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Outlet Velocity of Steam through a Valve
3

Outlet Velocity of Steam through a Valve

Outlet Velocity of Steam through a Valve

(OP)
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

RE: Outlet Velocity of Steam through a Valve

The usual practice in this situation is to assume an isenthalpic process.  T and P are known in the upstream condition.  Calculate the enthalpy.  Assume the same enthalpy for the downstream where P is known.  Calculate T downstream.

Good luck,
Latexman

RE: Outlet Velocity of Steam through a Valve

Latexman is right, use his recommendation.

Unless you use a Self-drag valve, the flow will be acoustically choked and a shock wave would occur at the valve outlet.

RE: Outlet Velocity of Steam through a Valve

(OP)
Thanks Latexman and Davefitz! really appreciate your help, i've got it now!

Cheers

RE: Outlet Velocity of Steam through a Valve

2
Just to add a little consideration.

Being the pressure reduction throughout a valve an isenthalpic process (as already pointed out above), steam downstream the valve should have a temperature higher than that a saturated steam would have at the same pressure, and more precisely the temperature increase deltaT would be:

deltaT = (h1 – h2)/cp2

where:

h1 = saturated steam enthalpy at pressure P1 (upstream  the valve)
h2 = saturated steam enthalpy at pressure P2 (downstream  the valve)
cp2 = saturated steam specific heat at pressure P2 (downstream the valve).

So the final temperature Tf of steam downstream the valve would be

Tf = T2 + deltaT

Where:

T2 = saturated steam temperature at pressure P2

So one should have a superheated steam. This is correct for saturated dry steam.
In the real world steam has always a degree of dryness which lower than 100%. In this case the heat in excess will be used to increase the steam dryness.
 

RE: Outlet Velocity of Steam through a Valve

ione (Mechanical)
The formula      deltaT = (h1 – h2)/cp2

is for a perfect gas, constant specific heat. Yet, the entalpies you are using are for a real gas.

If you know the sat pressure at conditions 1 and 2 , then the sat temperatures and therefore delta T is easily calculated.

RE: Outlet Velocity of Steam through a Valve

sailoday28,

Maybe I am wrong, but I can't see where the problem is.

The definition of specific heat stands still for saturated steam downstream the valve at the reduced pressure P2 (I think it is undeniable). The process throughout the valve is isenthalpic (undeniable too). Now steam at pressure P2 (downstream the valve) has got an enthalpy, it wouldn't have if it was at the saturation temperature at pressure P2 (just take a table with steam properties). So there is an excess of enthalpy (difference between enthalpy at upstream pressure P1 and enthalpy at downstream pressure P2) that is used during the transformation to increase the steam temperature.

cp2 x deltaT = h1-h2

You will get exactly the same result plotting an isenthalpic transformation from P1 to P2 on a Mollier diagram (h vs s).

I am pretty sure about what I have stated, but in this forum there many members more authoritative than me that can put a light on this matter.
 

RE: Outlet Velocity of Steam through a Valve


ione (Mechanical
Having an equation of state and specific heat data, one may obtain expressions for enthalpy or changes in enthalpy.
integ[dH]= integ[CpdT] - integ[d(v*tau)/dtau|p]dp
for non two phase states. Where tau=1/T
Note  there are 2 integrals on the right hand side
Your integration assumes the second integral on the RHS is zero and Cp is a constant giving--  delta H= Cp delta T

Typically for a throttling process, delta H is approx zero and therefore for a perfect gas deltaT is zero.

Note, once obtaining H from the integration, Cp may be obtained from  DH/DT at constant pressure.

Specific heat (Cp) for a saturated liquid or sat gas is not DH/DT at constant pressure and if you have an old Keenan and Keyes  C' and C" are defined as specific hts for sat liq and vapor respectifully.



 

RE: Outlet Velocity of Steam through a Valve

Maybe I have been not too clear with index.

