Physics Division

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Introduction

Reference Material

Policy and Requirements

Methods of Compliance

Operational Requirements

Appendices

1.

Properties of Liquids

2.

Relief Valve Sizing

2a.

Convective Heat Transfer

2b.

Relief Vent Pressure Drops

3.

Oxygen Deficiency Hazards

 

APPENDIX 2b:  RELIEF VENT PRESSURE DROPS

We calculate the pressure drops caused by the flow of gas through tubes and orifices using (CGS units) the following:

Define

    G = mass flow (grams/(sec.cm2))

    ρ = density (g/cm3)

    η = viscosity (poise)

For flow through a tube or channel

               (1)

    where
    Dn = effective diameter = 4 x   

             ℓ = length of tube

    and

             ψ = 0.326 (GDn/η) -0.25 (turbulent flow)

We also include a “minor loss” term

         ∆P =                        (2)

for the pressure drop occurring at any orifice or sharp bend in the flow channel.  For most relief path geometries term (2) will be larger than term (1).  Application of expression (2) is not always straightforward, a number of examples are discussed in ref. [8].  In what follows we assume the density of helium to scale according to the ideal gas law.

Example I.  Vacuum Vessel Relief Vents

                    These are of the form

We treat this as an orifice with area = the lesser of (πDih) or (πD/4) and we take the pressure drop to be given by (2).  We calculate the flow of helium gas through these vents for a pressure drop equal to the cracking pressure Pc plus one PSI.  Under these conditions, the dimensions are:


                  Unit                        Pc                     Di                     h                      Area


A.          ATLAS Linac          1.1 psi           15.24 cm            2.5 cm              121.6 cm2

B.          Rebuncher               3    psi              5.08                 0.64                    10.2

C           PII*                         1    psi           15.24                  2.5                   121.6 cm2


*for PII the overpressure = 2 psi

We assume (conservatively) that

     (ideal gas) at 1 ATM.


For ATLAS:  He at 15.1 K, DP = 2 psi, Vent Area = 121.6 cm2

Using ∆P =    (1 psi = 6.9 x 104 dyne/cm2)

then G = 29.2 g/cm2 sec

thus Vent Capacity = 29.9 x 121.6 = 3630 g/s

 

Example II.  Helium Vessel Vent – Rebuncher

 

 

 We assume helium gas at 5 K, the dominant term is the “minor loss” term we take

    ∆P =

at each 90º bend, and at the outlet orifice – working backward from the outlet, and assuming r = 9.8x10-3 P (g/cc) where P is in atmosphere (ideal gas density), for a total flow of 420 g/sec, G = 420/9.62 = 43.7 g/sec cm2

 

Location

Exit Pressure

Density

DP

 

 

 

 

Outlet

15 pisa

9.8 x 10-3

1.4 psi

2nd bend

16.4

1.07 x 10-2

1.3

1st bend

17.7

1.16 x 10-2

1.2

 

 

We note that the frictional loss gives DP1 = 0.2 psi for this case, and

 

 

 

 

 

 

 

DP      =  4.1 psi

 

    Total