Calculating Choked Flow of Propane Through a Valve

In summary: 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 BLoop gain = A x B x CClosed 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 .
  • #1
cps.13
40
1
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
 
  • #3
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 .
 
  • #4
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.
 
  • #5
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 ?
 
  • #6
Nearly. Find the full formula for qvf. What are the terms?

Specifically the 1. what term is there?
 
  • #7
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
 
  • #8
yes, so what reference gas are we using? is it 1 or 0.65
 
  • #9
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
 

Attachments

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1. How do you determine the choke flow rate of propane through a valve?

The choke flow rate of propane through a valve can be determined by using the Choke Flow Equation, which takes into account the pressure drop across the valve, the specific gravity of the propane, and the valve's flow coefficient. This equation can be found in many engineering references or can be calculated using specialized software.

2. What factors affect the choked flow rate of propane through a valve?

The choked flow rate of propane through a valve is affected by several factors, including the size and shape of the valve, the pressure drop across the valve, the specific gravity of the propane, and the temperature and pressure of the gas upstream of the valve. Additionally, the type and condition of the valve can also impact the choked flow rate.

3. Can the choke flow rate of propane through a valve be calculated for any valve size?

Yes, the choke flow rate of propane through a valve can be calculated for any valve size as long as the necessary information is known, such as pressure drop, specific gravity, and flow coefficient. However, it is important to note that the Choke Flow Equation is typically only valid for valves with a diameter greater than 0.1 inches.

4. How does the specific gravity of propane affect the choked flow rate through a valve?

The specific gravity of propane plays a significant role in determining the choked flow rate through a valve. Propane, which has a specific gravity of around 1.5, is heavier than air and will require a larger pressure drop to achieve choked flow. This means that for a given valve, the choked flow rate of propane will be lower than that of a gas with a lower specific gravity.

5. Can the choke flow rate of propane through a valve change over time?

Yes, the choke flow rate of propane through a valve can change over time due to factors such as wear and tear on the valve, changes in gas temperature and pressure, or changes in the specific gravity of the propane. It is important to regularly monitor and maintain valves to ensure accurate and safe flow rates.

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