Calculating Choked Flow of Propane Through a Valve

AI Thread Summary
The discussion focuses on calculating the choked flow of propane through a valve, given the choked flow of natural gas and their respective relative densities. The primary equation used is Qvf = Qvr * sqrt(1/d), where Qvr is the flow rate of natural gas, and d is the relative density of propane. Participants express confusion over the necessity of the relative density of natural gas and whether both densities are needed for the calculation. Ultimately, the correct approach is clarified, emphasizing that only the reference gas's density (0.65) is required for the calculation, leading to the final flow rate of propane being determined as 6.7 L/s. The discussion concludes with a consensus on the correct formula and the importance of using the appropriate reference density.
cps.13
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Hi,

Homework Statement


I have the following question in my course work. What I do not understand is why it provides the relative density of natural gas?!

Calculate the equivalent choked flow of propane through a particular valve if the choked flow of natural gas through it is 10 litres s–1.

The relative density* of propane is 1.45 and that natural gas is 0.65. [*with reference to air]

Homework Equations


I believe the equation I need to use is:

Qvf = Qvr * sqrt(1/d)

Where

Qvf = volumetric flow of the actual fluid (the propane)
Qvr = volumetric flow of the reference fluid (the natural gas) = 10l/s-1
d = relative density of the actual fluid = 1.45

The Attempt at a Solution


So

Qvf = 10 * sqrt (1/1.45) = 8.304L/s-1

Am I going wrong somewhere?

Thanks
 
cps.13 said:
Hi,

Homework Statement


I have the following question in my course work. What I do not understand is why it provides the relative density of natural gas?!

Calculate the equivalent choked flow of propane through a particular valve if the choked flow of natural gas through it is 10 litres s–1.

The relative density* of propane is 1.45 and that natural gas is 0.65. [*with reference to air]

Homework Equations


I believe the equation I need to use is:

Qvf = Qvr * sqrt(1/d)

Where

Qvf = volumetric flow of the actual fluid (the propane)
Qvr = volumetric flow of the reference fluid (the natural gas) = 10l/s-1
d = relative density of the actual fluid = 1.45

The Attempt at a Solution


So

Qvf = 10 * sqrt (1/1.45) = 8.304L/s-1

Am I going wrong somewhere?

Thanks

Hi CPS.13

As I have the same question as you got .
I used the same solution as you posted on here but I don't understand when it said "that natural gas is 0.65 [with reference to air] "
does the question ask to find both of density of propane and natural gas ?

QVF = 10 x sqrt (1/0,65) = 12.4 l/s-1

So this question will have 2 answers . This is what I understand .
 
jaff90110 said:
Hi CPS.13QVF = 10 x sqrt (1/0,65) = 12.4 l/s-1

The relative density of the reference gas is 0.65. The relative density of the actual gas is 1.45.
 
cjm181 said:
The relative density of the reference gas is 0.65. The relative density of the actual gas is 1.45.
Thank you cjm181for reply this thread agai,
So we only need to find the equivalent choked flow by using the relative density of the reference gas which is 0.65 . There is nothing to do with the actual gas ? From equation qvf = qvr x sqrt 1/ 0.65 is that right ?
 
Nearly. Find the full formula for qvf. What are the terms?

Specifically the 1. what term is there?
 
cjm181 said:
Nearly. Find the full formula for qvf. What are the terms?

Specifically the 1. what term is there?

The full formula is qvf =qvr x √ρR/ρF
 
yes, so what reference gas are we using? is it 1 or 0.65
 
cjm181 said:
yes, so what reference gas are we using? is it 1 or 0.65

For reference gas, using 0.65 .
then
qvf = 10 x √0.65/1.45
= 6.7 L s^-1
is that correct?
 
  • #10
That's what i got!
 
  • #11
cjm181 said:
That's what i got!

So that is final formula that we need to use with that answer right?
But why we could not apply for the qvf = qvr x √1/d ?
 
  • #12
Could you please help me with this question again?

1. Write down expression for the open-loop , loop , closed loop gains of the system shown in below , in term of the stage gains A,B,C .
Here is my work .

a)
Open Loop gain = A x B

Loop gain = A x B x C

Closed Loop gain = θo/θi = A x B x C / (1+(A x B) x C)b) The table below gives the magnitude and phase angle of the stage gains of the system represented by block diagram shown in below . Magnitude of gain Phase angle
Controller A 0.5 13°
Process B 2 113°
Measuring System C 0.5 47°- Determine if the system is stable ?
here is my work ,
Loop gain = A x B x C
= 0.5 x 2 x 0.5
= 0.5
As the loop gain of a system is less than unity , this is stable system.

- Determine the new value of controller gain that will cause the system output oscillate with constant amplitude.

Controller gain x 2 x 0.5 = 1

Controller gain = 1/2 x 0.5 = 1

Could you please check this out for me , as I'm not confidence these are the right answer. Thank you
 

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