Outlet Velocity of Steam through a Valve 3

[MattC1234]

I am having problems calcualting the outlet velocity of Steam across a valve and am looking for help.

The valve is 24" ANSI 600# (With a body bore of 22") my process conditions are causing some what of a problem to em and are giving very high outlet velocities, but these do not take into acount the temperature change of the steam during the pressure reduction stage.

The flow rates for the steam are 165000 and 235000 kg/hr and require pressure control from 64.7 Barg to 0.24 Barg with an inlet temperature of 282 Deg C. The ratio of specific heats is 1.86. Can anyone help me with the calculation for the temperature of the fluid at outlet and thus help me with the velocity calculation?

Any help would be appreciated. I am assuming using steam tables that the steam would be superheated as from an enthalpy entropy diagram it would be above the saturation line.

Cheers

 

[zekeman]

"V = 353.7* (Q* vs)/d^2"

Rule of thumb? Why?

Doesn't this presume a specific volume, so how can it always work.

My take on the overall problem of velocity is that the valve is opened until he gets the throttled pressure from which you you first use conservation of h leading to the estimated output state and then specific volume for his case and then velocity.Then from Q you get V.Then you correct the "error" in assuming constant h, get a new output h and redo the problem iteratively.

 

[zekeman]

Latexman,


"To me, it sounds like he needs help with Tout and he'll use that to calculate velocity, which he knows how to do. Just my interpretation."

Yes, of course you are right.With the estimated T he gets the thermodynamic state and knowing Q, the mass flow rate, he gets V But to get T he will have to iterate the solution as I pointed out previously.

 

[ione]

It is not a rule of thumb, with the units I have reported it is exactly what it is:

V = Volumetric flow rate/ outlet area

I have specified "IMO" making reference to the whole context.

What is the problem with the specific volume once you’ve identified the state downstream the valve?

Anyway my approach was just a “manual” approach to the problem (I still consider a 22” reduction valve too small). When dealing with critical applications it is always opportune to rely on much more complicated equations and sizing criteria, which cannot be handled handy. It is worth to use a specific software (the suppliers of the reducing valve should/must have this) which complies to standards such as IEC 60534.

 

[Latexman]

But to get T he will have to iterate the solution as I pointed out previously.

I think ione has proven that an acceptable Engineering solution for this case can be obtained without iterating. There is usually more than one way to get an acceptable answer to most problems, and I think I am a better Engineer today for learning from others when the opportunity presented itself. The key is to know the technology so the fastest method which gives an acceptable answer can be selected once the problem is understood; that comes from knowledge and experience.

Good luck,
Latexman

 

[zekeman]

"V = 353.7* (Q* vs)/d^2"

If Q is mass flow rate, show me how this facr\tor,353.7 makes this always true.

 

[zekeman]

"d I think I am a better Engineer today for learning from others when the opportunity presented itself. The key is to know the technology so the fastest method which gives an acceptable answer can be selected once the problem is understood; that comes from knowledge and experience."

You would be a better engineer if you did NOT make assumptions as to this application which may not be the "technology" you speak of.
Maybe, given your knowledge of the industry, you could tell me in this case, if the velocity term is or is not significant.

To say that h1=h2 without examining the energy equation proves that your grasp of the general problem is wanting.

 

[Latexman]

Even if the outlet velocity of the 24" valve is very fast, say Mach 0.5 to 0.75, it's contained energy is at most about 3% of the incoming steam's enthalpy. The OP and the steam's downstream usage (which we do not know) will determine whether this is significant or not, but IMO, it is more insignificant, than it is significant. This may be a good application for a drag valve. CCI makes some good ones.

Good luck,
Latexman

 

[zekeman]

So by your logic, tell me how step b step he gets a handle on tha velocity.
If h1=h2 then the velocity is near zero. And the temperature he gets will be meaningless to get velocity. What next?

 

[ione]

Zekeman,

Q = mass flow rate [kg/h=kg/(3600 s)]

vs = specific volume[m3/kg]

d = body diameter [mm]

V = steam velocity [m/s]

V = Volumetric flow rate/ outlet area

Volumetric flow rate = mass flow rate * specific volume = Q*vs [kg/(3600 s)*m3/kg] = (m3/3600 s)

Outlet area = pi*d^2/4 ? (3.14159/4)*d^2 [mm2]=(0.7853975/1000000)*d^2 [m2]

V = (1273240.62/3600)* Q*d^2 [m/s] ? 353.7*Q*d^2

 

[ione]

As I have stated above the transformation through a PRV should not be named isenthalpic, as the transformation is not composed of a succession of state characterized by the same enthalpy level. Anyway the enthalpies upstream and downstream the valve can be considered the same.