I should have written

deltaT = (h2 – h2')/cp2'

where
h2 =     enthalpy of steam at pressure P2 (downstream  the valve) which is equal to h1 (dry steam enthalpy at pressure P1, upstream  the valve)

h2' = enthalpy of dry saturated steam at pressure P2

cp2' = specific heat of dry saturated steam at pressure P2


Let's make a practical example for 1 kg/hr of dry saturated steam from

P1 = 64.7 bar(g) to P2= 0.24 bar(g)

At P1, the steam enthalpy is h1 = 2777.28 kJ/kg

Downstream the valve at P2 the enthalpy is the same (isenthalpic process) h2 = 2777.28 kJ/kg

For a saturated dry steam at pressure P2 the enthalpy should be h2' = 2685.26 kJ/kg and the specific heat cp2' = 2.0664 kJ/(kg*K)

Downstream the valve we have a steam which is superheated and its temperature is higher than that of saturation at P2 = 0.24 bar(g) which is be T2' = 106.069 K.

deltaT = (2777.28 – 2685.26)/2.0664 = 44.531 °C

So the temperature of the steam downstream the valve is

T2 = T2' + deltaT = 106.069 + 44.531 = 151.221 °C

The same result can be found from a Mollier chart.
 

RE: Outlet Velocity of Steam through a Valve

`The process is not isenthalpic since it is a flow situation where the energy , i.e. the sum of K.E and enthalpy are conserved, so  the difference in the enthalpies is  equal to the difference in the kinetic term, V^2/2g.
You have to use the steam tables for this and the possibility of choked flow where the mass flow rate is reduced.
In either case you can  iteratively get the differences in h, and go back and get the density ratio or V ratio , etc. differences to meet the energy conservation requirement.

 

RE: Outlet Velocity of Steam through a Valve

For an adiabatic steadystate process, zekeman gets a star.

RE: Outlet Velocity of Steam through a Valve

I agree with Zekeman that the transformation should not be named isenthalpic. During the process throughout the reduction valve dynamic phenomena take place, and so in a Mollier diagram (h vs s) the transformation should not be represented with a straight horizontal line from initial to final state (isenthalpic transformation).
The transformation in a Mollier diagram could be better depicted by a broken line (first step vertical isentropic line from pressure P1 to pressure P2 and second step isobaric line at pressure P2 to the final state at enthalpy h2 = h1). The thermodynamic transformation which takes place in a pressure reduction valve is the resulting of expansion phenomena with energy transfer due to conversions of static and dynamic pressures. In a throttling valve we have an irreversible expansion with the downstream enthalpy equal to the upstream enthalpy, and this is what matters. For a DRY (ideal) steam, at the end of the process the steam is superheated (and this, from an engineering point of view makes it comparable to a perfect gas). I still consider as valid the approach I've described in my previous posts and also the consideration that in the real world depending on the "wetness" of steam (upstream the valve), the superheating effect is minimized.
 

RE: Outlet Velocity of Steam through a Valve

ione.  For the throttling valve, chances are that the KE upstream can be neglected. If the KE downstream ( and not heat transfer) can't be neglected, then h1= stagnation enthalpy2 at pressure2.  If and only if, KE2 can be neglectged then h2=h1
And IF the gas is perfect, T2=T1.  

RE: Outlet Velocity of Steam through a Valve

I don't want to be boring and repetitive or appear arrogant, since I have always something to learn, anyway this time I think I will stay with my conviction.

Just one last thing.
Please take a look at the paragraph "What use is the temperature - entropy diagram (or T - S diagram)?" from the link below: there are described the transformations that happen during throttling.
I think (or at least hope) that those guys know their job, dealing with steam from day to night.

http://www.spiraxsarco.com/resources/steam-engineering-tutorials/steam-engineering-principles-and-heat-transfer/entropy-a-basic-understanding.asp
 

RE: Outlet Velocity of Steam through a Valve

I have found by chance on the web this calculator of steam (saturated and superheated) properties.

http://www.spiraxsarco.com/resources/steam-tables/superheated-steam.asp

If you enter the following data for a superheated steam:

Pressure: 0.24 barg
Henthalpy: 2777.28 kJ/kg (the same as that of saturated steam at 64.7 barg)

You get a temperature T = 151.526 °C (practically equal to the value I've posted on 16 Feb 2010 3:36).
 

RE: Outlet Velocity of Steam through a Valve

ione,

Touché!

Often as Engineers, we make judgements between what is practical ("that's close enough") and what is academically correct.  It's always good to hear both sides, as we have in this thread, lest we forget to use the academically correct method when conditions warrant it.

