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Determining Isentropic Expansion Coefficient

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phosty

Chemical
May 13, 2009
4
I have a copy of:

Shackelford, Aubry, "Using the Ideal Gas Specific Heat Ratio for Relief Valve Sizing", Chemical Engineering, pp. 54-59, November 2003.

I am struggling to understand how, in the first worked example he presents (Case I), the isentropic expansion coefficient (n) of 0.746 is arrived at.

The last sentence of the second box on page 58 titled 'Changed to API-RP 520 and their implications' reads
Alternatively, the isentropic expansion coefficient can be calculated directly using the real gas specific heat ratio and the fluid properties at the inlet conditions using Equation 3 - yielding an even better estimate of the isentropic expansion behavior of the fluid than the ideal gas specific heat ratio.

Eqn 3 is given as:

n = (-[ν]/P ) . ([∂]P/[∂][ν])T . (Cp/Cv)

The example calculation provides the following physical properties for the relieving fluid (saturated isobutylene vapor):

Relief pressure: 26.62 barg (2763 kPaa)
Relief temperature: 395.5 K
Real Gas specific heat ratio: 1.887
Vapor density: 81.98 kg/m[³]
Compressibility: 0.575

The first worked example (Case I) then simply reads
Using the actual isentropic expansion coefficient for the fluid at relieving conditions, 0.746.......

How is 0.746 calculated from the data provided? Or is there a few missing steps?
 
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