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Mass flow quality 2-phase calculation in ISO 4126-10 (2010) - Low amount of vapour rate through psv

Mass flow quality 2-phase calculation in ISO 4126-10 (2010) - Low amount of vapour rate through psv

Mass flow quality 2-phase calculation in ISO 4126-10 (2010) - Low amount of vapour rate through psv

(OP)
Hello to all,

We are calculating PSV for two phase scenario and we have noticed that the mass quality fraction (x0) is too low, about 0.0006, that mean that from 100 kg/h 99.9994 kg/h are liquid and only 0.0006 kg/h are vapor.

On Fundamental Equations for Two-Phase Flow in Tubes - Masahiro Kawaji the definition of the mass flow quality is:

x0= mg / (mg + ml)

where:
mg is gas mass flow
ml is liquid mass flow

On the other hand, the ISO 4126-10 (sorry I can't put the link there) the definition of the mass flow quality is literally: Ratio of gas mass flow rate to total mass flow rate and the equation is:

x0 = e0· sv-l / ((1 - e0) · sv-g + e0 · sv-l)

where:
e0 is the void fraction in the pressurized system at sizing conditions - in other words, the percentage of the reactor's upper empty zone. e0=(1-filling level)
sv-l is the specific volume of the liquid
sv-g is the specific volume of the gas

The first we notice is that is that the equation is reversed from the one explained in the book and also the text from the ISO Ratio of gas mass flow rate to total mass flow rate. In the ecuation of the ISO is referred to the liquid phase.

And this is even more puzzling, they multiply the sv-l with the void fraction. We do not undestant that because the void fraction has to be multiplied by the gas phase...

All of this give us a x0=0.0006 for a 99% full reactor for Ethyl Acetate, we think it makes no sense.


We propose change the equation for this one referred to the literature:

x0 = e0 · sv-g / (e0 · sv-g + (1 - e0) · sv-l)

That have more sense, whit that change we calculated that we have when the reactor is 99% full a 65% gas phase flow rate and when is 80% full a 98% gas phase flow rate, that is logical becouse if the reactor is 20% empty the liquid will not arrive to the PSV...

So, what do you think about that??? do you have any recommendation? do you have the same results?

RE: Mass flow quality 2-phase calculation in ISO 4126-10 (2010) - Low amount of vapour rate through psv

HI,
I'm not familiar with your standard ! Based on your mass balance why don't you size based on liquid instead of a mixture (2 phase), which I believe is not going to make a lot of difference .

I've added reference for your calculation (Leser), you can download their engineering booklet using your favorite search engine , in particular chapter 7 related to PSV sizing with standards' comparison.
https://www.leser.com/en/support-and-tools/enginee...

Pierre

RE: Mass flow quality 2-phase calculation in ISO 4126-10 (2010) - Low amount of vapour rate through psv

(OP)
Hi pierreick,

I have to desing it for two phase, also we think that the ISO mass balance is not accurate

RE: Mass flow quality 2-phase calculation in ISO 4126-10 (2010) - Low amount of vapour rate through psv

SF6-146,
streams with operating conditions close to bubble point can show large variations for two-phase speed of sound / dischargeable mass flow rates,
the reason being the growth of vapour bubbles inside liquid, the same is not true for streams close to dew point,
see this example (stream n-butane 0.7 n-heptane 0.3 mol.fract, P 30 Bar.a model GERG 2008) speed of sound calculated in Excel with StrMSS() macro included in Prode Properties library,

liquid molar fract. T(K) speed of sound m/s
1.0 441.9 289.27
0.999999 441.992 139.7543795
0.999 442.012 42.860
0.9 444.018 55.327
0.5 454.14 101.603
0.1 466.51 141.08
0.001 469.565 149.63
0.0 469.6 150.002

as you see the minimum speed of sound / dischargeable mass flow rate appears with very low vapor fractions..
there are specific two-phase models for these conditions, see for example HNE-DS (available in Prode) and others...

To answer your question, my understanding is that ISO converts from volume fractions to mass fractions

x0 = E0 * Vl / ((1-E0)*Vg + E0*Vl)
where E0 = 1 - Vl / V

where Vl, Vg are liquid, vapor volumes and V total volume (see table B.2 for details)

RE: Mass flow quality 2-phase calculation in ISO 4126-10 (2010) - Low amount of vapour rate through psv

(OP)
Here is the problem,

Quote (PaoloPemi)

x0 = E0 * Vl / ((1-E0)*Vg + E0*Vl)
where E0 = 1 - Vl / V

If you define E0 this way, E0 corresponds to vapor phase (void fraction) so u cant multiply it by the Liquid... becouse you are saying that the liquid correspont to the void fraction.

RE: Mass flow quality 2-phase calculation in ISO 4126-10 (2010) - Low amount of vapour rate through psv

there are no examples for E0 = 1-Vl/V where V ?
however they say that x0 is the mass flow quality at sizing conditions, i.e. the gas mass flow rate related to the total mass flow rate of a two-phase mixture, adding that mean value of the specific volume of the homogeneous two-phase mixture at sizing conditions in the pressurized system is given as
v0 = xo*Vg + (1-x0)*Vl
which should give sufficient details about the procedure (I base my interpretation on these points, see my previous post)
by the way, if you are solving this specific case (i.e. stream condition close to bubble point) take care to select a suitable model (see my first post for details)

RE: Mass flow quality 2-phase calculation in ISO 4126-10 (2010) - Low amount of vapour rate through psv

(OP)
Hi to all, but in special to PaoloPemi,

Here is the example they show, is used to calculate the specific volume at the pressure and temperature of sizing conditions (the escenario is a runaway reaction):



In the paper I attach in the figures 2, 3 and 4 is shown what you explain about the flow quality:

Quote (PaoloPemi)

as you see the minimum speed of sound / dischargeable mass flow rate appears with very low vapor fractions..

the problem is: is difficult to imagine that in a reactor 90% filled of a solvent that is exposed to fire and is uncontrolled heated, when the pressure arrive to the sizing pressure of the PSV you will only have 0.0006 kg/h of the vapor phase.

Also, I have to sum that in the equation: x0 = e0· sv-l / ((1 - e0) · sv-g + e0 · sv-l) they multiply the sv-l with the void fraction. It leave logicals values accordint to the critical flow? yes...

But for example if you change the equation to x0 = e0 · sv-g / (e0 · sv-g + (1 - e0) · sv-l) you obtain a phase of 0.35 vs the 0.0006 that seems more logical in the escenario I propose

Attached document: Two-Component Two-Phase Critical Flow

Edit:

P.S. I have seen the Prode web, it is free??

RE: Mass flow quality 2-phase calculation in ISO 4126-10 (2010) - Low amount of vapour rate through psv

they define x0 as the mass flow quality, i.e. the ratio of the gas mass flow rate to the total mass flow rate of a two-phase mixture, if you know total volume (of vessel, see also the definition of V) and (vapor and liquid) volumes you should be able to get x0 value,
the paper presents a homogeneous model "modified by taking into account the void fraction and two-phase mixture density dependence on velocity slip", the authors say "the better agreement of the model data is achieved by appropriate modelling of the slip between liquid and gas phase velocities" which is the goal of the models mentioned in my first post, take care that these model require specific parameters not easy to estimate (for best accuracy you should regress from experimental data),
the standard HEM model (liquid and vapor in equilibrium, traveling at same speed) is more conservative with low vapor fractions as shown in my previous post but it is much easier to solve (you do not need to estimate specific parameters).
Yes, there is a free version of Prode Properties available for students, it has a limited database.

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