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clh99
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- Homework Statement
- I have the question
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 ( at STP) 2.8x10^6 litres per min
Density (at stp) 0.752 kg per m^3
specific heat ratio 1.31
inlet pressure 20bar
outlet pressure 6 bar
inlet temperature 20 degrees Celsius
- Relevant Equations
- Cv = alternative flow coefficient (gallons/min)
Kv = flow coefficient (M^3h^-1)
Y = expansion factor
Qm = mass flow rate (kg/hour)
xT = critical pressure differential ratio
x = pressure differential ratio
Fl= liquid pressure recovery coefficient
Fk = specific hear ratio factor
k = specific heat ratio
x=(p1-p2)/p1
xT = Fk * x
xT < x
Y = 1 - x/( 3 x Fk x xT )
Kv= Qm / ((2/3) x 31.6 x √ Fk x xT x P1 x p1)
Working out as follows
x= (20-6)/20 = 0.7
Fk = k / air specific heat ratio
Fk = 1.31/1.4
Fk = 0.94
xT = Fk * x
0.94 x 0.7 = 0.658
because xT < x = 0.658<0.7 flow through the valve is choked
Expansion factor Y
Y = 1 - x/( 3 x Fk x xT )
Y= 1- 0.7/( 3 x 0.94 x 0.658)
Y = 0.623
Although this is the expansion factor because the flow is choked a value of 2/3 or 0.667 is used.
I am now struggling to kind a value for Qm which is mass flow rate. I believe once i have this value i can then incorporate it into Kv= Qm / ((2/3) x 31.6 x √ Fk x xT x P1 x p1) to find the valve flow coefficient
x= (20-6)/20 = 0.7
Fk = k / air specific heat ratio
Fk = 1.31/1.4
Fk = 0.94
xT = Fk * x
0.94 x 0.7 = 0.658
because xT < x = 0.658<0.7 flow through the valve is choked
Expansion factor Y
Y = 1 - x/( 3 x Fk x xT )
Y= 1- 0.7/( 3 x 0.94 x 0.658)
Y = 0.623
Although this is the expansion factor because the flow is choked a value of 2/3 or 0.667 is used.
I am now struggling to kind a value for Qm which is mass flow rate. I believe once i have this value i can then incorporate it into Kv= Qm / ((2/3) x 31.6 x √ Fk x xT x P1 x p1) to find the valve flow coefficient
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