helium interface area. The total length of
cold mass in a Q2a&b is 11 m with a single 70 mm bore tube, yielding 2.4
m^2^ of coil/helium interface area. The resulting estimated heat rate
for a Q2a&b is 48 kW.

At the relieving pressure of 73 psia, the minimum value of the specific
heat input is 22.6 J/g at 6.5 K. The discharge flow rate resulting from
a 48 kW heat rate is 2.12 kg/s. The standard volumetric flow rate is
1500 scfm air. This calculation includes a 3 psi pressure drop along the
relief piping.

This calculation assumes a 9 K temperature in the piping leading to the
relief valve. The flow rate is 2.12 kg/s with a temperature rise from
6.5 K (the temperature at which the specific heat input at 73 psia is a
minimum) to 9 K. This corresponds to a required heat input of 52 kW from
the relief piping. Using 3.8 W/cm^2^ as the heat input from bare relief
piping \[4\], a 52 kW heat input would require 20 ft of 2-1/2 IPS pipe.
The installed piping includes 18 ft of vacuum jacketed 2-1/2 IPS pipe
and 6 ft of bare 2 IPS pipe. The heat input is expected to be less than
52 kW, so the assumed 9 K temperature is therefore conservatively high.

# Combined Quenching and Loss of Vacuum

Loss of insulating vacuum on the test stand while a magnet is powered
would also result in a magnet quench, so the relief system must also
account for these events occurring simultaneously. Adding the heat rates
for the two individual events results in an equivalent flow rate of 3638
scfm air. This calculation includes an estimated 10 psi pressure drop
along the relief piping. This value was arrived at by scaling the
pressure drop for loss of vacuum with the square of the mass flow rate.

## Rate of pressure rise in test stand

The estimated rate of pressure rise in the test stand was used to choose
the burst pressure of the rupture disk relative to the setpoints of the
relief valves. The relief sizing calculations look at steady-state flow
from the test stand at maximum flow rates for each situation, with the
relief device inlet conditions at the specified temperature and
pressure. However, the buildup of pressure and cool down of vent lines
is a dynamic process. The experience at MTF is that vent lines drip
liquid air within a second or two of a quench on the horizontal test
stands. The settings of the relief valves were chosen such that the
quench relief will open first, followed by the main system relief,
followed by the rupture disk. The maximum rupture disk flow occurs 1.5
seconds after the system relief begins venting and 0.5 seconds after the
rupture disk bursts. These setpoints will allow the piping to the
rupture disk to be precooled before maximum rupture disk flow is
required.

Adding the 148 kW of heat for loss of vacuum to air and 48 kW of heat
for quenching gives a maximum of 196 kW of heat added to the test stand.
Figure 2 shows a plot of test stand pressure versus time with a
continuous addition of 196 kW of heat. The test stand is treated as a
closed volume of helium (i.e., no lead flow or addition of LHe). The
calculations are started with the feed can containing 19 psia saturated
liquid above the lambda plate at the normal operating liquid level and
19 psia/1.9 K subcooled liquid below the lambda plate. The helium gas
between the liquid and the top plate is assumed to be saturated vapor at
the same conditions as the liquid. In reality, there is a temperature
gradient from the liquid temperature up to room temperature. The mass of
vapor in this space is small relative to the total mass of helium in the
test stand and magnet, so treating it as saturated vapor is reasonable.

Based on helium vessel and magnet dimensions, the estimated helium
volumes are 750 l at 1.9 K below the lambda plate, 90 l at 4.5 K above
the lambda plate, and 120 l of gas space. This yields a total helium
volume of 960 l. The 196 kW heat load will quickly raise the temperature
of the 1.9 K helium. This temperature rise is accompanied by a
significant reduction in helium density. When the 1.9 K volume reaches
4.2 K, the entire helium space is filled with 19 psia/4.2 K subcooled
liquid. As heat continues to be added to the helium, the subcooled
liquid eventually makes the transition to supercritical fluid.

