Pressure Regulator Cv for Natural Gas Train

I'm not likely the ultimate one to help you, but it would be better if you define a couple of terms:

Cv could mean lots of different things to different people.

A train is a vehicle that runs on rails. Other multiple things connected one after the other are also referred to as trains, but trains of (whatever).

Santy

The Cv of a valve is the flow of water at 60 degF in US gallons per minute which will result in a one psi pressure drop with the valve wide open.

Cv = Q x (g/delta P)^0.5 Where Q is in usgpm, g is secific gravity and delata P is the pressure drop across an open valve.

You should be able to look up the Cv of your valve quite easily.

The rated Cv for your regulators should be available if you know the manufacturer. Check out the Fisher regulator/control valve site, you can get to it through www.emerson.com and they have the equations you need to figure this out. You'll probably need the temperature of the gas as well at a minimum. I would not use the liquid equation for this application.

"I would not use the liquid equation for this application"

Oops - really should read the question properly and apply mind not a***.

Flow Coefficient - C_v - for Air and other Gases

For critical pressure drop the outlet pressure - p_o - from the control valve is less than 53% of the inlet pressure - p_i. The flow coefficient can be expressed as:

C_v = q [SG (T + 460)]^1/2/ 660 p_i (5)

where

q = free gas per hour, standard cubic feet per hour (Cu.ft/h)

SG = specific gravity of flowing gas gas relative to air at 14.7 psia and 60^oF

T = flowing air or gas temperature (^oF)

p_i = inlet gas absolute pressure (psia)

For non critical pressure drop the outlet pressure - p_o - from the control valve is greater than 53% of the inlet pressure - p_i. The flow coefficient can be expressed as:

C_v = q [SG (T + 460)]^1/2/ [1360 (dp p_o)^1/2] (5b)

where

dp = (p_i - p_o)

p_o = outlet gas absolute pressure (psia)

You will find an online calculator here

Regulators control pressure not flow. Cv is related to flow. You have stated that your flow is 1250 Nm³/hr. Bernoulli said flow in equals flow out. Your flow before the first regulator, between regulators, after the second regulator would equal the equivalent of 1250 Nm³/hr at .5 kg/cm². You know the ID of the pipe or tube. The velocity in any section of the system is the specific density (Nm/meter) at the existing pressure divided into the flow (eg. if one meter contains (for example only) 4 Nm/m @ .5 kg/cm² then 1250/4 = 312.5 meter/hr. The flow will be slower at higher pressue. Tom

Dear All,

Thank you for your prompt reply and support.

@dkwarner.... lol!! i dot know if you are trying to pull my leg....but with due respect, i guess everyone else who replies understood what i meant by Cv & a gas train.

@Kaisan.....thanks for the formula, had got that one from some reference books and earlier forums posted on this website, i did work out some calculations based on it, am pasting the same below can you please if possible check them for me if i got the concept right??

@Tom.... got that one.....can you please check my below calculations and let me know if am right / wrong or need to tweak a little......

Thanks in advance all.

My calculations below:

Inlet pressure at PRS is 25 kg/cm2

First stage reduction is 25 kg/cm2 to 4 kg/cm2

Second stage reduction is 4 kg/cm2 to 0.5 kg/cm2

Process requirement is 0.5 kg/cm2 with 1250 Nm3/hr. This also becomes the outlet condition of 2^nd stage regulator.

=========================================================================

Now to get flow requirement at outlet of 1^st stage regulator

P1V1=P2V2 considering constant temperature at inlet and outlet of regulator

Where,

P1 = the absolute inlet pressure of 2^nd stage regulator

P2 = the absolute outlet pressure of 2^nd stage regulator

V1 = the flow of gas at inlet of 2^nd stage regulator

V2 = the flow of gas at outlet of 2^nd stage regulator

Hence,

(4 + 1.033 kg/cm2) x V1 = (0.5 + 1.033 kg/cm2) x 1250 Nm3/hr

5.033 x V1 = 1.533 x 1250

V1 = 381 Nm3/hr

Conditions at outlet of 1^st stage regulator

Outlet pressure: 4 kg/cm2

Flow: 381 Nm3/hr

=========================================================================

Cv of 1^st stage regulator

Since, Inlet pressure is greater than 2 times outlet pressure we use the following formula

Cv = {Q [SG (T + 460)]^0.5} / 660 (Pi)

Where,

Q = Flow in CFH

SG = Specific gravity

T = Temperature in (°F)

Pi = Absolute inlet pressure (psia) which is actual pressure + atmospheric pressure

Converting 381 Nm3/hr to SCFH

Now 381 Nm3/hr is at standard condition of 1 atm and 0 deg.C

Converting Nm3/hr to m3/hr

Q(act) = Qn x T(act)/(273 x P(act))

Where,

Q(act) = m3/hr

Qn = Nm3/hr

T(act) = Temperature in K

P (act) = Pressure actual (atm)

