SAFETY VALVE CAPACITY DETERMINATION FOR GASES AND VAPOURS AND VALVE SELECTION
The Seetru safety valve capacity determination expressions below are based on BS 6759:
Part 2: 1984. They apply to critical flow conditions, which occur when
(1)
r=\frac{P_b}{P}\leq \left ( \frac{2}{k+1}\right )^{\frac{k}{k-1}}
where
P
= Actual flowing pressure in bar absolute
Pb
= Back pressure in bar absolute
k
= Isentropic exponent of expansion at actual flowing inlet conditions
For air, k = 1.4, and r becomes
r\leq 0.53
Hence critical flow for air occurs broadly above 1 bar gauge set pressure, when discharging to atmosphere.
The mass flow of gas, qmg, being discharged by the safety valve is then given by
(2)
q_{mg}=A\cdot K_{dr}\cdot P\cdot C\cdot \sqrt{\frac{M}{Z\cdot T}}\;\;\;\;\;\;\;\;\;\;(kg/hr)
where
A
= Flow area (i.e. valve seat bore) in mm2
Kdr
= Derated coefficient of discharge declared on valve datasheets. (For type tests according to German TÜV rules, the working coefficient of discharge designated αw and is also declared)
P
= Actual flowing (inlet) pressure (in bar absolute). P = (1.1p+1) where p is set pressure in bar gauge, the actual flow pressure being given at 10% pressure accumulation.
C
= Function of k (isentropic exponent of expansion) taken from BS 6759, and varying from a maximum of 2.8 for monatomic gases down to 2.4 for polyatomic vapours.
M
= Molecular mass of gas in kg/k mol
T
= Absolute flow temperature in °K given by (273 + t°C)
Z
= Compressibility factor, given by
(3)
Z=\frac{v\cdot P\cdot M}{R\cdot T}
where
v
= Specific volume of gas P & T in (m3/kg)
R
= Universal gas constant equal to 0.08314 for units of measure employed here.
Any departure of the value Z from 1 is a measure of the departure of the vapour properties
from that of a gas. Z can be calculated from vapour tables, but may be taken as approximately
1 for significantly susuperheated vapours. BS 6759 provides curves of Z as a function of
vapour conditions.
By substituting (3) in (2) a modified expression is obtained
(2a)
q_{mg}=0.2883\cdot A\cdot K_{dr}\cdot C\cdot \sqrt{\frac{P}{v}}\;\;\;\;\;\;\;\;\;\;(kg/hr)
which can be convenient if the specific volume, v, of a vapour is known.
For air M = 29, C = 2.7, Z = 1. Thus from (2) the mass flow of air, qma, is given by
(4)
q_{ma}=14.54\cdot K_{dr}\cdot A\cdot \frac{(1.1p+1)}{\sqrt{273+t}}\;\;\;\;\;\;\;\;\;\;(kg/hr)
For standard rating flow temperature of 15°C and a valve seat bore diameter of do mm
(5)
q_{ma}=0.673\cdot K_{dr}\cdot d_{o}^{2}\cdot (1.1p+1)\;\;\;\;\;\;\;\;\;\;(kg/hr)
This is the mass flow given on the Seetru safety valve data sheets.
To convert a critical mass flow of air qma at 15°C for a given valve to that of another gas of molecular mass M, value Zgas, and flowing at t°C, this follows as
(6)
q_{mg}=\sqrt{\frac{M}{29}}\cdot \frac{C_{gas}}{2.7}\cdot \sqrt{\frac{288}{273+t}}\cdot \sqrt{Z_{gas}}\cdot q_{ma}\;\;\;\;\;\;\;\;\;\;(kg/hr)
(7)
=F\cdot \sqrt{\frac{288}{273+t}}\cdot \sqrt{Z_{gas}}\cdot q_{ma}
where
(6a)
F=0.0688\cdot C_{gas}\cdot \sqrt{M}
The factor F is tabulated for a range of gases see table
(6) can be transposed
q_{ma}=\sqrt{\frac{273+t}{288}}\cdot \frac{q_{mg}}{F\cdot \sqrt{Z_{gas}}}
This form is useful to give the mass flow of air equivalent to a known mass flow of gas. It
will allow a safety valve sizing to be checked against tabulation of air mass flow in kg/hr
from (5) and (6), for a given mass flow qmg of a gas, the required safety bore do can be found using
d_o\geq \sqrt{\sqrt{\frac{273+t}{288}}\cdot \frac{q_{mg}}{F\cdot {\sqrt{Z_{gas}}\cdot 0.673\cdot K_{dr}\cdot (1.1p+1)}}}\;\;\;\;\;\;\;\;\;\;(mm)
taking Zgas=1
(8)
d_o\geq \sqrt{\frac{0.08756}{F\cdot K_{dr} \cdot (1.1p+1)}\cdot \sqrt{273+t}\cdot q_{mg}}\;\;\;\;\;\;\;\;\;\;(mm)
As an alternative, the rated flow of gases is often given in standard litres per second,
which is the volume occupied by the mass per second at 1 atmosphere pressure (1.013
bar absolute) and 15°C.