Figure 2 shows the test stand pressure as a function of time. Time = 0
in this plot is the point at which the helium volume is filled with 19
psia/4.2 K subcooled liquid. After 1 second, the first relief valve will
open. After 1.3 seconds, the second relief valve will open. After 2.5
seconds, the rupture disk will burst. It is important to remember that
the pressure charted in Figure 2 assumes a closed volume. Opening a
relief valve will slow the rate of pressure rise, allowing additional
time for the vent lines to cool.

![](media/image2.wmf){width="4.1875in" height="2.7916666666666665in"}

**[Figure 2]{.underline}** Test stand pressure vs. time for simultaneous
magnet quench and loss of insulating vacuum.

# Summary

+-------------+-----------+--------------+----------------+----------+
| Requirement | Required  | Device(s)    | Size/capacity  | Setpoint |
| or event    | capacity  | satisfying   | of installed   | of       |
|             | \[scfm    | requirement  | device(s)      | i        |
|             | air\]     |              |                | nstalled |
|             |           |              | \[scfm air\]   | d        |
|             |           |              |                | evice(s) |
|             |           |              |                |          |
|             |           |              |                | \[psig\] |
+-------------+-----------+--------------+----------------+----------+
| Primary     | 224 at 55 | AG relief    | 838 scfm at 55 | 55       |
| relief (CGA | psig      | valve        | psig (0.785    |          |
| 5.2.2)      |           |              | in^2^ orifice) |          |
+-------------+-----------+--------------+----------------+----------+
| Quenching   | 1500 at   | 2 AG relief  | 2 @ 838 scfm   | 50, 55   |
|             | 55 psig   | valves       | at 55 psig     |          |
|             | inlet     |              | inlet (2 @     |          |
|             |           |              | 0.785 in^2^    |          |
|             |           |              | orifice)       |          |
+-------------+-----------+--------------+----------------+----------+
| Fire        | 1169 at   | Rupture disk | 2658 scfm at   | 100      |
| (secondary) | 115 psig  |              | 115 psig       |          |
| relief (CGA |           |              |                |          |
| 5.3.3)      |           |              | (1.5 in.       |          |
|             |           |              | diameter       |          |
|             |           |              | rupture disk)  |          |
+-------------+-----------+--------------+----------------+----------+
| Air         | 2691 at   | Rupture disk | 2808 scfm at   | 100      |
| c           | 122 psig  | (1.47 in.    | 122 psig inlet |          |
| ondensation | inlet     | diameter     | (1.5 in.       |          |
| (secondary  |           | min.)        | diameter       |          |
| relief)     |           |              | rupture disk)  |          |
+-------------+-----------+--------------+----------------+----------+
| Air         | 3638 at   | 1 AG relief  | 1581 scfm at   | 50 (AG   |
| c           | 117 psig  | valve and    | 117 psig inlet | valve)   |
| ondensation | inlet     | rupture disk | (0.785 in^2^   |          |
| plus quench |           |              | orifice) and   | 100      |
|             |           |              | 2706 scfm at   | (rupture |
|             |           |              | 117 psig inlet | disk)    |
|             |           |              | (1.5 in.       |          |
|             |           |              | diameter       |          |
|             |           |              | rupture disk)  |          |
+-------------+-----------+--------------+----------------+----------+

# 

# References

1.  ASME Boiler and Pressure Vessel Code Section VIII, Div. 1, UG-125
    through UG-136;

2.  CGA S-1.3 - 1995 (Compressed Gas Association Pressure Relief Device
    Standards\--Part 3\--Stationary Storage Containers for Compressed
    Gases);

3.  Gerin, G., Vullierme, B., and van Weelderen R., "Measurement of the
    Thermo Hydraulic Behaviour of LHC Dipole Prototypes After Quench",
    CEC42, Columbus, OH, 1995.

4.  Lehmann and Zahn, \"Safety Aspects for LHe Cryostats and LHe
    Transport Containers\", ICEC7, London, 1978.

5.  Miller, Richard W., Flow Measurement Engineering Handbook, 3^rd^
    ed., p. 13.4.