The sizing of safety valves designed to discharge gases or vapours, according to lspesl Collection “E”, requires knowledge of the isoentropic exponent k at discharge conditions.
Careless application of the lspesl Collection “E” chapter “E.1”, concerning the sizing of safety valves, can lead to an overestimation of the discharge capacity of valves and rupture discs.
This article gives some guidelines to estimate the value of k for real gases and
highlights the mistake by considering k equal to the ratio of specific heats Cp/Cv
A first and gross mistake to be avoided is to use the formula in Collection ‘E’, valid for gases or vapours, in situations where a two-phase discharge of liquid and gas/vapour takes place. In such cases, in fact, the calculated diameters will undoubtedly be undersized compared to the real need.
A second error, which in many cases can lead to the undersizing the safety system, is to give the isoentropic exponent k the value of the Cp/Cv ratio. While the first point will be the subject of a series of subsequent articles, here we would like to give some useful hints for calculating the isoentropic exponent and show, in concrete cases, the size of the error that can be made.
Isoentropic outflow through a nozzle
The formula  that is used in the collection “E”, as well as in other Italian  and foreign  standards, for the calculation of safety valves that must discharge gases or vapours, is that of the isoentropic outflow through a nozzle under critical jump conditions, which for an ideal gas is:
where the expansion coefficient C is given by:
being k the exponent of the isoentropic expansion equation:
(q’/q) x 100
q’= flow rate calculated with k = Cp/Cv (20 °C, 1 atm)
q = flow rate calculated with k = (Cp/Cv) • (Z/Zp)
By introducing the experimental coefficient k of safety valve outflow, which globally considers the real outflow performance of the valve, a safety coefficient of 0.9 and the compressibility factor Z1 for the real fluid, we arrive at the formulation of the collection “E”:
where Z is the compressibility factor defined by Z=P x V / R x T and Zp is the “derived compressibility factor”. When applying formula , according to collection “E”, the values of Cp/Cv, Z and Zp must be evaluated at discharge conditions P1 and T1.
The derived compressibility factor Zp is defined in formula  as:
Where Pr^SAT is the reduced vapour pressure corresponding to a reduced temperature value Tr=T/Tc=0,7. Appendix A shows the Ω values of some fluids. Z e Zp can also be derived directly from an analytical equation of state.
A numerical example
Turning to a numerical example, suppose we need to calculate the discharge capacity of a safety valve under the following conditions:
the discharge pressure is given by:
being for n-Butane: Tc=425,18 K and Pc=37,96 bar, we have:
and using the tables in Appendix A, we have:
Knowing the specific volume of the vapour at the discharge conditions (P1, T1) equal to 0,01634 m^3/kg (0,0009498 m^3/g-mole), we could also have calculated Z from the:
Given the ratio of the specific heats at constant pressure and volume, at discharge conditions (P1, T1), equal to 1,36, from formula  we have:
Applying formula , with calculation of the flow rate
Applying formula , which was solved for the calculation of the flow rate, we have a discharge flow rate value of 147.060 kg/h.
Applying formula , using the value of Cp/Cv at 1 atm and 20 °C
If we had instead used the value of Cp/Cv at 1 atm and 20 °C, we would have had k = 1,19 and from formula  an discharge flow rate of 174.848 kg/h.
This would have led us to overestimate the discharge capacity of the safety valve by around 19%
The error that can be made by assigning the value Cp/Cv to k can be much higher than in this example.
To give an idea, the following table shows the flow rates of an 18-mm orifice for other saturated hydrocarbons, calculated in the two cases. The calculations were performed with specially developed software.
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