Engineering Calculating valve sizes for a compressible liquid

AI Thread Summary
The discussion focuses on calculating valve sizes for a compressible liquid, emphasizing the conversion of pressures from bar gauge to absolute values. The pressure drop is calculated, indicating that the flow is likely choked. The user expresses uncertainty about selecting a valve size due to the lack of mass flow rate information. Another participant suggests using the product of volumetric flow rate and density to progress, while also noting the complexity of relationships between mass and volumetric flow rates for compressible fluids. The conversation concludes with acknowledgment of potential discrepancies in volumetric flow rate values provided in different sources.
shocked_prawn
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Homework Statement
A valve is required to regulate the flow of natural gas. Details of the gas are given in table below. It can be assumed that the pipe size and the valve size are the same [i.e. piping geometry does not have to be allowed for]. The chosen valve type is the V250 rotary ball valve by Fisher
Controls International [the V250 data sheet is available on Blackboard].

- Determine the required valve size in inches from the range of sizes given in the data sheet [the data sheet gives a range of nominal pipe sizes (NPS) in inches].
- Estimate the percentage the chosen valve would have to be open to carry the required flow.

REquired volumetric flow rate [@STP] 2.6 X 10^6 litres per minute
Density [@ STP] 0.752kg per m^3
Specific Heat Ratio 1.31
Inlet pressure 20 bar gauge
Outlet pressure 6 bar gauge
Inlet temp 20ºC
Relevant Equations
Kv=qm/31.6YSQRT(xp1P1) .........for non choked flow
Kv=qm/21.1SQRT(F_yXtp1P1) ..... for choked flow
where kv is valve coefficient
qm is mass flow in kg/h
Y is expansion factor
Fy is the heat ratio factor
p1 density
P1 inlet pressure

Y=1-(x/(3FyXt)) ..........2

Fy = y/1.4
Screenshot 2021-12-18 at 22.33.44.png

I have converted given pressures from bar gauge to bar abs
Inlet pressure = 20 - 1.013 = 21.013 bar abs
Outlet pressure = 6 + 1.013 = 7.013 bar abs

therefore pressure drop (x)
x = {p_1 -p_2 \over p_1 } = {21.013 - 7.013 \over 21.013} = 0.6662542236 bar abs
Comparing to the chart values for X_T I have determined flow is likely choked.

However, in the information given I don't have the mass flow rate (qm) so I am unsure how to progress to select a vale size.
Any help would be appreciated
 
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Hello @shocked_prawn ,
:welcome: ##\qquad## !​
So you found the same (*) exercise you are working on, and, like yours, there was no answer :nb) !

Did you notice the given vol flow rate and the density ? The product would at least have the right dimension :wink: !

[edit] (*) to my surprise the two Fv are different :oops: !?

##\ ##
 
BvU said:
Hello @shocked_prawn ,
:welcome: ##\qquad## !​
So you found the same (*) exercise you are working on, and, like yours, there was no answer :nb) !

Did you notice the given vol flow rate and the density ? The product would at least have the right dimension :wink: !

[edit] (*) to my surprise the two Fv are different :oops: !?

##\ ##
Hi @BvU

Thanks for the help :)
Whilst working on it over the weekend I have used the product of volumetric flow rate and density for now (thank you!)

In the textbooks, it says that the relationship between mass and volumetric flow rates for a compressible fluid is more complex than that for a liquid and quote the combined gas law and ideal gas law.

As these values have been given at STP has the "hard work" already been done for these calculations?

I am sorry I don't know what you mean by the two Fv are different?:smile:
 
Last edited:
shocked_prawn said:
As these values have been given at STP has the "hard work" already been done for these calculations?
Correct !

I am sorry I don't know what you mean by the two Fv are different?:smile:
You have 2.6 106 dm3/s; the old thread has
clh99 said:
Required volumetric flow rate ( at STP) 2.8x10^6 litres per min
but probably one of the two is a typo :smile:

##\ ##
 
@BvU Brilliant thank you for your help :cool:
 
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