Considering operating temp as 30 deg.C = 303.15 Deg.K

4 kg/cm2 = 3.871 atm

Q(act) = 381 x 303.15 / (273 x (3.871 +1)

= 115500.15 / 1329.783

= 86.85 m3/hr

Now,

1 m3/hr = 35.314 scfh

Therefore,

86.85 m3/hr = 3067.079 scfh

Specific gravity of Natural gas is 0.7

1 deg. C = 33.8 Deg.F

Back to Cv formula

Cv = 3067.079 [0.7(86+460)]^0.5 / 660 (355.58 + 14.7)

= 59961.198 / 244384.8

Cv = 0.25

=======================================================================

Cv of 2^nd stage regulator

Since, Inlet pressure is greater than 2 times outlet pressure we use the following formula

Cv = {Q [SG (T + 460)]^0.5} / 660 (Pi)

Where,

Q = Flow in CFH

SG = Specific gravity

T = Temperature in (°F)

Pi = Absolute inlet pressure (psia) which is actual pressure + atmospheric pressure

Converting 1250 Nm3/hr to SCFH

Now 1250 Nm3/hr is at standard condition of 1 atm and 0 deg.C

Converting Nm3/hr to m3/hr

Q(act) = Qn x T(act)/(273 x P(act))

Where,

Q(act) = m3/hr

Qn = Nm3/hr

T(act) = Temperature in K

P (act) = Pressure actual (atm)

Considering operating temp as 30 deg.C = 303.15 Deg.K

0.5 kg/cm2 = 0.484 atm

Q(act) = 1250 x 303.15 / (273 x (0.484 +1)

= 378937.5 / 323.232

= 1172.33 m3/hr

Now,

1 m3/hr = 35.314 scfh

Therefore,

1172.33 m3/hr = 41400.44 scfh

Specific gravity of Natural gas is 0.7

1 deg. C = 33.8 Deg.F

Back to Cv formula

Cv = 41400.44 [0.7(86+460)]^0.5 / 660 (56.89 + 14.7)

= 809375.95 / 47249.4

Cv = 17.13

==============================================================

Santy

Have no time at the moment but a quick look indicates that you are mixing things up a bit. V1 and V2 are volumes, not flows.

Nm3/hr is at a set of known conditions, the actual volume flow will vary with pressure, but the normalised volume flow remains the same across the sytem.

As Tom says, the flow of 1250 Nm3/hr (44143 Ncuft/hr) is the same across the system.

I agree with Kaisan and others, you should have a constant 1250 Nm3/hr passing through each regulator as the system needs flow in = flow out unless your regulators are physically unable to do that (possible but not likely in this case). I would convert the 1250 Nm3/hr to your actual cubic meters and then repeat the calculations.

Oh and thank you for responding to our questions and keeping the thread alive, that can be somewhat of a rarity.

Sorry for that goof up. Thanks again Kaisan, Betomachine and everyone else.

That means the flow rate remains the same at the inlet and outlet of a pressure regulator. Is it under some defined conditions or it always like that?

I am confused. As per the conditions in my question the process requirement is 0.5 kg/cm2 with 1250 Nm3/hr flow rate. How do i ensure that the inlet flow of the second stage regulator and the inlet flow of the first stage regulator is also 1250 Nm3/hr.

I may be asking too many dumb questions here, but would like to get my this thoroughly cleared. Referring a lot of books and self learning, but actual people like you can make it easy to understand with examples and experience.

Regards,

Perhaps I can help after all. The N in 'Nm^3/hr' means normalized, or converted to standard pressure. One Nm^3 is the same amount (mass, weight, number of molecules) of gas at any pressure. At the higher inlet pressure, the actual volume will be smaller than the actual volume at the reduced outlet pressure, but the flow in molecules per second MUST be the same at the input as the output. Otherwise you have a leak, or are magically producing new molecules!

thanks dkwarner.

So should i consider the flow rate for both the pressure regulators as 1250 Nm3/hr?

What if the flow receiving at the inlet of 1st stage regulator say from an MRS/DRS (or any other source) is higher than 1250 Nm3/hr.

Could you also help me on how to size the pressure regulator?

Regards,

IF your process actually uses 1250Nm^3/hr of gas, then that quantity (mass, not volume) must flow through every stage of the system, including the regulators, from whatever source is used up to the manifold or other point where the flow splits into more than one stream.

Under steady state conditions (after initial surges at turn-on), the flow at the input will automatically adjust to conditions at the output. If you stop the flow after the regulators by turning off the low pressure valve, the flow at the input must also stop (a very short time after the output is stopped)! The flow at the input can never be different than the flow at the output except briefly during more-or-less sudden changes in either the input pressure or the output pressure, such as initial start-up.

As you saw in my first post, I am not experienced in fluid flow calculations, so no, I can't help in sizing the regulators. I would expect the regulator vendors to be willing to help with that, although there is a good possibility that they might recommend a larger unit than needed, to maximize their sales. But then oversized is definitely preferable to undersized!