We can distinguish two macro phases:

1. Steam accelerates as it passes through the orifice and the gain in kinetic energy is realized at expenses of the steam enthalpy. This first step could be considered as an isentropic transformation from the starting state at pressure P1 to pressure P2.
2. Steam expands in the lower pressure region and here occurs the heat recovery at expenses of the steam kinetic energy. Friction contributes to the lost of a small amount of energy, which is considered negligible from an engineering point of view. This second step is an isobaric process at pressure P2.

 

[Latexman]

zekeman,

Meaningless? No. Close enough for Engineering purposes? It depends what happens downstream to the steam, and I repeat, we don't know that. The OP didn't say. If the steam is going to be condensed and the condensate recycled, it will be close enough for a satisfactory condenser design because the vast majority of the heat duty is the latent heat. I have assumed that accounting for > 97% of the steams energy will lead to a conclusion indicative of all it's energy. IMO without any information on what's happening downstream, it's simple and fast and close enough.

I get the feeling you are not going to be satisfied with anything but a rigorous, academic-worthy solution, so I will not entertain you with the simple, classical, approach. I would thoroughly enjoy seeing your solution though.

Good luck,
Latexman

 

[Latexman]

MattC1234,

Are you sure about "The ratio of specific heats is 1.86"? My reference shows it to be in the 1.26 to 1.32 range.

Good luck,
Latexman

 

[sailoday28]

Lateman and ione
Neglecting change in elevation and heat transfer, for a steady state process the fact is:
Vdownstream= sqrt[Vup^2+2*(Hupstream-Hdownstream)]....1
Make what ever further reasonable assumptions, shocks, isentropic, homogeneous two phase etc, BUT eq 1 remains true.

 

[ione]

Latexman,

The link below (NIST) shows that the value reported by the OP for specific heat ratio = 1.86 (dry saturated steam at 64.7 barg)is ok

http://webbook.nist.gov/chemistry/fluid/

 

[zekeman]

IOne,
"The correlation below works (IMO) quite good.

V = 353.7* (Q* vs)/d^2"

Sorry to put you through this , but your use of the term correlation and my not reading through, made me think "rule of thumb"

My bad.

 

[Latexman]

ione,

Yes, I must have looked at the wrong graph. I see now that is true for ~ 65 bar, but down at the low pressure side of his problem, 0.24 bar, Cp/Cv is ~ 1.34.

Good luck,
Latexman

 

[ione]

Zekeman,

The use of the word "correlation" was misleading. My fault due to my not good command of English

 

[katmar]

Regarding the Specific Heat Ratio

The specific heat ratio may well be 1.86, but that information is usually not terribly useful to an engineer. We more often need the isentropic exponent which might be the same as the specific heat ratio for an ideal gas, but for for a real-world, non-ideal gas like steam it is quite different. The values of 1.26 to 1.32 mentioned by Latexman are correct for the isentropic exponent. At the conditions given by the OP my steam tables (WASP) give a value of 1.266.

For more info see

http://www.personal.utulsa.edu/~kenneth-weston/appC2.pdf

http://twt.mpei.ac.ru/MCS/Worksheets/WSP/WKDiag15.xmcd

http://www.globalspec.com/reference...ids-Specific-Heat-and-Ratio-of-Specific-Heats

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[Latexman]

katmar,

Thanks! I still get confused on Cp/Cv versus k. Your references made it clear for an ideal gas; they are the same. With a non-ideal gas where they are not the same, use k in compressible flow equations?

I searched and found the following thread:

thread135-240728

Is there a better reference?

Good luck,
Latexman

 

[ione]

As katmar pointed out isentropic exponent is the parameter of interest to define whether an application is critical or not. Katmar’s value is precise (I got k = 1.254 for isentropic exponent).
No doubt here with the data passed on to us by the OP the application is critical.

Pcrit = P1 * [2/(k+1)]^[k/(k-1)] = P1*0.55418 = 35.85 barg

The definition of isentropic exponent states that it is equal to the partial derivative of enthalpy with respect to internal energy at constant entropy:

k = (?h/?U)s

For a perfect gas enthalpy and internal energy depends on temperature, and this leads to an isentropic exponent equal to the ratio of specific heat:

k = dH/dT = (dH/dT)/(dU/dT) = cp/cv