The general expression according to the gas laws for the volume V litres of a mass W kg
of gas of molecular mass M at pressure P bar absolute and temperature T °K absolute is
(9)
V\geq 83.14\cdot\frac{W\cdot T}{M\cdot P}\;\;\;\;\;\;\;\;\;\;(litres)
For standard litres T=288°K,P=1.013 bar
(10)
\therefore V=\frac{23637\cdot W}{M}\;\;\;\;\;\;\;\;\;\;(std. litres/s)
It may be noted that for European standards the standard litre is defined at 0°C, and hence it must be reduced in the ratio
\frac{273}{288}=0.948
Multiplied by 3.6, this standard litre per second converts to European normal cubic metre per hour: N m³/hr.
Hence a mass flow of qmg kg/hr is equivalent to
(11)
Q_{gas}=\frac{23637}{3600\cdot M}\cdot q_{mg}=\frac{6.566}{M}\cdot q_{mg}\;\;\;\;\;\;\;\;\;\;(std.litres/s)
where Qgas = Volumetric flow at standard conditions
By substituting from (2)
(12)
Q_{gas}=6.566\cdot \frac{K_{dr}\cdot A\cdot C_{gas}\cdot (1.1p+1)}{\sqrt{M\cdot Z_{gas}\cdot (273+t)}}\;\;\;\;\;\;\;\;\;\;(std.litres/s)
For air flowing at t=15°C
Q_{air}=\frac{6.566\cdot 2.7}{\sqrt{29\cdot 288}}\cdot K_{dr}\cdot A\cdot (1.1p+1)\;\;\;\;\;\;\;\;\;\;(std.litres/s)
(13)
=0.193\cdot K_{dr}\cdot A\cdot (1.1p+1)\;\;\;\;\;\;\;\;\;\;(std.litres/s)
In terms of flow diameter do
Q_{air}=0.1524\cdot K_{dr}\cdot d_{o}^{2}\cdot (1.1p+1)\;\;\;\;\;\;\;\;\;\;(std.litres/s)
Hence to convert to Qgas std. litres/s flowing at t°C from tabulated values of Qair (e.g. In Seetru datasheets)
Q_{gas}=\sqrt{\frac{29}{M}}\cdot \frac{C_{gas}}{2.7}\cdot \sqrt{\frac{288}{273+t}}\cdot \frac{Q_{air}}{\sqrt{Z_{gas}}}
=1.9945\cdot \frac{C_{gas}}{\sqrt{M}}\cdot \sqrt{\frac{288}{273+t}}\cdot \frac{Q_{air}}{\sqrt{Z_{gas}}}
(14)
=f\cdot \sqrt{\frac{288}{273+t}}\cdot \frac{Q_{air}}{\sqrt{Z_{gas}}}\;\;\;\;\;\;\;\;\;\;(std.litres/s)
where
(15)
f=\frac{1.9945\cdot C_{gas}}{\sqrt{M}}
is also shown from the gas properties table that (14) can also be transposed;
(14a)
Q_{air}=\sqrt{\frac{273+t}{288}}\cdot \frac{\sqrt{Z_{gas}}\cdot Q_{gas}}{f}
This form is useful to give the air flow in std. litres/sec equivalent to a known gas flow in
std. litres/s. It will allow a safety valve sizing to be checked against tabulation of standard
volumetric air flow.
Using (12) and (14), for a given flow of gas Qgas Std. litres/s, the required safety valve do can be found;
d_{o}\geq \sqrt{\sqrt{\frac{273+t}{288}}\cdot \frac{\sqrt{Z_{gas}}\cdot Q_{gas}}{0.1524\cdot f\cdot K_{dr}\cdot (1.1p+1)}}\;\;\;\;\;\;\;\;\;\;(mm)
For Zgas=1;
(16)
d_o\geq \sqrt{\frac{0.38665\cdot \sqrt{273+t}}{f\cdot K_{dr}\cdot (1.1p+1)}}\cdot Q_{gas}\;\;\;\;\;\;\;\;\;\;(mm)
If for a given range of safety valves the value of the coefficient of discharge is not
uniform, then by putting the lowest value into (16), the minimum safe seat diameter will be
obtained.