Good luck,
Latexman

RE: Outlet Velocity of Steam through a Valve

Latexman,

Thank you very much for your post.
I was not looking for approval at all costs, but I have dealt with this matter more than once and always did the way I've described. It was a bit sad and also frustrating to hear my approach was completely wrong. As I have already said above, I have always something to learn but I also consider some things I've learnt as a pillar of my (little) knowledge (especially if they've worked well till now).

RE: Outlet Velocity of Steam through a Valve

Ione,
 
Yes, for steam it turns out that it is far from ideal and a classical throttling process ignores the effects of velocity, since  it is usually low and Mollier diagrams are very convenient.

But the OP wants the velocity, not the temperature.

BTW, your example showing a decrease in temperature  seems correct to me,although I don't understand why you introduced Cp*delta T , unless you have no access to a Mollier diagram and steam tables.

 

RE: Outlet Velocity of Steam through a Valve

Quote:

Can anyone help me with the calculation for the temperature of the fluid at outlet and thus help me with the velocity calculation?

zekeman,

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.

Good luck,
Latexman

RE: Outlet Velocity of Steam through a Valve

Zekeman,

In all my previous posts I have made considerations concerning the superheating effect on dry saturated steam due to pressure reduction, and outlined a different behaviour between real wet steam and dry steam.

I take now the opportunity to add some considerations on velocity (just describe the way I usually proceed in these cases).

The velocity of steam downstream the valve, especially with a pressure reduction as that described by the OP (from 64.7 barg to 0.27 barg,) asks for particular care. It is recommended to keep velocity below a threshold of 200 m/s. Higher velocities produce bothering noise and vibration issues.

The correlation below works (IMO) quite good.

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

Where:

V = steam velocity downstream the valve [m/s]
Q= steam flow rate [kg/h]
vs = steam specific volume at downstream pressure P2 [m3/kg]
d = diameter of the body section valve [mm]

With the data given by the OP

Q = 235000 kg/hr (worst case)
vs = 1.54755 kg/m3 (from superheated steam table at P=0.24 barg and h=2777.28 kJ/kg)
d = 610 mm (this value must be checked as only the OP effectively knows it)

V = 353.7*235000*1.54755/372100 = 345.7 m/s (this is a too high value)

Conversely we can calculate the min diameter (dmin) of the reducing valve to ensure a max value
for the steam velocity V = 200 m/s

dmin = SQRT(353.7*Q*vs/V) ≈ 802 mm ≈ 31.5" (It seems a valve of 22" is little)
 

RE: Outlet Velocity of Steam through a Valve

"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.

  
 

 

RE: Outlet Velocity of Steam through a Valve

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.  

RE: Outlet Velocity of Steam through a Valve

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.
 

RE: Outlet Velocity of Steam through a Valve

Quote:

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

RE: Outlet Velocity of Steam through a Valve

"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.

RE: Outlet Velocity of Steam through a Valve

"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.


 

RE: Outlet Velocity of Steam through a Valve

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

RE: Outlet Velocity of Steam through a Valve

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?

RE: Outlet Velocity of Steam through a Valve

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
 

RE: Outlet Velocity of Steam through a Valve

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.
 

RE: Outlet Velocity of Steam through a Valve

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

RE: Outlet Velocity of Steam through a Valve

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

RE: Outlet Velocity of Steam through a Valve

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.


 

RE: Outlet Velocity of Steam through a Valve

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/

RE: Outlet Velocity of Steam through a Valve

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.
 

RE: Outlet Velocity of Steam through a Valve

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

RE: Outlet Velocity of Steam through a Valve

Zekeman,

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

RE: Outlet Velocity of Steam through a Valve


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/10737/179909/Chapter-3-Physical-Properties-of-Fluids-Specific-Heat-and-Ratio-of-Specific-Heats

Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com

RE: Outlet Velocity of Steam through a Valve

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: Isentropic exponent and specific heat ratio

Is there a better reference?

Good luck,
Latexman

RE: Outlet Velocity of Steam through a Valve

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 = (δh/δU)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
 

RE: Outlet Velocity of Steam through a Valve

The link below is for a calculator of the isentropic exponent of water and steam.

http://twt.mpei.ac.ru/OCHKOV/WSPHB/WKDiag15.html

The isentropic exponent is described as:

k = - V/P *(dP/dV) at constant entropy

and it derives from:

P*V^k = constant

Passing to the logarithm

LogP + k*logV = constant

Differentiating

dP/P + k* dV/V = 0

k = - V/P *(dP/dV)
 

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